source stringlengths 620 29.3k | target stringlengths 12 1.24k |
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Unable to change package footprint solder pad size in eagle as the image shows, I want to change the BGA soldering dots from current size to something larger. Only size changes, or shall we say diameter changes, no shape or anything, but I couldn't. I clicked change tool, entered the new diameter (and I'm sure it's different than the original) and then clicked the pads I wanted to change, nothing happens. Can someone please help? Thank you. <Q> Looking at it your footprint uses a circle drawn on the tstop layer which is why you see diameter setting when you look at info. <S> The pad itself is actually an SMD pad, which does not have a diameter setting. <S> Instead it has two key properties: <S> SMD Size which changes the overall size of the copper pad roundness which will be set to 100% (i.e. circle - 0% would be square) <A> Using the Info tool (the button that is the letter i), click on the PTH <S> you are interested in and adjust the Diameter from auto to another value that is greater than your Drill size. <S> Click Apply and visually inspect the hole - if it's still not as much copper as you want, increase the Diameter again until it's as thick as you want. <S> For reference see this link... <S> Reference link <A> it is impossible to make changes to a particular part in pcb layout... <S> edit the part in its library and make your changes then update this library in the pcb layout...
| So to change the size, leave the roundness set at 100%, and change the SMD size setting, making sure to keep both the width and length of the pad the same.
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Problem with testing 74ls173AN D flip-flop I'm building a simple 4-bit computer (a Nibbler) from 74 series ICs on a breadboard. Right now I'm working on 74LS173AN IC registers. I can't make it work. I have connected the input enable (pin 9,10), output control (1,2), and clear (15) to ground and some data inputs to vcc. The output pins (Q1-Q4) are connected to ground via LEDs. I'm simulating the clock signal by swithing a tact-switch on/off (with ground connected by 10k resistor and direct connection to VCC. I tried different connections and searched web (many hopes with https://www.youtube.com/watch?v=9WE3Obdjtv0&t=14s , but it didn't help). Nothing works. I suspect there might be a problem with clock signal, but I would really appreciate the opinion from an experienced electronic specialist. <Q> A number of tips about dealing with (LS)TTL: 1) Inputs should be either grounded or pulled high with a 1k pullup resistor. <S> 2) <S> Output LEDs should be connected to +5 with a 1k pullup resistor. <S> And yes, this will produce a signal inversion in the sense that a LOW output will turn on the LED. <S> simulate this circuit – <S> Schematic created using CircuitLab 3) <S> You must decouple the power pin. <S> Connect a 0.1 uF <S> ceramic cap from pin 16 to pin 8, and don't use jumpers - plug the leads directly into the breadboard at the IC. <S> Make sure you test the LED polarity by unplugging the IC, applying power, and grounding the output contact. <S> If the LED doesn't light up, it's bad or you've got it backwards. <S> When you've checked out the LEDs, turn off power and plug in the IC. <S> Otherwise your circuit looks OK with one exception. <S> You've missed the part in the Youtube video where it is pointed out that TTL inputs float high if no connection is made, so your circuit really ought to be turning on ALL the LEDs, not just the 2 that are connected to +5. <S> With proper LED connections, leaving the inputs floating or pulled high should result in all LEDs being off. <S> Finally, using a switch to generate a clock to a flip-flop is fine for what you're doing now, but anything more complicated will require debouncing of the switch. <A> <A> "The output pins (Q1-Q4) are connected to ground via LEDs." <S> There is your problem. <S> By the spec sheet that device does not drive high.. <S> IOH is -1.0 mA.. <S> ( weird spec that is...)
| When you push the tact-switch, the metal contacts might bounce off each other more than once triggering more than a single clock pulse which will lead to an unexpected output.
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Why speed of the wind turbine permanant magnet synchronous generator speed reduces when connected to the load I have 100 W vertical axis wind turbine which is mounted on PMSG. And its generated 3 phase voltage is converted to dc using H bridg rectifier. When I connect the load, the turbine speed will reduce. Why it happen so? <Q> This is more an aerodynamics than electronics question. <S> When a turbine aerofoil is gliding through the air, it settles down to a speed where the 'propulsive' force, the lift * aerofoil angle, is equal to the total load, comprising aerodynamic drag, and frictional and load resistance from the output shaft. <S> If the load is increased, then the excess of load over the propulsive force will slow the turbine down initially. <S> What this does is, for constant wind velocity, and a constant mechanical blade angle, it increases the angle of attack, increases the aerodynamic blade angle, which increases the propulsive force to match the new load. <S> This is the stable operating region of the turbine, where the wind power input can cope with the load, and small load increases result in small speed decreases. <S> If you increase the load too much, then the blades will slow down so much that the angle of attack increases too much, and the blades enter an aerodynamic stall. <S> This is an unstable operating region. <S> The lift now decreases with angle of attack. <A> It's a very basic behavior of any generator. <S> Load creates torque reverse to one that is driven by the power source. <S> Imagine a small generator, like your own, connected to an electrical locomotive. <S> No chance it will move! <S> Another way to look is by power. <S> Power it limited at some point. <S> Mechanically it is rotation speed times torque, electrically it's voltage <S> (speed times a factor) times current. <S> When no load applied, mechanical power is completely wasted on mechanical losses. <S> If load is applied, electrical power is no longer zero, so mechanical lisses must be reduced, and it means less speed. <A> The turbine slows down because the electrical load on the generator results in torque that resists the rotation of the wind turbine. <S> Since your permanent-magnet AC generator drives a bridge rectifier, the complex behavior of a generator interacting with an AC line does not apply here -- this speed reduction is simply a matter of the increased load on the generator. <S> Conservation of energy requires that if the generator's electrical power output increases (amps times volts), then the mechanical power input to the generator (torque times angular velocity) must also increase. <S> This is why the torque at the generator's input shaft increases when you connect the load. <A> The turbine slows down because you're increasing the load on it. <S> This is the same basic effect as when a table-saw blade bites into wood and slows down a bit (you can hear it)... or when you're riding a bicycle and hit a hill and <S> you slow down. <S> Part of good generator design is assuring that the generator doesn't put such a load on the turbine as to make it "bog". <S> You want it to slow down enough to get maximum useful power (torque x revs or volts x amps), but not too much that it bogs. <S> This problem is not unlike the MPPT problem for solar panels. <A> For a motor or generator the general rule is: speed \$ \propto \$ voltage and torque \$ \propto <S> \$ current. <S> When current starts to flow torque will increase and the speed will drop. <S> For a specific wind speed there is a speed - torque line for the turbine. <S> This becomes a voltage - current line on the electrical side. <S> See Figure 120 in this reference for a sample torque - speed diagram. <S> Good inverter design can ensure that the generator runs a maximum power. <A> Other answers have answered the immediate question : why does it slow down? <S> But I'm sensing an underlying question : <S> what can I do about it? <S> The answer to that is, tune the load to get the best performance from the generator at all windspeeds. <S> At no load, the generator spins happily but delivers no power. <S> As you add load, you extract power, slowing the generator down, until the turbine blades are stalled and produce very little power. <S> Somewhere in between, you will extract maximum power from the turbine - the ideal load will vary according to the windspeed. <S> I did some working out for a specific generator in this answer which may help. <S> For initial experiments you can simply connect in different load resistances and measure voltage across them, to evaluate performance. <S> But a practical way of automatically tuning the load may be similar to the "MPPT" tracking systems used in solar power.
| A small increase in load results in a decrease in speed and a decrease in lift, which results in the turbine stopping.
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Keep individual LED brightness steady when other LEDs are added or removed I'd like to use several LEDs on a project, with brightness control (it's a scale model of a street with houses, with LEDs of different brightness in each house, which I've been working on with my son). However, I'd want the brightness of the LEDs not to change when I turn some of them on and off (or equivalently, if I add or remove LEDs). I know I can control dimness of LEDs either by directly varying the current applied to them as in this project from someone else or by using PWM as suggested here -- but in both cases (I implemented the two projects listed above as a test), the current source is the same so their brightness will vary depending on the number of connected LEDs (which make sense, since the full LED set is plugged on a single transistor's collector). What would be the simplest way to achieve what I'd want in this case? I'd rather avoid using, for example, an Arduino (would be too physically large and would involve programming -- I'd like this to be as simple, small, cheap as possible, without microcontrollers). Is there some simple way to decouple each LED from the others, and still have central control of their brightness? Connecting the LEDs in serial would not be good either, since I would not be able to remove those in the middle (and this would require a too high voltage to work, because of the accumulated voltage drop of the LED string). Thank you! <Q> Update: <S> Op has noted his issue is with using a NPN transistor as a high-side controller. <S> It would not be working in a saturated mode and as the load changes it will have odd voltage and current properties. <S> The proper simple solution is to use a PNP transistor suitable for the load. <S> Or switch to a NPN low side setup, noting that the PWM period will be inverted. <S> I'm looking at the PWM circuit, a simple high-side transistor + 555 timer setup. <S> I'm not seeing why this would change the brightness if you remove a series string from it. <S> Unless you are removing a single led from a series string. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Here, all the leds and resistors are the same. <S> So the voltage between <S> Node1 and Node2 are the same, as is the voltage and current going through all the strings. <S> If you turn off string 2 (R2, D3, D4 via SW1), the other strings should not see any change in voltage, and thus, current and brightness. <S> If you turn off D6 in string 3, via SW2, then the voltage across D5 goes up, and thus current and brightness (if it doesn't burn out right away). <S> But String 1 and String 2 are still the same . <S> If you add leds to any of the strings, you will see a brightness change as the voltage is divided among them. <S> You would have to adjust the resistor value to make sure the same amount of current still goes through. <S> The only problem you may face is if your voltage source is a constant current source instead. <S> Or if your voltage source is very load sensitive. <S> If the source has a high equivalent series resistance, then as your load increases in current, a higher voltage drop is seen, affecting everything. <S> Use a high current capacity and/or quality power supply. <S> Also keep in mind that your transistor needs to be sized for your load. <S> A 2n3906 will only handle 200mA, so if your load is more than that you will see problems. <A> Put all of the LEDs in series. <S> Drive the series string of LEDs with a current source (or sink). <S> All of the LEDs will light. <S> Put a short across any LED (s) that you want to turn off. <A> The starting point should be a common anode LED matrix and just on and off. <S> Once you have the on off working, add dimming. <S> And if desired in the future add a micro. <S> This is the matrix used in signs. <S> Typically driven with a micro-controller, but could be done with out one. <S> Rather than a micro-controler build a state machine with a PLD. <S> The PLD would also have registers holding the on off values State machine logic is a good way to learn how a micro controller works. <S> After all a micro is just a large state machine. <S> To "program" the on off use two 8 position switches, one for address and one for data. <S> Once the on off is working add another dimension to the on off registers and up the frequency of the on off matrix and tweaking the state machine logic. <S> Registers come cheap in PLCs. <S> When, if, you want, remove the two sets of switches and replace them with a micro controller. <S> When you are done, if there is not a chip already out there doing this the state machine PLC could be a salable product. <S> I do not know how well you will understand this. <S> In the 1980s I was designing data networking ICs for IBM (token ring 802.5) and GE (token bus 802.4). <S> This project as described is how ICs are designed. <S> You build your model with off the shelf logic then migrate it to custom silicon.
| You will need a set of on off registers for each level of dimming. Because the string of LEDs is driven with a constant current, the brightness of the remaining LEDs will remain the same.
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How to add power terminal pins in PCBNew in Kicad? I am trying my hands-on with Kicad recently. I have created an LED astable multi-vibrator circuit in the Eeschema as follows. Then I have created a PCB out of it as below. I wanted to add the two power supply pins on the PCB. So, I used two SolderWirePad_single_0-8mmDrill s. But then I needed to connect these two drill points (+, GND) to the actual tracks of the circuit. I did this before successfully for the POS(+) terminal encircled in PINK colour. But now I cannot add a track for the (GND) terminal encircled in RED colour (seems like I am missing something). Whenever I am adding a track starting either from one of the emitters of the BC548 or from the GND drill pin, the track is not added after the double clicking on the ending terminal. (Please note: I am able to draw the track, but, it seems something is not validating, may be, the connection.) How to get the power supply tracks when there is no connection in the schematic? Is there any better way to handle this? Thanks in advance. <Q> The power flags aren't placeable parts. <S> To place connectors for power, you should use connector components in the schematic for those connections. <S> Here's your schematic with power connectors and power flags: <S> The purple text by the power flags are the names of the footprints I assigned them. <S> Here's the completed layout: <S> Note that there aren't any components placed for the power flags although I did have footprints set for them in the schematic. <S> To repeat a comment I left earlier:DO NOT turn off the design rules check. <S> It is there to help you. <S> When you turn it off, you downgrade KiCAD from an electronics design package to a drawing program that happens to have a library of electronics parts. <S> You can see this in your own layout: That white line from R3 to C2 says that the trace you drew to connect those parts isn't really connected at one end. <S> If you make the board as is with that line still in place, you may find that there is a gap in the trace - no connection, circuit doesn't work. <S> You picked a nice, simple, common circuit to start with. <S> That's good. <S> Use this opportunity to learn how your tools can help you. <S> Turning off rules check and forcing a part onto the board from the layout editor worked this time - but you worked against your tools. <S> Don't do that. <S> Let the program help you, and learn to work with it. <S> It will make things easier in the long run. <A> The "+5V" and "GND" symbols you used are really just net labels - they don't represent physical parts. <S> You have to place connector symbols of some sort to get a pad to connect your tracks to. <A> You should post a picture of your schematic too. <S> Pin 1 in that red circle has a No Connect X through it, so you won't be able to connect anything to it and it will just be floating. <S> EDIT: <S> After having my vision adjusted, I see the schematic now. <S> I don't see any component in the schematic that would associate with those +6V and GND pads? <S> pcbnew is wondering what they represent in the schematic. <S> Add a 2x1 jumper for them in the schematic, then associate them with a single footprint in cvpcb, then generate the netlist and re-read it in pcbnew <S> (you probably know the routine). <S> Or split them and do two 1x1 jumpers. <S> Either way, they have to have a representation in eeschema. <S> Have you tried switching to OpenGL mode (hit F11 and F9 to get back)? <S> OpenGL mode is a lot better at routing and since it does push and shove routing by default, it will "show" you if you have a clearance issue by routing your trace away from pads. <S> Also, as to the strange issue of your cap not being connected even though it clearly looks like the trace is connected: Did you drop the cap's pad on top of the trace after you had already routed the trace? <S> The only thing I can think of is that KiCad is just being finicky about your pad placement relative to the track. <S> Try hitting Backspace over the track segment that is connecting the cap. <S> Then go to Preferences, General and set both options for Magnetic Tracks and Magnetic Vias for "Always. <S> " This will guarantee that your track is being placed right by snapping to the pad. <S> Then try routing it again.
| I have often used a one-pin schematic symbol associated with a footprint containing a single pad to show a wire connection point.
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What is details meaning of output and input Impedance? I was studying the CC, CB and CE Amplifier. And I have found a sentence somewhere that, "Maximum power transfer occurs when load impedance is equal to source impedance." As much as I understand impedance is something that opposes current. But, what are this load impedance, output impedance, input impedance? I mean how are the related to. In the case of CC, CE, CB amplifiers, how are they related to each other and what are the meaning in this case? Thanks in Advanced. Hope somebody will help. <Q> To give a mathematical definition, if we were to excite the input of a two-port with a sinusoidal current source <S> $$i_{in}(t) = <S> I_0 <S> + <S> I \sin(\omega t)$$ with \$I\$ small enough to not cause any nonlinear behavior, we'd find $$v_{in}(t) = <S> V_0 <S> + V_i \sin(\omega t) <S> + <S> V_q \cos(\omega <S> t)$$ <S> Then we could define $$R_{in} = <S> \frac{V_i}{I}$$ and $$X_{in} = <S> \frac{V_q}{I}.$$ <S> Then the input impedance would be \$Z_{in}\$ if we defined $$Z_{in} <S> = R_{in} + j X_{in}.$$ <S> In words, this means that the real part of \$Z_{in}\$ tells us how the in-phase component of the input voltage depends on the input current. <S> And the imaginary part of \$Z_{in}\$ tells us how the quadrature (90 degree phase shifted) part of the input voltage depends on the input current. <S> We would also find that $$R_{in} <S> = \frac{{\rm{d}}V_0}{{\rm{d}}I_0}.$$ <S> The output impedance is defined the same way, but as a relationship between the output current and output voltage. <S> All the other things we know about input and output impedance and how they relate to circuit behavior can be connected back to these definitions. <S> "Maximum power transfer occurs when load impedance is equal to source impedance." <S> This isn't quite correct. <S> First, when talking about possibly complex-valued impedances, maximum power transfer occurs when the source impedance is the complex conjugate of the load impedance. <S> Second, this maximum power transfer condition is for choosing the load impedance when the source impedance is fixed. <S> If source impedance can be controlled, then either very high impedance (with a fixed-current source) or very low impedance (for a fixed-voltage source) should be chosen to deliver maximum power to a fixed impedance load. <A> Input impedance is the impedance that is "seen" by a device when connecting to a circuit's input. <S> For example if I connect a source to the input of any of the amplifiers you just mentioned, the input impedance would be Vin/Iin <S> And the load impedance is simply the impedance of the load itself that you attach to the output. <S> It's also important to note the impedance is actually a complex value that includes the magnitude and phase shift that will result at a given frequency. <S> As for the line you quoted.. it can be mathematically proven that Power is maximized when Rs = RL... Take for example a simple circuit with source with Rs and RL in series. <S> VL = Vs*RL/(RL+Rs) <S> PL = <S> VL^2/RL <S> if we differentiate PL with respect to RL we can prove that: d(PL)/d(RL) <S> = 0 <S> When RL = <S> Rs <S> Therefore showing that the maximum value of PL occurs at RL = <S> Rs <S> But another thing you should realize is that the quote you posted isn't entirely correct. <S> In fact it's wrong... <S> The power is maximized when source impedance equals the conjugate of the load impedance .. <S> The key word is conjugate <S> This is because impedance as I stated before is a complex value that represents magnitude and phase shift. <S> The phase shift (imaginary part) serves to place the voltage and current out of phase and therefore reduce power.... <S> If the source and load impedance are conjugates: Zs = R + jX & ZL = R - jX <S> Then the imaginary parts (phase shift) will cancel and you will be back at the derivation I defined previously <A> Be aware that maximum power transfer is a 6dB loss of voltage. <S> Thus broadband amplifier design does not use matched interfaces. <S> I trained a team of highly motivated RFIC designers. <S> They were concerned about onchip transfer of signal between "stages". <S> I explained "At these distances (100 microns), we are implementing broad-band operational amplifiers that happen to amplify 250MHz sinusoids. <S> There is no matching possible nor needed."
| The output impedance on the other hand is the impedance seen when looking into the circuit from the output.
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Water on Li-Ion battery fire: good idea, bad idea, or neutral? I always thought (like this guy ) that putting out a Li-Ion battery fire with water was a bad idea because of the reaction between water and lithium. But now I read from one source : Lithium-ion batteries contain little lithium metal and in case of a fire they can be dowsed with water. Only lithium-metal batteries require a Class D fire extinguisher. Is this accurate? Can I really use water on Li-Ion battery fires? And if so, is this safe for batteries of any capacity, or is it dangerous beyond some mAh? <Q> Oregon State University published this Lithium Fire Prevention Fact Sheet . <S> For Li-Ion batteries <S> : If formally trained, you may use a st andard ABC fire extinguisher or water to put out a lithium ion battery fire. <S> For batteries containing elemental Lithium: <S> Only Class D fire extinguishers that contain a copper powder are approved for combating a lithium fire. <S> DO <S> NOT USE WATER OR ANY OTHER TYPE OF EXTINGUISHER BECAUSE <S> ORGANIC & INORGANIC LITHIUM METAL FIRES REACT HIGHLY WITH WATER AND COMBUSTIBLE SUBSTANCES. <S> The paper also strongly recommends that you "Let the fire department fight fires. <S> " <S> In other words, you should probably only fight them yourself if you absolutely need to. <S> For more information, take a look at the Lithium Battery Safety and Handling Guide . <A> It will prevent the surrounding material from catching fire and maybe even preventing the battery from exploding! <A> The free lithium in lithium ion batteries travels between the graphite cathode and cobalt (or manganese) <S> oxide anode both of which are soaked in a solution of lithium hexaflourophosphate (or other lithium salts) in ethylene carbonate (or other organic solvents). <S> None of these react dangerously with water. <S> When you dump water on this, it won't soak in quickly enough to explode.
| From a practical point of view, if I have a Li-Ion battery fire, and all I have is a water hose, I most certainly will try to put out the fire by dousing it with water.
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A question about coupling differential signals safely and Input Common Mode Voltage Range This may sound a weird question but couldnt resist to ask here. Lets say I have a differential signalling mode floating battery powered transducer(where the outputs are mirrored) and the differential voltage across its outputs Vd at a moment is 1V. Lets call the output wire pairs as V+ and V-. At this moment I only know the differential voltage between V+ and V- by using a voltmeter or a scope as 1V. Below illustrates the transducer and its differential outputs: But in the above situation I dont know the voltage across V+ and the earth ground and similarly I dont know the voltage across V- and the earth ground. Lets say I measure the voltage across the V+ and the earth and V- and the earth and I measure them as 101V and 100V respectively keeping the differential voltage as 1V again. Imagine I hook up this transducer outputs to a differential amplifier where its system analog ground is earth grounded. I draw the equivalent circuit as: In this case, the common mode voltage Vcm = (100 + 101)/2 = 100.5V and the differential voltage remains Vd = 1V If it is true so far so good.. But in data-sheets of the amplifiers there is something called input common mode voltage range. I think it is the max voltage an amplifier can handle with respect its system ground. And imagine this amplifier has +/-15V input common mode voltage range. My questions are: 1-) Would the amplifier get damaged in this case? 2-) If it would, does that mean that I have to measure the differential output pairs of the transducer one by one with respect to system ground of the amplifier each time before I couple them? Is there a common practice for that? <Q> If you have a 'floating battery powered transducer' , and you measure 100v between one of its outputs and ground, then you do not have a 'floating' battery powered transducer, you have a transducer which is for some reason 100v above ground. <S> If you connect that to an amplifier whose permissible common mode input voltage ranges +/-15v about ground, then it certainly would not work linearly, and it would probably get damaged. <S> If your transducer is truly floating, then it would take only a nominal resistor, say 100k, connected between ground and some terminal on the transducer, say one of the outputs, or a battery terminal, to bring it down to within the amplifier common mode voltage. <S> If your transducer is elevated to 100v above ground by, say, a connection to the system you are measuring, then you will have to use an amplifier with a larger common mode voltage input. <S> Attenuating both outputs 10:1 with respect to ground will bring them both down to 10v, within your common mode range. <S> You would need to take care to match the two attenuators very carefully, any mismatch will result in some common mode signal appearing incorrectly as differential. <A> 1) Not unless ESD event occurred which being CM insulated, makes it a susceptible target. <S> Otherwise floating battery powered Diff Amp works fine, then use optical or digital or magnetic isolation for output to data collection system. <S> The CM gain can be computed by the (series leakage to input shunt ) <S> resistance ratio * the tolerance error of the matched R's * differential gain. <S> Coupling from Common Mode <S> noise sources can be a problem if cables and R matching is not perfect as CM noise tends to be orders of magnitude large in E fields <S> e.g. 10V/m from powerlines than the differential signal but also much higher impedance. <S> So a CM shunt connection of grounds reduces this somewhat. <S> The environmental reasons for isolation depend on each application affect the solution. <S> e.g. HV, safety leakage, RF immunity etc. <S> Sometimes a choke is used to get low f grounding but improved RF isolation. <S> Sometimes an RC connection is used for ground connections. <S> Sometimes an active CM signal is used to active guard the shield or use as a remote ground and only receiver ground connection is used for a shield. <S> This is standardized as "Right Leg Drive" in patient monitoring. <A> The short answer is YES. <S> That said there are solutions to measurements in very high Vcm environments. <S> Amplifiers such as this optically isolated instrumentation amplifier or <S> this capacitively isolated amplifier can stand off many kV.
| If you have common mode voltages in excess of the Vcm range of an amplifier you will damage either the measurement system or the source sensor.
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Electrical resistance of two conductors of equal surface area in contact with another? Resistance of a wire can be determined by the formula R = resistivity*(length/area). What if you have a wire composed of two parallel metals that are in contact with one another? Will the resistivity be a weighted average of the two values of each metal? <Q> That one's a bit harder than it seems. <S> So, first of all, let's assume the conductivities of the two metals are in the same order of magnitude, so that a simplification that "all significant current is carried by the lower-resistance wire" isn't justified. <S> In a first approach, you could model the wires as sum of small resistors that are contacted regularly: x·₁ <S> x·₁ <S> x·₁ <S> x·₁A ---+-[===]-+-[===]-+-[===]-+-[===]-+-- <S> | | | | <S> |B ---+-[===]-+-[===]-+-[===]-+-[===]-+-- <S> x·₂ <S> x·₂ <S> x·₂ <S> x·₂ <S> With conductor A having the specific resistance (ohm per meter) ₁, and B having ₂. Each resistor than has a resistance x· ; x is the small wire length between contacts. <S> We set the total length \$X= <S> N\cdot\Delta x\$. <S> A single "parallel element" has the resistance <S> $$\begin{align}R_{\Delta x} &= <S> (\Delta x\cdot\rho_1)||(\Delta x\cdot\rho_2)\\&= <S> \frac{\Delta <S> x\cdot\rho_1\,\cdot\,\Delta <S> x\cdot\rho_2}{\Delta x\cdot\rho_1+\Delta <S> x\cdot\rho_2}\\&=\frac{(\Delta <S> x)^2\rho_1\rho_2}{\Delta x\cdot(\rho_1+\rho_2)}\\&=\frac{\Delta x\,\rho_1\rho_2}{\rho_1+\rho_2}\end{align}$$ <S> The total resistance of A||B then becomes: $$\begin{align}R_X &= N\cdot <S> R_{\Delta <S> x}\\&= N\frac{\Delta x\,\rho_1\rho_2}{\rho_1+\rho_2}&\text{with <S> $N=\frac{X}{\Delta <S> x}$:}\\&= X\frac{\rho_1\rho_2}{\rho_1+\rho_2}\end{align}$$ <S> In other words, it's the same formula as if you put two resistors in parallel, just applied to the specific resistances (times the length of the conductors). <S> Notice that this construct will also be subject to the Seebeck Effect , which means that if you have a temperature difference between the ends of your composite wire, you might see a current flowing – a small one. <A> No. <S> You have two parallel resistors. <S> $$R_1=\rho_1\cdot\frac{l_1}{A_1}$$ $$R_2=\rho_2\cdot\frac{l_2}{A_2}$$ Parallelizing resistors follows the formula $$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots$$ For two resistors, this can be rewritten as $$R=\frac{R_1\cdot R_2}{R_1+R_2}$$ Putting in \$R_1\$ and \$R_2\$ from above, with \$l <S> = l_1 = l_2\$ and \$A = <S> A_1 = <S> A_2\$ <S> $$R=\frac{\rho_1\frac{l}{A}\cdot\rho_2\frac{l}{A}}{\rho_1\frac{l}{A}+\rho_2\frac{l}{A}}$$ $$\cdots$$ <S> $$R=\frac{\rho_1\cdot\rho_2}{\rho_1+\rho_2}\cdot\frac{l}{A}$$ <S> When \$A_1 \neq A_2\$ things get more complicated. <A> Assuming the two metals which make up the wire have a constant cross section, then you can calculate the resistance of the wire this way: <S> Work out the cross section of each metal. <S> Use the resistivity of that metal and the length of the wire to work out the resistance of the two metals as if they were separate wires. <S> The total resistance of the combined wire is the resistance of the two sub-wires in parallel. <A> Assuming you have two exactly parallel uniformly thick wires contacting each other: <S> There's no sideways current <S> (* You can calculate the resistances separately and the combined resistance by R <S> =R1*R2/(R1+R2). <S> The equivalent resistivity is R*Area/Length <S> The result <S> Note: we have doubled the area. <S> *) <S> How can I know that? <S> - <S> Simply there's as many volts/millimeter in both wires. <S> Theres in every point <S> zero volts at the sideways direction.
| : The equivalent resistivity for the combination of geometrically equal wires is 2* r1*r2/(r1+r2) where r1 and r2 ase the individual resistivities.
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How to mount electronics on a vehicle chassis safely I want to implement a control system I designed for a light electric motorbike. I would like to mount it as professionally as possible on the chassis. It consists of a power system pcb and a SAMC21 development board which I'll merge later once this all works properly. I have tried screwing it directly on the aluminium chassis but have found that shocks and vibrations from the horn disturbed the system or even destroyed it upon large shocks like a few hammer hits. Is there a specific way car and motorcycle manufacturers mount their electronics to make their system as robust as possible?Thank you for your help! <Q> If you can't avoid them, choose components that use materials or constructions techniques targeted to minimise the so-called "microphonics". <S> Also, use microphonics minimisation design strategies on your own. <S> Microphonics is an unwanted behavior due to the piezoelectric effect of ceramic materials. <S> Usually, it will manifest itself as spurious voltage burst when the component is subject to mechanical vibrations or shocks. <S> Obviously, it can wreak havoc in your circuit, upsetting digital circuits and/or triggering switching in analog circuits. <S> Apart from microphonics, electromechanical devices (relays, swtches, etc.) can suffer upsetting under vibration/shock. <S> Also, mechanical stress can dramatically affect reliability and precipitate failures unless the PCB and its enclosure is specifically design to withstand those vibration and shock levels: <S> Solder joints may break up due to tension/compression forces. <S> Solder joints may not break up but can apply tension force to some components and break them up (ceramic chip capacitor are a classic). <S> Glued components may loose (if glue is too stiff). <S> Electromechanical components may latch-up (permanent damage), especially relays. <S> PCBs may break up, either externally or internally and traces may get damaged. <S> etc. <S> All these issues are usually addressed at design level, having the environmental specifications in mind. <S> The specifications drive all the following: <S> The selection of components and materials (reliable under vibration, no microphonics). <S> The electrical design of the circuit (tolerance to upsets). <S> The mechanical design of the PCB (higher clearance to PCB edges, etc.). <S> The footprints in the PCB (bigger footprints for better solder joints wetting, etc.). <S> The soldering process (reliable joints with good wetting, no cold joint, etc.). <S> The mechanical interface between the circuit and its enclosure. <S> (stiffening mechanical parts may be required to avoid resonant vibration modes). <S> The mechanical design of the enclosure (to avoid vibration amplification). <S> The mechanical interface between the enclosure and the place where it has to be mounted (to avoid vibration amplification and to damp vibrations and shock as much as possible). <S> As you can imagine, if you don't address all these things from the beginning, then you can only pray for it to work under vibration/shock and/or try to mitigate it by replacing some parts (ceramics) and trying to reduce the mechanical energy your board is getting. <A> For prototype devices (which should last several months and are OK to fail occasionally), it is usually enough to mount the PCB on rubber inserts instead of bare screws). <S> This greatly reduces peak accelerations your PCB sees, while keeping your prototype accessible and requiring minimum design changes: For commercial products which should last 10 years or more, you want something more robust. <S> I have seen automotive electronics where parts are assembled (or even the whole device is potted) with RTV silicone , which greatly helps with both mechanical sturdiness and heat transfer. <A> Board flex will crack ceramic capacitors and other brittle parts, like SMD resistors (which have a ceramic body). <S> It will also crack BGA solder joints, if any. <S> Lower-intensity board flex and vibrations will make X7R and related high-K ceramics act piezoelectric. <S> This means any high impedance nodes (like a sensor, or a voltage reference) filtered by a High-K ceramic capacitor will be very noisy. <S> Using other capacitor types solves this. <S> Now, there are several strategies to mitigate this: Use flex-resistant components (like J-lead caps, flex-termination caps, SMD parts with leads, or even thru-hole parts) Minimize flex and shock reaching the board itself <S> You can make your board stiffer by mounting it inside a solid enclosure or on a metal plate, with lots of screws. <S> Thermal management issues must be considered. <S> You can orient the board in such a direction that the main shock force will be parallel to it. <S> Forces perpendicular to the board will make it flex more than forces parallel to its length. <S> Now that you have a rigid metal-backed board, you can mount it (or its enclosure) to the chassis using rubber shock-mounts, or even rubber shock washers. <S> The idea is to avoid resonance, and create a mechanical lowpass filter without a peak in its frequency response. <A> One thing you'll notice in many cars is that the electronics are often mounted in an enclosure, and the enclosure is mounted to the vehicle. <S> The enclosure may be metal or plastic, metal enclosures are connected to chassis ground and used for devices sensitive to electrical noise (e.g. radios, GPS/tracking devices). <S> A pcb may for example be mounted on rubber bushes into a box that's attached with plastic clips. <S> Some retrofit devices use sticky foam pads to mount the PCB. <S> The box also absorbs some of the large-scale flexing in the PCB. <S> Prototypes are often worse than production devices, as they may use bigger boards and heavier components/connectors (even through-hole).
| Quite often both the mounting of the electronics to the enclosure, and the mounting of the enclosure to the chassis have some degree of damped flexibilty in them. Avoid ceramic capacitors and ceramic components for applications under environmental conditions that include vibrations and shocks.
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Using a dc power supply instead of batteries I have a device which takes 4xD batteries which I would like to instead hook up to a spare wall charger. I have looked at the voltage of the batteries and it seems most batteries are 1.5V. My device has 4 of them (seemingly in series - but with 2 down one side and 2 down the next. Ie: simulate this circuit – Schematic created using CircuitLab There is 1 wire (black) connected to the negative terminal on the top. A Red wire connected to the positive terminal on the top, and then an orange wire connected to the negative terminal at the bottom (although this terminal is just a plate of metal so I guess it's neither positive nor negative as they are both connected? Whereas the positive and negative at the top are separated.) My wiring diagram seems to indicate this orange wire is ground (I guess that makes sense if both positive and negative terminals are connected then the voltage on this plate is 0 correct? If that is the case - then am I correct in thinking 4 batteries in series this way can be substituted for a 6V dc charger? (the wiring diagram shows the terminals as -3V and 3V). Also if I substitute the batteries for a dc charger, how do I connect ground? <Q> Unfortunately, just a 6 V supply won't do it. <S> You really need a ±3 V supply. <S> Or, you can get two 3 V supplies and connect the + of one to the - of the other. <S> That common connection will be the ground of your circuit. <S> The remaining + and - ends are the +3 V and -3 V ends. <S> The latter is probably easier to source. <A> You should also look at the current requirements for that device. <S> Two small 3VDC plug-in wall transformers will work if the current they supply is adequate for your device. <S> Since you're using a cell as large as the "D" cell it is possible to find small transformers with the correct voltage but lacking the needed current. <S> Not to be picky but to help with terminology, those are 4 "cells" and not batteries. <S> All 4 cells would make a "battery." <S> Again, my only purpose with that is to expose you to correct terminology even though cells are often referred to as batteries. <A> It is entirely possible that the device needs a split supply (±3 V) and that replacing with a simple 6V "wall wart" will be insufficient.
| Basically, get two cheap 3 V wall warts.
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The role of magnetism in gravity Ok, I haven't done any experiment and please don't get all technical "what type of scales are used" But here is my question Does the weight of and electromagnet change when it on? Here is my though! If you place an electronic magnet on a weighing scale and weighed it. Would the weight change when you turned it on. Would the magnetic field generated by the electromagnet attract or repel against the earth magnetic field changing its weight in a Positive or Negative direction What's your thoughts. Dose magnatisum play a role in gravity? <Q> This has nothing to do with magnetism. <S> However, the total force between the earth and the electromagnet can change when the magnet is turned on. <S> This is because the earth produces its own magnetic field. <S> The two magnetic objects will have magnetic forces between them, depending on alignment. <S> This force on the electromagnet due to it interacting with the earth's magnetic field is in addition to, and independent of, the force between the magnet and the earth due to gravity (the magnet's weight). <A> Does the weight of and electromagnet change when it on? <S> Yes, the weight of an electromagnet changes when it is energized (but only by an extremely small amount). <S> When you put current through an electromagnet to turn it on, you are storing energy in its magnetic field. <S> As you probably know, all mass produces a gravitational field around it. <S> Energy <S> also produces a gravitational field. <S> This includes the energy stored in the solenoid (or stored in a capacitor, etc.) <S> So increasing the energy increases the gravitational field around the magnet, which increases its weight in Earth's gravity. <S> As I said, though, it's an extremely small amount. <S> The effective increase in mass is determined by the formula \$e= <S> mc^2\$. <S> Even the 340 MJ of energy stored in a big MRI magnet only weighs 4 micrograms . <S> I think the forces due to the magnetic field itself would completely swamp any attempt to measure it. <S> Here's a paper proposing to use this effect to produce man-made gravitational waves (though I would think a giant centrifuge would be cheaper). <S> (Note that the mass of the solenoid does not change when current is going through it, since the total number of electrons stays the same, whether they are moving or not.) <A> The MASS of an electromagnet does not change when it is energized. <S> There is, as far as my understanding of physics knows, no relationship between magnetism and gravity. <S> I can imagine that an experiment could be constructed where the apparent weight of the electromagnet changes, based on interaction with the Earth's magnetic field, or objects in its environment. <A> Hah, Shameless plug. <S> You can measure the magnetic force on "something" and determine it's magnetic susceptibility. <S> So we are weighing a magnet and the seeing how the weight changes when magnetic material is brought near. <S> http://www.teachspin.com/foundational-magnetic-susceptibility.html
| The weight of the electromagnet is only a function of its mass and the pull of gravity.
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How to a build a payphone controlled iPhone? I'm trying to create a project for my kitchen that involves using the shell of an old payphone and a modern cell phone ( perhaps iPhone ) to actually make calls. Basically, I would like the user to be able to pick up the receiver, press one of the number buttons, and have the unit auto-dail one of nine pre-defined numbers. With the microphone and speaker in the payphone being functional. As I try to conceptualize a way to build this, here is what I came up with. Use an external iPhone microphone and speaker and place them in the handset. Use an actual iPhone as the unit that makes the call since I don't have anologue service at my house. The part I'm a little unclear on is, how do I use the payphone button pad to somehow trigger the iPhone to make a call to 1 of 9 numbers in memory, and how do I terminate the call once the user hangs the phone up. <Q> I doubt an mobile phone would do it. <S> A GPRS Modem would definitely do the job, if it has microphone speaker terminals. <S> The key is the microphone and speaker connections. <S> Not a simple solution, the GPRS modem has a learning curve. <S> At first glance it appears to be very similar to a regular land line modem, do not be fooled. <S> it is nothing like a regular land line modem. <S> I have an old Multitech GPRS modem where the serial cable broke out to a separate phone jack. <S> Multitech no longer makes it. <S> I was also buying knock offs of the Multitech out of China with a 20 piece minimum. <S> There may be some of them floating around still. <S> Ericsson Telecom made an inexpensive unit <S> This is NOT the one I have, but the connector is 15 pin video connector not a DB-9 <S> as most GPRS Modems have. <S> Connector Pin out, notice mic and speaker . <S> PDF Manual GPRS Modem <A> The listed method will work, but is a lot of hassle when it doesn't need to be. <S> You can order a keypad for a Raspberry Pi that has the same keys a phone keypad would. <S> From there, you plug it in and set your Pi up to make phone calls as outlined here . <S> What I would do is remove the payphones keypad and attach one connected to the PI. <A> What you want to do is already on the market, usually called "retro handset". <S> Those are typically just Bluetooth keyboards + audio devices which your smartphone pairs to, sometimes with a charging connector for the smartphone itself. <S> There are also iPhone models which seem to work via the connector rather than Bluetooth. <S> You may need to modify the keypad of your payphone to make it compatible with the handset, but it shouldn't be too difficult.
| You could probably buy one and fit the electronics inside your payphone, rewiring the microphone, the speaker and the keypad to the ones the payphone provides.
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How to build a self-made Qi compatible (wireless) charger So I recently discovered that my phone, a Nexus 5, has Qi wireless charging capabilities. I was wondering if it is possible to make my own charging coil with 24/26awg wire that would look something like this (but without the driver circuit). Then, I would use a micro controller/build my own circuit to drive it at the correct frequency with a DC to AC converter. I was wondering if this would work, how precisely I would have to make the coil, and if I am missing something about Qi charging completely. I ask this because there are very few internet resources on homemade Qi coils. <Q> The Qi wireless charging protocol involves a complicated communication handshake from receiver (target device) to transmitter in order to enable charging. <S> It is not trivial and requires strict timing. <S> The standard is open but you need to be a member of the wireless power consortium to get full access. <S> See http://www.low-powerdesign.com/article_TI-Qi.html for some details. <A> Simply powering the coil is unlikely to work properly. <S> Information on the QI standard is available at the Wireless Power Consortium web site . <S> However, keep in mind that some information is only available to members; you may have a hard time finding enough information to create your own implementation. <A> You can build your own without the full spec, just need the right parts... <S> The BQ500412 is at NRND state but Digi-Key still sells them.
| QI charging requires the coil to have certain electrical properties, and involves some active communications from the device to the charger. You 100% could make your own coil and driving circuit but it would be an endeavor for the sake of bragging to do it.
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How can I add high and low voltage signals to a high voltage DC offset? It's only a theorethical question, since I'm not really going to build this circuit... I'm just curious, that's all. Anyway, thats my question! I have these signals: 3 kHz, 1V (peak) 20 Hz, 100V (peak) (DC offset is 70 V) Since I doubt it would be impossible to add them (only impractical, I guess), I was wondering: how can I create a single signal out of those parameters?... Or better: Is there a "standard" method do such things when op-amps cannot be used? EDIT: Thanks for your answers: each one is a different approach to solve the problem, so thank you, really! <Q> The answer depends on how acceptable it is for one signal to back-drive the others. <S> In the worst case, you isolate the AC signals each with their own series inductor and capacitor. <S> These would be narrow band filters to pass only the specific frequency (3 kHz and 20 Hz in your case). <S> The DC would be isolated with a "large" series inductor. <S> This would let the DC pass, but have sufficiently high impedance at 20 Hz to not attenuate that signal much. <S> The 3 kHz signal will see much higher impedance, so anything that works for the 20 Hz signal will work for the 3 kHz signal. <A> This will superimpose an AC signal on top of the DC level. <S> You can feed the primary from a ground referenced AC signal too. <A> While the other answers are perfectly correct, you should keep in mind that building a high-voltage amplifier to do the job is not Real Hard as long as the load currents aren't too high. <S> Your relatively low frequencies help in this regard. <S> A simple high-voltage MOSFET and a normal op amp will do the job if you're willing to do some work on stability and frequency response. <S> And for 14 bucks you can get an op amp from Apex Microtechnology which will give you 300 volt swings. <S> In your case you'd need asymmetric power supplies, like +250 and -50, but that's hardly impossible.
| You can use a transformer with the secondary in series with the DC source.
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Can I use two 7805 ICs in parallel to get double current capacity? I use a 7805 for a project where the circuit needs a higher current (~2.8 A) at 5 V. So I assume that if I use both ICs in parallel I can increase the maximum current capacity. Would it work? <Q> As others have already said, paralleling multiple linear voltage regulators is a bad idea. <S> However here is a way to effectively increase the current capability of a single linear regulator: <S> At low currents, there is little voltage across R1. <S> This keeps Q1 off, and things work as before. <S> When the current builds up to around 700 mA, there will be enough voltage across R1 to start turning on Q1. <S> This dumps some current onto the output. <S> The regulator now needs to pass less current itself. <S> Most additional current demand will be taken up by the transistor, not the regulator. <S> The regulator still provides the regulation and acts as the voltage reference for the circuit to work. <S> The drawback of this is the extra voltage drop across R1. <S> This might be 750 mV <S> or so at full output current of the combined regulator circuit. <S> If IC1 has a minimum input voltage of 7.5 V, then IN must now be at 8.3 V or so minimum. <S> A Better Way Use a buck regulator already! <S> Consider the power dissipated by this circuit, even in the best case scenario. <S> Let's say the input voltage is only 8.5 V. <S> That means the total linear regulator drops 3.5 V. <S> That times the 2.8 A output current is 9.8 W. Getting rid of 10 W of heat is going to be more expensive and take more space than a buck switcher that makes 5 V from the input voltage directly. <S> Let's say the buck switcher is 90% efficient. <S> It is putting out (2.8 A)(5 V) = <S> 14 W. <S> That means it requires 15.6 W as input, and will dissipate 1.6 W as heat. <S> That can probably be handled just by good part choice and placement without explicit heat sinking or forced air cooling. <A> With two voltage regulators in parallel, one might want to naturally produce 4.99 volts whilst the other will want to produce maybe 5.01 volts. <S> The "winning" regulator will be the one that produces 5.01 volts and the losing regulator <S> will basically switch itself off in an attempt to lower the output voltage <S> but, the output voltage won't lower because the 5.01 volt regulator has "won" and will provide all the current to the load until it overheats. <S> Then the "cold" regulator will take over and then it will overheat and really it ends in a bit of a power struggle (no pun intended). <S> Here's a decent looking circuit that adds two transistors around a 7805 to give significantly more current and short circuit protection: - <S> Normally, as current approaches the limit for the 7805, the presence of the 6R8 resistor drops enough voltage for the MJ2955 PNP BJT to turn on and start supplying more output current. <S> If that current reaches about 3 amps, the NPN BJT will shunt the 6R8 thus turning the PNP off. <S> Circuit taken from here and there appear to be several variants of this on the web such as this: - Taken from here . <S> Or just build a little 5A switching regulator like this but make sure your target application doesn't require a particularly low noise and low ripple voltage supply: - <A> If you need that kind of current, linear regulators are usually not the answer, as they will dissipate quite a lot of heat. <S> A ready-made, integrated switcher will stay cool and use less space. <S> Here is a selection of switching converters for 5V, 3-5A output. <S> Another one And another... <A> If you want a 7805 that can handle more current, use a STS LD1085V50, 5V, 3A <S> On Semi KA378R05TUTI LM1085IT-5.0Exar <S> SPX29300T-L-5-0 <A> Yes you can, however, you need to isolate them from each other which will lower the output by about .707 volts each, the voltage drop of the silicon blocking diode that you would need to install on the output of each one, before putting the output of the diodes in parallel. <S> It is simpler to use a bypass transistor or even a higher output regulator. <S> Just keep in mind that the filtered but unregulated input current to the regulator circuit must be greater in amperage than your desired output, in order to maintain regulation, if the input current drops below the set output current, there is no telling what can happen, from circuit damage, to oscillation of the output, which would be the equivalent of feeding 5v AC to your circuit being powered. <S> And Yes I have seen this happen when this exact same thing was tried, in lab, when I was a student, and another student tried this exact same setup, feeding the regulator circuit from a 12 volt 1 amp regulated power supply. <S> The input voltage was monitored on an Oscope and never changed more than a handful of millivolt downward, but the output of his circuit was a flat topped high frequency pulse in the 1000 Hz range
| Short story is that you can't reliably or cleanly get twice the current from two paralleled voltage regulators that ostensibly produce the same output.
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How can 16 buttons be connected with only 8 wires? Here is the product. I understand the idea: these 16 buttons use a 4x4 matrix. 4 lines for rows, 4 lines for columns, and we have 8 cables. But: How can this work without any multiplexer ? Can this detect accurately presses of multiple buttons? Even if buttons use same row or same column? Example: buttons at position (2,2) (2,3), (3,2), and (3,3) pressed at the same time. How does it work? <Q> How can this work without any multiplexer? <S> It doesn't. <S> The keypad board seems to have just switches, with maybe some diodes we can't see. <S> However, the left board looks like it has a processor on it. <S> Almost certainly, multiplexing is being done in firmware. <S> The multiplexing algorithm works something like this: <S> Drive one row high, the others low. <S> Enable passive pulldowns on the column lines. <S> See which column lines are high. <S> The buttons at the intersections of those column lines with the one asserted row line are pressed. <S> The other buttons on that row are released. <S> Repeat back to step one, asserting the next row in sequence. <S> The above process is repeated fast enough so that all the buttons are checked within a time that still feels instantaneous to a human observer. <S> The human limit of "instantaneous" in this context is about 50 ms. <S> Even a low end microcontroller can scan a 4x4 keypad in much less time than that. <S> Can this detect accurately presses of multiple buttons? <S> Probably yes. <S> Those could be on the bottom side of the board you show. <S> With the diodes, the algorithm described above just works. <S> Another way is by putting resistors in series with lines and measuring their analog voltage. <S> Even with multiple buttons shorting multiple row/column lines together, you can eventually figure out which buttons are pressed. <S> This requires A/D inputs in the micro, not just digital inputs as when diodes are used. <A> Without diodes... you'll get phantom rectangles. <S> For instance in your example (2,2) (2,3), (3,2), and (3,3); you chose an ironic example by choosing a rectangle. <S> If you actually press all four, it will work. <S> But if you press any three corners of the rectangle, the fourth corner will also appear to be pressed, even though it's not. <S> That keypad is clearly meant for keyboard data entry, where the user convention is one keypress at a time. <S> You notice there are no "shift" keys on a gas station credit card entry keypad, for instance. <S> However, if you made S1 a shift key, and S3 an alt-shift key such that users might reasonably hold both while also pressing S13, then S15 would also appear to be pressed. <A> Here's how to do it without continuous scan: <S> Set all drivers to high Enable passive pulldowns on receiving lines Set pin change interrupt to detect one of the lines going high (or changing in level) <S> When this happens, enable keypad scan, say every 10ms Scan normally and process key presses After a timeout without events, stop scan and restart at first bullet point <A> I have done this with Assembly Language on bare Atmel AVR microcontrollers. <S> For simplicity:Keypad Rows: use Arduino Pins 4,5,6 and 7 <S> (AVR pins PD4, PD5, PD6 and PD7) <S> Keypad Columns: use Arduino Pins 8, 9, 10 and 11 (AVR pins PB0, PB1, PB2 and PB3)Make Row pins inputs with pull-ups enabled. <S> Make Column pins Output and output zeroes to them. <S> Enable Pin Change Interrupts on all row and column pins. <S> http://playground.arduino.cc/Main/PinChangeInterrupt Pressing a button on the keypad will pull a Row pin low. <S> The interrupt routine needs to read the row pins and find which pin is low. <S> The upper 4 bits should be three ones and one zero. <S> If you use a uint8_t 8 bit variable you can divide it by 16 or (var >> 4) to put the 4 bits in the lower bits of the number. <S> You can use a bitwise OR | operation with 240 to set the upper 4 bits to 1s and a bitwise NOT ~ operation to invert all the bits so you only have one bit set representing the row the button press was on. <S> Store this number as 0, 1, 2 or 3. <S> Switch the pin configuration: <S> Make Column pins inputs with pull-ups enabled. <S> Make Row pins Output and output zeroes to them. <S> Read the column pins and find which pin is low. <S> Do a similar manipulation with this value except you don't have to shift the number to the right. <S> After the var ^ 240 <S> and ^var you have a single bit to represent the column the key press was on. <S> Remember to disable interrupts before reconfiguring ports as it will likely trigger unwanted interrupts. <S> You may have to clear interrupt flags when leaving the interrupt routine to prevent duplicate interrupt handling.
| One way is to put a diode in series with each button.
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How do you find out if a MUX is bidirectional I've used this MC74VHC157 in a project thinking this would be bidirectional but I'm not sure. How do you find out if any MUX is bidirectional? <Q> Digital mulltiplexers are almost certainly uni-directional. <S> Analog multiplexers probably are bidirectional. <S> However, studying the datasheets should tell you for sure. <A> Its unidirectional, anything with gates is one way. <S> There are some digital logic ic's with pull down logic that is bidirectional, ususally they will advertise the bus that it works for (like I2C) <S> Not all analog muxes are bidirectional, the ones that aren't show input and output and usually have a pmos and nmos. <S> Bidirectional muxes (switches) will usually have a picture of a switch on the block diagram in the datasheet. <A> The datasheet shows that the pins are distinct inputs/outputs: What you search is usually called an analog switch , and the datasheet would show a symbolic switch (e.g., NLAS1053 ) or the actual implementation, <S> an N-channel/P-channel FET combination (e.g., SN74LVC1G3157 ):
| The 74xx157 is clearly a uni-directional digital multiplexer, as shown by the logic diagram in the datasheet.
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Can a time-varying magnetic field pass a metal sheet? I'm using CT (Current Transformer) to measure the current flowing on the power line. The frequency of the current is 13.56 MHz. CT appeared to be metal-shielded (This metal case of CT may be to be grounded). CT has a toroidal shape and the power line under the measurement passes through the center hole of the CT, so azimuthal magnetic field around the power line induces EMF (ElectroMotive Force) on the coil inside the CT. This EMF is measured so the current on the power line is measured. This is the basic story of how CT works. If the current frequency is low like 10 Hz, then I fully accept that CT really works. But I'm now confused at a high-frequency operation. The current is the source of the magnetic field and when the current oscillates, there is not only the magnetic field but also the electric field. The combined field is what we call an EM (ElectroMagnetic) wave. My frequency is 13.56 MHz which is a rather high-frequency, so the power line should play as an EM wave emitting antenna. EM wave generated from the power line propagates to CT first. However, CT has a rather thick metal case (much thicker than the skin depth at this frequency) so EM wave will be reflected from it. It means the magnetic field as a part of the EM wave fails to reach the coil inside the CT so CT should not work! I think the only way for CT to work is that the magnetic field alone pass through the metal so they get in touch with the coil. But..Is it really possible? The time-varying magnetic field can exist alone without the electric field? Of course, there is a number of commercial CT in the same type which work well even for higher frequency. Could you please give me some idea of breaking this confusion? <Q> The time-varying field around an AC conductor wire is in the circumferentialdirection. <S> So, it induces current in metal that is perpendicular to thatcircumferential direction, i.e. in the radial/Z axis plane. <S> The currenttransformer ought to have no such metal in its shield 'container'. <S> With a sufficiently low burden resistor, AC performance of transformers caneasily be quite good; that burden resistor and its secondary currentact against the B field penetrating the core, so magnetic core lossesare kept small (because internal B field excursion is small). <S> As for 'electric field', a good conductor can get quite high inducedcurrent from nearby AC wires, but a good conductor completely blockselectric field very well: that's why electromagnetic waves (light) reflect well from the thin aluminum layer of a common mirror. <S> The field lines are an abstraction for visualizationpurposes, not trajectories of particles that can ... bounce. <A> It is the losses in the core at high frequency that will make the CT not work at all. <S> Refer to Eddy current losses . <S> Measuring large 14 MHz currents is not trivial and you will you will need a well designed axial current shunt, CT or transducer. <S> (By the way, the magnetic field in normal CT operation is confined to the toroidal core and the metal casing plays no part.) <A> A single conductor generates a circumferential magnetic field around it. <S> If a single conductor is looped through a magnetic core, then this magnetic field will induce flux in the core. <S> As the first wire links the core, this will look like a series inductance to the current in the wire. <S> If a further wire is now looped through the magnetic core, and short-circuited, the emf generated in the loop will cause a current to flow in this second wire, which will tend to reduce the inductance seen by the first wire. <S> This is the basis of a current transformer. <S> In practice, the second loop is replaced by a large number of turns, 1000 is a popular figure, which reduces the current flowing in the secondary by the same factor. <S> It also allows the secondary to be 'short-circuited' <S> by a much higher value resistor, 50 ohms is a popular choice so the secondary can be measured with a normal oscilloscope. <S> If there is a further loop of conductor around the core, the secondary current will divide between the loops, compromising the gain accuracy of the intended output. <S> If a conductive tape is placed over the secondary for electrostatic shielding, it will be insulated from itself (in a professionally made current transformer at least) so that it does not form a complete conductive turn through the magnetic core, so that even though an emf is generated across the facing edges, no current flows through it. <A> I've gotten a lot of comments and answers on this question. <S> Thank you so much. <S> I've read all of these <S> and I tried to summarize them. <S> I posted my own summary here, so if there is a need to correct it, please give me some comments. <S> CT (Current Transformer) typically has a metal case <S> so you might wonder how it works at a high-frequency current. <S> The primary conductor carrying the high-frequency current plays as an EM (Electromagnetic) wave antenna so B-field (magnetic field) becomes a part of the EM wave. <S> In order for CT to work, the B-field must reach the coil (secondary winding in CT diagram) of CT. <S> But the metal case which is much thicker than the skin depth at this high frequency will reflect the EM wave so it appears that the B-field doesn't touch the coil! <S> If the metal case has no opening, then, yes, there is no way for CT to work at high frequency. <S> However, if the metal case has the opening, the situation changes. <S> Think about toroidal CT which metal case is split into two equal pieces (the plane of splitting is the plane of the largest circumference of CT) and the separation between the pieces is fairly small. <S> The EMF (Electromotive force) <S> E-field, which is a dual to B-field in Maxwell-Faraday equation, around the surface of the metal case induces the current running the case surface in the plane of the primary conductor and radial vector from it. <S> This current flows into the opening, running the inner surface of the case, and returns to the outer surface, to form a closed loop. <S> The current on the inner surface is equal in magnitude <S> but opposite in direction to the current on the outer surface. <S> The B-field induced by the inner surface current has the same direction to the B-field outside. <S> The field magnitude may also be the same (exact confirmation of this requires rigorous thinking of applying the Maxwell-Faraday and Maxwell-Ampere equation on both sides). <S> As a result, the B-field from the primary conductor with high-frequency current reaches the coil of the CT via induced current on the metal case which flows into the inner surface of the case through its opening, so CT works.
| Magnetic field does not 'pass through' anything, it is not a flowof some material.
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Can I use aluminium containers to transport ESD sensitive PCBs? I want to transport PCBs mounted with LEDs which are ESD sensitive, in knocked down condition. Standard storage bins are not available in my specific size, hence was thinking to fabricate aluminium bins as per my specification. I dont have the option of grounding the bins, cause I have to transfer the PCBs regularly form one section to another. Please help!! <Q> Conductive bins are just fine for transporting ESD sensitive boards. <S> Any protective packing in the bin to avoid mechanical damage to the boards must be conductive, for instance conductive foam. <S> The ground conductor(s) of the board should be in electrical contact with the bin. <S> You must also make sure you load and unload the bins correctly. <S> Do not touch the boards in the bin without taking the same precautions as for loading and unloading. <S> When you load a bin, the board and the bin must be at the same potential. <S> This means electrically connecting the bin to the board or board handling kit before, and during, the transfer. <S> This is perhaps most easily done by putting the bin on your anti-static workstation mat, or briefly connecting a ground lead to it. <S> The same goes for unloading a bin. <S> The bin must be at the same potential as the surface or kit that is to receive the board. <S> Once the bin is loaded, it can be disconnected from ground, and freely taken to any potential. <A> I did my own manufacturing of a telecom product of 20 years. <S> When I started out telecommunications manufacturing was regulated by Bellcore. <S> Manufacturing floors had Bellcore Quality Control Inspectors on site. <S> ESD was a major part of quality control. <S> Some products were manufactured at Siemens Telecom in Lake Mary FL, which met the most extensive of Bellcore requirements. <S> My product was housed in an anodized aluminum case. <S> The answer to your question is: <S> Almost all aluminum products are anodized to prevent oxidation and is inexpensive. <S> You are not the first to ask this question. <S> There was a study done on ESD on Anodized aluminum for ESD transport container used in automated handling equipment to see if anodized aluminum passes tests set by ANSI ESD standards . <S> LINK to Study: Anodized Aluminum Alloys Insulator or <S> Not Regarding some of the comments about ESD handling, they made my skin crawl. <S> Procedures were not touch this or that first then touch that. <S> Procedures are specified by organizations like ANSI and Electrostatic Discharge Association. <S> ESD Association has a very general basic procedures. <S> Keeping in mind these guys sell the ANSI documents so they are not going to give it all away. <S> A guide for implementing the standards is a better and more practical thing to know. <S> NASA has it own ESD manual which follows ANSI procedures <S> and they do not charge it. <S> It is an overall comprehensive handbook for implementing <S> ANSI/ESD S20.20 Chapter 8 <S> , ESDS ITEM HANDLING, is all you should need to know. <S> NASA HANDBOOK, WORKMANSHIP MANUAL FOR ELECTROSTATIC DISCHARGE CONTROL <S> LINK: Basic ESD Control Procedures and Materials <S> , EOS/ESD Association <S> The above should cover the topic, if for some reason it does not, there are volumes of ANSI/ESD standards. <S> The two most applicable would be Wrist straps and Handling Equipment. <S> LINK: <S> ESDA Document List <S> LINK: Discharge Susceptible Items, Wrist Straps 2006 <S> LINK: Discharge Susceptible Items,Automated Handling Equipment 2007 <A> You can safely load and unload PCBs to /from bins, by holding the GND trace of the PCB ----- with your hand ----- and then touching the metal bin <S> ----with <S> your other hand ---- <S> before placing the PCB into the bin. <S> Key is to discharge any static buildup between you/PCB/bin before moving the PCB. <S> This same procedure works on removal from the bin: touch the bin with one hand, touch GND of PCB with other, then remove the PCB. <S> The ESDstrap is just a way of getting a high-value-resistor between you and the bin, so the static does not build up. <S> Of course, if you go shuffling your shoes on carpet as you touch the PCB, that ESDstrap may not keep the voltage to ZERO as you shuffle. <S> There are reasons to think about these methods, not just blindly trust.
| Yes. , if you anodize the aluminum with Type II or Type III (hard coat, thicker coating), which uses sulfuric acid, the most common methods.
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How to Boost an AC Voltage signal from 18 Vpp to 30Vpp range I am interested in creating a portable diagnostic tool, and as such would like to passively (no wall plug) amplify the 18Vpp sine wave signal that I have (4-7 kHz range) to about 30 Vpp to adequately power a disc style PZT on my device. I have searched extensively and have not been able to come up with a good solution thus far. Alternative ideas require using some sort of MOSFET circuit still requiring a larger external voltage. This is not ideal, since I was hoping to use the setup and signal I have right now to power my device. Thanks. <Q> I suggest you to find a suitable transformer, maybe with an amorphous core. <S> Start to compute anything at lowest possible frequency and maximum voltage amplitude, because the flux will be largest at low frequency. <S> Then, if you put a signal with higher frequency or smaller amplitude, anything will still work normally. <A> If the frequency was fixed, you could have used a LC resonant tank, but frequency is variable. <S> Another solution would be to use switching converters to get higher DC voltages, and power your signal amplifier from these. <S> Both have merits in terms of size, weight, efficiency and cost. <A> A normal audio transformer (20 Hz to 20 kHz) will be fine as long as it it properly derated. <S> $$ <S> V_1 <S> \cdot <S> f_1 = <S> V_2\cdot f_2$$ <S> This means that you can maybe use a 15 V p-p to 9 V p-p audio transformer or even something like a 5 V p-p to 3 V p-p transformer.
| So, if you want to passively increase voltage on a signal, pretty much the only solution you have is a transformer.
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Purpose and explanation of resistor near output of LM317, high-current, adjustable-regulator circuit I'm trying to make a high-current, adjustable-voltage regulator using LM317 data sheets. Specifically, I am using the schematic below. So, what is the point of the resistor related to Note A? Is it supposed to represent the load? What does it mean that the minimum load current is 30 mA, and can or should that value change? Also, what determines that value? Basically, I'd be happy to learn as much as possible about that resistor. Edit: The figure above is incorrect in that the TIP73 is depicted as a PNP transistor. The figure below is from a more recent LM317 data sheet, correctly showing the TIP73 as an NPN transistor. <Q> Note A says the minimum load current for this circuit is 30 mA. <S> Without the 30 mA minimum load, the the circuit will not regulate correctly - the output voltage will probably rise. <S> The LM317 by itself (without the additional transistors shown here) has a minimum load requirement of 5 mA, which the recommended voltage adjustment resistors will draw, so no "extra" load resistor is required in that case. <A> In case your load might draw less than 30 mA, you add this resistor to make sure the regulator is always outputting at least 30 mA. <S> If your load might be totally disconnected, you choose the resistor to draw 30 mA at your (minimum) output voltage. <S> If the load might at minimum draw only 27 mA, you choose the resistor to draw 3 mA at your (minimum) output voltage. <A> The technical reason why 30mA must be drawn is to bias both transistors to the starting threshold of conduction so that they can begin to be active to bypass the regulator with a current sharing ratio defined by the other resistors. <S> This begins when the voltage drop across the 22 Ohm R <S> *30mA <S> = 660mV. <S> Plan B Works better than original for variable output voltage. <S> Change TIP73 from PNP to NPN like 2N3055. <S> V drop & I determines power dissipation. <S> I have simulated the LM317 with an emitter follower. <S> I also simulate two input options a) fixed 5V out and b) adjustable 2.5 to 7.5 triangle sweep <S> I used an ideal transistor as a dummy active load. <S> Now the 330 Ohm pre-load can be removed completely (still shown) as the bypass PNP is inactive with no load until the load current exceeds 20mA <S> then both bypass transistor take all the extra current. <S> Does this closed loop have a name? <A> I am using the schematic below. <S> I'm pretty sure the schematic is wrong. <S> as is, it is wired for a npn power transistor. <S> it can work with a pnp but <S> the wiring is slightly different. <S> So, what is the point of the resistor related to Note A? <S> I guess they are trying to get a minimum amount of current so the pnp is conducting -> <S> 22R * 30ma = ~0.7v. <S> Not necessary in my view.
| If the circuit will not always have a load of 30 mA or more, you need a resistor there to draw 30 mA and satisfy the minimum load requirement.
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How do I properly shield wires to my stepper motors? I have a project where I need to run stepper motors within a few centimeters of wireless RF transmitters and receivers. On the breadboard I have experienced some strong interference. I am purchasing shielded wire so that I can avoid it but I am concerned about how effective my shielding will be. I will have fully shielded wire running from the stepper to the driver but I am concerned about the open soldered connections to the board. Will this also cause interference? I also will have to attach this shielded wire to the unshielded wire coming out the the stepper. I will make sure I do this as close to the stepper as possible but there will inevitably be some un shielded wire. Is there anything I can do to further avoid interference at these connection points or am I being nitpicky and it won't matter to much? <Q> The radios will perform well as long as you do not overload the first transistor amplifier (the LNA). <S> That means run lots of current through that transistor. <S> Also, do not share GROUNDS with the motors or power drivers, even inadvertently; thus avoid Efields and Hfields coupling into Radio VDD or Radio Ground. <S> Another trick is to use LOW Data Rates, so motor transients are short compared to a radio data-symbol duration. <S> The shielding effectiveness of cable strongly depends on the percent-coverage of the cable-shield; thus a foil-shield should capture more of the electric field. <S> For magnetic shielding, you need the magnetic fluxes to CANCEL; that means the stepper-motor current must have NO OTHER PATH than the cable. <S> Since there will be parasitic capacitances inside your motors, there will be currents/charges flowing through wiring-insulation to the CASE of the motor. <S> You need to capture that current. <S> I suggest you evaluate attaching the CASE to the cable shield, while electrically isolating the CASE; in a plastic drone, that may be simple. <S> EDIT Standard PCB foil is 35 microns thick (0.035 millimeters, or 1.4 mils) and weighs 1 ounce/foot^2. <S> Frequencies at 4MHz ( <S> 125 nanosecond Trise, Tfall) will be somewhat attenuated; frequencies at 16MHz (30nanosecond Trise, Tfall) should be attenuated by 2 Nepers (2 <S> * 8.6dB); that atten of 17.2dB is not much to protect your RF Receiver front end circuits from the overloading magnetic flux of 1 amp being switched in 30 nanoseconds. <S> PI matching circuits should be chosen to provide High Pass Filter behavior, not Low Pass Filter behavior. <S> That means the components both before and after the PI matching are contributing to reduction of induced Magnetic field overload energy of that first RX amplifier. <S> Thus capacitors into the PI and Out of the PI are guaranteeing High Pass Filtering; if you see a DC path from antenna to the LNA base or gate, you have failed. <S> One of you questions is "the open connections soldered to the board". <S> Yes, those matter, because the distance and area from RTN to HOT wire are the loop area creating a flux that will overload the radios. <S> Make that area as small as possible. <S> Something like this is good: <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Use twisted pair shielded cable. <S> Connect each coil with respective pair conductor and connect shield to earth/ground. <S> I would be more concerned about the EMI in your PSU that can interfere with power supply of the RF receiver/transmitter . <A> A stepper motor and its drivers do not generate RF nor sense it. <S> Theintelligence behind the driver transistors, on the other hand, might be sensitive. <S> A common-mode choke (one of those big ferrite bead things) with thestepper drive wiring shoved through it can be put next to the drive transistors. <S> That keeps the motor and wiring from being an antenna. <S> Ground theconduit, of course.
| If the cable to the stepper is loose wires (not twisted) there can be somedifferential pickup, so keep the wires together (or fit them intoa flexible metal conduit, available at hardware stores).
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Is better to have holes on different sides of the shield instead of the same side? I suppose the best is to have no hole on the shield, but that is practially impossible if you have to let cables come out of it. So I read somewhere that having the holes on different sides of the shield (assuming that a shield is like a box around some electrical equipment, like a chassis around PC hardware) is better as opposed to having them on the same side. If we have a box shield, with 6 sides, drilling 3 holes on 3 different sides is better than drilling 3 holes on the say upper side? <Q> I'm not sure it makes much difference. <S> But if the holes remove a majority proportion of the side it's probably not good. <S> However, if it's cables coming out as opposed to holes for plugs to go through, having cables on multi-sides can be an assembly nightmare. <A> If you are serious about keeping trash separated by that shield, you'll need filter networks (bypass caps, or even "L" filters) that bolt into the shield, allocated to EACH WIRE. <A> They are generally pretty small, so at the frequencies of greatest interest (usually < 1 GHz), they have very low leakage. <S> Long seams are a problem (even if extremely thin), but not your average 1/2" - 3/4" diameter hole. <S> There are higher frequency signals (wifi radios, PCIe/SATA signaling, cpu cores), but those are usually well confined, individually shielded -- for instance, most wifi radios come as shielded modules with an antenna output. <S> I am not saying you can't have EMI problems at 5 GHz, but more often you get into trouble in the 100s of MHz. <S> The problem with holes is that you usually have wires going through them. <S> Wires act as antennas if they are not properly shielded, grounded, and filtered. <S> So I don't know if holes on the same side vs. opposite side are technically worse, but it is a relatively minor issue compared to what you do with your cables and their shields.
| Holes per-se are not usually a big problem. In particular, you really want to have your cable shields electrically connected to the enclosure right at the point of penetration, otherwise it can conduct noise from inside the enclosure to the outside.
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Changes when \$V_{be}\$ is increased in npn transistor For the given circuit, a statement has been given in my book, saying It is obvious that if Vbe is increased by a small amount, both hole current from emitter region and the electron current from the base region will increase. As a consequence both Ib and Ic will increase proportionally I believe there is a mistake and it should be rather, electron current from the emitter region Because BE seems to be forward biased and thus the majority carriers, that is, electrons from Emitter should increase and go into the base and further in collector. Am I right or wrong? <Q> When you increase the base-emitter junction voltage the electron current injected into the base (and thus the collector current) increases but also the hole current injected from the base into the emitter increases similar to an isolated pn-junction. <S> This hole current is usually the dominant cause for the base current flowing into the base. <S> Thus the collector current and the base current increase proportionally to each other when you increase the base emitter-voltage. <A> You are correct and the book is wrong if interpreted as follows. <S> The emitter majority carriers in an NPN BJT are electrons that are injected into the base. <S> Holes from the base cross the b-e junction and recombine as minority carriers with electrons in the emitter just as electrons recombine as minority carriers in the base with the majority-carrier holes, resulting in base current and finite beta. <S> However, the book is referring to the minority carriers in both emitter and base and as such, they are holes in the emitter and electrons in the base. <S> So both you and the book are right if each is referring to the right currents, minority or majority, in base and emitter. <A> You can take into consideration the formula <S> Ic= <S> Is exp(Vbe/Vt), so obviously if Vbe rises the collector current rises, and if transistor is directly active, you can use the formula Ic=beta Ib, so obviously if Ic rises, Ib also rises.
| The statement of the book is correct.
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What's an Air Gap Layer in a PCB? At work I've inherited a multilayer PCB design that I need to send out for quote and eventual fabrication. It contains two inner layers that are labeled "AIRGAP". What is the purpose of these air gap layers? The board stackup is as follows: 1. Top Silkscreen 2. Top Soldermask 3. Top Copper 4. Ground Layer 5. Ground Layer Airgap 6. VCC Layer 7. VCC Layer Airgap 8. Bottom Copper 9. Bottom Solder mask The highest voltage on the board is about 40 volts, so I wouldn't think it's a high-voltage design. Would this be considered a four-layer board, or more? Some of the board houses we've sent it to are confused as well. <Q> As Peter Bennett said, the air gap layer is probably a Gerber containing areas to be milled out of the layers, possibly the top and bottom prepreg, leaving the core intact. <S> Since there are only 4 copper layers, this would likely leave open cavities on the top and bottom with copper potentially exposed on the power/ground layers. <S> In some cases, components are completely embedded into the PCB. <S> I believe this process typically would have the (in this case) core run through a pick and place machine, soldered, cleaned and then laminated and the holes plated through with the top and bottom prepreg. <S> Here is an example of a stackup with completely embedded components from Altium : <A> An air gap is for creepage an clearance for high voltages to meet regulatory. <S> I'll bet the designers have a different depth on the PCB for the milling trace and they use that distance in the stack up to achieve a custom depth. <S> This is probably so the depth will show up in the 3D design or for manufacturing, and a milling track could be created with a custom depth in the PCB. <S> So if the design is for a power supply or something with creep-age and clearance then that's what it is. <S> If it actually is an air gap layer I'd be shocked. <S> Edit: <S> One other place I have seen air gaps (which this probably is) is in rigid flat flex PCB's which have kapton inner layers and FR4 outer layers. <A> An air gap is a physical less than conductive distance between two sections of a electronic circuit. <S> It is intended to enforce a non conductive section between two points using non conductive (in normal circumstances) material. <S> This air gap is chosen based on the typical working voltage of the circuit. <S> A mains voltage air gap will be smaller than an air gap for 1k volt or higher circuits, for example. <S> The spacing between two multi killivolt paths will be much larger than the spacing between two bare mains voltage paths. <S> The typical air gap is calculated based on the conductivity of atmosphere (a mix of various gases). <S> Of atmosphere would conduct at that voltage at a given distance, the air gap is not enough.
| The air gap is to promote flexibility if you have more than 2 kapton inner layers as shown in the 8 layer stackup. This could be used to recess components into the PCB.
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Does Shielding "Electrically" shield "Magnetically" too? I know it sounds like a newbie question but I can't wrap my mind around it. An electromagnetic field is electric + magnetic field. So this means that when shielding an equipment offensively, for example from avoiding causing interference with other electronics, we need to shield the electromagnetic waves, that means both electric and magnetic shielding. So if we put say a radio inside an aluminium box, aluminium is pretty much the most cost effective material you can find. Some may use copper but aluminum is more cost effective. Now an aluminum box will shield the electrical field very efficiently given if the box has no holes or seams, or if the cables coming out of holes are properly shielded and grounded. But what about the magnetic field? Aluminum has very low permeability. So how can the aluminum box shield nearby equipment from the magnetic field of the radio inside it? It does shield the electrical field, but not the magnetic? Can somebody explain to me how shielding works with electrical/magnetic waves? Because I can't wrap my head around it how can it shield the electrical part but not the magnetic? Does magnetic field leakage pose any noise danger to the nearby equipment from this theory perspective? <Q> You would not be alone in this one. <S> This is an often misunderstood phenomenon. <S> Static magnetic fields can not be shielded. <S> They can be re-directed using ferrous materials but even those will not block them. <S> Electric fields on the other hand can be. <S> Since an electric field is basically a voltage in space, they can not pass through a conductive plate that is held at a fixed potential. <S> Space is shorted out as it were. <S> Alternating magnetic fields of sufficient frequency however, will not pass through a metal plate. <S> The alternating field generates an eddy current in the plate which generates a cancelling magnetic field. <S> This is all explained in much better detail here.. <S> Wikipedia <A> In a box, the distance from circuit-to-shield may not be adequate to develop an electro-magnetic-wave. <S> In that case, you can validly consider the Efield separate from the Hfield. <S> The sea-of-mobile-electrons in metal is very effective for Efield shielding; the electrons roam to where needed on the metal's surface, to oppose the incoming Efield flux lines, coercing that flux to only impinge on the shield metal at exactly 90 degrees. <S> The ratio of Magnetic Permeability to Electric Permittivity hints at dramatically different effects between Hfield and Efield shielding. <S> Standard 1 ounce <S> /foot^2 copper foil of 35 micron thickness gives some attenuation (a few dB) at 5MHz. <S> At 50 MHz, that same 35 micron provides <S> sqrt(10) <S> * dB/Neper, or 3.14 <S> * 8.9dB = 28dB attenuation. <S> At 500 MHz, that 35 micron provides 10.0 * dB/Nepers, or 89dB attenuation. <S> To begin to shield against 60Hz, you need sqrt(5,000,000/60) <S> ~~ sqrt(100,000) = <S> 316X more thickness; thus 35micron * 316, about 10,000 microns, or about 1 cm. <S> For magnetic fields, aluminum and copper have nearly the same behavior. <S> Mu is same for both; differences appear from their different conductivity. <S> Aluminum instantly tarnishes, so you cannot solder to it. <S> Copper is easily soldered, using a big hot iron. <S> Regarding your question about noise-danger to nearby equipment, the answer is YES. <S> Signals can interfere with each other. <S> Check out my answer to "Distance between SPI traces..... <S> " question. <S> {edit} High voltage Efields cause lots of charge movement. <S> If the frequency is low, you'll get detectable EXTERNAL movement of charges due to the Efield. <S> In other words, the SkinEffect is your friend but SkinEffect only predicts attenuation; SkinEffect does not prevent external charge movement. <A> I do not know the theory that <S> well <S> but I can tell you what I saw being practiced at Qualcomm when I used to work there about 15 years back. <S> So while conducting tests on the phones/chips (such as reference sensitivity tests) we placed the phone in a metal box about 50cm x <S> 35cm x 20cm. <S> From the color of the box it seemed more like copper than aluminium <S> but I guess you can put on artificial colors. <S> There was a wire that carried the signal to and from the outside world. <S> For more sensitive testing the phone along with other test equipment was placed inside a metallic cage, the size of a small room. <S> There were all kind of other precautions that we took so as not to influence the test results. <S> Just to clarify the signals the phones carried were GSM/GPRS/WCDMA signals in the range from about 900MHz to a few GHz.
| Magnetic shielding varies with frequency.
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Smart Light Switch power supply solution for 2-wire (without neutral) scenarios Considering a smart switch based on a power greedy WiFi ESP8266 ( 800mA spikes and 100mA on idle, @ 3.3V ), there is the common situation to have the neutral wire missing from the wall switch box (left with 2 hot wires, maybe some earth wire that I don't want to consider in any way). Powering the electronics in the smart switch from the 2 hot wires will draw current through the light bulbs, making them either flash or light up (maybe except for some incandescent ones, but I can't rely on the light bulb type in any way; all usual types ranging from 1W to 100W, 110/230VAC should be considered). At this point you might have already guessed the problem: what kind of smart switch design can bypass this issue? Some suggestions I read were to use an extra capacitor in parallel with the light bulb so that the current drawn by the switch won't get through it. Although I'm a beginner with electronics and I don't fully understand how exactly should that work, or how to scale the capacitor (maybe someone could explain me), I only wonder why didn't the LED/CFL bulbs manufacturers just include this feature in their products, since this seems to be an universal issue. Are there any reliable solutions to this problem? <Q> What kind of smart switch design can bypass this issue? <S> I have not seen a Wi-Fi switch that doesn’t need neutral. <S> Perhaps this can be done, but I guess the power consumption is such that manufacturers choose different protocols. <S> There are popular generic RF (433 mhz, etc.), z-wave and ZigBee switches that work without neutral. <S> All those protocols are designed around low-energy standards. <S> Some can be used with a generic X to Wi-Fi gateway which is powered from mains. <S> Why do some bulbs require a capacitor in parallel for non-neutral-wire switches to work? <S> The switch requires some current to pass even in “off” state. <S> If the bulb doesn’t let the current pass or if it starts to flicker as a result of this current, then you will be instructed to install a capacitor in parallel to the bulb. <S> Why does it solve the problem? <S> The capacitor forms a capacitive dropper (together with some components in the switch) that bypasses the bulb. <S> A capacitor in an AC circuit forms a current limiter because when current flows one way the capacitor lets it pass until the capacitor is charged in one polarity and then when the current is reversed the capacitor discharged and lets the current flow the other way. <S> The amount of current depends on the capacity of the capacitor and the frequency of the AC. <S> Also see this relevant discussion . <S> Why isn’t this standard in all bulbs? <S> Because it would waste energy. <A> When the switch is off, we can't allow any current through, or LED light bulbs will flicker. <S> Therefore, we need a low standby current. <S> WiFi is therefore out of the question. <S> This leaves: Infrared Bluetooth Low Energy <S> Infrared would be very nice, because standby power is ridiculously low. <S> However, we might want to place the command button in another room, and it doesn't go through walls. <S> When the lamp is on, the circuit can be powered. <S> For example, make the switch drop 3.3-3.5V or so, which is easily done with a triac and a couple diodes. <S> This voltage will be used to charge a small NiMH coin battery. <S> BTLE is great for stuff that transmits whenever it wants, but here, the device will need to periodically wake up and ask its master if it needs to turn on. <S> It can't receive orders while sleeping. <S> Therefore, its power consumption will be higher than expected. <S> OK, let's say your Bluetooth Low Energy device needs about 1mW to connect 10x per second in order to know if it needs to flip the switch or not. <S> I fudged the figure from this paper . <S> Now, how to power the device at 1mW average... <S> CR2032 cell (3V 225 mAh) : 1 month. <S> Too short. <S> 2x AAA batteries (1Ah) : 4 months, why not... <S> Still sucks. <S> Would fit in Euro size switch, probably not in US size. <S> Supercap 1 Farad 5V: 1.5 hours. <S> No go. <S> Rechargeable 100mAh 3.6V cell: 15 days. <S> All these options suck. <S> The rechargeables would need the switch to be turned on periodically to recharge. <S> Lowering the polling frequency 10x would lower power draw 10x also, but make the switch sluggish. <S> Another option would be to bite the bullet and allow some leakage current... <S> but how much? <S> It only needs 300µA at 3V3. <S> That's enough to make a LED lightbulb go blink. <S> Therefore, I propose to change the problem: If you put the received and controller inside the light fixture, then it can have its phase and neutral wires, and be powered constantly, as needed. <S> This leaves the problem of the old switch now being a hole in the wall. <S> Simply replace it with a coin cell powered transmitter, which will have the same role as the other wireless switches in the house. <S> No need to connect it to the wires at all. <A> While the switch is off the voltage across the switch can be used to light up a LED and charge a supercap. <S> A triac can be used to close the switch. <S> Once the triac runs out of power the triac will switch off, allowing the supercap to be charged again and presto... <S> The circuit can also be designed to switch off for half-a-cycle every now and then to re-charge the supercap before it runs out of power.
| Unless you need some current to pass constantly to power this special kind of switch you shouldn’t install such a capacitor with the bulb.
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Is the ATtiny85 actually used in any consumer electronics? I was wondering if anyone here knows of any products that utilize the ATtiny85. It appears to me that it's solely used by makers/electronics enthusiasts. I'm asking because I think it would be interesting to see to what products its limited IO and storage can be applied. Sorry if this is the wrong place to ask a question like this - I couldn't find a more applicable SE site. <Q> The answer is yes. <S> Is that surprising? <S> no. <S> Why? <S> Because Atmel can't care at all about a couple thousand hobbyists buying a couple thousand low-cost chips per month. <S> What they aim at with low-cost chip is mass-produced low-cost devices, selling millions of chips instead. <S> Most products in this world need some microcontroller logic, but not much. <S> You really don't need much IO or RAM to make a microwave oven beep. <S> Or to control the charging in a electric toothbrush. <S> Or to set up the sequence in which to start the power supplies inside a complex product¹. <S> Of course the hobbyist market creates visibility and allows to sell a couple products that are profitable in small numbers; but really, in the greater scheme of things, hobbyists can't sustain a large semiconductor manufacturer. <S> ¹ fun fact, doing power sequencing is something very often done by a microcontroller reserved for that purpose only – you just can't turn on your complex main MCU before power runs. <S> And that's pretty much the upper limit of what an Attiny85 can do – a bit of I²C and a bit of voltage/ADC math, and the memory's full. <S> With a well-known environment to rapidly write down the firmware for that, plus competetive prices, especially for the non-DIP form factors, why not go <S> ATTiny? <S> I hate 'em, personally, but if you're more used to Atmel than to 8051... <A> Is the ATtiny85 actually used in any consumer electronics? <S> It appears to me that it's solely used by makers/electronics enthusiasts. <S> Outside of consumer electronics and enthusiasts markets, there is a huge universe out there. <S> The fact you only see them in one tiny segment doesn't mean they aren't used in other places. <S> I have used them to process buttons, and as data aggregators in industrial control projects. <A> The ATTINY85 is used in Dame Product's flagship FIN product... <S> https://www.dameproducts.com/products/fin <S> It handles the user interaction, motor control, and battery monitoring. <S> The A/D converter, internal voltage reference, extremely low power sleep with wake on pin, and ability to direct drive LEDs, small physical size, wide supply voltage range, deep & wide supply chain availability, and low cost make it perfect for products like this. <A> I had to repair an automatic car battery tender ( Car battery tender) <S> and it had an ATtiny in an 8-pin package <S> - I can't remember which family member. <S> The market for those far outweighs that of hobbyists. <S> Its capabilities are limited but it is very cheap to build into a product.
| I'm sure they, and similar devices, are used in billions of products these days wherever a small amount of control or intelligence is required; remote controls, thermostats etc.
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Can a "static" magnetic field carry Information? Information is electronically represented as some energy that is changing, therefore we measure an on state (1) and an off state (0), to create a (0,1) binary representation of the information. Now a static magnetic field by definition has a frequency of 0 Hz, so frequency modulation can't happen. But what if we modulate the amplitude? (Tesla), would that consist as a static field? Can an amplitude modulation work to send information from an electronic device that creates a magnetic field with constant frequency? What if the device is already electrically shielded, as this is a follow up to my previous question . If we already have a device that is sufficiently electrically shielded, can information leak out from the magnetic field via amplitude modulation? I am specifically interested how can / if possible / information leak out from an electrically shielded device by modulating a magnetic field. EDIT: I made the question more precise. Referring specifically to the relationship between amplitude modulation / frequency modulation, and a static magnetic field. I believe a magnetic field with 0 frequency is only static if the amplitude is constant as well, reading the answers below. <Q> A static field (indeed, any static symbol) conveys no information, byShannon's theorem. <S> This is because the probability of the state is1, there is only one 'symbol', and if it is unchanging, the frequencybandwidth is zero. <S> Once one talks of amplitude modulation, however, the field is NOT static,there is a bandwidth, and a signal that can carry information. <S> Every loop-shaped antenna picks up modulated magneticfield, so we know that kind of receiver is workable. <S> In a sense, a permanent magnet with a shutter (driven vanes of softsteel, perhaps?) makes such an amplitude modulated signal. <S> Such amodulator, in my auto, senses crank position from the passage ofa notched rotor next to a magnet and sense coil. <S> The crank sensor might be less sensitive if electrically shielded, butwould still work. <S> Magnetic fields are hard to shield. <A> Commonest information carried by a static magnetic field : this way is North, that way is South. <A> Depends on your definition of static. <S> The presence of a magnetic field itself can carry information. <S> It's permeation through space-time can also carry information . <S> If you believe in monopoles and their interaction / entanglement, there are more possibilities then... <A> Of course information can be encoded in static magnetic fields — that's the whole idea behind all forms of magnetic recording, from the stripe on your credit card, to audio and video tapes, to the trillions of bits on a hard drive. <S> Remember, a magnetic field is bipolar, which means that it has a definite orientation, and information can be encoded in how that orientation relates to some other reference (frequently another magnetic field). <S> In most cases, it's easiest to read out the data electronically by introducing motion between the recorded information and a sensor of some sort — swiping your card, moving a tape past a read/write head, spinning a hard disk — but it isn't necessary. <S> For example, you can read out magnetic patterns optically using a "developer" fluid that contains tiny magnetic particles. <S> There are also the mechanisms used in magnetic core memories and Hall-effect devices (such as shaft angle encoders, which can be read out regardless of whether the shaft is moving).
| A static magnetic field can carry information, but being static, that information cannot change.
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How to drive a linear actuator in both directions I would like to drive in both directions my linear actuator (12VDC, 13A max).My electronic board outputs two signals (obviously, I also have GND signal): Forward : +12VDC (2A max) Backward : +12VDC (2A max) When I press the forward button, the forward output is +12V, on the contrary, when I press the backward button, I have +12V on the backward output line. The battery for this system is an AGM battery, 12VDC, 200Ah. I know that the correct way would be to use a driver, but it is expensive and I do not need lots of functions, I only need to open and close the actuator. I was thinking to trigger two relays by using the forward and backward signals, but I can't figure out how to connect them in a correct way. Can you give me some suggestions, please? <Q> Use an H-bridge. <S> It is cheap and straightforward to build. <S> http://www.bristolwatch.com/k150/port4.htm <S> I use this for high current switching: http://www.bristolwatch.com/k150/pics1/mosfet_hb2.png <S> These N- and P-channel MOSFETs have a fairly low V_DS(ON) resistance, and don't run very hot at 2A. P = <S> (I^2) <S> x R = 4 x 0.8 = 3.2W for the IRF9630, less for the IRF630. <S> I don't personally like BJT transistors in H-bridges except to control the MOSFETs, but perhaps there are scenarios where they're better. <S> If so, others may chime in with their cluebats. <S> Hope it helps. <S> Edit: the linked schematic is not for PWM (I used it for 12V actuators as well). <S> Cost was low, <5USD <A> I was thinking to trigger two relays by using the forward and backward signals, but I can't figure out how to connect them in a correct way. <S> Can you give me some suggestions, please? <S> Try this for a relay reverser: - A motor is shown but that can be your linear actuator. <S> Contacts required in relay: two single pole changeover (or double pole, double throw). <S> You might need to put back-to-back zener diodes across the actuator to "catch" any back-emfs. <A> Use two contactors with a interlocking mechanism. <S> You will find such contactors in a electical specialized shops. <S> Try to google: reversing contactor. <S> This kind of contactor has the mechanical lock between them, preventing to be engage both of them at the same time - this would cause short circuit. <S> https://www.youtube.com/watch?v=HRv_LhhMofE http://www.cesco.com/b2c/product/Schneider-Electric-LC2K0901B7-Reversing-Contactor/614180 <A> In response to the comments on both the question as well as Andy's answer I've decided to post my own answer. <S> From what I've gathered, you want the operation to be normally stationary. <S> Then either forward or reverse but only while the button is pressed. <S> How the circuit below works, with no buttons pressed there is no positive voltage on the terminal of the relay, no voltage no motor movement, simple stuff. <S> When the REV button is pressed, it connects \$V_{S}\$ to the relay terminal. <S> Now there is \$V_{S}\$ on your \$MOT-\$ and \$0V\$ on your \$MOT+\$. <S> So your motor will move in reverse whilst the button is held. <S> When the FWD button is pressed, like before it still connects \$V_{S}\$ to the relay terminal. <S> However, this time it also uses \$V_{S}\$ to saturate the NPN transistor which will cause current to flow through the relay coil and the relay will actuate changing the poles. <S> Now you have \$V_{S}\$ on your \$MOT+\$ and \$0V\$ on your \$MOT-\$, so your motor will move forward whilst the button is held. <S> The purpose behind the diode on the FWD button line is so that when you press the REV button, it doesn't saturate the transistor, there will be an ~0.7V drop when your motor is running in the FWD direction <S> but it shouldn't make much difference to the speed. <S> Depending on your motor though, you will need quite a high current diode. <S> As for the transistor and base resistor values, without all the voltages and currents I can't really give any recommendation. <S> simulate this circuit – <S> Schematic created using CircuitLab
| There are also ready made kits, with motor protection and reversing contactors.
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Op-amps, their impedance and current flow I'm an undergratuate student and I have a question about op-amps. As I understand it you need infinite input impedance so there is voltage drop across the op-amp and not the signal device. However, doesn't infinite resistance means that no current will flow through the op-amp? Do you get any form of current at the output? Thank you for taking your time to read this. <Q> Infinite input <S> impedance means that no current flows into the input terminals of an ideal op amp. <S> The image above shows a non ideal op amp in an inverting configuration. <S> To idealize this, \$Z_{in1}\$ <S> and \$Z_{in2}\$ <S> are equal to \$\infty\$ , and \$Z_{out}=0\$ , making \$ <S> e_{out}=v_{out}\$ . <S> To finish off the ideal assumptions, \$A_{OL}\$ is the open loop gain of the op amp, and is equal to \$\infty\$ <A> An OpAmp can be considered a voltage-controlled voltage source. <S> You apply a voltage at both inputs, the OpAmp 'measures' the differential voltage \$v_D\$ and applies a voltage proportional to <S> \$v_D\$ at the output. <S> The proportionality is determined by the open loop gain of the OpAmp. <S> It is usually very high, about 1E5, infinitely high for an ideal OpAmp. <S> The energy to drive the output comes from the supply rails, not from the input. <S> This is the trick with an ideal OpAmp: It has inputs where no current flows in (aka high impedance input), but a voltage appears at the output that can supply some current (aka low impedance output). <S> So it is an impedance converter. <S> As the open loop gain of an OpAmp is so high, you usually do not use it in this configuration for amplifier designs. <S> Instead, you apply negative feedback (output somehow connected to inverting input). <S> If there is negative feedback, you can make another important assumption of an ideal OpAmp: <S> \$v_D\$ is zero. <S> This means, the OpAmp will drive the output to whatever value is needed to achieve \$v_D=0\$. Additionally, still no current is flowing into the OpAmp. <S> So, the inputs of an ideal OpAmp with negative feedback employed show no current inflow and no voltage across them and are therefore neither a short (no voltage, maximum current) nor an open loop (maximum voltage, no current). <S> Keep in mind, that applying negative feedback inserts a connection from input side to output side and therefore influences input and output impedance. <S> So the OpAmp is not anymore the near-to-ideal impedance converter mentioned above. <S> Instead you have to analyze input and output impedance depending on the feedback circuitry. <A> The output current of an OpAmp provides surprises. <S> Over much of the useful range of frequencies, there is a 90 degree phaseshift between Vin (the difference between Vin+ and VIn-) and the Vout. <S> You can see this phaseshift, in the left plot shown below. <S> What does this phaseshift do? <S> It makes the Vout appear inductive, particularly with the output current dropping with frequency. <S> Add on a capacitor, and you get another surprise: peaking of frequency response, shown above in the right plot. <S> In this next set of plots, we see the OpAmps INDUCTIVE Zout on left (looking back into the OpAmp stage); then we look back into the Cload stage and see the combined effect with the sharp resonance of Lout and Cload. <S> The output current, and output voltage, have other surprises: thermal noise and power supply ( <S> deterministic) noise. <S> Those cause a wiggling of the output, even when Vin is fixed. <S> In this next screenshot, look on the lower right, to readthe error from ThermalNoise and from Aggressors (the only activated Aggressor is PSI --- power supply interferer --- show in topright checkbox). <S> Notice the 22.8uV of Thermal Noise and the 15uV of (60Hz, 120Hz) <S> Power Supply noise. <S> Here is what a 25uV peak signal, at 200Hz, with added 1:1 (0dB) noise power and signal noise, into a 200Hz LC filter. <S> Notice most of the OpAmp noise is gone; we see some wandering of the sinusoid and some "distortion" which is just the noise not being totally removed so that energy upsets the sinusoid shape. <S> The Operational Amplifier is very useful, on PCBs or on silicon, but the physics, and math, involved are worth learning so you do not expect too much yet can get excellent performance. <S> Here is a plot of Zout versus frequency, for an opamp with UGBW of 100MHz. <S> The plot is particularly interesting because the opamp has a ChipSelect pin, thus we see Zout with output transistors controlling the output <S> pin and with those transistors disabled. <S> Up near 500MHz, the Zout nears 30 Ohms, which is impedance of 10pF. Also note the dip in impedance; 10pF and 10nH (Vout pin and VDD pin inductances) resonate at 500MHz. <S> I'm thinking the 30 ohms Zout is the NPN reac in parallel with PNP reac, where reac is 0.026/Ie_ma thus 0.5ma produces 52 ohms in both emitters, which in parallel become 26 ohms. <A> The model of the ideal op amp shows a dependent voltage source with no output resistance as the output of the device. <S> Remember on practical devices you see 2 power supply pins. <A> Ideally, an opamp has an infinite input impedance (and thus zero input current) and an infinite open loop gain , along with infinite output voltage range, bandwidth and slew rate and zero impedance at output (and thus infinite output current) with zero noise. <S> But in reality that is far from being true. <S> An real opamp has: High, but finite input impedance, so a little input current (a few tens nanoapms to only some picoamps I think); Low, but non-zero output impedance, with an output current of a few miniamps to hundreds of miliamps (like 200 mA) for power opamps; <S> Finite bandwidth <S> (it is easy to find one with over 2 MHz bandwidth for a small price); Finite slew rate of, commonly, 5 to 100 V / microsecond; <S> Output voltage limited to input voltage. <S> Always check the datasheets before buying any component. <S> That can even help you to understand what are the important parameters for a type of electrical or electronic component.
| The ideal op amp also has zero output impedance , and most certainly provides current.
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Excessive noise in amplifier circuit when switching from DC power supply to AC-DC adapter I'm working on breadboarding this amplifier circuit before transferring to a perf board. When we power it from an adjustable DC power supply set to 12V, the audio output sounds clear. When we switch to a 12V 5Amp AC-DC adapter, we are getting almost nothing but noise. We've already switched from one adapter which was providing an uneven output between 10V and 12V, to an adapter which we've measured at a constant 12.6V. We're using the negative power terminal as our ground. I would appreciate some suggestions for ways to troubleshoot this issue or any ideas as to why we're only getting clear sound from the adjustable power supply. <Q> Based on your description, I am assuming your schematic should be reading +12v <S> and 0v, rather than <S> +/-12v which, as mentioned above, would yield a total of 24v. <S> I am also going to assume that the noise is a buzz or a hum, rather than a hissing, or white noise. <S> If that is the case, the first thing I would suspect is that the wall adapters are switching power supplies, and your adjustable supply is linear. <S> So, I believe you may also be amplifying the noise generated by your adapter. <A> looking at the datasheet, your dc power supply is not sufficient. <S> I would size it about twice the non repetitive peak Amperage for power headroom, but twice its repetitive peak output would be my recommended minimum. <S> If you go by typical practices, the maximum current draw should be 80% of the power supply so the rectifier filters in the power supply can recover. <S> But that is only the 1/3 of the issue, <S> the main issue it has is its design. <S> Unbalanced = 100% of power supply noise injected onto the signal. <S> Now if you create isolated power, or create a ground reference that has the inverse of the power supply noise on the ground, the noise would get subtracted by the output circuit. <A> Could be several things: 1) <S> The wall adapter is junk. <S> Older SMPS are known to be noisy. <S> Newer SMSP like Mean <S> Well LED supplies are known to be very good and can be very quiet. <S> 2) <S> Even if the SMPS is the modern good variety, they may not "like" large capacitors directly on the output. <S> The ideal scenario is to use a common mode choke that is as large as you can find for the current required and then follow that with whatever size cap you need to filter down below 20kHz. <S> For example, you should be able to find at least a 1mH choke which in combination with 10uF cap will filter out everything above ~1.5kHz. <S> 3) <S> Even if the SMPS is the modern good variety, they may throttle switching which causes moduation at low frequencies if the output is not loaded with at least 10-20% of the load capacity. <S> So you might need to just add a smallish resistor of appropriate wattage. <S> But unless the SMPS is very large, my guess is that the quiescent current of your amp is enough to load properly. <S> So my guess would be that the issue is 1 <S> - the adapter is some junk pulled out of a bin. <S> There is no recovering from that. <S> You can get a proper 24V (or 12V) <S> Mean <S> Well constant current LED supply for under $20 USD on Mouser.
| From my understanding, certain amplifier circuits can be more sensitive to the noise generated by the switching action, whereas linear supplies do not produce such noise.
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Matched and Identical Concepts of Transistors I realize some expressions like "Q1 and Q2 are matched" or "M1 and M2 are identical" about BJT and MOSFET configurations. I have some questions related to this. Are "matched" and "identical" concepts' meanings same? Which parameters of transistors that i mentioned above are same if transistors are identical? Thank you. <Q> Identical transistors are very easy to find....... in your SPICE simulator. <S> This can sometimes be misleading, when your circuit has superb performance, because everything matches perfectly, distortions cancel, etc. <S> If you pick two transistors from the bag, they will have different characteristics (Vbe, hFe, capacitances, Vgs, RdsON, etc). <S> Also, your transistors will be at different temperatures. <S> Vbe changes 2mV/ <S> °C for bipolars <S> , hFe changes too, and FETs are also temperature-dependent. <S> This will create offset, distortion, and drift in your current mirror. <S> If you want a better current mirror (or differential pair), you can: <S> Add degeneration resistors, to swamp differences in Vbe. <S> Use matched transistors, like DMMT3904 which have matching guarantees, and they are in the same package, ensuring close (but not identical) temperature, at least if dissipation isn't too imbalanced. <S> These are two independent transistor chips, so they are isolated. <S> Use dual monolithic transistors, which are made on the same chip, therefore much better matched, also better temperature tracking, and very expensive. <S> Or, in an IC, you can have excellent matching, because transistors are made next to each other on the wafer, in the same fab run, same process, etc, so they will have very close characteristics. <A> There is no such thing as an identical transistor. <S> Variations in manufacture dictate that each and every one is slightly different. <S> Identical is a mathematical 'ideal' used for math purposes only. <S> However, you can never achieve identical in all parameters. <A> Your circuit shows transistors operated with DIFFERENT VCEs. <S> Thus the currents will be different[EarlyEffect], and the heat dissipation will be different[Ic*Vce], thus the Vbe will be different[~2mv/degree Cent], causing ~ 5% Ic mismatch. <S> Check out current-mirror topologies that use stacked transistors, so <S> the Vce (and the heating) are identical.
| Matched means the transistors were selected from a pile by measuring them to find the two that "match" closest for a particular variable, presumably the one you care about.
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Master switch to turn on and off independent lights i am adding some led lighting to a dolls house where each room can be turned on and off independently but i would like a master switch which turns all lights off(simple) but should turn on all lights even if they were off before. I just can't figure how to get this to work. Example With 2 LEDs in the system. When the master switch is turned on all LEDs turn on. Led1 is switched off, led2 is kept on.-When the master switch is turned off all LEDs turn off. And then when the master switch is turned on all LEDs are turned on. <Q> You will not be able to do the last step without resetting relays (costly, not easy to find), complex logic circuit (hardware intensive) or a microcontroller. <S> A microcontroller solution with push buttons will be super easy. <S> On/off switches will require slightly more code but pretty easy too. <S> The microcontroller solution can be cheap, and consist of nothing more than the switches and some resistor + Transistor pairs (and your leds of course). <S> Depending on how many lights you could skip the transistors too. <S> The internal Pull-Ups are enough for the inputs, but external ones can be used too. <S> simulate this circuit – <S> Schematic created using CircuitLab Code would be simple: <S> Boot/Power <S> On{ turn all lights on, set inputs; Wait for button press{ <S> At button press toggle light; At master button turn all lights off; At second master button, go to boot. <S> }} Or skip the master button and use a hard power switch turning the microcontroller off. <S> When turned on it starts at boot. <S> Cuts down on some of the coding. <S> simulate this circuit <A> Your description is kind of vague, but I understand it to mean that you want a 3-position master switch: " <S> All Off", "All On" and the third state in which all of the LEDs are controlled individually. <S> You'll need some diodes: simulate this circuit – <S> Schematic created using CircuitLab <A> Despite I'm using buttons rather than switches , you can achieve this using a 4013 dual D flip-flop with set and reset, where you have a flip-flop per LED configured as a T flip-flop to toggle the LED's state with a single button. <S> But the main point we're using this is for the SET and RESET functions which can turn all LEDs on and off. <S> Note that I put debouncing circuitry to prevent accidental extra triggering of the clock. <S> The SET and RESET do not need this. <A> simulate this circuit – <S> Schematic created using CircuitLab <S> In Figure 1, when the master-on switch is open, then each of the individual switches has one pole on and one pole off. <S> When the master-on switch is closed, then each of the individual switches has both poles "on", effectively overriding the individual switches. <S> Of course, the separate off switch prevents any current flowing, so it overrides everything (including the master-on switch). <S> Figure 2 uses a single center-off DPDT switch as the master. <S> This works mostly the same way, where having the switch in the "up" position powers both terminals of each switch, "center" isolates ground and prevents all current from flowing, and "down" connects ground and exactly one pole of each individual switch (allowing for normal operation).
| If you're comfortable with each of your individual light switches being SPDT, and you're comfortable having separate master switches (or one center-off DPDT master switch), you can do this without any additional components.
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what does the c stand for when charging li ion batters ie Max. Discharge rate: 2C (6.8A) looking to charge a 4 pack of li-ion battery but dont understand what the 2c means know the rest. i.e. Max. Discharge rate: 2C (6.8A) <Q> C in this context means the battery capacity in Ampere-Hours but divided by hours. <S> If battery's capacity is 3.4 <S> Ah, then C would be 3.4A, and 2C would be 6.8. <A> This would mean 2C discharge rating for the battery is 6.8 Amp <S> , the battery would be rated at 3.4 Ampere-hours. <A> A battery has a capacity rating usually in mA hours or Amp Hours. <S> The C or C-Rate is the C harge and dis C harge rate, expressed as a ratio of the battery capacity. <S> A 1Ah battery should provide 1A for one hour. <S> If it is powering something that is draining 2 Amps that is a 2C discharge rate. <S> The rate is used when the specifics of the battery such as voltage and capacity are not an issue. <S> It is a generalize term applying to all batteries or the type of battery being addressed. <S> For example: Li-Ion batteries are recommend to be charged at an 0.8C rate. <S> This means the 1Ah battery should be charged at 800mA. <S> A 2Ah battery would be charged at 1.6 Amp. <S> The C rate can be applied to any battery regardless of it capacity or voltage. <A> Charge and discharge rates of a battery are governed by C-rates. <S> The capacity of a battery is commonly rated at 1C, meaning that a fully charged battery rated at 1Ah should provide 1A for one hour. <S> The same battery discharging at 0.5C should provide 500mA for two hours, and at 2C it delivers 2A for 30 minutes. <S> Losses at fast discharges reduce the discharge time and these losses also affect charge times. <S> A C-rate of 1C is also known as a one-hour discharge; 0.5C or C/2 is a two-hour discharge and 0.2C or C/5 is a 5-hour discharge. <S> Some high-performance batteries can be charged and discharged above 1C with moderate stress. <S> Table 1 illustrates typical times at various C-rates Credit https://batteryuniversity.com/learn/article/what_is_the_c_rate
| "C" is the ampere-hour rating of the battery.
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Relay VS Push Button: blind test (how to tell the difference) I have a classic push button that closes the circuit when pressed down.This push button was replaced by a simple relay, the relay is activated when 3V goes through its trigger points, and the circuit is closed.The result is the same, the circuit is being closed. Now, if I was on the other side of a wall and someone was to replace the push button with the relay (case 2), how can I tell the difference from my side? I do not see anything, so I don't know when the other person pushes the button or when he triggers the 3V for the relay.If I have access to an oscilloscope would I be able to see a different when the circuit is closed in case 1 VS case 2? Is there any simpler test? See my beautiful drawing: <Q> Contacts bounce , and pushbuttons usually do this more than relay contacts and in an unpredictable way. <S> If you happen to know the specs of the relay, or at least, the type (e.g. reed relay) you can make an educated guess by looking at a scope print of that bounce. <A> And I am now able to tell the difference! <S> The relay always shows a last bounce that seems to be always at the same period from previous bounce (in the case of that particular relay: ~460 μs). <S> The push button in the other hand shows random bounces for each push which seems to be the "factor in the human finger issue" as suggested by @Trevor. <S> Cool experience, thanks for your help guys! <A> An O-Scope (overkill) or a simple meter can be used to detect a change in potential in your circuit on the right side of the wall, the same will hold true regardless of how the circuit is completed. <S> Open circuit, there is a difference in potential between the two wires, closed circuit, no difference in potential. <S> Ah, after reading the other response, I think I misinterpreted the question. <S> I like the contact bounce idea. <S> It's unlikely that the contact resistance is going to be consistently different enough unless maybe you had a lot of empirical data to compare to.
| I compared the contact bounce for both cases. Nothing would be different in either case.
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Measuring weight on a gym machine We're doing a school project where we have to measure weight on a fitness machine. The user can select how many weight plates to use by moving a rod like shown in the picture. How would you do it? We've talked about the following solutions: a strain gauge measurement unit beneath the stack, be we doubt that it will be robust, and it will be expensive. an IR distance sensor that counts the spaces between each plate. inductive proximity sensors Some kind of electromagnetic solution? <Q> What not to do <S> That would probably work well, albeit with some extra peaks and troughs in the signal as the weights move up and down. <S> But: the wire and it's connection to the weights may be safety critical. <S> If the wire parts, it will lash around, and the weights would drop with quite a bang. <S> So don't interfere with the wire unless you're very sure it's remaining strong. <S> Some Ideas Put a microswitch in each hole into which the weight selector can be placed. <S> Use a microcontroller to monitor the switches. <S> Watch out that only the lowest one which is being lifted will definitely dissapear from the sensor, as lower weights may still block higher sensors. <S> As above with a LDR and LED for each weight. <A> Switches activated by a cam on a vertical mounted bar next to the stack. <S> As the plates move upward the top switch will remain closed until the last plate being lifted passes. <S> That is the point in time to read the switches states. <A> Do you have a lot of students who exercise in the dark? <S> No, you say ? <S> Optodetectors <S> Position them <S> so they are quite close to the weight, and have them report back an analog reading of light level. <S> You can try using ambient light level, and some software to normalize and remove false readings from people walking by etc. <S> If this problem proves difficult, provide your own light by putting an IR LED right next to the IR optodetector. <S> If this continues to produce falsing, switch the lights on and off semi-randomly (so you don't contend with other machines using the same system), a sensor which changes dramatically with your pulsing is on a weight. <S> It may be possible to economize on sensors by having a sensor one position above <S> the top weight and then a sensor at every other weight below. <S> If the above-sensor detects whilst all the weight sensors are still detecting, then the starting weight is not on a sensor. <S> Given that most gyms kit out a variety of machines from one manufacturer, and those makers have only 1 or 2 standards for weight packs, it should be possible to fab a PCB that positions the components correctly. <S> Another option is to junk all that and go with pneumatic resistance machines. <S> There are weight machines which do exactly that; they need an external air supply which you manually feed or bleed off with two pushbuttons on the machine. <S> I have seen machines with no electrical connection, only an air line. <S> They have an electronic display reading out weight level, which takes minimal power, and presumably uses batteries, solar or a microturbine. <S> the machine automatically pumps/bleeds to your weight setting, while reporting via WiFi your performance <A> Ok... <S> a strain gauge measurement unit beneath the stack, be we doubt that it will be robust, and it will be expensive. <S> A load cells under the stack or in the cable would be the ultimate solution. <S> AS you have noted though, normal load cells are not cheap. <S> However, it may be possible to "arrange" something that adds measureable deformation under the plates, perhaps a spring, or a slab of rubber or some such and measure that deformation. <S> One COULD even create a simple weight detecting capacitor... <S> And detect the change in the capacitance. <S> That would be a cool little project too. <S> As JRE Mentioned, better to weigh what you are NOT lifting than when you are lifting though and do the math. <S> an IR distance sensor that counts the spaces between each plate.inductive proximity sensors Although these are easy to do, they really do not answer the challenge since they do not measure weight. <S> The weight exists without moving the plates too far. <S> Some kind of electromagnetic solution? <S> No idea <S> what that would be. <S> BY THE WAY: Stack-Exchange is usually NOT the forum for answering school projects and <S> such so I wont give you much more details. <S> However, I personally am giving you help because I believe your are doing an appropriate amount of research and utilizing the resources the web has to offer to you, HOWEVER <S> I expect you to publish the fact that you used this help in your development. <S> You do not have to mention me personally, but you should include the fact that your consulted with on-line "experts". <A> This started out as a comment and got too long. <S> Scrap the load cell concept. <S> Even if it was under the entire stack trying to measure unused weights, if Joe Sixpack lifts a substantial weight as high as it goes and drops it (happens all the time), the load cell will be exposed to huge forces. <S> Surviving that is going to take a load cell rated for huge forces, or it wil be out of calibration within a couple weeks. <S> The result will be pathetic dynamic range and SNR, and a super expensive solution. <S> Pretty much any other concept will work better for less than 10% of the cost of a sufficient load cell.
| One obvious solution would be to put a load cell in-line with the wire, above the weights. Put an ultrasonic sensor near each weight, and watch for them dissappearing. When the plates are lifted the and the top most switch is activated (on) then you have a valid count. Put a load cell under the unused weights, and subtract that from the total. As above with a microswitch on a stick for each weight. It wouldn't be hard to have an NFC smartcard which you swipe and
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Telling which way a KY-040 Rotary Encoder has been turned I am stripping this problem down to the bare minimum. I have KY-040 Rotary Encode, When it is turned clockwise I want a blue LED to blink for every increment it is turned.When it is turned anti clockwise I want a red LED to blink for every incremented is turned. I can have anything between them (but not a PI or arduino type barebone computer) I have looked at flip flop chips but with the wave signal that comes out of the KY-040 Rotary Encoder I got quickly confused. <Q> probably the 2 outputs have a quarter pulse lenght shift. <S> Use one output as the known amount rotation indicator (=every rising edge means a known angle increment) and the other is the direction; checked when the angle increment occurs. <S> Your led flashing circuit: ADDENDUM: <S> The fi2 output of the rotation sensor states how long the led light pulse is ON. <S> Of course you might want to limit that in case the rotation stops just when fi2=High. <S> For that you can add a monostable multivibrator between fi2 and the nand <S> gates to limit the on-state lenght. <A> simulate this circuit – Schematic created using CircuitLab R! <S> is negative logic input. <S> A Leads B for Up <S> Alternatively using S with positive logic and Q bar outputs. <S> SIM <A> I don't think it'll come as a surprise when I tell you that you'd often just write a bit of software that runs on a microcontroller to do exactly that. <S> However, the encoder logic is so simple that you can write it down as boolean combination of the sensors. <S> So, with a very minimal amount of discrete logic gates, you can do that. <S> So, the mechanism for this kind of problem is to write down your problem as state machine (you're a CS person – you should know state machines) and then implement these states in hardware, simplifying where possible. <A> Bi-Directional encoders usually give out two signals in quadrature. <S> That is, two square waves 90 degrees out of phase. <S> BTW: Here is the Spec Re your LEDS.... <S> Simply feed one phase into the clock of a D-Type latch and the other phase into the D-Pin. <S> The output Q will change with direction of rotation. <S> Use that to enable power to high end of the appropriate LED (Resisitor). <S> Use one phase to drive the low end of both LEDS.
| You can tell the direction by the state of the second line on the rising or falling edge of the first line.
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On what condition does Harmonic components of power generated in 50/60 Hz Grid Generators depend on? I have power-recording data from 11 countries (USA-60Hz mains frequency / other nation 50Hz). These data were collected using a audio recorder (in a computer) connected directly to the power mains via a step-down transformer. So, the 50hz/60Hz electrical voltage sine wave is basically stored in the computer as a .wav file. Now if I perform a Fast Fourier Transform on these data, a distinction arises between data collected from one generator (from a grid in a country) to another in the magnitudes of their harmonics. I am including the images after performing the FFT. Grid A Here, only the main 60Hz component is prominent. Grid B Here, the main 50Hz component is prominent, but the odd multiple components also are present with decreasing significance. Grid C I can't post 3 images due to lower than 10 reps. Thus, Grid C has no image, but the description still holds. Here, the main 50Hz component is prominent, but the odd multiple components also are present, but not with decreasing significance. The 3rd component has the most prominence than 1st, 5th or 7th. My question is what makes this harmonic components arise in the first place (may be some aspect of the generator). What are they dependent upon or are they arbitary? So far, from the large amount of data in my possesion, It seems that this should be a pattern of the generators it self because data from the same grid showcases this trait consistently. Can a grid (including the generators and the load) suddenly stop producing harmonics or alter their magnitudes of different harmonics? Moreover, I can see that in Grid A, which is from USA, the harmonics are nearly absent. Where as Grid B (Lebanon) and Grid C (Turkey) have significant harmonic components. Could this be because of the better Generator and load control mechanisms of the networks of USA in comparison to those of Lebanon and Turkey? P.S.Any information about this would be great. It would also be great if you could point out some relevant literature. <Q> It could be the generator, but it's more likely to be the type of loads being driven. <S> A simple rectifier/capacitor DC supply draws current only around the peaks of the supply voltage waveform, giving rise to a current with lots of odd harmonics, and a 'clipped sine' voltage waveform. <S> This type of supply was prevalent in computers, TVs etc for a long time, but is now being replace in new equipment above some power level like 300watts or so with 'power factor corrected' DC supplies. <S> These use a programmable current boost converter to force the supply input current to be as if it was going to a resistive load. <S> It's quite plausible that there are more PFC corrected supplies in the US, and more of the old type in Turkey and Lebanon. <S> Let me google a reference for you ... https://en.wikipedia.org/wiki/Harmonics_(electrical_power) <A> When you consider the 'Grid' you are talking about a complex of both generators and loads. <S> Your AC supply may consist (locally) of either a single generator or multiples in parallel. <S> With a single generator you will see waveform distortion mostly based on it's construction and lower levels of harmonics for larger gensets. <S> With multiple generators in parallel you will see harmonics based on construction and synchronization offsets Your loads may also create local waveform distortion (harmonics) based on their current consumption periods. <S> One typical load that creates high power line harmonics are large power VFD's Read <S> this for some ideas. <S> In general for your US supply, there will be fewer and possible better controlled generators on line giving low harmonic distortion. <S> In areas where the generator load and supply is much more variable, you may see more harmonic distortion. <S> You may also see droop influence synchronization between different generator sizes paralleled. <S> For harmonic generation based on alternator construction this may help. <S> For harmonic generation based on paralleling this may help. <A> From wikipedia: Current harmonics are caused by non-linear loads. <S> When a non-linear load, such as a rectifier, is connected to the system, it draws a current that is not necessarily sinusoidal. <S> The current waveform can become quite complex, depending on the type of load and its interaction with other components of the system. <S> So diodes and transformers with a ferrite core (hysteresis in core) generate the harmonics. <S> The generator doesn't really add harmonics. <S> Its the load. <S> Here is a great diagram from an article on Harmonics in power systems . <S> It should also be noted that the line impedance also has a big factor on harmonics, because the impedance blocks the source (a big generator) from 'seeing' the load. <A> In the wealthy industrial countries there has popped up every kind of pricing mechanisms that are designed to quide how people behave. <S> One of them is to make the reactive power cause the price increase for the electricity. <S> Nonlinear loads cause harmonic reactive power that only cause losses just like the normal reactive power. <S> But it's worse. <S> Harmonic reactive power can't be compensated by reactances or equivalent synchronours motors by the local distributor. <S> Thus there's a good reason to quide mass consumers to have linear loads (=filter out the harmonics that are caused by electronic power supplies and triac controls) <S> In the developig countries there are much bigger problems still to be solved. <S> If one has got the line into his house and can pay the bill or the bribes, he can connect virtually anything as long as the wires or fuses do not go down.
| If you are in a local supply condition with multiple different sized generators in parallel you may see poor synchronization and hence potentially higher levels of harmonics.
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Purpose of series resistors in this bipolar rectifier circuit? The following circuit uses two series resistors, one for each polarity of the supply. I understand function of smoothing caps and overvoltage zener diodes, but the resistors I do not. I would like to understand the function of these resistors and how to calculate what wattage these parts should be. Also, how critical is their resistance value? <Q> It's a series resistor for the zener diode. <S> One is needed to make the zenerdiode able to cut off the exessive voltage. <S> The zenerdiode draws just that current which is needed (in addition to the load current) to drop the voltage to the zener voltage. <S> This is the way how zenerdiodes are used as voltage regulators. <S> This principle is practical only for low power systems due the continuous dissipation in the zenerdiodes. <S> The wattage of the resistor <S> : Let Ux = the maximun DC voltage in capacitors after the rectifier, for example 20V. <S> It's <S> about 1.4 x <S> the output voltage of the transformer. <S> Calculate the voltage drop in the series resistor <S> : Ux - Uout. <S> If Ux = 20V <S> then the drop is 8V. NOTE the zenerdiode draws all the current that's needed for 8V drop in addition to the load current. <S> The resistance is detemined in accordance with that. <S> The dissipation in the resistor = <S> (Ux - Uout)^2/R = <S> (8V)^2/270Ohm = 0.24W <S> Theoretically <S> 0.25W resistor is ok, but in practice it easily gets hot due poor cooling. <S> Take at least 0.5W, if Ux=20V happens to be your case. <A> Voltage regulation, and to a lesser extent filtering. <S> The resistors and the zener diodes together form a simple linear regulator. <S> The zeners set the output voltage, and the resistors set the current which flows through them. <S> If the resistors are too small, then the current through the zeners may be too big power is wasted heating them up. <S> If the resistors are too big, then they restrict the total current available. <S> The current drawn by the load could reduce the current through the zeners to zero, then the voltage would begin to droop. <S> If you look at the zener datasheets, and start working out the performance of the regulator for different resistor values, you'll quickly see that this approach is only suitable for crude regulation and small currents. <S> The resistors will also form something of an RC filter with the caps, though that effect is not really separable from their task as part of a regulator. <A> The Zener diodes can only dissipate a certain amount of power. <S> Also, their actual voltage drop varies from their rated voltage drop as the current through them varies. <S> And in the resistors, for that matter. <S> This kind of regulator, a shunt regulator, is one of the least efficient. <S> The load current causes a drop in the resistors and the Zeners are conducting more current to clamp the voltage. <S> If the load current drops, the Zeners must conduct more current to counterbalance it. <S> It's an OK circuit if you can afford the wasted power through inefficiency or have a very small load. <S> Otherwise a series regulator, linear or switching, is more efficient, though each comes with its own pros and cons.
| The resistors are therefore to limit the current through the Zeners, attempting to keep the current to one that produces a tolerable voltage drop range and that doesn't dissipate excessive power in the Zeners.
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PIC's output pin not turning LED on I've been trying to test my PIC just by turning a LED on, I verified that the PIC is energized and that the MCLR pin is HIGH. The PIC I'm using is the PIC18F45K22 , running at 3.3 volts. The LED has a Vf of 1.8 volts and I have verified that its polarity is correct. It is connected to pin RB4 with a 1 kΩ resistor in series. Not sure what else to try here. I have only 1 LED connected and that's RB4, is there something that might be blocking RB4 from giving the output? Is there anything else that I have to setup? This is my complete source code: #include <p18f45k22.h>void main(void){ ANSELB = 0; TRISB = 0; while(1) { LATBbits.LATB4 = 1; } return;} <Q> Sounds to me as thought your oscillator settings are incorrect. <S> Interestingly enough, I've been able to program and verify a pic32 with a broken oscillator. <S> Make sure you set your configuration bits corresponding to the board you have. <S> Verify that the board has proper bypass capacitors. <S> If you have access to an oscilloscope, verify that the oscillator is running by probing on the driven pin (refer to the oscillator section of the PICs family datasheets). <A> These what i have in mind try any of the following: <S> 1- try to use a different pin say <S> portB pin32- reduce the resistor value make it 450 ohm. <S> 3- add delay to your code. <A> Two things. <S> 1 set the fuse. <S> Initialize the pins as goio. <S> edit: <S> your code, when it is done right.
| Also try running with the internal oscillator. Some modes will have your PIC just stall waiting for a correct clock. 4- make sure that you are connecting the LED in an active high connection (anode to the PIC and cathode to GND)
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Why are thermionic valve/tube heaters generally designed to be 6.3 V? Why are the heaters generally 6.3 V (or multiples thereof)? Does it make for an easy transformer winding ratio when used in countries with 120 VAC or 240 VAC mains? They often pull quite significant current, and as there's generally a higher voltage supply available, a higher voltage and thinner wire could have been used. What are the advantages of having a low voltage supply for the heaters? <Q> When vacuum tube radios were invented, only a fraction of houses had mains electricity, therefore, first radios (and their tubes) were battery powered, they used three batteries: "A" battery for heaters. <S> As heaters require a lot of power, this was a rechargeable battery. <S> "B" battery for anodes. <S> This was a high voltage non-rechargeable battery, it lasted longer than the "A" battery though. <S> "C" battery for negative grid bias. <S> As grids do not really use any current, this battery lasted very long. <S> I guess the 6.3V heaters continued to be used just because there was no real reason to change the voltage. <S> Using a high voltage (220V) <S> heater would be problematic because you would need a very thin wire for the heater (220V 9mA heater would need a really thin and long wire) and the high voltage may affect the signal in the tube. <S> Later tubes intended for battery operation used 1.2V or 2.4V heaters which is a multiple of a NiCd battery voltage. <A> 6.3V tubes became common at the time the first car radios (and, I suspect, other vehicle-mounted electronics, usually not for personal civilian use) were developed. <S> 6V was the standard for car batteries back then; 12V applications were easily dealt with by intelligently building series circuits of heaters - heater currents @6V were specified in the datasheets. <S> DC voltage converters were awkward and costly to build in these days (though often needed for the anode voltage anyway - but why make them bigger or more complicated than necessary), so designing a valve series from the ground up for car use was the most economic solution. <A> The voltages were selected to minimise the current from available batteries to gain the longest duration of heating. <S> However as the voltage increased the effects of the voltage on directly heated filaments would affect the bias point at the non grounded filament end related to the grid voltage. <S> It would cause a spread in the bias and gain if DC heated with possible problems in reaching cut off. <S> It would also cause a massive AC component in the cathode current that would be amplified if the filament was AC heated. <S> Some of this noise was removed by using centre grounded AC filaments to have the opposite ends cancel some of the problem and also indirect heating of the cathode when practical to hide the filament potentials. <S> A voltage of 6.3V was a compromise that took as much of the limitations into account as possible. <S> It was close to the 2V and 1.5V cell chemistry multiples <S> ,that let one use 3 lead acid cells or 4 zinc chloride cells.
| A 6V lead-acid battery usually is at 6.3V, so this voltage was chosen as standard. Some tubes were designed to powered from the mains, their heaters were designed so they all draw the same current (at different voltages).
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Why isn't line to line voltage zero? If voltage is measured relative to something. For instance, voltage-meters measure voltage differences not individual voltage levels, or by analogy altitude is measured relative to somewhere. Mount Everest is so-and-so compared to sea level. But zero compared to itself. How come line to line voltage is not zero when both lines have the same value? For instance, a three phase three wire Y connection of 120V with an angle of 120 degrees. Line to line voltage here is $$120V \times \sqrt[]{3} \approx 208V$$ not zero. <Q> The difference between +1 and -1 is 2 yet both have the same magnitude of 1. <S> If you connect two identical 9 volt batteries together with a single wire and measure across the unconnected terminals you might measure 0 volts or you might measure 18 volts depending on how you connected the single wire. <S> Polarity matters and it matters in 3 phase systems just the same. <S> As you can see, although the 3 individual phase voltages are rising and falling identically, they are displaced in time and therefore there is a voltage between any two. <S> Picture stolen from here <A> ...because direction matters AC signals are time varying and they have both an amplitude and a phase. <A> The very short answer is... <S> If there is zero volts there is zero current . <S> So power is zero too, which is not very useful. <S> By having a phase difference you have a voltage and so current can flow if the load allows it. <S> Now we have voltage and current = power. <S> Now that is useful.
| The phase is the angle between the voltages so they are not both "the same value" at the same time .
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What does "dV/dt" mean for TRIACs? This is probably obvious, but since I still don't have an engineering education, I ran into this problem: What does dV/dt mean? What does it affect on a TRIAC? <Q> When the current across the triac falls under \$I_H\$, which is the holding current, the triac stops conducting. <S> With a pure resistive load this happens at the very end of the sine wave cycle, and voltage and current are in phase. <S> When the load has an inductive component (e.g. a motor) <S> then there is a lag between current and voltage. <S> At the moment when the current drops below \$I_H\$, the voltage has already risen with the opposite polarity. <S> Therefore when the triac turns off there is a big dV/dt on the triac - " <S> voltage is cut off immediately". <S> This situation can lead to self triggering the triac, and it starts to conduct uncontrolled. <S> The remedy is to use a snubber circuit, i.e. an RC in parallel with the triac. <A> dV/dt is the derivative of the voltage with respect to time. <S> In other words, it's the change in voltage (delta V, or ΔV) divided by the change in time (delta t, or Δt), or the rate at which the voltage changes over time. <A> If you had a curve of \$y = <S> x^2\$ like below: - The slope at when x = 3 (y = 9) can be estimated by calculating how much <S> y changes divided by how much x changes. <S> The change is called "delta" hence the slope is \$\Delta y/\Delta x\$. <S> Taken to infinitesimal changes <S> , mathematically it gets "renamed" to dy <S> /dx. <S> It can even be algebraically proven by adding dy and dx to the original formula: - \$y + dy = <S> (x + dx)^2\$ \$y + <S> dy = x^2 <S> +2xdx +dx^2\$ <S> Subtracting y ( = \$x^2\$) from both sides gives: - \$dy = 2xdx <S> + dx^2\$ <S> Then noting that if \$dx\$ is very small then \$dx^2\$ can be ignored hence: - \$\dfrac{dy}{dx} = <S> 2x\$ <S> In other words at any point on the curve \$y = <S> x^2\$ the slope is 2x <S> In terms of voltage changing with time it is dv <S> /dt. <S> It has significance for triacs and mosfets and can cause such devices to trigger or partially activate if the rate of change of voltage with time is too great. <A> So far everyone has explained what \$\frac{\Delta v}{\Delta t}\$ means ( rate of change of voltage, its gradient, the 1st derivative of a voltage w.r.t. time ) <S> But what does this have todo with TRIAC's? <S> Triacs, like Thyristors/SCR's can be re-gated if there is high dv/dt across the device http://class.ece.iastate.edu/ee330/miscHandouts/AN_GOLDEN_RULES.pdf <S> This is most likely to occur when driving a highly reactive load where there is substantial phase shift between the load voltage and current waveforms. <S> When the triac commutates as the load current passes through zero, the voltage will not be zero because of the phase shift (see Fig. <S> 6). <S> The triac is then suddenly required to block this voltage. <S> The resulting rate of change of commutating voltage can force the triac back into conduction if it exceeds the permitted dVCOM/dt. <S> This is because the mobile charge carriers have not been given time to clear the junction. <A> the energy mechanism Q = <S> C*V, when we make incremental changes and see what happens, becomes <S> dQ/dT= C * dV/dT + V <S> * dC/dT. <S> After chosing to ignore the 2nd part, and recognizing current = <S> dQ/dT, we are left with $$ <S> I = <S> C <S> * dV/ <S> dT$$ <S> whereupon we discover high rates of change of voltage will trigger the Triac. <S> The charge injection of dV/dT also puts FETs at risk. <S> Unless sufficient Source contacts and Well contacts are in place, the charges will pursue ALL POSSIBLE PATHS; current crowding into contacts may cause I*R drops large enough to turn on emitter-base junctions of parasitic bipolars, in which case the bipolar adds to the current flows. <S> In many cases, that brings about Gain > 1 positive feedback, and the FET/bipolar try to discharge the entire VDD charge storage network down to ZERO VOLTS. <S> With that mere attempt, silicon and aluminum melt. <S> How to avoid? <S> Design the Source and <S> Well contacts for transient charge tasks, not just for DC leakage control. <S> Here is microphotograph of high voltage under transient conditions (1volt per nanosecond) injecting charge, with that charge then crowding around a Well Contact.
| Dv/dt is the expression for charge injected into the Triac's internals (the silicon);
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Calculating CCA of a starter battery I'm searching for some kinda of formula that can be used to calculate the Cold Cranking Amps (which seems to be an industry term for a 30 sec discharge pulse current at a specific temperature, 0 degrees F) of a battery. I work in design, what I'm trying to do figure out the required CCA to start an engine and compare it to the CCA of the battery installed. The idea being to make sure the batteries speced aren't oversized. I came across the equation: E = V* I* t = V* A-hr Which makes sense to me intuitively, but doesn't account for the effect of low temperature battery. Any help would be greatly appreciated. Thanks <Q> The problem is that the current you can get from a battery is based on the exact temperature and details of the battery. <S> Even if temperature is specified, the details of the battery construction vary. <S> Different batteries that have the same voltage and same amp-hour capacity may have different cold cranking amp values. <S> Consider for example this: you may have 12V 45Ah deep cycle battery and 12V 45Ah starter battery. <S> Which one of these would you expect to have higher CCA value? <S> You probably guessed it correctly: the 12V 45Ah starter battery has a higher CCA value. <S> So, if the only thing you know is that the battery is 12V and 45Ah, you can't deduce the CCA. <S> You really need the datasheet of the battery! <A> The CCA of a battery is merely a rating that manufacturers give it. <S> If you want a formula to calculate the theoretical, temperature dependent CCA of an individual battery it would require extensive differential equations and a very in-depth knowledge of the battery's chemistry, and it would be even more difficult to accurately predict the actual performance of a specific battery, as there are a massive amount of variables that can affect CCA. <S> So no, there is no formula to calculate CCA. <S> It is simply a rating. <A> CCA cannot be “measured,” but it can be “estimated” and the process can take a week per battery. <S> A full CCA test is tedious and is seldom done. <S> SAE Test Fully charge battery accordingto SAE J537 and cool to -18 <S> °C (0°F) for 24 hours. <S> While at subfreezing temperature, apply a discharge current equal to the specified CCA. <S> (500 CCA battery discharges at 500A.) <S> To pass, the voltage must stay above 7.2V (1.2V/cell) for 30 seconds. <S> IEC Test Fully charge battery accordingto SAE J537 and cool to -18 <S> °C (0°F) for 24 hours. <S> While at subfreezing temperature, apply a discharge current equal to the specified CCA. <S> (500 CCA battery discharges at 500A.) <S> To pass, the voltage must stay above 8.4V for 60 seconds. <S> DIN Test Fully charge battery accordingto SAE J537 and cool to -18 <S> °C (0°F) for 24 hours. <S> While at subfreezing temperature, apply a discharge current equal to the specified CCA. <S> (500 CCA battery discharges at 500A.) <S> To pass, the voltage must stay above 9V for 30s and 6V for 150s . <A> You can't. <S> A starter battery specs list <S> both 20-hour discharge capacity (in Ah) AND CCA in Ampere, which is also a capacity, but for some pretty high current discharge where lead-acid batteries are profoundly non-linear. <S> The capacity in both modes usually differs 10..30-fold. <S> Both of them can only be tested and not really measured (except by testing multiple times and approaching the right value). <S> Both of them are usually guaranteed only for a new battery and should be derated for aging AND for using a partially-charged battery. <S> They are related to different modes of operation and a battery used in some car (or other battery-started engine) is selected to fulfil both. <S> A battery of a given size is a compromise between 20-hour capacity, CCA, cycle life, calendar life, price and other objectives. <S> Each one of these can be tweaked against others - more active mass gives more 20-hour capacity but leaves less room for conducting paths so less CCA and so on. <S> ... <S> and, a sane car w/ 2.5l gasoline engine in a moderate weather can be started w/ a new, fully charged 9Ah UPS battery. <S> I had to try once <S> and it worked. <S> It doesn't mean that the original 60Ah/600CCA battery is over-engineered. <S> I had to start it in this exotic manner exactly because the original battery was too smal for my (rather exotic, but pretty much legitimate) use case. <A> No way to calculate it properly afaik. <S> However, some of the cheaper Chinese battery testers on the market (Biltema, Topdon etc) simply seem to estimate it by measuring the battery's internal resistance (R), then present the CCA as: CCA = 2985,6204 / R (with R in mOhm) <S> , CCA rounded to nearest multiple of 5. <S> Very strange. <S> I have tried to find out why they do it this way.
| To test CCA, apply different discharge currents to see which amperage keeps the battery above a set voltage while cold. No such formula exists.
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Why do lead-acid batteries fail after about five years of operation? I had some time ago a failure in the batteries of my UPS. The UPS periodically performs a self-test of the battery, and it failed, causing all the computers behind the UPS to shut down. The batteries were gel cells, and failed five years after I purchased them. What is the physical/chemical reason for the batteries failing after about five years of operation? I do understand that lead-acid batteries suffer from sulfation if not kept continuously full, but in this application they were continuously full and in fact never saw a full discharge cycle. I have adjusted the float charging voltage to be 13.65 V. Is the float charging voltage incorrect, as the batteries failed? According to Wikipedia, starting batteries suffer early damage if kept in continuous charge due to corrosion, but these are not starting batteries. <Q> Pulse charging can also reduce sulphation rate but not replace need to periodically equalize cells with full charge voltage. <S> ( e.g. every 6 mos) <S> Ambient temperature rise greatly accelerates sulphation. <S> https://googlegalaxychemistry.blogspot.ca/2017/03/leadacid-battery-preventive-maintenance.html#content http://batteryuniversity.com/learn/archive/can_the_lead_acid_battery_compete_in_modern_times <A> Five years is not that bad for a non-AGM battery <S> Float charge voltage for lead acid batteries is about 2.26v per cell plus or minus 0.1v <S> NEVER over charge a sealed battery. <S> Set the float at 2.25v/cell. <S> Temperature should be kept near 20°C ambient. <S> While charging the charge the internal temperature should be held at 25°C or charge voltage should be reduced by 3mV per cell for each degree above and increased by 3mV per cell for each degree below. <S> The charge voltage ripple should be limited to 5%. <S> And you have no deep discharge issues or water and electrolyte to maintain. <S> Lead Acid batteries use a CC/CV (constant current/voltage) charge method. <S> Lead Acid needs a slow charge, 12-18 hrs, in two stages, CC and CV. <S> A stationary battery <S> even a slower charge is beneficial, around 3x slower. <S> For your stationary sealed battery, 36-48 hours. <S> The first stage, is a constant current which lasts for about 3-4 hours until the battery is about 70% charged. <S> Stage two is a "topping charge" a constant voltage, as the battery starts to saturate the current gradually slows. <S> When the charge reaches 3–5% of the its rated capacity, stage two is over. <S> This usually takes about 7–10 hrs (20-30 hrs for yours). <S> The battery should not stay at the topping voltage for more than 48 hrs. <S> This is critical for sealed batteries. <S> Sealed batteries are not overcharge tolerant. <S> Setting the correct full charge voltage (2.30 - 2.35v per cell) is critical and debatable. <S> Keeping the battery at topping voltage, this maximum capacity avoids sulfation on the negative plate. <S> On the the hand over charge saturation causes grid corrosion on the positive plate. <S> It's a balancing act. <S> For a stationary battery under load, after stage two, the battery would then go into a float charge at about 2.26v/cell. <S> For a UPS, not being under load, rather than a float charge a hysteresis charge is recommended. <S> When the battery goes to standby after the full charge, the float disconnects. <S> When the voltage drops due to self-discharge <S> a CV topping-charge is used to bring it back to full charge. <S> Pulse charging should not be used. <S> Sulfation can not be measured so pulse charging results are inconclusive. <S> Why fix what may not be broke, at the risk of doing harm? <S> The topping charge does a good job at preventing sulfation. <A> With wet batteries you have options. <S> I had a car battery last 11 years by adding EDTA and also doing an equalizing charge a few times a year: 4 amp solar panel connected for a few hours, lots of bubbling. <S> Golf Cart batteries lasted years also, with EDTA <S> and they were connected to the solar panel through a charge controller. <S> But I had some problems with the controller at first (long wires to panel made it oscillate) so the golf cart batteries overcharged sometimes (smelled like a swamp in the radio room). <S> If you want to make a small solar - battery system for emergencies or hobby, a modest panel and two golf cart batteries with EDTA added is your cheapest and best bet.
| Monitoring specific gravity and electrolyte levels can lead to longer life with periodic full equalization charging. There is not much you can do to extend the life of sealed / gel batteries.
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Measuring noise on DC rails with active probe I have a Tektronix P6245 active probe (allows only DC input) and wish to use it to see noise on various power supply rails. I've done this already on a +3.3V rail, and can get sufficient DC offset on the scope (Tek TDS744a) to see millivolts. However, it gets tricky with higher voltages (within the probe limits). For example, a +12V DC rail is tough. Bringing the vertical resolution to a level that will provide meaningful data is simply not possible. I'm considering a dc blocking capacitor between the probe and the rail under test. Is this a reasonable option? Any suggestions for best way to do this? <Q> Options: Money <S> no object? <S> The cost of a new R&S scope to use it with may be prohibitive, though. <S> I'd be surprised if Tek, Keysight, LeCroy, etc., don't come up with similar probes in the near future. <S> Use a different probe that allows for ac-coupling. <S> Forget the fancy probe and just contact the power rail through a 50-ohm resistor with the center conductor of a chopped off piece of coax (and contact the o.c. to ground nearby), and ac-couple the other end to your 'scope. <A> Yes, go ahead and use a DC blocking cap. <S> Active probes do have high input impedance, so peculiarities of capacitor actual R-L-C do not matter much. <S> If you need more accuracy along the spectrum, you can always calibrate your tip. <S> I usually use the standard Tektronix tip (plug), shave the sharp edge flat (with Dremel), and solder a 0603 ceramic capacitor as a new tip. <S> A 0.1uF <S> 25V MLCC cap should give you a bandwidth from about 1.6Hz and up. <S> CAUTION: <S> However, be aware that the single-ended probe must have good grounding, so be careful not to pick up noise on the ground lead over the air, or from improper grounding point. <S> Instead of long lead, use a pogo-pin (spring-loaded) plug into ground receptacle of the P6245 probe, and have an open ground surface near your point of measurement. <S> You will need to properly align both tips to make good connection. <A> Your probe seems to have a gain of x1, therefore it provides only a dubious benefit in this case. <S> Since a power supply has low impedance (hopefully it has decoupling capacitors!) <S> Set scope to AC coupling. <S> You don't need to enable internal 50R termination, since you have 50R source termination. <S> Soldering the coax frees up your hand to perform other tests, if you want to measure your supply noise with your microcontroller/board/DSP/whatever doing various things, for example. <S> I like MELF resistors for this purpose, because they are small, have good HF behavior, and unlike SMD chip resistors, won't break when the cable pulls slightly. <S> This will give you the best measurement possible. <S> The place you solder the coax ground to is also important. <A> So the best way would be to just use a normal probe lower bandwidth probe (if you don't have one maybe you should get one) and run the probe in AC mode at a higher gain than 1x and take out the hassle of adding capacitors. <S> I also find it hard to believe that you can't set the scope to AC coupling.
| the easiest (and also best) way to measure its noise is to solder a 50 ohms resistor to your test point, and connect this to the scope using a piece of 50 ohms coax with a BNC at the end. R&S has a new probe specifically for this application, with separate AC and DC signal paths so that large offsets can be accomodated while measuring small AC signals.
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A motor with high holding torque? Is there a motor which has regular torque, but extremely high holding torque? I mean, the motor probably holds by some mechanical lock rather than by the torque generated by the magnetic field. Is there some such motor, either linear or regular? <Q> A worm gear is what you are looking for. <A> The usual type of brake used is a spring-set brake. <S> A spring applies the brake and a electric solenoid releases the brake by compressing the spring. <S> That way, power is not required for the brake to hold. <S> AC motors are readily available with a spring set brake mounted on one end. <A> Depends on your definition of holding. <S> A stepper motor will have a high holding torque within plus minus half the pole angle. <S> If it's a servo motor with positional feedback and proper controls you can also lock a motor "around" one spot with a lot of torque. <S> If you need it rock solid in both directions, or need it to hold when the lights go out, a worm gear or mechanical lock / brake will be required. <A> One common type for industrial purposes is referred to as a "Brake Motor" or "Conical Brake Motor". <S> This type of motor is arranged such that the magnetic forces that create the motor rotation also cause the armature to move laterally such that when the motor is unpowered a spring forces the armature into its resting position and engages a brake. <S> It can therefore slow-down the load and lock it in place.
| What you seem to be describing is a motor with a friction brake.
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Why Circuit Breaker has neutral and no ground? My understanding is circuit breaker only need to trip when the hot end current jumps up (e.g. short circuit), but I have seen circuit breaker has dual (two) poles with both red cable and blue cable going in and out - what is the reason of connecting neutral to circuit breaker? Also I do not see anywhere around the circuit breaker where there are cables connected to the ground - without ground, how could any voltage have a reference? Sorry for the newbie question, but I have read a few on circuit breaker but did not find the answer. <Q> Its not about voltage it's about current. <S> Old style circuit breakers toggle off when the current going through them is enough to trip a relay and work like a fuse. <S> "Earth leakage circuit breakers" also watch the neutral line. <S> Theory being <S> what ever go's UP <S> must come Down. <S> That is the current going out of the live wire does not match the current returning on the neutral wire, there must be a leak so ground somewhere... <S> However it's a bit of a misnomer <S> , the current could actually 'LEAK' back to the panel through some other neutral path too if the wiring is not correct. <S> The newest standard also detects electrical arcing in the system. <A> Dual pole circuit breakers are used in biphasic circuits. <S> In some distribution circuits the demands are supplied using two phases and a neutral cable instead of one phase and one neutral cable. <S> This has advantages for both utilities (it is easier to balance loads) and end-users (sometimes when a fault occurs only one phase ends damaged and the other keeps working). <S> There are also some devices that require high amount of power and are designed to be fed with two phase cables. <S> The red and blue cables you mentioned are probably the two phases of the circuit, not a phase and the neutral (neutral cables are typically white). <S> The circuit breaker does not need connection to any reference, as it acts only as a switch. <S> For that reason there is no ground nor neutral cable connected to it. <S> Finally, you yourself said it "circuit breaker only need to trip when the hot end current jumps up", the breaker trips with high current, not high voltage. <A> (Assuming 120V split phase wiring) <S> The second wire is not neutral, but an additional "hot" wire that is 180deg out of phase with the red. <S> This will appear as 240V at the outlet. <A> In your observation, are you assuming the blue wire is Neutral? <S> Seeing that you are in the US, that would NOT be the case. <S> Neutral wires are required to be White in North America. <S> But elsewhere in the world, blue is used as Neutral. <S> Blue conductors here generally are Hot wires from a different phase or pole, depending on your system feed. <S> Different rules in different parts of the world dictate whether or not you switch the neutral circuit or not. <S> Here in the North America, we do NOT switch the neutral wire for the most part. <S> You CAN if you want to as long as you follow other rules about doing so <S> but in general, that is almost never done. <S> So in that case, the answer would be "in depends". <A> In my country (New Zealand), breakers for caravans, mobile homes etc require a true two pole Phase & Neutral breaker with current imbalance tripping as well as current overload. <S> This makes more sense in many cases as it ensures that both connections are broken regardless of how the input is wired. <A> Here's a French example... <S> Large breaker on the left is the RCD which feeds standard breakers. <S> They cut both live (red) and neutral (blue) for extra safety. <S> Normally touching the neutral should be safe, but there are corner cases where it isn't... <S> so it is mandatory to cut both, at least in new installations. <S> Note this is not two-phase (120V+120V=240V) <S> like in North America, this one is just one phase and neutral. <S> Protection Earth is never interrupted, obviously.
| In countries conforming to IEC wiring and grounding conventions, there are different types of power systems that REQUIRE you to switch the neutral conductors along with the "hot" conductors in a circuit. It makes especial sense in the case of equipment fed via flexible leads or temporary circuits which have a much higher risk of having connections swapped than do fixed wiring.
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Given the diagram, where do I add/replace a larger capacitor for continued power while bike is stopped? Following the circuit diagram, included as an image in this guide , where do I replace/add a larger capacitor for continued power while the bike is stopped? My guess is that it should replace the capacitor between the rectifier and the voltage regulator, but I'd like some input from someone who knows . I'm planning to insert a 2200uF capacitor. <Q> Your guess is right. <S> Placing it on the "left" side of the regulator works well. <S> You can also place it on the "right" side, but then you would need a few extra components (at least one extra diode). <S> It's easier on the "right". <S> Note: I have to use stupid words like "left" and "right" because the person who drew that schematic failed to add reference designators to the components. <S> Don't be that guy. <S> A few caveats: You have to make sure that the capacitor has a sufficient voltage rating. <S> Since you're taking the AC voltage from a dynamo, it's hard to say how high the maximum voltage will ever be. <S> I have very little experience with them. <S> I assume that 6 V is just a "nominal" value, and the output can increase. <S> The peak voltage will be the AC voltage times 1.4, then subtract about 1.6 for two diodes in series (your bridge). <S> $$6 \cdot 1.4-1.6=6.8$$Now double this (13.6 V), and buy a capacitor rated for at least this much. <S> I suggest at least 16 V, but 25 V would feel safer. <S> As brhans pointed out in a comment, unless you seriously beef up the capacitance, chances are that you will hardly notice the increased time. <S> This is hard to say without knowing the expected load. <A> The generator undoubtedly has a current limit, so a 'capacitor' after thediode bridge and before the regulator will work, but a battery in thesame place will work better. <S> 6V gel cell would be appropriate, andyour wheels won't go dark at an intersection on a rainy night. <S> A battery will be damaged if it discharges too far, so a disconnectswitch would be prudent. <A> If your generator is like the "dynamo" (sic) on my bike, as your road speed increases it tends towards a constant (AC) current device. <S> Reason: <S> both its EMF and its internal reactance increase in proportion to its rotation frequency. <S> I have for several years used a stack of either 5 or 6 1.2V AA size NiMH cells in this position. <S> They sometimes need recharging offline if I do too many miles with the lights on.
| The drawback of using a silly amount of capacitance (part from the obvious: cost and size), is that you will get a slower start-up time; it will take time to charge up the capacitors.
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Can I operate a 12vdc valve using a capacitor and AA batteries? First, I need to specify I'm very noob, so maybe in my situation a vague awnser is better than something very specific, I don't know. I want to make an airbow (airgun that shoots arrows) like the one Joerg Sprave made on his YouTube channel. Except, he powered it with a 12 dc battery, whitch is very straight-forward, but also quite expensive and heavy, compared to AA or maybe higher batteries, boosted with a capacitor. The thing is I know nothing of what to expect about the output of the capacitor. I know it can be quite high, because once I made fun of shocking myself with one made for a flash, with friends, and it gave quite a jolt. Anyway. I want to have a switch button, once pressed it would give power to the valve just long enough to deliver the compressed air and shoot the arrow, so it doesnt need long, once the arrow it shot, any excess air is futile. If it's simple to install, I would like a LED to signal when the valve/gun is ready to fire, but it's optional. I can go a bit with trial and error, but I wouldn't fry my valve in the process since it's the most expensive part of the project yet. I dont mind a schematic for the circuit if you can provide, but (ok I know I sound very helpless) if there was at least some clear note (e.g.: your valve, power source:aa battery), it would help me a lot to understand the big picture. I'm sure I can, but the last time I worked with electronics was more than 20 years ago, and I'm 37. I hope my question is not too boring or too easy or too vague, thanks for any help. <Q> 8 AA in series will produce 12V. <S> It should trigger the air valve at probably 9V (A 9V battery has a high ESR and a low capacity and current delivering power, so stick to AA or C cells for multiple triggering). <S> Unless the valve is incredibly power hungry. <S> The capacitor is to make it a bit easier on the batteries. <A> The first thing you need to do is find how much current the valve needs, and for how long. <S> Then you can calculate what size capacitor can deliver that current for that time, and not drop too much voltage in the process. <S> Once you have that, you have to decide how to charge up the capacitor. <S> You could put enough batteries in series to get 12 V, and then just connect the capacitor across them. <S> Since the capacitor by itself can operate the solenoid, you don't need a battery with very low internal resistance, like a car battery. <S> Since the equivalent resistance of the voltage source charging the cap doesn't matter, you can use just a few cells or even a single cell and a boost converter. <S> This is exactly what goes on in the camera flash you mentioned. <S> The boost converter takes the low voltage from the battery to make high voltage, and a cap gets charged with that high voltage. <S> This can take a few seconds because the energy required to activate the solenoid is provided by the battery slowly, stored in the cap, then released from the cap suddenly. <S> Boosting a single 3-4 V rechargeable lithium cell to 12 V isn't hard. <S> The time you have to wait for the cap to charge up depends on the size of the cap, and the current the boost converter can provide. <A> If the air valve operates on 12 volts, you just need a 12 volt battery. <S> The battery must be large enough (enough stored energy and current capability) to supply the current the valve requires. <S> I'm guessing the battery Joerg Sprave used <S> was a lead-acid car battery <S> (but I haven't searched for his YouTube video). <S> Eight AA or C cells in series will produce 12 volts and may produce enough current to operate the valve (but a datasheet for the valve would help to confirm this). <S> A capacitor will only charge to the voltage of its charging source - if you connect the capacitor across a 12 volt battery, it will only charge up to 12 volts. <A>
| If you don't want to use several batteries you can use a step-up switching regulator to generate 12V from lower voltages.
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Need help understanding this op amp/BJT configuration to troubleshoot why I'm reading different voltages than expected I am building this VCO in an attempt to study it and I'm having trouble understanding the references section seen bellow. I have used a KIA431 that I 'stole' from a broken switching power supply on the breadboard in place of the specified LM4041CZ-ADJ and it is giving me the +5V I was expecting. I am using the LM336Z-5.0 as a substitute in Multisim because I couldn't find a part model. Besides this, the rest of the circuit is quadruply checked to be as specified in the schematics and as seen in the Multisim snippet. To make this as clear as possible, I need the following things:a) to know what this op amp/bjt configuration is meant to achieve and why a simple buffer (or amplifier with a gain of 2) wasn't used to acquire the +-10V referenceb) the name of the bjt configurationsc) why I am reading different voltages than what is specified (+-10V). I am reading +13.85V and -13.18V on the breadboard and +12V and -11.3V on the simulation One thing that's also really bugging me to understand is that in both cases we see one diode drop difference between the positive and negative output voltages. P.S. I am assuming that the only reason this op amp/BJT configuration is used is to feed the CA3280Es at the Tri to sine converter on page 6 of the linked PDF. I would appreciate a confirmation on my assumption (or the opposite) and a link to any resources that might help me get my head around this. (This is not part of the question in the title so it will not be considered in my answer acceptance.) <Q> a) <S> You'll have a higher output current capability with this configuration compared to the one without the op amp alone. <S> b) I don't know if this configuration has a specific name. <S> I would call it "A BJT with current limitation" c) <S> Are you sure about R7/R5 and R9/R4 combinations? <S> Those look like gains of 4. <S> I'm guessing that the op amp can't output more (less) than a defined voltage (which is lesser than +15V or higher than -15V depending on which op amp you're looking at), not high/ <S> low enough to output the full +15V/ -15V of your rail. <S> That being said, you should make R7=R5 and 2xR4=R9 to get the targeted gain of 2. <S> Regarding the diode voltage drop you talked about, all I'm seeing on the simulation is an oscillating voltage (Vpp=11.3V and Vdc=-). <S> Let us know if the problem persists after the gain correction to think about it. <A> "I have used a KIA431 that I 'stole' from a broken switching power supply on the breadboard in place of the specified LM4041CZ-ADJ <S> and it is giving me the +5V <S> I was expecting. <S> I am using the LM336Z-5.0 as a substitute in Multisim" <S> That's the root of your problem. <S> I've never heard of the KIA431, but that LM336 is quite a different part to the LM4041. <S> The latter allows you to use resistors to set the reference output voltage from 1.2V to 10V, while the LM336Z-5 generates a 5V reference that you can trim by only a few hundred millivolts either way. <S> With the pot in the midpoint the LM4041 will generate a reference of about 2.5V, which the opamp circuits will multiply by 4 to get +10 and -10. <S> With the LM336Z-5 in its place, you'll get a 5V reference, which the opamps will try to multiply to +20 and -20 <S> but fail because they can't get anywhere near it given the supply voltage and the circuit configuration. <S> Try using an LM336Z-2.5 instead if it's available in your simulation. <S> Oh, and the transistors provide current limiting on the output. <S> The first is a current amplifier for the opamp output and it drives the load through the 10 Ohm resistor. <S> If the current flowing through the resistor reaches about 600mA the second transistor will start to turn on, reducing Vbe of the first one and limiting the output. <A> I also suggest you add a feedback cap, to change the phaseshifts of your opamp;a load resistor may also help, by reducing Rout of the emitter follower: simulate this circuit – <S> Schematic created using CircuitLab <S> Let's examine the world of stability. <S> We have two delays in this circuit: the opamp, and the transistor. <S> First we'll operate the transistor with 1uA current, because 1uA is a likely leakage current through an electrolytic on that emitter. <S> The Rout (from the emitter) of a bipolar is 0.026V/Ie_amps, thus 0.026/1uA = 26,000 ohms. <S> The delay of 10uF and 26,000 ohms is 0.26 seconds. <S> This very slow behavior, being monitored by the OpAmp through R3, produces an oscillator, unless we provide a sneaky way to monitor the OpAmp's output, through C2. <S> What if you add an intentional current through that emitter, to greatly reduce the emitter's Rout? <S> Make R4 be 10K ohms. <S> Rout is 0.026v/0.001 = <S> 26 ohms. <S> We've sped up the delay by 1,000x to 0.26 milliSeconds. <S> But the OpAmp is out at 1MHz, and acting very impatient with such long delays. <S> Again, use C2; if R2 and R are each 20,000 Ohms, the Requivalent on (-) <S> OpAmp pin is 20K||20K or 10Kohm. <S> Strive to get the feedback response up to 1MHz, or 160nanoSeconds. <S> Using $$Omega <S> * Tau <S> = 1$$, with $$Omega = <S> 2*pi*frequency$$, and $$Tau = R*C$$, <S> I know 10Kohm & 100pF - <S> > 1uS, thus 10Kohm and 16pF -> 160nanosecond. <S> Bigger caps are OK. <S> Here is BODE plot, along with OpAmp with NO Cfeedback and a large Cload. <S> You can alter the params, and rerun yourself.
| I agree with HatimB the opamp VDDs are a problem.
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Feedback in combinational digital circuits It is recommended that one should not use feedback in a combinational circuit and the reason is pretty clear because, the delay of the output that feeds to the input causes malfunction. However, the ACT-1 cell from ACTEL (see Fig. 2 from datasheet ) is a basic cell consists of some multiplexers (which is combinational) which is capable to implement even flip-flops! See this example which uses feedback to implement an RS latch. So, the question is, should we use feedback or not? <Q> As mentioned, combinatoric logic with feedback is used to implement flip flops and latches, so it isn't intrinsically bad. <S> However, there are some specific problems when using it, especially in an FPGA (where you are presumably eventually going to drive flip flips). <S> The first problem is that the settling time of a circuit with feedback is not statically determinable. <S> In the absence of feedback, the synthesis tool can simply sum the propagation delays of all the gates to determine if the outputs will settle by the next clock cycle, and thus determine the maximum operating frequency of a circuit. <S> If there is feedback, that is not possible. <S> Second, the type of feedback you are describing normally generates latches, not flip-flops. <S> This presents a few problems. <S> A latch can 'see' temporary / intermediate states and respond to them in a way that you don't expect. <S> For instance, if you have a counter that goes through the states 00 , 01 , 10 , 11 , <S> and you want a latch to fire when you see the state 11 . <S> During the transition between 01 and 10 <S> you may go through the state 11 . <S> Then, if you recompile the supply voltage changes, or the chip temperature changes the behavior can shift so <S> instead you go through an intermediate state. <S> This instance could be fixed by using Gray codes, but that won't always be possible. <A> Any flip-flop is built around combinational logic with feedback. <S> And nobody can stop you from building a completely asynchronous (and functional!) <S> design. <S> However, those designs are really complicated and certainly not for beginners. <S> For most designs, synchronous logic is preferred since many design and debug tools (like Static Timing Analysis) can be used only on the environment of a synchronous design. <S> So the answer to your question is (as already said in same comments), <A> Judging from your tags, you're suggesting to use VHDL and an FPGA to design combinatorial memory elements. <S> However, although you can simply design gates and multiplexers in VHDL, their logical function will actually be implemented into lookup tables (LUTs) on the FPGA. <S> The timing behavior will differ depending on how the functions are mapped onto these LUTs: what inputs are used, how many inputs are used, etc. <S> In addition to this, the size of the LUTs is limited (usually to 4-6 inputs), so for more complex functions multiple LUTs need to be connected together. <S> The connections between the LUTs (routing elements) each introduce another timing elements that all are very dependent on the final implementation in the FPGA. <S> You can configure your design's timing using design constraints: this tells the synthesis, place and routing tool what timing you want to achieve on your signals. <S> It is used a lot in complex high-performance applications. <S> However, it is very labor-intensive and quite difficult: the tool will not see the bigger picture for you, so you have to know what you are doing. <S> It also likely increase run-times of the synthesis, place and routing, as these tool will have to put a lot of effort into achieving your goals.
| unless you know very well what you are doing, do not use feedback on combinatorial logic.
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Why do some wire gauge ampacity charts state vastly different current ratings? This is a question that I've had for a while but have somehow managed to go without asking. Looking at all the wire gauge ampacity charts online, some of them appear to provide vastly different ratings for the same wire gauge. I understand this may be due to different standards, temperature, conductor types, strand count, etc. However, lets looking for the current rating associated with 10AWG copper conductor at a low operating temperature (60c): Chart 1: 15 amps Chart 2: 30 amps I've only cited two sources, however, I've come across many other charts that provide this 15 to 30 amp difference. What gives? <Q> Notice that your first link gives two ampacity values for AWG 10 wire. <S> For power transmission applications, the rating is 15 A. <S> For chassis wiring applications, the rating is 55 A . <S> These ratings (and the third one at your other link) are based on different assumptions of convective cooling available and allowances for wire self-heating. <S> For example, the first link says The Maximum Amps for Power Transmission uses the 700 circular mils per amp rule, which is very very conservative. <S> The Maximum Amps for Chassis Wiring is also a conservative rating, but is meant for wiring in air, and not in a bundle. <S> You should also consider resistive voltage drop along the wire if you find one of the higher ratings appropriate for your thermal environment. <A> The temperature a wire will reach when carrying a certain amount of current depends upon the gauge, the ambient air temperature, and the amount of thermal insulation between the wire and the air. <S> Passing a certain amount of current through a piece of bare copper wire which is surrounded on all sides by air will not cause its temperature to rise nearly as much as as passing that same current through a like-gauge wire which is buried under 4 inches of fiberglass insulation. <S> The difference between the 55A rating in one table and the 10A rating in the other is likely because of this. <S> The 55A and 30A figures assume the limiting factor is heat dissipation. <S> The 15A rating takes into account another factor: any power that gets converted into heat won't be usefully delivered to its destination. <S> If one were trying to send 2400W of power over a cable, one wouldn't want to lose hundreds of watts in the cable even if it could dissipate that much heat safely. <S> While the required gauge to usefully send a certain amount of current with a certain fraction of loss will depend upon the voltage and transmission distance, there are many situations where it will exceed the gauge that would required to avoid overheating. <A> The two charts are based on different criteria. <S> The first chart uses a rule which says that a wire should carry no more than 1 amp per 700 circular mils of cross-sectional area. <S> The second chart gives the maximum current for several different temperature increase. <S> (The environmental conditions are not specified.) <S> You'll notice that the first chart says 15 amps for "power transmission" (long runs in enclosed spaces) but 55 amps for "chassis wiring". <S> So there's a wide range depending on the thermal conditions. <S> I suggest staying closer to the low end unless you know what you're doing. <A> Probably more importantly; The first chart is a listing for the COPPER wire itself and if you notice, they also list it as 2600Hz with regard to skin effect. <S> This is a chart for engineers who are going to make cable to be used in high frequency applications, not use it in the real world. <S> it even says that in the beginning: "This data is useful for high frequency AC engineering." <S> details matter when reading things on the net. <S> The second chart is based on the National Electric Code for insulated wire, as in what you would buy and install in an every day application. <A> The type of insulation used determines the maximum ampacity of a wire. <S> In reality, this maximum must also be derated for safety (normal load should be 80% of rated ampacity), ambient temperature, armoured sheaths, raceways or adjacent power conductors. <S> The first site is more: As you might guess, the rated ampacities are just a rule of thumb. <S> Long power runs, assume 15A for #10. <S> The Maximum Amps for Power Transmission uses the 700 circular mils per amp rule, which is very very conservative. <S> Short power runs, assume 55A. <S> The Maximum Amps for Chassis Wiring is also a conservative rating, but is meant for wiring in air, and not in a bundle. <S> They protect their butt by: NOTE: <S> For installations that need to conform to the National Electrical Code, you must use their guidelines. <S> Contact your local electrician to find out what is legal! <S> If you figure a maximum of 3% voltage drop to a feeder, this means a specific maximum distance for #10. <S> At 120V, 3% means 3.6V. <S> At 30A, this equates to a feeder resistance of \$0.12\Omega\$ or feeder length of 120ft or 60ft from panel. <S> 30A with a load 60+ft from panel violates the 3% maximum voltage drop. <S> Too much current for the insulation, which would go through irreversible deterioration and possible fire, death and destruction. <S> You could do the math and figure out the correct size of wire for the wire distance or using their rule of thumb and half the current. <S> To get 30A, double the area or #7.
| You should use a rating consistent with how you're using the wire and the cooling environment around the wire in your application.
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Can I use a resistor to slow down a DC motor? I have a few DC motors of 1.5 V and 9 V and need them for some applications. These motors run at a certain speed (rpm). But I need to slow them down. Will it be a good idea to connect a resistor in series with the motor and the battery in order to reduce the motor speed? Will doing so create any bad effects to the motor? I know there are other methods I can employ to reduce the rpm of the motor such as using gears or a PWM circuit, but I'm interested in knowing whether using resistor is a good idea or not. <Q> Using gears is always best because that is the method that makes available the highest percentage to the motor power capability. <S> Since power is torque multiplied by speed, keeping most of the motor's power capability increases torque capability while reducing speed. <S> Pulse width or pulse amplitude modulation is second best. <S> That can preserve the highest torque capability while reducing operating speed and power capability. <S> The best way to maintain torque capability is to have an inner control loop that regulates torque by regulating current. <S> An outer control loop regulates speed and provides the current reference. <S> Because a switching regulator is used, there is less power wasted than would be wasted by a linear control technique or series resistance. <S> A series resistor can certainly be used. <S> It can be a variable resistance (rheostat) or one or more fixed resistors. <S> With a series resistor, the wasted power is directly proportional to the percentage speed reduction. <S> Since there is no current control, adding resistance increases the speed change that results from any load change. <S> The speed variation due to load variation increases in proportion to the amount of speed reduction. <S> There is also speed variation caused by resistance change due to resistor temperature variation. <S> Additional Considerations <S> If not only "slowing down," but also variable speed is required, it is best to select gearing for the proper maximum speed and then use electronic control to provide variation. <S> Larger DC motors are seldom used for fixed speed operation, but electronic control or series resistance may be required just for starting a larger DC motor to avoid excessive starting current. <S> For very small DC motors, series resistance may be perfectly adequate in some situations depending on various factors such as power source, duty cycle, cost analysis, maximum speed reduction, load stability, etc. <A> The speed of a permanent magnet DC motor is primarily determined by voltage. <S> The problem with using a series resistor is that its voltage drop is proportional to current. <S> If the motor draws constant current then the resistor will drop a constant voltage and the motor will run at a fixed (lower) speed. <S> However the motor will draw more current when starting up and as the load increases, causing higher voltage drop and reducing speed even more. <S> If it is driving something whose loading increases as speed increases (eg. <S> a propeller) the speed will eventually stabilize, but if the load is variable then speed regulation will be poor. <S> If the resistance is increased too much the motor may stall under heavy load, or not even start up if the initial load is too much for it. <S> This is good for some applications because the resistor limits torque and protects the motor from being damaged by high current, but if you want to maintain relatively constant speed under varying load then it's bad. <S> If you want to reduce rpm without compromising speed regulation then use a voltage regulator or PWM controller to lower the motor voltage. <S> If you also want higher torque then use a gearbox (which increases torque by the same proportion as it reduces shaft rpm). <S> Most small DC motors are designed to run at high rpm, so for applications that need low rpm a gearbox is usually the best option. <A> If you want to go really slow, the resistor method will probably cause the motor to stall way before you reach your desired RPM. <S> Using PWM ensures you get pulses of full torque, which allows you to drive the motor to really slow speeds. <A> Yes, using resistors were the first method of controlling motor speed. <S> Back in the early days of electricity, street trolleys were driven by electricity and the speed controlling was done by moving mercury in/out of tungsten pipes. <S> The more mercury filled the pipes the faster the cart was running. <S> The speed of a carousel was also controlled by a bucket of salty water, deepening electrodes in it. <S> Back those days electric wires were kept as straight as they could not to loose high speed electrons by missing the curves and leaving the conductors forever. :) <A> This assumes you've just got a little motor such as is common in battery-powered toys or fans, running at low voltage, and you just want to slow it down a little. <S> It also assumes that you really want a cheap and cheerful solution rather than a control loop, and are prepared to tolerate some inefficiency. <S> One or more series diodes may be used to drop the voltage the motor sees while still allowing it to draw more current under load than a resistor. <S> This may be too much for the 1.5V motor (though it may be possible with a careful choice of diode taking into account the forward voltage drop at the current you will use) but works well in the 3--12V range, for example for slowing PC fans to make them quieter. <A> Simplifying matters to some degree, the motor speed is proportional to the voltage applied to it, while the torque is proportional to the current. <S> A series resistor will only limit that voltage reliably when the current is kept constant, meaning the motor is running at more or less constant load. <S> If this is your case, a rheostatic speed control can be a viable solution (lots of toy slot cars use exactly that): <S> If your load will vary significantly, so will the current. <S> As the voltage drop on the resistor is proportional to the current, you will observe that the motor speed gradually drops as the load increases. <S> This behavior is not harmful to the motor, but it may be not what you want.
| You CAN use a resistor, but understand all you are doing is dumping power out the resistor to drop the voltage to the motor.
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Is it safe to drive I/O pins of an unpowered AVR ATmega328P? Is it safe to drive the pins of this AVR chip while it's unpowered? For example, if I were to share a motor driver such that it can be driven by two microcontrollers, would either microcontroller be damaged while the other is unpowered, but is seeing the signal voltages on its pins from the other microcontroller? EDIT: Because this was marked as a duplicate, here is some clarification. The MCU specified here is different from a different manufacturer. Though I'm not sure, their operation and limits would most certainly differ. This question was not about wether it'll 'weaken' the MCU but about if its safe to do so in the first place. IMO weaken would mean that it is possible but over time it'll degrade the performance of the MCU. <Q> You may be interested in EEVBlog's video about powering an MCU without connecting the power pins . <S> If you drive the <S> I/O pins on the MCU, you may end up powering up the MCU through its protection diodes. <S> If, after the diode drop, your VCC ends up being below or above the minimum voltage required to drive your MCU (or draw more current than the internal diodes can handle), then it will not operate correctly. <S> Of course, this is assuming your VCC is open circuit and not shorted to ground (or zero volts). <S> Here is a diagram of the equivalent circuit for your Atmega328p: <S> Note the diodes from ground to VCC. <S> Either way, the datasheet does not explicitly specify the exact behaviour of driving the Atmega328p from an I/O pin. <S> However, it does say the absolute maximum voltage you can apply to an (non-nRESET) I/O pin is Vcc+0.5V. As mbrig pointed out in the comments, if you phantom power the chip, the diode drop will very likely cause the voltage on the I/O pin to be larger than Vcc+0.5V. <A> From the datasheet: The maximum voltage on any pin other than _RESET is specified as "Vcc + 0.5V". <S> If the chip is unpowered, Vcc is 0, so no -- <S> you cannot safely apply any voltage above 0.5V. <S> If you want to share an <S> I/O between two parts, where one of them may be unpowered, you need some kind of switch. <A> No. <S> At the input of most pins on a device such as the AVR ATmega328 there are diodes to protect the chip from electrostatic discharge (ESD). <S> These diodes connect to the supply rail of the device. <S> This may not cause damage, but often it will affect the power-on-reset circuitry, so the device will not power up correctly when power is available. <S> It is often a cause of unexpected power consumption in battery powered circuits. <S> You need to put in a multiplexer to select which MCU drives your motor driver or arrange that both MCUs are powered, but the one that is not intended to operate the motor keeps its I/O ports in a high-impedance state (e.g. as an input). <A> Unsafe for both AVR and the other MCU. <S> The AVR will try to power up when the other MCU drives its I/O pin "high". <S> Not good for AVR <S> - it is meant to power-up from its dedicated supply pin. <S> Bad things can happen otherwise. <S> Since the manufacturer explicitly says "don't do this", the manufacturer is not obliged to say what "bad things" you might encounter. <S> My experience is that "bad" ranges from very strange to smoke . <S> Since the AVR may look like a short-circuit to ground, the other MCU (trying to pull its output pin high) would encounter a heavy load. <S> It may fail to pull as high as it should, and cause too much current to flow (putting it at risk). <S> At risk also is your motor driver, especially if it is PWM. <S> Not knowing if it is to be "high" or "low" (somewhere in-between), it can overheat. <S> So you risk everything . <A> According to the spec V ih = <S> V CC + <S> 0.5 V so from that alone it's no. <S> However, if V CC is really "open", feeding the I <S> /O pins will apply a phantom V CC to the device <S> so it won't really be off, so that kind of negates the above issue, but it opens a whole new can of worms. <S> Even if the above were not an issue, there is no guarantee you will actually be ABLE to drive an I/O pin high when the device is powered off. <S> You may find the output is extremely low resistance to ground. <S> The latter issue also applies to input pins too by the way. <S> Assuming the chips V CC has decayed to ground, the input will represent a diode to a capacitor to ground on any signal you are trying to measure with the other device. <S> Basically... don't do it. <A> Although it is not in the datasheet, the rule of thumb is that the internal ESD protection diodes can have 1mA for a 8-bit AVR chip. <S> The diodes allow more current, but the 1mA is the recommended maximum. <S> When a voltage divider is made with two resistors, this 1mA can be used to calculate the maximum allowed voltage for the voltage divider. <S> The Atmel Application Note AVR182 mentions the 1mA. A positive current into a pin flows to VCC and could power the AVR chip and/or the circuit. <S> That could result into unpredictable behaviour when the AVR chip is running near its lower voltage limit. <S> With enough voltage and current it could even raise the voltage far above 5.5V, and destroying the AVR chip. <S> Especially when the AVR chip is in sleep mode. <S> It is therefor dangerous to use, but it can be done. <S> Always be careful when applying voltages to a pin. <S> Use a protection resistor to limit the current. <S> That is also safer for normal operation. <S> That means that when two AVR 5V chips are connected to each other, there should be 4k7 resistors between them to limit the current to about 1mA when one AVR chip has no power. <S> Depending on the circuit, that 4k7 might influence the signals. <S> There might be even more consequences, I'm not sure that mentioned them all.
| If you drive a logic signal into a pin when the device is unpowered it will attempt to power via the diode. The data sheet states that operating the ATmega328p outside of it's absolute maximum ratings can cause permanent damage so I would recommend against it.
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Can floating cat6 shield be worse than unshielded? I'm having an electrician fish an Ethernet cable through a wall. I can put shielded or unshielded cat6. However, the electrician doesn't know how to crimp Ethernet cable so I agreed to do that part myself, and I only have materials on hand and experience for plastic crimps. Thus if I put shielded cable, the shield will be left floating. Can a floating shield potentially have worse characteristics than an unshielded cable? <Q> Usually the cable goes to patch panel where you wrap the shield wire on the metal case that is grounded. <S> The other end goes to metal socket and similarly the shield is connected on metal case. <S> If you don't connect the shield, then there is no benefit using FTP over UTP <A> Worse, an ungrounded shield becomes an antenna that will resonate at a frequency that is dictated by the length and geometry of the wire. <S> That frequency will of course be nicely transferred to your signal lines. <S> So yes, ungrounded shield can make things worse. <A> I would rather specify HDPE sheath. <S> It will protect the Ethernet pairs as it is much harder than the normal PVC. <S> Make sure that you properly test the cable. <S> (If it is damaged, redo the pull. :-)
| If the shield is not grounded, at least at one end, it is not really a shield so much as a wall, and noise can get around walls.
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Why transformerless power supply need a capacitor to decrease the current? Assuming we have a 12 V load which needs 1 A current, when we connect this load to 60 Ah 12 V car battery it does not need to resistor and it will ony draw 1 A. But on the other side, assuming we have a transformerless power supply circuit connected to 220 V so the DC voltage after diode bridge will be 310 V. Why can we not connect 100 LEDs in series (each 3.1 V) directly without using a capacitor or resistor? I know this capacitor is used as reactance (\$X_C\$) to limit the current, but why is it needed since the voltage of LEDs is equal to the output voltage of the power supply? Shouldn't the LEDs should draw the required current? According to this equation R=(Vin-VLED)/I =(310-310)/I = zero resistor needed <Q> How confident are you that your rectified mains voltage will be exactly 310V? <S> How confident are you that 100 LEDs in series will have a voltage drop of exactly 310V? <S> Have you read the datasheet? <S> What is the Forward Voltage tolerance specification? <S> What if your mains is actually 315V and/or your string of LEDs adds up to 308V? <A> If you short the capacitor the total resistance in the circuit will be about 100 \$\Omega\$. <S> The total impedance with the capacitor not shorted at 50 Hz <S> is $$|100 + <S> {1\over j \omega C}| =|100 + <S> j~3185| = <S> 3186~\Omega.$$ <S> At 60 Hz it will be <S> about \${5\over6} \times 3186~\Omega = 2653~\Omega\$. <A> It is called "capacitor dropper". <S> Note that you are talking about AC current (230V AC). <S> Capacitors and inductors has some impedance, it decreases current. <S> Lower capacitance = <S> higher impedance = <S> lower current. <S> So, current is limited. <S> If you jump it, it means that you remove that impedance and current jumps very high and burns LEDs. <A> Assuming we have a 12 V load which needs 1 A current, when we connect this load to 60 Ah 12 V car battery it does not need to resistor and <S> it will ony <S> draw <S> 1 A. <S> You started out with a fallacy. <S> To have a load on 12v drawing 1 <S> Amp you would have to have a 12Ω load. <S> Because the load would be resistive. <S> You could not connect a string of LEDs directly to a battery with out them burning, <S> so why would you think you can connect them directly to an AC source and have a different result? <S> They required something to limit the current. <A> Your mistake is assuming that a raw LED is a "normal load", it is not. <S> Normal loads have a relatively gentle voltage/current relationship. <S> Small changes in conditions, either voltage of the power supply or temperature of the load, result in small changes in the current. <S> LEDs have a very steep voltage/current relationship, <S> a small change in voltage can result in a very large change in current (so much so that a common first approximation of a LED is a component that maintains a fixed voltage for any forward current). <S> They also have significant variability with temperature and manufacturing. <S> When LED manufacturers give a range of forward voltages they are NOT talking about a range of voltages that the LED can be fed with blindly and operate correctly, they are talking about a range of voltages you might encounter when you drive the LED with a specified current. <S> Even with the capacitor the design of your light is awful. <S> Leaving asside the problems of flicker that come from using rectified mains, the current drawn will vary massively over the range of mains voltages you are likely to encounter and likely also with the temperature of the LEDs.
| LEDs need current limiting - they are not voltage-driven devices. LEDs have a dynamic resistance.
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Send simple signal over AC power line for Arduino / ESP8266 I am looking for the simplest electronic solution to be contained in boxes B1 and B2 so that when the switch SW is toggled, a signal (free choice) should be carried across the AC power lines (residential 110/220V) to B2 where it should be detected and converted to a small voltage meant to feed some input GPIO pin of a microcontroller. In other words, the uC should be aware whenever the switch gets toggled, provided the configuration. B1 and B2 are some feet apart and the circuit is also used for other consumers, therefore no huge voltage or current spikes can be used unless filtered by B2. The carried signal can be of any kind as far as B1 can generate it and B2 can detect (and further cancel) it. I hope for an easy solution, not involving additional uC or heavy signal modulation processing, if possible. LE: based on the provided answers I want to stress out the need of the simplest solution to send/receive a "sign" of any kind over the power lines . I don't need to de/modulate data, a fair voltage spike generated by B1 that can be detected and filtered out in B2 would be enough for the "sign" I need, if possible. I'd rather avoid uC and heavy processing in B1, maybe just in B2. <Q> You were not clear if you wanted to design your own or buy an off the shelf product. <S> Either way I doubt you will do better than this cost wise. <S> LINK <S> Product Information <S> LINK to buy Shield $50 <S> They also have a Powerline Communication Module $29 Or build your own from the schematics (PDF) <A> Unfortunately they seem to be black magic with regard to tracking down a good schematic, however the process was described as: "Such products are based on a standard called HPAV - HomePlug AV. <S> They essentially couple the AC line with the digital baseband signal. <S> The AC line has a frequency of 50-60Hz, while the digital communication happens between 2Mhz & 50Mhz. <S> The upper limit is determined by the flavor of HPAV used. <S> On the receiver, the AC interference is filtered out and the digital signal is demodulated. <S> - https://www.quora.com/In-a-nutshell-how-do-ethernet-over-power-line-adapters-work <S> This is appears to be what you are trying to do. <S> In searching I found the following site about making a WiFi switch for an electronic device, so depending on how tight your constraints are either solution may work... <S> http://www.esp8266-projects.com/2015/03/esp8266-cbdb-web-power-switch-for-mains.html <S> Whatever you end up going with, be careful since it is mains and try to have someone else around just in case. <A> If you are willing to take a different approach, then there is a simpler and inexpensive way to do all this. <S> You can install two esp8266 modules - one each in B1 and B2. <S> Esp in B1 will detect the switch state and transfer the information to B2 over wifi. <S> Follow this answer to learn how to sense ac using a micro-controller: https://electronics.stackexchange.com/a/208596/68606 <S> The answer was written for R-Pi but it will hold true for any micro-controller.
| Powerline Communication Shield for Arduino So the first thing that came to mind was those ethernet over power boxes (not to be confused with Power over Ethernet).
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Failure mode of TTL device A question came up about common failure mode of a transistor. Searching past posts I actually found varying answers, either open or short. Maybe it depends on the application, and I'm missing something? "The output of a TTL NAND gate is being used to drive a relay ON by pulling the source low. The diode for the relay was installed backwards, effectively shorting the gate's output to the power supply. Which component has most likely failed? The gate's output transistor has failed open The gate's output transistor has failed shorted" simulate this circuit – Schematic created using CircuitLab The book provides that 'open' is the correct answer, not 'short', but doesn't explain why . <Q> If the transistor failed short, it (and the diode) would form a direct path from power to ground, and would experience a large current flow. <S> Between the two of them, one would eventually fail, and the transistor is probably the more likely. <A> In my experience, your specific example will have two distinct failures. <S> I'm assuming that the power supply can supply several Amps of current when I describe the following: <S> First: the transistor fails SHORT Second: <S> the bonding wire that runs from the die to the package pin burns open. <S> Depending upon the energy levels involved, the second part of that failure could be silent, smelly, noisy, or both smelly and noisy. <A> Failures can be analyzed, and a lot of different mechanisms exist. <S> The particular properties of this circuit (the diode carries excess current, asdoes the NAND, if the gate goes LOW) <S> don't really determine if the diodewould fail open or short, or if the NAND would fail open or short, butbecause currents are unlimited (except by the power supply impedance)a fault IS expected. <S> If the diode were MISSING, a current surge from therelay might dump enough heat into the chip to make it fail short (andthe melted aluminum might pool to connect the pin to that short). <S> That's only 'likely', though: the surge of current could alsodamage the power source, or trip a protection circuit that hasn'tbeen diagrammed. <S> Or, the diode (also stressed beyond its specifications)could prove less robust than expected. <A> I would think the NAND gate failed since it was sinking 50 mA which exceeds the the max <S> I OL <S> for any TTL logic gate. <S> Logic output transistors are not the same as a bus driver like an 74xS244 <S> I would expect the output transistor to fail open as the transistor likely burned. <S> When the current would flow through the reversed diode the TTL gate would fail. <S> The diode can take 300mA where the logic gate is 10mA or less. <S> There is no way a logic gate failure could take out a power supply. <S> Not likely to take out a 300 mA diode either.
| A likely scenario is an open gate; the aluminum metallization atop the gate chip would melt before the diode (assuming it is a plug-typeconstruction) would overheat, or the dopants in the gate would rediffuse.
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Open-collector comparator with hysteresis and LED indicator at output I am trying to choose the correct resistor values to make Vout go high when Vbatt falls below 18V and go low when Vbatt rises above 20V. I've built the actual circuit and my measured values are 18.0V for the lower threshold and 19.6V for the high end. I used TI's reference design to choose the values for Rx, Ry, and Rh. Since the LM293 is open collector at the output, I read through the post here trying to account for the discrepancy between my measured upper threshold and my theoretical threshold. I believe D1 is the cause of my calculations being incorrect, but I am unsure how to analyze the circuit while accounting for it. Using the NSPW500BS model in LTspice for D1 gives me very accurate simulated values compared to the measured values of the actual circuit. This circuit is a modification of an existing circuit, and I am restricted to only changing the five resistor values. How do I need to modify my calculations to achieve a desired range of hysteresis? simulate this circuit – Schematic created using CircuitLab <Q> You need to learn to do error budgets. <S> 1) Are your 24 and 5 volt supplies exactly 24 and 5 volts? <S> If not, what will this do to your trip points? <S> 2) I'll assume your resistors are 1% units. <S> Go through the various permutations of each resistor being 1% off (99% or 101% of nominal) and look at the results. <S> Combine this with 1) above. <S> 3) <S> Your speculation about D1 is probably correct (at least as a factor), but there is no excuse for not determining its effect rather than speculating. <S> With the 293 output high, exactly what voltage do you see on the output? <S> How about when the output is low? <A> The 5V must be exact or else use a bandgap ref Voltage and just add another 1k pullup R to get 5V out. <S> Your desired Hysteresis is +/-1V <S> or +/- <S> 5.26% about mean of 19V. <S> Other notes <S> Your ref from 24V seems to be 2V for 19Vin <S> thus it swings 5.26% for input limits which is what the feedback thresholds must be set to from 5V. Except the comparator saturates around 150mV instead of 0 which adds considerable error, but not as much as the diode. <S> If 5V sags then your Vbat thresholds drop, which ought to be fixed with a 2.5V band gap ref. <S> VOL for comparator is also 150mV nom. <S> which also affects thresholds. <A> When the output of the open collector comparator is OFF then output <S> high voltage of the comparator is determined by the LED bias resistor, the forward voltage drop of the LED by the small current through the feedback resistor into the voltage divider. <S> To make your analysis simpler do consider taking the LED mostly out of the equation by placing another resistor in parallel with the BIAS resistor and the LED. <S> This additional resistor would be from the +5V to the comparator output. <S> At the low current that the 715K feedback resistor takes I would guess that the voltage drop across the added 1K resistor is small enough that the LED never biases on at all and can be neglected in the analysis of the high level output mode.
| A good starting value would be to use 1K ohm as long as the comparator is capable of sinking the current from both the added 1K branch and the LED branch when the output goes low.
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What are some ways to improve a voltage multiplier? I am working on a Nixie power supply, but I would like to improve it. I have 4x9V batteries in series for a total of 36V to be switched across a multiplier. A (TTL) 555 timer is running astable from only the first 9V battery to generate a 8.5-ish volt square wave, 10kHz (or any frequency you want, I guess), approx. 50% duty. The 555 output drives the gate of an N-channel BS170 MOSFET . The MOSFET drain is connected up to 36V through an approx 1.2kΩ resistor. This resistance needs to be as low as possible to push current into: a 6-stage Cockcroft-Walton multiplier , which produces a nice ~220VDC output under no load. Unfortunately, it sags to about 155VDC when loaded by a 47kΩ resistor in series with the tube. Things I like about this circuit: It Works™ It can be built by extremely common parts I'm likely to have on-hand, e.g.: It requires no inductors. It requires no specialized IC's such as boost converters. It requires only capacitors and diodes with voltage ratings to handle each stage, not the full shebang. It crashes Multisim. Things I don't like about this circuit: The output voltage sags to ~155VDC under only ~600μA load. I'm too stupid to think of a better way to switch 36V across the multiplier: While the 555 timer output is high, I'm wasting over 1W across the drain resistor just to drive the multiplier. The multiplier input voltage is hampered by the drain resistor. How can I: make improvements that can enable ~10mA to be sourced with less than 40V drop in supply output? I have tried: Replacing the MOSFET driver section with something like this: simulate this circuit – Schematic created using CircuitLab I toasted quite a few transistors trying this inverter. As shown, the gates of the inverter are pulled up to 36V by the 10kΩ resistor. Is it possible that the gate charge time is what destroyed the transistors? EDIT: I just realized that the maximum ratings for gate-source voltage on both inverter FETs is ±20V. That would explain why they fried. Hmm, maybe instead of a single 10kΩ, I could make a voltage divider to drive each gate separately? reading the Wikipedia article regarding improvement methods: For these reasons, CW multipliers with large number of stages are used only where relatively low output current is required. These effects can be partially compensated by increasing the capacitance in the lower stages, by increasing the frequency of the input power and by using an AC power source with a square or triangular waveform. studying other popular Nixie power supply designs, such as these . I suspect that switching the 36V across the multiplier more efficiently would go a long way toward improving the performance. EDIT/SUMMARY: Switching the 36V across the multiplier more efficiently went a long way toward improving the performance. As several people suggested, something called "push-pull" was a quick fix here. A CMOS inverter with separately-driven gates makes the charge pump much more effective: The supply now stands at ~216VDC when loaded with two tubes, a huge improvement: <Q> You need to ditch Rd from your first schematic, and use a low impedance push-pull output as in your second schematic. <S> However, as you correctly say, 36v will toast the gates of 20v Vgs FETs. <S> There are few fets with Vgsmax greater than 20v, and none to my knowledge with more than 30v. <S> Amongst the options are to use a) suitable level shifters to control the FET gates, small bipolars would work well here <S> b) a gate drive transformer (though usually only used for higher power applications) c) <S> how about 18v push-pull drive from two batteries, but in push-pull, like this ... simulate this circuit – <S> Schematic created using CircuitLab <S> I've illustrated 4 stages here, the extension to more stages is obvious. <S> Now, I've not connected the upper capacitor. <S> There are two options a) <S> Cockcroft Walton stylee, where you are limited by maximum voltage. <S> Here, you'd connect C5 to the D1/D2 junction. <S> This allows low voltage across each capacitor, but results in high output impedance. <S> Also known as a Villard cascade, though invented by Greinacher. <S> b) Dickson charge pump stylee, which results in a much lower output impedance. <S> C5 connects back to the driven end of C2. <S> This means C5 needs a higher voltage rating, but if you can get caps with a suitable voltage rating cheaply, 250v or even 400v are commonly available, then this configuration has a much lower voltage droop with current. <A> That way you start the voltage multiplication already with voltage \$\gg 36V\$. <S> But make sure that the MOSFET can hable the voltage (for BS170 \$V_{DS, <S> max}\$ is 60V) and the MOSFET can handle the max. <S> current (mainly depending on operating voltage, switching frequency and inductance); for BS170 \$I_{D,max}\$ is 0.5A. <A> 36 V on the gate will destroy the devices. <S> You need to find proper MOSFET driving circuits. <S> Absolute maximum rating for BS170 Gate-Source voltage: \$|V_{GSS}| < 20~V\$. <S> The simplest way to fix the circuit will be to use a transistor push-pull stage. <S> The pull-up resistor \$R_1\$ will only have to be a little bit smaller. <S> Say \$3.3 <S> ~k\Omega\$? <A> as a voltage multiplier, its current output is inversely related to its voltage output. <S> so to increase the current output, you have two choices, without stepping out the topology: 1) increase drive current: the 555 can deliver 200ma and your bs170 a few ma. <S> you could try an emitter follower as a buffer; or a dedicated driver; 2) increase drive voltage: run the whole thing at as high of a voltage as you can; if i were you, i would try driving the multiplier directly with the 555 first. <S> if that doesn't deliver enough current, think about a different approach, like a step-up converter. <A> What kind of batteries do you use? <S> The internal resistance of a 9V battery can be quite high. <S> I think a normal alkaline can only provide around 3 Amps because of that. <S> 36V * 3A / 220V is about 500mA at the output without considering any losses in the circuit. <S> I think rechargeables could perform better.
| Replace resistor \$R_D\$ by an inductor which gives you a boost converter.
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Mixer spurs and IF output filters I'm designing a simple diode ring mixer for down-conversion from 1.5G to 200k. I want a band-pass LC pi-filter on the mixer IF output. I've designed the filter for 50R source and load impedance by using a low-pass butterworth prototype and used the usual equations to scale and transform into a band-pass at my center frequency of 200k. My LTSpice simulation shows the input impedance is only 50R at the center frequency, dropping to 0R in the stop band. Should I be concerned that the filter wont absorb mixer spurs and harmonics in the stop band and will instead reflect them back into the mixer? What is best practice for terminating an IF output with a narrow bandwidth? <Q> Here is one way to do it: Terminate the mixer into a common-base (or common-gate ) amplifier, using a resistor in series with the input to increase the input impedance to 50 Ohms. <S> The output of this amplifier can drive your filter. <S> Increase the bias current if you need less distortion. <S> Instead of terminating L1 and C1 to ground, connect them to the positive supply voltage. <S> Add a 50 Ohm resistor in parallel with L1 and C1 to provide a 50 Ohm output impedance. <S> The current to feed the collector of the common-base amplifier will go through the inductor. <S> If you increase the filter input impedance (and the collector resistor) to a value above 50 Ohms, there will be gain. <S> The value of this gain can be adjusted to maximize dynamic range, improve sensitivity, or reduce distortion. <S> The advantage of having L1 and C1 in parallel with the resistor is that the resistor value can be made larger without reducing the collector voltage. <S> Usually, the worst problem with an amplifier in this position is that the second-harmonic distortion of the amplifier will cause a signal at 100k to be converted to 200k, which is in your passband. <S> A 100kHz notch filter on the input side of the amplifier will help get rid of this. <S> To stabilize the amplifier, put a ferrite bead or 10 Ohm resistor between the base terminal and ground. <A> This is unfortunately an unavoidable problem. <S> The best course of action is to build a mixer with decent port to port isolation and terminations. <S> Have something like a 3dB pad at all outputs (between the mixer and filter etc) to offer some attenuation to the reflections. <S> This way repeated reflections will be avoided as the signal incurs additional attentuation with each round trip. <A> The low-pass zobel will block RF from reaching the opamp while presenting a nice wideband 50R impedance to the mixer. <S> Then use a MFB filter to bandpass at 200k. <S> My LO and RF inputs are already driven by well matched RF amplifier outputs, so I'm hoping no 3dB pads will be required here. <S> Thanks for the comments.
| I think what I'm going to do is pass the IF output through a low-pass zobel bridged tee and terminate in 50R at an opamp input.
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Type of capacitor for crystal resonator The Atmega328p datasheet recommends the following circuit to work with its internal crystal resonator: simulate this circuit – Schematic created using CircuitLab I can't find any recommendations on the type of capacitors to use. I think the reasonable options are mica and thin film. Maybe also ceramic NP0 (emphasis on the NP0 type), which is a lot less expensive. Any advice? I tend to believe that either type is ok, mica and thin film probably being the absolutely safe bet? Again, any comments will be appreciated. <Q> Unless you need absolute frequency stability, any capacitor should be just fine. <S> The crystal’s own temperature coefficient may affect the frequency more than the capacitor’s temperature coefficient. <S> If you need absolute frequency stability, you need to find capacitors whose temperature coefficient will compensate the crystal’s (that’s a TCXO) or make a circuit whose temperature is controlled (that’s an OCXO). <A> So choosing an NP0/COG is wisest for several reasons. <S> (low-microphonics, zero temp error (<50ppm) and stable accuracy with V, aging etc.) <S> But if your tolerance is high , you can use the internal RC oscillator. <S> For ceramic caps with stable temperature curves, N250 is -250 ppm/'C while <S> P150 is +150ppm/'C and NP0 is 0 ppm <S> +/-50ppm/'C ( aka COG) <S> The value is twice the rated crystal load capacitance (in series becomes half) <S> minus stray load proximity to ground plane and input capacitance <S> (e.g. 2*C + <S> 2~3pF = CL ) <S> But the other answers are also correct, in a benign environment, any cap will help it oscillate. <S> AT cut Crystals have a 3rd order temperature curve that are cut for different tolerances and temperature,T, ranges to control the peak frequency f(T) <S> swings which crossover at room temp. <S> So the value of these caps shifts to Xtal curve at 25'C temp to the precise frequency specified for rated load within the specified ppm tolerance. <S> (25~100ppm) <S> Parallal Cap Xtals are always specified by f, Tolerance/ Stability/ aging and temp range. <S> e.g. XTAL, 10.000 MHz 50/50/5 ppm, <S> -40 ~ <S> +70'C xxyy case <S> While Tuning fork resonators have a 2nd order (parabolic) temperature curve centered at room temp. <S> so a cap. <S> selection cannot compensate for this in a TCXO, only a varicap diode with a custom correction curve. <A> Worked with a gal who designed crystal oscillator circuits for MCUs. <S> She called me into the lab one day because "My amplitude at 40MHz is so high we are afraid the crystal will be overheated and break. <S> Can you figure out why the VoltPP is so high?" <S> After a couple hours of lab observation and listening, and couple days of musing as worked on other tasks, I realized the ESD structures were consuming so much energy, at 40MHz, the stable amplitude was indeed higher than simulated. <S> The ESD losses needed to be in the simulation. <S> Only recently, 10 years later, is that company learning to characterize their IOpads/circuits with Network Analyzers. <S> Summary: the capacitor losses matter. <S> There is no difference between PCB capacitors and onchip ESD-diode-junction-capacitance. <S> ESR matters.
| One might choose an X7R cap if was essential for profitability to shave a penny in small sizes or for gen. purpose use but in important applications where accuracy and quality is important, always choose NP0/COG 1% if the value is <1000 pF. ( low k density)
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Difference between a memory cell and a memory chip? I have gotten very confused on the fundamentals of computer memory in regards to memory cells and chips. I have been reading Assembly Language Step By Step Programming with Linux by Jeff Duntemann and in Chapter 3 "Lifting the Hood" he dives into what a computer is, starting with how memory in a computer works (RAM, memeory chips, and cells). The confusion starts when he uses memory chip and memory cell interchangeably. It states that transistors act as switches and have either a state of on or off which represent 1 or 0 respectively in binary. Then it states ...the transistor switch and its support components are called a memory cell. A single computer memory cell, such as the transistor-based one we’re speaking of here, holds one binary digit, either a 1 or a 0. So from the above quotes I have the impression that a cell holds one bit and has a transistor. The confusion starts here: Whereas in the beginning one chip held one transistor, in time semiconductor designers crisscrossed the chip into four equal areas and made each area an independent transistor. Not it is stating a chip is holding the transistor but before it said the cell held a transistor. It also says the chip was redesigned to hold four transistors so in essence now a memory cell/chip holds four bits of memory not one as it said previously. Then it goes back to using them as separate objects. ...256 memory cells could occupy one chip of silicon, usually in an array of 8 cells by 32. Now it says chips hold cells. The question I have is what are the differences between the cell and chip? Also if you could help clarify my confusion of how memory works. Follow up question is how does this relate to RAM? Does RAM hold multiple chips or is the chip actually RAM in this context? <Q> Old school: <S> A memory cell holds one bit of information, a 1 or 0. <S> A bit and a cell can be used interchangeably. <S> Memory chips are made up of one or more cells. <S> In modern hardware, memory chips contain millions or billions of cells. <S> Modern terminology: 8-bit per cell memory? <S> RAM, or Random Access Memory, is a type of memory that allows random access. <S> Here's an example of some RAM: <A> Back in the day you could see each memory "cell", made of a magnetic core, holding one bit of data. <S> Cost about $1 per bit. <S> They put them on plug in cards. <S> Then in more modern times they put them in "chips". <S> Cost about $0.01 per bit. <S> Then in 1972 the Intel 1103 DRAM chip was release. <S> Cost, $0.01 per bit. <S> And you could no longer see the memory cells. <A> I would not rely too much on a book about software to describe how hardware works. <S> The RAM memory in a modern computer or computing platform can be either DRAM or SRAM. <S> Both memory types loose their data when power goes off. <S> A DRAM cell is actually a small capacitor with its own transistor, it can store 1 bit. <S> The transistor is needed to read out that bit. <S> The data is lost when it is read and also it leaks away over time. <S> So DRAM needs to be refreshed every few thousandths seconds or so. <S> An SRAM cell consists of 6 transistors as far as I know. <S> 4 are needed to remember the data (again one bit) and 2 extra for read and write operations. <S> Chips always contain many cells of memory in a matrix like structure. <S> I think you should ignore the text: <S> Whereas in the beginning one chip held one transistor, in time semiconductor designers crisscrossed the chip into four equal areas and made each area an independent transistor. <S> as I think it makes no sense at all . <S> And concerning: ... <S> 256 memory cells could occupy one chip of silicon, usually in an array of 8 cells by 32. <S> almost any configuration is possible so he's basically saying "chips contain arrays (matrices) of memory cells." <A> Memory chips contain memory cells. <S> Each memory cell can hold a single bit of data. <S> Each memory cell may be made of one or more transistors. <S> The number of memory cells = <S> The number of bits the memory can store. <S> The transistor per cell count determines the type of memory (SRAM, DRAM, flip-flop based etc). <S> If you're talking about SRAM based memory, each cell contains 4 transistors. <S> Thus, a 128 byte (or 1024-bit) SRAM contains 128*8=1024 cells which turns out of be 4096 transistors. <S> DRAM memory on the other hand is essentially 1 transistor = 1 cell, thus, a 128 byte DRAM would contain 1024 transistors. <S> Here's a representation of a single SRAM memory cell... <S> Here's the waveform of the latch in action... <S> When Enable i.e., E=1, Q tracks the D pin and when E = 0, Q retains its value despite D changing. <S> Note that Q' is the logical opposite of Q. Put together N latches <S> and you have N-bit memory. <S> If you tie all the N enables together as a single pin, you have an N-bit wide memory bank. <S> If you construct another N-bit bank in a similar fashion, then overall you get 2 enable pins (Call them E[1] and E[0]), each pin controlling one set of latches (or bank) and thus you now have N-bit wide and 2 locations <S> deep memory <S> (Totally N*2 bits organized as N-bit x 2). <S> The final E pin can be seen as a 1-hot address (E[1]=0 E[0]=1 selects bank 0 ; <S> E[1]=1 E[0]=0 selects bank 1). <S> Note that each bank can store N-bit. <S> To confuse matters a little, some people consider each bank to be a cell so in that case, the memory you constructed has 2 cells, each cell capable of storing N-bit. <S> However, from the perspective of a integrated circuit design engineer, each memory cell is treated as 1-bit (and that should be the way to describe it). <S> Here is the logic gate level implementation (5 gates) of a latch. <S> Note that a logic gate is built using transistors. <S> The chip is actually RAM. <S> Of course, you could string together several RAM chips on a circuit board and create even more RAM (or you could just buy bigger RAM chips). <A> In DRAM, memory is a capacitor. <S> Each bit, each cell, is a capacitor. <S> The presence of many electrons in a cell defines the 1/0 value. <S> These capacitors are built on the silicon in rectangular grids. <S> Thus a memory chip is covered with rectangular patterns. <S> To read out, or to write to, just one bit at a time, requires too much circuitry. <S> Thus entire rows of capacitors are read out at a time <S> , the value 1/0 of electron charge in each capacitor is stored in a group of FlopFlops/latches, and a multiplexor selects just the ONE BIT you wished. <S> A portion of the address is used to specify the row, and rest of address tells the multiplexor exactly which BIT you wish. <S> To write, part of the address is used to select the Row, that entire Row is read into the FF/latches, and a multiplexor changes JUST THE ONE BIT you planned to write. <S> Then, that entire Row of data is copied from the (many) latches into the Row of capacitors. <S> This dual-task ---- read, then write ---- <S> makes writing the slower of the 2 memory activities.
| RAM can be made up of one or more memory chips.
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Implementing LM339 comparator to count analog peaks Updates after applying Trevor's advice:I have attenuated the output of the CPS to between 2 and 2.5 volts, as shown in the scope pic: , channel 2 is the constant "high" output from the schmitt trigger, which is not connected to any input in that screenshot. the issue now is that when I connect the trigger input it drops the voltage way down and the scope looks like this: I have absolutely no idea why it is dropping the voltage so much, any advice on what to change? see Trevor's answer for schematic. original question below: I am an ME lost in a EE world... Overall problem is need RPM input from an engine to an arduino for PID control. Went to the shack and picked up some stuff to experiment with. Seems that the LM339 should be able to pickup the peaks of the analog signal from the crank position sensor and output a logic signal that the arduino can count and convert to rpm. I have referenced the TI datasheet and application guide [see figure 6] in my efforts so far. See attached schematic for my starting point. Also attached the scope reading of the analog signal from the cps. My question is basically can this work or am I missing some fundamental EE reasoning? Input/advice on resistor values, decoupling caps, and the proper Vref value would also be greatly appreciated. would not let me post links to the application guide or my search results, found some info on decoupling caps for IC's but not sure how to apply it to my circuit. <Q> The circuit below should do the trick. <S> I A.C. coupled the input signal and biased it to 2.5V using C1 and R1, R2. <S> Shotky diodes D1, D2, <S> with R7 afford you protection in case the signal ventures over 2.5V peak to peak. <S> C2 provides some noise filtering. <S> You may need to adjust this size to best suit your application <S> The reference is set on the plus pin at 2.5V with about +/- <S> .44V hysteresis. <S> If the signal has a ripple or noise larger than that decrease the size of R5. <S> R6 provides the final pull-up. <S> simulate this circuit – <S> Schematic created using CircuitLab EDIT: Since your input signal is a lot larger than you originally intimated, You could also use the following circuit to simply half-wave rectify it and attenuate it a little. <S> Benefit of this method is you pretty much guarantee full swing on the input of the comparator. <S> That allowed me to decrease R5 giving you more hysteresis and lowered C2. <S> simulate this circuit <A> The LM339 won't like inputs that drop below 0 volts. <S> The absolute maximum negative input voltage is -0.3 volts below the negative power supply rail. <S> It might be an issue if the RPM is low <S> but I don't think you have much of an option. <S> Some comparators (like the MAX999) are designed to be used with a negative input signal but this still has to be restricted and I would consider using a potential divider to reduce the input signal to <S> +/- <S> 75 <S> mV. <S> Your pull up resistor is also far too low and implies an amp of current into the device when driving a low output signal. <S> The MAX999 (and others) have a push pull output and don't require this resistor. <A> Suitable for signals 2Vpp with 500 Ohm load to reduce stray noise coupling. <S> Use twisted pair , pref shielded from Arduino gnd side only. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Simplest solution is to stick a resistor in the input, like 10k, and let the input protection diodes shunt the current to GND. <S> A much better solution, especially near motors which may create lots of noise, is: Use hysteresis Filter input signal <S> Here's a quick example. <S> Here, I set hysteresis high enough to reject the noise on the 1kHz waveform. <S> You can use LM339, I used LT1018 here because the model comes with LTSpice. <S> See how comparator ignores multiple zero crossings due to hysteresis? <S> A small RC filter on the input signal to get rid of HF noise is also a good idea. <A> can this work... <S> the basic concept is sound, aside from excessive positive feedback. <S> in cases like that, the challenge often is a signal with varying degrees of "average". <S> so you cannot "pre-set" the reference voltage. <S> if your application calls for that, you may think about how to take that on.
| If you feed a voltage below about -0.6V to LM339 inputs, then a high current will flow out (limited by your sensor) and LM339 input stage may not survive. I would consider biasing the input signal to somewhere between the power rails. Looking at the hysteresis resistors I think you have too much hysteresis, so much that the input signal won't be able to switch the output from one state to the other.
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What capacitor(s) will be best replacement Living and working in West Africa. We cannot purchase capacitors here, we can only beg (and pay) used parts from the workshops which rewind electrical motors. Our gray-water-pump is broken, the always-connected run capacitor is dead. It is a simple 400W single phase induction motor. I can start it by hand in either direction. But it will not start any longer by itself. The dead capacitor is marked 6 uF. I found no 6 uF capacitor locally, have already asked in three different towns at several workshops. I found a used 5 uF (cylindrical shape, same size as the dead one) and a 1,25 uF (small box-shaped unit). Because of the keeping-the-water-out-seals, I cannot open the pump many more times, because the rubber rings suffer with each opening and closing of the case. Question: Should I install the combination of 5 uF + 1,25 uF in parallel or just install the 5 uF? What are advantages and disadvantages to consider please? Are there formulas which I could apply, having very little documentation? (It is just a dirt-water-pump from a garden-center labelled 400W.) If I could calculate the "nominally needed capacitance" then I could better decide whether to go 1 uF too low or rather 0,25 uF too high with the added complcation of installing two different capacitors in parallel. What do the users think please, who got real-life repair experience? I can tell you that the motor without any load (dry on my work bench) does start on the 5 uF alone. But I wonder whether it will be "enough" with the added load of having the blades submerged in the water (physical friction). The local repair-veteran (without any formal education) has recommended to just install the 5 uF. But I would be 17% off the original capacitance. Why do I not just order a new 6 uF capacitor from Germany? I did, but mail-order and shipment to West Africa takes several months. And I do love my family and want to repair our toilet-flushing system. So your help is appreciated please. This is not a theoretical question but meant to help me fix a real problem. Thank you. (This is my first question here. If I got it wrong, please do not hate me, rather please help me edit to get it right. I love Stackexchange.) <Q> I needed a 6 uF replacement and only found a used 5 uF and a used 1.25 uF. <S> With the help of several answers, I decided to only install the 5 uF capacitor. <S> The motor started fine on its own. <S> Having assembled the pump with its pumping "blades" (extra mass) and the outer case it still started fine in my workshop. <S> Then I re-installed the 1-inch hose and the supporting rope and tested inside our gray-water tank. <S> I learned that I now have less starting torque. <S> But in the wild, the pump is starting just fine, even suspended almost 1 meter down in gray water, i.e. having an actual "load" in the outgoing hose. <S> It is not mud, just water from our washing machine and shower. <S> So, nominally, I am 17% low on capacitance, and the used capa might even be somewhat lower than what its label says. <S> But my pump is repaired and we no longer need to run with buckets to flush our toilet. <S> Thank you all. <S> Maybe this will help another user... <S> Update2: It is now May 18th and the pump is still starting every single time with load (in the water) at first flip of the switch. <S> It never hesitated with my replacment capacitor with 5 uF. <S> In the meantime a brand-new 6 uF capacitor has arrived in the mail; but I do not even want to open the pump again. <S> I will keep it as a spare part for when my second-hand fix will die of the heat some day. <S> Update3 <S> : <S> Today Oct 11, 2017; pump still starting fine every time. <S> And keeping the replacement capacitor on standby. <A> First I must repeat that I know nothing about motors! <S> However, knowing that most capacitors like that have a tolerance larger than 17%, I would just use the 5 µF. <S> When they originally selected a capacitor with 6 µF of capacitance they knew that it could have a large manufacturing spread. <S> Of course, maybe the original one had more than 6 µF, and maybe the new one have less than 5 µF, but it's likely still within the accepted range. <A> The previous capacitor was both start and run. <S> So, choose a non-electrolytic capacitor with a high temperature rating, such as 105 C. <S> If the two capacitors you have are non-electrolytic, are rated for high temperature, are rated for motor applications, and rated for the absolute maximum AC Voltage that may get applied to the motor, then connect them in parallel. <S> Using two capacitors increases the outside, heat dissipating area, which may increase the time to the next failure. <S> You may need to ask your local electricity supply authority what maximum AC Voltage is provided - not the standard, but the actual. <S> If you run your motor from a local generator, then use a multi-meter to check the maximum Voltage - you may need to measure at many times during the day; if your multi-meter uses a moving needle to indicate Voltage, then you can leave it connected all the time, and just note the Voltage each time you walk past.
| Do not use an electrolytic capacitor for this AC application. It may have lost its capacity through overheating.
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Can an alternating magnetic field be "informationally" shielded? I have asked previously about the properties & shielding of the "static" magnetic field previously , where I have determined that static fields are hard to shield. However what I am really interested in, is that is it possible to shield against an alternating magnetic field, in the sense that can the information itself be blocked? When I refer to shielding I don't just refer to EM shielding but also information shielding. For example: We put a radio transciever a metalic box to act as a shield. The metalic box can easily shield against the electric field going out or going into the box, therefore we can't send out signals from the box, since the electric field is blocked by the shield. So far so good, the electric field doesn't leak information, however that electric field can have an alternating magnetic field. (static field doesn't matter as pointed out above) So if we put a magnetometer ouside the shield, can the magnetometer pickup the signal from the transciever? It may be electrically shielded, but the magnetometer looks for the magnetic field, and therefore the information can go out via the magnetic field? Some people have pointed out that eddy currents could be generated by the alternating magnetic field , that would manifest itself in the shield. So the eddy currents would cancel out the change in the alternating magnetic field, and make it static. So I don't quite understand this phenomena, my point is, I don't care if a static magnetic field goes out from the box, which will inevitably do according to the answers on my earlier question. So a static magnetic field will go out, but that's no issue (since it can't carry information) But will the alternating magnetic field succesfully be stopped by the eddy currents, or will the magnetometer pickup the information leaking out from the box? In other words, can information be sent out from an electrically shielded environment via alternating magnetic fields? How effective is an electrical shield against information blocking that may be carried by magnetic waves? <Q> can information be sent out from an electrically shielded environment via alternating magnetic fields? <S> Yes. <S> How effective is an electrical shield against information blocking that may be carried by magnetic waves? <S> Depending on the thickness of the shielding and the frequency, anywhere from very effective to totally useless. <S> The thinner the shield and the lower the frequency, the less effective it is at attenuating the magnetic field. <S> Whether it will be enough to 'block' the signal also depends on distance and sensitivity of the receiver, and the nature of the signal. <S> Real shields don't completely block the electric field either. <S> If you put a sensitive receiver close enough to a 'totally' shielded high frequency transceiver you could probably pick up enough rf to get some information from it. <A> A superconducting material will block both DC and AC fields totally, because it allows a current to flow that exactly cancels the impinging field. <S> Materials with nonzero resistivity will be less effective, because the energy of the induced current gets dissipated as heat, causing the current to die out. <S> This is why conductive boxes provide better shielding at higher frequencies. <S> Unless the permeability is infinite, there will still be some field that escapes, but modern practical materials offer very high attenuation values. <A> Because they are each intertwined you can think of magnetic signals in the same way as electric signals, but in a 90 degree rotation. <S> You can literally understand a shield as very much like an inductor in a signal line. <S> As you are aware, an inductor "BLOCKS" sudden transitions or high frequency electrical voltage signals. <S> In the same way, a shield BLOCKS sudden changed in magnetic field and attenuates the level of alternating fields. <S> Just like a step voltage "decays" across an inductor in a signal line, a step magnetic field will also decay through a shield. <S> In an electrical inductor you have a coil of wire around a ferrous material and the inductance effect is caused by the energy you need to store to build up that magnetic field. <S> Similarly, in a magnetic inductor, you need to build up an electric field around the "magnetic Current". <S> Literally electrons spinning around in imaginary wires buried inside the shield material. <S> This is what we call "eddy currents". <S> So, to answer your question fully. <S> Yes, low frequency magnetic "Signals" can be sent out of a metal box.
| Magnetic materials (high permeability ) block magnetic fields because they readily form an internal field that cancels the impinging field.
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Is there a light sensor that varies output with wavelength? What components are sensitive to the wavelength (energy) of impinging light? I am wondering if there are any that have a reliable and detectable response to wavelength instead of just intensity. For example, if some hypothetical component responded to light in the range of 400-1000nm, then when exposed to say 10 photons at 400nm perhaps its voltage (or impedance) is a significant multiple of the voltage when exposed to 10 photons of 800nm? Clarification: I know that it's possible to do spectral imaging using prisms and/or filters so that a sensor can say, "I was just hit by this many photons," and based on the filters we know what those photons' wavelengths could be. What I'm wondering is whether there is any component that can say something like, "A photon hit me this hard," where this is a relatively precise measure of the photon's energy. <Q> Basically all bandgap devices (read: semiconductor devices) are specifically sensitive to the wavelength for which photons have exactly the energy needed bring an electron from lower to higher energy band. <S> So, a blue LED will be significantly more sensitive to blue light than to red light. <S> Technically, it's much harder to do the opposite of what you want: have an output that depends only on intensity, not on wavelength of the incident light. <S> This is really simple to test <S> : Get a monocolor LED. <S> Take Blue. <S> Now, there's different ways to use an LED as photodiode, but I think the easiest for you will be to use the LED, and connect it in reverse to a (stable) voltage source (e.g. 4.5V coming from three AA batteries). <S> Use a sensitive multimeter in series to measure how much current passes through the LED in reverse direction. <S> First, cover your LED in darkness, measure current, then expose it to bright blue light – another LED (operated in forward to actually emit light) of the same kind, and compare the current readings. <S> Then, compare current readings when subjected to other colors of light – from a RED LED, for example. <S> If your ammeter isn't sensitive enough for µA to mA, you might want to feed the current to the base of a transistor pair in Darlington configuration. <A> All opto-detection devices have a light frequency sensitivity profile. <S> However, they also all produce a single variable, e.g. voltage, resistance, or current depending on how it is configured. <S> As such, it is impossible to tell what changed, the impinging light intensity, or the light's frequency. <S> What you are asking for is a chromatic light sensor. <S> These do exist as equipment and rely on using prisms or lenses to "split" the light onto an array or line of light sensitive semiconductor devices or to focus a particular wavelength onto a single sensor. <S> However, the technology called " Foveon X3 " uses a newer method to make a single spot on a silicon substrate sensitive to different wavelengths. <S> However, unless someone else can find one, I know of no single "diode like" product that uses this technology. <S> If you have a project in mind though, you may be able to use a Foveon X3 camera sensor. <A> For example, if some hypothetical component responded to light in the range of 400-1000nm, then when exposed to say 10 photons at 400nm perhaps its voltage (or impedance) is a significant multiple of the voltage when exposed to 10 photons of 800nm? <S> You can do this two ways: Choose a semiconductor material for your sensor with a bandgap energy somewhere between the photon energy associated with 400 nm wavelength and the photon energy associated with 800 nm wavelength. <S> Unfortuntely I'm not familiar with the materials with bandgaps in this wavelength range <S> so I can't suggest any particular material off the top of my head. <S> As Marko's answer suggests, a red LED might be a good candidate. <S> Also, this approach won't get you a detector that's sensitive at 800 nm but not at 400 nm. <S> Start with a detector that's sensitive to both wavelengths and put a reflective coating on its surface (or on a piece of glass in front of it) <S> that is highly reflective at 800 nm, but highly transmissive at 400 nm. <S> Using multiple thin film layers, it's possible to design a coating with essentially any profile you like of reflectivity vs wavelength. <S> This way you could also make a detector sensitive at 800 nm but not at 400 nm. <S> Cost of course goes down if you choose a profile that can be achieved with a small number of layers. <S> You could, of course, also think of using things like absorptive filters in front of your sensor to vary its response to different wavelengths. <A> In order to measure the wavelength of a photon, you need to measure its energy. <S> This is common for example with x-rays or gamma rays because they are ionizing radition -- they have enough energy to knock electrons free from atoms, and we can measure the energy of the electron. <S> It is not normally possible with visible light because the energy it too small, but there are detectors, called transition edge sensors (TES) <S> that are energy resolving photon counters. <S> They measure the energy of a photon by the change in temperature of a thin metal film right at the superconducting transition. <S> Unfortunately, they typically work at 0.1 K and require complicated electronics and optics to use.
| So, for example, LEDs (light emitting diodes) work pretty well as selective photodetectors.
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easy to construct FM low-pass or band-pass filter Suppose I wanted to filter a Raspberry Pi's (or other similar) GPIO digital output (running pifm or rpitx or similar) to be clean enough to not splatter RF harmonics outside of FCC Part 15 (or equivalent for other countries) unlicensed RF broadcast allowances. What might be some suitable FM broadcast band low-pass filter or band-pass filter designs that could be constructed out of the easiest-to-procure and manipulate components and materials (e.g. something suitable for a kid's student project, nothing tiny surface mount, etc.) Is this possible? <Q> If you dont care about bulky and messy construction, a DIY filter made with hand wound coils and copper clad is the way to go. <S> Get yourself some enamel wire and a ceramic capacitor kit from ebay. <S> However, in my opinion do not bulid the doubly tuned band pass filter suggested by one of the answers, atleast not the form shown. <S> The filter shown, couples the resonators using capacitors only, this introduces a "zero" in the frequency response. <S> Simply put, the filter roll off for frequencies greater than the resonant frequency is terrible, so it is not a sensible choice for harmonic suppression. <S> Furthermore, aligning it without variable (expensive!) <S> capacitors is difficult. <S> (The filter is however an excellent choice for an IF filter) <S> Build <S> a 7th or 9th order Butterworth Low Pass filter using similar techniques <S> , they are much more well behaved and easier to construct. <S> Here a 9th order LPF I built. <S> The bandwidth is 100Mhz and at 200Mhz I get > 70dB attenuation. <S> Use the calculator found here: http://www.wa4dsy.net/filter/hp_lp_filter.html EDIT: How to wind the inductors: At these frequencies the inductors are small enough to be air core and hand wound. <S> So start by some enamel wire, wind a few turns around something like a pen and measure the inductance produced. <S> This will give you a standard to refer to in terms of inductance per turn for a certain diameter. <S> Now take it from there. <S> You do not need to be spot on, just get close enough and then tweak the inductance by varying the coil separation. <S> This is how I did it. <S> You might not have a LCR meter capable of measuring such small inductances (I didnt have one either). <S> So simply, build a fast edge square wave (or use the TTL output of your cheap signal generator) and feed it into a test circuit with a known capacitor and unknown inductor. <S> Probe it using a X10 probe and measure the ringing frequency and calculate the inductance. <S> Careful about the length of coax you use in the above setup as that is going to have comparable capacitance to your test circuit. <S> EDIT: <S> The wire I've used has a 0.6mm diameter and the coil diameter is about 7mm. <S> So this is a good starting point for you to use for building inductors for this frequency range. <A> that is, air-core coils wound from simple copper wire, and standard film/ceramic or air gap caps (air gap primarily if you need to make things adjustable). <S> devttys0 made a video on those; his filters look like this: I've personally got no doubt you could, with a bit of RF design knowledge, also design filters that work very well on standard dual-layer PCBs with SMD ceramic caps and small commercially available inductors <S> – I don't fully buy the "you can't buy inductors with high Q" Craig's telling me, considering the Q factor of an inductivity is just the ratio of reactance to ohmic resistance – so, <S> \$\frac{\omega L}{R}\$ at the frequency <S> \$f=\frac{\omega}{2\pi}\$ <S> that you're interested in, and inductivities with low resistance can very well be bought. <S> For the single-digit GHz range, SAW filters can be purchased that are really, really awesome, also in terms of suppression per buck you buy. <S> For higher frequency or stronger suppression requirements, things often get a bit hairy – look into cavity resonators. <S> So yes, there's plenty of viable choices for filters that can suppress harmonics of a square wave (which are only at every odd multiple of the fundamental frequency, so you got twice as much bandwidth between pass- and stopband), and you can probably build something that works well enough for the law yourself! <A> You'd want a filter with passband in the FM band (approximately 100 MHz.) <S> with a bandwidth of about 75 - 150 kHz. <S> While it would be easy for a manufacturer to fabricate a surface acoustic wave (SAW) filter, or a ceramic filter, a quick scan of suppliers shows no hits. <S> An alternative would be LC filters that you build and align yourself. <S> The high-Q required of a filter makes this option difficult. <S> Building is quite possible, but aligning requires complex procedures and test equipment: VHF bandpass filter
| For the frequencies you're mostly interested in ( FM Broadcast bands probably means something like 87 MHz – 108 MHz, depending on where you live), passive filters made of discrete inductances and capacitors should work pretty well –
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Why does the stator field rotate at the same speed as the rotor field in a synchronous generator? In the case of the induction motor the rotor never catches up with the rotating field of the stator because if it did the induced voltage would be zero as there is no relative movement between the rotor and the stator field. What changes in the synchronous generator that makes the stator field rotate as fast as the rotor field? (!)If they rotate at the same speed,there is no relative movement.So how is the voltage induced and the current that creates the revolving stator field produced?(!) However, there is relative moment between the rotor and the windings. Is this what causes the current? Edit: I completely understand how the induction motor works. What I'm trying to work out is the synchronous generator and why isn't there a problem if the rotor and stator field are synchronized as there is in the case of an induction motor leading to the 'slip'. Why don't we have a slip in the synchronous generator? <Q> In an induction motor, the speed of the rotor structure is always less than the speed of the stator field. <S> However the rotor field rotates faster than the rotor structure so that the rotor and stator fields are synchronized with each other. <S> In a synchronous motor, the rotor magnetic field is produced by permanent magnets or by DC current in the rotor winding. <S> In either case, the rotation of the magnetic field of the rotor is mechanically fixed to the motion of the rotor. <S> For uniform torque to be produced, the both the rotor structure and the rotor field must move synchronously with the rotor field. <S> In other words, both synchronous and induction motors have synchronously turning magnetic field with torque produced in proportion to the angular displacement between the stator and rotor magnetic fields. <S> In the induction motor, the rotor structure must turn at a slower speed than the magnetic fields while in a synchronous motor, the rotor structure must move synchronously. <S> Re: Question Edit <S> In a synchronous generator, the stator magnetic field rotates behind the rotor magnetic field with respect to torque angle. <S> The current produced produces a rotating magnetic field in the stator that is synchronous with the rotor magnetic field but has a torque angle displacement. <A> The synchronous machine has a field winding or permanent magnets to generate the d-axis flux, where the induction machine relies on the changing magnetic field to induce flux in the rotor. <S> If the rotor speed in an induction machine equals the synchronous speed <S> then there's no induction and no flux. <S> In the synchronous machine the windings or magnets provide flux regardless of rotor speed. <A> The synchronous and the asynchonous both need a magnetic field to make and keep the rotor rotating thereby transforming electric energy into mechanical energy. <S> In the asynchronous motor the magnetic field is created due to the slip of the rotor in respect to the stator. <S> This slip creates a magnetic field in the rotor thereby creating the magnetic field required. <S> Such a magnetic field can not be generated without the slip as is the case with the synchronous motor. <S> Never the less a magnetic field is required. <S> Therefore the only remaining option is to create a magnetic field with a field winding or permanent magnets. <S> Thereafter the only task of the rotating field magnets is to remain in sync with the rotating field in the stator. <S> Here you also find a complication. <S> A synchronous motor normally can not start without additional measures.
| It is the relative motion between the rotor magnetic field and the stator windings that allows the magnetic field of the rotor to produce current in the stator.
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Why doesn't the LM7805 circuit short circuit? I'm new to electronics and have made some simple projects using a LM7805. I'm using the following schematics: I don't understand why this circuit works. Why doesn't it short circuit? I thought electricity always takes the "easiest" path which in my opinion would be the following: Why does the electricity even bother going through the 7805 to begin with while it can just fill up the 0.33uf capacitor and then go right through it back to the battery? <Q> Capacitors do not present a short circuit path. <S> The Capacitor will charge then stop when full, acting as a parallel voltage source when the Input voltage source drops for whatever reason. <S> And electricity does not just take the easiest path, or path of least resistance. <S> Electricity will flow through all parallel paths. <S> You may see a greater current through one path due to its lower resistance, but it will go through other paths too. <A> Your thinking is actually 100% correct! <S> The "electricity" (current) does "fill up" the 0.33uF capacitor before flowing to the 7505 regulator. <S> However, the capacitor "fills up" in a fraction of a second . <S> After that, flow to the capacitor essentially stops and current flows to the regulator only. <S> Note - in practice, things are slightly more convoluted due to parasitic effects: https://en.wikipedia.org/wiki/Parasitic_element_(electrical_networks) <A> Let's have a look inside a capacitor to see what prevent the short circuit. <S> A capacitor consists of two conducting plates. <S> And there's an isolating plate between the two conducting plates. <S> How is current able to pass through the the isolating plate? <S> When there's a change in electrons (charge) on one conducting plate, a change in the charge of the other conducting plate occurs. <S> The two conducting plates affect each other because the isolating plate is very thin. <S> This way current can pass through a capacitor. <S> I consider it as a virtual current. <S> When you apply an AC signal, a short circuit will happen because there's a rapid change in the charge of the plates. <S> When the frequency is decreased, lower current can pass. <S> Capacitor is considered as a resistor and this resistance is called "capacitive reactance". <S> When you apply a DC voltage (zero frequency): <S> At the beginning, the capacitor will charge because applying DC voltage itself is considered a change in charge. <S> After completing charging, current will stop flowing because there's no change in the charges. <A> When you apply a voltage to the 7805 the input capacitor is charged and then the DC current through this capacitor stops. <S> At the same time the other way for the current is,through the 7805 to the connected load at the output and a little bit to the ground connection. <S> So you are right. <S> There would be a current through the capacitor but only during the charging of the capacitor. <S> This current is limited because the capacitor has an internal resistance in series with the actual capacitor. <S> Once the capacitor has internally the same voltage level as the input voltage the charging stops.
| Another way to view it is that the capacitor goes from a "short circuit" when it's fully discharged to an "open circuit" when it's fully charged.
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Advantages of using a current loop in driving a BLDC motor I've been reading the answers to the question: How to realize constant acceleration control for BLDC motor? However, it is not specified what are the advantages of adding a current control loop to drive the motor. Why not just drive the motor PWM from the output of the velocity loop PI controller? <Q> Yes, you can just drive BLDC motor only with velocity loop. <S> But please note that without current loop you can overload current of stator coils <S> when torque and power at the shaft will be higher than rated. <S> When the load on the shaft is being increased - it's faster to detect increasing current than decrease of velocity of motor <S> so response time of regulation control is faster. <A> Do you need a current loop to control an electrical machine? <S> no <S> However... there is a specific point in the question you linked: ACCELERATION, more to the point controlled acceleration Acceleration is proportional to torque and the constant of proportionality is Inertial \$T = J \dot{\omega} \$ <S> Torque is proportional to current ( <S> \$T = K_t I\$ ) <S> Thus, if you want to control acceleration you MUST control the current. <S> Equally for complete system stability bounding hte current loop to a known bandwidth & ensuring a decade difference between the other loop bandwidth is good practice <A> Actually, though the other answers here are correct, in fact YOU DO NOT HAVE TO modulate the current to control the acceleration. <S> The torque produced is a factor of current and the offset angle between the coil and the magnets. <S> You can in fact use a constant current driver, or open loop current driver and adjust the phase angle instead. <S> Like a car engine, you can adjust the acceleration by advancing or retarding the switching time. <S> Which method is better/easier is debateable.
| Secondly, if you control current - it's possible to control torque directly.
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Overcharging in regenerative braking Depending on the system used in a electric car there may be a situation were more energy is being regenerated than can be safely used to charge the battery. What happens to this excess energy? Can the motor controller just briefly disconnect the motor and the battery until the rate of energy transfer subsides to a suitable level? I'd imagine the energy has to go somewhere but I'm not sure where. <Q> One way or another the energy must be converted to heat. <S> It can be converted electrically by regenerating it to resistors. <S> If the motor is briefly disconnected, but the vehicle still needs to slow down, conventional friction brakes need to be used, <A> On many electric hybrid electric vehicles the braking force is both from regeneration and from friction braking. <S> Regeneration is generally used in preference to friction brakes up to the limits imposed by battery state of charge and charging rate. <S> Beyond that point the friction brakes are blended with regenerative braking to provide the driver commanded braking. <S> The Tesla models are a bit different in that the brake pedal does not increase regenerative braking <S> it only adds friction braking to the already existing regenerative braking that happens as soon as the throttle pedal is lifted. <S> This avoids the tricky process of blending the regenerative and friction braking - in many hybrids this can be noticeable and affects the drivers feel of the braking process. <S> If the tires lose traction the ABS, regenerative and friction brakes have to be coordinated to safely slow down the vehicle - maintaining control of the vehicle <S> is more important than recovering kinetic energy in this case. <A> To add to the other answers.. <S> Regenerative braking is when the motors attached to the wheels are being driven by the inertia of the vehicle and are acting as generators. <S> How much torque those generators create to brake the vehicle is dependent on how much current is drawn from the generator. <S> The control system draws an amount of current necessary to perform the designed amount of braking required depending on how far the brake pedal is depressed, or when the accelerator pedal is released. <S> What happens to that current will vary. <S> If the battery needs charging some, or all, may be used by the charger. <S> All these systems also include regular friction braking systems which either work in tandem with the regeneration system or when demand requires, i.e. heavy braking or electrical system malfunction. <A> You have to dump the energy somewhere else. <S> Usually braking resistors, that are switched on/off at a certain voltage level. <S> If the voltage further rises by means of exces regenerative power, then the only way to prevent damage is to disconnect the motor. <S> I guess, by just inhibiting H-bridge transistors won't help since they have freewheeling diodes.
| The rest will either be dissipated into a resistive load and dumped as heat, or used to charge a storage capacitor for retrieval when required.
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Can an inductor/capacitor/resistor combo emit the same frequency as a given crystal oscillator? Beginner in electronics and radio frequencies in particular interest me. That said I'm having trouble understanding what components are used when. When I initially searched about radio frequencies, most knowledge articles said those frequencies are produced based off of a crystal oscillator that'll oscillate at a specific frequency. However, in some DIY FM transmitter tutorials I've looked at a lot of the circuits don't incorporate any crystal. Rather, the circuits incorporate certain capacitors, and a coil (inductor?). In this case are they doing more or less the same thing? For example, if I had a crystal oscillator generating a frequency at 88.1 MHz.. could I create an equivalent circuit using capacitors & a coil that also produces an 88.1 MHz signal? If no, why not? If yes, is there a reason to use one over the other? Appreciate any responses. <Q> Using inductors and capacitors, you can reach frequencies much higher or much lower than you can reach with crystal oscillator. <S> The reason why crystal oscillators are used is because they have a much better frequency stability. <S> Frequency stability does not matter much for an experimental DIY toy FM transmitter, but it does matter for many other uses. <A> Crystal oscillators are nice if you want to broadcast from a precise frequency. <S> You may achieve the same goal with a coil & capacitor (called an LC tank circuit).Probably in the circuits you have seen so far utilizes this tank circuit, with their values chosen to resonate at your preferred frequency. <S> It is all nice with capacitors, but most of the time, you may not be able to find a good inductor, so you will have to wind it by yourself. <S> However, I know of no way to measure the inductance without the help of at least a frequency generator. <S> So you may follow a tutorial, wind up your own air core inductor, and it may still not work. <S> And you cannot be sure if the fault is inductor related or something else. <S> Crystal oscillator saves you from this pain. <A> Time to bring servo-loops or negative-feedback into your skillset. <S> OpAmps are examples of negative-feedback, to achieve precise ratios of voltage-gain. <S> The key bit of math, given G <S> = opamp openloop gain, H= <S> the feedback ratio: <S> $$Avcl = <S> G/(1 <S> + G*H)$$ defines <S> the input/output ratio(gain). <S> Huge G values -> very precise gain. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> We use the same feedback methods, but instead of making voltages be a precise ratio, we make the ratio of zero-crossings a precise ratio, using FlipFlops and an Edge_Comparison box, the PhaseFrequencyDetector: <S> The VCO (the 16MHz +-1Mhz) might be 7MHz to 89MHz, but the Phase Frequency Detector ---PFD--- will handle that enormous range, even as the inductors and capacitors and transistors (providing power gain, so the LCs will have a growing amplitude, required by an oscillator) change properties over temperature and with VDD and simply because inductors and capacitors are never exact. <S> A proper PFD is a very interesting beast, and as long as the oscillator is oscillating (some stop oscillating if the VTUNE is requesting too high a frequency) and the PFD functions over the entire range of the VCO, the UP and Down pulses push the VTUNE[the control voltage out of the 2C/1R filter] to the needed voltage; simply by comparing the timing of edges, this feedback system performs frequency lock and phase lock. <S> Key to precise frequency ratio is the charge-storage performed by C1. <S> Any residual error, caused by slight up/down charge imbalance, caused by the 8th edge out of our 1/2 <S> *1/2 <S> *1/2 divider not exactly aligning with same polarity edge from the Fref, will build up and up (or down and down) on C1, even if just tiny charges. <S> Thus over time, the 8th edge from the variable-oscillator is more and more exactly aligned with Fref from the crystal reference. <A> An oscillator is an amplifier with positive feeback, and a "frequency determining element". <S> This can be a crystal, or an LC resonant "tank". <S> The LC oscillator isn't usually stable enough for most applications. <S> Here is what typically happens when you build your first LC oscillator. <S> First, the mechanical properties of the tuning element (usually a variable capacitor) are such that after you adjust it, it "relaxes" slightly, causing the frequency to change. <S> So you end up adjusting it past the desired frequency, and letting it "settle" back to the target frequency. <S> Next, you find that as you move your hand away from the circuit, the frequency changes again. <S> The stray capacitance added by your body affects the frequency. <S> You end up moving your hand towards or away from the circuit in order to keep it on frequency. <S> Next, you find that even if you hold your hand steady, the frequency is drifting and you have to move your hand again to correct for this. <S> This drift may be from a change in temperature, voltage, or more mechanical changes. <S> So, the answer is "yes", but you eventually realize life might be a lot easier if you used a crystal. <S> Crystals aren't perfect, but they are magnitudes better. <S> They also require less power to keep the oscillations going. <S> Good LC oscillators can be built; they will feature a well-regulated voltage supply, rigidly mounted components, and a metal shield enclosing the circuit. <S> There will also be temperature compensation (or temperature control). <S> There were some remarks about frequency multiplication. <S> This could be accomplished by introducing distortion into the oscillator, which produces harmonics. <S> Then, another LC tank circuit is tuned to a harmonic of the original frequency. <S> Another amplifier may be used to boost the resulting signal. <S> This is how it was done in the days of pure analog circuitry. <A> The crystal is at least much easier if you need a precise frequency. <S> Tuning an LC group and keeping it stable if the environment changes is an annoying task. <S> You can get both in the same frequency range (crystals can be bought off the shelf in 100+ MHz which would be right in your FM range)On <S> the other hand, if you really intend to produce FM radio signals, the LC group might be the better way because, contrary to a crystal, an LC group can be easily (de)tuned via a capacitive diode or similar methods, which makes it quite easy to modulate your frequency. <S> A crystal will oscillate at its cut frequency and not far outside of that, so it's better suited for applications that require stability, like for clock generation.
| Yes, you can build a oscillator with inductors and capacitors that will have the same frequency as your crystal oscillator.
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Is it possible to power an arduino-like device from soil? I am exploring some IoT applications I could work on. One of them is a sensor that does some periodical measurements of soil and saves it into inner memory or broadcasts over low-power radio channel. I'd make the device as low in power consumption as possible, e.g. small size, low-voltage transmitter, long periods when the device runs measurement (say 2 times a day). I know there are many factors that can impact it, but is there a chance that soil around the device could give it enough power to function or is it impossible to count on it as a source of energy in a real world project and I should abandon this idea and use batteries? The idea comes from the article about "earth batteries" , which creates a battery by burying two electrodes in the soil. <Q> This isn't going to work. <S> Think about it. <S> If dirt to power worked, it would be done regularly. <S> There are some ways that soil can help to get some energy, but these will be difficult to extract or the amounts would not be useful for most purposes: <S> Sufficiently wet soil can be the electrolyte of a battery. <S> It wouldn't be a very good electrolyte, and the energy would come more from the plates that you'd insert into the soil than the soil itself. <S> Therefore saying you're getting power from soil is a bit of a stretch. <S> From thermal differences. <S> You can use soil as a big temperature averager, especially if you go a few feet or more down. <S> The instantaneous temperature difference between the ground and the air represent power that can in theory be extracted. <S> Note that the Carnot efficiency is not your friend here. <S> Carnot says that the maximum possible theoretical efficiency of a heat engine is T diff /T hot , when the temperatures are measured on some linear absolute scale, like Kelvin. <S> For example, Lets say the soil is 68 °F and the air 90 °F. <S> That's 293 °K to 305 °K, for a Carnot efficiency of only 4%. <S> That means you have to have a large thermal connection to both the soil and the air to extract meaningful power levels. <S> I would look into solar cells. <S> Perhaps these could keep a battery charged so that the device can run whenever it wants to, whether the sun is shining at that moment or not. <A> In principle, yes. <S> The available power from such a battery is highly dependent on the design and install location, so I suggest that you research and experiment with earth batteries and determine how much power you can reliably achieve. <S> From that data you can determine a power budget and therefore the hardware and software requirements. <S> I imagine that any such battery will require regular inspection and maintenance, and that the performance will vary substantially with weather. <S> Reliability and predictability are not going to be easy to achieve, I suspect. <S> I seriously, seriously, doubt that you'll be able to take this beyond "interesting demonstration of the idea" stage. <S> It's not going to work as a product. <S> Maybe an art installation or something. <A> The electrodes themselves are sacrificial (they store the energy as refined metal), and over time your battery internal resistance will rise as the waste products build up. <S> You'll probably also need to use a super-cap (maybe only a small one) to maintain the voltage whilst the mcu is active. <S> A small solar cell and a small battery is probably a better combination. <S> Keep the battery at the coldest part of the enclosure, particularly if you use LiPo technology.
| Although it's possible to construct a battery to extract energy from the bare electrodes, this is unlikely to provide a viable long-term solution.
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What role does the capacitor play in an AM demodulating circuit? The goal of this question. Demodulate the AM signal. (Please assume a diode demodulating AM radio of the most imaginable simplicity.) I am assuming an alternating current sine wave is coming into a wire from the antenna? In order for an envelope to coexist with the sine wave the amplitude is varying I assume? If the antenna is the correct length then resonance of some fundamental frequency of the EM wave allows the signal to increase in strength in the antenna before it enters the diode? An assumption on my part. The signal is then converted to DC but since the strength of the signal determines the fluctuations in the "envelope" the voice information is in "duplicate" so the bottom half is cut out of the rectifier , no? From this point I am lost. The sketch shows a capacitor in parallel. I know what a capacitor does in principle. Is it attempting to maintain the voltage as the current changes since capacitors don't like to have their voltage changed and resist it, no? Please complete my picture to demodulate the AM signal. <Q> The capacitor is not required for the demodulator to work. <S> However it does improve its operation is two ways:- <S> By charging up to the peak signal amplitude and holding most of that charge between peaks, it increases the audio output voltage. <S> Without any capacitance after the diode, the average output voltage of a half wave rectifier is Vpeak*0.318. <S> With a suitably sized capacitor the audio output voltage increases closer to the peak value. <S> The result is a stronger audio signal, potentially as much as 10*log(3.18 2 ) = <S> 10dB louder. <S> In practice audio circuits have some capacitance of their own, which will affect the output to varying degrees depending on the rf frequency and complex impedance of the load. <S> For example a magnetic headphone (commonly used in 'crystal' sets) has high inductance which resonates with stray capacitance at some unspecified frequency. <S> An audio amplifier may be connected through a signal cable whose capacitance depends on length, and the amp's input capacitance may vary depending on the volume control setting. <S> Having a parallel capacitor in the detector circuit reduces the effect of varying load capacitance and inductance, making the audio signal amplitude more stable and predictable. <S> However if the capacitance is too high then it will fill in the gaps between peaks enough to distort the troughs in the lower part of the audio waveform and reduce the amplitude at higher frequencies. <S> To faithfully reproduce the original modulation there needs to be a balance between filter capacitance and load impedance. <S> You can't hear rf frequencies but <S> your amplifier might. <S> Without low pass filtering the rf component of the signal is higher than the audio component. <S> If the amplifier doesn't filter this out internally then its power consumption will be increased and it could produce distortion and spurious frequencies (which could be audible). <S> Effective power output will also be reduced because the amp will overload at a lower audio signal level. <S> Even if the amp can handle this rf leakage without distortion, you now have high power rf signals in the audio circuit, which could feed back into sensitive rf stages to cause distortion or even oscillation. <S> Since only a small amount of capacitance couples rf signals quite well, the amount of feedback could change depending on the position of speaker or headphone wires or the operator's body. <A> We need to start at the beginning. <S> MODULATION <S> At the source the music signal (represented by a sine wave) is combined (used to modulate) a high frequency carrier signal at the transmitter. <S> The music signal contains low frequencies (as humans hear from about 20hz to 20Khz) <S> The carrier signal is much higher frequency (say 100 KHz). <S> note: the carrier signal is a fixed frequency as each radio station operates at a different frequency. <S> TRANSMISSION <S> This modulated signal is now transmitted a long way over the air. <S> RECEPTION <S> Somewhere far away, say 20 km, we now need a way of recovering that original music signal from the 'received signal' we receive at the antenna. <S> To do this we are going to use a circuit called an envelope detector. <S> The envelope detector is first going to convert the AC signal into a DC signal. <S> RECTIFICATION <S> Imagine the received signal (the modulated signal) is a 'messy' sine wave that varies 10V peak to peak (it goes from -5V to +5V). <S> We are going to ignore the negative half of this 'messy' sine wave. <S> A diode only allows current to flow one way, and hence has the effect of producing a constant voltage the -5V to 5V signal has an average voltage of 0V the 0V to 5V signal has an average voltage of 2.5V <S> DEMODULATION <S> Now imagine that this 'messy' sine wave is not a 'true sine wave' but is actually made up by a more complicated waveform that contains the original message signal (at low frequencies) and the 'messy' bits at high frequencies. <S> The envelope detector is now going to remove the high frequencies from the low frequencies. <S> This is done with a lowpass filter (resistor and capacitor). <S> A lowpass filter lets low frequencies pass but stops high frequencies from passing. <S> Now we have recovered the original signal! <S> note: <S> Sometimes a picture can really help understanding <S> : I have tried to give you the bigger picture and have skipped some details here. <A> This circuit is a peak detector, with decay rate fast enough to allow the audio/music to appear on output. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Here are some useful waveforms, showing a charge-fast, discharge-slow behavior
| The capacitor filters out the rf frequency.
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Reading 4.2v with the ADC on an ESP8266-12E The ESP8266-12E is a low power, blazing fast, MCU with WiFi capabilities making it the best option for small robotics projects. I am trying to measure the voltage of a 3.7v lipo in order to determine the battery life of the robot. The battery, when fully charged, is 4.2v but my regulator drops out at less than 1v. I obviously can not pass the 4.2v from the battery directly into the MCU's onboard ADC, and I know I need a resistor in order to drop the voltage to a certain level under 3.6v (it's maximum voltage rating) and generate a percent on that new scale, but to calculate the value of said resistor, I need to know how much current the onboard ADC for the ESP8266-12E draws. How much current does the ESP8266-12E's onboard ADC draw? This is what I came up with: This should allow me to drop the voltage from the maximum possible 4.2v of a fully charged lipo down to solid 3v. I also added a MOSFET to keep the voltage divider from constantly draining the battery. Anyone know if this will work alright? I am using a barebone ESP that I purchased here . @mkeith, I remember seeing somewhere on a datasheet that the ADC can take up to the maximum voltage of the unit itself, which is 3.6v. I could always redo my calculations and drop the voltage down to under 1v to be on the safe side. I just don't want to lose any resolution in my final percentage if I don't have too. I trust this community more than I trust some poorly translated Chinese PDF. <Q> The raw ESP ADC has a full scale range of 0-1V, so you would need a divider of about 4.2:1 to capture the 4.2V input. <S> Since you're switching it, the resistance value is not critical and can be made reasonably low. <S> Perhaps something like this: simulate this circuit – Schematic created using CircuitLab <S> The ADC sees a source impedance of about R2||R3 = 11.5K when measuring (15K when off, but that doesn't matter) and the 0-1VADC input range translates into 0-4.32V. <S> You might want to make the divider ratioa bit more conservative since the ADC in that part is so very loosely specified. <S> This circuit draws less than 110uA when measuring and almost nothing when 'off'. <A> Using just a resistor will not work. <S> However, I am assuming you need to do that in a way that will not consume much battery current. <S> As such you will need large resistors and possibly some kind if MOSFET switching circuit to turn it off when not needed as @mkeith suggested in the comments. <S> If you use the latter you will also need to be careful that the signal line does not go over Vcc when the switch is closed. <A> You should realize that ADC pin also has got an internal resistance which is ~ 220K as per my calculation. <S> So if you use large resistors like R1=220k, R2=100K for the voltage divider then you will see quite significant difference in drop while you have connected the middle pin (across R2) to ADC pin and while its not connected to that pin. <S> So you should take that into account as well while you are calculating the resistors for voltage divider
| You need to use a resistor divider to bring the voltage down to a range the device can accept.
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My IC keeps blowing up - PAC1720 current sense IC I am using a PAC1720 current measuring IC to measure the current/voltage/power being delivered to a lead acid motorcycle battery that is being charged. Here is a newer datasheet and an older datasheet (with different info in them!) Everything seems to work okay until I remove the power supply to the PAC1720 chip. When Vdd to the PAC1720 is removed (i.e. my PCB power supply is off), it appears that the chip begins conducting current from its sense pins to the Vdd or GND pin (or both, not sure), which ultimately fries the chip. This is a critical flaw because it means that if someone hooks up a battery to my product without having the power supply turned on (highly likely to happen), the product will destroy itself. Here is a schematic of how the product is laid out: simulate this circuit – Schematic created using CircuitLab I'm guessing it has something to do with the internal ESD strucutre of the PAC1720. One datasheet for the PAC1720 shows a typical layout like this: The phrases at the bottom of that picture disturb me. It says: Note 1: The device MUST be biased PRIOR to applying VSOURCE. Failureto do so will result in damage. 2: The unpowered bias on VDD must begreater than VDD, if VDD will be removed prior to VSOURCE. I don't understand what they mean by "biasing the device"... sounds like they are talking about biasing the internal clamping diodes? The second note doesn't make grammatical sense does it? I don't understand what they are trying to convey in the second note. Now, an older datasheet shows the internal ESD structure of the PAC1720: This isn't very helpful since it just shows a generic box labelled "Edge-Triggered ESD absorbtion circuit" between the upper clamping diode on Sense1 +/- and GND. So, my main question is: what is happening when I disconnect Vsupply while a battery/charger is connected and the PAC1720 starts smoking? Any way that I can solve this problem now that the product is in production? <Q> I think with bias they mean the combination of the 5.2 V zener diode and the 100k to 1M resistor. <S> So the zener diode is providing a voltage to the chip making those parts work which protect the chip from the voltage across the sense pins. <S> So if you don't have that combination in your unit, you will fry the chip as soon as VDD is removed and nothing is left to power those parts. <A> It appears the chip needs a VDD applied before the SENSE pins are connected to the battery. <S> Since its IOs are open collector (or open drain) it can be powered while the rest of the circuit (SMBUS side) is unpowered. <S> It will not send voltage to your unpowered micro. <S> Now, when the microcontroller is powered but there is no battery, the SMBUS pullups will let some current enter the inputs of the unpowered chip. <S> I don't know if this would be a problem. <S> When in doubt... <S> My solution would be to use a Diode-OR from both the battery and your 3V3 supply. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> The biasing thing means :- You MUST have the power turned on to the device before you attach the battery <S> OR turn on the charger. <S> It also means don't turn off the device before you disconnect the battery AND the charger. <S> In other words. <S> when connected as shown SW1 literally = KILL SWITCH. <S> How do you fix it at this point.... <S> maybe add a relay to disconnect the battery circuit both sides of the sensor resistor when vSupply is disconnected and enough bulk capacitance to keep the chip biased for the duration of the relay opening time.
| Your only option is thus to power it from the battery side, using a resistor and zener as shown in the schematics you posted, or even a voltage regulator.
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Trying to regulate 24V from 24-30V 3A input I am trying to convert 24-30V input to 24V 3A regulated output. I tried using ti's 5A adjustable regulator LM338 but I faced heatsink problem which seems like impossible to keep cool while trying to supply that much. Do I have to buy a DC/DC converter or is there any other 'diy' way of doing it? This regulator is going to be used for powering a water pump which requires 24V 3 Amps. Heat dissipation calculation: lm338 has a characteristic of 'Junction to Case, RθJC(top) 15.7°C/W' The load will be dissipating 24v*3A=72 Watts of power, and the regulator will be dissipating (30v-24v)*3A = 18 Watts of power. So total is 90 Watts. 90*15.7= 1413°C seems too much isn't it? <Q> The real question is: why the hell do you want to feed regulated DC to a motor?.... <S> If it needs to be run at a constant RPM, use a PWM speed controller with a speed sensor. <S> LM338 has 2-3V dropout voltage anyway, so with 24V in, you'll only get 22V out maximum. <A> You will not be connecting the top of the case to a heat sink. <S> Instead, look at the number for the bottom of the case (the metal tab). <S> You'll notice a number of 0.7 deg/W. Add another 1 deg/W for a good heat sink, and you'll get a temperature rise of of about (1.7 x 18), or 30 degrees above ambient. <S> This brings up a question which you have not addressed: exactly what sort of heat sink were you using when you found it "impossible to keep cool"? <S> If you used a pcb heat sink such as <S> well, there's your problem. <S> You need something a lot beefier. <S> That particular heatsink has a thermal resistance of about 24 deg/W under natural convection. <S> Instead, you need something like which has 1 deg/W. <S> Of course, it's 4 1/4 wide by 5 1/2 long, and will run you more than 20 bucks new, but that's what it takes. <S> You can, of course, go with a smaller heatsink and get a bigger rise, or you can mount a fan to it to greatly improve airflow and cooling. <S> But really, 20 watts on a TO220 is no big deal. <A> You are partially correct with your calculations. <S> The load power has nothing to do with the power dissipation of the LM338. <S> 30V - 24V <S> = 6V <S> 6V \$\times\$ <S> 3A = <S> 18W <S> $$P_D = <S> \frac { <S> T_{J\ Max} - T_A}{\theta_{JA}} $$$$T_{J\ Max} = <S> P_D \times \theta_{JA} + T_A $$$$18W \times <S> 22.9^{\circ}C/W + 20^{\circ}C = 412.2^{\circ}C <S> $$ <S> Assuming ambient temperature is \$20^{\circ}\$. <S> The maximum for the TO220 package is \$260^{\circ}C\$ (at least the one I reference). <S> So thermal shutdown. <S> If you decrease input voltage to 27V. <S> You meet the minimum Input-to-output voltage differential. <S> The LM338 will not work below 27V to provide 24V. To minimize losses and to keep parts cooler, it is best to operate linear regulators at the minimum input voltage. <S> 27V - 24V <S> = 3V <S> 3V \$\times\$ <S> 3A = 9W <S> $$9W <S> \times 22.9^{\circ}C <S> /W + <S> 20^{\circ}C = 226.1^{\circ}$$ <S> Below \$260^{\circ}C\$, so this should work with a heatsink. <S> It will be hot and waste power, but it will work. <S> LM338 datasheet
| If it will overheat and burn at 30V (but not at 24V) use a PWM to keep the average voltage on the motor at 24V. Most likely the motor will run just fine on 24 - 30VDC, and you don't need any regulation at all. You are using entirely the wrong thermal resistance.
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Questions on AND gates I have been looking at AND gates and noticed there are a couple different ways to build them. I am wonder when each of the circuits is better than the other: or Also with the first circuit I am a little confused about how it works. If x and y are 5v then q3 will turn off. Why is this? Also, if X and Y are 0v what prevents the current from just flowing directly from the power source into Z making it 5v? <Q> Really wide question and neither is really great. <S> Option A is better if you want to drive something low.. like pulling current down through an LED connected to 5v. <S> simulate this circuit – <S> Schematic created using CircuitLab Option B is better for the converse if you want to supply current to some load. <S> simulate this circuit <A> To answer your last question, when X and Y are both on, there becomes a low-impedance path from the Q3 base resistor to ground. <S> You can think of this as replacing Q1 and Q2 with a short to ground. <S> Therefore, the base resistor is pulled low and Q3 is switched off. <S> When X and Y are off, current flows into the base of Q3 which switches it on. <S> Again, this creates a low-impedance path to ground so Z must be at 0v. <S> As an aside, current will flow through R4 but it will drop 5V. Therefore, Z will still be at 0V. <A> With both Q1 and Q2 on, the base voltage of Q3 should be low enough to be off. <S> It could be fairly close, though. <S> it would be better if there was a voltage divider feeding the base of Q3. <S> That way the threshold voltage for turning on Q3 would be higher, which would give you more noise immunity. <S> Option B is just plain wrong if you are expecting the AND function. <S> Even with the top transistor off, the bottom can drive its emitter high just from the B-E diode action. <S> Therefore, when B is high, the output will be high, whether A is high or not.
| Option A is better because it has a chance of actually working.
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VHDL that can damage FPGA I read somewhere that bad VHDL code can lead to FPGA damage. Is it even possible to damage a FPGA with VHDL code? What kind of conditions would cause this and what are the worst case scenarios? <Q> Adding to @Anonymous's answer, there are designs you can build which can damage the fabric of an FPGA. <S> For starters if you build a very large design consisting of huge quantities of registers (e.g. 70% of the FPGA) all clocked at nearing the FPGAs maximum frequency, it is possible to heat the silicon considerably. <S> Without sufficient cooling this can cause physical damage. <S> We lost a $13k FPGA because it overheated due to the dev-kit having a terrible cooling system. <S> Another simpler case can be combinational loops. <S> For example if you instantiate three not gates chained together in a ring, and disable or ignore the synthesizers warnings about such a structure, you can form something which is very bad for an FPGA. <S> In this example you'd make a multi-GHz oscillator which could produce a lot of heat in a very small area, probably damaging the ALM and surrounding logic. <A> Code is not a right word in this context. <S> While Verilog or VHDL look like program, the output of the compiler is a configuration which is loaded into the FPGA chip forming electronic circuit within it. <S> Current flows - might be excessive current - which eventually damages the gate(s); logical damage: circuit may handle flash chip, or configuration device improperly, and corrupt data image in it, this whole device eventually malfunctions. <A> Misconfiguring a block of input pins as outputs might do it if whatever else is driving them is stiff enough. <S> I don't know if configuring some pins for LVDS or one of the LVCMOS standards while the IO bank is powered from an overly high voltage (3.3V power with a 1.8V IO standard for example, or the opposite on an input) would do it? <S> Obviously thermal problems may be a possibility by doing something silly like instantiating many, many, ring oscillators. <A> FPGAs can be reconfigured at runtime with a new (partial) bitstream. <S> Normally, this stream is loaded from an external source, but you can also create it by your self in the FPGA (e.g. by an embedded softcore CPU). <S> Using such a solution for e.g. dynamically relocating subdesigns, doesn't provide all the consistency checks as done by the vendor tools. <S> So if your algorithm is broken, you might enable the false path transistors in an FPGA and burn them. <S> You could also chose false operations modes for FPGA primitives like PLLs or transceivers. <S> Dynamic reconfiguration is like self modifying code in software.
| Two types come to my mind: physical damage: for example, several FPGA pins are connected together (or to another device) and start outputting different logical voltage at the same time.
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Push-pull power amplifier on complementary darlingtons I decided to make a power DC-amplifier. I used complementary darlington pair TIP142 and TIP147. Also TL082 used for negative feedback. The schematic is as below: I gave 100 kHz sine signal and under oscilloscope I noticed the crossover distortion. It seems opamp is too slow to compensate ±1.4V "dead zone" (distorion on oscilloscope above). Then I decided to bias darlingtons bases using four 1n4148 diodes as follow: Unfortunately after 5 seconds TIP147 blew up although transistors were mounted on a heat sink and the fun was over. :D I was reflecting what actually gone wrong. I suppose that: I didn't put diodes on the same heat sink as darlingtons.Consequently BE voltage dropped under the diodes bias. In Horowitz's "The Art of Electronics" I have read if I we are usingemiter resistors, then four diodes biasing are insufficient. I would to know are my conclusions correct and also how to effectively get rid of crossover distortion. <Q> Four diodes provide too strong bias. <S> Together with positive temperature coefficient of output transistors and lack of thermal coupling, the result is sadly expected. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Better solution is to use transistor. <S> It must be mounted on the same heatsink, preferably on top of one darlington. <S> Quiescent current should be adjusted to about 20mA (start with maximum resistence, measure mV on emitter resistor). <A> On 1 you are correct. <S> The diodes are really temperature sensors. <S> On 2 the answer depends on your definition of "sufficient". <S> Generally you want to have the same number of pn junctions in the bias section. <S> As the idle current is small typically in the ma range the emitter resistor isn't really a factor. <S> In your case I think issue 1 is a far bigger driver . <S> insufficient biasing would have saved you, everything else being equal. <S> Btw, oscillation many times kills amps too. <S> So keep an eye on that. <S> Edit as your rail is single ended you will need a capacitor on the output to block dc. <A> Place a 1 microfarad ceramic cap between the darlington bases and look for waveform improvement on the scope .This <S> improvement should be more visible at say 20KHz. <S> This is because loop gain decreases with frequency paticularly for opamps. <S> Increase the value of the emitter resistors .A <S> rule of thumb is up to 10% of the load resistance .It <S> is a trade off between output swing and thermal stability .Your diode circuit is adequate but the thermal coupling should be adressed .What <S> I have done <S> is make my own diodes out of TO126 transistors like BD139 or <S> what ever is handy .Bolting <S> this to the same heatsink has made for reliable Audio amplifiers operating well into class A .I <S> have tied the base to the collector to make a diode .Also <S> I have jacked it up with BE and BC resistors making a VBE multiplier. <S> I have seen this VBE multiplier called a Rubber Diode .The <S> 1 mic cap is still just as useful . <S> The key here is that the thermal resistance of the TO126 package to the heatsink is really low compared to your 4148 or other peoples <S> BC547 <S> .There is little point in designing fancy bias circuits before your thermal impedance is dealt to .
| An option is to use three diodes, but the circuit is still thermally unstable until diodes are on the heatsink, and it has smaller but still very high crossover distortion.
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What to do with short-curcuit 18650 I was soldering wires to 18650 batteries, and while I was soldering the + tip at one of them, it suddenly became very hot. I thought that maybe some drip of tin get under the tip and shorted it, so I took voltmeter and it measured no voltage, so the cell must be shorted. As it was getting hot and I wasn't able to find any drip of tin which shorted it, I just throwed the cell into iron bucket filled with water. Is this what I should do? Or what am I supposed to do if this will happend again?Thanks <Q> No, you are not supposed to do any of that. <S> It is not good idea to solder directly to li-ion battery, as li-ion batteries really hate heat. <S> And it is not good idea to extinguish li-ion battery by water. <S> You can use sand. <A> This process takes only a fraction of a second (to weld the wires to the terminals) and the battery would not be heated up as opposed to soldering. <S> In your case buying a holder would be the safest solution. <A> Throwing a single cell into a bucket full of cold water isn't actually that bad of a thing to do. <S> Because lithium battery powered vehicles are becoming more and more relevant, there have been investigations <S> (German document, sorry) on how to extinguish a burning car filled with lithium ion batteries. <S> You actually cannot extinguish a burning cell, as the oxygen required for the exothermal reaction <S> is generated inside the cell due to thermal runaway and decomposition of the materials inside the cell, so even putting it in sand will not extinguish the fire. <S> It will just prevent the spread of the fire. <S> What is the benefit of using water then? <S> For a single cell it's probably not as relevant as in a battery pack. <S> That way some cells remain "intact", at least they won't burn. <S> Another point for water is that stuff getting out of the cell (fumes, electrolyte) is bound in the water. <S> But it was noted that you need a lot of water. <S> Do not solder lithium batteries which don't have a solder contact. <S> You will introduce heat very close to the internal chemistry of the cell. <S> It is very likely to damage the separator with excessive heat which occurs during soldering, thus creating an internal short circuit which leads to thermal runaway. <S> Even if you have cells with solder contacts, make sure to be quick about soldering it.
| You were probably lucky that the cell was not fully charged when that happend, otherwise the reaction would have been more furious. The proper way to connect wires to the battery terminal would be using a spot welding machine. In a battery pack the water will cool down the other cells and prevent decomposition in those cells. The easiest way would be to buy a simple 18650 battery holder, cheap enough.
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attenuators volts to micorvolts I am a software engineer and have limited knowledge of electronics. I am using a Raspberry PI to control a AD5360 ADC eval board ( http://www.analog.com/en/products/digital-to-analog-converters/da-converters/ad5360.html#product-overview ). I need to attenuate +-5 volt output form the DAC to +-50 Microvolts on 16 independent channels. I am thinking that I need an on chip attenuator. I am still prototyping so and part needs to be used on a bread board or with pins that can be soldered to a solder board. I need a fast response time because I change voltage on each of the 16 channels between 256 to 2048 time each second, and a high level of precision. If required I could build them, however because I need 16 I would prefer to buy them. I have tried to use a simple voltage divider however, this introduces too much variance at the microvolt range. I am open to other ideas if they can solve the problem more efficiently or are more cost effective. <Q> They add little noise and drift. <S> For example, you could use a 1M:10 ohm divider, using 0.1% or 0.05% resistors. <S> Or lower accuracy if that meets your (unstated) requirements. <S> Added noise due to the resistors will be down around the theoretical minimum of 0.41nV/sqrt(Hz), so over a 10kHz BW only 41nV RMS. <S> You do need to take proper care preventing errors due to thermal EMFs if you expect to get high accuracy with only 50uV full scale, but that's a construction detail and outside of the scope of this answer as it applies to any possible method of producing the voltages. <S> In general you would minimize asymmetry, minimize material choices that result in high uV/°C EMFs and keep everything isothermal. <S> If you need lower than 5-10 ohms you may have to add amplifiers, but that will greatly increase noise and could add large errors depending on the choice of amplifier. <S> As far as 'speed', such a divider will be limited by the DAC output- it would settle in nanoseconds with an ideal step input (microseconds at most)- <S> so your kHz should not really be a worry if the DAC is up to the task. <S> If you were looking at really high speed (above hundreds of kHz) <S> you would frequency-compensate the divider by matching the ratio of 1/capacitances across the resistors. <A> I'd still go with voltage dividers, just use 1% (or even 0.1%) precision resistors. <S> Note that "typical" resistors in hobbyist kits are usually 5% precision. <S> And then buffer each voltage divider output with precision op-amp, because if you take the output straight from voltage divider to load then your voltage will definitely jump all over the place. <S> So build 16 of 10MΩ/100Ω voltage dividers, and connect simple op-amp followers to them. <S> Input resistance of the op-amp is very high, so it will not affect the output voltage of the divider, and then output resistance of the op-amp is quite low, giving enough current for your load. <S> As for op-amp selection, look for "precision op-amps": TI , Linear , others. <S> " <S> 2048 time each second" is not quite fast. <S> It's in the middle of sound spectrum, so any op-amp should be capable of working at that frequency. <A> +1. <S> Opamps are unnecessary. <S> DAC has pretty good on-chip buffers, so 100k/1ohm dividers will provide low output Z. 10k/0.1ohm could even be used...
| If your output impedance specification is reasonably high (say 10 ohms) then JUST using resistors is going to be your best solution.
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Advantage of series vs parallel battery in handheld product There are many handheld product like laptop or an Handheld measurement equipment that all parts would works fine with 3.3V or lower VCC. However these devices still comes with a 11.1ٰV battery or higher voltage. My question is, doesn't it more efficient to use 3 cell in parallel instead of series in order to reduce power dissipation on voltage regulator? As an example in fanless laptop (like Asus UX305 ), 5v seems to be sufficient for most parts, so probably 2 cell 7.4v battery suits better than 3 cell 11.1V. Where am I going wrong? <Q> The decision to use a particular voltage input is not necessarily made on the major supply voltage used in an appliance. <S> Consider your question around the ASUS laptop: <S> Its input is 19 V; a fairly standard laptop supply voltage, but not likely used for anything internally compute or peripheral related. <S> @19 <S> V and 45 W you can expect about 2.3 A maximum line current. <S> @12 <S> V, that would rise to about 3.75 A. @5 V ... <S> 9 A. @3.7 V (1 cell voltage) ... 12 A. Voltage loss would become more critical in the wires and connectors at lower voltages making overall design much more challenging. <S> Internally they may not even charge the batteries at full current when the laptop is on at the same time to control both the power dissipation within the shell and current magnitude in the input supply line. <S> You can see from the specs that the compute/display side of the laptop is only 5 -7 W depending on display usage (resolution/brightness). <S> The battery in the ASUS is likely a 4 cell 14.8 V battery with active capacity management but no cell balancing. <S> I don't see it specified, but would assume about 1 - 2 hours maximum charge time for the battery at say 2C. <S> Most of Intel's reference designs are based around a 12 V supply (Core level compute elements). <S> To abandon the reference design (which is fully debugged) and design a new way to do it would be a risky undertaking. <S> You rarely see OEMs such as ASUS, Apple or Microsoft stray far from the reference platform in anything other than peripheral devices. <S> So back to your question ...could the laptop be powered by a single cell <S> paralleled battery pack .... <S> sure, <S> but it would be significantly harder to design the appliance. <A> I would put the cells in series. <S> The buck converter can be designed to work well over a fairly large input voltage range, so you can draw the cells down to such a low voltage that there is no energy left. <S> With multiple cells in parallel, I'd worry about a small mismatch and one cells trying to drive other cells backwards when everything is supposed to be off. <S> This is assuming everything is at a "low" voltage, like under 20 V or at least under 30 V. <S> You'd rather not have 100 V in to drive your 3.3 V device, but at 11 V versus 7.5 V, the 11 V is probably easier to use efficiently. <A> A common arrangement, especially in power intensive applications like modern CPUs, is called point of load regulation . <S> This feeds a higher voltage at a lower current around thinner traces on the main board to feed buck regulators at the places where the lower voltages are needed. <S> At that point it becomes lower voltage with higher current and thicker traces. <S> This has the advantage that the power routing is easier, takes less trace room, the traces are thinner, and you get reduced losses in the power circuitry. <S> It's the exact same idea as having high voltage power lines to transfer power around the country and then transform it down to lower voltages for consumption: less transmission losses and thinner cables. <S> And of course for that you want a higher voltage, not a lower voltage. <S> So you start with a high voltage battery arrangement to make life easier for you. <S> After all, boosting to 12v from 3.7v to then buck back to 3.3v or 1.8v <S> would be very wasteful.
| Generally, higher voltage at lower current is easier to deal with and will have less loss than lower voltage at higher current. It's also much more challenging to design an internal battery charger since you now have to separate input/battery at the same voltage with FETs to allow a SM charger to operate.
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What does "AC frequency response : (60-1000)Hz" mean for multimeters? I'm considering to buy a multimeter. The following is its specs as listed at Amazon: Size:KT6666 Featuring a wealth of functions, it is intended for troubleshooting various electrical issues arising in household, industrial and automotive devices. Features: Fast, accurate measurements at 3 readings per second update speed; Backlit LCD Screen with perspicuous readings display; Overload protection in all ranges; Heavy duty build, suitable for indoor & outdoor use; 6000 counts display with TRMS , Large LCD size at 6.2*3.8CM Rubber holster to protect the meter dropping when use. Multi meter Specifications: DC Voltage: 600mV ±(1.0%+10) DC Voltage : 6V/60V/600V/1000V ±(0.5%+3) AC Voltage (True RMS): 600mV ±(3.0%+3) AC Voltage: 6V/60V/600V/700V ±(1.0%+3) DC Current: 600uA/6000uA/60mA /600mA /10A/20A ±(1.5%+3) AC Current: 600uA/6000uA/60mA /600mA/10A/20A ±(1.5%+3) Resistance: 600/6k/60k/600k/6M/60 MΩ ±(0.5%+2) Resistance: 60MΩ ±(1.5%+3) Temperature: -20~1000℃ (-4~1832℉) ±(1.0%+5) Capacitance: 10nF ±(5.0%+20) Capacitance: 100nF/1uF/10uF/100uF/1000uF ±(2.0%+5) Capacitance: 10000uF ±(5.0%+5) Frequency:99.99/999.9/9.999k/99.99k/999.9k/9.999MHz,±(0.08%+2) Duty Cycle:1-99%,±(0.8%+2) Range Selection: Auto / Manual Ranging Max Display: 6000 counts Sample Rate: 3 times/s Diode test: Yes Continuity buzzer: Yes Low battery indication:Yes Data hold:Yes Auto power off: Yes Restorable Fuse protection: Yes Shock proof protection: Yes AC frequency response: (60-1000)Hz Battery: 2x1.5V (INCLUDED) Dimension: 180*90.5*45mm Product Net Weight: 343g Working environment: 0~40℃, relative humidity < 80%" From: https://www.amazon.com/KASUNTEST-Multimeter-Capacitance-Resistance-Transistor/dp/B01L9F09NM/ref=s9_simh_gw_g469_i1_r?_encoding=UTF8&fpl=fresh&pf_rd_m=ATVPDKIKX0DER&pf_rd_s=&pf_rd_r=QCNF28ZDGZC93PVZTEE3&pf_rd_t=36701&pf_rd_p=1cded295-23b4-40b1-8da6-7c1c9eb81d33&pf_rd_i=desktop ==================================================================== Question: This meter is advertized as a true RMS multimeter. What does "AC frequency response (60-1000)Hz" in the specs mean? Is it this meter's sampling rate to calculate RMS? If so, how does it compare to other true RMS multimeters'? After writing the above, I just noticed that there's "Sample Rate: 3 times/s" in the specs. Is this the sampling rate to calculate RMS? Isn't this pretty low? <Q> This seems like a poor choice in case you someday want to use this meter on a 50 Hz AC system. <S> But if you live in North America or the right parts of Japan, maybe you don't care about that. <A> Three readings per second is typical for digital displays. <S> Faster than that is difficult to mentally absorb (for many users). <S> Slower than 3-per-second makes some users have that "hurry-up-and-update" feeling. <S> Display rate is usually not associated with the RMS calculation of AC signals. <S> Between 60 Hz and 1000 Hz, this meter has +/- <S> 3% error. <S> For AC waves of other fundamental frequency, you will get a reading with higher error. <S> For very low-frequency waves, you will often get readings that jump around, making an amplitude determination difficult. <S> For readings higher than 1000 Hz, error will be greater, often on the low side. <S> You should determine if error is 3% of full scale , or 3% of the reading . <S> (3% of the reading is probably a better spec). <S> A 3% error is not very good for a digital meter, especially one that has a four-digit display. <A> It means that you will only get accurate RMS readings within those frequency ranges. <S> You can only use it as a frequency counter above those frequency ranges. <S> If you want to have true RMS readings to a high frequency (such as 10 MHz) <S> you are going to need to buy a more sophisticated meter such as a Fluke Scopemeter which is expensive.
| It means that for AC signals with frequency below 60 Hz or above 1000 Hz, you can't expect this meter to give an accurate AC rms voltage reading.
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What are the effects of trimming the diameter of a wire down to half its gauge? I have a three-phase, 4 AWG 259/28 stranded power transmission cable (3 conductors total) running ~200m into a slip ring. The only way to connect to the slip ring is by using its 8 AWG pig tails. These pig tails are rated 600V and 40A, which is sufficient for our needs. I would like to trim down each of the 4 AWG conductors so they can fit into an 8 AWG butt connector or c compression clamp to splice to the pig tails. As far as I can tell, when trimming a conductor down you are simply changing the resistance/length of the cable in that area and its rated amps. When trimming the conductor from 4 AWG to 8 AWG, would the mΩ/m of the trimmed section shift from 0.8152 to 2.061, as shown in wikipedia's table on wire gauges , or is there more going on here than I think? Since the diameter of the 4 AWG conductors will suddenly change, will there be power transfer or heat issues that I am unaware of? <Q> If you simply cut strands from the 4 AWG wire, you will have an unpredictable resistance between those strands and the remaining strands. <S> If you don't use all of the strands for the majority remaining length of the full wire size, the voltage drop will be the voltage drop will be somewhere between the voltage drop of 8 AWG and 4 AWG for that length. <S> I would attach ring lugs for 8 AWG on the leads form the slip rings and lugs rated for 4 AWG on that wire. <S> Then, in a junction box, connect each phase of 8 AWG to 4 AWG using short bolts through the ring lugs. <S> Tape up the lugs with appropriate electrical tape. <A> If you are affected by the electrical codes, you will violate code doing that. <S> You may also affect the cable's strength in unforeseen ways. <S> Any chance you can obtain different lugs for the machine? <S> Otherwise my approach would be to run 8 AWG from the slip rings to some convenient location for a junction box, and then splice the 8 to the 4 there. <S> I would use a lug multi-tap connector, <S> e.g. One is insulated, the other naked. <S> (the insulated one's wires enter from the same side, you can get any variation). <S> Both under $10 at your local electrical supply. <S> All these can handle a wide range of wire. <S> For insulated ones, I select where its maximum size equals my largest wire; that way that wire's insulation fits snug into the rubber. <S> The plastic plugs cover setscrews and unused wire holes. <A> You have no problem. <S> You say the amps will be less than 40. <S> 8 gauge wire will handle 40-55 amps. <S> There will be no heat build up. <S> Yeah <S> sure, the resistance will increase by a negligible amount, like 4µΩ for a couple of millimeters. <S> This is not necessary and may not be an option depending on your circumstances. <S> If you want you can not cut the extra strands but peel them back about 3 inches. <S> Then cut the strands going into the CTAP so the peel back strands are about 2" longer than the other strands. <S> The once the connector is crimped you can take the peeled back strands and twist them around the outside of the crimp. <S> And solder them if you want(if you use the copper CTAP).
| Depending on the method of supporting the wire, vibration or just the weight of the heavier wire could put excessive strain on the short length of reduced-size wire.
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How to Measure Power Consumption on Extremely Low Power Devices? This might be old news in a half a decade or two but by today's means, I am referring to electronic prototypes and designs which would draw in a μA (uA) and even nA range of current. Some recent MCUs, such as SAMD21 that I am using atm are armed with internal clocks such as ,always on, Ultra Low Power Internal 32kHz RC Oscillators which would draw only 125nA, and the whole microcontroller is capable of consuming only 6.2μA on STANDBY mode with a live RTC. In these type of quiescent current and power consumption levels the smallest limitations in the internal machinery of bench measurement devices such as multimeters and oscilloscopes could add a fair bit of error to the overall measurement or even measure a flat out wrong value in situations like a different relay kicking in when changing the resolution from 6 to 8 decimal places accuracy on your multimeter. What is the most precise method of measuring the overall quiescent current/power consumption for such applications? Update: As I mentioned in one of the replies, measuring low currents is hard but very possible, however, making conclusions on the integrated amount of current consumption to come up with numbers for the realistic over all power consumption is more what I had in mind. I have bumped into some solutions such as wide range current to frequency converter , however the wide range in this application note is only limited to the max of 200uA and in my case, my max current can rise to milliamps when my radio is transmitting and could drop to as low as 3uA when the whole system goes to sleep. <Q> These are designed to offer an extremely high input impedance to both inputs of the amplifier (in excess of 1 giga-ohm), while allowing you to amplify this signal by relatively large factors (1000x is not uncommon). <S> Note that the fact that there is a really high input impedance isn't too terribly important for this particular application, however the high amplification factor is. <S> The basic schematic looks like this (I'm using IA <S> is a self-contained package for an instrumentation amplifier; often, these have an external gain resistor so you can choose whatever gain you want): simulate this circuit – <S> Schematic created using CircuitLab <S> The large amplification factor allows you to use a relatively small sense resistor, mitigating a large portion of the effect of the burden voltage on your DUT. <S> If you're just looking to buy an off-the-shelf solution which does effectively this, you could look into something like the uCurrent . <S> There are probably also specific IC's designed for this current range. <S> Since the outputs of these type of current sensors is just a relatively isolated analog voltage, you can use any standard oscilloscope or voltage meter to measure the current. <S> These very simple devices are good enough for things in the nano and micro ampere ranges and are relatively easy to use. <S> For even smaller currents (pico or fempto ampere ranges), there are specially designed chips such as the LMP7721 , along with a few pages of application notes on low current design. <S> It's unlikely you'll want something like this for measuring power current draw. <S> These are typically used by the scientific community for measuring sensor outputs (photodiodes/other very low current sensors). <A> The Microchip AN1416: <S> Low Power Design Guide, on page 6 specifies a very interesting and simple solution to measure very low current static consumption, using what it called 'the capacitor method'. <S> A known charge is set on a known capacitor. <S> This charge is then used to supply power for the Device under Test. <S> After a known time, you disconnect the capacitor from the Dut and measure their residual voltage. <S> With this delta and with a formula supplied by the same document, you can estimate how much current your device consumes over a period of time. <S> The document also points out which types of capacitors to use and to how to account the leakage current of the capacitor. <S> Below the document from Microchip. <S> http://ww1.microchip.com/downloads/en/AppNotes/01416a.pdf#utm_source=Facebook&utm_medium=Social&utm_term=Post&utm_content=MCU8&utm_campaign=Low+Power+Design+Guide <A> The professional solution is to use a sufficiently good bench multimeter. <S> I've met people who measured the average current consumption (< 10µA) as part of their software development routine, using something like a Keysight 34465A with the 50000 measurements/s option. <A> I easily measured 1uA currents. <S> With a scope I can see when different functions of my application are running. <S> You can connect it to a scope or a benchtop voltmeter. <A> I've been developing battery powered IoT devices now for over 10 years, and have found multiple methods to do this depending on what I'm trying to accomplish. <S> If simply trying to find the low sleep current of a static system, I like to keep my setup relatively simple, and use common items you can find in most labs, and use basic electrical concepts. <S> Referencing the image below, choose a sense resistor (R1) value that gives roughly a few hundred milivolts with the expected current draw. <S> This will allow a standard DMM to get a relatively accurate measurement while still providing adequate voltage to the DUT, even at low supply voltages. <S> Using Ohms's Law, you can calculate the current: <S> I = <S> V/R. <S> Using the expected current value from the original post of 6.2 uA, a sense resistor value of 20k-30k (0.1 to 1%) would be sufficient. <S> In a case where the DUT needs to be initialized to a low power state, a shorting jumper could be placed across the sense resistor R1 until the low power state is maintained. <S> This would allow the DUT to draw as much current as it needs without causing an excessive voltage drop. <S> Once the DUT gets to the expected low power state, the shorting jumper can be removed, and the idle current measurement can be taken. <S> While the above method works well under static conditions, it will not work under dynamic conditions, especially with the peak currents typically seen in battery powered devices due to the high impedance that the measurement method presents. <S> For these more real world operating conditions as you describe in your update, you will need a device that accurately measures and records the current over a very wide dynamic range, possibly up to 100,000:1 (100mA down to 1uA), do it with enough speed to capture the quick turn on and off transitions, and continuously integrate the results. <S> This was something that always took a great deal of time and effort in my early days. <S> So much so that I decided to create a device that was purpose built to handle this for me. <S> Check out the link below: Better Embedded Engineering Battery Energy Estimator 300
| One solution is to use an instrumentation amplifier to measure the voltage drop across a shunt resistor. An off-the-shelf solution is a uCurrent from CMicrotek , worth the price.
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Reason for USB power bank LED to be always on? I hope this is somewhat on topic. I received a free 2200 mAh USB power bank today and noticed that inside its casing it has a blue LED that is always on, even when not giving charge or being charged. Are there any common design reasons for such behavior in a device of this kind? To me it just seemed peculiar for a device meant to retain energy. (I don't have much experience in electronics) <Q> Leakage current. <S> Older simpler designs leak current through the boost inductor and diodes. <S> Ten bucks says that the led is dim when not in use and brighter when in use? <S> That's because it's tied directly to VUSB instead of a dedicated control IC like newer designs. <S> Since the battery is not disconnected, it leaks and powers the led. <S> When the current kicks up, the voltage of the boost circuit will rise to the 5V it is set for, and the led gets a higher voltage and hence brightness. <S> Basic Switching Circuit: simulate this circuit – Schematic created using CircuitLab <S> When the switching circuit is off, the battery's 3.7V still leaks through L1 and D1. <S> The battery typically has a protection circuit ( DW01 IC ) + mosfets that will disconnect it when the voltage hits the Over Voltage and Under Voltage thresholds, but in normal voltage ranges, are not affected. <S> The battery is never disconnected. <S> It's not a minimum load for regulation, as that only matters when the current to a load is significant. <S> While leaking it will be a milliamp or less. <S> It is actually discharging the battery unnecessarily, just very slowly. <S> It's a bug in the design. <A> This might be a "leakage", but also it might be the method how some (all?) <S> chargers detect the presence of load. <S> When phone gets charged and its power consumption goes below certain level (50-100mA), most chargers switche into some pulsed mode, to reduce waste (quiescent conversion current) when either the charge is done, or when nothing is connected. <S> The duty cycle could be 1:100, or something. <S> On every short ON cycle the LED goes full bright, so on average you see it glowing. <S> When there is a load above threshold, the charger goes 100% on, charge begins, and the LED shines full scale. <S> Some chargers IC do have a dedicated status LED , which keeps it separate from output, so no light comes out after 15-20 seconds timeout. <S> Apparently your charger is really on a cheap end, and has the LED simply on the output. <A> Some cheaper battery banks require a small current output all the time or at least once in a while for them to maintain a charge in the battery pack. <S> That LED barely draws anything but it's not helping neither.
| This LED is apparently connected to the charger output.
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Cable joins on wiring LEDS in parallel So I am a newbie. I want to wire up a bunch of LEDS in parallel, and I have it all working fine on a bread board. I can find loads of nice drawing of the schematicsBut what I cant find a good example of is how to attach LED/Resistors to the wires.I have seen wire nuts and people just removing the insulation on the wire and soldering on to the bare wire, and then not covering it up. All these just smell bad. What would is considered best practice. The project is about a square meter in size. <Q> This is the best way to avoid short circuit connections. <S> Note that it may be time consuming to actually have to do this for a meter square of LED connections. <S> Here is an example below of what I am talking about: You would have to place the heat-shrinking element before you solder your LED to your resistor and when the solder connection is done, move the heat-shrink tube down over the solder joint. <S> Using a solder iron or a lighter, you can shrink the tubing material very easily and make it stick to the solder joint. <S> That way your circuit will be protected from potential short circuits and no LED will go bad. <S> Be careful to not place the solder iron or the lighter over the same place for too long since it might cause the solder joint to melt and detach itself. <A> Use one of these PC boards from RadioShack; https://www.radioshack.com/collections/breadboards-ic-sockets <S> You can also get a breadboard which has the same layout as the PC board making this a simple transfer. <A> There are sockets for LEDS, such as the one shown below. <S> Digi-Key part <S> no 2914-ND <S> However, that will get VERY expensive if your board is a meter square. <S> You would be much better to break up the board into a number of smaller, identical, panels make from perf-board or a PCB, that interlock together, like "Leggo" using a suitable inter-board connector. <S> On another note: You do not mention the spacing between LEDs. <S> I have the feeling that you intend to have hundreds if not thousands of LEDs in this panel. <S> If so you are going to have current and power issues. <S> Normally, arrays of LEDs are not driven all at once, but rather they are driven as a matrix. <S> Each row is turned on individually in sequence under the control of a sequencing circuit. <S> This all happens fast enough that the human eye can not tell the difference. <S> Sometimes the individual LEDS are also augmented with a small capacitor in parallel to keep the LED turned on for a short period after the row is "scanned" so <S> they work more like a phosphor TV. <S> Either way, the module method helps with power distribution too. <S> You could even have each module have <S> it's own, inbuilt power convertor.
| What you can do is solder the LED to the resistor with the bare wire obviously as you mentioned but cover it up with heat-shrink tubing insulation material.
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How to automotive/boat devices limit voltage to 12 volts? I live on a boat and get all my electricity from 12 volt battery arrays. The batteries are charged using either 1) alternator on the diesel engine or 2) charger/ inverter powered from shore power at a dock or a portable generator. Yeah, some 120v power while on a dock or running the generator but not usually. Most everything onboard is powered 12 volt DC, but in reality it's actually fluctuating from 12.4v to 14.6v. So how do electronics like chart plotters, SSB radio, am/fm car radio tolerate these changes? Obviously they do it, and hopefully efficiently. Is there a simple device to do this? So far all I could find is " consider a simple PWM driver with mosfets controlled by a microcontroller, and figure out the correct pwm voltage correction empirically and be done with it. " which sounds interesting but doesn't do it for me. <Q> Devices designed for automotive environments have a wide input range that covers the typical automotive 12V rail. <S> They spec the input at 15V on average. <S> Most cars have a nominal voltage of 14.6V while on and charging via the alt. <S> They work fine at these raw fluctuating voltages. <S> Some like motors or fans do not need regulation. <S> Others, like radios or computers, require cleaner signals so they internally regulate down/up and filter the noisy 12V rail as best they can. <A> Automotive voltage is often referred to as 12 Volts DC when it is actually 13.2 volts DC. <S> The voltage is determined by the chemical make of each cell. <S> The chemical make-up produces 2.2 volts DC per cell multiplied by 6 cells. <S> Many auto alternators charge at the rate of 15 volts. <S> Equipment designed for automotive operation is also designed to deal with these variations. <S> A bigger problem to consider would be the damage that can be caused when a battery is allowed to fall below 12 volts DC. <A> There are multiple ways you can achieve 12V regulation: <S> Buck-Boost Converter LDO (Voltage Regulators) <S> Voltage divider (That is never used... <S> Very inefficient) <S> Let's focus on the buck-boost converter since this is probably the best way to supply 12V continuously over a range of time. <S> A buck converter will convert high input voltages to a fix output voltage whereas a boost converter will convert low input voltages to a fix output voltage. <S> A buck-boost converter is a combination of both converters described in the past couple sentences. <S> As you can see in the circuit above, there is a switch controlled by PWM that will control the voltage on the load. <S> Here is how the current will flow depending whether or not the switch is open or close in the circuit. <S> Note that the 12V can also be regulated using a LDO. <S> In this case, it can be useful since there is not a major voltage drop across Vin and Vout. <S> Therefore, the power dissipation through the LDO is much less. <S> Sometimes, dropping the voltage from 12V to 5V, care must be taken since the LDO may overheat and destroy itself. <A> This question is easy to mis-understand since there are so many possible devices. <S> For the parts of the devices that care (mostly the digital electronics) the internal circuits will be operating at typically 5 volts, or maybe even 3.3 volts for modern self-contained circuits. <S> You might also find some parts of the circuits running off 9 volts. <S> All of these are easy to generate from your battery bank (nominally 11V to 16V) with either linear or switching regulators. <S> Since the regulators are inefficient, it makes sense to only regulate the parts of the circuit which are sensitive. <S> This means that the backlights, etc, might run off the unregulated inputs. <S> Amplifiers (and to some extent radio transmitters) can effectively act as regulators themselves so might not require such a carefully controlled supply rail.
| If you need a regulated 12V input, use a low dropout voltage regulator, or a switching regulator.
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does 8085 CPU have an extra register within ALU? From the 8085 CPU architecture, when ALU done calculation, the result is clocked back to accumulator A on next clock edge. But accumulator A is directly wired as ALU input, what if the clock edge didn't raise fast enough to cause A is being added twice or many more times, it would be extreme difficult to detect such an error, such a design is very "fragile" to me. Unless, there is an extra register within ALU to temporary save ALU results? https://en.wikipedia.org/wiki/Intel_8085#/media/File:Intel_8085_arch.svg https://en.wikipedia.org/wiki/Intel_8085 <Q> As @duskwuff suspected, I've looked into this. <S> To answer the question, the 8085 has two extra registers in the ALU. <S> WZ is part of the register file, while ACT, A (accumulator) and TMP are located in the ALU circuitry itself. <S> Here's a diagram of how the ALU works: <S> The ACT register has several important functions. <S> First, it holds the input to the ALU. <S> This allows the results from the ALU to be written back to the accumulator without disturbing the input, which would cause instability. <S> Second, the ACT can hold constant values (e.g. for incrementing or decrementing, or decimal adjustment) without affecting the accumulator. <S> Finally, the ACT allows ALU operations that don't use the accumulator. <S> One interesting consequence of the 8085's ALU setup is that a value can be loaded into the accumulator only after passing through the ALU. <S> Details on the 8085 register set are here and details of the ALU are here . <A> In synchronous designs it is an important task of designer to ensure such things do not happen. <S> Register, which is having data being "clocked" into, is having specific dynamic properties like clock raise time, clock hold time, data stable prior and after clock signal change. <S> If timing is violated, resulting state is not guaranteed. <S> In your particular case <S> ALU is having its propagation delay, and to add A twice there should be a time until new A is added to previous A within the adder and result appears at its output. <S> Most probably it was simulated and calculated that such thing will not happen within defined allowed clock frequency range for the device. <S> That's why datasheet explicitly has minimal and maximal clock ratings. <S> For 8085A-2 it says: Minimal CLK cycle period: 320 ns Maximal CLK cycle period: 2000 ns <A> The accumulator is the ALU's output register. <S> The 8085 has a two phase clock. <S> Where single clock instruction like a NOP took 2 clock cycles. <S> Similar to the 8088 used in the original IBM PC, the 8088 had a 4Mhz four phase clock and executed instruction at a rate of 1Mhz. <S> With the two phase clock you have two oscillator cycles for each instruction cycle. <S> Internally, for timing, any edge of either clock can be used. <S> The clocks are inverted <S> so there are actually four clocks available for timing. <S> Then you have transparent latches that latch on the falling edge and D-Flip Flops that latch on the rising clock edge. <S> A transparent latch allows the input data to propagate through to the output beginning at the clock's rising edge, and the values are latch on the falling edge of the clock. <S> The 8085 had many options when it came to avoiding propagation race conditions.
| The 8085 has several "hidden" registers: a 16-bit WZ pair and two 8-bit ALU helper registers: ACT and TMP.
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5v power supply from 18650s - high power, low profile I am trying to use 2 18650 batteries to power 288 APA102 LEDs. The LEDs require 5V, and have a max current draw of 60mA/LED for a total of 17.28A. I will also have a microcontroller attached (currently arduino micro), powering some other low-draw devices. So let's say I'm aiming to support up to 18A. Most of the time though, the current draw will be much lower. My understanding is that I can use a buck or boost converter (depending how the batteries are wired), and outboard bypass transistor(s) to shunt additional current around the converter (with proper heat-sinking). I am curious though, are there considerations for the buck vs boost configurations? Is my stated understanding even correct? And lastly, will I need to protect my arduino somehow if I'm drastically changing the power coming out of the batteries (e.g. Strobing the LEDs from fully on to fully off)? I'm open to alternative battery suggestions as well, but I need everything to fit in a cylinder of ~20mm, and energy density is very relevant for this project. <Q> The LEDs require 5V, for a total of 17.28A. OK, let's round that to 100W. <S> One 3000mAh 18650 rated for 20A continuous current, as used in e-cigarettes, contains 11 <S> Wh. <S> With two, that's 22 <S> Wh. <S> With 100W power draw you cannot expect more than 15 minutes battery life. <S> Factoring in converter losses, this will most likely be 10 minutes. <S> I assume you're okay with that, since you say "I need everything to fit in a cylinder of ~20mm". <S> I assume you did your homework, and if you needed more than 10 minutes battery life, you would have specified an adequate size and weight for batteries. <S> It is your problem. <S> My understanding is that I can use a buck or boost converter (depending how the batteries are wired) <S> For this level of current, you do not strictly need a multiphase buck converter, but it would make the job easier, use smaller inductors, reduce I2R losses... <S> Ex-National Semiconductors has several chips which will do what you want. <S> and outboard bypass transistor(s) to shunt additional current around the converter (with proper heat-sinking). <S> Uhhh? <S> What? <S> I am curious though, are there considerations for the buck vs boost configurations? <S> At high output currents, buck converters tend to have higher efficiencies. <S> They are also easier on the batteries, since drawing power from a higher voltage source will require less current in the batteries. <A> This isn't a direct answer to your question, but pointing out something you should look at. <S> The LEDs require 5V <S> No, they don't. <S> LEDs that produce visible wavelengths require roughly in the range of 1.6 to 2.5 V. <S> Some "white" LEDs really emit UV and use phosphors to convert that to visible colors, and can require up to 3.5 V. <S> Maybe you have LED modules that require 5 V, but the LEDs certainly don't natively. <S> Since power requirement is a serious problem in this project, you should look carefully at how to require less power in the first place. <S> If possible, drive them with constant current directly from a switcher. <S> Regulating voltage instead of current forces you to put resistor in series with each LED. <S> That drops voltage and therefore wastes power. <S> Even if you do try to regulate voltage, use a separate rail for each type of LED. <S> Each rail can then be tuned to be only a few 100 mV above the LED nominal voltage so that the series resistor doesn't waste much power. <S> With a fixed 5 V in, there is either another conversion going on, or a considerable fraction of the power is wasted in series resistors. <A> Have you built this project? <S> I was thinking about the drivers used in high power flashlights. <S> There are 22mm, round cut pcbs that crank off a lot of power to run CREE, COB and SMD leds. <S> Another idea I'd consider is a less compact setup using a Lipo battery like those used in RC helicopters, cars and planes. <S> I use them and you can actually start a car with one. <S> 3S lipos are 11.1 volts and are rated from 10C all the way to above 100C. <S> I'm thinking if you were to combine one with a buck converter, you could make it work. <S> Also I think you're best off to provide more power than required and bring it down to your needs than up. <S> This allows for a bank of power that wont drain out fast. <S> Like 5v 250ma. <S> So I am sure you could achieve your goal if you haven't already. <S> Also they sell battery alarms for lipos when not in protected circuits. <S> I have some and they are insanely loud but a nice thing is they are adjustable in volume and as well as can be set to alert at whatever voltage you prefer as low as 3v per cell. <S> If any cell on a lipo reaches the set voltage alert level the alarm will go off. <S> The safe level for a lipo is 3.2 min per cell. <S> That's why LiIon batteries sometimes explode and get on the 6-o-clock news. <S> here's a Google link of these battery alarms. <S> If you wanted you could easily wire each terminal on the alarm to the right spot on a series of 18650's. <S> Just start with the positive and work your way down the line. <S> https://www.google.ca/search?q=lipo+alarm&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjMvoj6nezZAhVIR6wKHZ0uB70Q_AUICigB&biw=1920&bih=888
| We use BEC's (battery eliminator circuits which are basically just buck converters) to reduce the insanely aggressive power coming out of these batteries to provide power for much more delicate systems like cameras, flight controllers and telemetry units which run on much lower softer power. Anything under 3v will damage the battery and make it puff up and if the packaging tears, the lithium inside will react with the oxygen in the air and combust. Start by defining what kind of light at what intensity you really need, then work backwards to find the sets of LEDs to emit that directly.
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Two grounds when using transistor switching Background I'm designing a small circuit in which an MCU and all its peripherals should be shut down unless a certain condition is met, which is represented by a switch in this simplified schematic. The MCU requires a 5V power source from which it supplies 3.3 Volts for the peripherals. In its simplest form I came up with the two solutions below. The measurements reflect reality -- I probed both circuits on a breadboard. Question Which alternative is more suitable for this: Observations: In case 'A' the MCU receives 4.3 V, which should be sufficient for the internal regulator to operate and output a proper stabilised output, however, two separate grounds will result. Digital ground will be comparatively higher than 'real' or analogue ground. Peripherals will have to stick to using the digital ground, everything else (i.e. not connected to the MCU) may use the real one. Observations: I case 'B' digital and analogue grounds are the same but the MCU only receives 3.7 V instead of 5 V. It still operates and establishes the 3.3 V output, but I think this it is really pushing it and is bound to fail intermittently. Is either of these a viable solution, or should I do something along the lines of this (which will increase the part count): <Q> If you have analog signals crossing the boundary, and that is your "analog ground" in the sense that it is the reference for all of your analog signals, then I think the high side switch is less likely to give you problems. <S> The low side switch will introduce a potentially nasty voltage between the two references due to its finite on-resistance and need to handle the switching current from the digital logic. <S> The "digital" ground could also be the reference for internal ADCs, for example. <S> It is a good idea to decouple the MCU side to provide a low impedance source for the switching current. <A> In this situation, I would use an N-channel MOSFET on the ground. <S> Alternatively, as previously mentioned use a P-channel MOSFET on VCC. <S> Either way works, as long as you are keen on ensuring there are no sneak paths. <S> Why? <S> Using an NPN will result in voltage drops Vbe and Vce. <S> Using an N-channel results in Rds(ON). <S> If Rds(ON) is low (e.g. milliohms) and your circuit only draws milliamps, your voltage drop would be very small and wouldn't effect your circuit. <A> First I'd use a mosfet rather than a transistor so your voltage losses will be much less if you are not pulling huge currents. <S> Whether to switch the high side or low side is really dependent on what's happening in the digital circuitry block on the right and anything else in the system. <S> You can use either method. <S> HIGH side is usually preferred though to maintain the integrity of the grounding system. <S> However, great care has to be taken to ensure that no input or output on any gate of the micro or other devices can exceed the maximum and minimum voltage specifications when their power is turned off. <S> Your circuit seems to indicate there is an analog side to this design. <S> As such, I have to assume there are signals leading from the analog side to the digital side. <S> Appropriate circuitry needs to be added to ensure those signals do not fry your digital circuits. <S> Similarly, additional elements me be required to ensure that the digital circuitry does not affect the analog circuits when powered down.
| As others have commented, a P-channel MOSFET is a better choice for a high side switch.
|
Are DC-DC converters designed with fixed output or fixed output/input ratio? When I was learning DC-DC converters, I think they are a "DC version" of transformer (which is for AC-AC). So when I was designing a 100-30V flyback converter, my suggestion is: if a user inputs a DC 100V source, he gets a 30V output; if he inputs a DC 50V source, he gets a 15V. Just like how transformers act with AC, which have a fixed output/input ratio. And someone then told me: no. If he inputs a 100V he gets 30V output; if he inputs a 50V, he should still get a 30V; even if he inputs a 15V, he still have to get a 30V. (We were talking about flyback converter. I'm not sure what about buck/boost converter. ) This is like a DC power supply (or it is), with which you want a fixed output. So I'm curious and a little confused now about the actual situation. Do manufacturers always make DC-DC converters with fixed output voltage? Or with fixed output/input ratio? Or both are possible? In this case, is one of them are more often to use then the other? The "someone" was like "How dumb are you! How can you even don't know this! " while telling me the story. I think it's more possible what he said makes more sense, because I'm really stupid. But what confused me is: if the actual situation is we never design DC-DC converters, like buck or boost or flyback or any other DC-DC converters, with a fixed output/input ratio, then how do I do if I want a "DC transformer" which give me output = f(input) = input * 3? Or, I don't know, maybe we will never want a DC transformer like this? <Q> The search term for what you are describing is "proportional dc-dc converter". <S> For example, you can buy high-voltage DC-DC converters, where you put in 5-12V and get 1000 times the voltage out (5K to 12K volts). <A> Yes, there are converters that have a fixed ratio of input to output such as 48v down to 12 or 4:1. <S> They are normally referred to as "unregulated". <S> One application is in telecom servers where they are used to provide galvanic isolation and convert to a lower voltage for distribution in the rack. <S> A switching regulator close to the point of use would then convert the voltage to one that the load needs - for example 1v for a CPU. <S> This is an example Synqor unregulated converter <S> You could change a normal converter design act as a ratio converter by using the input voltage as a reference rather than a zener diode or equivalent. <A> Most DC-Dc converters are used as regulators. <S> In places which require a fixed voltage supply irrespective of the changes at the source. <S> So having a transformer like action is not really the agenda. <S> Also, the inductors and the capacitors in the converter are designed for a particular voltage and current ratings. <S> So you will not be able to have a converter work satisfactorily in any type of input condition.
| But for most DC-DC converter uses, the purpose is to get a fixed output voltage even if the input varies.
|
Is it healthier for a capacitor to be discharged to itself or to "ground"? A practical-use question from a non-EE who enjoys simple repairs of home electronics. Am curious about whether "where" to discharge makes a difference to the continued health of an electrolytic capacitor - and for caps still in-circuit on a PCB also to the health of its neighbors. A brief web-search yielded: some discharge 1) directly across the terminals, 2) from cap to "ground" (metal case, or ground pin of the power plug), or rarely 3) from cap to "ground" (the earth). (And most do use a resistor to slow things down.) <Q> It is true that in most cases one side of the capacitor will be grounded and the other attached to some rail, <S> HOWEVER this is NOT TRUE in all designs . <S> There is no guarantee that grounding either pin of the capacitor to frame ground will discharge the capacitor. <S> Further, by doing so you may actually be applying power to some circuit that does not expect it and can potentially damage it. <S> It is also prudent to re-test after discharge with a suitable delay to make sure that some circuit on the board is not bleeding, or pumping, more charge into the capacitor. <A> To discharge a capacitor there must be a circuit, a loop, that passes through both terminals of the capacitor. <S> Whether there is also a connection to some "ground" makes no difference to the process. <S> What you might be seeing is that in a specific circuit, one terminal of the capacitor is already connected to ground (or any other bus / voltage reference), meaning that if you then connect the other terminal to the same bus you've created a circuit to discharge the capacitor. <S> Depending on the exact physical wiring there might be measurably more resistance and inductance in that connection than "shorting the terminals", but if you're going to include an actual series resistor in the discharge path (a good practice) then that deliberate resistance will dominate and the rest of the wiring can be whatever. <S> Fun fact <S> : I wrote the above thinking you were talking about designing a circuit with a discharge feature, not discharging manually for service — but everything I said applies to both cases just the same! <A> Furthermore, if you accidentally connect the ground terminal of the capacitor to the metal case, the capacitor will not discharge. <S> Because of this, it is a good idea to discharge all capacitors by connecting the terminals together (either with a conductive material or a resistor) until the capacitor is discharged. <S> (You can check with a multimeter.) <S> In regard to the "health" of the capacitor, high discharge currents can damage it or reduce its lifespan, so it is favorable to discharge through a resistor.
| The only GUARANTEED safe answer is to discharge the capacitor, through a suitable resistor, across the capacitor terminals. In some cases, one terminal is connected to "ground" (the metal case of the device) but it is not always easy to determine which terminal it is.
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Can I Run An Appliance From a 24v Battery? First, a little backstory. My wife is planning to operate a booth at some craft fairs and we require power for a piece of equipment (Silhouette Cameo, a computer controlled vinyl cutting machine). We've done this once before with success at a show that had power provided. However, other shows do not provide power. The Cameo’s AC to DC pack states that it provides 24v power at 1.25 amps. My initial thinking is that you could run two 12v batteries in series to provide the power, then at the end of the day, let them recharge overnight and have them ready for the next day. I have a few questions on the practicality and feasibility of doing this. First off, what kind of battery can I use? I’ve read a few things on the web stating that certain batteries are not good for this application (ie: car batteries). Is this an applicable battery? https://www.amazon.com/ExpertPower-EXP1270-Rechargeable-Lead-Battery/dp/B003S1RQ2S/ref=pd_lpo_23_bs_tr_t_2?_encoding=UTF8&psc=1&refRID=NV9MYPMB3E5Q6JT65SZS Secondly, as batteries don’t provide a straight up 12v as they describe, and the voltage differs based on the amount of charge they are holding, can this damage the appliance? If so, is this what a regulator can help with? Pardon if anything I’m saying comes off as incredibly stupid as I only understand electronics at a very basic level. My hope is that connecting two 12v batteries directly to the Cameo’s DC input will work without damaging it, and if not, a little regulator chip ( like this: http://www.jameco.com/z/7824T-Major-Brands-Standard-Regulator-24-Volt-2-2-Amp-3-Pin-3-Tab-TO-220_51414.html ) would be enough to supply a steady current to the machine with no risk of damage. Thanks in advance for any help. <Q> That should be close enough to 24 V to run the device. <S> It is quite unlikely that ±1 V will cause any trouble, although there is no guarantee of that. <S> Since the power supply is rated for 1.25 A, you know the device won't draw more than that. <S> Even assuming the full current draw for 8 hours, that comes out to 10 <S> Ah. <S> You don't want to run car batteries down very far, but even small car batteries can do well more than that. <S> 30 <S> Ah is a "small" car battery, and two of those should be just fine with 10 <S> Ah drained from them. <S> Overnight should also be plenty of time to charge them back to full. <A> First, instead of regular car batteries to get that ~24VDC, use deep discharge batteries. <S> Second, 12V lead acid batteries are fully charged at between 14.4 and 14.8 VDC gassing point. <S> Third, yes, two 12VDC batteries in series can properly supply this instance with plenty of power to get you through the day. <S> You only have to supply the current requirements. <S> Fourth, your goal would still be better realized with one 12VDC battery and a buck/boost supply or three 12VDC batteries regulated down to 24.Fifth option, use 18650 batteries stacked in series to make between 23 to around 26 VDC. <S> Stack parallel banks to obtain the current you need for the day or use multiple sets.@user253751 <S> , how do you get <S> +2.4 VDC from two 12 volt series-connected lead acid batteries? <S> 2.24VDC each from 12 cells results in 26.8 VDC. <S> A 12 volt lead acid battery produces barely any voltage above 12.2 after having been removed from a charger for 15 minutes. <A> It's a moot point anyway because the voltage going in on the power connector is regulated in all instances unless the regulator itself can't handle the added watt or two of dissipation or there are capacitors directly across the input rated for 25 volts, either scenario of which is highly unlikely. <S> https://batteryuniversity.com/learn/article/charging_the_lead_acid_battery
| Just get two car batteries and put them in series.
|
LDR working differently than usual I observed something different in LDR (Light Dependent Resistor) today. I made a "magic" LED circuit such that when lights are off, the LED glows, else when the lights are on, LDR creates resistance and the LED stops glowing. Later when I connected just the LDR and LED (in series) to the battery, LED glowed very dim when there was light in the room, and stops glowing when no light reaches the LDR. Why does this happen? I know just the basics of electronics. I request you to provide an answer I can understand. <Q> Typical LDRs (light-dependent resistors) increase their resistance when dark. <S> Therefore connecting one in series with a LED should give you exactly what you observed. <S> When dark, there is too much resistance to light the LED noticeably. <S> I go into more detail, include a complete circuit, at https://electronics.stackexchange.com/a/53681/4512 . <A> LDR creates resistance and the LED stops glowing. <S> You didn't share your circuit or your LDR part number and datasheet, but you are probably wrong about this. <S> LDR's generally decrease in resistance when light is present. <S> Using some kind of transistor circuit, this can easily made to decrease the current flowing to an LED, giving your "magic LED" circuit. <S> Later when I connected just the LDR and LED (in series) to the battery, LED glowed very dim when there was light in the room, and stops glowing when no light reaches the LDR. <S> Why does this happen? <S> Because the LDR resistance increases when light is removed. <A> As Mr. Lathrop stated, most LDRs have a quite high (a few hundreds of kOhms) dark resistance. <S> When it comes to light, this value can drop to a few kOhms depending on the light intensity (e.g. you can find a value in datasheet as \$R_{100}\$ meaning resistance under 100lux). <S> Even this "low" value can be high to drive a LED directly from a 9V battery. <S> Suppose you have a 5mm red LED and an LDR having \$R_{100}=5k\$. <S> Under 100 lux of light intensity, LED current will be (9V-2V)/5k=1.4mA. <S> This amount of current will cause the LED to glow dimmed.
| It takes a active circuit to use the resistance of a LDR to turn something on when it gets dark.
|
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