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12V solenoid drawing more current than it needs I'm fairly new to practical electronics so I apologize if any of this is obvious. I am currently trying to use a microcontroller to control the energizing of several 12V solenoid valves using a PC power supply to provide the 12V for the valves and 5V to power the microcontroller. The issue I'm getting is that the valves appear to be drawing a lot more current than they require to operate. I've measured the voltage and current going through the valves and they read as 4V and 12A. They are rated at 12V and 0.5A by the manufacturer. I can't explain what the problem is or where it resides. I have verified that the proper 5V is being output from the microcontroller and the NPN BJT used for the switching has a 300-ohm resistor on its base. Any help would be appreciated, this project has definitely lagged as a result of this issue. <Q> I would guess that either you have a defective solenoid with a partially-shorted activation coil, or you've made a major wiring error somewhere, with suspicion pointing toward the latter. <S> The reason I'd guess you've done something very wrong is that 12 amp current. <S> Assuming your base drive is 5 volts, a 300 ohm resistor will give you a base current in the ballpark of 10 mA. <S> A reasonable (although slightly optimistic) gain for an NPN transistor <S> (unless you're using a Darlington) is about 100, so I'd expect a current of 1 amp. <S> EDIT - <S> And, we have a winner. <S> Two, actually. <S> Well, three. <S> First, you are completely misusing your Fluke. <S> It is incapable of measuring current directly. <S> This suggests (since your usage makes no sense) that the "12" you are reading is actually the 12 volt supply. <S> And I have no idea what the 4 volts is. <S> Consequently, there is no way to tell what the circuit is actually doing. <S> Certainly if you're trying to measure current by putting your Fluke in series with the solenoid, that will explain why it doesn't work. <S> Second, as has been pointed out, by Bruce Abbott among others, a 2N5551 is not suited to your needs. <S> Third, assuming you do get a decent transistor, it can't be a single transistor, or at least not a BJT. <S> Assuming your Arduino can supply 10 mA of current, you need to be aware that for switching purposes (such as your application) you should assume a gain of 10 to ensure good switching. <S> This puts an upper limit of about 100 mA on your solenoid drive. <S> You might conceivably try for 200 mA, but not much more. <S> The solution? <S> Use either a MOSFET (n-type in this case) or a Darlington NPN such as a TIP140 or TIP141. <S> END EDIT <A> If you're using a 12V supply and you're using the BJT common emitter, (with the solenoid between the supply+ and the collector) then there should be close to 12 volts across the solenoid when the transistor's turned ON. <S> Since there's only 4 volts across it <S> and there's 12 amperes being taken from the supply when the BJT is ON <S> , it sounds to me like you're either trying to shunt control the solenoid by wiring it and the BJT across the supply in parallel, or you've made that wiring error. <S> How hot does the BJT get when it's turned ON? <A> The 2N5551 is a high voltage low current transistor, not suitable for switching 0.5A. <S> Depending on the manufacturer it may be rated as high as 600mA absolute maximum , but above 40mA its current gain drops off rapidly (to less than 10 at 500mA). <S> Assuming the solenoid is a basic single coil type, to draw 0.5A at 12V <S> it should have a resistance of 24 Ohms (12V/0.5A). <S> You measured 4V across it, which suggests an actual current draw of ~167mA (4V/24Ω). <S> The transistor is limiting the current because it doesn't have enough gain to turn on fully. <S> The solenoid cannot have been drawing 12A because the transistor won't let it. <S> Perhaps the meter was set up wrong and you were measuring voltage instead of current?
| The fact that you're seeing 12 amps indicates that something, somewhere, is very wrong, and on the basis of what you've told us I can't be any more specific.
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Removing spikes from multiple onboard power supplies I have a board that is based around an STM32F303 and am using the onboard ADC/DAC, fed by a 3V power supply derived from a USB 5V input. The 5V also feeds a module, RD0515, which creates +/- 15V for opamps on the board. My problem is that the RD0515 also creates quite substantial spikes of about 2uS width and repetition rate of 20uS. These are finding their way through to the analog supply pins on the STM32 at around 100mV amplitude and screwing up measurements. The board is 4 layer (GND and 3V internal planes) and has plenty of capacitor decoupling. Is there a better way of getting the +/- 15V that will not cause such problems? <Q> The RD0515 is a switching type converter. <S> As such it is not surprising that it produces spikes in an unoptimized design. <S> There are several things you can do to help with this situation. <S> At the input side to the converter make sure to apply a very low impedance high value capacitor to supply the input current surges as the internal circuit switches on and off. <S> It may also be necessary to place some inductor or power type ferrite bead in series with the 5V supply and the input of the converter. <S> The added capacitor should be as close to the converter as possible and would be on the converter side of any series inductance added. <S> On the outputs you may need to add a low pass filter to clean out spikes before the +/- <S> 15V goes to other circuitry. <S> This may need to be a multi stage filter with several series inductors/ferrite power beads with intermediate capacitors to GND. <S> An alternate for the outputs would be to run your analogue circuits on +/- <S> 12V and place low current linear regulators (7812/7912 types) between the converter and the analogue circuits. <S> I can add the comment that the RD0515 is part of RECOM's "econoline". <S> The parts are small, convenient and have attractive prices. <S> But for this you do give up some on how much filtering that the manufacturer puts inside the converter. <S> The econoline is most certainly designed to be adequate for many applications that do not care about some switching spikes. <S> On the other hand if you select this type part then you take on the burden of cleaning things up to be suitable for your specific design. <S> As such the output voltages can change with load current variation and input voltage variations. <S> If your analogue circuit designs have a dependency on the accuracy of the DC voltages and have low tolerance to power supply variations you may very well want to look at the option (3) above. <S> Lastly be aware that this part should have a minimum load of 10% of its rated capacity in order to meet the data sheet performance specifications. <A> Separate the supply lines using inductors. <S> Inductors block AC, their impedance increases with frequency. <S> The technique is sometimes used to create, to separate analogue and digital grounds to ensure that high frequency noise (voltage spikes) doesn't propagate on to the AC ground. <A> Summarizing your needs: <S> +5V <S> In <S> +15V Out <S> VERY CLEAN <S> -15V <S> Out VERY CLEAN <S> If I was starting from scratch and NOT forced to use the RD0515, then I would do the following concept for both +15V and -15V power rails (the following only shows the +15V): 1) <S> +5V IN to 2) <S> Boost/Inverter Switching Voltage Regulator (or Module) with +16V to +18V OUT (depends on "drop out" voltage of LDO in step#4). <S> 18V is fairly common, but if you have a variable output then tweak the voltage a little closer to the LDO <S> drop out. <S> to 3) power filter (possibly a pi-filter and ferrite bead) to 4) <S> low drop out (LDO) <S> linear voltage regulator. <S> to 5 <S> ) +15V OUT to 6) <S> capacitor(s) <S> and analog rail
| The noise is being coupled along the supply lines to the power line that feeds the device, so an inductor in that line will block the AC. Also be aware that the RD0515 part does not have regulated outputs.
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PCB, Plated hole with annular ring on only one side I need my pads for connectors to be plated so that the pins from the connectors can be soldered. But the bottom side cannot have copper pad/annular ring. I have made the pads as shown in the figure below. Bottom pad size is same as the hole size and the hole is Plated. Is this the proper way? Will there be risk that the holes will be unplated my the manufacturer? <Q> Well as comments have said only your fab house can tell you for sure. <S> You should try to keep a close relationship with them as they're basically your partner in all your designs :) <S> Now this question just happened to come up for me as well a few months ago on a really dense board. <S> I use Via Systems when in the US, a very large and capable manufacturer. <S> I asked them a similar question <S> , could I make vias that only had an annular ring on one layer. <S> Their reply was <S> no don't do that <S> , we can't plate the via properly that way. <S> Your shop may tell you it's ok <S> but I would expect they won't like it. <S> There's always surface mount connectors. <S> Or and you'd have to ask about this to <S> but I imagine you could just not plate those vias and only route to the top layer where the pad is. <S> There really wouldn't be much holding the connectors on the board at that point <S> and I'd look at mechanically securing them to the board. <S> Maybe epoxy or screws or something like that. <A> Will there be risk that the holes will be unplated my the manufacturer? <S> Yes. <S> Most PCB plating operations occur after drilling and before etching. <S> A board with copper on both sides drilled for all the holes requiring plating. <S> Then a conductive solution is sprayed (or dipped) <S> such that all the holes now have a conductive surface - this is the activator. <S> Using electrolytic plating, copper is plated - this thickens the existing copper on the sides, and puts copper over the activator inside the holes. <S> The holes are now plated. <S> The PCB then gets cleaned and an etching mask is applied that will protect the copper that is to be left on the board. <S> However, there's always a degree of mis-registration in this process, and it's possible to end up with a damaged plating in the hole if the mask isn't placed precisely. <S> This is why most PCB fabricators specify a "minimum annular ring". <S> You'll need to have a discussion with your PCB fabricator about this. <S> If you can live with a small annular ring, then the minimum may be sufficient. <S> If you cannot have any annular ring, the PCB fabricator will likely be able to make your board with special instructions and an additional cost. <S> I'd start by carefully examining exactly what you need when you say <S> , "bottom side cannot have copper pad/annular ring" because there's always a little error in everything, and I have a hard time believing you're using a standard component <S> and it doesn't allow an annular ring. <S> There are probably other solutions as well, but without a better description of the part and how it's meant to be attached, it'll be hard to resolve this for you. <A> I think you have to provide a convincing reason why you think you cannot have pads on the bottom of the board. <S> If you have pads on the top side among the connector pins there is certainly no reason that it is not possible to have pads on the bottom side too. <S> Through hole connectors are generally soldered from the bottom side and would certainly be done that way at a production fabrication shop. <S> I suspect that wave soldering would also tend to be unreliable without pads.
| Hand soldering from the bottom without pads would be tricky at best and likely unreliable. In theory the mask could be applied just at the hole, and thus leave you with the finish you want - no annular ring.
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How to find potential difference at A As shown below the schematic diagram is given.Point B is grounded. I did take the potential at A is 2V since it is on the same node where the 2V batteries are located , am I correct? . I want to verify the accuracy of my answer !. What is your opinion? <Q> Another approach is to take each one of the branches with batteries and substitute them with their Norton equivalent, i.e. a \$5\Omega\$ resistor in parallel with a \$ \dfrac{2V}{5\Omega}=0.4A\$ current source. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Then you get three paralleled \$5\Omega\$ resistors driven by two paralleled \$0.4A\$ current sources. <S> This means a total \$0.8A\$ flowing into an equivalent \$\dfrac{5\Omega}{3}\$ resistor. <S> simulate this circuit <S> By Ohms'law you get: \$ V_{AB} = <S> \dfrac{5\Omega}{3} <S> \times 0.8A <S> = \dfrac{4}{3}V \approx <S> 1.33V \$ <S> EDIT <S> (Since it seems you have not grasped the concept of circuit ground). <S> A voltage is an electrical quantity always measured between two points in a circuit. <S> Since it is cumbersome to draw schematics full of arrows (or +- symbols) <S> specifying where are all the voltages you need, it is customary to take a common reference point (node, to be precise) for all voltages: <S> that point (that node) is called circuit ground . <S> Then you specify only node voltages , i.e. voltages between a node and the ground node. <S> That is, for each node k , \$V_k = V_{kG} \$ where G indicates ground. <S> This is a definition! <S> So, by marking a node (B in your case) with the ground symbol you are saying that any other node voltage is measured/specified with respect to that node (B). <S> This allows specifying node voltages with only one subscript. <S> Moreover this doesn't lead to a loss of generality, because it can be easily shown that a voltage \$V_{XY}\$ between any two points X and Y in the circuit can be written as a difference between two node voltages, i.e. \$V_{XY} = V_X - V_Y\$ (if either X or Y is ground, the corresponding node voltage is of course 0). <A> You can solve this by stepwise reduction. <S> 1) <S> The left side of the two batteries is connected, hence the right sidse of the batteries have the same potentional. <S> Connect those two point with a wire. <S> 2) <S> Now you have two same-voltage batteries in parallel. <S> Remove one. <S> 3) <S> Now you have two 5 Ohm resistors in parallel. <S> Replace them by a single suitable resistor. <S> 4) <S> Now you have a simple circuit with one battery and two resistors in series. <S> Calculate the current and use it to determine the voltage at the junction of the two resistors. <S> Optional step: <S> The fact that the batteries do not have one side grounded might make you a little nervous. <S> But two elements in series can always be swapped without influencing the rest of the circuit, so you could swap both battteries with their series resistors. <S> (But do note that this will influence the voltage at the junction of the battery and its series resistor!) <A> Your outright guess that node A is 2V with respect to node B <S> is not correct. <S> There are resistors in series with the two batteries that must be taken into account. <A> Ground is just a reference point. <S> Just move the ground to point A and calculate the voltage at B. <S> The relationship between A and B will always be the same, regardless of where you put ground.
| Or alternatively, regard the two resistors as a voltage divider and use that to calculate the voltage at the junction.
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What is equivalent Resistor different Resistors put them in series or in parallel if they have a different wattage and different resistance? What is equivalent Resistor of two this different Resistor put them in series or in parallel if they have a different wattage and different resistance? In practice somtimes we have a resistor with \$\frac{1}{4}\$watt , \$\frac{1}{2}\$watt, \$1\$watt or etc... But let`s say that wattage of resistor is \$\frac{3}{4}\$watt or \$\frac{3}{5}\$watt, or etc... So what is the equivalent Resistor of two this "\$w_1\$ watt \$r_1\$ resistance" and "\$w_2\$ watt \$r_2\$ resistance" resistor put them in series or in parallel? We know that if they have the same wattage \$w\$ watt and \$r_1\$, \$r_2\$ resistance then the equivalent Resistor might be: 1) in Series \$w\$ watt \$r_1\$ + \$r_2\$ resistance 2) in parallel \$2w\$ watt \$\frac{1}{r_1}\$ + \$\frac{1}{r_2}\$ resistance But what about my question? <Q> Forget the wattage of the resistors to begin with, that's causing confusion. <S> Do the analysis, combine resistors together first. <S> You can combine resistors in series or parallel. <S> When in series the resistors add together. <S> When in parallel, the calculation is a little more complex. <S> Here are the two formulae you need. <S> When you combine resistance values together, the power doesn't come into it. <S> In practise, it's your job as a designer to ensure that you don't execeed the power rating for the resistor <S> so you have to calculate the power dissipation of the resistor and ensure it doesn't exceed that stated for the resistor <S> , so you need to work out the current through the resistor and what voltage is dropped across the resistor. <S> You need to know Ohms Law and Power formula <S> (R=V/I , W =VI), re-arrange, combine if necessary and apply. <S> If you have a problem with actual power exceeding the maximum power rating for the resistor then there are things you can do to get around that: you can use multiple resistors in series, the current through each is the same but the voltage is shared across the resistors, thereby reducing the actual power dissipated by each resistor. <S> Alternatively putting resistors in parallel will reduce the current but keep the voltage the same. <S> But you have to ensure the the combined resistance of the multiple resistors (whether in series or parallel) <S> achieves the value you want to achieve. <A> Two 1 watt resistors in series can be regarded as a 2 watt resistor if the individual resistance values are the same. <S> If the resistance values are different then this is not true. <S> If one resistance is much bigger than the other then the power rating of both combined in series tends towards the power rating of the higher resistance resistor because it will always have a bigger volt drop across it. <S> For resistors in parallel, the power rating of the combined parallel pair tends towards the one with the least resistance because this resistor always takes most of the current. <A> First and foremost, resistors have resistance. <S> Given a Resistor \$R_1\$ and another one \$R_2\$, the total resistance of both is $$R_{series} = <S> R_1 + R_2$$$$R_{parallel} = <S> R_1 || <S> R_2 = <S> \frac{R_1 <S> R_2}{R_1 + R_2}$$ <S> You can show that \$R_{series} <S> > <S> R_{parallel}\$:$$R_1 <S> + R_2 <S> > \frac{R_1 R_2}{R_1 + <S> R_2}$$$$(R_1 <S> + R_2)² <S> > <S> R_1 R_2$$$$R_1² + <S> 2 R_1 <S> R_2 + <S> R_2² > <S> R_1 R_2$$$$R_1² + <S> R_1 R_2 <S> + <S> R_2² > 0$$ <S> The amount of power that they consume depends on the voltage that you apply or the current that runs through them. <S> $$R = <S> \frac{U}{I}$$ <S> $$P = <S> U <S> * I = R <S> * I² = \frac{U²}{R}$$ <A> In series, the current is the same through each resistor so you can use P = \$I^2R\$ for each resistor Ri, and of course the total resistance is the sum of the Ri. <S> Power for each resistor must not exceed the rating for each resistor, but it's perfectly legitimate to mix wattages. <S> For example, a 1/4-W 10 ohm resistor can handle 158mA but at that current, a 100 ohm resistor would dissipate 2.5W. <S> In parallel, the voltage is the same across each resistor so you can use P = <S> \$E^2/R\$, and ensure that each resistor power rating is not exceeded. <S> A 100 ohm 1/4W resistor can handle 5V, but a 10 ohm resistor would dissipate 2.5W. Of course the parallel resistance is found from the reciprocal of the sum of the reciprocals of each value. <S> Rx = <S> \$\frac{1}{1 <S> /R1+..+1/Ri}\$
| Only resistors of equal resistance can have their individual wattages added to yield the combined wattage.
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Altium: Surface Mount Test Pad Passing Through PCB I'm working on a PCB in Altium that will sit inside an assembly above another board. The board on the bottom has a test pad, but once everything is assembled it will be inaccessible. Therefore, I am attempting to add two pads to the top PCB (one on the bottom layer, one on the top) that are connected through vias around the edge (No, I cannot simply use a through-hole pad). The issue is this: In order to connect the vias to the top and bottom pads, they must all be connected to the same net. However, there is not a designated net for these pads, as they are simply fed through the PCB and are not used by it. It is a PCB-only feature and is not part of the circuit, so I don't want to add it to the schematic. How might I go about connecting these top and bottom layer pads through vias? Here's an image showing the top side of the PCB: The bottom layer pad is directly beneath the one shown in the image. <Q> You can manually add a net using Design->Netlist->Edit nets, then select each pad and assign that net to it. <S> But I don't see why you don't want to do this in the schematic. <A> I see, it's not a hole but more two pads on top of each other. <S> Two options: Option 1: <S> Design->Netlist->Edit Nets (add a dummy net - e.g. DUM1 and DUM2) <S> ; You can then use this net in the PCB to assign it to your pads and tracks and vias. <S> Option 2: <A> The way I would do this would be to create a double-sided testpoint library component, and add it to the schematic for the board. <S> By doing that you can assign both pads to be the same net in the schematic. <S> Alternately, you could move the test point to be accessible, if that's an option. <S> Lastly, if you select the pad and look at PCB Inspector, you can change the object specific property "Net" to the desired net.
| Draw two dummy objects in the schematic, connect them and assign pads as footprints; then transfer them to the PCB
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How to wire screw type banana jacks I recently purchased some of these: Digi-Key Banana Jack . Unfortunately, I'm new and there isn't the supporting documentation I need to be able to determine how they're to be used. Unlike other banana jacks, there is not a hole to feed the wire through and screw down on. I understand the large nut is for holding it against the enclosure, but I don't know what the intention is with the threaded bottom section and the two nuts. I'm guessing that the wire is simply looped around and a nut is tightened down against it, but this doesn't seem very durable. Why the double nuts, what is supposed to be done with them? Edit To clarify, I understand that they are the female piece of a banana connector. I have male banana connectors that I am using these with. My lack of understanding comes from how to wire this female piece to the... for lack of a better word "connection wire". I.e. how to wire this female portion into a circuit so that the circuit is accessible through it. <Q> As the data says it mates with a "standard banana". <S> Standard is 4mm. <S> Something like this : <S> Or this: <S> Not this (but perhaps you can see the resemblance in the springy bits): <S> The part you have (female) is just a hollow tube with no springy parts, so it depends on the male to supply the compliance. <S> Some plugs have a cross-drilled hole so another banana plug can be inserted sideways. <S> In the case of the PCB, a toothed lockwasher is a good idea. <S> In the case of the terminal, a toothed "star" terminal can be used (photo from here ). <S> You drill a hole as the datasheet shows and attach the jack to a panel with the large nut. <S> The datasheet indicates that the jack also accepts shrouded plugs (insulated on the outside of the metal and often with plastic at the end so that a human finger cannot contact the metal). <S> This is commonly used on the meter end of test leads. <A> The threaded post on the back is intended for a ring lug. <S> The diameter of the ring can be 4mm or US screw #8. <S> To make a better electrical contact, add a star washer between the nut and the rung lug. <S> Source: Datasheet for the banana jack. <S> US screw sizes , <S> Metric screw sizes p.s. <S> Pomona is a reputable supplier. <S> I would expect a better drawing from them. <S> The datasheet claims " IEC " and " 4mm ". <S> Yet, the dimensions are only in inches, including dimensions of metric threads. <A> Run down to your local hardware store and get matching ring crimp terminals and non-slip washers. <S> Crimp the terminals onto wires, then slip them in between the nuts on the bottom with a non-slip. <S> This will provide a fairly secure connection, dependent on how well you crimped the terminal of course.
| The screw and nuts are to fasten onto a ring terminal or onto a through-hole PCB.
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Using GPIO in Altera I'm trying to test the GPIO functionality of Altera (DE1, Cyclone II) with this simple program. If the GPIO_0[0] gets a high (1) signal, LEDG[0] will light up. If it receives a low (0) signal, LEDG[0] will turn off. For the input signal, I am asserting 5 and 0 V DC . Here's my code: module gpio_test (CLOCK_50, GPIO_0, LEDG);input CLOCK_50;input [35:0] GPIO_0;output [7:0] LEDG;reg [7:0] LED;assign LEDG = LED;always @ (posedge CLOCK_50)if (GPIO_0[0] == 1) // if GPIO received a high signal LED <= 1; // turn LED onelse if (GPIO_0[0] == 0) // if GPIO received a low signal LED <= 0; // turn LED offendmodule The problem is, it seems that GPIO_0[0] is always receiving a high signal (even though I assert a high or low signal) because the LED is always on. Assuming that my pin assignments are correct, what went wrong? <Q> Posting the pin assignment here might help us find the problem. <S> Anyway, Please test it without a clock and let us know. <S> module gpio_test (CLOCK_50, GPIO_0, LEDG);input CLOCK_50;input <S> [35:0] GPIO_0;output <S> [7:0] LEDG;assign LEDG[0] = <S> GPIO_0[0];endmodule <A> Sorry for posting this late. <S> I have finished the project I'm working on months ago. <S> The code above worked, there is nothing wrong in my code and in the pin assignments. <S> The problem is on my hardware connection. <S> During my testing, wherein I used the code above, I connected the GPIO pin for input to an LDR (light dependent resistor). <S> That being said, since the output of LDR is not really digital , the input GPIO is not getting proper signals. <S> That's why it is not producing the desired output. <S> Garbage in, garbage out. <S> The code above worked when I introduced Arduino to perform as my ADC. <S> I would appreciate if someone backs up my explanation with in a more technical manner. <A> Considering your self-answer, the issue was the input signal which didn't have proper logic levels. <S> A typical approach to deal with signals with significant noise, intermediate voltage levels or slow edges is to use a Schmitt trigger. <S> Those are build-in in most MCUs, notably, in Arduino digital pins and Raspberry Pi GPIO, so they are often perceived as a given by hobbyists. <S> In FPGAs, Schmitt triggers are often not implemented because they would prevent the pins from being used at their maximum speed with proper digital signals. <S> If you need those, you can pick an FPGA which has them (in case of Altera/Intel that would be MAX II and MAX V devices), or add an external Schmitt trigger or ADC IC.
| Most likely the problem is the pins assignment, make sure they are assigned correctly.
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What proximity detection technologies can work through glass? I'm looking for sensor options for proximity detection through a pane of glass. Something that will sit 6-24" on one side of a pane of glass, and be able to detect presence the same (roughly) distance on the other? I assume ultrasound won't work, but am not sure about infrared. If not, are there other options? Edit per this link from the comments, most glass is opaque to IR, so no, it won't work. That leaves the second part of the question - are there ways to detect presence / motion on the other side of a pane of glass. Video-based motion detection won't work because I need to be able to restrict distance (i.e. something present/moving within a fairly tight distance of the glass). Use case is something along the lines of a home security motion sensor looking out the window. <Q> You could use a microwave presence sensor <S> e.g. <S> this one which mentions glass specifically In general <S> , they will work through glass, without a big reduction in range. <S> They are often combined with PIR detectors to reduce false triggering (from sources of heat), but usually configurable so you can only use the microwave part. <A> Soda-lime glass: <A> This kind of situation is usually dealt with using video processing. <S> A camera pointing out through the "window" and comparing successive frames. <S> You can either do "advanced" object detection (there's a big open source project just for this kind of thing: http://opencv.org/ ) or take a more simplistic approach which I have used myself: <S> Maintain a "buffer" of the past (for example) <S> 10 frames Identify any pixels that have changed between frames Count the percentage of pixels that have changed Calculate the number of successive frames that have changed pixels over a certain threshold. <S> If enough has changed then you have motion. <S> You can also split the image up into "zones" so that you have only one area that is sensitive to motion, or make some areas more sensitive to motion than others. <A> Induction would work, I believe. <S> IR doesn't work because of the glass, yes, but close range coupled coils can still sense each other, and the voltage drop across the primary is inversely proportional to the distance of the secondary. <S> Coupling gives you a means of differentiating receivers, voltage drop gives you a means of measuring distance. <S> If it's for security though, it probably not going to be the case that some robber will carry a tuned receiver coil, so perhaps have an IR section hidden on the outside, fed power by induction and sending data back by induction too. <S> Food for thought. <A> It is glass. <S> You can use visible range, right?If your purpose is to detect someone's presence, have a camera which does it for you.
| IR goes through glass if it’s anywhere near the visible region. Motion is a great project to achieve this.
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How would you solve this resistive network? simulate this circuit – Schematic created using CircuitLab I'm not really sure how to solve this. I can't see any series or parallel connections. I need to find the Total resistance, V1 and V2, and finally I3 with a direction. Any help would be much appreciated. (edit) Okay so now I have taken your suggestions and redrawn it and worked out the values. it is much easier now to work out, so thanks for all the great suggestions. Feel free to correct me on any faults in my calculations. <Q> This is a really easy circuit to solve. <S> The reason it appears difficult is because the schematic is obfuscated by deliberately using unintuitive layout. <S> First, this is a good lesson on why logical schematic layout matters. <S> Second, you'll see this circuit is actually very simply once you draw it logically. <S> In general, draw schematics with power and ground voltages being horizontal lines sorted by ascending voltage bottom to top. <S> Then show the rest of the circuit with logical flow (nothing to do with current flow) left to right. <S> In your case, draw the battery at left with the positive node going up from the top, then across the sheet to the right. <S> Similarly, the negative node should go down from the battery, then across the sheet left to right. <S> Now fill in the resistors vertically between the two horizontal power lines that run across the top and bottom of the sheet. <S> Once you do this, the circuit will be obvious and you'll wonder how you didn't see it before. <S> Hint <S> : Look at how many nodes this circuit actually has. <S> Yes, it really is that simple. <A> This is obviously a homework question. <S> Please do us all a favor next time and actually tell us that right up front. <S> We won't GIVE you the answer <S> but we'll help you figure it out yourself. <S> I'll start off by making a suggestion. <S> Use the letter A to keep it simple. <S> Now do the same for the battery (-) and use the letter <S> B. <S> Now re-draw the schematic with that information. <S> Modify your question if you are still having problems. <S> [Edit] <S> I'll add one more suggestion: <S> re-draw the schematic with the battery on the left and all of the resistors on the right. <S> Place all of the resistors vertically with all of the same letter at the top. <A> As Olin said - Yes its really that simple. <A> If you pretty up your circuit it'll look something like the schematic below, so now all that's left is to figure out - knowing that V1 and V2 are merely the the battery voltage across the array - is I3, then the resistance of all the resistors in parallel and, finally, Is.
| Follow the line from the battery (+) terminal and mark each end of every resistor that connects to that (+) terminal.
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The role of reference voltage in buck converters? I asked a question here which made it quite clear that it's probably not a good start to make something that I plug directly into the mains. Instead, I've decided to build a low-voltage buck converter for starters. But I've run into something I don't understand. 1) You need a stable voltage to power your switching IC, or external circuit, or what have you. Usually lower voltage than the current you're trying to lower. 2) You need a stable reference voltage for your switching mechanism so it knows when it's duty-cycle is right on. With these two things required in a buck converter, it seems you've already accomplished your end just to build the buck converter. I.E. you have to have pre-established stable, lower voltages to run a buck converter who's purpose is to produce a stable, lower voltage. Why would you want a buck converter, and how does one establish a reference voltage and a voltage to power the switching mechanism? EDIT Reading both the answers I think I understand. For the switching mechanism, we KNOW the load it's going to draw and can plan accordingly. We can produce a very stable linear regulator if we know how much the load will be. We can even account for little fluctuations with a zener diode. However, using the same process for the overall power supply would be inefficient, and if we drew too much current, the heat emanating from the components would be uneconomical. this is without mentioning that we would be unable to switch the load without getting a different voltage. Correct? <Q> The reference voltage is required in the feedback loop of the switch mode supply so that the switching chip has something to compare the output voltage to - <S> it needs to know whether it has to raise or lower the output voltage. <S> The reference voltage is usually a high-impedance, low output current reference, so it does NOT replace the entire switch-mode supply (which by definition has a low output impedance and high current capability). <S> A simple (but not ideal) way to produce the reference is with a reverse-biased zener diode. <A> A voltage reference is at its heart a very precise linear regulator composed of a diode and a resistor. <S> The higher the load on the reference, the more power wasted as heat. <S> This of course defeats the purpose of using a switching regulator in the first place so it is recommended to use as little current from it as possible. <A> I will try to answer your points assuming I have read them correctly. <S> 1) <S> The driver for the buck circuit will generally need some kind of regulation, but this can be a low power regulator. <S> 2) <S> If your have either a dynamic supply voltage or load then the buck converter will need to be regulated to maintain a stable voltage. <S> This is achieved by varying the duty cycle to the power MOSFET and feedback is required to the driver circuit to determine how the duty cycle is adjusted. <S> You can use voltage feedback and then some closed loop control circuit to alter the duty cycle pending on the feedback signal. <S> To summarise a buck circuit is a good way to step down and regulate a large voltage to a smaller voltage, this can then provide power to a small, medium or large load. <S> Additional regulation is generally required for control circuitry such as a microcontroller and driver circuit, but these circuits generally have very small power requirements.
| You then use the buck circuit to step down your supply voltage and regulate it, this will provide a lower stable voltage with more current for your load.
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Would it be possible to make a solid state transistor by hand? I am curious, would it be possible to fabricate a modern BJT/FET transistor with hobby/garage tools only if size and aesthetics was not important? Which materials would one use and how would one acquire them? <Q> Well, yes and no. <S> You can make a transistor by hand in your kitchen. <S> It will involve some nasty chemicals, but it is doable. <S> A friend of mine made a LED a couple of years ago. <S> Not a transistor, sure, but the same process was involved. <S> (She was studying semiconductors at that time, so she had help. <S> The process itself was quite similar to producing transistors). <S> However, It is very unlikely that you'll ever be able to make a transistor in your kitchen that even comes close to modern mass produced transistors. <S> You wouldn't even come close to what has been produced in the 70th. <S> Nonetheless you may end up with a working transistor that does amplification. <S> If you want to do this, and you're willing to invest some time and money into it, please do so! <S> That would be really cool. <A> Sure it can be done, if you have ninja skillz: Homebrew NMOS <S> Transistor Step by Step - <S> So Easy <S> Even Jeri <S> Can Do It <S> Making Microchips at Home - Cooking with Jeri Part 1 <S> I wanted to post more links <S> but I don't have enough reputation points to allow that. <S> Just search for the followup videos to the 2 links I have provided. <A> A very similar question has been asked before. <S> My answer there included this link to the website of a someone who has actually made thin film transistors out of zinc oxide. <S> That site also includes links to otehr people doing similar work - at home and in the garage. <A> It's possible, but not so easy and the result is not going to be anywhere near the quality of components that you can buy. <S> The main issue is that in order to actually form the semiconductor junctions inside of the transistor, you need to carefully change the properties of a piece of semiconductor in a controlled manner. <S> This requires rather specialized tools as well as some pretty nasty chemicals. <S> Namely, you at least need a quite high temperature (1000C) <S> oven. <S> See <S> https://www.youtube.com/watch?v=w_znRopGtbE <S> for a high level explanation of what you would need to do to make a FET. <S> To make something even remotely 'modern', you basically have to build your own clean room and acquire used semiconductor manufacturing equipment. <A> If I recall it right, Jeri Ellsworth made a few transistors, depletion-mode NMOS be exact, and recorded videos logging the process of making it. <S> You can check YouTube for how she made it and her tests.
| So, it is possible to make transistors at home, but from what I see it is more of a learning experience as the produced transistors have really poor performance. It is possible to make some very primitive diodes, such as what is used in a crystal radio, but this is more 'finding' a diode than making one as the actual junction is already in the crystal, you just need to go poking around with a piece of wire until you find one.
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Power consumption when DC motor runs on battery I have small doubt about power supply selection for a running motor. I have a small dc motor, which is rated for 12V , 3A(rated ). When the motor runs with a load 4000N, the current consumption is 1.5A. So I have to choose a 12V, 3A = 12 * 3= 36W power supply to run the motor.This is because DC power supply can supply continous 3A current without any disturbance. Now I wanted to run same motor on battery. I would like to know how much power should be supplied by the battery to run the motor theoretically. When motor runs on battery, it takes full current from the battery; as per formula (\$e= l \frac{di}{dt}\$). It said that current required by motor = 3 \$\times\$ current required while running on starting . When we run the motor on battery eventually battery voltage got dropped, taking more current. I would like to know what is the percentage of current it takes. If we consider above system 12V, 75Ah battery is it efficient? Can someone explain with proper calculation here? While running on battery;motor will turn on every 5 minutes and after that it will be turned off. The motor is made to run 11 hours on daily basis. <Q> Before calculations you need to be clear about some fundamental concepts about motors. <S> "It said that current required by motor = 3 × current required while running on starting. <S> " This statement can not be true, because motor draws more current at start up than it rotates in operational angular velocity. <S> "When we run the motor on battery eventually battery voltage got dropped, taking more current." <S> This statement is also false. <S> Motor speed <S> ---- <S> > Voltage, Motor torque ----- <S> > <S> current. <S> My advice is as follows. <S> Your application does not require detailed calculations with motor specs. <S> What you need to do is measure the current when your motor rotates under the load, which you choose. <S> Then compare it with battery capacity. <S> About voltage drop of battery, it does not draw more current when battery gets low on Voltage. <S> Voltage is about speed(Theoretically). <S> Thus, as long as your load is the same, theoretically, motor draws same current. <S> NOTE: <S> Of course current change when the Armature voltage changes, but this is negligible. <A> 75A/h means "75A for one hour." <S> Maximum. <S> So if your motor, on average, draws 1.5A, and ran continuously, then it should run for 75/1.5 = 50 hours. <S> If it only operated one second out of every minute, then it's duty cycle would be 1/60, so would last up to <S> 50h * 60 = 300 hours. <S> The motor will slow down some as the battery gets depleted, so it's a good idea to de-rate this at least 15%. <S> Some battery types (such as lithium-polymer) can even be destroyed by over-discharging them. <A> It's not 75 A/h, it's 75 <S> Ah. <S> That means you can draw at 75 amperes for an hour <S> , then the battery would be drained - theoretically - this represents the amount of charge in a battery. <S> If you multiply that with the (battery) voltage, the resulting value gives you an energy measured in Joules, the SI (IS) unit for energy. <S> J and Wh are both units of energy.$$1kW h = \bigg(\frac{1kJ}{1s}\bigg)*3600s = <S> 3600 <S> kJ$$ <S> For a brief reminder: <S> \$P = IV\$, where \$P\$ is power, \$V\$ is voltage, \$I\$ is current <S> intensity.and \$E <S> =Pt=IVt\$ where \$E\$ is the energy and \$t\$ is time. <A> I am interested in spinning a PMG and kick the wind out of equation. <S> The torque demanded would shrink some of that 25 hours. <S> I suggest Varta Silver or gel kind of batteries. <S> Succes !
| Keep in mind most batteries self-discharge a little. If the DC motor is 12 v and has 3 amps, then is 36watts and a 12 volt battery on 75 amps/h will give you 12*75=900watts/36 = 25 hours.
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What is the optimal battery charging speed? I have 3.7V 4000mAh Li-ion battery . I also have 0-30V/6A power supply which can limit output current.Provided that I charge the battery with voltage limited to 4.2V , I can change the maximum current output of the power supply. Let's say that have set these parameters on the PSU -voltage limit = 4.2V -current limit = 500mA On the powerIf current is higher than 500mA , then the voltage will drop to a value where the current is 500mA . Assuming the battery is @ 5% , I can set the current limit to 4A and charge the battery less than 1 hour. My questions are > What is the optimal charging speed at which the battery will be safely charged? What is the fastest charging speed (safe charge)? What is the fastest charging speed (unsafe charge)? unsafe= can damage battery life or charging cycles Temperature may and may not be considered in the answer. <Q> Lithium Ion batteries have an upper charge limit of 1C. <S> This means once its capacity, or in your case 4000mA. <S> This does not mean that is the best, though. <S> It is good to keep a margin. <S> But the margin doesn't have to be large, as healthy cells of the LiIon type do not have a chemical recombination and produce nearly no heat while charging at an appropriate current. <S> Usually I take 0.5C, or in your case 2000mA. <S> It'll get the job done in about 2 hours and your cell suffer nearly as little as when you charge it with 400mA or less. <S> Of course, if you want to squeeze ten more charge/discharge cycles out of it, you can limit your charging and discharging to 1/20th C, i.e. 200mA and your battery will love you for a few days more. <S> Other than <S> that there's some guidelines to the best charging curves (which are a bit more complex than what a PSU does limited to current and peak voltage), but they will again only add a few percent at best to the lifetime. <S> The best advise is, charge them to 4.2V or even 4.15V instead of <S> 4.25V or more and don't keep them connected to the power too long after they are full and to never discharge them below 10%, or more conservatively 20%. <S> Or put in a voltage: Shut off at about 2.85V instead of 2.55V <A> It depends on battery, the exact rules change all the time. <S> Remember, that once you had to charge your cell phone for 24 hours first time? <S> Now you don't have to. <A> Li-Ion batteries are designed to be charged via a battery management IC, such as this one . <S> They charge the battery using the following charging profile: which is constant current (CC) followed by constant voltage (CV). <S> That said, it is possible to charge a Li-Ion battery with a good lab bench supply, which apparently you have. <S> It is important that the power supply have a meter that can monitor the current being drawn. <S> The charging voltage should be set to 4.2V or slightly below. <S> The current limit should be set to C/2 or C/3 for manual charging ( <S> 2A or 1.33A in your case). <S> Charge the battery until the measured current draw drops down to 0.03C (120 mA in your case). <S> If you let your power supply current go down to 0 mA, this "trickle charging" phase will oxidize the electrolyte inside the cell which is not good. <S> Li-Ion batteries don't like to be trickle charged. <S> Periodically check the temperature of the battery with your finger to make sure it is not getting hot.
| Still, i would say, based on my experience with various cell phones qnd chargers, that i wouldn't exceed 2A.
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How to default logic high after a voltage divider? I have a 5V logic pin coming in that I want to cut down to 3.3V. When the 5V pin is detached, however, I want the pin to default logic high (3.3V). How can I choose R_pu appropriately? V_in = 5VVCC 3.3VR1 = 20kR2 = 33kRpu = ? <Q> There is no universal answer, since the strength of the pullup depends on the current from the high voltage side. <S> That's why a diode translator is more effective: simulate this circuit – <S> Schematic created using CircuitLab <S> A low on the input pulls the output down to V f . <S> A high on the input is blocked by the diode, and the pullup pulls it up to 3.3V. <S> And since there is a pullup, a floating input lets it take over to pull the output up. <A> Ignacio gives you an excellent solution. <S> This has the advantage of having symmetrical rise & fall times if you add a noise-suppression capacitor to the 3.3V node. <S> There is one other advantage of a purely-resistive divider: if the input of the diode-coupled level translator shown above should ever have a significant transient that goes below ground, the diode will happily couple that transient directly into the 3.3V logic input. <S> That doesn't happen with the resistive divider shown here. <A> Both are good solutions. <S> For bidirectional lines it is common to use a levelshifter circuit.
| There is also another method: resistive voltage divider to ground plus a pullup-up resistor to +5V. simulate this circuit – Schematic created using CircuitLab
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Order of data bits In general when labelling data bits, say for a binary number. If I have four bits to represent such a number and I label them \$\ b1, b2, b3, b4 \$. Should \$\ b1\$ or \$\ b4\$ be the most significant bit? Which side do I start counting from? Is there a general rule for this or does it rely upon specification? <Q> In principle the agreed upon style of writing is as suggested by Gregory, we start with 0 and count up, as such representing the power-of number: <S> b0 has a value of 2 to the power of 0 --> 1 <S> b1 has a value of 2 to the power of 1 -- <S> > <S> 2 b2 has a value of 2 to the power of 2 --> 4 <S> b3 has a value of 2 to the power of 3 --> <S> 8 <S> So, by convention of use the b0 would be the Least Significant Bit and b3 would be the Most Significant in a 4 bit number. <S> Whether those bits come first or last on a transmission medium or if the one is on the left or the right side in a data storage device, that's entirely up to the hardware and protocols used. <S> But it is a very general convention for a b0 to be Least Significant. <S> Keeping that will let your developments be understood by the rest of the world. <A> Depends on the context. <S> If b1 - b4 is referring to the order in which the bits are received over a digital communication channel, the MSB and LSB must be specified and agreed upon by both sides ahead of time. <S> There is no universal standard and neither order is more correct than the other. <S> However, if you're representing a binary number in writing, the convention is standardized by grammar. <S> A binary number is still just a number. <S> We read numbers (and words) left to write. <S> The left-most digit is universally understood to be the most significant digit. <S> For example, the binary representation of one-hundred twenty-eight is 10000000 . <S> Here, the 1 would be considered the MSB. <A> And, start nubering from 0.
| For me it's easiest to declare that the bit number is the power of 2 that is represented by the bit.
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Are there downsides to using resistors with too high power ratings? I'm hoping to measure the current draw to my ESP8266 with a scope and a shunt resistor. I think a 1ohm resistor might drop the voltage too much, so I'm thinking to use a 0.5ohm (or even a 0.1ohm) instead. However, most 0.5ohm (and 0.1ohm) resistors tend to have at least 1W, 2W power ratings. Are there issues with using resistors that have power ratings much higher than required? <Q> The only usual downsides that come to mind are size, cost and sometimes resistors with a higher power rating use technologies such as wire wound that can have higher tolerance rating than some others. <S> Other factors such as inductance that may be an issue with RF circuits won't be an issue for a current shunt. <S> So in general if you're happy with the size, cost and tolerance of the resistors you've found <S> then there's no reason not to use a resistor with a higher power rating than you require. <A> Are there issues with using resistors that have power ratings much higher than required? <S> V across an inductor = <S> L\$\dfrac{di}{dt}\$ and this is the problem. <S> Typically, a 3cm long piece of wire of <S> 0.4mm diameter will have an inductance of 30nH (irrespective of its resistance) and if the current changes at 1 amp per microsecond (quite normal in a buck or boost converter), the volt drop due to di/dt will be 30mV. <S> If the resistor is 3cm long and <S> 0.03 Ohms, there will be a total volt drop of 60 mV for a sudden change of 1 amp flowing i.e. 50% bigger than what would be expected from a pure resistor. <S> Bigger sized resistors also have more self-capacitance (and capacitance to ground) and means another error when measuring high-speed signals. <S> Here is a table from Vishay that shows tha parasitic inductance and capacitance of small SMT resistors - note the final columns: <S> - You can see that capacitance (parallel) and inductance (series) reduce with size and if you are operating above 1GHz resistor sizes become a big deal. <S> Also, as frequencies of RF transmissions rise there <S> comes a point when the physical length of a resistor can cause standing wave problems. <S> At 300MHz, RF has a wavelength of 1m and it is fairly unfeasible for anything significantly less than one-tenth of a meter will cause problems but <S> what about wifi frequencies at 2.45 GHz - this has a wavelength of 12cm and conceivably, at this sort of frequency resistors that are about 1cm long will start to show problems. <S> WiFi is now occupying the 5 GHz band and some are starting to emerge at 60 GHz. <S> 60 GHz has a wavelength of 5mm so realistically only 0201 resistors are not going to cause a significant problem. <A> Usually larger resistors have higher ESL values, these are not listed for most resistors, but if your switching or using them for a DC to DC converter this could make a difference. <S> Larger resistors could have a different temperature coefficent.
| Bigger sized resistors may have higher self-inductance and this self-inductance in a component that is used to measure fast currents can lead to errors.
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Newbie question: How do pots limit current? From my understanding, a series circuit has a constant current, and when resistance is placed in the circuit it is voltage that changes across the resistors and not current. Potentiometers, from what I understand, are simply a variable resistors. However, when placed in a circuit they are usually used to control current. I am struggling to understand how what is essentially another type of resistor is suddenly breaking the constant-current rule I have been told. How is this possible? Thanks for the dumbed-down explanation, I have an engineering background but not in electronics, and I can't find a good explanation for this anywhere else. <Q> A pot doesn't break KCL or KVL, and as you say it's just a variable resistor. <S> Usually it controls current by converting a voltage on the pot to a current. <S> That's just Ohm's law- <S> The current is the voltage across the resistance divided by the (variable) resistance. <S> There's no law that says a series circuit has a constant current, only that elements in series carry the same current. <A> A lot of problems in the question: - A series circuit has whatever current it demands from the supply and is not normally constant by any means. <S> When a resistor is added to a circuit, voltages can rise or fall and current may do the same. <S> Potentiometers ARE NOT simply variable resistors. <S> As the name implies they control "potential" i.e. voltage. <S> Ohms Law prevails as always - there is NO constant current rule <S> All this is possible because what you have been taught (or learned) is incorrect. <A> That is not what happens. <S> If you have a resistor 10ohms , with another resistor 10ohms , and a voltage of 20V then a current of 1A flows through the circuit, and 1A will be generated because of I = <S> V/R = 20 / (10+10) in this case. <S> Here 1A flows through both, and current in both is same. <S> However if you replace one of the resistors with a pot, then if the value of resistor on being changed becomes 20ohm , then new total resistance R = 30ohm <S> So new I = <S> V/R = 20/30 = <S> 0.66 <S> A . <S> Now every time you vary the pot resistance, the entire series resistance will change, so total current will change. <S> But both the Pot resistance and the other resistor will still follow the rule, and both will see 0.66 Amps. <A> Pots are variable resistors and if the resistance is higher than the rest of the circuit loop <S> it reduces the current. <S> but you must choose wisely so that the power and voltage ratings are not exceeded. <S> other info related to current limits <S> Also, Pots can be made with Wire-wound (WW) for higher current, Cermet ( a resistive carbon-ceramic) material and metal film. <S> (MF) <S> More commonly today with MF plastic and a metallized coating embedded are used for reduced cost and smooth wiper action. <S> These MF pots cannot handle much current or power at all and especially if you apply a constant voltage near 0 Ohms where the wiper almost reaches an end terminal. <S> There are also many kinds of physical pots. <S> Including; Linear, audio-log, sin-cos rotary, sliders and trimmers or trim pots. <S> With Volume Controls on consumer goods and cars, Pots have been replaced with continuous rotary encoder dials using dual rotary tracks and wipers that are staggered by 90 deg to sense small angular motion digitally. <S> As always, the current limit is defined by the datasheet specification by the power level or wiper current so that the P=I^2R = <S> V^2/R does not melt the conductor.
| If you need to regulate high current then a transistor(s) with high current gain or a FET with voltage controlled transconductance or ON resistance in a Power FET is biased by a pot to become the higher current conductor.
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How to program an ATMega328 using a Serial Port without using programmers such as USBasp I am currently learning electronics. I recently bought an ATMega328 and an USBasp programmer to begin my learning. But I couldn't use the USBasp to program my chip. So I researched on other ways to program a AVR and I stumbled upon the idea of using a serial port to program it. I have a RS-232 port on my old desktop. I figured I could use it to program. But I couldn't figure how to do it. I have searched through the Internet. Some advise me to use a Max232 chip. I couldn't understand all of them. Below is a list of sites I have researched on this question: Programming Atmega with serial port http://www.hobbytronics.co.uk/arduino-atmega328-hardcore I want to know how to program the chip. Answers are appreciated. <Q> If you can afford buying an Arduino UNO board (about 20EUR, original, but you may find compatible ones which are cheaper), you can use it to program another Atmel MCU, in particular an ATMega328. <S> See this article on Arduino as ICSP and this one (programming breadboarded MCUs with an Arduino) . <S> I've personally done that <S> and it is seamless ( <S> Arduino UNO programming a breadboarded ATMega328P, exactly as shown in that article). <S> BTW, I was also able to burn a bootloader on that chip, since I bought some ATMega328 chips with no preprogrammed bootloader. <A> If your old computer has an LPT port, then all you need is literally 5 wires and <S> the program called AVReal: http://real.kiev.ua/avreal/langswitch_lang/en/ ; http://real.kiev.ua/old/avreal/en/adapters.html#FBPRG <A> You may use AVR910 programmer. <S> It uses serial port on PC side and ISP port on microcontroller side. <S> How to program: <S> write code, compile, and program the chip. <S> Make sure AVR910 selected in programmer setting.
| For compiler, you may use any compiler that support Atmel AVRProg (AVR910) programmer, for example CodeVisionAVR.
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What is the difference between clock and pulse? What is the difference between clock and pulse ?For example: Is the CLK-signal a pulse ? By the way, why do they invented the term frequence ? In my point of view, they could have said 20 times that period in a second ? <Q> 1.) <S> Generally, when referring to a "clock", the signal in question is a never-ending pulse train with known frequency, amplitude, and edge rates. <S> However, a single pulse used, for instance, to initiate the propagation of a data signal through a "D" type flip-flop is often referred to as a "clock pulse" and, in fact, many logic chip diagrams label the clock input "CP". <S> 2.) <S> The term isn't "frequence", it's "frequency", and it was invented in order to indicate the number of generally regular occurrences of an event in a particular unit of time. <S> In the scientific community, the frequency of an occurrence is measured in hertz, with one hertz being equal to one cycle per second. <A> By the way, why do they invented the term frequenc[y] ? <S> In my point of view, they could have said 20 times that period in a second? <S> In fact if you look at old writing, we did used to say "50 cycles per second" or "50 cps" to describe a frequency. <S> You would see radio frequencies described in terms of "kilocycles" and "megacycles" as a shorthand for thousands or millions of cycles per second. <S> Only relatively recently did it become very common to say "50 Hz" instead. <S> According to Wikipedia , the unit of hertz was introduced in 1930, but only became widely adopted after 1960. <S> Of course the quantity that these units measure has been called frequency for a very long time. <S> According to Etymonline , the sense of the word meaning "rate of recurrence" dates form 1831. <S> We use frequency rather than "rate of recurrence" or "times the period occurs in one second" because it's much shorter to say, and having shorter terms for frequently used concepts helps us to study and think about physics. <A> By the way, why do they invented the term frequence (sic)? <S> In my point of view, they could have said 20 times that period in a second? <S> Well, that would be the definition of frequency (with a "y", not an "e" at the end). <S> Frequency is not a human invention, it exists everywhere in nature and has existed long before humans. <S> Frequency is the rate at which an event occurs. <S> The rotation frequency of earth around its axis is 1 rotation a day, or 0.0000116 Hz. <S> Your microcontroller may have a 16 MHz clock, then the clock generates 16 million pulses per second. <S> So a clock signal consists of an indefinite series of pulses. <S> A pulse is sudden change in signal level, in a digital signal from low to high or vice versa, and after some time a return to the original level. <S> A pulse is most often relatively short, but your digital signal may consist of a 1 second pulse, so that's not a requirement. <S> In a digital clock signal the pulses are often half the period wide, what is called 50% duty cycle. <S> A 1 MHz signal has a 1 µs period, then the clock pulses will often be 0.5 µs wide. <S> Note that a repetitive signal with a given frequency consists of pulses, but you can also have pulses in a non-repetitive signal. <S> A single pulse will have zero frequency. <S> This question has some good answers explaining frequency.
| Talking about clocks, frequency will normally be defined as the number of clock pulses per second.
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Are these resistors in parallel? Suppose I have the following circuit: Are resistors R1 and R2 in parallel, assuming that through resistor R1 flows current because this circuit is connected as a feedback to another circuit? And if so, why? <Q> No, R1 and R2 are not in parallel unless the load was 0 Ω (so vout = 0). <S> Assuming the load is non-zero, but it is not a simple resistive load that can be separately measured, you can get an equivalent resistance by measuring the current through the load and the voltage across it. <S> $$R_{load} = <S> \frac{vout}{I_{load}}$$ <S> Then, the resistance from the junction of R1 and R2 to ground would be: $$R_{parallel} = <S> \frac{R1 + R2 + R_{load}}{R1 <S> \times (R2 + R_{load})}$$ <S> Note that if you plug in 0 for \$R_{load}\$ in the above equation, you get the formula for just R1 and R2 in parallel, as stated in the first paragraph. <A> R2 and R1 are not in parallel. <S> R2 and the sum of R1 plus whatever resistance is placed across vout <S> are in parallel: R2 || (R1 + Rvout) <S> The calculation of the resistance at vout may be a non-trivial calculation, but in every case, the voltage across R2 will be the same as the voltage across R1 plus vout: VR2 = <S> VR1 + vout <A> However, if it was, then there would be no voltage at vout--It would be measuring between ground and ground (0 volts difference). <A> To the base of the transistor (only) they look like they're in parallel.
| Think about it this way: In order for R1 to be in parallel with R2, its right-side would need to be connected directly to ground. No, they are not in parallel.
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Does current flow on an open transmission line? Please take a moment to look at the diagram below: The question is if the lightbulb will momentarily flash when the switch is closed. I think it will but I get the feeling I am wrong. The reason why I think it will flash is because when the switch is closed, the transmission line wire electric potential should become the same as the electric potential found on the battery terminal and in order for that to happen, electrons will need to flow through the wire until electric potential balance is reached. As electrons flow through the wire, they will need to go through the light bulb filament causing the light to turn on. By the way, I realize that the light bulb will not light up a room or that it will even light up at all, I am only using a light bulb here to illustrate my question and not mean to represent some sort of real life experiment. Thanks. <Q> Yes, there will be a brief pulse of current through the bulb as the portion of the transmission line (i.e., its capacitance) to the right of the bulb charges to the supply voltage. <A> There will be a slight current pulse at switch on even if you consider the circuit a lumped element circuit, i.e. without resorting to transmission line theory. <S> Just keep in mind that in a real circuit stray capacitance is always present, therefore you might model the open end of the transmission line as a capacitor (with tiny capacitance, say ~1-10pF). <S> Therefore you have a lumped RC circuit, where R is the filament. <S> Thus when you close the switch you are charging that tiny capacitor through the filament. <S> Assuming some ballpark figures like \$R=100\Omega, \; C=10pF\$ and \$V_{DC}=10V\$ <S> you get an initial current \$I= <S> V_{DC}/R=100mA\$ exponentially decreasing with a time constant \$\tau <S> =RC=100\Omega <S> \times 10pF = 1ns\$. <S> Of course the energy transferred to the filament before the current dies out is so tiny that a normal lamp won't be able to emit any detectable light. <A> No <S> and yes <S> and it depends on your view. <S> No if you view it as a schematic symbol representation. <S> This is typically what most engineers view when doing calculations or designing most of their work around - the schematic. <S> In this view, current flows when you have a continuous connection driven by a voltage, but there is no continuous connection here, so no current flows. <S> Yes if you view it as a transmission line. <S> As @Andyaka and @DaveTweet mention, a change in voltage will propagate through the transmission line and every point in the transmission where there is a voltage change, you will have current flow (displacement current). <S> However it will stabilize itself relatively quickly, until there is no longer any change in voltage. <S> As a crude analogy, you can think of it as, if you stand still and don't move, are you moving ? <S> If its relative to the earth, then no you are not, but relative to the sun, you are moving - quite fast actually. <A> Electricity has to flow or else <S> how could the power source know there wasn't a load at the end. <S> This is all embodied in transmission line theory. <S> It's called characteristic impedance by the way. <S> Another interesting side effect is that an infinitely long lossless transmission line will conduct current indefinitely based upon the voltage supplied and the cable's characteristic impedance.
| The current that flows is based on the input impedance of the transmission line.
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Run a program on a PICKit 2 chip I've got a PICKit 2 with a PIC16F690 chip plugged in, and installed MPLAB and made a sample program in C, which compiles fine. However, there's only the option to "Program to target device", "Read from target device", or some other options to erase memory, all of which work fine, but I can't work out how to actually make the program run on the chip. MPLAB says I can't debug the chip: "PK2Error0028: Unable to enter debug mode" "NOTE: This device requires an ICD Header for debug. See "Header Specification" DS51292."After some research, apparently I need a different interface board to debug it. When debugging, the option to run the program comes up, but when set to "Release" mode, it only gives the option to read/write/erase from the board. How can I run it? The code is as follows: #include <htc.h>#include <stdio.h>#include <stdlib.h>int main(int argc, char** argv) { return (EXIT_SUCCESS);} Which doesn't actually do anything yet <Q> "PK2Error0028: Unable to enter debug mode" <S> Is it a genuine PICkit2? <S> Many cheap clones cannot debug. <S> When you program the chip you place the code onto the chip. <S> When you aren't programming the chip, if MCLR isn't held low, the program is running. <S> You don't "run" the program, it just runs unless told specifically not to. <A> You need what Microchip calls a "Debug Header" in order to debug this chip. <S> The Debug Header brings out all of the chip's <S> I/O pins for your use and includes extra pins that connect to the debugger (PICkit or ICD). <S> Be sure to hold MCLR HI when testing your blink code. <A> What's there to run? <S> Your program starts and immediately exits. <S> Embedded program should never exit main. <S> Instead there should always be some sort of infinite loop. <S> Something like this <S> : #include <htc.h>void init_io <S> (){ // init timer(s) <S> here // <S> init <S> an I/ <S> O pint for output here}int main(){ init_io <S> (); while (1) { <S> // <S> set <S> i/ <S> O pin <S> high <S> // wait <S> 1/2 second <S> // <S> set <S> i/ <S> o <S> low <S> // wait <S> 1/2 second }} <S> You can then look at the I/O pin with a scope, or connect it to a multimeter, or light an LED.
| The easiest way to see if your chip is working is to include some very simple LED blink routines and see what happens when you run the chip with just power supply and the LEDs.
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JFET in place of digipot for gain control? I'm hoping to build a sound sensor using this schematic ( http://cdn.sparkfun.com/datasheets/Sensors/Sound/sound-detector.pdf ) To programmatically adjust the gain, I could use a digipot. My (naive) understanding of the main issue with using a digipot is that it's linear, but gain control should be logarithmic for constant change. That said, I guess the MCU could do the conversion. Suppose I'd like to save the MCU from the conversion, is JFET a good alternative? If so, which JFETs specifically are good for this? Are there other better alternatives? Thanks <Q> You can use a J-FET but there are challenges. <S> The main problem is distortion. <S> Here's what happens <S> : the resistance of the channel (D-S) changes as Vgs changes. <S> If the J-FET is in series with the signal, Vgs changes with the signal. <S> There are techniques that minimize this problem but it is something to be aware of. <S> The next problem is repeatability. <S> Each J-FET has different sensitivity in term of how the channel resistance changes as Vgs is varied. <S> One place where J-FETs are used very successfully is as the gain-control element in an audio AGC or compressor / limiter circuit. <S> Because the Vgs control voltage is created by sampling the output signal, variations in Vgs sensitivity basically drop out of the equation - these circuits have a classic negative-feedback control circuit and the negative feedback simply compensates for the Vgs sensitivity between different parts. <S> For what it's worth, I'm currently working on a design that uses a logarithmic-taper digital pot (32k, 100 steps) from Catalyst / <S> On Semiconductor. <S> Although this part is scheduled for end-of-life this year, we will simply purchase enough parts to see the product through its manufacturing lifetime. <S> And, yes: using a digital pot with log taper makes my hardware design dramatically less complex and expensive. <A> For volume control, yes, a logarithmic taper pot is desirable. <S> That said, there are log taper digital pots. <S> They are not nearly as cheap or abundant as linear, but they are out there. <S> If it were my project, I would probably tackle the problem with some sort of VCA, either built with an analog multiplier, or OTAs. <S> If you want to keep your control entirely in the digital domain, another option is a 4 quadrant multiplying DAC. <S> Feed your signal into the Ref pin, and DAC's output code sets the attenuation (ie. <S> on a 10-bit DAC, setting the output to 512 would attenuate the signal to half amplitude). <A> the old LDRs were made of cadmium sulphide BUT new ones don't have cadmium I have shown excellent linearity on a actice <S> PFC prototype <S> THE LDR is a lowspeed device and has been used in faders decades ago <A> You can use a logarithmic digipot: you specify the attenuation in dB (logarithmic obviously) and it does the conversion. <S> http://www.maximintegrated.com/en/products/analog/data-converters/digital-potentiometers/DS1882.html
| ONE way of implementing volume control without distorsion is to use a LDR that is driven by a LED
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Do conductors age under reasonable electric load? Construction steels are prone to fatigue failure - they get structurally weaker when exposed to repeated mechanical stress. Is there any similar process for conductors which makes them less usable once they conduct current for long enough? In other words, suppose I have an old transformer which has served for say 30 years. It still functions fine (didn't burn down, short or anything like that) but it is rather old already. I expect that magnet wire insulation will get weaker because of being heated for all these years. I also expect that core plates insulation will get weaker. What about the wires themselves? Will aluminum or copper in them be any different from what I'd find in newer wires? <Q> If the question is about small transformers, the above answers are adequate. <S> If the question is about large transformers, i.e. oil-immersed power transformers - that is a whole different game. <S> Power transformers are constructed with paper insulation, which has a limited lifetime. <S> The cellulose in the paper degrades with normal heating and becomes mechanically weak; additionally, the degradation of cellulose releases water, which lowers the di-electric strength of the oil. <S> The rate of degradation is exponential with increasing temperature (see 1 for some discussion.) <S> I believe transformers are usually designed to achieve 25 or 40 year lifetime under rated load. <S> All of which is to say - the metallic conductors in an old power transformer will be fine - but beware old insulation . <A> Not in a transformer - there should be no change unless the transformer was abused. <S> However, a reasonable load in the context of an IC is considerably higher than a wound coil. <S> At about \$10^6\$ <S> ~ <S> \$10^ <S> 7 \$ A/\$cm^2\$ (depending on material etc.) <S> electromigration becomes a serious limitation on life. <S> Thin wires can sometimes deteriorate due to chemicals in the environment- <S> I've seen fine copper coils fail in the presence of (gaseous) sulfur or sulfur compounds getting into pinholes in the insulation (kind of a black plague). <A> The following is an interesting document about the ageing of conductor cables in nuclear plants . <S> At pages 20 and 21, inside a table reporting cable components stress factors ( stressors ), you find possible causes of conductors failures. <S> You may notice that in no way the conductor material (aluminium, copper) "ages" just by itself, but it develops problems when interacting with other "factors". <S> The simple act of conducting current is not harmful, unless it causes thermal stress or mechanical stress on the cable (not on the material). <S> As for the material itself, copper used in building conductors is extremely pure, so it doesn't degrade unless exposed to factors that either trigger a chemical reaction (e.g., oxidation, corrosion) or a nuclear reaction (metals exposed for long time to heavy radiations, especially neutrons, undergo a change in their physical properties, since atoms in the crystal trellis may be mutated into other isotopes or even in other elements). <A> As mentioned above, your transformer should be fine unless it is overloaded. <S> A minor overload, generating excessive heat will degrade the insulation, and potentially lead to turns shorting. <S> A major overload will burn open one of the windings entirely. <S> The conductors themselves do not degrade in their ability to carry current in any capacity as long as they operate within their ratings. <S> Again, passing 1000 amps through a piece of #30 copper isn't going to work for long, but that is because the heating will melt the conductor. <S> The primary failure mode of conductors, be they aluminum, copper, or whatever is an enviornmental breakdown of the insulation systems, or in the case of medium to high voltage (15KV to 345KV) insulated cables improper installation, typically not observing the minimum allowed bend radius. <S> Environmentally, the most common cause of insulated cable failure is exposure to ultraviolet light, with a close second of exposure to oils or other chemicals that break down the actual insulation jacket. <S> Very rarely there can be failures due to water intrusion, but again, that isn't actually a failure of the conductor, but rather the insulation system. <S> To be complete, there is a possibility of actual conductor failure, but I don't consider it to be a true failure of the conductor personally, from improper installation leading to galvanic action or corrosion (think wires on an automotive battery terminal corroding from the acid, or copper developing <S> it's typical green patina when exposed to salt water). <S> Those failures almost always happen at a connection point, and are the result of poor or improper connections.
| The metallic conductors in a transformer don't deteriorate with age, but the insulation does.
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Can I use a supercapacitor (like Maxwell boostCAP 3000F) on a coilgun? Can someone tell me if I can use a supercapacitor (like Maxwell boostCAP 3000F) on a coilgun? I want to obtain a very intense magnetic field for a project, and I'm not sure what capacitors should I use <Q> Let's try it! <S> From the datasheet: <S> Capacity : <S> 3000F <S> Absolute Maximum Current : 1900A <S> Now the charge stored, Q, = <S> C <S> * V = <S> 3000 <S> * 2.7 <S> = 8100 <S> Coulombs. <S> So at Abs Max Current (1900A) discharge time can be as low as ...8100 / 1900 = <S> 4.26 seconds. <S> So divide the length of the gun by 4.26 seconds to get the muzzle velocity. <S> For any reasonable size of gun, you could walk faster than the bullet. <S> (Edit : I am being a bit unfair : if 1900A is a useful current in launching a small, er, mass at high velocity, then the capacitor can launch a number of small masses over a 4 second burst. <S> But as far as dumping all its energy (10.9kj) into one mass is concerned, that would take 4 seconds, for an average power about 2.5 kw. <S> That's really not much by projectile launching standards.) <A> Short answer: <S> a super capacitor will not be useful at all in this scenario. <S> If you are trying to induce a large magnetic field the key is having a large amount of current dumped into the coil in a very short time. <S> Having a large capacitance can actually make this process take much longer and you get a much smaller field out. <S> The magnetic field from a coil is proportional to the rate of change of the current through the coil. <S> The key here is the balancing act between how much current you want and how much current you can use before the coil burns up. <S> As always, be cautious and safe when using capacitors, especially at high voltages, as they can be very dangerous. <S> PS: calling anything a "gun" may draw unwanted attention, use "projectile launcher." <S> ;) <A> You need to focus more on high-voltage , low-capacitance for higher energy projectiles; not the other way around. <S> Although this is a new-guy mistake, so I would advise you read a bit more on designs like this, and basic electronics. <S> The long-term limit is overheating, not energy. <S> Remember that <S> and it gets clearer that you can put a hell of a lot of power in quite easily, but controlling it is not as easy. <S> Also rail guns are way cooler. <A> Using a boostcap CAN work, I started working on one myself but never got around to finishing it unfortunately and due to divorce <S> I now live in a small apt. <S> where testing such a thing isn't feasible heh. <S> Anyway, the first thing you need to know is <S> you CANNOT use a SCR to fire it. <S> It will take too long to discharge and melt your coil and plastic firing tube along with it. <S> My plan was to fire short bursts so I could fire multiple shots without a recharge. <S> Unfortunately, the firing timing circuit malfunctioned and left it on and that's exactly what happened lol. <S> What I did was used 50 MOSFETS in parallel connected together via 1" thick aluminum rails for the power rails and smaller ones for the gates. <S> You need the thickness because you are conducting a LOT of current, you want to keep resistance low so most of the voltage drops across your coil, plus it helps to heat sink your MOSFETS too. <S> You will need a timer circuit or an IR LED/detector setup to ensure you only activate the MOSFETS when needed. <S> You'll need a beefy current source for the MOSFETS plus high power bleed resistors since you want the MOSFETS to turn on and off quickly because unlike BJTs, MOSFETS have a gate capacitor that needs charged/discharged to turn it on/off. <S> If it helps, I used 5 boostcaps connected together with 1/4" thick aluminum plates and 10awg wire for the coil. <S> I had dual 4awg jumper wires to take the power from the cap bank to the the coil gun, which was sorta table <S> mounted on a big piece of plastic. <S> The 10awg wire was selected because with low voltage, you need to keep resistance low too <S> so you want thicker wire with less turns. <S> Unfortunately, even with that you won't get the shot power a single shot high voltage coil gun has. <S> BUT, you do get to fire it probably a few dozen times! <S> Also, it should have more power than battery versions due to the far lower current limitations of batteries.
| Do a little bit of research on the discharge rates of capacitors and you will see they are dependent on the initial voltage (which 2.7v from a super capacitor is too low to use for almost anything), the size of the capacitor, and the impedance of the load that is discharging the capacitor.
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Zener diode voltage too low I'm trying to regulate 12V to 4.7V with a zener diode. (I'm doing this to protect an AVR's gpio pin, so the current is very low.) I tried the simplest circuit with the zener, but it does not work as expected. simulate this circuit – Schematic created using CircuitLab I expect the multimeter to measure 4.7 Volts on the zener diode, but it only shows about 3.2-3.3 Volts. I replaced the diode with other ones with the same model, and the results are the same. Are my diodes faulty, or am I doing something (very) wrong? <Q> The test current shown in the datasheet is 45 mA. <S> You are only allowing 2.5 mA, which is apparently not enough to get over the diode's "knee". <A> You are using a 1W max zener diode at far below its rated current: <S> the datasheet shows the zener voltage as 4.4 .. 5V at 45 mA. <S> You are using it at ~ 7V / 4.7kOhm = <S> 1.5 mA. <S> The same datasheet shows the Zz as 13 Ohm, so the 'missing' 43.5 mA would account for a missing ~ 0.6V drop. <S> The effect you see is larger, probably because you are operating the zener outside the current where the stated Zz is a valid approximation. <S> Solutions: <S> use a smaller (400mW) zener <S> use a smaller resistor use a diode clamp to Vcc instead of a zener (but make sure the input can handle the extra 0.6v, and that Vcc can absorb the extra current!) <A> If you're just trying to drop 12 volt signal to around 4.7 volts, you ought to be able to use a simple voltage divider, like this: <A> The zener voltage is guaranteed to be within 4.4V and 5.0V at 45mA. <S> You are putting about 1.8mA through it. <S> It will have a lower voltage at such a low current.
| If you want a more predictable result you can use a zener diode that has a lower test current. That is a fairly husky Zener diode.
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Does USB always supply 5.0V I was wondering about this part of a circuit diagram. I believe the component marked as JP12 is a DC barrel connector, which I thought had three pins. One to detect input, one with V+ and one with ground. However I thought this device was also receiving 5.0V from USB. The diagram shows pin 1 of the DC connector directly connected to the VBUS line, does this mean the DC connection is also powering the USB device. If so, what is pin 2 being used for. Also, I have no idea which of the three pins in JP12 is supposed to be ground. Am I mistaken, and this DC connector does not supply ground? <Q> CN2 is probably the barrel connector you are looking for. <S> It is labelled as a "DC_2.0mm" which would be a 2mm diameter DC-jack, which can be known as barrel connectors. <S> CN2 is connected to the circuit GND and the VCC net. <S> JP12 looks like a jumber to allow you to set the power source which U2 regulator is fed from to generate 3.3V. <S> If pins 1 and 2 are shorted together U2 is fed from the 5.0V net, where as if pins 2 and 3 are shorted together <S> U2 is fed from the VCC net. <S> The jumper will usually either be a set of 2.54mm posts which you can short together using a special shunt piece that fits on top like this: Or surface-mount pads close together that you can short with a solder blob or a surface-mount resistor like this: <S> So in this circuit, the USB connector CN1 always provides 5V when connected, but the current available will be limited by the USB host (usually to 500mA). <S> You can choose where to supply your circuit from by shorting the appropriate pins of jumper JP12 . <A> JP12 is a three-pin male header. <S> A jumper connected to two adjacent pins is used to select which of 5.0V and VCC will be used as the supply for the 3.3V regulator. <S> The barrel jack is CN2 , on the bottom left of the diagram. <A> USB provides up to 0.5A, so if you need more, you put external power supply. <S> Usually external side of the plug is minus. <S> Use DVM to check it. <A> JP12 is a three-position post header, used as an SPDT switch to select power from either USB or CN2 (presumably a coaxial power connector).
| You would use a shorting plug on JP12 to connect between pins 1 and 2 to use USB power, or between 2 and 3 to use power from CN2. The designer has provided an optional way of supplying your own supply (which could be greater than 5V as long as U2 does not overheat) if the power drawn by the 3.3V circuit is too great for USB.
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pull down resistor in PIR circuit I have an motion sensor circuit. The circuit calls for a voltage of 2-15 dc volts. On the PIR data sheet the source pins calls for a pull down resistor of 47K at 5 volts. My question is if I use 8 volts on the circuit do I need to recalculate the source pin resistor? I have searched this forum and others for the usage of higher or lower voltage, but cannot seem to find any examples. All examples are made around 5 volt. Sorry . I do not have a data sheet on the circuit. I am using PIR325 on the circuit I have now and it calls for a 100R resistor and it seems to be rather slow. I am using 8 volts on it and I have the circuit tied into a PICAXE 08m2. I also have a regulator and a diode for the voltage on the picaxe. I am posting a picture of the schematic for the circuit without the picaxe. The picaxe is tied into the circuit a the large dot next to the led. I want to change PIR to a PIR LHI878. Here is the data on the PIR. excelitas.com/downloads/dts_lhi778_lhi878_pyd1388.pdf – <Q> Well, to calculate it the mathematical way, set it up as a ratio and solve for X: <S> 47k / 5v = <S> x / 8v <S> Get rid of "/8 <S> " by doing the opposite to the left: (47 * 8) / 5 = <S> x <S> 376 / 5 = <S> x <S> x = 73k. <S> That will provide a scaled pull-down resistor value for 8v instead of 5v. <S> We can check the actual current by using Ohm's Law . <S> Assuming the full supply voltage was being imposed across the pulldown resistor, I = <S> E / R <S> I = <S> E / <S> RI = 5v / 47k <S> I = <S> 8v / <S> 73kI = <S> 0.000106A <S> I = 0.000109AI <S> = <S> 106uA <S> I = <S> 109uA <S> Close enough. <S> Give it a try, should work just fine. <S> Test a +/-20% variance and see what effect it has on the circuit. <S> I doubt the value is super-critical, but one particular value may work best. <S> 73k * 0.20 + 73k = <S> 87.6k <S> 73k - 73k * 0.20 = <S> 58.4k <S> If a particular pull-down current is specified as best, then calculate the resistor value for the given operating voltage by: E = I * <S> R R = <S> E / <S> I <S> R = 12v / 100uA <S> R = 120k <A> The simple answer, given all the information (including the datasheet you linked in the comments: Pro-tip, add it to your post, preferably at the first posting, to get the best answers the most quickly), is: No need to change the resistor. <S> If you change the supply voltage of the Sensor to between 5V and 12V, anywhere, the response will be in large orders the same. <S> The "at 5V" in the datasheet is sort of to make it clear that the sensor will not achieve that output when you power it with only 2V. <S> (The point about the op-amps is that not all op-amps in the world will work with a single-ended supply and/or one above 5V, but the LM3xx-series op-amps will do you perfectly fine in most use cases for this schematic.) <S> So, no, you need not change anything in this schematic. <S> The changes in response from the sensor will be minimal and the effect of the changes is easily caught with the trim potentiometers "RV1" and "RV2". <S> You are allowed to experiment with other resistors, such as 22k or 68k or 91k to see if the response and filtering will match your application more, but it should not be necessary. <A> Update on what I have tested. <S> Changed to 47K and 12 volts but the sensor would not settle down. <S> Just stayed on the blink. <S> Changed to 12 volts 75K and the same thing happened. <S> Changed to 8 volts and 47k and the sensor worked <S> well sensed up to 70 feet and was responsive. <S> Changed to 8 volts and 75K and worked very well. <S> sensed up the 90 feet and was very responsive. <S> Before I decided on what to use I will continue to test other resistors that were suggested. <S> I will update when completed. <S> Thank You
| The only thing that should change in this schematic, provided all the op-amps used are actually the ones in the schematic, is the set-points of the potentiometers, but you are probably going to empirically find them anyway.
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Why is transformer higher voltage primary usually closest to the core and has the shortest loops? AFAIK the typical transformer design includes the primary being wrapped closer to the core and the secondary being wrapped atop of primary and so further from the core. Why this way and not the other way around? This question presumably mentions the design where the secondary is closer to the core and made of thick wide aluminum tape. It's a last mile distribution grid transformer with primary being fed with something like 6 kilovolts and secondary producing something like 110-230 volts (consumer voltage), so the secondary has 30-60 times higher current than the primary and with that current I think it's reasonable to place the secondary closer to the center so that each turn is shorter and the secondary itself is shorter and therefore has lower resistance and lower losses. The same reasoning should apply to all transformers which lower the voltage - the secondary voltage is usually 10-20 times lower than the primary voltage (110-230 volts primary vs 12 volts secondary is typical) - it's reasonable to keep the secondary wire as short as possible to lower the losses (and save some thick wire). Yet I've never seen a transformer with a secondary located closer to the core (except in the linked to question). Why is primary winding usually closer to the transformer center and has smaller coil loops? <Q> For "concentric" wound transformers, I find both ways of them to be used, and the factors influencing it are most likely (may not all apply to your case): <S> Cost of the material involved. <S> Higher current needs thicker wires, but the same number of turns (and even thicker if its on the outside because it is longer there). <S> Just calculate what is needed for the number of turns required in either inner or outer layer, and then do a comparison. <S> It seems to me the higher the step down ratio, the more beneficial it is to keep the high current ones inside. <S> Tapping. <S> A lot of transformers have multiple tappings. <S> Precise positioning and space for them is a lot easier on the outside than it is on the inside. <S> Serviceability and failure modes. <S> Bigger transformers are actually so expensive that it might be feasible to repair them. <S> Depending on the expected failure modes putting one or the other winding on the outside is more useful. <S> Aluminum might be better suited here than the softer copper. <S> This applies of course only at rather high currents (that are distributed unevenly in the conductor). <S> Metal fatigue considerations however might drive you away from aluminum here. <S> Note that I don't call them secondary/primary but low/high voltage sides <S> , I think those factors are more influential than the direction they are used in. <A> There are probably several reasons. <S> First off the manufacturer may have primary winding coils wound up for a whole family of transformers that use the same turns count and wire size. <S> Secondaries are added as needed according to output specs. <S> Secondly it is probably always a simple calculation of wire total resistance versus total length. <S> Primaries of the type of step-down transformers that you mention have many more turns of smaller gauge wire. <S> Putting it near the core has a greater advantage for lesser total length and thus less total winding resistance. <S> So it comes down to a simple trade off computation between secondary and primary. <S> I could imagine that there is a computation that involves wire size loss versus total winding length that resolves to the lowest physical size. <S> The reason we see most step down transformers with the primary on the inside is because the computation most often wins out that way. <S> I look back to my early teen age years when I was very glad that old TV power transformers had secondary windings that I could easily remove and replace with my own to achieve the voltage and current ratings I wanted without having to touch the primary windings. <A> I googled "power transformer construction" and copied various images: - I think most small - medium power transformers have primary and secondary stacked side-by-side: <S> - Here's <S> one where the primary (HV side) is on the outside: <S> - <S> And another: - <S> Considering winding losses and turns ratio, it make sense to wind the low-voltage coil on the inside. <S> Broadly speaking, the average length of 1 "inside" turn might be half of that when wound on the "outside". <S> This means the outer winding is probably double the resistance of the inner winding for the same number of turns. <A> Two reasons : Safety and efficiency. <S> Safety : if the primary winding insulation breaks down, it is better protected from the external environment (your hands!) <S> and closer to components that are likely to be earthed, namely the inter-winding screen (where a screen is employed). <S> Thus a primary breakdown should connect live to earth, tripping breakers or fuses. <S> Efficiency. <S> (1) <S> The winding closest to the core is most tightly coupled to it, producing the magnetising field with lower losses. <S> (2) <S> As more power is dissipated in the primary, it makes sense to wind this with the shortest turns to minimise its resistance and I^2*R losses. <S> Both of these are relatively minor advantages in the overall scheme.
| The low voltage winding resistance is "seen" on the high voltage winding by the ratio of the turns ratio squared and, given that the current drawn from the secondary is reduced by the turns ratio (not squared) onto the primary it makes sense to keep the low voltage winding as short as possible. Rectangular cores might need different material strenghts and properties since internal forces are higher than in circular cores.
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Why are some generic door alarm magnetic switches built using Normally Open reed switches instead of NC type? Some generic, low-cost "door alarm" magnetic switch components are made of using "Normally Open (N/O)" type reed switches, ie, when the magnet is in proximity, the current stops flowing. If I were building such a system, I think I would choose a Normally Closed type so that the current would flow when the magnet is pulled away, eg, when the door is open, and I could sense it and ring the alarm. My question: Why are generic, low-cost magnetic switch components go for Normally Open reed switches instead of the opposite? <Q> N/C magnetic switches are unfortunately nearly useless for security. <S> Since their "unalarmed" state is for them to be open, if the cable going to them is cut then the alarm will not trigger, and of course not be triggerable anymore. <S> With a N/O magnetic switch the circuit is closed in the unalarmed state, and cutting the cable will trigger the alarm. <A> I think that a clarification is in order. <S> When a switch, relay, transistor, etc. is NO or NC, it is in this state/condition, when the device is not activated .What <S> this means, is that a NO switch (for example) will be closed when it is activated (a magnet is in proximity; door closed). <S> With this kind of switch, you make a "closed loop," which will be "sensed" if the loop is broken (the door opened; wire cut). <S> This type of application, is the reason why the switch has to be NO , when not activated! <A> Normally Open magnetic switches can be put them in parallel rather than series when you want one wire pair to monitor several openings. <S> Unfortunately if the wire is cut, it doesn't alarm, but this is taken care of by having each switch in parallel with a large resistor. <S> If the wire is cut, the resistance changes upward, and you can tell that the wire was cut, rather than a door opening (which reduces the total resistance to zero) or closed (which has a maximum resistance based on the number of openings in the string). <S> It's a different topology, and is more useful when you are mixing sensor types. <S> For instance, a floor pressure switch is normally open, and closes when someone steps on it. <S> If you wanted a door and a floor switch to be on the same circuit, it's easier and cheaper to use an NO switch at the door, than to find a NC floor pressure pad. <S> Most alarm systems today have enough zones, though, that mixing various types in one zone isn't needed. <A> It is called Fault Condition. <S> Both conditions: Open door and Power Supply failure are covered. <A>
| The reason NO switches are used is simply that the MB circuits are cheaper to produce, monitor, service and rectify faults because they are simple circuits without spurious parts such as resistors to go wrong than with NC switches.
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does SPI master always receive data in same time he send data I try to communicate with EPPROM using SPI. the image illustrate the read status register sequence. the program I'm using checks the receive data register each time SPI finish sending (send data register empty). my question is, did SPI master receives data on MISO when he send the instruction data as shown in image below. and in general did SPI always receive data on MISO in same time he send data on MOSI? in other words, to get status register out data, should I send other NULL data (0x00)?? <Q> That said every SPI device will work differently with regard to what bits are used in each path. <S> It is up to you to read the DATASHEET for your device to figure out what bits required are on the MOSI side to get the device to do what you want and provide back what you need on the MISO side. <A> SPI is a full duplex bus. <S> That means that communication in both ways happen simultaneously. <S> In hardware it may be implemented by 2 shift registers that are clocked by the SPI clock. <S> The transmit register will clock data out, and the receive register will clock data in. <S> Most microcontrollers will access the transmit register on write, and receive register on read. <S> Both shift registers are clocked with the same clock. <S> So if you want to clock data into the receive register, you need to send a 00h byte. <S> This is often considered "safe" to send. <S> In the code bases I have worked on, we have typically implemented SpiTxRx routines. <S> It takes a byte to transmit, and also returns the byte read. <S> In my opinion this is good practice, because if you don't read the SPI receive buffer (e.g. you think can optimize by writing a faster SpiTx routine) at all the hardware may flag a buffer overrun and lock up the SPI peripheral in an error state. <A> SPI data transfer in either direction means the master sends a clock and whoever wants to say something (including the master itself) puts it onto the data line with the clock cycles. <S> In principle, the master could just generate the clock without sending data out, but this is the same as sending a 0x00 . <A> Some SPI libraries include only one routine, which sends and receives a byte simultaneously. <S> Others have a separate routine to send a byte and ignore the result. <S> Including the latter function in a library may allow performance to be improved significantly (sometimes almost doubled) on some platforms since the transmit-and-ignore-result routine can return as soon as the hardware starts transmitting the byte in question, even though it wouldn't be able to return data until the action finishes . <S> To allow high speed reception, some libraries may offer a "get a byte and start receiving another" function. <S> I haven't noticed much consistency in how the higher-performance libraries operate, and avoiding data corruption while overlapping operations can be tricky, but the performance improvements that can be obtained by overlapping I/ <S> O and computations can be very significant and worth the effort.
| The general concept of SPI is that it can send and receive bits at each clock time.
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White noise generator for audio jamming ( from 250 Hz to 4.5 KHz ) I want to implement a white noise generator device for the purpose of audio recording jamming , as I can make the recorder record the noise and the speech together so any one try to record the speech , he/she will hear a noisy speech , But from where I should begin ? Thanks <Q> Seems like the optimum spectrum for masking speech yet not sounding too unpleasant is a 'haystack' spectrum that mimics speech. <S> White noise is more unpleasant than pink noise but neither apparently work all that well. <S> Cambridge Sound Management makes their opinion fairly clear: Both white noise and pink noise are anything but unobtrusive, and neither is very effective at blocking speech. <S> Should a vendor attempt to convince you that his system is better than another because it uses “pink” noise rather than “white” noise, run the other way, don’t walk. <S> Such a vendor is at best naïve, and more likely a borderline charlatan. <S> Keep in mind that the document <S> I lifted this from is intended to promote their products, but I think the principle is sound (no pun intended). <S> If you want to make white noise (which can be shaped into pink noise with a filter), the usual method is with a pseudo-random number generator. <S> This can be a few CMOS chips or a small microcontroller. <S> You might be able to do even better if you have a better processor by playing with speech samples. <S> Of course if the samples or "white noise" are predictable it might be possible for someone to remove the noise. <A> One or both of those frequencies was frequency-modulated with a noise signal. <S> The idea is that the frequencies mix together on the mic's diaphragm and thus create an audible masking noise that makes it impossible to record anything else that is a lower level. <S> This worked well for both dynamic and electret microphones. <S> I have no idea how well (or if at all) with modern MEMS microphones. <A> Speech intelligibility is best measured around the SPL of the voice relative to the background noise at specific frequencies. <S> In order to block speech, the best way is to blast noise ranging from 500Hz to 2kHz. <S> Turns out, we're really good at understanding someone even with high amounts of noise (or a fair amount of downsampling, etc). <S> But if those specific frequencies are below the background noise (of what threshold I can't remember off the top of my head) <S> then understanding the words spoken becomes much much harder.
| One of the techniques that used to work well for jamming microphones used for recording was to use two- high-frequency signals above 20 KHz (about 1 KHz apart) that beat together.
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Print PCB on copper using Altium I've been trying to print and make my own circuit using Altium (print it on film paper, iron it on copper-coated plastic and then dip it in Ferric Chloride etc.), using Gerber files. Only problem is, Gerber files don't show the holes (and obviously I can't print the drill file) and I can't print the PCB directly since that I can't separate the layers while printing the PCB. Any idea on how to print holes normally? I've been looking for an answer for a long time but couldn't find anything that worked for me (looked mostly at this ). <Q> So you have got a drill file, in txt format. <S> You can drag this txt and drop into the Altium Designer. <S> You will get the following dialog: <S> Click OK and your drill file will be loaded like this: <S> Now, you can select \$ <S> \textbf{File} \longrightarrow \textbf{Print}\$, set Print Scalling to \$ \textbf{User Scale = <S> 1.0}\$. <S> The page orientation can be set under the Printer Setup option. <S> As a demonstration I have just printed it as a PDF document, here is the output on an A4 sheet: the board dimensions: 67.6mm <S> x 149.25mm <S> the A4 sheet dimensions: <S> 210mm <S> x 297mm <S> According to this: \$ \frac{297mm}{149.25mm} = <S> 1.9899 \$, there must be enough space along the height of the page for almost two printing. <S> It can be seen on the picture above that this condition is fulfilled(, if we place the printing at the edge of the sheet). <S> Consider it as a workaround, there could be much more sophisticated solutions. <A> You have not stated how many layers you use, but I suppose that you use 2 layers. <S> Top and bottom. <S> It is actually pretty easy to print the PCB on paper, through printer. <S> Here is an example, we will print top layer 1st, then bottom layer. <S> > <S> Open the PCB which needs to be printed. <S> File <S> >Page Setup... <S> (Composite Properties window should pop up) <S> On the Composite Properties set your paper and quality setting, also set these> scale mode = <S> Scaled Print scale = <S> 1.00 corrections x = <S> y <S> = 1.00 <S> color set Mono <S> if you experiment with offset settings you can get both layers on one paper(!) <S> Go to Advanced... <S> Delete all layers except top and multi layer (right click>delete) Check holes and mirror <S> (because top needs to be mirrored) > <S> OK <S> File > Print Preview... - make sure it looks like it need to be, all holes should be there (if not, reset Altium Designer) <S> Print... <S> > <S> select your printer settings and print to the paper. <S> We finished top layer. <S> To print bottom layer, do all steps from 4 to 8, but make sure you add bottom layer (right click>insert), delete top layer and mirror is unchecked. <A> Altium won't print the drill holes on it's output options as anything other than white, so if you want a negative you need to place a small dot in the center of each through hole pad on an unused layer, set it to print that layer in black and move it to the topmost print.
| You do not have to use Gerber files to output the PCB, but use print option and select printer to print on.
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How do I get my Arduino to reset another board? Let's say I have another micro controller board. It has a 2 pin header on it. If you momentarily connect those two pins together -- either shorting them out with a jumper cap, or hooking them up to a button and then pressing said button, and so on -- the board will reset. Now let's say I want to set up my Arduino so that it can trigger this reset when a certain condition is met -- received a command over a wireless link, a counter passed a certain value, whatever. How would I hook this up? What additional components would I need, if any? Do I just hook both pins of the reset header to Arduino pins and digitalWrite(PIN, 1) them? Sorry for this admittedly newbie-sounding question, but I am still fairly new to electronics. <Q> MCU reset pins are normally active-low with pull-ups, which means you ground them to trigger the reset and leave them open to run normally. <S> There's a special kind of output called open drain that's perfect for driving this sort of pin. <S> Unfortunately, it seems like Arduinos don't support open drain outputs directly. <S> This page suggests putting the pin in input mode to produce the open state and output mode to produce the grounded state. <S> All this assumes that the MCU board is using the same 5V or 3.3V supply as the Arduino. <S> If this is correct, it's safe to do what I've described. <S> But if the MCU board uses a higher voltage than the Arduino, you'll need to use a transistor (NPN or NMOS) to pull down the reset line. <S> If the reset pin is active high, you'll have to drive it with a normal output, and the voltages probably have to be the same. <S> You can figure out the polarity of the reset by measuring the voltage on the shorted pins with a multimeter. <S> You can get the board voltage by measuring both pins separately. <S> You'll also need to connect the grounds of the two boards together to provide a return path for the current. <A> Something like this would work nicely. <S> It's a digitally-controlled, analog switch. <S> Best of all it's less than $1! <S> Also, there are many "reset supervisor" chips out there that can do the same thing, but are a bit more complex to implement. <A> I would think the easiest solution (without needing to worry about linking grounds, etc) would be to simply use a normally-open relay. <S> Something like this would do nicely: http://www.digikey.com/product-detail/en/9007-05-00/306-1062-ND/301696 <S> All you need to do is connect the relay coil pins (2 and 3) between the Arduino output and ground, and connect the other two relay pins (1 and 4) across the header contacts used to reset the other micro. <S> It is the equivalent of a pushbutton switch connected to the header, except that it's controlled electronically rather than mechanically.
| Alternately, if the reset is totally controlled by the Arduino (i.e. it never toggles itself), you can drive the pin directly with a normal output.
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How is a semiconductor electrically neutral? I'm in the process of learning how transistors works, which starts with understanding how doping is used to create n-type and p-type semiconductor materials. All the resources I've read sort of explain this the same way, and I'm missing something. P-type semiconductors have extra holes and are predisposed to accept electrons, whereas n-type semiconductors have extra free electrons and are predisposed to donate them. This is the fundamental principle of how transistors work, as I understand it. But every resource emphasizes that in spite of this both n-type and p-type semiconductors are electrically neutral, which is where I'm lost.If one has extra electrons, and one is missing electrons, how are they electrically neutral and not charged? I seem to have a block about this or something, I just don't get it. <Q> Take silicon as an example. <S> Silicon has four valence electrons, and silicon atoms in a crystal lattice form four bonds with neighbouring atoms. <S> Transistors, and other semiconductors, are made of silicon crystal with small amounts of dopants added. <S> These dopants change the electrical properties because of the way they interact with the crystal lattice. <S> Phosphorous, for example, has 5 valence electrons. <S> It's still electrically neutral <S> (number of protons = number of electrons) <S> but since the silicon crystal structure only requires 4 bonds per atom, there is an 'extra' electron that isn't really participating in the crystal structure. <S> With a bit of extra energy, that electron will go into the conduction band and freely roam around the crystal lattice. <S> This corresponds to an n-type semiconductor. <S> There is a similar process for p-type semiconductors - boron, for example, only has 3 valence electrons. <A> Good answer can be found here, taken from a physical standpoint. <S> The terms n- and p-type doped do only refer to the majority charge carriers. <S> Each positive or negative charge carrier belongs to a fixed negative or positive charged dopant. <S> p and n type materials are NOT positively and negatively charged. <S> An n-type material by itself has mainly negative charge carriers (electrons) which are able to move freely, but it is still neutral because the fixed donor atoms, having donated electrons, are positive. <S> Similarly p-type material by itself has mainly positive charge carrier (holes) which are able to move relatively freely, but it is still neutral because the fixed acceptor atoms, having accepted electrons, are negative. https://physics.stackexchange.com/questions/81488/how-can-doped-semiconductor-be-neutral <A> When a donor (for example) is ionized, it creates a free electron, but also it creates a positively ionized donor atom. <S> The charge on the free electron and the ionized donor are equal and opposite. <S> So as long as the electron doesn't go anywhere, the net charge remains zero. <A> They are not always electrically neutral. <S> An n-type semiconductor has an excess of 'free' electrons -- electrons that can move about freely in the semiconductor (very similar to electrons in a metal). <S> These electrons are 'donated' by immobile donor impurities doped in to the semiconductor. <S> If you imagine starting from that state, then the result is still neutral. <S> However since the electrons can move, they have a tendency to diffuse away from regions of high concentration. <S> If you connect another material (e.g. p-type) to the n-type (forming a pn junction), electrons will diffuse from the high concentration region to the low-concentration region. <S> This won't continue forever (unless you have a power source connected), because in leaving the n-type region, they leave a + charge behind. <S> This creates a restoring electric field, and at some point this restoring field will balance the diffusion process and an equilibrium will be obtained. <S> The specifics of this depend on the materials, the doping and temperature, as well as any external voltage applied between the 2 materials forming the p-n junction. <S> Since (starting from neutral), electrons (negative charge) have left the n-type region, it will become net positively charged, and the p-type negatively charged. <S> In a similar way holes ('anti-electrons') from the p-type diffuse over to the n-type, further charging it positively. <S> A similar behaviour would occur if you connected a heavily doped n-type to a lightly-doped (in fact it occurs any time there is a concentration (or temperature) gradient). <S> The material as a whole isn't charged (just polarized), but if you connected it to another conductor (e.g. a wire), charge would move between the cloud of free electrons in the wire to the semiconductor, putting a net negative charge on it. <S> Although it is small, it could in principle be detected by observing electrostatic forces. <S> It cannot be measured by (e.g.) connecting a voltmeter to the semiconductor and the metal because charges would also flow into the leads of the voltmeter, exactly canceling and leaving no net voltage. <S> If in fact there was a temperature difference, you would be able to measure a voltage -- this is the Seebeck (thermocouple) effect. <A> Simple version: <S> When n-doping by adding phosphorus, we're actually adding a positive phosphorus ion , plus a mobile electron. <S> When p-doping by adding Boron, we're actually adding a negative boron ion , as well as a mobile "hole."
| The semiconductor has both free charge (electrons and holes) and immobile charge (lower band electrons, nuclear protons, and ionized donors and acceptors).
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Designing an ethernet isolator I'm trying to galvanically isolate a device which is controlled by ethernet (1000base t) via computer. I've been looking and apparently the ethernet standard provides some isolation, but after I connect the cable to my otherwise isolated device to my computer, my multimeter reads short between its ground and earth ground. So clearly it's not isolating enough. I'm wondering how I should proceed; according to wikipedia, gigabit ethernet is 8-bits, so a digital isolator won't work. Can anyone suggest a kind of chip that would work? They don't seem to make 8 bit digital isolators. In any case, it seems the ethernet signal can go below ground, so I'm not sure opto couplers would be a (direct) option, although I suppose I could bias them before and remove the bias after with a cap? It also seems like they are not fast enough for gigabit ethernet... <Q> Buy a pair of gigabit fiber media converters and a few meters of fiber. <S> That will give you more than enough isolation. <A> Ethernet does provide 'transient' isolation (to around 1.5kV but don't quote me on that) as distinct from functional isolation (a permanent ac or dc bias, which is a different spec), via a transformer which may be a discrete transformer on the PCB, or integrated into the RJ45 connector ("integrated magnetics"). <S> This provides a floating differential pair driven/received by the transformer at each end - 2 of them for 10/100BaseT, or all 4 pairs (of Cat5/6) for 1000BaseT. <S> Unless this is a PoE scenario, I would expect open-circuit between the "ground" of the main circuitry at each end. <S> If you don't, I think there's something wrong somewhere. <S> A separate issue is whether the "ground" of the circuitry of your PC & mains-earth are also connected, which varies - desktop PCs usually are, laptops usually aren't. <S> This has nothing to do with the 'galvanic isolation' you're seeking between PC & peripheral. <S> GigE isn't "8 bits", it uses all 8 wires of Cat5/6 cabling, 4 (floating) differential pairs. <S> Forget about digital isolators or optocouplers, this isn't your problem or your path forward :) <A> Well, this was a lot simpler than I thought. <S> Thanks The Photon for the comments giving me the idea; it turns out the cat6 cable I was using <S> is metal shielded. <S> The metal shield is connected to ground at the computer end, and to the ground on my microcontroller. <S> Using a cable with no metal shield breaks the connection.
| True 'Ethernet Isolators' can provide true 'functional isolation' between each end, for (rare) situations where the "ground" at either end has a permanent ac or dc differential.
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Can you use diodes instead of OR gate? I am building a computer and thought if I could simply use diodes instead of OR gates. Will this work with logic ICs such as the 74LS, 74HC and 75HCT series? Sorry if this is a stupid question, I am new to electronics. <Q> The conditions are: 1) <S> The circuit is low speed. <S> 2) <S> The load impedance is high, such as a CMOS input. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This can be expanded to as many inputs as needed by adding more diodes. <A> Yes, that is called diode logic, but you have to be aware of some constrains. <S> You will need a pull down resistor to give a path to the current. <A> Yes you can. <S> For glue logic on slow events (like power supply sequencing or connecting a pushbutton switch), this kind of logic should not cause problems. <S> But If you use a negative supply like in davidrojas's answer, then you would be exciting the esd protection diodes of the downstream chip, which could cause unexpected behavior if there's nowhere to draw the pull-down current from. <S> Also, you might not have a negative supply voltage available. <S> To make the speed fast, you need to make the pull-down resistor value small, and that will increase the power consumption whenever the logic is high. <S> You lose noise margin in the high state due to the drop through the diodes. <S> It's more difficult to guarantee timing specs compared to using a dedicated OR gate chip. <S> Some alternatives that are often better are: Use a dedicated logic gate. <S> One-gate chips only cost a few cents, so that the cost of the pick-and-place assembly might be higher than the actual chip cost. <S> Invert the logic and use open-drain outputs in wired-AND configuration.
| There is a really cool trick that you can use under specific conditions that allows you to build a 2-input OR gate or AND gate using only 1 diode and 1 resistor.
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How to get started with microcontrollers? My point is not to start a flame war, or have an opinion based fight. I really just need advice based off of my experience. I am pretty experienced with using an arduino, I understand C, & C++ fluently. What is a 8 bit microcontroller I could start with that would give me a solid foundation and understanding? I understand the programming and theory, but I am lacking practical experience. Besides the chip, do I need an IDE? or an IDE & a compiler? <Q> "Arduino" as a hardware platform is useful because of the range available for almost every niche, even if you choose not to use the Arduino IDE & toolchain & bootloader - you can install your own standard AVR8 toolchain (i.e. GCC + AVRDUDE + avrlibc + linker + etc, there's usually packages available to make installing all this easay), or <S> Atmel's Windows-only-based 'Atmel Studio', and other options, & use an AVR device programmer, and your own preferred text editor. <S> With this you can take off the Adruino 'training wheels', get dirty with the MCU's registers, and really understand how to drive it. <S> The same can be said of just about every other mainstream MCU family, there's usually cheap development boards available for virtually all of them, though without so much of the community & hobbyist-level <S> hardware add-ons that surrounds the Arduino. <S> Each will have a toolchain of their own. <S> I would recommend sticking with 8-bit MCUs for a while - get to know how to really drive them without the Arduino training wheels, because this will hold you in good stead to deal with the substantially greater complexity you'll encounter in 32bit MCUs. <S> The IDE is just a pretty GUI layer on top of the underlying toolchain, and is optional, but ideal for beginners - setting up a toolchain & make system is not easy unless you're already familiar with it from other programming work. <S> Opinions vary greatly on which IDEs are 'best' (from those available within a given brand/family of MCU), but reality is that few if any are 'best'. <S> Although if you start using a JTAG debugger device (as distinct from a simpler in-circuit device programmer) to step through you code & set break-points & inspect registers & memory etc, then an IDE really helps here. <A> I made a comment above that expressed my opinion that 8-bit (or simple) microcontrollers are a better jumping off point than ARM Cortex for beginners. <S> Despite that earlier comment, I just ran a student through an independent study using an STM32F4 as his very first controller, and it went OK. <S> The main problem is the relative sizes of the user bases. <S> If you go with a PIC or AVR, (dev board and compiler choice make little difference) and you don't know what you're doing <S> , you get on the web and poof , there's the answer you're looking for, and more help than you could possibly need to get you to the right place. <S> Not so with an ARM Cortex. <S> Even selecting a compiler and IDE, and then getting the toolchain running can be very tricky. <S> There are enough online resources such that someone <S> who knows what's going on can get needed guidance. <S> IMO, someone who does not know what is going on can have a very hard time. <S> If you have buddies who can help, do what you want. <S> If you're a novice, and trying to make a go of it using just online support tools, go with a PIC or AVR toolchain. <S> You're just shooting yourself in the foot if you're a novice in it alone and unsupported trying to start with an ARM Cortex. <S> When you're ready, the tools are out there for you to make the switch. <A> These kits also has debug. <S> For ide, you can download ST Visual Develop and Cosmic CxSTM8 32K 4.3.13. <S> Cosmic licance is free for 32k memory. <S> This is ide and linker. <A> My first microcontroller was the PIC18F1330, which is a very decent chip and great for beginners. <S> I had to buy the chip ($2) and the PICkit (2 or 3) for $35-$45, but then I was ready to go. <S> I used MPLab 8.6 and the C18 compiler, though I think I heard that C18 is being phased out. <S> XC8 is taking its place. <S> All of the software is available from the Microchip website for free, and there is a lot of help documentation out there. <S> The library references built into the compiler/ <S> MPLab are also very thorough. <S> MPLab v8.92 (I recommend staying away from X at this point--it's still very buggy) <S> : http://ww1.microchip.com/downloads/en/DeviceDoc/MPLAB_IDE_8_92.zip XC8 v1.34: http://www.microchip.com/mplabxc8windows
| in my opinion, you can start smt8 discovery kit then you can try smt32.
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Current source or current mirror? I've been told that using current sources in simulation programs may produce "too optimistic" results compared to using current mirrors. Could someone please elaborate? I'm still a bit confused about this topic. <Q> Your question is very specific, but I think what you're getting at is a more general concept. <S> An ideal current source will provide the exact amount of current requested regardless of external influences. <S> It will adjust its voltage instantaneously as needed to continue to provide that exact current flow <S> no matter what. <S> Realistic? <S> Not at all. <S> Any real current source will always be influenced by external forces. <S> A sudden step change in load resistance, for example, may cause the current source to spike up or down. <S> Or what if the load becomes open circuit? <S> The realistic current source may attempt to raise its voltage until it fails catastrophically (or simply powers down if it's smart enough). <S> Using a current mirror as a realistic current source is one of many ways to model a more realistic current source. <S> Current mirrors are very simply to build - just two transistors and a resistor - which is probably why this person suggested using them. <S> But current mirrors exhibit properties that may not be adequate for your simulation. <S> Sometimes an ideal current source will work just fine. <S> Other times a more sophisticated model is necessary. <S> As with all things electronics, it depends on your application. <A> I've been told that using current sources in simulation programs may produce 'too optimistic' results compared to using current mirrors. <S> I wouldn't have thought so <S> but, if you can find some quote on the internet, that would be a better approach for this question. <S> We don't know the validity of the person who said this AND "simulation" could mean anything. <S> Some sim tools (and models) might do a very good job of predicting the actual current produced whereas some others may be poor. <S> From my own experience of using micro-cap and having designed several current sources for strain gauge bridges, pressure bridges, RTCs and individual gauges I would say my simulation was accurate to within 0.1%. <S> Of course it may be argued that 0.1% is still inaccurate. <S> I say this because the current sources that I use are all based around op-amp feedback systems and, at DC (up to 100s of <S> kHz) <S> the op-amp would ensure pretty tight control of the current generated. <S> Current mirrors tend not to use a high-gain, high-precision components like an op-amp to regulate current so, they are bound to be more flaky in both real life and simulation. <A> If you're talking about biasing in an analog IC, the use of ideal current sources will give you biasing that never runs out of headroom (the source will work even if the voltage across it reaches zero or goes negative) and infinite output impedance. <S> Real current sources need a bit of voltage to work and have finite output impedance (and less at higher frequencies). <S> Using a current mirror can model those effects, so <S> your amplifier stops working at low supply voltage and has a more realistic lower gain than you'd expect with a perfect current source. <S> Better current mirrors (for example Wilson mirror vs. a simple current mirror) may more closely approach the ideal current source. <S> If you are modelling the other transistors accurately, then using ideal current sources in your circuits will tend to give better results than reality. <A> Simple current mirrors (2 transistors) are prone to some undesired effects, such as not exactly matched pairs, dependency of the collector current on Vcc etc. <S> Some can be mitigated by adding components, some cannot.
| In typical general-use simulators, "current sources" are modeled as ideal. As for current mirrors I would say that these are likely to be more problematic than the general term "current source" AND therefore more likely to be flaky in simulation.
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Pull-down resistor ; why is it crucial These 3 diagrams represent 3 test cases. Dia 1: When I perform an analogRead of analog port 0, then I get a high value obviously. According to the manual, the analogRead method obtains the resistance. I expected it would say that it returns the voltage. Anyway, in this case both are directly proportional. Dia 2: With a resistor + pull-down resistor, the reading of the analog port 0 decreased. Dia 3: However, I would have expected that dia3 would also result in a decreased value. My reasoning was: The supplied voltage of 5V is constant. So, it must be divided between the resistor and analog port 0. A bigger resistor should result in a higher voltage for the resistor, and a lower one for the analog input. --> But that seems to be incorrect. Second attempt: The analog port reads resistance and not voltage. Maybe that's important. R = V / I . Apparently the resistance remains the same. So, applying this formula: if resistance remains constant and voltage decreases, then it could be caused by an increasing current. But it doesn't make sense that the current would increase by adding a resistor. PS: if I add the analog pin to ground, then I get a low reading. Where exactly is my reasoning flawed ? EDIT: As many people have pointed out. I have indeed wrongfully interpreted the following phrase in this tutorial . Thank you for pointing that out to me. in the main loop of your code, you need to establish a variable to store the resistance value. <Q> Analog inputs measure voltages, not resistance. <S> It is possible to infer the value of a resistor by setting up a circuit in a certain way and measuring voltage on an analog input. <S> But analog inputs on their own only measure voltage. <S> Can you show the context from the manual you're referring to that claims an analog input obtains resistance? <S> The results you're getting for your three experiments sound perfectly correct. <S> In Dia 1, as you seem to understand, the analog pin sees 5V and reports it. <S> Analog inputs draw extremely little current. <S> So little, in fact, that the current can be ignored in most cases <S> *. <S> In Dia 2, you've created a resistor divider between 5V and GND. <S> It's not really appropriate to call the 2nd resistor a "pull-down" resistor in this context. <S> Redrawn, Dia 2 looks like: simulate this circuit – <S> Schematic created using CircuitLab <S> The analog input pin draws no current, but there is a current path from 5V to GND. <S> This causes a voltage drop across both resistors. <S> The voltage at the middle is easily solved using Ohm's Law. <S> The equation for the divided voltage looks like:$$V_{divide}=5V\frac{R2}{R1+R2}$$The "decreased value" you observed was in fact the divided voltage value. <S> Choosing different values for R1 and R2 will give you a different analog voltage readings. <S> For Dia 3, consider the effect of Ohm's Law on that resistor. <S> Remember, virtually no current is consumed by the analog input pin. <S> If there's no current, then the I in \$V= <S> IR\$ is zero. <S> If the <S> I is zero <S> , then the voltage drop <S> V is zero. <S> If the voltage drop is zero, then there's no change in voltage on either side of the resistor. <S> Therefore, both sides of the resistor read 5V and the analog pin will read a high value, exactly as you observed. <S> * <S> There are times when this assumption is not correct and will result in measurement errors. <S> But those cases can be safely ignored for introductory purposes. <A> According to the manual, the analogRead method obtains the resistance. <S> Either the manual is incorrect or you have misinterpreted this statement. <S> Any analog read will return a digital number that represent the voltage at that input. <S> It does not convey any value of resistance that might be fitted at that input. <S> I expected it would say that it returns the voltage. <S> So would I. <S> I would have expected that dia3 would also result in a decreased value. <S> For values below 10kohm (generalism alert) <S> I would expect it to be 1024 (10 bit ADC resolution) because the input impedance of the ADC is so much higher than the 10kohm and therefore does not form a significant potential divider. <S> Note than an input impedance of (say) <S> 10 Mohms would result in the 5V source being lowered (due to the 10k and 10M) by about 0.1% - maybe you would read 1023 instead of 1024: - <S> Rtop would be 10k and Rbottom would be 10M in this example. <S> On the other hand if you were scanning around those inputs in software and the ADC inputs were fed (internally) to a ccommon ADC via a multiplexer you might see variations in the read value. <S> This is due to the finite time it takes to charge internal capacitors at the ADC input (after the multiplexer): - The sample and hold capacitors need to charge up to within 0.1% of the output of the multiplexer and, if the multiplexer is scanning around very quickly, an error can result on some inputs because there might not be enough time to charge the capacitors. <A> As far as I know analogRead return a value (10bit wide by default) which represents the fraction of the ref. <S> voltage being read. <S> For example if you have 5V hooked up to the ADC and your ref. <S> voltage is 5V <S> you will get a value of 1023 (or something just below it). <S> Your resolution is 5V / 2^10 = about 4.9mV per ADC code. <S> so 1023 <S> * <S> 4.9mV is about 5V. <S> When you are shorting 5v to the ADC you are causing 5V to drop over the ADC <S> so you get max. <S> code for 10bit ADC = 1023. <S> When you hook up 5v through a resistor you will still get about 5V on the ADC input because the input impedance of the ADC is very large so almost all the voltage drops on the ADC and almost none on the resistor. <S> Your ADC resolution is not fine enough to show these differences. <S> When you hook up two resistors with ends going to 5V and GND you have created a voltage divider which according to the resistors value relation will be some fraction of the 5V voltage. <S> For example if you have 2 10K resistors you will get 2.5V at their junction. <S> This voltage is sampled by the ADC <S> so you will get a code about 512 (512 * 4.9mV = <S> about 2.5V right?). <S> It would not be correct to call the resistors pull up or pull down as that term is usually used for resistors that introduce a certain voltage level to a digital pin which is floating.
| It all depends on the value of the resistor and "other things".
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Discrete Schmitt Trigger: make output logic level at 0V, not 3V I designed a Schmitt Trigger with discrete components (using an emitter-coupled pair) that operates with a Low Disable Trigger of around 3.5V and a High Enable Trigger of around 4V on circuit simulation. The circuit simulation uses a sine wave function generator to simulate a changing analog waveform. The desired output of the Schmitt Trigger for high and low are 5V and 0V respectively, but this Schmitt Trigger outputs a logic high of 5V and a logic low of 3V, which is not suitable for the rest of my circuit, as they require 5V-0V logic levels to work. The question is how can I make the Schmitt Trigger output a logic level of 5V-0V from a logic level of 5V-3V? What circuit can I "bolt on" to this design to make it output that logic level, or even better, an entirely different circuit altogether that achieves similar functionality (a discrete-component comparator that can achieve a 5V-0V logic level is also cool with me). Keep in mind that I cannot use any ICs due to project requirements, and I also cannot use any other power supply other than 5VDC. <Q> The basic problem is how the positive feedback in your Schmitt trigger robs low output voltage. <S> The feedback voltage is generated across RE, which is always added to the low level output voltage. <S> A Schmitt trigger is basically a amplifier with a little bit of positive DC feedback. <S> Can you think of other ways of using two transistors to make positive gain? <S> Perhaps some of those configurations allow for a feedback path that does not limit the output voltage swing. <S> Hint 1 <S> : Your exising amplifier topology can be used with a different feedback path. <S> Hint 2 <S> : The two transistors can be a mix of NPN and PNP. <S> That may open your thinking to more amplifier topologies. <A> Or, since the specs for TTL allow a 400mV output voltage with full noise margin you could use a few hundred mV of that up for the hysteresis and divide the input voltage down and probably get to your goal, but that's a bit ugly. <S> Even unloaded you can't get to 5V/0V output swing because a transistor will have some 10's of mV across it when saturated, so you need some slightly more relaxed requirements. <S> You can get close to 0 and 4.9x V (unloaded) with a fairly simple addition on the output side. <A> Logic levels for 5v CMOS are not necessarily 0v and 5v; 1v-3.5v will usually work fine. <S> You can however get 0v-4.5v with two extra transistors and resistors.
| In theory, yes, you could use a negative supply.
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If we put a led light into a oscillating current, will it blink? Sorry for the misleading title cause I have no electronic knowledge. It is a simple question. This is the circuit diagram. In case you don't see clearly, it is XOR GATE.My question is what will happen with the LED light, will it turn on/off or blink ? <Q> It's a shame that we don't better encourage these learning questions on the stack, because this is a good one. <S> What you have stumbled across is an indeterminate state. <S> While it is possible that the circuit may oscillate it would happen at an incredibly high speed. <S> Whether or not the LED appears to be on would depend on how much time it spends forward biased each cycle, and the characteristic of the diode. <S> I'm guessing it probably wouldn't be visible, but this depends on so many part specific factors it really is just a guess. <S> The problem is that combinational logic circuitry is purpose built to respond to input as fast as possible, and you've shorted the output to the input. <S> This problem is the whole reason sequential, clocked logic exists. <A> I'm going to answer the intended question, and not analyze the circuit. <S> These may be too fast to see, but they can be picked up on a photodetector and translated back to an electrical signal. <S> It is possible to transmit data this way. <S> You circuit as shown may not work as intended, but that's another matter (and should be a separate question). <A> In the massless, frictionless, noiseless Perfect World (beware of the spherical cows) where DC is pure and logic signals only have two states, you will have made a high speed oscillator. <S> The LED will blink, with a 50% duty cycle, at half the switching frequency of the gate. <S> Since perfect gates run at ∞Hz it will be rather difficult to measure the pulse rate. <S> In a less perfect world, the gate will probably be unable to switch fast enough and will settle down into a noisy state, the final output probably varying between components. <S> They aren't designed to be wired like that. <S> Add a buffer and a simple RC delay. <S> In reality, the circuit you drew will burn out almost immediately. <S> You've drawn a short circuit from the gate to ground via the LED. <S> Hard to say which component will release the Magic Smoke first, but one of the two is going to have a short service life. <A> A CMOS xor gate with low enough operating voltage can oscillate, the LED can get a square wave voltage and nothing puffs the smoke out. <S> I think this is possible if your LED has high enough forward voltage. <S> Let's assume the system oscillates. <S> It must happen at several MHz, so no blinking can be seen by eye. <S> It can be detected with fast enough instruments - except if the LED use fluorescent wavelength conversion to make short wavelength (blue or UV) radiation usable for lights. <S> LEDs of this type are common in lamps. <S> Their light decays so slowly that the light is virtually continuous if the voltage oscillates faster than at few hundred Hz.
| If you connect an LED to an oscillating (or pulsing) circuit, it will blink as fast as the oscillations.
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What makes cables suitable or unsuitable for a given purpose? I've been working on a homebrew set of VR Goggles w/ head tracking, and am just beginning to plan the cable that will connect the parts on the user's head to the electronics on the main unit (which will be mounted under & behind the seat the user will be seated in) I'm expecting about a 4ft cable run between the two units, and the connection will need somewhere between 6 and 10 wires (depending on how many ground wires I share): Manditory wires: +12v for the video display in the goggles GND I2C SDA (3.3v) I2C SCL (3.3v) Composite Video Mono AudioOptional wires: +3.3V for the headtracker sensors (Could be obviated with an LDO regulator from the 12v GND (paired with +3.3v) GND (Video-) GND (Audio-) Other requirements: the cable needs to be flexible, light weight, and possibly coiled to handle head movement without discomfort or tangling. the cable needs to be commodity (this is a personal project) the cable needs to be reliably disconnectable/reconnectable I think I understand the various factors in play (crosstalk, emf, capacitance, etc.) but I don't have any perspective on their importance for a 4ft cable run. My back-of-napkin guess would be to use an ethernet cable (CAT 5), wired like this:Pair 1: Audio+/-Pair 2: Video+/-Pair 3: I2C SDA/SCLPair 4: Power +12V/GND (Use an LDO for the 3.3v on the sensors.) Is this sensible? What factors should I be most concerned with? Which signals will be most impacted by connectors and wiring choices? <Q> Your main considerations should be as follows for your cable: <S> Shielding/grounding for the composite video portion and other low-level signals Impedance for signals with content above 16 MHz (see calculation below) <S> For a cable run of 4 feet and assuming a typical velocity of propagation of ~66% for the cable, the one-tenth wavelength rule says you can ignore transmission line effects below about 16 MHz. <S> I don't know enough about the frequency content of your I2C lines to give a recommendation there, but analog baseband composite video will have frequency content below 8 MHz in all regions of the world. <S> If you decide to go with a Cat 5E Ethernet product, try to find one designed for continuous flexing. <S> Those types of products are used in industrial robotics products. <S> For the composite video portion, you have two choices: <S> Make a composite cable (cable made up of multiple elements) yourself with a coaxial element and your Ethernet element. <S> Use a 100-ohm twisted pair from the Ethernet with a balun. <S> If you decide to make a composite cable yourself, find a 75-ohm video product designed for continuous flexing (long flex life). <S> Combine it with the Ethernet element in a braid. <S> To maximize your flexibility and the flex life of your finished product, you want to use flexible materials (soft plastic and many strands of copper) and construct things in such a way as to minimize the twisting and untwisting of the internal elements. <S> I'm a former applications engineer for a wire and cable company. <S> Here is a link to a general-audience presentation about wire and cable: http://www.belden.com/resourcecenter/cablebasics/upload/Cable-101.pdf <S> And here is a link to industrial products in a wire and cable catalog: <S> https://www.belden.com/docs/upload/Cabling-Solutions-for-Industrial-Applications.pdf <S> This 2002 book is also a good overview of the audio/video cable industry, even if it is somewhat outdated: http://www.amazon.com/Audio-Video-Installers-Pocket-Reference/dp/0071386211 <A> The video's most likely to pickup noise from the digital signals, audio at headphone power level won't pick up or interfere with any of those signals noticeably. <S> using a separate shielded cable for the video could be a win but you may find that an off-the-shelf video balun works well enough. <A> To be honest I don't think you will have much of an issue with simply using the Cat 5 other than flexibility. <S> The wires inside are solid copper so they are rigid and prone to wear after too much flexing. <S> The frequencies those signals (like the I2C) will likely be running at shouldn't cause an issue. <S> I'd only be concerned about power draw which I'm assuming will probably be less than 250mA in this case? <S> That's about the max depending on the gauge of the strands in your Cat 5 cable in particular. <S> Also, the twisted pairs provide isolation because they are somewhat perpendicular, which means less magnetic field is induced in its fellow pair.
| Flex life, i.e. how long the cable can be flexed before it breaks(very important for your application)
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What is this on power supply unit? I need to power a large 180W LED lamp that runs on DC so I got this transformer that is 12v × 15amp. But I don't understand the labeling on the terminals. There are 2 +V terminals but no -V terminal, why? What is the 'com' terminal for and why 2 of them? What is that terminal between 'com' and 'N'? I'm guessing ground. What grade/gauge/type of wiring must I use for safe and efficientoperation? Any additional equipment devices needed for safe operations? What is the purpose of 'output adjuster' knob? There is a knob inthe corner but it doesn't look like it can be adjusted by hand. Whatkind of tool is needed here? See image below. My aim is to use AC from the wall and turn it into DC to supply the LED. Please show me the correct path lest I get electrocuted. Somebody told me the 'com' terminals are -V output and said there are 2 of each -V/+V so you don't have to jam a thick wire in and instead use two smaller wires. This is a bit confusing. Please elaborate and whether this step is even necessary. If 2x -V/+V terminals present, is total current/voltage splitbetween the two? Can I use just one set of terminals to get fullvoltage & current, or must I use both? What happens if I use justone set? Please help. I don't want to hire an electrician for a job that may be very easy. Neither the LED nor the power supply came with any manuals or supporting documents. LED manufacturer said to use the above power supply. Additional info on power supply, Link to actual product page Item description is found at this link <Q> Disclaimer: <S> If you don't know what you're working with, I suggest you hire someone to help. <S> I know it's not what you want to hear <S> but it's better safe than injured or dead. <S> You only get once, maybe twice to seriously mess up. <S> From left to right on the back: The AC "L" and "N" probably mean "Live" and "Neutral" respectively. <S> The ground next to it is the Earth ground. <S> Just some clarification: <S> What you're referring to when you say V- is probably "ground" or to be precise, 0 Volts. <S> With respect to the circuit with the LED anyways. <S> COM and V- and Ground can all mean 0 Volts. <S> However, V- is typically reserved for actually negative voltages, like -10V, etc. <S> The reason why there is a set of two is probably because there are two "rails." <S> A rail acts as an independent supply so that you can easily isolate or distrubute loads so that two separate circuits don't interfere with each other or rob each other of too much power. <S> This is just a guess since there's zero documentation for this supply. <S> Also without documentation we don't know if it's possible to run those outputs in such a manner you can power the one 180W LED without some other external circuitry. <S> A lot of high power LED's I've seen run at ~30V <S> or so. <S> Power = <S> Current x Voltage, <S> so 180W = <S> Current x 30V, solved: 6A of current. <S> You'd set the output voltage of the power supply by using the adjust potentiometer "+V ADJ" before connecting anything. <S> According to the following table you probably need 12 gauge or larger wire (allowing more amperage to over-rate so the wires don't heat up easily). <S> http://www.powerstream.com/Wire_Size.htm <S> Also another consideration: <S> Be sure the LED has plenty of heat sinking otherwise it will burn up immediately. <S> Even 20W - 50W LED's can require heatsinks as large as a typical desktop CPU's (including a fan). <A> 1) "I don't want to hire an electrician for a job that may be very easy." <S> On the one hand, you can probably go to any electrician, bringing your power supply and a power cord, and he'll show you where the wires go. <S> For free. <S> On the other hand, how much is your life worth to you? <S> Seriously. <S> Mess this up and you can do serious damage. <S> Respect your limits and don't try for a Darwin Award. <S> 2) <S> If your system is really pulling 180 watts, you are asking for trouble by using a cheap 12 volt / 15 amp supply. <S> You can give it a try (and since you have it, I expect you will). <S> Be prepared for the supply to overheat or otherwise give you trouble. <S> Make very sure you have good ventilation. <S> Driving a power supply right at its rated output is asking for trouble, but you may get away with it. <A> It is quite common on power supplies to have a remote sense connection adjacent to output. <S> In most cases it may be simply strapped together at supply. <S> On a single voltage supply the output is between +V and com (common). <S> It sounds as though your common is also referenced to ground. <S> Measure -with no power connected- <S> with ohm meter to chassis to verify. <S> Do not have ANY connections between output and power line! <S> Doc
| The V+ and COM are the DC output terminals, which you could also say are V+ and V-, respectively.
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Why are high values of capacitance in low pass active filters not recommended? I am designing low pass active filter (Sallen-Key structure) and I have come to a point where I have to decide what should be the value of C (C1 = C2 = C) and then calculate values of resistances. I know that capacitance can not be very low because then I would have to take into consideration parasitic capacitances of other elements on the PCB. I also know that it is not recommended to take high value of C but I do not know why. <Q> High values of C would imply low values of R, therefore high signal currents. <S> In simple circuits (e.g. emitter followers) when the signal current is a large proportion of the quiescent current, you have to consider the device's nonlinearity (i.e. distortion). <S> In op-amp based circuits, as long as the opamp has sufficient bandwidth and drive capability, this is a lesser concern. <S> In either case, high signal current means high supply current, i.e. heat and wasted power. <S> However, go too far in the other direction, and low signal currents mean high resistor values, and higher noise ( <S> both shot noise from the lower currents, and Johnson noise from the higher resistances). <S> The right balance depends on the application. <A> Define high. <S> Anyway, there is a thumb rule: <S> the bigger is the capacitor, the bigger ESR and ESL it has. <S> Of course, ceramic caps are always better than electrolytes of same capacity, but the rule still works. <S> Maybe i could tell more if i had more info on the application. <A> High value capacitors have lower leakage resistance associated with them. <S> Certainly this depends on the type of capacitor used. <S> If you use paper/mica capacitors their leakage is relatively lesser but they are often too bulky. <S> So if you want to generate 1 Hz, it is not difficult, but generating 0.001 <S> Hz will be. <S> So the method used to use a digital divider. <S> Similarly row frequency (0.001hz) ramp is generated by digital means. <S> Some of the newer Farad capacitors may possibly solve many problems. <S> I have not tried these. <S> But repeatability of an oscillating frequency will still be a problem due to their tolerance and temperature dependence. <S> It will be a good exercise to attempt to make a ramp growing from 0 to 10 volt in say 1000 secs. <S> You will have to have a lot of patience to determine the repeatability if you change the temperature!
| Active filters can amplify the capacitor by miller effect and effectively solve lower frequency issues with not so high a capacitor.
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Low leakage current methods of initializing pin inputs at power up Suppose I have an IC that has a standard level triggered input pin. I need this pin's state to be setup immediately at power on at a time when my MCU's GPIOs may not have established levels. What are my options for accomplishing this while still being able to control the pin level at a later time? My way that of course comes to mind is using a pull up/down resistor however that can have high leakage current. During the times my MCU is online I will have no use for the resistors as the pin state can be set by the MCU so this leakage is highly undesiable. Are there any options i'm not thinking of? simulate this circuit – Schematic created using CircuitLab <Q> Also i think actel's igloo does the job with megaohm pull resistors. <A> It could probably be done, but the details would be very dependent on the actual circuit. <S> For example, suppose the microcontroller is reset by a supervisory circuit- <S> you could have the pullup enabled by a hardware delay that stretches the reset signal. <S> After the delay is over the pullup resistor is disconnected and the pin controls the input directly. <S> Another trick that is a bit more general and may not require any extra port pins is to use a keeper circuit- a positive feedback non-inverting gate that has small enough phase delay it won't oscillate. <S> Also called a bus hold circuit... some parts ABTH type have it built into the input. <S> As you can see the resistor draws negligible current if the input is open or tied to the high or low rails statically. simulate this circuit – Schematic created using CircuitLab <A> If the peripheral has a reset line, you could connect that to I/ <S> O on the micro (you'd probably want to anyway), and then you could assert the state of these setup lines from the micro while performing a reset on the peripheral, at some point after power-up. <S> Are you absolutely sure the consumption via these pull-ups actually matters in your application?
| I once use megaohm pull downs and very low leakage mosfets as a cmos buffer.
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What destroys a LED in the reverse direction? When a LED is included in a circuit that applies a reverse voltage that exceeds the reverse breakdown, a reverse current can flow and the LED might be destroyed. But what is it that actually destroys the LED: is it the reverse voltage itself, or is it the reverse current that is made to flow, or is it simply the overall power dissipation caused by the reverse current and voltage exceeding the device rating? Or something else? So, for instance, if I connect a 12 volt source in reverse to a LED that breaks down at 5 volts, via a resistor, the passage of reverse current will cause a voltage drop across the resistor which in turn can limit the voltage on the device to its reverse breakdown value (and thereby define the current that would flow) - rather similar to what happens in the forward direction. Would this in itself destroy the LED as long as the total power was within the LED rating? Normally of course, one would place a regular diode in parallel with the LED in the reverse direction to limit the LED reverse voltage to 0.7 volts or so, but there may be situations where this might not be possible or economic to do. I am just trying to understand how much circuit design flexibility I might have to meet different requirements. And if it is possible to expose a LED to a reverse voltage, what precautions should be taken to avoid damage, and which spec parameters are relevant? <Q> I have seen heterojunction LED failures that appear to be partial. <S> Failure with DC current in reverse is probably related to power dissipation, but it might be unwise to depend on it. <S> Breakdown could be quite high, hence the allowable current could be quite low (maybe less than 1mA). <S> Safest is to follow the LED data sheet recommendation- <S> usually 5V or so reverse is guaranteed. <S> Many types of LEDs have much higher actual reverse voltage breakdown (perhaps 15V to 70V), but it's unwise to depend on it- the LED maker could change the chip supplier or process or purchasing could go to a different vendor. <S> It's not really a good idea for efficiency to make the supply voltage much higher than the sum of the series LED forward voltages (often, but not always, just one LED is used). <S> For example, you might use 5V for a single 2-3V LED array, or 12V for an array of series strings with 6-9V per string. <S> Since the individual LEDs can take 5V each (usually guaranteed), you'd be fine in either case. <S> See this nice instructables gif : <A> Most datasheets from famous and respected manufacturers including Vishay and others, show breakdown voltage of 5V and reverse current of 10 to 50uA. <S> That is not true. <S> I just tested white, red and green leds, with a 0-30V power supply and a 1k resistor in series, measuring voltage across the led, and current in the circuit. <S> This is the result: White 9V=0.4uA, 13V=1uA / Red 5.3V=0.3uA, 6.7V=0.5uA, 12.5V=1uA / Green 5.11V=0.3uA, 6.5V=0.5uA, 9.9V=1uA. <S> So, no led presented anything close to half a microAmper at 5V as stated the manufacturers, and only right after 10V they presented ONE microAmpere. <S> That is not enough to damage the component. <S> According to what I read, it would need a minimum of 1mA reversed to increase the depletion area. <S> Wagnerlip <A> Diodes have properties which create what is known as the depletion region. <S> This is the barrier which prevents current from flowing across it with a forward bias until the depletion region is minimized (forward voltage dropout). <S> Reverse biasing a diode increases the depletion region, acting like a one way door. <S> However if you apply enough voltage to it, the mechanism breaks down and current flows either way, usually after the PN junction is shorted and the diode is effectively destroyed. <S> Basically what destroys the diode is power dissipation, or whatever causes the diode to be physically altered. <S> The reverse bais case may typically exceed the normal power dissipation of the forward bias case before it damages the device. <S> However there are certain types of diodes like zener diodes which are made to break down in reverse at a specific voltage, which makes them useful as voltage references and limiters. <S> For the 12v case on a 5v LED in theory using a limiting resistor to reduce the current (and drop the reverse bias voltage anyways) shouldn't, in theory, destroy the LED. <S> Some are more forgiving than others. <S> For your last question I'm wondering what scenario you'd have reverse voltage applied to it? <S> Usually there is protection applied before it gets to the LED. <A> LED's are still diodes, in reverse bias they will avalanche. <S> (Though I've not done anything like an exhaustive search, I've never found an LED that breaks down in reverse at less than 20V.) <S> It's my guess, that it's the heat/ power dissipation that would kill an LED in reverese. <S> So as long as you limit the current such that the power is less than ~10 <S> mW <S> LED's can handle reverse bias, IME <A> In my experience the standard 3mm and 5mm red LEDs can block 12V easily <S> so I have used them as reverse polarity protection in 12V systems where current is about 10 mA DONT use them on a 24 Volt system some LEDs died at approx. <S> 30V
| The typical situation where LEDs are exposed to reverse voltage is when they are operated in a multiplexed configuration- they will see up to the supply voltage in reverse. ESD appears to cause damage due to hot spots or some other localized damage.
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Computer peripherals to immediately (<5ms) detect digital input state change in C# programs The C# application I'm working on needs to be able to be able to call a function within 5 milliseconds of a digital input signal's state change (0V = off, 5V = on). Rather than using software to repeatedly poll that input's state, I would like some hardware to simply call a software function when the hardware detects a state change (like how hardware interrupts work). Does anyone know how to do this (what ports to use) on a standard PC tower? Or does anyone know of any hardware peripherals that allow this functionality? Occasional latencies are fine. I need <5ms average. I basically need to take as many samples as possible of the input digital waveform. (preferably in the C# program, rather than in a peripheral micro-controller, although I am open to suggestions). For instance, I found out the Labjack U3 can poll its digital input in less than a millisecond. This is a valid answer to the question, but I am interesting in more answers, especially ones that don't use polling or poll faster than 1ms. High input impedance would be nice (digital inputs). <Q> Do you need to perform a specific real-time action, or just record the event? <S> If the former, you will never accomplish this using commodity hardware and consumer-grade interfaces. <S> A user-centric OS won't interrupt a disk write just because the USB bus wants some attention. <S> If the latter, get yourself a signal recorder and post-process the buffer output at your convenience. <A> If your signal does not have too much of DC (long pauses with no data), use the sound card. <A> I ended up using an Arduino Uno's Timer Input Capture Interrupt Service Routine to set data over USB to a C# SerialPort object immediately upon input capture. <S> The SerialPort fires a serialdatareceivedeventhandler immediately upon detecting an End Of File Character (0x1A), and a new thread launches to process the input data (such as assigning a timestamp to it). <S> To all the neighsayers, <S> What the Heck? <S> Obviously you were wrong that it couldn't be done.
| You will need both a laboratory-grade signal processor ( it will be a card that can access hardware interrupts directly) and a real-time operating system like VxWorks.
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find VCC, GND and IN in relay I`m new to electronic stuff, I was reading an artice on how control AC light using arduino, In part of solution there is a relay module which has a VCC and GND and IN and COM and NO and NC, but I don`t have that module instead, I have a relay like this: can you please help me to find VCC, GND and IN in my relay? Thanks <Q> To find out which is which, use a ohmmeter. <S> There will be some finite resistance between the two coil leads, probably a few 10s of Ohms, maybe up to 100 or two Ohms. <S> The resistance between each of these pins and any of the other three will be infinite. <S> The COM and NC pins will be shorted together (well under 1 Ohm). <S> The NO pin will have infinite reistance to all other pins. <S> Until you activate the relay, you won't be able to tell the differnence between the COM and NC pins. <S> Guessing from the layout of the pins shown in your picture, the coil pins are the two at right. <S> This guess is based on the fact that there are two of them grouped separately from the others. <S> Relays are sometimes used for their electrical isolation between their input (the coil) and their output (the switch contacts). <S> As such, there is often a larger physical distance between these two sets of pins. <S> Flip the relay over and look at what is printed on the top. <S> This will often tell you the voltage or current the coil requires to activate the switch, and what the switch is rated for. <S> You aren't going to be able to drive this relay directly from the digital output of a microcontroller. <S> A simple low side NPN switch is usually the easiest way to control a relay from a digital output. <S> There are many questions here already on that topic, so no need to repeat the details in this answer. <A> Here's your relay's datasheet: <A> Ouch. <S> Your Arduino most likely can't drive that relay directly. <S> You are probably better off purchasing a ready-made module that has the appropriate driver transistor and <S> back-EMF diode already installed. <S> You <S> CAN do it yourself <S> but I'm worried about you not being able to work with AC Mains power safely. <S> If you don't have somebody local to you to help you through this project, you are better off with the pre-manufactured module. <A> You have made an understandable mistake. <S> The relays in the link you provided do not have a Vcc connection, and they have only 5 pins. <S> The relay MODULE (the complete PC board with relay intstalled) does use 6 pins, and does use Vcc. <S> The MODULE has, in addition to the two relays, two ICs (8 pins each). <S> These ICs are driver chips, and they are what actually drive the relays. <S> The VCC input provides power for the drivers and the relay coils. <S> As a matter of fact, if you look at the Arduino connection block, you'll see that it has VCC, GND, and 2 inputs, 1 for each relay. <S> So the module as a whole has 10 pins. <S> And I have no idea why your module can run without connecting ground. <S> It may be that you are connecting it to the power supply ground that also connects to the Arduino, but in that case you are, in fact, electrically connecting it to the Arduino.
| Your relay has 5 pins, which are almost certainly the two connections to the coil, and the common (COM), normally-open (NO), and normally-closed (NC) connections of the switch. What you have in your picture is a plain old relay.
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Do batteries in series equalize charge If I connect 8 batteries in series and two of the batteries are fully charged, while the others are partially charged, do the batteries eventually reach some sort of equilibrium? <Q> Not generally a good idea, but it depends on the battery type. <S> NiCd is very tolerant of overcharging, and will equalise ok if you keep the current below C/10. <S> Other technologies will not work. <S> Worse than charging an unbalanced pack is discharging it. <S> The empty cells will be forcibly drained far past their safe minimum voltage, possibly until they are inverted. <S> This will cause permanent damage, even if done at low current. <A> No, a single series string will not equalise the cells in the normal course of charge/discharge, although the degree to which this is a problem and the degree to which it can be dealt with varies by battery chemistry - from not at all, to a reasonable amount. <S> Installing unequally charged cells in a series string is just plain bad practise - you'll hit over- or under-voltage limits on the over-/under-charged cells way before the others do, and you'll likely damage them. <S> Multiple single cells can be parallel connected, and they will equalise in voltage, although the current that can flow when they're initially parallel connected can be extreme, and may need to be mitigated with series resistors (and if you're talking about big batteries, then you're talking about beefy big resistors). <S> What's more, due to manufacturing & chemical & temperature variations between one cell to another, you can't expect all the cells in a string to stay at the same state-of-charge / voltage after multiple charge/discharge cycles. <S> That's when cell equalisation becomes a necessity. <S> Without knowing anything more specific about the chemistry & size of the cells you're referring to, that's about all that can be said. <A> It's best to not do this because the charged cells will get stressed out while the flat ones are being brought up. <S> From a theory viewpoint charging the cells independently is always best. <S> If you must do this, keep currents very low and don't even think about lithium.
| Multiple strings of series-connected cells that are connected in parallel will equalise with their peers on the overall series-string length (i.e. String-cells-1+2+3 in parallel with String-cells-4+5+6 will eventually equalise to the same voltage), BUT within the series string the cells will not equalise to each other by themselves, and you have the same original problem that some cells will reach full when charging, or flat when discharging, before their peers within the series string, and thus be damaged. Lithium batteries even have separate 'equalising wires" connected to each cell, so the charger can adjust the current individually.
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Forgot to put a via near bypass capacitor, and now the boards have been fab'd - what can I do? I goofed and didn't notice this until the board was already fab'd and assembled. The board is an RF amplifier; the portion I have pictured is a part of the DC control pat (so no RF is nearby but we're talking 100MHz-1GHz so it's surely floating everywhere). See the possibly disastrous screenshot marked by 'missing a via here'. (The fab removed by hand the huge trace to nowhere, before anyone asks). I really need to be more careful with altium's polygon pours... I'm really kicking myself right now about this stupid error, it's a 20 board run and money is really tight; I'm in academia, so these boards are not getting remade. The problem is C18 is a 100nF bypass cap for a high speed op-amp. It seems to me that without a via to the ground plane, theres only that ultra tiny slice of the pour connecting it to a via 'very far' away. I might be wrong, but from everything I've read, the cap might as well not even be there because the inductance will be so large. I don't have the boards back yet, so the fab might even have eliminated that small trace entirely! It's only a few mils thick. Maybe I'm being overly worried, as I don't yet know if this will cause issues. But is there anything I can do 'by hand' to improve the decoupling? Would soldering a small wire to ground be effective? I guess my main concern is oscillation with RF signal floating around everywhere; the op amp I'm decoupling is the LME49990, and I've seen this thing oscillate when bypass caps arent placed right. <Q> Stripped wire-wrap wire works well. <S> It's 30 AWG, so <S> the hole that you drill can be very small. <S> Use a Dremel tool in a drill press if you have such available - I routinely drill #78 holes reliably without breaking the bits. <S> Your can modify all of your boards in fairly short order. <A> I think you can most likely add a little fly wire from the existing GND via (or a closer one above) to the capacitor. <A> Given how close the nearest ground region is, you probably don't need to use any wire or drill holes. <S> In the picture below, the points in the green circle are both Ground, in which case all you need do <S> is scratch off the solder mask nearby and add a solder bridge. <S> This will create a nice large (low inductance!) <S> connection between the C18 ground pad and the rest of the plane. <A> The inductance between a bypass capacitor and the power supply is nearly irrelevant. <S> The entire point of a bypass capacitor is to supply brief bursts of energy to some IC through a low-inductance connection. <S> It's the inductance between the IC <S> and it's associated capacitor that is important. <S> The suggestion to "mount a through-hole capacitor (with leads cut to the minimum) on top of the chip." <S> (-- Wouter van Ooijen) gives close to the minimum loop area, and will be the best you can do -- it appears that will give even less inductance than adding a via to your design and refabricating your board. <S> I see there is already a good connection between the high side of the capacitor and one power pin of your <S> IC.I agree with Spehro Pefhany that a short bit of wire on the low side of the capacitor will almost certainly fix your problem, but I would connect the other end of that wire to the GND pin of the IC to minimize the inductance between the capacitor and the IC.Such jumper wires are extremely common in commercial PCBs and in space-qualified hardware; one hopes that if it's good enough for NASA, it's good enough for your application --see <S> "Is Rework Unacceptable?" andp. <S> 1 of "NASA Workmanship: jumper wires" <S> .It is unclear if piggybacked parts are good enough for NASA --p. <S> 16 of "NASA Workmanship Standards: through-hole soldering" seems to say no,whilep. <S> 3 of "NASA Workmanship Standards: Deadbugs" seems to say yes. <S> If you are lucky, that GND pin on that IC already has a connection to the power supply that is more than adequate for slowly recharging the bypass capacitor between bursts; if not, a second wire attached according to (a) Spehro Pefhany's suggestion or (b) directly between that GND pin on that IC to some nearby GND point may be needed -- the difference between (a) and (b) is nearly irrelevant. <S> EDIT: <S> Many chips don't have a GND pin, or even when some pin happens to be connected to GND, they get their power from some other pin. <S> For such chips -- for example, an opamp that gets its power from +15VDC <S> and -15VDC -- the bypass capacitor should go directly across the power pins -- in this example, the capacitor should go directly from the +15VDC pin to the -15VDC pin. <A> First, try as it is. <S> If you have problems, content a wire from other pad. <S> The dremmel advice is cool, but before you ruin your board, try patching conventional patches. <S> I think all you may need is a wire from other gnd pad and maybe additional capacitor. <S> But probably it will just work. <S> Many designers put there a serial ferrite or inductor, so just close your eyes and imagine this is what you did.
| Drill a small hole that will allow you to simply run a free thin piece of wire from the ground plane on the opposite side of the board to the capacitor. If you're having problems, do what Dwayne suggests and drill a hole and scrape the resist off, but you've got little to lose in trying the easy way first, especially if this op-amp is not directly handling the high frequency signals.
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Insulation resistance and leakage current in capacitor I am using the capacitor in a battery powered device but unable to find the leakage current. They have mentioned the insulation resistance minimum as 10 GOhms. Can I take this into ohms law to find out the leakage current? Capacitor link <Q> Yes you can but that would be useless in my opinion. <S> The 10 GOhms is what the manufacturer guarantees, so it can be 20 GOhms butalso 100 GOhms. <S> In general your battery will have significantly larger internal leakage ! <A> Insulation Resistance is a measure of the ability of the dielectric to withstand the passage of electrons through itself. <S> Insulation resistance is usually expressed as a value of "ohms x farads". <S> Please be sure that whether the given value is "10Gohm.uF" rather than simply 10Gohm. <S> Then you can use these equations to calculate the leakage current. <S> -- <S> > DCL = <S> (Voltage(V))/(Insulation Resistance(IR)) <S> = <S> V/IR IR = <S> Internal Resistance Spec./Capacitance Take this example,47µF commercial X7R MLCCs for power-rail filtering, at a working voltage of 5V. Insulation resistance specification of 500MΩ.µF IR = <S> 500MΩ.µF/47uF = 10.6MΩ <S> DCL <S> = 5v/10.6MΩ = .47uA <A> Yes, you can assume R > 10G ohms, so the leakage current is less than <S> Vtest/10G. <S> You should make sure that the voltage you care about is equal or less than the test voltage the manufacturer uses. <S> If your circuit is sensitive enough to care about that small a leakage, it will likely be troublesome in other ways.
| In general you do not need to worry about capacitor leakage current provided you observe the polarity (for electrolytic types).
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Lab power supply to breadboard Recently I purchased a lab power supply, with banana jack output. Are there any specialized products that snap onto the breadboard on which you can connect banana jacks to deliver power to the board? Any suggestions to products or DIY solutions are appreciated. I have looked at the MB102 breadboard power supplies, but they use a USB or 2.1mm jack connection to power the board. I am specifically looking to power it from a lab power supply. Thank you! Edit: I must add that I have a simple 830 pins breadboard without any binding posts, so I am looking for something like a MB102 module but for banana plugs. <Q> Nothing great. <S> For breadboards with female bananas binding posts on them, you just run the wire to a bus. <S> https://cdn.sparkfun.com/assets/4/e/b/7/3/518c07b8ce395fea62000001.jpg <S> The latter are not extremely robust arrangements, but solderless breadboard circuits, in general, are not the most robust of prototypes. <A> One solution I've used in the past to robustly connect to a breadboard is to fit a Molex KK connector with locking ramp using superglue. <S> You can then make up a cable from the KK connector to whatever you want at the other end <S> (banana plugs in your case). <S> You need the superglue otherwise, whenever you try to unplug the cable from the breadboard, the connector comes out instead. <A> You can get banana leads to alligator clips online. <S> Then also get breadboard pins (also known as PCB pins) which plug into the breadboard. <S> https://www.jaycar.com.au/0-9mm-pcb-pins-pk-50/p/HP1250 <S> You can then clip the alligators onto those. <S> Also a good idea if you need leads for an oscilloscope or other equipment which uses BNC plugs. <S> In that case, get BNC to alligator clip leads and use the same method of clipping them onto the gold pcb pins which you plug into the board. <S> The parrot clips and banana binding posts running to rails mentioned above are also good ideas <S> (might try it myself sometime). <S> There's many ways to do this, it's just about personal preference and what works best for you. <S> These options give you plugs which go into an individual breadboard socket of course, not any kind of module. <S> Can be useful for op-amp projects when you want to have multiple input values. <S> Hope that helps.
| If you want to make up your own leads to avoid breaks in the connections, you can also buy breadboard jumper leads with plugs on both ends, cut off one side and crimp alligator clips on the other ends. Otherwise, Male banana to test clips are probably the best bet/ from www.apogeekits.com/images/banana_to_grabber_leads.jpg
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The best way to judge the capacitive and inductive load I've a ADC sampling system. What I can get from the connected load is the sampled data of the output voltage and output current. I want to know if the load is capacitve or inductive from these datas. For a pure AC wave, I can get the information from the time of the 'zero crossover' point. But in my situation the AC may contain high frequency harmonics. Does the 'zero crossover' method still work? Is there other (faster) method to do this work? <Q> The most popular method to obtain harmonics from a signal is digital filtering. <S> Most method use Fast Fourier Transformation (FFT). <S> You can google it, because the theory is quite broad. <S> The discipline which covers transforming analogue to digital signals and use them later is called Digital Signal Processing (DSP). <S> The literature in this topic is also broad, there are some free books available online (you may start with this one ). <S> The main idea of this approach is to use Taylor series to obtain all harmonics, including DC component (the 0th harmonic). <S> The highest harmonic one can measure is determined by sampling rate (see Shannon theorem ). <S> You should convert samples of voltage and current to their complex form (see phasor ) and divide one by another, so you can get the complex impedance, which is easy to calculate the angle. <S> If the phase shift is the only thing you are interested in, you can just subtract the phase angles of voltage and current. <S> For any n-harmonic (for $$n \geq 1$$) the characteristic of load does not change. <S> So for any capacitor, for any harmonic, current is always leading, and for any inductor the current is always lagging. <S> There will be differences between the phase shifts for each harmonic. <S> The reactance of an inductor, for n-th harmonic is $ <S> $X = n <S> \omega <S> L $$, <S> while for a capacitor it is always $$X = \frac <S> {1}{n \omega <S> C}$$. <S> Because it is numerically easier to handle higher values than lower ones, you may check for each harmonic (for example first 10), for which one the voltage and current are the largest and then divide them. <S> However, you should get correct results even for the 1st harmonic. <S> UPDATE <S> In general, there is a possibility that each harmonic has different phase angle. <S> If they are combined, there can be many zero crossings in a single period. <S> You will not know which one crosses the zero in which moment, if you do not decompose them. <S> You can also miss the zero crossing, for example there is large DC component (say 10 Volts), and low 1st harmonic (say 1 Volt). <S> You will never get a zero-crossing in this case, but it is possible to measure the impedance for the 1st harmonic (and thus its angle). <A> If you have both current and voltage waveforms, you could filter the output digitally (or analog, but it's difficult to get two filters which are exactly the same), and remove any high frequency components. <S> There are plenty of digital lowpass filter design sites on the 'net, which even should you the code you have to use in the program. <S> You can even use one of the many audio processing programs which can apply filters. <S> After the filtering you use the phase difference to see if the load is capacitive (current is leading the voltage) or inductive (current is lagging), and by how much. <A> Use the load voltage signal to produce a -90 degree (lagging) shifted reference waveform (maybe with a couple of low pass filters). <S> The lag produces a reference waveform that "looks" like the load current of a pure inductor. <S> Take this reference and convert to a logic square wave using a comparator/schmitt trigger. <S> Use this to control an analogue switch. <S> The signal input to the analogue switch is a voltage representation of the actual load current. <S> The analogue switch is now performing synchronous rectification of the load current representative waveform. <S> If the output waveform averages at a negative value then the current through the load is capacitive. <S> If the average output is a positive voltage then the load is more dominated by inductance. <S> Here's a picture of an input sinewave and how it can be "synchronously rectified" by a square wave: <S> - The example above is for a -90 degree shifted reference sq wave - the output has an average value of zero and the load is therefore resistive.
| In your case the zero-cross method is not valid.
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Is it possible to measure energy consumption from power socket? I wanted to know if it is possible to measure how much energy is being consumed in the entire house by plugging in a power meter in any power outlet in the house? I am aware of some portable meters but they can only measure energy consumed that specific appliances but not of entire house. Sorry if the question sounds stupid, i am new here <Q> Energy is the accumulation (or integration) of power over time. <S> Power is voltage x current. <S> To measure current you need to have something that is "in-line" with the current flow. <S> You can't do that from any particular outlet. <S> If you can't determine what the current taken is you can't calculate power. <S> If you can't calculate power you can't determine energy consumption. <A> I once heard someone describing such a device and was very confused. <S> The usual method is something like the Kill-A-Watt, which goes between the appliance you're measuring, and the wall. <S> But this can only measure the consumption of something plugged into it! <S> It turned out to be possible. <S> A smart meter (mounted at the main board) communicates over the power lines with an indoor terminal, plugged into any outlet in the house. <S> Here is a picture of one: <S> They are most commonly used for prepaid meters, where the metering and control is done outside on the pole, and the user just plugs the indoor terminal into any wall outlet. <S> It then displays the power consumption of the whole house, and allows credit to be added to the meter. <S> In some cases it might be possible that even though you have a regular billing electricity meter, you have the option of buying a plug-in unit which reads the consumption from the meter. <A> Unfortunately not. <S> It's just like the plumbing in your house - you can't tell how much water is flowing out of the shower by looking at the tap in the kitchen. <S> It is exactly the same with electricity. <A> the better and safety way to measure current is to use a coil surrounding the wires comming from outside. <S> You can found split transformers (i have forgotten the right name) on ebay for less than 10$
| You have to either put a meter on the outlet you want to measure or install one on the main pipe into the house.
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Do LED have an unproductive base power consumption like incandescent bulbs? To generate light from an incandescent body, like a tungsten filament, the body needs to be heated up to a temperature minimum needed to emmit light from that material. Up to that minimum temperature, only heat is emitted. Above it, heat and light is emmitted. What is the related behaviour of LED? When the current increases from 0 A to the specified standard current, I assume there is some range where the LED does not emmit any light. What happens in that range - I would guess it's not behaving just as a resistor? <Q> Early LEDs had a kind of knee below which you didn't get much light out of them, and thus a sort of threshold current (maybe 1mA). <S> Most modern LEDs will output light with microamperes of current, and are reasonable efficient at low currents. <S> As current increases the efficiency will peak at some current then decrease as I^2R losses increase, but I think it's pretty constant from (say) <S> 10% (maybe even 1%) to 100% of the recommended operating current. <S> Here's an early (2001) white LED with a relatively high minimum current (from this web page ) <S> And another that is better (2007 Nichia) <S> Modern LEDs are almost never operated at the optimum level for efficiency - the main thing is to get many lumens out of the expensive semiconductor as practical. <S> You'd get more attractive looking curves if you plotted lumens/mA <S> but that's not representative of energy efficiency- <S> the forward voltage drop increase with current. <A> Let's take a look at a typical LED datasheet . <S> Figure 4 showers relative luminous intensity versus applied current. <S> It is assumed that the voltage is fixed here and the diode is forward biased. <S> We see then that any amount of current creates visible light. <S> So an LED does not have the same dead area of an incandescent light bulb. <S> However, seeing the non-linear curve, we recognize that there is an optimal current for lighting efficiency. <A> You get this kind of graph. <S> It's pretty linear all the way down. <S> It behaves like a semiconductor junction. <S> A particular voltage level is needed to push charge across the band gap. <S> However, every electron-hole recombination emits a photon. <S> This is independent of temperature and everything else, so there's no real minimum. <A> Old LED technology had surface leakage issues and was inefficient so we would always drive them <S> well I have used standard general purpose optocouplers in the microamp range proving that the LED must be making light
| There is no real minimum LED current needed to get light output in modern LEDs due to their extremely low leakage
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Required subjects to understand FPGAs I would like to get into FPGAs. I'm a computer engineer student and I have knowledge of electronic, electromagnetism, circuit, architecture, microcontroller, software development... but I studied them only in a summary way. What are relevant and preliminary topic I must know to understand FPGAs? e.g. I found that Digital Design is an important aspect of the story. NOTE : I'm NOT asking about FPGAs itself. I'm asking about things AROUND FPGAs, that are requirements to understand it. <Q> You mentioned that you have some background in electronics. <S> Did you ever have to study transistor logic? <S> Did you ever make <S> AND OR NOT and other logic gates out of transistors? <S> Did you work with theoretical logic circuits where you were given several gates, inputs, and you had to calculate what will be on the output? <S> This is what an FPGA is all about - field programmable gate array. <S> Thus, it's just an array of logic gates which you could utilize to build a system off of them, which will do some simple or complex function. <S> To make it easier, you can use hardware description language, which can be then synthesized and implemented as the transistor logic on an FPGA. <S> Thus, you need to know logic design techniques. <S> What are the counters, flipflops, memories, types of logical operations, etc. <S> , and most important <S> - how can you use all that to create a system that can perform desired function. <S> As mentioned in the other answers, the FPGA is a pure logic device and you can not work with analog signals directly. <S> Instead, FPGA is usually accompanied with different peripherals including ADCs and DACs, and certainly an oscillator that provides the FPGA with the logical clock - that you need to know as well. <S> There is certainly more to it than just digital logic design, however I would suggest to get a strong grip on it first. <S> Eventually, the rest of the aspects will come as you learn more. <A> Have a read of FPGA programming, where to begin ; it does what it says on the tin. <S> There is some ambiguity in your question, and I'm not sure if your question is too broad - "I'm asking about things AROUND FPGAs" <S> - what do you mean by 'around'? <S> An FPGA by itself is somewhat useless. <S> It's not a microcontroller, so it doesn't come with practical things like Analogue-to-Digital convertors, Digital-to-Analogue convertors, PWM controllers, UART/USART/SPI/I2C/I2S/CAN communication modules, etc., <S> and there's no real memory/storage on board. <S> Some of them can be written in code, but some will always require external circuitry. <S> ADC is (probably) the most popular of add-ons that people need. <S> As an example, FPGA development boards seem to always have a Wolfson audio IC [example datasheet] included which allows you to connect a line-in source, line-out/headphones, and PC-type microphone to the FPGA via a choice of comminucation protocols. <S> Have a read through the above datasheet and it will help you to understand the concept of integrating an external device, and just how simple it is. <S> From there it's only a small jump to interfacing the FPGA with another type of IC, like a FLASH memory or EEPROM IC - it's all about the communication between the two. <S> You read the datasheet and then implement a controller in software on the FPGA. <S> I hope the above has been in the direction of your question. <S> If not, please edit your question to be more specific. <A> [GOOD COURSE ON THIS TOPIC] , hardware programming languages such as VHDL, Verilog and a solid knowledge in electronics generally. <S> Here is the list of most famous FPGAs manufacturers. <S> So, choose&buy one, and start with the research. <A> I know how the website is against posting links, but I don't think there is a short answer to your question. <S> Basically you would want to go over this class: Introductory Digital Systems Laboratory . <S> I'm sure your university also offers a class (or two) that covers the same topics. <S> During my studies I had 3 classes covering the core materials required - Intro to digital systems, Digital systems & VHDL. <S> Intro is about Boolean algebra and simple logic gates and how to implement and minimize simple functions. <S> Digital systems is about more advanced topics, such as advanced implementations and components. <S> VHDL is where you bring almost all of this material together to create something useful in today's terms. <S> Good luck!
| Things you need to know to get started with FPGAs are digital electronics circuits and their working principles
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Use a single-output floating supply as a dual relative to ground I have a single output 15V supply that is isolated with respect to any ground. I'd like to use it to produce something like +12V and -2V with respect to the ground of my circuit. The current required is less than 500 mA. Is this possible, and if so, what's the best way to tie it to my circuit ground? <Q> I'd like to use it to produce something like +12V and -2V <S> with respect to the ground of my circuit. <S> Depending on how much current you might be expecting to take from the 12V (or the -2V) to 0V (ground) <S> a beafier version of the following may be needed. <S> From the 15VDC uses a voltage regulator to produce 2V relative to the most negative lead on the 15V DC supply. <S> Connect this output to 0V on the rest of the system. <S> You now have a -2V rail (formerly the most negative lead on the isolated 15V supply). <S> By inference you also have a +13V rail that can be regulated down to +12V with a low drop-out regulator. <S> Or maybe you use one of these: - <A> Although the schematic above is nice, I think it's too complicated. <S> Also there's a transformer <S> and I only use those when I have to <S> ;-) <S> I would be looking at something like this: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> But this simple configuration would only work if the current in and out of the ground on the right was a few mA. <S> If you need to be able to supply more current through the ground one or more transistors will need to be added ending up in something like: BTW, in case you have an old fashioned audio power amplifier IC lying around (TDA 2003 etc) then you could use that instead of the opamps + push-pull stage. <S> As a bonus you will get short circuit and temperature protection :-) <A> Of course the LM7905 will dissipate about 1.5W worst-case, so a smallish heat sink would be required.
| Assuming most of the current flows from the +12V to GND, you could use an LM7912 to give you the GND rail and then the minus input would be at -3V, which you could regulate to -2V with another negative regulator (perhaps a TLV431 shunt regulator if you only need a few tens of mA on the -2V rail). And then a negative voltage regulator to get -2V
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Making prototypes with high speed SMD components Even today we have breadboard and strip board on which one can make "quick and dirty" circuits and also test prototypes. However, we have now moved into an era of predominantly surface mount components many of which work at very high frequencies. If one wants to make a circuit one may first need to design some simpler prototypes of fully understand how the complex off the shelf ICs work and determine their performance, sometimes we may find that the datasheets are wrong too. In this process if we find that a connection was wrong or the chip was fried, we can rewire/replace them easily when we have a strip board or veroboard, we can always use a breakout board for surface mount devices to generate pins from them. However, for circuits that work and high speeds, doing so is not possible anymore. How does one follow the prototype stage in this case? It seems that the process will become longer and more expensive and rather inconvenient if PCBs have to be respinned many times. <Q> There are techniques that can be used for prototyping. <S> If you're prototyping a whole system you're probably not doing it right- but to test smallish bits of analogish circuitry, it's practical. <S> It's good to go to a PCB layout early, but not necessarily as the first step. <S> Here's one hacked together circuit by a fellow I happen to know John Larkin- <S> (he's since moved to gold-plated for the boards) <S> And another (the high-speed section is kept very small) <S> This is done with shears, dental burrs etc. <S> The ground planes under everything mean that the circuits are fairly quiet. <S> You can also stack breakout boards for SMT chips on the ground planes (you can buy them or have your own made). <A> Well once you start designing real things you need to do it the real way :) <S> At a professional level, and even for home projects we spin boards for proto all the time. <S> You're not going to get clean power, clean signaling from a wired up breadboard. <S> As an example I recently made something very similar as a side project <S> PIC32 and some audio circuitry. <S> The 4-layer bare boards cost me about $200 or less and I bought a stencil too. <S> Then hotplate re-flowed the whole thing, including some very tiny QFN components. <S> It's much easier to make a board and blue wire <S> your mistakes then it is to blue wire a whole board and try to decipher the rats nest of problems in front of you :) <S> Another approach of course is to buy a pic32 eval board, and an eval board or breakout board from say sparkfun for your ADC. <S> Then connect those together as a prototype. <S> For tiny simple circuits I sometimes use surfboards (check digi for those). <S> Otherwise many people have come before you looking for ways to make cheap pcb boards faster for in house proto. <S> There's everything from etching it yourself, to cnc routing, and now even conductive ink printing. <S> In a case like yours I find just shipping it out and waiting 5 days is the most convenient. <S> Of course you can't get around the time it takes to design a schematic and layout the board. <S> But carefully designing your schematic should be your first step anyway :) <S> If you find that the expense of these boards is too much for you or your project <S> I'd suggest trying to simulate as much as possible, especially spice your analog input section. <S> Review your design early and often, and have someone else review it as well before sending it of to be built. <A> I'm in the same boat as you - I want to be able to quickly prototype high-speed circuits, similar to the ease and speed of a breadboard. <S> They're expensive and take time to design/make/order, so this obviously isn't ideal. <S> You just have to look at the requirements of a high-speed circuit to realize that it isn't feasible to "prototype" in the conventional sense: <S> Device footprints - <S> only PCBs give you the ability to mount small-pitch parts properly, with a soldermask and such. <S> Higher-speed parts lead to more difficult pagckages, like BGA Ground plane requirements - high-speed circuits <S> must have proper grounding, usually in the form of a plane. <S> In addition, your audio circuit example is a mixed-signal device, meaning the grounds need to be isolated to work properly. <S> Trace lengths <S> - Usually clock sources need to be as close as possible to the relevant IC. <S> This is also true of data buses and other signal paths that require strong integrity. <S> There may be one option for you, one that I have used in the past. <S> It's called Manhattan construction. <S> This technique is commonly used by RF tinkerers to build circuits. <S> Basically, you build everything over a piece of copper-PCB, which acts as a ground plane. <S> You connect discrete components by directly soldering leads together. <S> Here is a good post documenting the subject (you can also find more online). <S> You can then buy these pads for attaching SMD IC's. <S> I used it to build a simple DDS circuit that had a 50MHz clock and interfaced to an FPGA. <S> It worked pretty well, but you usually have to use through-hole components, which gives you some noise. <S> I'm like you; always looking for a way to prototype my high-speed designs fast and cheap. <S> But due to the nature of these circuits, it's difficult to do without a proper PCB.
| Unfortunately, I think PCB's are pretty much the only way to do things.
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BJT Biasing with negative voltage I got this question as a school assignment (1st year electrical engineering). I have to calculate the 3 unknown resistor values. The values which were given are in the schematic. Further \$\beta = 100\$ and the transistor is an NPN BC547B. The problem for me is the negative \$V_{EE}\$ voltage and the \$V_x = 0\$V. I don't really know where to start (already stuck for an hour). Can you give me some hints or an example so I know how to go further? I have already looked at a lot of BJT biasing examples but can't find one (I think) which really fits my problem. <Q> In all cases it's not the "absolute" voltage which matters, it's the voltage across the resistors that matters. <S> \$R_1\$ <S> \$R_1\$ and \$R_2\$ form a voltage divider with a small load <S> \$I_B = I_C/\beta <S> = 10\mu\$A. <S> You're given \$V_x = 0\$, so <S> the voltage across \$R_2\$ is <S> \$0 - V_{EE} = 15\$V. <S> By Ohm's Law the current through \$R_2\$ is \$15\text{V}/R_2\$, which is \$3\$mA. <S> The current through the divider is much larger than the load current, so if you just need an approximation you can ignore \$I_B\$. 1 <S> In that case the voltage across and current through \$R_1\$ is the same as \$R_2\$ <S> and therefore the two are equal. <S> Alternatively, you can use the voltage divider equation, which can be simplified by noticing that the voltage divider output is halfway between \$V_{CC} = 15\$V and \$V_{EE <S> } = -15\$V <S> and therefore the voltage is divided in half. <S> The simplified equation is $$\frac{R_1}{R_1 + R_2} = <S> \frac{1}{2}$$ <S> which gives you the same result, <S> \$R_1 = <S> 5\text{k}\Omega\$. <S> If you can't approximate, note by KCL that the current through \$R_1\$ is the current through \$R_2\$ (\$3\$mA) plus \$I_B\$, or <S> \$3.01\$mA. <S> Now you have the voltage across \$R_1\$ (\$15\$V) and the current through it, so you can use Ohm's Law to calculate it exactly. <S> \$R_C\$ <S> You're given \$I_C = 1\$mA and \$V_Y = <S> 5\$V, <S> so you know you have \$15-5=10\$V across \$R_C\$. <S> \$R_E\$ <S> The transistor is on since \$I_C = 1\$mA, so assume a \$0.6\$V (or \$0.7\$V) <S> \$V_{BE}\$ drop. <S> Since \$V_x = V_B = 0\$V, that means \$V_E = -0.6\$V. <S> This gives you the voltage across the resistor: \$V_E - V_{EE} = -0.6 - (-15) <S> = <S> 14.4\$V. Remember $$I_E = \frac{\beta + 1}{\beta}I_C$$ and both \$\beta\$ <S> and \$I_C\$ are given, so you know \$I_E\$. <S> Alternatively, you already know \$I_B\$ <S> so you can also use the fact that $$I_E = <S> (\beta <S> + 1)I_B$$ <S> Again, you have the voltage across and current through the resistor so you can use Ohm's Law to calculate the required resistance. <S> 1 <S> Typically the rule of thumb is that if the unloaded divider current is 10 times the load current it is safe to make an approximation. <A> Negative voltages are nothing to be afraid of. <S> They work exactly like positive voltages, just with a minus sign. <S> I'll walk you through the first part, which is to figure out \$R_1\$. <S> You know the voltage across \$R_1\$ is 15V. <S> If you find the current through \$R_1\$, you can calculate the resistance with Ohm's Law. <S> The current through \$R_1\$ is equal to the current through \$R_2\$ plus the base current. <S> Ohm's Law gives you the current through \$R_2\$: $$ <S> I_{R2} = \frac{0\ \mathrm <S> V - -15\ \mathrm V}{5\ \mathrm{k\Omega}} = <S> 3\ <S> \mathrm{mA}$$ <S> You're given the collector current and \$\beta\$ <S> , so you can easily calculate the base current: $$I_B = <S> \frac {I_C} {\beta} = \frac { <S> 1\ \mathrm{mA}} {100} = <S> 0.01\ <S> \mathrm{mA}$$ <S> So the current through \$R_1\$ is: $ <S> $I_{R1} = I_{R2} <S> + <S> I_B = 3\ \mathrm{mA} + 0.01\ \mathrm{mA} = <S> 3.01\ <S> \mathrm{mA}$$ <S> This is almost ( <S> but not quite!) <S> equal to \$I_{R2}\$ by itself. <S> In real life, you would probably ignore \$I_B\$ unless \$\beta\$ were lower, but this is homework, so let's do it the hard way. :-) <S> Now you can calculate \$R_1\$: $$R_1 = <S> \frac <S> {15\ \mathrm V - 0\ \mathrm V} {3.01\ \mathrm {mA}} = <S> 4.98\ \mathrm{k\Omega}$$ <S> I'll let you handle \$R_C\$ <S> (trivial) and \$R_E\$ <S> (harder) on your own. <S> Please be sure to post the work you've done if you have any follow-up questions. <A> I will just give you some hints since it is course work. <S> You can calculate Rc with the information provided, because you know the voltage on both sides of Rc, and you know the current flowing through it. <S> You can calculate Ib with the information provided. <S> After you do that, you can calculate the current flowing through R1, so then you can calculate R1 using the voltage and current. <S> In order to calculate RE, you need to assume a certain voltage drop across base-emitter junction of Q1. <S> You can pull that from the datasheet if you want, or you can use a typical value such as 0.6 or 0.7V. <S> But since the type of transistor is actually specified in the problem, I would probably go look at the datasheet to see if it shows a graph of typical Vbe vs Ib.
| Since you know the voltage across and current through the resistor, Ohm's Law gives you the required resistance.
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Testing batteries by measuring their short circuit current, why it doesn't work? I've experimented a little bit with some of my old, rechargeable batteries. (They "slept" around 5 years in a cabinet. I've long forgot their previous charged status.) What I did: I've measured their voltage. It was okay, maybe a little bit lower as their nominal voltage. I recharged them: I gave them around 12 hours in a battery charger. Unfortunately, this battery charger didn't give me any information, with the exception of a led showing, some type of charging actually happens. I've measured their voltage again. This time, it was better. Earlier I experienced that completely (or near completely) exhausted batteries give their nominal voltage measured by a multimeter, but it doesn't mean, that they could produce usable power on load. My idea would be, that if measuring the voltage doesn't show anything about their power, maybe measuring their short circuit current would. Yes, I know it harms the battery, but I think, for the some seconds while I read the current on the display of my multimeter, it is practically negligible. And knowing the voltage, and their short circuit current will show, if they are really usable or not. One of my friends (professional electric engineer) said, it is a bad idea, but he didn't explain clearly, why. So, why is it a bad idea? How could I test a battery, if I have only a multimeter? (P.s. they are NiMH batteries, if it matters.) <Q> As a secondary concern it may damage the battery. <S> That said, I do this myself briefly with alkaline batteries because I know that the cell or battery can produce only a few amperes into the 10A multimeter input. <S> Some battery types can produce much larger currents- enough to damage themselves <S> (heat up, vent electrolyte, explode, even), melt the test leads causing severe burns etc. <S> The proper way would be to apply a reasonable load (reasonable depends on the battery chemistry, size, design etc) and measure the voltage. <S> NiMH is a bit thorny since it's difficult to determine the charge condition from the terminal voltage. <S> It's better to discharge with a moderate load to a known discharge condition (perhaps 1V per cell or a bit less) then charge from there. <S> Overcharging NiMH batteries causes irreversible cumulative damage. <S> Once the battery has charged properly you can discharge it into a reasonable load and measure how long it lasts (taking care not to discharge it too much, which can also cause permanent damage). <A> Short circuit test of batteries is a bad idea because it can damage the batteries. <S> Lithium polymer battery have a large discharge current on short circuit it may explode. <S> You can discharge the battery using a proper dummy load for testing the capacity of the battery. <A> The proper way is to measure internal resistance. <S> You do that by drawing an AC current, which can be small. <S> Simply switch a resistor across the battery with a FET, or use a switchable current source... <S> Now, an AC current will cause an AC voltage to appear on the battery, depending on its internal resistance, <S> U=RI after all. <S> If your ADC has tons of bits you can acquire this voltage directly. <S> Or you can use a cap to AC-couple the battery voltage, then amplify the AC part with an opamp, and acquire it with a standard microcontroller ADC. <S> Synchronizing the ADC with the current pulses will give you two voltages (with and without current applied) and the difference will give you internal resistance. <S> You can average over several cycles to reduce noise if needed. <A> A few seconds can cause irrecoverable harm depending on the cell type/chemistry. <S> Using a reasonable load (power resistor) is a more appropriate approach. <S> Stepping the DC load (eg, 1A -> 2A) is a common method for gaining insight into cell health and determining equivalent DC resistance of the cell. <A> If you really want to delve into how good your battery is working, then you need to monitor the voltage and current over time and find the charging cycle and then compare it to curves for healthy or bad cells. <S> Comparison of the discharge voltage of an alkaline battery (red) and a NiMH battery (blue). <S> The green line is the voltage at which the battery is considered dead. <S> Source: <S> Using Nimh <S> (This link will tell you that you might need a better charger) <S> You could do this with a multi meter and a variable power supply. <S> However it would require you to log the voltage and current every so often and put it in a table and then graph it out.
| It's generally a bad idea, especially if you don't know with 100% certainty a lower bound on the internal resistance of the battery. I would not do it with any other kind of battery, including NiMH.
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Why are these independent loops? According to my textbook, abca with 2 ohm resistor is independent. A second loop with 3 ohm resistor and the current source is independent. The third loop, with 2 ohm resistor in parallel with 3 ohm resistor is also independent. Now the definition of an independent loop is a loop that contains a branch that is not part of any other independent loop. Let's take the first loop, abca with 2 ohm resistor. Say the unique branch is 2 ohm resistor. now bc with parallel resistors 3 and 2 ohm is also said to be independant. but the latter contains the 2 ohm resistor, meaning 2 ohm resistor is not unique to a loop after all. Same goes for the current source and 3 ohm resistor loop, 3 ohm is also not unique. So according to the definition, why are these three loops independant? <Q> I teach circuits and use the same textbook that the figure is from. <S> My students have asked me about this and it took a while to come up with a sound definition. <S> The statement from the text "an independent loop is a loop that contains a branch that is not part of any other independent loop" is ambiguous and as was previously stated is necessary but not sufficient. <S> If you go strictly by this definition you are correct that you can form the left loop IL (abca) which leaves the 3 ohm resistor and the 2A source. <S> You can then form the right loop IR which starts at b goes through the 2A source to c and then back to b through the 3 ohm resistor. <S> At this point there are no more unique branches but you have not found all the independent loops. <S> What needs to be added to this definition is "Can any nodes be expanded creating a unique branch without breaking any independent loops? <S> If a unique branch is created without breaking an independent loop then a new loop must be formed which contains this unique branch." <S> If you expand node a into two points (a,d) then you will break the left loop, IL. <S> Therefore the unique branch created between points a and d must become part of loop IL so that IL can be closed. <S> No independent loop is added. <S> If you expand node b into two points (b,e) then neither the left loop, IL, nor the right loop, IR, are broken. <S> This creates a new branch between points b and e that must be part of a new independent loop, IM. <S> If you then expand node c into (c,f) <S> you create a new branch but break loop IM so the new branch must become part of IM to close that loop. <S> If you continue to expand any nodes no new unique branches are formed and you are left with the 3 independent loops, IL, IM, and IR. <A> Now the definition of an independent loop is a loop that contains a branch that is not part of any other independent loop. <S> If a loop has a branch that's not part of any other loop, that does guarantee independence, but I don't think it's required. <S> (Mathematically, it's sufficient but not necessary .) <S> In mesh analysis, you're trying to solve a system of equations. <S> For that, you need one equation per variable. <S> But the equations must be linearly independent -- if you can make one equation by adding, subtracting, and/or multiplying the other equations, it doesn't count. <S> For example: <S> $$x + <S> y = 5$$$$2x + <S> 2y = <S> 10$$ <S> The second equation can be produced by doubling every value in the first equation. <S> This doesn't give you any new information, so you can't solve for x and y. <S> But in this example: <S> $$x + y <S> = 5$$$$x <S> + <S> 2y = <S> 7$$ <S> you can't get the second equation by manipulating the first. <S> So you can find the solution: <S> x = 3 and y = 2. <S> Back to circuits. <S> Your system has three variables -- the mesh currents \$I_L\$ (on the left), \$I_M\$ (in the middle), and \$I_R\$ (on the right). <S> Here are the equations, assuming the mesh currents flow clockwise: $$10\mathrm V - I_L \cdot 5 <S> \Omega - (I_L - I_M) <S> \cdot 2 <S> \Omega = <S> 0$$$$-(I_M <S> - I_L) <S> \cdot 2 <S> \Omega - <S> (I_M - I_R) <S> \cdot 3 \Omega = <S> 0$$$$I_R = <S> -2\mathrm <S> A$$ <S> Grouping the variables gives: $$10 \mathrm <S> V - I_L \cdot 7 \Omega + <S> I_M \cdot 2 \Omega = <S> 0$$$$I_L \cdot <S> 2 \Omega -I_M \cdot 5 <S> \Omega + <S> I_R \cdot <S> 3 <S> \Omega= 0$$$$I_R = <S> -2 <S> \mathrm <S> A$$ <S> There's no way we can make one of these equations out of the other. <S> The first has a constant term, the second doesn't, and the third just gives us the value of one variable. <S> If we substitute \$-2\mathrm A\$ for \$I_R\$ and try to make the signs match, it's even more obvious: $$ <S> I_L \cdot <S> 7 <S> \Omega - I_M \cdot 2 <S> \Omega - 10 <S> \mathrm <S> V = 0$$$$I_L \cdot 2 <S> \Omega - I_M \cdot 5 <S> \Omega - 6 <S> \mathrm V = <S> 0$$ <S> The ratios of the coefficients and constants are totally different. <S> These equations are linearly independent. <A> So using your two loops, ABC and BC as examples: ABC contains branch 10V/5 Ohm that isn't in BC. <S> BC contains the 3 Ohm which isn't in ABC, therefore they are independent. <S> Additionally ABC through the 2A/5ohm/10V contains the 2A source which isn't in either of the first two loops, so that too is independent. <S> This not the only possible combination of three independent loops necessary to solve the problem.
| An independent loop contains at least one branch that doesn't belong to another loop.
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How to safely destroy an AC Adapter? I purchased an AC adapter for a nintendo wii. It hasn't worked and every fuse I put in to it appears to blow. The seller has offered to refund me and send me a new one if I send him a photo of the destroyed AC adapter. He's suggested submerging it in water which I feel uncomfortable with for obvious reasons. What is the safest way I can visibly destroy my AC adapter? <Q> Cut the cord off <S> right where it comes out of the box with some sturdy kitchen shears. <S> Take a photo- that will probably be good enough. <S> That method was used for 'destruction' of non-compliant adapters that were considered unsafe. <S> For example, .45ACP <S> slug, ceramic kiln, various machine tools etc. <A> There are many ways to safely destroy electronic devices. <S> Some of those include: Smashing with hammer or sledge hammer. <S> Cutting in pieces with a band saw or hacksaw. <S> Submerging a sealed device as your seller suggests is NOT a very good way of destroying your AC adapter. <S> There are more fun ways of doing this but all involve some hazards. <S> You could burn it up with a blow torch or do as Steve Ciarcia (of Circuit Cellar Ink fame) did with some of his old hard drives: take it out to a target range and blow holes in it with a gun. <S> And, of course, you could use a Blendtec blender to "See if it blends". <S> Search YouTube for "Blendtec" or <S> "Will It Blend" if you don't know what I'm talking about. <S> I'm kinda partial to applying large amounts of electrical energy to said devices <S> (it's amazing what you can do with a 480 Vac 100 Amp distro) but that is actively frowned upon where I come from. <A> The most expensive part of an AC adapter is usually the transformer. <S> In addition, many use a custom transformer that can only be ordered with a minimum run in the thousands or tens of thousands. <S> If you open up the adapter and tear apart the windings of the main transformer then it will render the adapter inoperable and not easily/cheaply repairable. <S> Don't forget to discharge all the capacitors first though, in order to prevent injury or death.
| I can think of other ways of destroying an AC adapter, but the seller might not readily recognize the remains.
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Microchip PIC Development with OTP Devices (12C672) I have to work on a system that uses 12C672 PICs from Microchip. These are one-time-programmable (OTP) devices. The board has been designed already without any in-circuit programming facility. How do people normally develop with these? Are the ICE systems good enough and affordable, or is there always a flash alternative part that people use? I am used to developing with flash AVR microcontrollers and just do in-circuit serial programming. This PIC project will be a small production run of 100 or so units, so I am also interested in recommendation for a programmer to use. <Q> We use the Microchip ICE2000 emulator with the appropriate module. <S> These are available on eBay for a reasonable price. <S> The processor module that you want to use for the 12c671 / 12c672 / 12ce673 / 12ce674 is PCM12XA0. <S> You will also need the device adapter which is part number DVA12XP081. <S> The cool thing about these tools is that they have essentially a lifetime warranty. <S> Microchip will repair or replace them when they die - which is almost never. <S> FWIW - we use the 12F675 instead of the 12C671 / 12C672 for all new designs. <S> Less expensive and contains eeprom as well as a better a/d converter. <S> You can debug them with the ICD2 / ICD3 / PICkit2 / PICkit3 with the appropriate debug header. <S> The only downside is that you get one breakpoint instead of unlimited breakpoints with the emulator. <S> Why do you want to use the 12C672 when better parts are now available? <S> These new parts are pin compatible with the older parts. <A> First, that's a ancient part with newer flash versions available that are pin-compatible, do more, and cost less. <S> Back in the pleistocene before flash parts, the usual way to develop with these things was to get a few of the JW variants. <S> These had a quartz window and were UV-erasable. <S> I usually required the customer to supply me 8 of them <S> so I would always have a fresh one ready with a batch of 4 in the eraser. <S> Keep in mind that erasing can take 20 minutes. <S> These were then often used in the first prototypes, since you expect those to go thru a few firmware revs. <S> Debugging with the ICE-2000 was the nicest environment, but you could get a lot done with the simulator too, especially on such a small part. <S> Often that's all that was required, and it wasn't worth getting the ICE-2000 module for little PICs like that one. <A> then I have small wire jumpers that I bring over to 0.100" headers which make testing easy. <S> Let's face it; code for a PIC12C672 is not going to be much different from code for a PIC12F683; you can develop on the "surrogate" part more easily and then when you feel you're about done and the code is correct you build for the real target and try it. <S> If it works, great. <S> If not you toss the part and go back to the flash device until you're confident again. <S> Much less waste and a HELL of a lot cheaper than an ICE.
| Generally speaking, if you purchase a used ICE2000 from eBay that comes with the proper processor module, it usually includes the device adapter as well. When I usually have to develop with OTP parts I will cheat; I will take the closest non-OTP version and develop on that, usually with a board that is built assembled with everything but the processor.
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Need to create a magnetic coil (no core) for low frequency field (20Hz-100Hz) I am doing some experiments with pulsed magnetic fields. I am using an audio amplifier as a power source, and a simple software as my frequency generator. I use mostly square waves and sometimes sine. Both straight forward, simple waves - sort of like a clean audio test signal of different shape. Nothing fancy. I can even hear them when I run them through my coil and when it is noisy around here, I can just put on headphones and hear clearly whether the coil is working. I am terribly uneducated when it comes to all things electromagnetic (I only know a bit about audio and that electricity going through a wire will create a magnetic field)… so I need some help. On the upside, I am a fast learner, given right direction. So far I have been experimenting with a coil I built, that is 4" in diameter, about 1" wide and has about 1lb of 22 gauge insulated copper wire on it (I think that is some 300ft but I don't remember any more) - as a result, it is 8 ohms resistance which was done on purpose as my amplifier is rated at 8 ohms at max power. In any case, the coil works quite well, down to about 300 Hz, and up to 5000 Hz although it tends to get hot around 300 Hz after 10 minutes or so. Haven't tested it lower than that, except today at 20 Hz when it overheated in a second and I had to shut it off. Since I am using an audio amplifier as a power source, I made sure that the coil is 8 Ohms resistance, as that's what the amplifier requires. Now, I have two questions: If I need a coil that can take low frequencies, like 20 Hz, the one that I have does not seem to like it, as it immediately gets extremely hot - within seconds literally. It is well varnished (I did it by hand, on every turn) and quite solid, but I imagine that such low frequency makes it vibrate enough to produce high heat. I tried even at very low power setting, such as 1/10 of the max that I normally use, but still it gets very hot very quickly and I am sure that if I let it run, it would melt in about a minute or two. Would using thicker copper wire be better for lower frequency? If so - instead of using 22 gauge wire, what should I use to be able to run that coil as if it is on higher frequency? And if thicker wire would work better, would that somehow weaken the magnetic field? (right now I can run 1000 Hz through it no problem, for hours on end, at max power which is around 1000 W maybe even a bit more). I imagine that since my power amplifier is rated at 8 Ohms I would just use thicker wire but with lower number of turns to achieve the same resistance. If it matters - I am using a Mackie professional power amplifier, the type that is used for concerts. It is rated at 1500 W when in bridged mode (both channels bridged together into one). How can one achieve an even higher power of the magnetic field using, say, an audio amplifier as a power source? I imagine, getting a more powerful amplifier would be a good start - perhaps a 2000 W or more. But then, how do I build a coil that will not overheat? Any general advice you can give me in that respect is much appreciated. I think, but I am not sure, that thicker wire is better suited for lower frequencies. Having lots of power, as in - strong magnetic field, is essential. I just don't know how to do it. I understand that there is a practical limit at which coil will necessarily melt, but I don't need to go that far. One more question: I made an observation that - the lower the frequency the stronger the magnetic field. Am I correct there? Namely, I run 1000 Hz, and I can hear some of the iron stuff on my desk vibrate - from my Luxo desk lamp to the box cutter knife, but it does not attract anything, and can run like that for hours… but at 20 Hz, that same knife stuck to the coil at 1/10th of the power (remember - the coil has no core - it is pure copper)! It was scary for a moment - I imagine that's what happens when someone brings metallic objects into an MRI machine room… <Q> Holy Cow! <S> You're running a kilowatt into a 4-inch coil and <S> you're wondering why it get's hot? <S> Please, rethink this one. <S> The reason you can run at higher frequencies is that the impedance of your coil is equal to the resistance plus a term that increases with frequency. <S> Look up impedance and inductance. <S> As your frequency increases, so does the coil impedance, and the total power drops. <S> At low frequencies this doesn't apply much, and that's why you're getting hot. <S> Increasing the wire size is not going to help. <S> What you need to do is something like this. <S> Then run water through the tubing to get rid of the heat. <S> Remember, if your amplifier is putting out 1 kilowatt of power, you have to get rid of that power. <S> If you don't heat something like cooling water or air, it all goes into the wire. <S> Just think about how hot a 1 kilowatt light bulb would get. <A> This is all expected behavior, because the coil has inductance as well as resistance. <S> At DC, you measure 8 Ohms. <S> But at AC, the inductance adds to this. <S> You have a series RL circuit: <S> \$Z_c = R + j <S> \omega <S> L\$ <S> I guess with the coil you describe, the inductive reactance probably dominates from 10 Hz or so. <S> After that, each doubling of the frequency will double the impedance of the coil, and the current will drop by two times too. <S> Try an experiment with your multimeter. <S> First, turn the audio level down, that amplifier sounds scary. <S> Connect the meter in series with the coil and drive it at 20 Hz, then increase the frequency to 30, 50, 70, 100, 150 Hz and observe the current decreasing. <S> This is the effect of the inductance increasing the impedance of the coil as the frequency increases. <S> If it's important to have the same current in the coil at different frequencies, you'll need to apply a correction to the input signal to compensate for the filter effect of the coil inductance. <S> If you are finding that the amplifier can't produce enough voltage to drive enough current you need through the coil, you could consider a few things. <S> A transformer would work to raise the voltage, or put another way, it would transform the high coil impedance done to something more suited to your amplifier. <S> A normal mains transformer might work, you're in roughly the right frequency range. <S> Use a series capacitor to compensate for some of the inductance. <S> This will make a much more narrow-band circuit, so it'll only work at one frequency, but you could completely remove the effect of the inductance at that frequency. <S> All of this assuming you can solve the overheating problem, with cooling or by keeping the pulses short. <S> One last thing to keep in mind: the circuit doesn't see square waves, it only sees sine waves. <S> When you apply a 50 Hz square wave, you're really applying a series of sine waves at 50, 150, 250, 350... <S> Hz. <S> The circuit responds differently to each of them, it filters out the higher ones. <S> So the current in the coil will be a very rounded-off square wave. <A> OP is clearly taking it back to 1990, where you use to be able to walk into any radioshack, and scoop up a "supercoil" for 10.00 each. <S> If memory serves me correctly, those things always burned up when you ran more than 500 watts through them. <S> That being said, how much power are you running through it? <S> I also feel your pain, due to the fact I just spent a decent amount of time on radioshacks site looking for one and if you query a search for one, 99% of the returns are for ignition coils. <S> lol Tho, I feel this might help you. <S> https://www.parts-express.com/cat/air-core-inductor-crossover-coils/297 <S> Since we're on old school setup, might want to consider making a bracket that allows a fan to be spaced off the MDF, about a half inch of space <S> and then the inductor coil above the fan, this will help keep those temps down... <S> Either way, good luck...
| Wind your coil on a piece of 4" diameter copper pipe, and solder copper tubing all the way around both ends of the coil form, as close to the windings as you can.
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3V6 Zener Diodes With Very Poor Accuracy I have a question about the regulation accuracy (or lack of) of a 3V6 Zener diode. I wanted to confirm that I wasn't doing anything stupid. I set up the standard test circuit shown below. No load except for the DMM. The two Zeners in question are: BZX55C3V6 (3V6 - 500mW) and 1N4729A (3V6 - 1W). Both Zeners came from separate suppliers. \$V_{CC} = 5\$V and is relatively stable (total variance no more than 40mV).I know that Zeners need a minimum current to regulate properly and I worked out the optimum current through the combination of \$I_{zt}\$ from the datasheet and the following "rule of thumb" calculation: $$I = \frac{(P/V) \times .7}{4}$$ Taking the 1N4729A as an example, the \$I_{zt}\$ is roughly 70mA.Working out the current limiting resistor we have: $$\frac{5 - 3.6}{70\text{mA}} = 20\Omega$$ When I tested the zener with a \$20\Omega\$ resistor I get a \$V_z\$ of just over 4V! In fact, as I increase and decrease the resistor value, the Zener voltage increases and decreases along with it. It's like the Zener resistance isn't decreasing as the current increases, which is not what I'd expect. What's odd is that the BZX55C is showing more-or-less the same behaviour although with different resistors. Is this normal? <Q> A "zener" diode with a breakdown voltage below ~5.6V are based on the Zener effect and will have a very soft knee, I personally never use them. <S> The ones with voltage of 5.6 and above exploit the avalanche effect and have a very much sharper knee. <S> Even so I expect that the diode you have does match its data sheet. <S> kevin <A> If you want decent accuracy throw the zeners away and use some low dropout 3.3V linear regulators. <S> Zeners are notoriously power inefficient even at no load. <S> At least at no load or very small load the LDO regulator will be way more efficient. <A> Rather than use rules of thumb which you may or may not understand, you should first look at the data sheet of the part you're using. <S> From http://www.mouser.com/ds/2/427/bzx55-se-58740.pdf <S> you will find that the BZX55C3V6 is tested at a current of 1-5 mA. <S> Since you are driving more than 10 times that amount of current through it, it is no wonder that your voltage is high. <A> The problem is how you are modeling your circuit, because the series resistance you are using is so close the zener resistance, it must included in the circuit model. <S> As shown in the circuit below which models the behavior that you are seeing of a voltage reading of about 4V from the zener diode simulate this circuit – Schematic created using CircuitLab <A> There are a couple of issues here. <S> First this stated in the question: - <S> In fact, as I increase and decrease the resistor value, the Zener voltage increases and decreases along with it. <S> If you increase the resistor the voltage across the zener should go down. <S> Because it rises (as per the quote) you must have an unstable power source of 5V and this is rising as you increase the load resistance. <S> Double check this supply. <S> The next problem is that at 70mA and 4 volts, the power is 0.28 watts and the self-heating may be causing the zener voltage to rise. <S> I'm not sure what the temperature coefficient is for this device and the data sheets are lacking in that respect. <A> If the 1W zeners were purchased from an unofficial channel they may, in fact, be something like 500mW units marked with the JEDEC 1N4729 number. <S> The 3.6V +/-5% @ 69mA is a JEDEC registered parameter and the voltage across the zener <S> should be in that range (at Tj <S> ~= <S> 30°C) with 69mA flowing. <S> 3.6V zeners typically have a negative temperature coefficient of -1.5 <S> +/-0.5mV/°C, so self-heating will not account for the anomaly. <S> I suspect that if you retest the "1N4729" parts with a test current of something like 20mA (70 ohms) <S> you may find that they're within spec voltage-wise. <S> The BZX55C3V6 is rated for 3.6V +/-0.2V <S> at 5mA, so your resistor would be more like 280 ohms. <S> You don't say what you saw there. <S> If you include the dynamic resistance in the model, you must use a lower voltage for the ideal zener. <S> The dynamic resistance describes the small-signal behavior near the test current. <S> Typically the dynamic resistance of the 1N4729A is about 5 ohms, so you could approximate it with an ideal 3.255V zener in series with 5\$\Omega\$ but only very near Izt = <S> 69mA. <S> are available.
| Zeners with nominal voltages less than 5V or so have very soft 'knees' (poor regulation) and are not really that useful as usually better alternative For low voltage shunt regulation either use a series regulator (LDO) or a device such as the TL431 which has a band gap reference, and an opamp in a 3-poin package to create an almost perfect "Zener diode". It's incorrect to add a voltage at the test current because of the dynamic resistance.
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Is it OK to put LEDs in parallel? I wonder if it's a good idea to put LEDs in parallel, as below: I've heard that it might not be because the voltage threshold won't be exactly the same for each LED, so they'll shine all with very different brightness and you don't have a way to balance that. Is that true? So does it mean it's a bad idea? <Q> Look how a (generic red) LED conducts current when you apply a voltage to it: - At 2 volts, the LED is taking 20 mA. <S> If the LED was manufactured slightly differently it might require 2.1 volts or maybe 1.9 volts to push 20 mA thru it. <S> Imagine what happens when two LEDs are in parallel - if they "suffer" from normal manufacturing variations, an LED that only needs 1.9 volts across it would hog all the current. <S> The device that needs 2.1 volts might only receive 5 mA <S> whilst the 1.9 volt device would take maybe 35mA. <S> This assumes a "common" current limiting resistor is used to provide about 2 x 20 mA to the pair. <S> Now multiply this problem out to 8 LEDs and the one that naturally has the lowest terminal voltage will turn into smoke taking the best part of over 150mA. <S> Then the next one dies then the next etc... <A> Not a good idea. <S> Not only will you have inconsistent brightness (which you may not care about) <S> you will have a cascade failure mode. <S> You've got 8 LEDs in your circuit with one resistor. <S> 8 <S> x 20mA is 160mA. <S> As long as each LED takes roughly the same current we are fine. <S> Now lets say one of them gets a bit warm, and it now pulls more current. <S> positive feedback increases the current until it burns out. <S> Now we have 7 LEDs receiving current meant for 8. <S> Rinse, lather and repeat until the final LED gets all 160mA and comes to a fiery end. <A> Yes you can do that, and to balance the brightnesses you can put a resistor in series with each individual LED (varying the value of the resistor to compensate for the difference in brightness). <S> Also, if you put a single current-limiting resistor in series with the complete LED group, you will need to take into account the fact that you are using multiple LEDs and calculate the total required current (allowing for their differences in forward voltage drop and the values of their individual resistors if you have put those in), and you cannot use the same value of resistor that you would use if you had just one LED. <S> EDIT: <S> As "Andy aka" has mentioned above, you need to watch out for such situations. <S> EDIT 2: <S> In short, no. <S> This is probably going to cause more problems than it is worth. <A> Yet another possible solution is this . <S> More expensive than resistors, but it pretty much guarantees equal brightness for each LED with much more relaxed voltage requirements. <S> Just be careful of power dissipation and temperature de-rating. <A> The best way to do this, if you have to, is to group the LEDS in series in groups of N = 1, 2, 3, 4, etc, depending on the maximum supply voltage and the LED voltage drop, with N giving the maximum voltage drop that will work with your supply voltage. <S> For example, let's say you have 2-volt red LEDs, and you are running them in a car, with a supply voltage between 12V and 15V more or less. <S> You can make strings of 5 LEDs in series, giving a voltage drop of 10V, and to anticipate the maximum voltage at 20ma current put the string in series with a (15 - 10)/.020 = 250 or 220 ohm resistor. <S> You can them connect as many of these modules in parallel as you want. <S> Of course as your supply voltage falls, the LEDs will dim. <S> But they wont all blow up if one fails. <S> For decent results you need to "bin" the LEDs. <S> Nowadays, the voltage drop from LEDs bought from the same source under the same part number will be within a tenth of a volt. <S> But it's always good to test.
| It's not a good idea. But if you use the correct current-limiting resistors for each individual LED then there shouldn't be a problem (although you will need to determine the correct value for the resistor by measuring the forward voltage drop across each LED, which is not a very elegant approach to circuit design).
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Why are NPN Darlington transistors used to sink current? I notice that NPN Darlington transistors are commonly used to sink current. Wouldn't it make more sense to use PNP for the purpose? This would avoid shunting the load current through both junctions at once. Granted, we might want to share the current between two transistors; but in which case, please note that the second transistor is still carrying the full load (half via the C-E path, and half via the B-E path). For that matter, why are transistors most commonly used for sinking current anyway; rather than driving it? I've never understood that. In the above example, it seems more sensible to either (1) place the load below the transistor; (2) use a PNP Darlington; or even better (3) use a complementary PNP pair as shown here: EDIT: To clarify, one of the questions I'm asking is: Why can't we place this NPN transistor as-is above the load? Or, for that matter, place a PNP Darlington below the load? And also, why do Darlingtons even exist, when a complementary pair looks to be a cleaner solution? <Q> Sinking load switches with an NPN darlington allow the control signal to be a GND referenced signal. <S> If you use high side sourcing switches it is most typical that then the control signal needs translation down to a GND referenced signal domain. <S> These days when MCUs are controlling almost everything the GPIO pins on such devices are GND referenced signals. <S> And so it should be obvious why many load switches use the syncing type components with a GND referenced input. <A> Regarding the use of NPN rather than PNP, Michael Karas answer is correct: you want ground-referenced control signals because the N-type transistors generally have better characteristics than the P-type equivalents. <S> Regarding other parts of your question: Darlingtons don't share the current between the two transistors 50-50. <S> The one where the input signal arrives on the base carries maybe 1% of the current through it (assuming a beta of 100; most integrated circuit NPNs have betas much higher (~250) <S> so the percentage is therefore even lower). <S> The other transistor is therefore carrying 99%+ of the driven current. <S> This is a good thing, not a bad thing. <S> This is without the need for pairing multiple transistors in parallel, which can cause uneven current splitting due to device differences, even on integrated circuits. <S> Lastly, NPN Darlingtons can be easily constructed on an integrated circuit effectively as a single meta-transistor; they share the same collector region but have different embedded base/emitter regions (with the size difference I mentioned earlier). <S> Connecting the emitter of the smaller to the base of the larger is pretty trivial. <S> I'm pretty sure this is what's done on the integrated multi-Darlington arrays <S> e.g. ULN2k series <S> (I don't have the details of access any more, but I did see some of this way back when doing my studies in this stuff). <A> In the Darlington configuration, the base current of the larger transistor helps to drive the load and is self-regulating. <S> If one needs to drive a 10 amp load and wants to avoid assuming a beta greater than 40, one will need to be able to drive the base of the large transistor with 250mA. <S> To get that 250mA, one would need to drive the base of the small transistor with 7mA. Using a Darlington configuration, if the load draws 10A, 9.75A will flow through the collector of the large transistor and 250mA will flow through the small transistor into the base of the large one. <S> The 7mA driven into the base of the small transistor will be "wasted". <S> If the load were to drop to 10mA, the base of the small transistor would still consume 7mA, which it would pass through the base of the large transistor, but little current would be required elsewhere. <S> In most other configurations, arranging for the large transistor to have 250mA available on its base when needed would imply that 250mA would be fed to the base of the large transistor even when it wasn't needed. <S> In cases where the load is known to require 10A, that wouldn't be a problem, but in cases where the load might require anything from 10uA to 10A, wasting 250mA at times when the load requires 10mA may be undesirable. <A> You should be able to see for yourself from your own diagrams that the lower circuit needs access to the power rail, whereas the pure low side switch can be pre-packaged without needing that connection.
| Integrated Darlington pairs are configured in physical layout with a significant size differential, such that the main drive transistor has a much larger junction area than the first, allowing for much lower C-E on resistance for lower drive currents and much higher max current handling capability.
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Why isn't this astable multivibrator oscillating? This is the schematic for a modified astable multivibrator I've built. I incorporated the diodes in order to get a sharper rise time, and therefore produce actual square waves. Principle of Operation (As I Understand It): One transistor will turn on first due to minute differences in the gain of the transistor. Assume Q1 turns on first. 1) Q1 Turns on.2) The left side of C1 is at .7V. The right side of C1 is at 0V.3) C1 charges through R3 at \$T = R_3C_1\$. 4) When C1's right plate achieves .7V Q2 turns on. Because Q2 was previously off, the right side of C2 was at rail voltage during the previous state. Since the right plate has dropped 5V the left plate also 5V. The negative voltage at C2's left plate is coupled to Q1 holding it off. 5) R4 charges C2 from negative rail voltage to +.7V at which point the states switch and the process repeats. Unfortunately, this oscillator is producing a standard voltage. This is confirmed by a simulation. What's the problem? What are the errors in my thinking? EDIT See related question here . <Q> The diodes are the problem. <S> As modified, the circuit has a stable state, with both transistors turned on, and about 0.7 volts on both sides of both capacitors. <S> Consider that the circuit was turned off for "long enough" for the caps to discharge to zero. <S> When you apply power, the caps start out at zero, and R2, R3, R4, and R5 all follow the power rail as it starts to go up. <S> When the transistor bases get up to about 0.7 Volts, the transistors turn on with the current through those resistors. <S> That clamps the collectors near zero. <S> The anodes of the diodes can get pulled up to 0.7 above the transistor collectors, and that's enough to turn on the transistor bases. <S> You could replace your 1N4007 diodes with lower-voltage Schottkey diodes <S> (maybe a BAT48, or 1N5817, 18, or 19), and that might allow the circuit to oscillate, but I'm not sure you'd really achieve your goal. <S> Here's a selection of likely Schottky diodes at Digikey: http://tinyurl.com/qat53yk <A> The basic circuit is OK but the resistors R3 and R4 are much too low in value. <S> I would expect them to be about about 10 times the collector resistor (500 ohms as it consists of 2 1K resistors in parallel) <S> so 5 k or so. <S> The transistors are so heavily biased into conduction that the cross-coupling (C1, C2) cannot turn them off. <A> Here's your circuit and one that works, so you can compare them, and here's the LTspice .asc file <S> so you can play with the circuits if you want to.
| There are two problems with your circuit, the diodes and the base bias resistors R15 and R16, the diode drops keeping the transistor bases above cutoff, no matter what, and the 330 ohm resistors putting too much current into the transistors' bases for each other's collectors to suck out and turn the transistors off.
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What is a Namespace in SSD? What is a namespace in NAND or NOR based Flash Memory? Is it a range of addresses of NVM? If yes, then is it specified by SSD Manufacturer? <Q> After formation each namespace contains “n” number of logical blocks with logical block addresses from 0 to (n-1). <S> Thus namespace is a collection of logical blocks. <S> Namespace ID is an identifier used by a controller to provide access to a namespace. <S> Private namespace <S> A namespace that is accessible by only one controller. <S> A host may determine whether a namespace is a private namespace or may be a shared namespace by the value of the Namespace Multi-path <S> I/O and Namespace Sharing Capabilities (NMIC) field in the Identify Namespace data structure. <S> Shared namespace A namespace that is accessible by two or more controllers. <S> A host may determine whether a namespace is a private namespace or may be a shared namespace by the value of the Namespace Multi-path <S> I/O and Namespace Sharing Capabilities (NMIC) field in the Identify Namespace data structure. <A> Namespace is actually the list of LBAs(Logical Block Address) in an NVMe Dev. <S> It is usually Vendor Specific & is embodied in the NVMe device. <S> Take an example of Intel® Solid-State Drive DC P3700 Series. <S> It says 781,422,768 total User Addressable Sectors in LBA Mode in 400GB capacity. <S> Also, a complete table is given for namespace identification. <A> When a SSD is manufacture which supporting namespaces, there should be no namespace in the SSD. <S> It is up to user to create and attach namespaces. <A> A quantity of non-volatile memory that may be formatted into logical blocks. <S> When formatted, a namespace of size n is a collection of logical blocks with logical block addresses from 0 to (n-1). <S> NVMe Specs Section 1.6.17
| Namespace is an area or quantity of non-volatile memory that is formatted into small logical blocks count.
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Significance of -1 slope in CMOS inverter transfer characteristics In the CMOS inverter transfer characteristics what is the significance of slope of \$-1\$ at the points where \$V_{IH}\$ and \$V_{IL}\$ have been shown? And how is this the occurrence of the values, \$V_{IL}\$ and \$V_{IH}\$ related to the point where the slope is \$-1\$? <Q> This seems to be the standard textbook definition of VIH and VIL. <S> Using a slope of -1 as the limit makes more sense if you think of inverters as amplifiers and remember that the input will have some noise: simulate this circuit – Schematic created using CircuitLab Going back to your graph, if the input voltage is between VIH and VIL, any noise will be amplified. <S> If the signal passes through many gates (which it probably will), the noise could be amplified enough to flip a bit. <S> This leads to the definition of the noise margins for the inverter: <S> $$NM_H = V_{OH} - V_{IH}$$$$NM_L = <S> V_{IL} - <S> V_{OL}$$ <S> Or at least, that's my understanding. <S> I found some sources that said choosing a -1 slope as the threshold maximizes <S> the sum of the noise margins \$(NM_H + NM_L)\$, <S> but I'm not sure how that works. <S> There seems to be a lot of debate among textbooks and academics about how best to define \$V_{OH}\$ and \$V_{OL}\$. <S> Some sources define them as the extreme limits of the output voltages, while some use the same -1 slope definition as the input thresholds. <S> (The latter seems to be more formally correct.) <S> I even found an IEEE paper which claims that all of the definitions are wrong, including the input thresholds! <S> I don't have a resolution to all this except to say that the significance of the -1 slope definitely has to do with noise margins. <S> I believe that answers the original question. <A> While I can't be sure, since you have provided no context or links to the original source, I would guess that this is part of a discussion on how to quantify 3 terms: input high, input low, and mid-voltage. <S> What the author seems to have done is to take the position that, for a logic circuit, there will be input/output high and low regimes, where changing the input has little or no effect on the output, and a middle area where the output is sensitive to changes in input. <S> Apparently the author has chosen to define the input high and low thresholds as those at which the change in input is equal to change in output - where the slope of input vs output is -1. <S> The mid-point, where the input equals the output, is obviously where the curve intersects the line Vin = Vout; <S> in otherwords where it intersects a line with slope = 1 and which intersects the origin. <S> This is a generalized approach to analyzing response curves, and has the advantage of dealing in a consistent way with curves which are not ideal. <S> An ideal inverter, for instance, has an input/output curve with flat in the high and low regions and a vertical transition region. <S> For such a devices, Vil, Vm, and Vih are the same. <S> But real devices don't have infinite gain, and your figure seems to be one way to deal with this. <S> Note that this is not the only way to deal with the problem. <S> You might, for instance, define the transition points as those which produce outputs of 10% and 90% of the output range, and Vm as the 50% point. <S> Or any other set of values you like. <A> The "slope=1" comment indicates that the diagonal line shows the point where Vin=Vout. <S> If one has two inverters in a back-to-back configuration to form a latching circuit (to change the latched state, use stronger transistors to overpower the latching transistors), both inverters could sit at that precise voltage indefinitely. <S> If one inverter's input was a little higher, that would cause its output (the other inverter's input) to go a little lower, which would in turn cause the that inverter's output (the first inverter's input) to go higher, etc. <S> eventually causing the latch to fully switch, but if the voltages are both infinitesimally close to the Vin=Vout voltage, switching could take an arbitrarily long time. <S> The places where the slope of the transfer function is -1 define Vil, Voh, and Vih, though it's not quite clear whether there's any special significance to the spot where the slop passes negative unity beyond the fact that on every gate with whose transfer function's second derivative is monotonic, and whose output voltage range is as large as its input voltage range, will have such points. <A> This property of the CMOS transfer curve is used to bias the CMOS unit and to use it as a "linear" amplifier because this point - more or less - is in the middle of the Quasi-linear part of the transfer curve. <S> Simple biasing is possible using a large resistor between input and output of the CMOS device. <S> This gives a stable bias point for amplification of an input signal - either connected via a coupling capacitor or via an additional resistor. <S> In the latter case, due to negative feedback the gain is smaller but has a value that can be selected (and fixed) by the resistor values (opamp principle). <A> The output remains relatively unaffected with noise if the small signal model gain (dVout/dvin) of the circuit remains below unity. <S> In the figure, the slope of the VTC begins from zero, becomes more negative above VIL, and it approaches zero again when Vin>VIH. <S> That is why some designers consider the limits of margin noise at slope=-1.
| I also see no specific significance in the points having a slope of -1,The most important point of this transfer curve is the point "VM" where input voltage=output voltage.
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microcontroller input protection with Schottky clipping diodes I regularly see the use of Schottky diodes in order to protect microcontroller inputs from overvoltages (>5V). The Schottky diodes are connected to the 5V power supply (cathode) and the microcontroller input (anode). I was wondering what happens when current flows through these diodes due to overvoltage? Where does this current flow into? It probably flows into the power supply unit...but what happens there? Isn't it so, in fact, that we are applying an overvoltage to the power supply output and that we expect the power supply to take care of it...? Anyone with bad experiences with this solution?Are there any better solutions? (In fact, the ESD protection diodes in the microcontroller itself apply the same trick for short spikes...the spike current flows into the power supply...) <Q> Your analysis is correct. <S> What happens to it in the power supply depends on the design of that part of the circuit. <S> If the current is a transient pulse (from EMI or something like that) then the circuit's decoupling capacitance and the output capacitance of the power supply will easily absorb it. <S> If it is a small constant current then it will be used by the chips in the circuit instead of drawing current from the power supply. <S> If it is excessive, it may cause the power rail to rise and destroy the rest of the circuit as most power supplies are not bi-directional. <S> The key thing is to ensure that the last situation cannot happen, and this is one of the reasons you normally see a series resistor before the protection diodes. <S> If the potential current is still too high it is better to use a disconnection FET or shunt clamp design instead. <A> The internal diodes are only there for ESD protection while handling the device (while it is not fitted on a PCB). <S> They are not rated for human body discharge model <S> 8kV/16kV. <S> They only satisfy some JEDEC standard. <S> In your final product, if your microcontroller's pins are exposed, external connectors for example, you fit ESD protection diodes to protect against more severe discharge events. <S> They can not protect for over-voltage! <S> you will need some series resistors if that what you are trying to achieve. <S> Also remember, regulators do not sink! <S> So if you increase the supply rail from another source, your 5V WILL go up. <S> 99.9% of the regulator designs do nothing about this situation. <S> An ESD event is a short pulse and gets absorbed by the decoupling capacitors (shorted), not by the regulator. <S> I hope this helps. <S> Cheers. <A> If I understand correctly: 1) <S> In most cases, the power supply will go up, due to current leaking through the schottky diode. <S> 2) <S> This means that, even with a series resistor at the input, power supply will go up. <S> How fast it goes up depends on the power supply design and the series resistor used. <S> 3) <S> So, if it (slowly) goes up, then not only all circuitry powered by this supply is in danger, but also the input that I wanted to protect, as it stays only at about 0.6V below power supply at that moment <S> (power supply minus schottky forward voltage)... <S> So, if inputs at the microcontroller input could rise @ <S> x2 the allowed input voltage and very slowly, one would advise not to use this kind of solution at all?
| The idea is that the current flows through the diode and into the power supply.
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Electrical symbols, what does a dot before a triangle mean? Good day. Trying to recreate a circuit, but I'm not sure what the dot BEFORE the triangle means. I know after it means invertor, does before just mean it's inverted BEFORE, instead of after? Rather confused, and not having luck searching. Really appreciate the help. <Q> Just another way of drawing a logic inverter ("NOT"). <S> The triangle-with-bubble is an inverter: logic high input yields logic low output, and vice versa. <S> Usually the bubble is shown on the output, but in a mixed-logic system, the bubble can be shown on the input instead. <S> When the bubble is on the input instead of the output, that indicates that the input is an active-low input. <A> Here you can find good explanation The “bubble” (o) present at the end of the NOT gate symbol above <S> denotes a signal inversion (complementation) of the output signal. <S> But this bubble can also be present at the gates input to indicate an active-LOW input. <S> This inversion of the input signal is not restricted to the NOT gate only but can be used on any digital circuit or gate as shown with the operation of inversion being exactly the same whether on the input or output terminal. <S> The easiest way is to think of the bubble as simply an inverter. <A> If someone's stumbled upon this question looking for a slightly different usage such as in this picture: <S> You might want to check this answer: https://electronics.stackexchange.com/a/452460
| It indicates an active-low input, active-low means you must apply 0V to make gate working.
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Replacing the Arduino voltage regulator I want to replace the onboard AMS1117-5.0 (broken) with a LM7805. However, the pinout is different between the two voltage regulators; AMS1117: Gnd, Vout, Vin vs LM7805: Vin, Gnd, Vout. I would like to know what is the best practice of how to approach this? Line-up the Vin and cross the Vout and Ground pins and add some insulation so they dont make contact. How do the experts at EESX do this? Follow-up: adding the lm7805 using jumper wires works fine with the 12v DC input i have. It is a temporary fix until replacement ams1117 arrive. <Q> Why do you need to change over to a 7805? <S> If you're gonna be bending pins up and using jumper wires, you're cruisin' for a bruisin'. <S> A quick look at the data sheets leads me to believe there is no good reason to switch, other than the possibility that this is all you've got on hand? <S> If it must be done, here's how I'd do it <S> : bend the large ground tab so that the LDO will sit at a large incline off the plane of the board. <S> Tack it on, then solder your jumper wires from the pads to the appropriate connections. <S> I do not recommend doing this, just sample a replacement part. <S> You'll get it for free, and have the satisfaction of a job well done. <A> The two devices may not be interchangeable in the target circuit (not shown). <S> The drop-out voltage for the AMS1117-5.0 is typically only 1.1V at a load current of 0.8 amps. <S> The LM7805 drop-out voltage is specified as typically 2V at 1A load current. <S> You may not find the replacement works correctly. <A> Given that the tab on the AMS1117 (SOT223 housing) is the output, it will have the largest copper area to connect to and will be mechanically more stable. <S> I'd clip the other two wires short and use pieces of insulated wire to make the other two connections. <S> That gives you a solid connection to the one solder pad best capable of supporting it, and flexible connections to the other two. <S> I'd then use some hot melt glue or epoxy to give it some mechanical stability. <S> All of this assumes you are using the 7805 because you've got one at hand and need to get the Arduino going again without waiting on parts. <S> As RYS says, the professionals would replace it with the correct part. <S> This also assumes nothing else died when the AMS1117 quit. <S> Also, be aware that the 7805 is not a low drop out part. <S> If you have been using a 6Volt power supply then it won't work right with the 7805. <A> I am using a 12V DC supply. <S> In any case, all answers given have made me reconsider and i am getting a replacement AMS1117-5v. <S> Given that you are using a 12V DC supply, and want 5V, you can leave broken AMS1117 and get a car usb charger. <S> 12V <S> in, regulated 5V out, better efficiency, less heat, neater package, <S> and you can use the Arduino usb connector.
| I think you need to justify the replacement. I would connect the 7805 output pin directly to where the AMS1117 output was (middle pin, large solder pad.)
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What are the basics I need to know to power LEDs with RF at a range of 30 feet? I would like to light up a path in the woods with hundreds of LEDs using RF. The idea is that I would carry a RF transmitter powerful enough to light LEDs. As I approach the LEDs would get brighter, showing the path, and fade out as I pass out of range. Similar to the effect of carrying a lantern. I would like the LEDs to become noticeable, though dim, at 15 to 30 feet. Is this possible? I have never attempted anything like this before and I would really appreciate any tips or help with this project. Update*Parameters: Transmitter weight as heavy as 20 lbs Forward voltage up to 3v No solar pannels No long wires linking LEDs Passive LEDs (no battery) Parabolic dish if necessary Also, is safety an issue when holding a transmitter so close? <Q> It feels intuitively like this might be possible. <S> If you carried a 10 watt omnidirectional light bulb, in a dark place, it would certainly light up a white object at 30 ft. <S> So why not at RF? <S> RF spreads out the same as visible light. <S> The capture area of a dipole is about 0.1*wavelength^2. <S> So I calculate that at 145 MHz, a 5 W transmitter will deliver 5 mW to a dipole antenna 10 m away. <S> The trick would be rectifying the RF without wasting too much of it. <S> If you manage 50% efficiency that will produce 1 mA in the LED, which should be visible. <S> 5 W is easily available from a handheld radio, but you'd need a decent antenna, not the rubber duck it ships with. <S> 145 MHz is legal if you have an amateur license. <S> SMD LEDs are bright enough at 1 mA. <S> So the hard parts are the rectifier, and getting the voltage high enough to drive the LED. <S> A full-wave dipole has a high impedance, which might generate enough voltage (5 mW into 1000 <S> Ohms is just over 2 V). <S> The finished path-light would be a string of wire perhaps 1m long, with a epoxy coated circuit in the middle. <S> LED, diode and matching circuit. <S> You could hang them from the trees. <A> It's doable, but there is a realness / hardness tradeoff which makes other methods more attractive. <S> Powering LEDS with RF at 30 feet is difficult and inefficient. <S> Cheating so that LEDs 30 feet away are power by RF is doable. <S> Activating LEDS with RF from 30 feet is a doddle. <S> Activating LEDS some other way may be even easier. <S> RF or Inductive power transfer over 30 feet requires high Q resonant circuits, generally needs larger than smaller antennas and is liable to be hard and expensive to do well at a size and cost that is acceptable. <S> Inductive power transfer only transfers power to a matching load when the load is "in range" and well designed circuitry draws far less power when not loaded. <S> (You could arrange an interactive system that determines proximty and raises power level, butr complexity rises rapidly). <S> If you MUST transfer power then having the receiver coils close to your path will reduce the transmit power and/or coil sizes. <S> You can then run wires to LEDs or other loads as desired. <S> Having the LEDs powered by eg batteries and detecting the proximity of a small portable transmitter is far easier than either of the above two choices. <S> Use of proximity detectors along the path is also easy. <S> These could be ultrasonic (modules with 5 to 10 metre range cost under $10). <S> Or light-beam-breaking or capacitive proximity or ... <S> As a guide to what can be achieved with range search for what MIT did sa few years ago. <S> Korea Advanced Institute of Science and Technology 2014 - 5 <S> metres(KAIST] <S> ( http://www.sciencedaily.com/releases/2014/04/140417124509.htm ) <S> Wikipedia - Inductive Charging Wikipedia - Wireless Power ) <A> A bit late answer.., but <S> some years ago I was able to light up a led at 1400 meters from an fm broadcaster. <S> Power was 20 kw erp. <S> The clue is that you have to transformer the energy to a very high impedance. <S> Antenna was a 1/2 dipole.
| RF transmitters "throw the power out into the aether" [tm] regardless of whether a load is present or not so must work at the maximum power level ever needed for the application.
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What happens when a power metal film resistor is exposed to higher power? What is the worst thing that could happen if a power metal film resistor rated for 2W is exposed to 15W due to a short in the circuit? I have seen smoke being emitted but switched off the power supply immediately. If I hadn't, what would have been the end-game? I am using this particular Vishay resistor: http://www.vishay.com/docs/28729/pr010203.pdf <Q> These resistors, according to the data sheet, have "Defined fusing characteristics". <S> This means that the components will act like a fuse at some level of overload and go to open circuit. <S> There are graphs, starting on page 12 of the linked data sheet, that show the time to interruption as a function of overload power. <S> You did not specify the ohms value of the resistor that that you are using in your circuit but <S> the graphs for the PR02 part series would show you the range of time to interruption (going open circuit). <S> So lets say for example if you were using a 1 ohm PR02 resistor the fusing time can range from about 0.35 seconds up to 20 seconds when the overload is at 55W. <S> At less than 55W overload the vendor is not willing to specify what the fusing characteristics would be. <S> The most likely scenario is that the resistor would get smoking hot and could very well allow damage to other parts of the circuit. <A> Well, sadly, you were already too late. <S> Once smoke is forming, the power rating and resistance value of the resistor have already changed. <S> By how much and if that is bad for your device is hard to say. <S> Depending on where it is located, related to other parts the heat of a resistor can already change or damage other components before it starts smoking, and even itself, though at the "before smoke or glowing" point usually the permanent changes to its own characteristics are within a very small range and will usually not affect much, unless it's a high precision part chosen for a very specific value. <S> (in stead of the most common power resistor use-cases of "between 1 and 2 ohm: so take 1.5") <S> Whether or not you have also affected anything in the neighbourhood of the resistor can only be measured or tested, but often if there hasn't been serious glowing or (near-) fire <S> , it's more likely than not that the only thing affected is the resistor. <S> Had you not turned it off, a beefy resistor like that could very well have caused an actual fire. <S> It's unlikely the resistor itself would have burnt (metal film on ceramic doesn't burn easily) very long or intense, but they can glow hot enough to make nearby plastics go "woosh". <S> I don't expect a Vishay to be made of shoddy ceramic core-material, which is good, because the real cheap "ceramics" that do exist out there may snap at high heat and actually burst a bit like a micro-grenade throwing off some shards in a few directions. <S> But as said, that's usually only a risk with the reaaaallly bad ones. <S> Normally a power resistor is made with materials that don't do too much bad stuff, including the very limited burning. <S> Just because if someone specs a power-R they usually expect there to be a lot of power, so there's more risk there, statistically. <A> It would have to be a very short 15W pulse for a 2W resistor to survive. <S> It mainly depends on the time it takes to heat-up the resistor. <S> My guess the 15W pulse should last less than a second before damage occurs. <S> You could desolder the resistor and measure it. <S> If its value is still what it should be then it might still be usable. <S> But if it has changed colour then I would replace it. <S> But note that there might be some other components damaged. <S> Power supplies are complex and dangerous beasts, you really have to know what you're doing ! <S> What if you repair it, it works, you use it for a day <S> but it bursts in flames when you're out of the office getting a coffee ?? <S> I would only use a powersupply that I can rely on.
| The resistor is supposed to be built in a manner that it will not sustain flaming (at least the coating is flame retardant) but the generated heat could very well cause adjacent non-flame retardant materials to catch fire.
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Why is serial EEPROM preferred over parallel EEPROM? In the wikipedia page for EEPROM: http://en.wikipedia.org/wiki/EEPROM it is given that "Parallel EEPROM devices typically have an 8-bit data bus and an address bus wide enough to cover the complete memory" and also "Operation of a parallel EEPROM is simple and fast when compared to serial EEPROM". In that case why are serial EEPROMs becoming more popular than the parallel EEPROM? <Q> It is very simple. <S> Number of pins and cost of packaging. <S> EEPROM devices are primarily used to store parametric data or characterization constants for a device. <S> The typical scenario is to write very seldom and read typically once each time the host device boots up. <S> For this type of application the relatively slow writing times of EEPROM are of little concern. <S> And the reading time to load at most a few K-bytes of data from a serial device (SPI or I2C) is not normally an excessive time impact. <S> There is another factor that has played into the popularity of serial devices over parallel devices. <S> That has been the migration of MCU devices from older microprocessor units with parallel busses to the much more prevalent modern types that have all their program storage memory and data memory built right on the chip. <S> Often there is no longer a parallel bus option directly available. <S> And in most applications there is very little interest in using up scads of pins to bit bang to a parallel peripheral. <A> In the early days, wires were cheap and transistors were expensive. <S> These days it's the reverse. <S> Hence why almost everything is done serially. <S> In the early days, chips weren't very sophisticated, and a CPU would power up and read the first thing it found on its memory bus at the starting address, so parallel EEPROMs effectively mimicked the DRAM that was hanging on the bus. <S> These days, DDR RAM is screaming away at gigahertz on huge wide buses, making a flash chip that could hang on the same bus would be prohibitively expensive and fairly pointless when modern CPUs have enough built-in intelligence (thanks to cheap small transistors) to boot from I²C / SPI flash. <S> With micros, these days the program flash and RAM is usually internal to the device. <S> External storage like EEPROM can hang on an I²C bus, saving I/ <S> O pins for other functions whilst maintaining acceptable throughput. <S> The fewer <S> I/ <S> O pins you use, the smaller, cheaper and more energy efficient you get. <S> Plus it's far easier to track two wires around a board than two 8/16/32-bit wide buses, with the associated EMC issues, etc., etc. <A> Don't forget there is a "half-way house" called SQI. <S> That is a multiple parallel bit serial interface (it stands for Serial Quad Interface ). <S> From a protocol point of view it is just the same as working with a normal serial interface, but instead of just one bit being transferred every clock, 4 bits can be transferred at once. <S> Instead of a single data/clock, or din/dout/clock arrangement it has 4 data pins and one clock. <S> This gives 4x the through put of a normal serial interface and doesn't require many more pins. <S> In fact many SPI flash chips can also run in SQI mode without requiring any more than the existing 8 pins they already have. <S> A significant increase in speed without any increase in real estate. <S> SQI is becoming a popular interface for faster loading of programs from external flash chips - not only used for simple microcontrollers, but also now often used for booting the BIOS of PCs, especially laptops, where space is a real concern. <A> The low pin count on the device itself is probably less important than the saving on the MCU or FPGA <S> you connect it up to. <S> Finding 8 data pins, plus many more address, select and enable pins means a much bigger package and probably more expense for the MCU too. <A> <A> Just for grins, let's say I have an old-timey 2-way radio in my airplane, with 16 frequencies available and selectable from the cockpit, where the control unit resides. <S> Aft, somewhere, is the transmitter-receiver unit with a cable running to the control unit containing, among other things, the 16 wires running to the cockpit selector switch required to do the frequency selection. <S> One day, when talking to a friend, I bring up the subject of the radio and ask him if it wouldn't be possible to encode the cockpit frequency settings into a four bit binary number and send that number over four wires (saving 12 wires) to the T/R unit where it would be decoded into the sixteen signals needed to do the frequency selection. <S> "Sure", he says, "but why stop there? <S> instead of sending the [four bit] number all at once, why not send it a bit at a time over a single wire and have the decoder in the T/R unit figure out the frequency to select, saving 15 wires in the cable and 15 pins each in the connectors connecting the units?" <A> Below are some reasons why serial EEPROM is preferred over parallel EEPROM. <S> Lower Current Consumption . <S> For example, operating currents for 16K serials are around 3 mA; the same for 16K parallel devices is approximately 30 mA and above. <S> So the lower the current, the lower the power consumption. <S> Lower Voltage - serial EEPROMs are available in the markets which operates on low voltages (1.8-2.5 V). <S> Low voltage operation also has a positive effect on power consumption. <S> Programmability - serial EEPROMs are easier to program compared to parallel. <S> Serial EEPROMs have the ability and ease of programming one byte at a time; Serial EEPROMs are available in smaller footprint <S> Lower pin count Available at a lower price compared to parallel ones <S> Low microcontroller overhead and support
| While parallel EEPROM chips are faster and less complicated to communicate with, serial ones are less expensive hardware-wise, as they require less pins, energy, and wires/circuits.
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Creating DPDT Solid State Relay with FETs I feel like this should be a very simple question but so far I have not found a solution. My goal is to make the following DPDT relay circuit using MOSFETs: The goal is to connect INX to OUTX or OUTY and INY to OUTY or OUTX depending on the state of an input signal (I have two data lines available for the switch control). Normally I would use a mechanical relay but due to the nature of my project this really won't work, I need a solid-state option. DPDT solid-state relays are very difficult to come by, and any I have found cannot be connected to work in the way shown in the image. I was looking at using two enhancement mode and two depletion mode N-channel MOSFETs, but since a FET needs a G-S voltage to switch on I would not be able to simply pass the data signals straight through the FETs as they need to be isolated from ground. So how might one construct a DPDT solid state relay entirely out of MOSFETs that would allow them to swap the two outputs (connecting INX to OUTY and INY to OUTX, or vice versa)? EDIT: I forgot to mention, I had been looking at using bilateral switches but most of them can only handle about 30mA. These particular data lines can carry up to 200mA, so I'm looking for a ~500mA solution (hence using discrete FETs). EDIT 2: I was looking at these: http://download.siliconexpert.com/pdfs/2014/7/28/1/10/31/351/aro_/manual/semi_eng_ge2a_aqw21_e.pdf My thought was to connect them as follows (please pardon the crudity of the drawing, it had to be done in MS Paint, and I didn't want to put detail of the FETs and all in it. Just know that each pair of wires on the output are effectively connected to a switch: <Q> You could combine multiple SSRs to do this. <S> Two form B and two form <S> A will give you a DPDT SSR. <S> Cost would be fairly high (about $20 in singles) but this is really a weird set of requirements. <S> Switching will be slow. <S> For example, two LCB710 and two CLA230 (IXYS) would handle 700mA. <S> You could also buy PV optoisolators and use them to drive back-to-back MOSFETs for each switch. <S> Since the former come in pairs you'd end up with 8 parts (plus maybe 4 resistors). <S> Again, slow switching. <S> Edit <S> Again you could use dual MOSFETs and PV drivers, but I don't see much point in that since the current requirements (and presumably your voltage requirements) are met by common parts. <A> As the OP said, it is very difficult to find DPDT SSR (solid state relays). <S> But there are a some ways around this; use two SPDT relays, like this one , or use one DPST-NO and one DPST-NC as proposed in another answer, or finally, use two DPST, with an inverter in front of the control for the second relay. <S> For this particular application, the need to handle relatively high currents (at least 500 mA) with a low on-resistance weere the critical specifications. <S> I found this DPST C347S SSR which is rated for 1A. 150 <S> mΩ. <S> So you would need two, and use an inverter in front of the second IC's control lead. <S> $8.09 @ <S> in singles , $4.72 in thousands. <A> this is the digital equiv of an h circuit, fwd rev motor drivers. <S> try the tlp222 optical isolated fet driver, and if the current demands are really high follow the design and use opto isolators to keep one side that is on from inverting the side that is off and reverse biasing the fets. <S> some diodes might come in handy during development <A> I would approach this as follows: Use a pair of photovoltaic optos to generate an isolated voltage source. <S> Connect enough N-channel MOSFETs (back-to-back in pairs) as you need. <S> That would be 4 pairs. <S> Finally, use opto-isolators to drive the gates of the MOSFETs. <S> You want to use the optos in a tptem-pole configuration so as to minimize the current from the photovoltaic source. <S> I'll draw up a quick schematic when I get in front of a computer. <S> This gives you the ability to switch very quickly and handle as much current as you need. <A> I think I found the simplest option: Instead of a "Solid-State Relay" <S> like I kept asking for, what I think I was actually looking for was <S> a DPDT analog switch like the NLAS44599 <S> (Datasheet: http://www.onsemi.com/pub_link/Collateral/NLAS44599-D.PDF ). <S> The functional diagram looks like this: <S> I hope this might help someone in the future who's looking for the same thing I was. <S> Sure, it's a 16-pin device, but it does exactly what I need it to.
| : Since you imply in the comment below that there is no requirement for power-off normally closed, the simplest method is probably to use two dual form A SSRs such as TLP222A-2.
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What does the arrow on VCC in a BJT bias circuit mean? At the top, the VCC has arrows on it. What does that mean? I thought it would be like in other circuits as empty holes that you could attach, but I'm not sure here. <Q> The arrows have no special pointing significance. <S> If you look around you will notice that there are a variety of different symbols used for power supply connections on schematics. <S> There are also various ways that the associated voltage rail are labeled. <S> Here is a sample of some of the common types. <S> Note that in my example there is no particular correspondence between the label type and the symbol used. <S> This is by no means an exhaustive list either and you can find other usages as well. <A> Those symbols just represent a source of voltage. <S> The arrow is a common representation, but other typical representations include bars, circles, or just a label. <A> That up pointing arrow is just a convention to say "I'm connecting here the positive voltage supply.". <S> You can actually find circuits where there's only the arrow and not the label: if you only have one supply that's clear enough. <A> The arrow just a symbol to represent this node VCC . <S> In Schematics you can have more than one VDC at your circuit, and you can use various symbols to designate each of them. <S> In this document you can see that have some symbols for the schematic documents that are not components, like hte ground symbol. <S> This is one that have more than one symbol too.
| The conventional symbols used on schematics include the use of the arrow as a symbol for a power supply connection.
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4-20mA to 0-3V3 Converter Looking for a simple design to convert from 4-20mA signals to 0-3V3, I'm thinkin about configuration shown at picture attached. And I would like to discuss its suitability. It is based on:- A follower configuration for a Rail-to-Rail OpAmp, powered at 3V3.- A 165Ohm resistor to get, 3V3 at 20mA, and 0V66 at 4mA. What do you think about? It is any improvement I can do?Or it would be better to work with another more precise configuration? It is thought for a low power applications (<5V), but it would like to know if it could work with 4-20mA transmitters powered at 24V too. <Q> First, that will not give 0V out at 4mA. <S> It will deliver 0V at 0mA. <S> For the full span, you will get your 3.3VDC out. <S> As for will it work with 24V powered 4-20mA transmitters... <S> yes. <S> It would work with 48VDC powered transmitters as well, as long as you sink the 4-20 signal to the transmitter common. <S> a 4-20mA loop is 4-20mA; it doesn't matter what voltage is driving it. <S> Your circuit would not work if it is not the least receiver in a string however. <S> It is common for 4-20mA loops to place the transmitter in series with several receivers. <S> For that reason, most receivers use a simple sense resistor to pass the loop, and a differential amplifier reading across the sense resistor. <A> It will work. <S> The offset zero allows you to detect underrange (this is useful for sensor break detection and such like). <S> Of course you lose a bit of resolution on the ADC- 20% on the bottom. <S> You could also use a stage with some gain rather than a voltage follower which would allow an even lower resistor (say 49.9 ohms) to be used and would allow a better performance op-amp to be used (relaxing the requirements to input common mode range includes ground and R-R on the output only). <S> Take care as to the layout- <S> use Kelvin connection to the sense resistor, especially if you want to have a high accuracy (< 0.1%) circuit and are using a lower value resistor. <S> The compliance of the transmitter will determine the minimum supply voltage, and your 3.3V subtracts from that, so a lower input voltage is desirable, however too low and you lose accuracy. <S> If several devices are connected in series on the loop then 12v or even 24V may not be enough (the transmitter may require 10V, an indicator another 10V and <S> your circuit 3.3V is getting close even at 24V). <S> Note that a short across the transmitter will damage the input resistor most likely and probably blow <S> the *** out of the op-amp and likely other stuff too. <S> This can be protected against, at some cost in input voltage and parts. <S> If you put a series resistor on the op-amp inputs it may limit the damage to just the input resistor. <A> would like to know if it could work with 4-20mA transmitters powered at 24V too. <S> Yes, even though a 4/20 mA transmitter may be powered at a much higher voltage it is the current that flows through the device that determines the voltage across the 165 ohm resistor. <S> Think of a device that controls the current through its terminals <S> - it is connected in series with a 24V battery. <S> Ov of the battery connects to your monitor 0V and the device feeds the input to your circuit producing a current between 4mA and 20mA. <S> If the device were powered from 20,000 volts DC (impractical of course), would you see a current that is any different? <S> No, because the device controls the current that feeds the 165 ohm resistor.
| You might want to use a somewhat lower value of resistor so you can detect overrange on the 4-20mA.
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Proton current? and its potential effects in cars and circuitry Is this statement true: "Within the (lead-acid) battery, the electric current is primarily due to proton (hydrogen ion) current which is in the same direction as the electric current." What are the implications of this statement? Does proton current exist at all in the rest of the car circuitry, outside of the battery? And what effects does it have, if any , on the wiring, components, fuses, connectors, circuitry, etc. in cars? References: How the Current Flows in a Car? answer mentioning different flow types: https://electronics.stackexchange.com/a/95049/66759 <Q> Mobile charge carriers (the flow of which we generally view are electricity) are only electrons. <S> There are 'holes' but those are just abstractions of electrons. <S> As far as battery chemistry goes, it's referring to ion's, which are atoms that have more or less electrons than normal, which makes them non-neutral, and so a flow of ions can cause a voltage and thus current. <S> The statement is referring to proton flow, again this is just an ion in this case (I'm not a fan of the term 'proton current'), but the ion is not flowing through the wires, it is only flowing within the battery chemistry. <S> Current is the flow of electrons, and electrons will not flow without a voltage. <S> A voltage is a potential difference between two areas. <S> So, if all positively charged ions (as they lack an electron) move to the positive side of the battery, and all negatively charged ions (as they have an extra electron) move to negative side of the battery, you get a difference where the one side is more negatively charged than the other, and thus you have a voltage. <S> Connect a wire between the two and you have a current. <S> Though, really, even internally in a battery it's still due to electrons if you think about. <A> The statement is - to the best of my knowledge - true, but only matters if you're concerned about the battery chemistry of lead-acid batteries; it has no impact on the rest of the system. <S> Electricity is electricity, whether it comes from a battery, an alternator, or the power grid. <A> The clause Within the (lead-acid) battery is important. <S> Outside the battery, current is carried by electrons in metal in the normal manner. <S> This has no implications outside the battery. <S> People are easily mislead by "current flows from A to B" when really we should say "current flows in a loop through A and B". <S> It's a lot like a drive belt, really.
| As far as anyone is concerned, outside of battery chemistry, all electricity is due to electrons.
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Solid state relay doesn't turn off I am trying to build a custom light setup for my computer desk. I want to use the scroll lock LED signal to control a pair of 12 volt DC CCFL lights run off the same inverter. The problem is when I plug in the CCFL's they just turn on and no matter what I run across the signal pins, they don't turn off. The SSR I have purchased is from Crydom part number CN048D05. This is a drawing of my circuit: I am not really an electronics guy, but I have some basic knowledge. The LED circuit that I am trying to use as a control works. I verified that. What am I missing here? <Q> From the diagram on the SSR (as given in the datasheet ), it uses a photo triac . <S> The SSR will not deactivate until the load current drops below the sustain current for the device, which is not given in the datasheet but is usually very close to 0 regardless. <S> This device is not suitable for controlling constant DC loads. <A> If you disconnect the input pins to the SSR and are sure you have not swapped pins 13 and 14 then the SSR is kauput. <S> Try measuring the resistance between pins 13 and 14 with all connections off.. if it measures low ohms, that would be a confirmation. <S> If it looks more like a diode (on the meter diode range) then it's probably okay. <S> Or maybe your 20mA is way low- <S> that does not seem to be enough for a CCFL tube- <S> I would expect some watts. <S> If the 20mA comes from the rating of the tube, note that the 20mA will be at 600V or something, so the current at 12V will be more like 1.2 or 1.3A, which will surely fry that wimpy SSR. <A> Thanks for the help guys, but I found my issue. <S> The relay I bought functions fine for this. <S> Apparently, when I was putting it together, I was supplying my load current backwards.
| 100ma is an extremely low maximum current and it's possible that you've fried the output, perhaps merely from charging a capacitor in the inverter (sometimes inverters draw high current whilst they are starting up).
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Increasing wire gauge by twisting pairs I have a cable like this with 4 wires inside. Each wire is 20 AWG. I've been told that I can twist the ends together (i.e. green and red, white and black) on both ends of the cable and this will effectively increase the wire gauge. This would make perfect sense to me if the wires were naked (no green/red/white/black plastic coat around it) so it would be copper twisted the full length instead of just the ends... does the coat affect the current? I'm not an electrician or EE so I'm not too sure about this. I want to use this for sprinkler valves. <Q> " <S> American Wire Gauges go down by about 10 for every factor of ten in cross-sectional area. <S> If you had ten #20 wires connected in parallel, they could carry as much power as one #10 wire. <S> With two #20 wires, you'd have the equivalent of one #17 wire. <S> (A handy "rule of thumb" value: #40 copper wire has about an Ohm of resistance for each foot. <S> By the rule above, #30 would have an Ohm for every ten feet, and #20 an Ohm for every 100 feet.) <S> Note that connecting wires in parallel may work at DC or low frequency AC. <S> For audio, RF, or other purposes, you'd just mess up the wire characteristics, and cause yourself problems. <A> Doubling the conductors has the effect of reducing the equivalent AWG by a factor of 3 (as in 1x 20AWG = 20 AWG; 2x 20 AWG = 17 AWG equivalent). <S> In order for the effect to continue with additional cable conductors, doubling is required each time (eg 2x 20 AWG = 17 AWG equivalent, to go down (larger) another 3 AWG would require doubling your 17 AWG equivalent once more; ie 4x 20 AWG = <S> 14 AWG equivalent, 8x 20 AWG = <S> 11 AWG equivalent; to go down another 3 AWG equivalent now requires 16 conductors, then 32, and so on). <S> Whether it's advisable to create a substitute cable in this way is debatable, and dependent on application, (it affects certain peramaters such as cable capacitance, inductance, and in AC frequency {audio, Radio, digital} applications may cause smearing of the signal if each conductor is not the exact same length physically and electrically), but if you want to use an existing installed cable, it may be adequate. <S> Bundling of cable (what you are effectively doing) reduces the heat dissipation capacity of the cable, so some safety factor should be considered (ie don't use a 2x 20 AWG cable for DC power if the application is power and requires a minimum or specified 17 AWG, but if the requirement is greater than 17 AWG but less than 20 AWG it might be OK, or if heat is not normally an issue, such as loudspeaker cable, heat dissipation issues can be ignored). <S> Again, all depending on application. <A> On a uniform wire the resistance is defined as: R=(p*L)/A p = Resistivity of material <S> L = length of wire A = area of cross section Increasing the area by twisting several wires together makes area larger = <S> the resistance smaller. <S> As the power dissipated by a resistance (your wire in this case) can be calculated with: Power= <S> Current^2 <S> * Resistance <S> That means that lowering the resistance will make your wire less hot. <S> If you have too high resistance the wire gets hot, it might burn off or burn soemthing else, in worst case cause fire. <S> If the current is small, probably the most annoying effect will be that at the end of the wire the voltage you put in will have decreased because of the resistance. <S> I don't know if a sprinkler in this case means a fire extinguishing sprinkler or just a gardening sprinkler. <S> If it is for safety equipment I would surely get the correct wire gauge from the start, to not risk that a twisted wire gets "untwisted" so the sprinkler won't be able to operate. <A> It'll effectively "increase the gauge" by lowering the resistance of each pair of two paralleled conductors to 1/2 the resistance of a single conductor. <S> From the table below, 20 AWG has a resistance of 10.13 ohms per thousand feet, so 1/2 of that would be 5.06 ohms per thousand feet, which corresponds, roughly, to 17 AWG at 5.054 ohms per thousand feet.
| If you twist two wires together, each would carry half the current, so you'd "effectively increase the gauge.
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What is a 3 wire type cross linked serial cable? I'm trying to use a controller which uses Serial communication with the PC. The documentation says I need to get a 3 wire type cross link serial cable. Is this a null modem cable? The following image is from the documentation. <Q> Yes - I would call that a Null Modem cable. <S> If the cable you get (or make) doesn't work, my first debugging step for serial communications is to swap connections on pins 2 and 3. <S> (Transmit and Receive data) <A> Normal serial cables have pins 2 and 3 connected straight through, and the DCE is responsible for cross-connecting them internally. <A> Whether it is a null-modem cable depends on your definition. <S> Yours can be described as a partial null-modem cable, in that it does not incorporate hardware flow control. <S> Or, since most PCs these days don't require flow control, it can be called a null-modem cable. <S> Your documentation is being correct by specifying a three-wire cable, but a full-up null-modem would also work. <S> See https://en.wikipedia.org/wiki/Null_modem for a description of the possibilities. <S> Additionally, as Nick Alexeyev has commented, a null-modem cable usually has connectors of the same gender, since standard RS232 distinguishes between a Data Terminal Equipment (DTE) box and a Data Circuit-Terminating Equipment (DCE) box, and these are distinguished <S> be different connector genders. <S> So the cable you show is not technically a null-modem cable, but rather a crossover cable without flow-control - in other words, a cross-linked 3-wire cable. <S> Add a gender changer, and it becomes what is nowadays called a null-modem in the PC world. <A> Some background. <S> RS-232 serial ports were originally designed to connect a modem (DCE) to a teleprinter (DTE). <S> There was a data line in each direction and also a number of lines for handshake and modem control. <S> Later computers, graphical terminals etc got serial ports too, again these were used for connecting to modems and so used DTE pinouts. <S> The DTE was connected to the DCE using a "straight through" cable.. <S> Sometimes it was desirable to connect a computer to a terminal or a computer to a computer or even a terminal to a terminal without using any modems. <S> So cables known as "null modems" got made-up. <S> These crossed over the data lines allowing the two DTE devices to be connected together. <S> What they did with the handshake lines varied. <S> Some looped them back, some crossed them in various ways, some didn't connect them at all. <S> In later years most null modem cables seemed to settle on connecting the RTS output on each end to the CTS input on the other and also connecting the DTR output on each end to both the DSR and DCD inputs on the other. <S> The cable described in your instructions can be regarded as a primitive form of null-modem cable with only the data connectors and ground present. <S> It also has an unusual combination of connector genders. <S> So the question then becomes: <S> can I use a regular null modem cable and a gender changer to connect this device to my PC? <S> or do I need to find/make a special cable to thier specs. <S> And the answer is most likely <S> yes <S> but there is the outside possibility that the manufacturer has done something stupid <S> and you really do need a cable that ONLY has the data pins and ground present.
| Since the TXD (3) of one side is connected with the RXD (2) of the other, this is indeed a null-modem cable.
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Why does the power company provide a neutral line? I understand that power is often transmitted in 3 phases (with no neutral). Then, when we get to a certain substation, the power company basically gives house 1, L1, house 2, L2, house 3 L3, and connects them all to a common neutral line. That is, each house gets one phase of power and a neutral shared between all 3 lines. Then, that neutral is grounded to earth at the substation. It is also my understanding that in the main panel in my home the neutral is tied to ground. It seems to me, if we are going to do this, why do we need the neutral wire at all. In fact, in any system, if the power company actually grounds the neutral wire at the substation, why can't each individual house simply provide its own neutral (i.e. each house has a single phase and a metal pole or two out in the back in the ground that serves as a neutral (current carrying earth ground) and a ground (for safety)). It seems to me this would save the electric company from having to provide a neutral wire. My point is is that the neutral wire is earth grounded at the substation and in each home, so why is it even necessary to provide it? In my setup, current would flow from the 1 wire coming into the home from the pole (single phase) and current would flow to an earth ground provided at each home. There is no reason for the power company to provide the neutral. Can you fix my misconceptions please? I have read many posts and I get conflicting or contradictory information about this. <Q> I think part of the reason you're seeing contradictory information is that practice differs in different countries. <S> So if you pick up part of the picture in one place and part in another it's not surprising <S> you'll see contradictory information. <S> Stick to one location and understand how its practice works, and you'll be in better shape. <S> In the UK, practice should follow BS7671 , formerly known as the "IEE Wiring Regulations", currently at the 17th edition. <S> The first edition dates from 1882. <S> Your first paragraph summarises it well <S> : each house gets a phase and neutral, and the neutral is grounded at the substation. <S> In this system, the ground at your house is isolated from neutral, not connected to it, and thus carries relatively little current. <S> However it should have low enough impedance to carry enough current to keep you safe in the event of a fault, until the main house fuse blows, or breakers trip. <S> This ground impedance is assured - traditionally, by connecting to the cold water pipe in the days when they were metallic - and it is tested - for new houses, before you move in. <S> In fact, in a house protected by modern breakers - RCDs in the UK, or GFCIs (US) - if you divert as little as 20mA from neutral to ground anywhere in the house, the breakers will detect the imbalance and disconnect power, in case that 20mA was flowing through you. <S> Consider what would happen if you economised and only used one conductor for both neutral current, and safety earth. <S> Now imagine excessive current flows because you overload the circuit, or a bad connection develops high resistance and overheats, and the neutral/earth conductor melts before the live... <S> Everything in your house (including the neutral/ground wire, and thus all exposed metalwork), is live, and you are wandering around in the dark trying to fix it. <S> How happy are you about that? <A> A voltage is a potential difference between conductors. <S> For instance the difference between the live and the neutral wire. <S> The power company delivers these two lines to ensure the potential difference to the customers. <S> If it would only supply the live wire and the ground of a local customer is used, it's unknown which voltage the customer receives. <S> It can be higher or lower. <S> And then the resulting voltage on your devices can also be higher or lower. <A> You could disconnect the neutral line and use the earth to carry the current back to the power company.... <S> as long as you only use a very little amount of current. <S> If you are brave, take a small LED night night and connect one of it's prongs to the hot side of a power outlet and the other to a rod driven into the ground. <S> It should light up. <S> (Be really careful if you are going to try this - it can be dangerous if you don't know what you are doing!) <S> Why only a little night light? <S> Why not disconnect the neutral from the power company from my house and run my blender using only a neutral connections from my house to the rod in the earth? <S> Because dirt has a much, much, much higher resistance than wire and the power station is far away. <S> The more current you try to send though the dirt, the higher the voltage drop there will be across the dirt. <S> There will not be enough power getting though the earth to run the blender. <S> Note that there actually are some "earth return" distributions systems that use the ground as one of the conductors- <S> they just typically run at much, much higher voltages than your house. <S> Because the voltage is so high, the current is low (ohms law) so the resistance of the soil is less of an issue. <S> https://en.wikipedia.org/wiki/Single-wire_earth_return <A> Anybody whom has ever seen what happens when a 12V car battery is cabled to two stakes driven in moist ground, will understand that wide-scale, high-power Earth conduction paths are not feasible. <S> (In fertile soil, the earthworms will wriggle right out of the ground - which is great for catching fishing bait, but bad for the soil.) <S> Using soil as a conductor on a large scale would disrupt many burrowing creatures. <A> Why do we need both ground and neutral in the US? <S> Well, in the US, the neutral is a current carrying conductor. <S> It is no different from the "hot" line except that it happens to be bonded to earth (in exactly one place). <S> The ground wires carry current ONLY in the event of a fault. <S> This scheme works well. <S> The ground wire is bonded to all externally accessibly metal of all line powered equipment. <S> Normally no current flows in the ground wire. <S> But if the hot wire comes loose inside the equipment and shorts to the metal case, it will cause a large fault current and trip the circuit breaker. <S> What if neutral is shorted to the case? <S> In principal, since current flows in the neutral wire, you could still get a shock in this scenario. <S> But most power outlets now have ground fault interrupt (GFI) circuit breakers. <S> What these do is compare the current flowing in the hot and neutral conductors. <S> If the currents are not exactly equal in magnitude (within a few mA) then the GFI trips. <S> This is possible ONLY because both ground and neutral are used. <S> And it is a great safety benefit. <S> So that is why we need both.
| It is because there is no knowledge about the potential difference of neutral between the power facility and your house.
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Using a 5V Optical Mouse USB Cable to Power a toy requiring 4 Size D batteries I have a baby swing which requires 4 Size D batteries to get power. Here's a picture of such a thing - a human baby goes inside. Googling the information on Size D batteries, I found that 4 Size D batteries equals 1.5 V * 4 = 6V I want to power this toy using an old optical mouse USB cable. The specification on the mouse says 5V and 100mA. Questions: 1) Can I use this cable to power my toy ? 2) Is this correct: Red Wire in the mouse cable connects to Positive Node of the toy and Black Wire in the mouse cable connects to Negative Node of the toy ? Thanks for reading! <Q> Normally, of course, such supplies work fine, and the baby isn't supposed to touch other grounded surfaces, but the swing wasn't really designed for it. <S> There are two areas of concern: <S> Electrical Safety <S> If your supply's isolation fails, you might have a dangerous situation as the baby is normally strapped into the swing. <S> A grounded adult touching such a swing might then electrocute themselves and/or the baby. <S> Physical Safety <S> The supply cord becomes a nice strangulation hazard. <S> Make sure that it's properly zip-tied to the supporting structure and routed out of reach of the baby! <S> When it comes to devices that are in forced, constant contact with the body, such as in a baby swing, I'd not use a random supply, but buy a medical grade power supply, such as this one . <S> It's only $15 at this time + shipping. <S> Before you spend on such a supply, it'd be worthwhile to calculate how much money you're going to save in batteries over the life of the device. <S> I'd also not randomly switch over to rechargeable batteries: they can be a real fire hazard in devices not designed (read: at least fused) to accommodate them. <A> TL;DR: <S> I don't recommend this. <S> You should use bigger wire. <S> The voltage will be fine; that's an insulation rating, which is really hard to make thin enough to be a problem at any low voltage without chafing through. <S> If you're completely re-purposing the cable with no connection to any of its original stuff, then the color code technically doesn't matter. <S> Just make sure that the right connections get made from one end to the other. <S> However, it's still good to follow a convention like Butzke mentioned. <S> I'd be more concerned about the current. <S> A D-size battery is probably capable of several amps continuous, and the cable is only designed for 100mA. <S> That rating is for the wire size. <S> It'll probably work though...for a while. <S> Then you might have hot wires oozing out the side of the cable. <A> A lot of devices will run until each battery drops to 1v <S> (See EEVBlog https://www.youtube.com/watch?v=4iEshd6izgk ), and a 5v phone charger can produce an output voltage from 4.75v to 5.25v, which should be more than enough to operate the toy. <S> You'll just want to make sure the current rating of the charger is high enough for the toy. <S> You could also try a few different chargers if they have the USB connection on them, if under 5v won't work.
| You may be able to get away with using a cell phone charger if you plan on stripping a USB cable. I would have a bit of a reservation about powering a baby swing from a random wall-wart supply.
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How to cut costs when fabricating large PCBs? I've looked online at PCB fabrication companies, and they invariably price their boards according to the size of the board as the primary factor. Why is this? The physical board itself isn't that expensive, is it? I'm guessing it's because the size of the boards dictates how many they can produce simultaneously, and that's the primary limiting factor in their profitability - Is that right? Anyway, is there a way to keep the costs down when fabricating large (~8x10 in.) but sparsely populated (~50 components) boards (other than just ordering from the cheapest Chinese factory I can find)? It seems silly to pay $50 for a board that's only gonna have $10 in parts on it. <Q> 10*8 isn't really large, but you may get only 2 per panel which will impact the cost. <S> Panel sizes will vary from fab to fab so it is worth talking to a few - and negotiating the details - as The Photon says, you may get 4 per panel and half your price that way. <S> And the economics of setting up a fab for small jobs dictate that buying 50 boards instead of 10 can half the price again, 100 or more even lower. <S> Single-sided PCB may be MUCH cheaper than 2-layer, especially since there's no through-hole plating stage, omitting silk screen and solder mask may save a little more money. <S> Some fabs may still offer phenolic material - much cheaper than FR4 fibreglass. <S> You might be able to use an Arduino-sized full spec PCB to hold the complicated stuff or customization, which you then plug into a much larger single-sided board which - because it omits the personalization for a specific project - you can buy in larger quantities and re-use for multiple projects. <A> @Josh: You might be able to reduce board fab costs if you could work out a way to use a number of smaller boards to achieve your goal. <S> For example, if you have either one long series string of LEDs, or, say, ten strings of five, and could use ten long, skinny boards (say, 8 inches x 0.25 inch). <S> That'd reduce the material cost, and allow for a variety of panelization alternatives, but with substantial impact on total assembly cost (as you'd have to connect the small boards together electrically and mechanically). <S> Perhaps this illustrates a more general rule: If something seems unexpectedly expensive or difficult, see if you can look at your problem differently to see if you can find an easier or cheaper approach. <A> Be sure to know what size panel you are building, and design your board to fit well on that panel. <S> For example, if you use an 18x24" panel (very common), and your vendors want 1" spacing around the edges, and 1/2" spacing between boards, a 7 x 10 board will be priced substantially lower tha <S> 8 x 10 board because you'll fit 4 per panel rather than 2. <S> Note the board spacing might be driven by your assembly vendor rather than your fabrication vendor. <S> Buy larger quantities. <S> If you are buying less than maybe 10 boards at a time, you are paying a relatively large amount per board for per-lot processes (programming machines, ordering materials, ...). <S> Your price per board will go down quickly as your lot size increases.
| In addition to The Photon's good suggestions, use the simplest process possible for the large board.
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How to avoid Relay buzzing noise? (LDR dependent) I have the following circuit where I'm switching a relay based on whether it's day or night. I made a voltage divider using a pot and an LDR, which leads to the NPN transistor. So when it's dark the NPN will switch, which then causes the relay to switch. Problem : I am getting a buzzing noise whenever there is a gradual decrease in light, which is a problem since I'm going to be placing this outside. My guess is there is a threshold where the voltage causes the relay to switch back and forth quickly, causing the buzzing noise. How do I ensure the relay only switches once when light slowly fades? Is there some latching device I could use? Also, my DC power source is via a full-wave rectifier without a capacitor, so there seems to be very small fluctuations in the DC voltage. Would that be the problem? EDIT: I read about relay hysteresis, but so far I only read about solutions using microcontrollers, which I want to avoid. simulate this circuit – Schematic created using CircuitLab <Q> You need what is called a Schmitt trigger. <S> And yes, you definitely need to put a capacitor on your power supply. <S> Try something on the order of 1000 uF to start. <S> Then measure the supply voltage, and it had better be more than 24 volts, although it almost certainly will be. <S> Then you need to get a comparator IC - an LM311 will do fine. <S> You also need to get a cheap DMM to read voltages and resistances with - eBay will get you a basic unit for less than $10. <S> Now measure the resistance of your LDR at the sort of light level where you want your lights to come on. <S> Let's call this 1 Megohm, just to have a number to work with. <S> This circuit simulate this circuit – <S> Schematic created using CircuitLab should work pretty well. <A> my DC power source is via a full-wave rectifier without a capacitor, <S> so there seems to be very small fluctuations in the DC voltage. <S> Would that be the problem? <S> Yes. <S> Without a capacitor to smooth out the rectified AC, the voltage will go up and down at twice the mains frequency. <S> This is why the relay 'buzzes' - it is turning on and off in time with the rectified AC. <S> How do I ensure the relay only switches once when light slowly fades? <S> Most relays have a large hysteresis (pick-up voltage > 3x drop-out voltage) which in your unity gain circuit <S> should be sufficient. <A> Maybe not so much a definite answer, but a nice step into the good direction allowing to gain yourself some knowledge. <S> The problem you face is two-fold: <S> Most of the buzzing will be solved by adding a capacitor to the power supply. <S> Rule of thumb is 2200µF <S> /A drawn by the load, so order magnitude 470µF should be sufficient, 4700µF is somewhat overdone (but not necessarily bad). <S> While the surrounding light slowly increases, the voltage on the relay coil is changing slowly with it. <S> You can partially solve this by moving the relay from emitter to collector line. <S> So connect emitter to ground and the relay goes between +20V and collector. <S> The relay may have enough hysteresis of its own to nicely switch on and off. <S> The advantage is that you actually use the transistor as (voltage) amplifier. <S> The advantage of 2. is that you already have all components required and it is easy to test. <S> The disadvantage is that the thresholds will probably change substantially. <S> You may want to increase R1 to 2k2 or so to protect the transistor base. <S> I agree with the other answer that you really need something like a schmitt trigger although I would personally probably opt for an all transistor solution. <S> See also: <S> Use bipolar transistor to power LED from a certain power on?
| The other cause is that the transistor as you use it delivers an approximately unity voltage gain.
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How use DMM thermocouple probe hands-free to monitor heatsink temperature? I suppose this is more of a lab practice question than electronics per se, but I wonder if there's a standard-ish way to mount the K-type thermocouple that comes with my Fluke DMM so I can monitor the temperature of a power transistor and/or heatsink over time, hands-free? I'm certainly open to a special-purpose thermocouple or other sensor that could be screwed or perhaps preferably clamped in place so I could follow the temperature changes without having to poke the little temperature probe tip in there each time I want to take a reading and wait the minute or so it takes to stabilize on the right temperature. Is there something the pros do for this sort of thing? <Q> Short of that <S> you can try and tape or clamp the flexible wire, I've had luck using a binder clip or some locking pliers like vice grips. <S> Then curve it so the bend in the wire provides natural holding force onto the heat sink. <S> That way what you're using to hold the probe doesn't affect the measurement. <S> But really glue is the way to go. <S> For more pro use you can use thermal loggers that use slightly cheaper thermal couples, there are some nice ones that help you measure air flow at the same time. <S> That said for another project I just built a little wind tunnel and started using the bent wire approach so it's up to you how accurate you need to be. <A> If you must mount the thermocouple on the heatsink then consider a silpad or mica washer to stop possibly high volts getting into the meter which is USER INTERFACE now <S> even if your volts are low you don't want the sensitive thermocouple meter to give errors Often the heatsinks are full of RF these days and there is enough coupling to make the meter give crazy readings turning the equipment off and seeing an abrupt change to a believable temp will confirm this Winding the thermocouple wire around a ferrite toroid has helped once or twice as has shielding the wire <S> BUT these days we use the IR temo probe <A> I like to use the self adhesive probes that have a flattened ribbon junction. <S> They require a bit of surface area of course. <S> You might have to add a bit of insulation to get an really accurate reading, especially if there is forced convection. <S> If the heatsink is thick and grounded you can also drill a small hole (at least 3x the diameter deep, preferably 10x, and epoxy in a small bead thermocouple. <S> This us potentially the most accurate method. <S> When you are done just snip the T/C off flush (they're cheap). <S> You can also use an IR probe, paint the heatsink black to increase the emissivity and calibrate the IR reading (manually or by fiddling with the emissivity adjustment) with a thermocouple as above. <S> This is useful if the heatsink happens to have hundreds of volts of high frequency AC on it and you can only get sensible readings with the circuit 'off'.
| Best way I think is thermal compound and glue it to the device, then just buy another thermal probe :)
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Are sealed lead-acid batteries safe to use indoors? I have a small, 12V sealed lead-acid battery. I know regular lead-acid batteries can be dangerous to use or charge indoors, due to the fumes they release and the potential for acid to leak out or spill. A sealed lead-acid battery wont release fumes or spill though, correct? Does this make it safe to use/charge indoors? Thank you! <Q> Actually SLA batteries have a vent... <S> so the name "sealed" is a bit of a misnomer . <S> VRLA (valve-regulated lead-acid battery) is actually a name for the same tech . <S> Practically every UPS (uninterruptible power supply) <S> I know of has one [or more] SLA[s] <S> inside, so it's generally safe for indoor use. <S> Here's a snippet from an APC white paper on the issue: <S> Valve regulated lead acid (VRLA) batteries [...] do not require special battery rooms and are suitable for use in an office environment. <S> Air changes designed for human occupancy normally exceed the requirements for VRLA [...] ventilation. <S> Vented (flooded) batteries, which release hydrogen gas continuously, require a dedicated battery room with ventilation separate from the rest of the building. <S> And bit later in the paper the difference in gas output is quantified as 60 times less for VRLA: <S> VRLA batteries are considered to be “sealed” because they normally do not allow for the addition or loss of liquid. <S> A vented battery can give off sixty times more gas than a VRLA battery in normal use. <S> And the reason for this is that in a "sealed"/VRLA battery: <S> hydrogen recombines under pressure with oxygen into water inside the battery. <S> Gas can only escape when internal pressure exceeds the rating of the pressure valve. <S> The fact that they're pressurized explains why in extreme cases of misuse or abuse they end up seriously bulged . <S> The plastic container is actually designed to cope with that scenario. <S> Also, small SLAs (almost certainly the one you have) use gel as electrolyte suspender so won't spill liquid[s] even if cracked. <S> The larger ones use a glass mat instead (gel is rather expensive). <A> With that in mind, it is possible (although unlikely) to encounter leakage if they are left untended for very long periods of time (just like dry cells). <A> I charged a flashlight that contained 2 6v lead acid batteries inside the flashlight ON my laminate kitchen counter. <S> The next day I noticed a pool of liquid below the flashlight. <S> The acid escaped from the batteries and leaked ON the counter. <S> It discolored the laminate to a lighter white color with no way to restore the original color pattern. <S> I now have a $3000 counter replacement ahead of me. <S> Was I safe? <S> Oh yea, definitely safe. <S> No physical harm at all.
| Yes, sealed-lead batteries are considered safe for indoor use -- they are no different from dry cells or NiCds in that regard, and can be found in emergency lights and other applications where low cost and relatively long livespan in float applications is critical. "Safe" is a relative term.
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How to test op amp stability? I've got a circuit I'm developing using an op amp to drive a power MOSFET for an electronic load: I'd like to test it for stability, but I don't have enough experience to know what perturbations are most likely to show any instabilities. I have the following ideas: Inject a square wave (offset such that low value is >= 0V) into the non-inverting input while monitoring the voltage on the sense resistor (load current waveform). This would simulate sudden changes in the control voltage (1V/A) likely to be supplied later by a DAC. Apply stepped input voltage to IN+, simulating connection of power supply under test. Are these sensible ideas that are likely to uncover any op-amp related instabilities? Also, are there LTSpice simulation exercises that might be worth trying? UPDATE: I revised the schematic to be more suitable for simulation. I found and used models for the specific parts I'm using and removed all compensation elements for a baseline simulation. I ran an analysis feeding a DC-offset square wave into the control pin (non-inverting input in this circuit). The result was dead stable. There's a teeeny-weeeny 1mV overshoot on the rise if you zoom way in, otherwise it absolutely mirrors the input (except it represents current flow of course). I even reduced the rise time to 1ns to see if I could get it to ring, but no luck :) I ran an .AC analysis like @Kevin White suggested, and found a 62 degree phase margin in the open loop gain. I built it up as @mkeith suggested, and unfortunately it oscillates like crazy on the breadboard :) I was able to make some progress stabilizing it a bit until I accidentally blew out my MOSFET. It's down to the store tomorrow to get a new one so I can carry on from there :) <Q> This is normally (all situations that I've witnessed or read about) <S> a stable configuration. <S> The op-amp would be stable with direct feedback <S> so the question is what does the MOSFET add in terms of gain or phase that might make the circuit unstable. <S> Well, in a source follower configuration the gain of the MOSFET is a little less than 1 so <S> on that score the circuit is still going to be stable. <S> As regards phase shift from gate to source there will be a little but given that the gain has probably reduced about a dB and that the MOSFET is going to be much faster (as a singular device) compared to the op-amp, I really don't think you would have any problems. <S> It's the sort of circuit that I wouldn't hesitate to build and not expect a decent result. <S> However, in the circuit you have shown I wouldn't use the op-07. <S> It cannot adequately drive its output down to 0V (read the data sheet) and this could mean that the FET is not turned off properly and that control is lost when trying to control small currents. <S> The same is true (if not more so) when looking at the input voltage range that device is capable of. <S> If you are trying to control 1A thru the 1R sense resistor, the voltage at the OP-07 input is expected to be 1V <S> and this is right on the limit of what the input range can be expected to handle (again read the data sheet). <S> So my conclusion is don't use an OP-07 or power it with a small negative supply instead of having its neg supply terminal at 0V. <A> Your methods are fine. <S> What you are looking for in the step response is overshoot or undershoot. <S> What you'd like to see is something approaching critical damping for fast response and good phase margin. <S> Graph borrowed from this SMPS app note. <S> Your circuit has a strong potential to oscillate- <S> I would definitely add a compensation loop around the amplifier. <A> Both of those are good ways to test the stability. <S> You can only break the feedback loop in simulation rather than the real world. <S> If you put two copies of the schematic in the LTSpice schematic you can do a Bode plot of the closed loop and open loop gains simultaneously.
| Another thing that can be useful is to measure the open-loop response by breaking the feedback loop by using an extremely large inductor (1Giga Henry for example) then injecting a sine wave at the non-inverting input (with a suitable DC bias) and measuring the voltage across the sense resistor.
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Can anyone identify this type of connector? My Sony CMT-X5CD (CD/DAB hifi soundbar thing) uses a very non-standard antenna connector, as shown in the photos. Can anyone identify it? It's used for the aerial(but unfortunately, the aerial supplied is absolutely terrible..) 2pence coin for scale (26mm dia) about the same size as a $1 American coin). Thanks! <Q> I believe it to be TE Connectivity, PN: 1470222-3 Open the datasheet and verify the pitch and other measurements. <A> Is it a Molex 3 pin connector? <S> It looks a bit small though... http://www.surplusgizmos.com/Molex-Style-3-Pin-Locking-Connector-w18-AWG-Wire_p_2516.html <A> I have worked with a connector that looks very similar to this. <S> Or google "3 pin cable dynamixel robotis". <S> Unfortunately, I can't find the name of such connector but perhaps contacting a supplier may help you. <A> This looks like the very small aerial/antenna connector plug used in Sony DAB/FM radios and probably in others too. <S> Having problems with DAB and sometimes FM radio reception with these aerials is very common. <S> Molex manufactures them in this size as the OP in this useful post on digital spy forums verifies <S> The post shows how the connections work. <S> Also here is a link to an informative radio and TV aerials site. <S> You can then attach to a basic FM aerial (half-wave dipole) for decent reception on FM & DAB, or to a DAB folded dipole for better DAB but worse FM. <S> These aerials are best positioned vertically as described in the 2nd link. <S> Hope <S> this is useful for some folks. <S> It took me a while to get the bottom of the DAB reception problems we have here.
| Look at the 3pin cable/connector for the Dynamixel servo motors from Robotis, here is a link http://www.trossenrobotics.com/p/200mm-3-pin-dynamixel-compatible-cables-10-pack . Essentially you need to connect the black wire in your plug to the centre wire of a 75 ohm co-axial cable either directly or via a co-ax connector.
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How can I evaluate a potentiometer susceptibility to change? I have a usage case where I would like a potentiometer for controlling the charge going to a lead-acid battery via a IC. It would be susceptible to some (although not extreme) vibration, such people carrying it around and various bumps. It would for most part be a set-and-forget thing hidden in a case that nobody will think of, unless the battery type was changed. I don't want the resistance to go shifting over time, it should ideally remain reasonably stable within a few percent. How can I evaluate if a potentiometer is stable enough to use for this purpose? <Q> Don't worry about it. <S> If trimpots shifted around a few percent with minor vibration they'd never be used. <S> Just about any trimpot from a reputable maker (and quite a few from disreputable ones) will be fine. <S> Minimize the range of the pot (don't require it to be set to 0.1%) and try to use it as a voltage divider rather than a rheostat. <S> But a few percent stability is not a high bar. <A> Solder female header to your board instead of the potentiometer. <S> If you're slick, make a series of small breakout boards with the resistor you need soldered on and silkscreen text that says what battery type and/or voltage it should be used for (i.e. - 12V lead-acid, 7.2V LiPO, etc.). <A> How can I evaluate if a potentiometer is stable enough to use for this purpose? <S> This is what data sheets are for. <S> Evaluate the data sheet first. <S> If you take one physical sample of a potentiometer and test it, you may get a very favorable result. <S> This might lead you to believe that this pot (or that pot) works really well and is stable in end-to-end resistance, ratiometrically pretty good and generally it makes you happy and confident. <S> Then, you take a look at the data sheet and it's a lot worse than what you thought. <S> So, don't bother testing until you have read the data sheet then <S> , you might realize that you don't need to physically test it - just buy a pot (from a reputable supplier) that does the job that you need. <S> Would you test every component individually <S> you fix to a circuit board?
| If you must use it as a rheostat, keep the setting away from the very ends of the range and try to use a relatively high value (avoid 10 ohm cermet for example, if you can) so that CRV (contact resistance variation) is not much of a factor. Then, if you're lazy, just stick a fixed-value resistor in the female header.
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Calculate voltage of Led driver required for a given amount of led's I have built some LED lights for my fish tank using six of these LED lamps: http://www.ebay.co.uk/itm/261813680886?_trksid=p2057872.m2749.l2649&ssPageName=STRK%3AMEBIDX%3AIT I have tried using this power supply: http://www.ebay.co.uk/itm/291074673667?_trksid=p2057872.m2749.l2649&var=590234643632&ssPageName=STRK%3AMEBIDX%3AIT When I connected the original six white LED lamps they flashed to indicate the 12v 60w driver was being overloaded. I changed it to four LED's two on each string so the circuit was well within the drivers current parameters; however They still flashed, at this point I took a voltage measurement of between 7 and 8 volts when the lights flashed. It did however work with only two LED's connected. The driver seems to have a fixed voltage at 12v and from what I have read I need a new driver one that compensates for the drop in voltage. I asked the person from who I bought the LED lamps what the power requirements if the chip were and he said it would be 12v per chip so around 66V? This left me a little confused because the 12v driver worked with two chips. Is this the correct way to calculate the voltage of the driver required? Please note I found another source for these chips who states the reserve voltage if 5v; however I believe this may be a typo from the Chinese chap and he probably meant reverse voltage. Also; constant current driver so no resistor suggestions please. Many Thanks, Ross. <Q> If it does this at 9V <S> and you apply 12V <S> you might be getting a forward current of 3 amps thru the LED. <S> There is no data sheet for the device (that I can find) <S> so it's impossible to be exact. <S> But, realistically you probably need to put in a current limiting resistor for each LED. <S> What makes you think it has a constant current circuit built into it? <S> The ebay link is really-really crappy at giving details - for example it states that the output power is 10 watts - this of course is rubbish! <A> The LEDs need a constant current feed. <S> The power supply is a constant voltage supply. <S> You can PROBABLY get an OK result as follows. <S> Operate ONE LED from the supply with a resistor between supply V+ and LED Vin+. <S> The power supply is rated at 5A max. <S> For 6 LEDs the current per LED is 5A/6 ~= <S> 830 <S> mA <S> Let's work on 800 mA = 0.8A. <S> LED spec says 9 to 12V. 9V is worst case so start there. <S> PSU = <S> 12V. LED = <S> 9V Resistor = <S> 12-9 = 3VR = <S> V/ <S> I = <S> 3V/.8A <S> = 3.75 Ohms. <S> A standard value is 3.9 Ohms. <S> R dissipation = <S> V <S> x I = 3V <S> x 0.8A = 2.4W R needs to be 3.8 Ohm and say 5W (10 W even better). <S> Wire PSU+ve - Resistor <S> - LED+ve <S> - LED- - PSU -ve. <S> ie current flows through resistor and through LED. <S> Set to 20V range. <S> Turn on as above. <S> Measure voltage across resistor. <S> Target is 3V. A bit more [tm] - say up to <S> maybe 3.3V is OK <S> Lower is better than higher. <S> Say V across R was 2.9V.Iactual = V_across_R / R = 2.9V <S> / 3.9 Ohms ~= <S> 745 <S> mA = <S> OK <S> If I is too high use <S> a larger R - say 4.7 Ohms <S> If I is too low use a smaller R BUT be careful - say 3.3Ohms. <S> Once you have this working OK repeat for all LEDs in parallle. <S> One R per LED.R's are cheap, LEDS are not (although yours are a good price) LEDs will vary in current somewhat. <S> You can fine tine as above if desired. <S> NBNBNBNB <S> LEDS MUST have heat sinks. <S> They will get very hot and die in no time with no heatsinks. <S> This can be a simple large metal plate of available. <S> Be sure not to short power supply when connecting LEDs to heat sink. <A> simulate this circuit – Schematic created using CircuitLab <S> The bulb does light up. <S> I took the measurements again and voltage fluctuated quite widely as the LED heated up. <S> I don't have my heat sinks wired up at the moment; the plan is to use the same power source if there is any current to spare. <S> I used the 10A setting on my multi-meter, swapping the leads over to the appropriate terminals. <S> I also bypassed the resistor with a wire and measured the voltage across the led as 11.87v, not sure if that helps. <S> Been looking at this ohms law solver: <S> http://www.ohmslawcalculator.com/led-resistor-calculator <S> and it confirms I need a 3.9 ohm resistor. <S> I think I will go buy a bunch of them, wire the whole thing up, then measure again hopefully that will take me in the right direction.
| Somewhere between 9V and 12V the LED will take a current of 900mA. Get voltmeter (eg in DMM)
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BLDC max speed KV or Hz? So from what I've understood, the KV rating for a motor gives us the rotor speed based on the voltage applied at the phase of the motor.Having looked a bit deeper in ESC / inverters, I've come up with the conclusion that it's rather the frequency of phase commuting that sets the motor's speed. But I guess there is some electromagnetic force (lorentz / laplace?) that defines the force exerted on a body knowing the voltage flowing through it. What I'm thinking is that the frequency commuting defines speed, but the current/volts going through each coil defines the maximum force/acceleration the rotor can be pulled with. And since commuting is done one zero-crossing of the floating phase, this max acceleration (thus time it gets to do the zero-xing) limits the commuting freq. Could anyone help me to figure this out ? <Q> You need to consider both frequency and Kv in driving a BLDC motor, and motor drivers differ in how they take this into account. <S> So, both points are correct. <S> The rotational speed is directly related to the commutation frequency and the pole count. <S> And Kv * rotational speed gives you the back-EMF. <S> If you define the motor speed by generating a fixed frequency, then as JonRB says, you must supply enough voltage to overcome the sum of: V1 = <S> Kv * actual motor speed V2 = <S> IR loss from the motor resistance and the torque required to overcome friction and load resistance V3 = <S> IR loss from the torque required to accelerate the rotor + load if actual speed falls short of driving frequency. <S> Just as Kv = <S> Speed / Voltage. <S> the torque constant Ki = Torque / Current. <S> In SI units, Ki is simply 1/Kv : <S> non-SI users have to remember a funny conversion number (which I've forgotten) as well as which of oz, lbs, feet, inches it refers to, or look it up in a textbook. <S> So, given Kv, you can compute Ki, and therefore V2 and V3 for both components of motor current I. Driving a BLDC from a fixed frequency <S> , V3 is critical : if the frequency is too high or the supply voltage too low, the motor simply won't start, so it's normal to ramp the frequency up until the desired speed is reached. <S> Then if the driving voltage is fixed, and higher than V1 + V2 combined, the motor will run, but inefficiently, i.e. waste power at steady state (at constant speed V3 = 0) so for efficient operation you need to reduce driving voltage to V1 + V2 (usually via PWM). <A> So from what I've understood, the KV rating for a motor gives us the rotor speed based on the voltage applied at the phase of the motor. <S> I think you mean Kv <S> (lower case 'v') <S> which is the motor's 'velocity constant'. <S> This is the rotational velocity required to generate a Back-EMF of 1 Volt. <S> If Kv is specified in rpm/V then multiplying Kv by the applied voltage gives you the rpm when Back-EMF equals the supply voltage, which is the fastest it can go on that voltage ( <S> any faster <S> and it would be acting as a generator, not a motor). <S> A perfect motor will always spin at this rpm. <S> A practical motor has resistance which reduces the effective supply voltage and causes it to run slower as current increases. <S> When the motor is running free it only draws a small current to overcome internal losses, so the no-load rpm is generally only slightly lower than Kv*Vs. <S> What I'm thinking is that the frequency commuting defines speed, but the current/volts going through each coil defines the maximum force/acceleration the rotor can be pulled with. <S> Primarily it is Volts <S> which determines speed and commutation frequency. <S> At start up the rotor is stationary and producing no Back-EMF, so current is limited only by resistance of the windings. <S> This current produces torque which accelerates the rotor. <S> As the rotor speeds up it produces Back-EMF, reducing voltage across the winding resistance and lowering current and torque. <S> rpm stabilizes when torque drops to match frictional losses and shaft loading (leaving no excess torque for acceleration). <S> A sensorless BLDC controller constantly monitors the Back-EMF waveform and commutates at each zero crossing. <S> Unlike an AC drive <S> It does <S> not set a frequency and force the motor to spin at that speed. <S> The ESC generates its commutation frequency as a reaction to the speed the motor is already doing. <S> Speed is controlled by varying the voltage applied to the motor (usually with PWM). <S> Commutation must necessarily be kept in sync with the rotation, so you could say that commutation frequency is 'defined' by force/acceleration of the rotor, which is in turn 'defined' by the motor constants Kv <S> (velocity constant), Rm (resistance), and Io (no-load current). <S> Using these constants you can calculate the expected rpm and current with any load. <A> Frequency is the vastly dominant quantity Frequency <S> A BLDC/BLAC machine is basically just Sync machines and thus the frequency of the rotor is proportional to the electrical stator frequency (constant of proportionality being the pole pair count. <S> Voltage <S> That said voltage is equally key because you need to be able to inject AC into the stator. <S> With an increase in electrical frequency, the mechanical frequency increases which in turn increases the backEMF presented by the machine (where the constant of proportionality is Ke). <S> At some point you will run out of supply voltage to overcome the backEMF, the resistive drop & the inductive drop. <S> With increase speed the amount of current required equally increases to overcome the drag torque. <S> Add some mechanical load into the system and <S> the inductive voltage drop is comparable to the backEMF <A> Every BLDC motor rotates <S> commuting (one of six) divided by number of poles divided by six. <S> A 6 poles motor makes one electrical rotation(six commuters) per one mechanical rotation. <S> A 12 pole motor makes two electrical rotation(six commuters) per one mechanical rotation. <S> A 24 pole motor makes four electrical rotation(six commuters) per one mechanical rotation
| This can get complicated, so most controllers operate as in Bruce's answer : motor speed is controlled by voltage, with the BLDC sensing either the phase of the back-EMF, or motor position separately via hall-effect sensors or rotary encoders, and controlling the commutation frequency to suit the actual speed.
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What's the actual difference between edge sensitive and level sensitive interrupts Currently I'm working on a C8051F120 MCU where external interrupts can be defined in two ways: Edge sensitive (falling) Level sensitive (low-level) In level-sensitive interrupts as soon as the MCU detects a low level at the external pin it will execute the ISR which is the same as detecting a falling edge. I know I'm wrong as both can't be the same. Hence I'm asking this question: what's the actual difference between the two, in their detection procedure or in the execution of the ISR? <Q> Its exactlly what is says. <S> If edge interrupt is set, the ISR will only get fired on falling/rising edge of a pulse. <S> While if level sensitive interrupt (as you say) is set the ISR will get fired everytime there is a low-level/high-level signal on the corresponding pin. <S> In short, edge interrupt gets fired only on changing edges, while level interrupts gets fired as long as the pulse is low or high. <S> So if you have low-level interrupt set, MCU will keep executing the ISR as long as the pin is low. <A> A level sensitive interrupt and an edge sensitive interrupt are actually two quite different things. <S> I'll try to give some general insights that might help you understand how other interrupts work too. <S> Let's assume that your CPU can execute code in two modes: normal mode, and interrupted mode. <S> To go from normal mode to interrupt mode an interrupt, whatever it is, must happen, while to come back the IRET instruction must be executed. <S> Let's also assume that if an interrupt happens while in interrupt mode it gets somehow saved but it is not immediately serviced, i.e. when in interrupt mode the CPU can not be interrupted. <S> So what is an interrupt? <S> I would say it is an event : <S> something that happens, a timer overflows, a pin goes low, whatever. <S> The CPU does something to respond to the event then resumes normal execution. <S> What happens if an event occurs while another is being serviced? <S> Usually a bit is set in some register and just after the IRET instruction the CPU is interrupted again, checks which bit is set and executes the correct interrupt service routine. <S> When your ISR on the level triggered interrupt is executed you probably clear the interrupt bit as first thing: if the level stays low the hardware immediately triggers another interrupt that will be serviced just when you are finished with this. <S> In an edge triggered interrupt you need the pin to go high and then low again to trigger the interrupt once more. <S> I can't think of a meaningful example of when you would need a level triggered interrupt <S> , edge triggered seems much more useful and what you'd usually need anyway. <A> On many systems, interrupts may be divided into three categories--not just two; many systems only support two of the three, but there may be some difference as to which two they support. <S> An edge-triggered interrupt will cause the CPU to switch to interrupt mode any time interrupts are enabled, the interrupt line has been in the inactive state some time after the interrupt was last reset, and has been in the active state some time after that. <S> A "pure" level-triggered interrupt will cause the CPU to switch to interrupt mode any time interrupts are enabled and the interrupting signal has been in its active state since it was last reset <S> (if it was in its active state when reset, it may have simply stayed in that state). <S> A major advantage of level-triggered interrupts is that if while a device is being serviced for one reason, another reason emerges that would cause it to require attention (or another device using the same line requires attention), the CPU will keep revisitng the device until it is completely satisfied and has no cause to hold the reset line. <S> The primary disadvantages of level-triggered interrupts are that they often require that the CPU take explicit action to reset them (edge-triggered interrupts are often implicitly reset by the interrupt controller when the interrupt is dispatched), and that an interrupt which gets enabled when the CPU has no idea how to service it can lock up the system, since the CPU will do nothing except repeatedly invoke the interrupt handler because the device will continuously need (but never receive) attention.
| A "pure" level-triggered interrupt will cause the CPU to switch to interrupt mode any time interrupts are enabled and the interrupting signal is currently active . You might now see why level triggered and edge triggered are two different things: they are two different events.
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How can I attach a heat-sink to legacy ICs? I work (play) with some legacy chips like the mighty TMS9918 . Which, if you didn't know, doesn't require a heat-sink but the thing does get really hot. Amazing they don't fail more often. Anyway, I actually have a heat-sink for it that is the perfect size (came out of an old Colecovision) but I don't know how to attach it. Modern CPU and heat-sinks have the mounting holes in the motherboard that clamp it on top of the CPU. This heat-sink or motherboard has nothing like that. It's literally a piece of metal that sits on top of the chip. I have thermal paste but that isn't glue. In fact, the reason I have this is because it was glued (I'm guessing) to the chip but over time fell off. So my question is, is there a thermal paste that can also act as glue to hold this on? I want to protect these legacy chips as much as possible. Thanks. <Q> From what you describe, you don't want the heatsink to be removable. <S> For that reason I would suggest using Arctic Silver binary thermal adhesive: http://www.arcticsilver.com/arctic_silver_thermal_adhesive.htm Note <S> this is the Thermal Adhesive , not the Thermal Compound . <S> It's good for more than 150*C, and is one of the most popular thermal adhesives I've come across. <A> I'm going to throw in a couple of cents, since I don't think Arctic will be very cheap bought as a one-off. <S> But I may be wrong. <S> You definitely will not need that level of heat conductance on the top of a ceramic chip package, let alone a plastic one. <S> You can do two other things: As Plasma suggest google for "Thermal Adhesive", many cheaper types exist, as an example of what I use, bought in bulk, for not-so-heat-critical things: Farnell Listing for Fischer Thermal Adhesive <S> May also be obtainable through easier mediums. <S> This does have the disadvantage of needing a precise measurement tool for the stuff. <S> If you need a small drop mixed, you're looking at miligrams of hardening liquid. <S> But, what you can also do is much simpler and household-budgetty: <S> Take a drop (or small stripe) of thermal compound on the heatsink, press it onto something made of glass (bit of alcohol and an old cloth <S> will clean it off, no worries) and see how much it spreads. <S> On glass with high pressure (don't break your glass!) <S> it will spread more than on a chip with a rough surface in comparison, but not by 50%. <S> Now find the drop that will cover about 80% of your heatsink. <S> Once you have found the drop-size you need (and over time you will get better and better at it in one go if you need to do it often): Apply the drop-size you need, put some high-strength glue to the small bits that you expect won't get covered. <S> Such as brand name super-glue, or usually better with plastic dips that get hot: "heat proof" (upto <S> 100degrees C is fine) epoxy or silicone. <S> Press on and hold in place with a decent clip until the glue is fully hardened. <S> Done, preso-magnifico and you will likely be using the glue elsewhere often enough to validate its purchase. <A> You should just use regular epoxy glue. <S> I can't do better than quote my old answer : <S> Thermal adhesives are better, but that's missing the point. <S> Some numbers from wikipedia :Air has a thermal conductivity of 0.025 <S> (W.m-1.K-1).Aluminium is about 200, so that's a ratio of 10,000 times. <S> If you can fill the gap with thermal grease (0.9), or a polymer like epoxy (0.3), or even water (0.5), the thermal resistance of the gap will decrease by about 100 times. <S> 3M claims a thermal conductive epoxy with 0.72, but this is not much better than average plastic around 0.3 to 0.5. <S> The important thing is that these are all 100 times more conductive than air, and they fill the gap eliminating the air. <S> So select an epoxy that meets your mechanical needs, something with the right temperature range and flexibility. <S> You can probably use it somewhat above <S> it's high temperature limit, as you don't need full strength. <S> Prepare the surfaces properly and fill the air gap. <A> As the others said, ultimate performance is unnecessary for the interface material, since the chip's dissipation is likely pretty low to begin with. <S> I would use a less permanent solution than Epoxy: good old thermally conductive adhesive tape . <S> It is also much less messy to apply than epoxy goop, since it behaves just like double sided tape. <S> It is also removable, if needed. <S> Also, I did not check if thermal expansion of the ceramic package is similar enough with Aluminium's so they can be bonded without risk. <A> Gluing the heatsink with the surface of SMD will work to some extent. <S> A few cases it will with stand for dissipation but in several cases it failed by days. <S> I would suggest a continuous air flow in that area will give good result. <S> Calculate the space near and around the SMD IC, fix a mini fan that is more efficient and low voltage operated(5v or 12v). <S> In addition to the heatsink. <S> (I think it is easily available and you can fix it near the PCB.)
| You can buy it on Amazon for a relatively low cost: Arctic Silver Thermal Adhesive Good luck! Almost any adhesive will be suitable.
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Will a RCBO 4P trip in case of neutral fault/failure/interrupt/float? We have 3 phase power supply in my home which is used to power single phase 240V 50 hz appliances with common neutral. (Diagram in the link at the bottom) The only protection we have as of now are MCBs down each phase. But we recently faced a big problem. Few days back, the neutral wire of the distribution line outside our home broke and fell on the ground. This happened probably due to impoper maintenance by company, summer heat and heavy load. Whatever may be the reason, this fault damaged our wasing machine, air condtioners, water filter, laptop charger and other appliences. This was a big setback for us. So, I am trying to figure out Why this damage happened and What can be done to prevent this in future? Are there some device which can be installed to protect the home circuit? I researched a bit around web and it seems this problem is called neutral fault. This decreses the volatage across some loads and increases across others. High voltage can reach up to 380 to 400 V. Solution I have found so far is to install a RCBO-4P at input from main lines. So I want to know will this solution work? Will a 4 pole residual current circuit breaker with over load protection trip, if the neutral input is disconnected? I have tried to simulate what happened here. Thanks a lot. Also I found one product which match my requirement, Neutral Loss Protection Relay.But I am doubtfull about it's current rating . Update Wikipedia says, "To provide some protection with an interrupted neutral, some RCDs and RCBOs are equipped with an auxiliary connection wire that must be connected to the earth busbar of the distribution board. This either enables the device to detect the missing neutral of the supply, causing the device to trip, or provides an alternative supply path for the tripping circuitry, enabling it to continue to function normally in the absence of the supply neutral." <Q> Unfortunately not. <S> An Residual Current Breaker trips only if the sum of the current on all four wires is nonzero. <S> This residual current means there's a leak to earth, possibly through a human, and the breaker trips. <S> In your case, the loss of neutral meant the supplied voltages became unequal, some too high, but this won't make the total current nonzero. <S> There may be other devices which can protect you, but the basic RCB won't. <A> I am looking for the same sort of protection, and so far the best I can do <S> is use an undervoltage/overvoltage relay PER PHASE to neutral such as http://www.acdc.co.za/rhomberg/docs/AP221mon.pdf which then connects to a shunt trip like this http://www.oez.com/uploads/oez/files/ks/3223-Z01-06_EN_PL.pdf which is attached to a breaker for the phase (i.e. before your RCB). <S> Your relay will connect via the Normally Open part to the shunt trip with a 5 to 10 second delay. <S> If voltage higher or lower than your preset values it will trip the phase. <S> I woud say go for your safe range as 200v to 255v. <S> Hope this helps. <S> p.s. <S> there is an option to use a 3 phase unit with one shunt trip to a 3 phase breaker, but if one phase is lost on the supply side, the relay will trip the whole house. <S> Not ideal! <A> <A> While 4P (3P+N) RCBO won't trip in case of floating neutral where the current returns through one of the other phases, a 2P (SP+N) RCBO on each of the three phases <S> could (but probably not a good idea as explained below). <S> This is because in a 4P RCD the incoming current equals the return current even in case of floating neutral condition where the current returns through another phase instead of neutral. <S> I1 + I2 + I3 <S> + In = 0 . <S> Whereas in a 2P RCD installed for each phase the equation would look like this: <S> I1 <S> + In = 0I2 + In = 0I3 <S> + In = 0 <S> While theoretically this sounds like a plausible solution, I'm not certain of its practical efficacy when faced with neutral breakage upstream of the house distribution panel, say at the neutral leg of transformer. <S> And the biggest case in point is the European Standard EN 50550 <S> whose conditions for a power frequency overvoltage protection (permanent overvoltage protection) device include: <S> Prohibition on using earth leakage or current differential as operating principles. <S> Unless you can get one of these Power frequency overvoltage protection (POP) devices (which I couldn't), the solution is to use Voltage Protection Relay (VPR) and shunt trip (or contactor) per phase to disconnect supply on overvoltage. <S> I wanted both transient overvoltage (TOV) protection and permanent overvoltage (POV) protection, so the wiring diagram below includes Surge Protection Device (SPD - 3P+N) and VPR+contactor for each phase. <S> In case of transient voltage spikes (lasting about 30 microseconds, very high voltage/current like 6kV/10kA), the SPD should protect everything downstream to it. <S> In case of permanent overvoltage (lasting indefinitely, higher than normal voltages in the range of 400V), SPD would probably fail after approx. <S> 5 seconds (as per its TOV withstand capacity). <S> The VPR and contactor would respond in <300 ms so it should protect everything downstream to it. <S> Contactor used: Schneider Electric A9C20862, datasheet Voltage protection relay (VPR) used: Selec 900VPR-2, datasheet <S> SPD: <S> Any 3P+N SPD would be suitable. <S> Type 1+2 or Type 2 <S> according to your lightning risk, they all have the same wiring characteristics if its 3P+N.
| A 4P RCBO/RCCB won't get tripped in the case of broken neutral fault but three DP RCCBs connected in each phase will trip in the said case of neutral fault protecting all downstream single phase equipments/appliances.
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Why does measuring the voltage drop across a thing not simply measure the battery voltage? Imagine a trivial circuit with battery and one resistor. To measure the "voltage drop" across the resistor, we stick a voltmeter in parallel with it. However, this means the voltmeter is also directly electrically connected to the terminals of the battery. Therefore, why doesn't it simply measure the battery voltage, regardless of the resistor? I hypothesized that the answer is because some of the available voltage/pressure/force/power/energy/magic/whatever-it-is from the battery is being drained through the resistor, it will not be passing through the voltmeter and will therefore not be measured by it, but this still doesn't fit, because that means the voltmeter is not measuring the voltage drop; it's measuring the voltage that wasn't dropped. If the battery supplies 5 V and voltmeter reads 4 V, how is 4 V possibly the voltage drop of the resistor, when the available voltage has actually dropped by 1 V and I still seem to have a non-dropped 4 V to play with? I guess I really do not understand what the voltage drop is supposed to mean. I don't know why the resistor is dropping voltage at all, when it is supposed to resist current. I'm so lost. <Q> [I'm ignoring non-ideal behavior since that doesn't seem to be what you're interested in.] <S> Your assumption is wrong. <S> Measuring the voltage drop across the resistor does measure the battery voltage. <S> The general rules are: Components in parallel share the same voltage Components in series share the same current Ideally, adding or removing the resistor doesn't change the voltmeter's measurement at all. <S> The battery, the resistor, and the meter are all in parallel, so they all share the same voltage. <S> If the battery voltage is 5 V, then the resistor voltage and the meter voltage must also be <S> 5 V. Voltage is basically a measurement of potential energy due to an electric force field. <S> If you go around the circuit in a loop, you end up back at the same potential, which means you lose whatever energy you gained along the way. <S> (Gravity also works this way.) <S> When you move from negative to positive through a battery, you gain energy. <S> When you move through a resistor, you lose energy. <S> If a battery and a resistor are in parallel and you move around that loop, the energy gained in the battery will equal the energy lost in the resistor. <S> In other words, their voltages are the same! <S> This principle is called Kirchhoff's Voltage Law . <S> More formally, it says that the sum of the voltages around a closed loop must equal zero . <S> Resistance describes a relationship between the voltage across the resistor and the current through it. <S> So resistors do "resist" the flow of current, but the way they do that is by dissipating energy. <S> It's similar to the way that friction resists the movement of an object. <S> Now as others have pointed out, in real life a battery is not an ideal voltage source. <S> The voltage of a real battery changes depending on how much current is being drawn and how much charge is left. <S> So in real life, adding a resistor can change the battery voltage. <S> But the battery voltage and the resistor voltage will still be (almost) equal. <S> (The parasitic resistance of the wires is normally very small.) <S> Hopefully this has clarified things. <S> Please feel free to post follow-up questions if you're still confused. <A> Because we are living in a non-ideal world, where batteries and wires have non-zero resistance and the actual circuit is as following: simulate this circuit – Schematic created using CircuitLab Voltage measured in such a circuit would be \$V=5\frac{R_1}{R_{wire}+R_{battery}+R_1}\$ which is less than 5V. <S> The non-ideality of the voltmeter is not taken in account here. <A> Lets simplify, the resistor or load in the circuit is consuming voltage. <S> If you were to remove (R1) creating a open circuit essentially you would be measuring battery voltage. <S> closing the circuit is putting the load to work. <S> in the first example r1 the meter should read 5v, showing the resister is using full battery voltage. <S> (KVL) voltage drop is an excellent way to identify undesirable loss in a circuit. <S> if in example 1, the meter reads 2.5 volts and a known good battery is <S> 5v would indicate unwanted resistance in the circuit. <S> (bad connections, long distance etc.)this is known due to KVL, and could be verified by doing a voltage drop from battery source to load, and after load to battery ground. <S> the total readings will equate 5v and identify the unwanted resistance. <S> in multi load circuits it can identify malfunctioning loads and voltage use from individual loads. <A> If you have an ideal battery you got 5v when you measure the voltage across the battery even if you put a resistor of 100 Ω Because it is ideal battery i.e. no internal resistance <S> But in your case you measured 4v instead of 5v which mean your battery have an internal resistorAnd by apply a kirchouve law you find 4v with neglecting the wire resistance which is in mili ohms VR=(V*R)/(R internal+R) <S> VR=(5 <S> *100)/(25+100) VR=4v <A> in your example, measuring the voltage drop across the resistor may not exactly matchthe results of measuring the voltage of the battery terminals. <S> think of very long leads between the battery terminals and resistive load which are also adding resistance in the mix. <S> in that scenario choosing where exactly to measure becomes practical. <S> Assuming a resistor is simply directly tied in parallel with the battery, there will be practically no voltage drop as the resistance will be extremely low. <S> that scenario also provides a few extra complications. <S> the battery is not an 'ideal' power source, and you are just essentially measuring the voltage of a battery as a resistor drains it down.
| The battery voltage and the resistor voltage are the same in your circuit.
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