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Capacitor exploding in a paper plant Two years ago I was in an internship in a paper plant in the Superintendence of Electrocontrol, in the area of maintenance of electrical equipment. Well, they had a problem regarding some capacitor banks that were installed to mantain the power factor at a certain level. The system was simple, it had multiple capacitor banks that were switched to change the reactive power and control the power factor, and they did have many switchings over a day. The problem they had is that, capacitors would not last long, they kind of exploded on a month basis, and this was expensive due to the maintenance and the need of more capacitors to replace the useless ones. I am kind of puzzled, because it was a 3-phase system, and the capacitors nominal voltage was higher than the voltage of the source they were connected (380 line voltage, and capacitors were approximately 600 V) My question is, why did capacitors "explode" so quickly? They should have lasted at least 3 - 6 months (1 year expected). I was thinking of over-voltage due to the transients when switching and producing like a "chain-reaction", less capacitors => higher voltage transient. However there is long time since I have not solved a circuit problem, could someome explain it in more mathematical terms? Am I correct in my statement? <Q> I would suspect harmonic distortion of the system voltage due to nonlinear loads. <S> The capacitors are sized to carry current due to the complex impedance of the capacitors at the line frequency. <S> If the actual voltage waveform has significant content of voltages at harmonic frequencies, that increases the capacitor current because the impedance is lower at those frequencies. <S> In addition, harmonic distortion can precipitate a harmonic resonance situation further increasing the capacitor current. <S> The nonlinear loads can be electronic motor speed controls or other power electronic equipment on the three-phase distribution system or even electronic equipment on single-phase distribution systems that are fed from the three-phase systems. <S> Many large motors in paper plants are powered through electronic power conversion speed controls. <A> Like anything, capacitors get hot when current flows through them. <S> The amount of current flow depends on the voltage and the complex impedance of the cap at the line frequency. <S> In the simplest possible analysis, RMS current will be \$V_{line} \times 2 \pi f C\$. <S> Typically, capacitors used for this sort of application will have an AC voltage spec, which takes all that into account. <S> Of course, you may have multiple capacitors in parallel which are not sharing current properly. <S> In that case the one with the lower impedance (higher capacitance) dominates. <S> Or there might be other effects contributing to capacitor currents besides the line voltage. <S> I've heard of cases where resonant effects between different capacitors in a plant expose them to higher currents than they would otherwise be expected to see. <S> I've even heard of cases where the failure of one capacitor changed the resonance, making the problem go away. <S> Until someone replaced the blown capacitor, of course... <A> Capacitors can fail early if their ripple current rating and ambient operating temperature rating is being violated. <S> If the ripple current rating for these 600V rated caps was 20A, but they were seeing peak AC current ripples of 30-40A (unit magnitudes are examples only), then the capacitors will wear down much faster than what they are rated for due to internal temperature rise thanks to ESR (equivalent series resistance). <S> Ambient temperatures being high will make this even worse. <S> The reality of contact resistance and parasitic inductance means no capacitor will share current exactly the same. <S> Some may bear just above their ripple current rating, while others down the chain (if you model parallel caps as a chain connected by parasitic resistances in the contacts and wires between them) have far less. <S> The ones closer to the input source will charge and discharge current more than the others, and will fail faster due to the above reasons. <S> I also found a useful document on PF correction capacitor banks and their failure modes here .
| The most likely cause of failure in this case is that the capacitor got too hot.
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How can I quickly and safely make defunct electronics safe and free of charged capacitors? I am planning a movie set which will contain old circuit boards from TVs, VCR's and other old electronics. The plan is to go to several Goodwill stores and buy up all of the TV's and such that they have, open them up and integrate them into the set. There is an electrical danger here due to charges still present on large capacitors. What is the best way to quickly and effectively remove these charges? Dunking these devices in a salt water bath seems like a good idea (we do not need for these devices to work.) If so, are there things we need to be aware of? If not, is there a better recommendation? We will be getting around 60 items, so finding and discharging all of the capacitors individually could be a big problem. <Q> However. <S> Do not dispose of the water carelessly afterwards. <S> There may be traces of lead and other nasties in it. <S> Find some sort of hazardous waste disposal company and get them to get rid of it (it may cost you a few $$ but it will be quick and fairly safe <S> if you are careful, rubber boots might be a good idea). <S> Also, I have no idea how well sealed up everything is inside them but some of the capacitors might not be shorted by the water properly, if you don't go digging through things that look like they may act as a sealant and you make sure you get all of the air pockets out when you give them a bath <S> I think you should be OK however. <A> They sell "wands" to do this sort of thing (Google for "electric discharge wand") but you can make your own. <S> It's just a wire with a clip on one end that you can touch to one end of the capacitor. <S> The wire goes through a resistor to a clip on the other end for hooking up to the other side of the capacitor. <S> Of course, you don't want to be handling that with your bare fingers so ... <S> And with that last line, I have to stop cause it's been far too long since I've done that. <A> I'm going to say that it is ok to dunk each component in a salt water bath. <S> It salt water will short out all of the caps and neutralize any lost over charges. <S> This is fast and generally effective against and charged component that a person may accidentally touch later. <S> This method is fast and effective for items that you don't want to function anymore. <S> Be sure to wear gloves and don't come in contact with the water at any point during the activity.
| The trick is to find the right resistor value and size but that can be calculated judging by the size of the capacitor and how quickly you want them to discharge. I feel that giving all of the items a salt water bath is probably the best option for your circumstances.
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Light effiency or efficacy and watt to heat output I am on a light forum and there are debates about watt heat output of lights. I'm using my common sense for the debate but can't find the data online to support my ideas. People are thinking that a 400 watt heater produces the same heat as a 30% efficient 400watt light. And further same as 400 watt of 60% efficient lights. Now my logic says that heat being the main by product of a heater will of course cause the 400 watt heater to produce the most heat A light that's 60% efficient at light production like new LEDs cannot produce the same amount of heat because its byproduct is photons or light. But still produces heat as its not 100% efficient. Can any of you geniuses explain to me smartly what heat production wold be produced from lights? HID lighting is about 30% efficient New led is getting to 60% and more But people think that 400 watts translated directly to heat update: I think I figured it out. The room is much cooler because: the 60% energy as light from LEDs is projected to concrete floor which goes for many meters outside the room and connected to the earth thus dissipating it effectively outside of my room. The remaining 40% heats the rooms air Where as a less efficient light uses 70% of its energy to heat the air and only 30% is sent to the concrete and this dissipated to earth and other rooms. Am I correct here? – <Q> Ok, so as a rule:$$Watts_{in} = Watts_{out}$$This is always true unless you're storing energy somewhere. <S> A 400W heater will produce 400W of thermal energy. <S> * A 400W incandescent will produce 400W of thermal energy and light energy combined . <S> Unfortunately if you measure the amount of light energy being produced, it is rather small compared to the total energy consumption of 400W. <S> A 400W LED fixture will produce 400W of thermal energy and light energy combined , but the amount of light energy is going to be a more significant figure in the process. <S> Now you will be able to tell that it's not 400W being lost to the heat sink, but some amount less than that. <S> (This ratio is the efficiency). <S> The rest is being emitted as mostly-visible light. <S> However, if you put the LED in a well-insulated black box, you would not be able to tell the difference between a 400W lightbulb and a 400W LED by the temperature of the box. <S> Efficacy is not the same as efficiency. <S> It takes into account the human eye response that determines how bright the light is, as produced from the source. <S> This is not in units of Watts/Watt like efficiency, but in Lumens/Watt. <S> For example, comparing a white LED to a UV or deep blue LED, the white LED would have much higher efficacy compared with the deep blue LED, even if the amount of energy released in photons is the same, because the human eye does not respond to deep blue very well. <S> * <S> As an aside, a 400W heat pump can present more than 400W of thermal power at its output, but it does that by moving heat rather than purely producing it. <A> A 400W heater and 400W, 60% efficient bulb will heat a room roughly the same amount. <S> Most of the energy that starts off as light will become heat once it's absorbed by items in the room. <S> The only losses would be out the window or by other energy conversions that somehow make it out of the room (like a solar panel). <A> If you're satisfied with monochromatic green light <S> it's more like 700lumens <S> /watt. <S> I don't think commercially available lights are much better than 100lumens/ <S> watt (though claims of higher efficiency for prototypes and for chips ignoring the power conversion circuitry float around). <S> Russell may have more to add on this. <S> So actual efficiencies are more in the 20% range and less when you take into account circuitry, ballasts, covers and so on. <S> It sounds like you are looking at actual bulbs. <S> There are a number of reasons why you may be seeing a discrepancy. <S> One is that electronic bulbs do not draw current evenly, they may have a lot of harmonic content so your wattmeter may not be accurate. <S> If there is actually light escaping through a window then there would be a slight effect, but you'd have to do an accurate well-controlled test to be sure about this- outside solar radiation and temperature (and adjacent room temperatures) would figure into it. <S> I think you'd need a fairly carefully controlled experiment to see the difference. <A> Thanks for your replies. <S> There's no windows. <S> It's mostly concrete <S> If you search Phillips is producing a 200 lumens per watt LED T8 light to replace the tubes. <S> Cree produces a cob led that when ran low power it is up to 70%efficient at 6500k at 3500k they're about 60% efficient with 25 watts. <S> Producing 200-240 lumens per watt depending on color of course. <S> I understand lumens and how it's really only for our eyes and different color lights effect the reading. <S> I was trying to understand why a 400w HID light at about 30% efficient heats <S> my room so damn much when 400 watts from very efficient led isn't close. <S> As discussed here it appears that because the efficient LEDs transfer that electricity to light more it is thus hitting my cool concrete where the less efficient light turns more of its electricity in to heat directly not light. <S> So I believe that's the answer. <S> I was thinking from knowledge of refrigeration aspects and energy needed to make gas cooler and the delta. <S> So it made me assume that the higher efficiency lights are less restrictive and cause lower heat and more light. <S> Wasn't thinking how light itself is heat or energy. <S> Still kind of confused.
| As far as I understand it the theoretical maximum efficiency converting light with reasonable color spectrum is something like 400 lumens/watt. In general, the actual efficiency of a light source (in Watts emitted as light) is not published.
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How to limit current in order to power small car radiator fan with laptop charger? I'm trying to do a makeshift small grinder/sander out of a car radiator fan, but I want to power it up with a laptop charger for convenience. Since it's a salvaged fan, I wasn't able to look up its schematics, I only know it takes 12VDC input (as most car radiator fans do).These fans typically go from 10-15 Amps while running, with up to 25 Amps inrush. The laptop power brick is rated at 18.5 Vout, 6.5 Amps, 120 Watts. I connected it as-is for a short period, and everything works fine, I get satisfactory RPM and torque on the fan. However, I wouldn't want to leave that setup running for a long time, for multiple obvious reasons. I'd like to be able to have some sort of potentiometer as well, in order to be able to regulate its speed. So far, I've only come up with this: Limit 16 Volts on the laptop output via triac voltage regulator or similar (at this point around 16v) Static resistor rated at 100 Watt / 3 Ohm maximum (at this point making sure it'd be around 16V 5A) Variable resistor (pot) rated at maybe 100 Watts / 100 Ohms (resistance is less important than wattage at this point) The problem with this setup, is that those kinds of resistors and pots are hard to come by, and usually expensive. Can anyone come up with a better / more intelligent solution and suggestions? Most answers I found are related to more common uses, at usually lower amperage. I have moderate experience with electrical appliances, what I don't know I'm willing to research, so don't hesitate to go wild. <Q> Is there a reason you're not controlling it via PWM? <S> The main argument against is the complexity, but would give you the effective reduction of voltage you're after. <S> Though there are potential issues with using PWM on an input to (what I assume is) <S> a DC brush-less motor <S> , you may want to end up smoothing it out with capacitance, but then it becomes a DC-DC converter... <S> Third option would be to put a known fixed resistor in series with the fan, and a lower rated pot across the pins of the motor. <S> So R1 is a high wattage (100W), low resistance device, R2 is a high resistance POT (you should really put a high value resistor to go in series with the POT to make sure you can't set it too low). <S> But his is a pretty wasteful way of controlling the fan, burning energy in the resistors rather that switching off as you would with a switched DC-DC or PWM option. <A> Cheap, Simple, but no over current protection. <S> Several resistors (resistor bank) in series. <S> Put a switch across each resistor bank. <S> Opening or closing the switches individually will change the overall resistance. <S> I will leave all the ohm's law calculations for resistance and power rating of resistors up to you. <S> EDIT : <S> Probably better to use the resistor banks in parallel (with switch then in series with each resistor bank. <S> This allows use of higher value resistors. <A> Just thinking a power transistor with a potentiometer as voltage divider/ base current control should work. <S> Just never fully open the transistor or something will die. <S> ;-)
| Alternatively, as your current plan is to change the voltage across the motor, it isn't insane to considering using a DC-DC converter with a potentiometer on the voltage setting pin.
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Buck converters- No Load condition How exactly do buck converters work under no load condition? The capacitor voltage will try to charge up to the peak input voltage, with no load connected across it to discharge. How to avoid this problem? ** I've included the simulation result. The output voltage (green curve) reaches 12V (peak input) slowly. I don't understand why the input voltage to filter has spikes. Even if the controlling circuit alters the duty cycle, wouldn't the output reach the peak input value if there is no load? <Q> A synchronous buck converter has no problem because it has two low impedance states in the push-pull output - it is either switch hard to the incoming supply voltage or switched hard to 0V. <S> In other words it's a voltage waveform generator and, a simple LC low pass filter then behaves as an averager: - However, it's different for the non-synchronous buck regulator because it only uses a single pass transistor and a flyback diode. <S> Under these circumstances, and assuming a perfectly efficient scenario, the duty cycle MUST drop when the load current decreases. <S> If the duty cycle doesn't drop, the output voltage will rise beyond acceptable limits. <S> Under near-no-load conditions a non-synchronous buck regulator must go into a different mode of operation where instead of operating at a fixed frequency of (say) <S> 100 kHz, it pulses at a much lower rate. <S> This is quite common to happen in many buck circuits. <A> Newer buck converter offer a "SKIP MODE" (with different trade mark names from company to company, <S> Linear Tech call theirs "Burst Mode") which basically makes the converter skips some of the MOSFET driving pulses, when low loads are detected by the controller. <S> This tutorial from Maxim Integrated explains skip mode in a bit more detail. <A> No, the capacitor won't charge to the input voltage. <S> That is because the switch stays off when the output is high enough. <S> This is of course assuming a buck converter rated for 0 load. <S> Most are, but some aren't.
| Most buck conveters are capable of going to 0 PWM duty cycle, or infinite time between pulses, depending on their design.
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What happens if there are many 120 ohms resistors as terminators on the CAN bus? I connected three devices in parallel on a CAN bus system. Every device has a \$120\,\mathrm{\Omega}\$ resistor built into the circuit. I am using two pairs of twisted cables of \$1.2\,\mathrm{m}\$ each for the communication. Will it have any effect on the communication? <Q> Assuming that the data receivers are capable of working with the smaller signal (which I'm sure they are), the only aspect that might cause problems are signal reflections. <S> Given that data rate is 19,600 bps, if you said this was a square wave of frequency 10 kHz, I'd be considering the wavelength of the 7th harmonic in order to justify that the cable length was OK. <S> 7th harmonic is 70 kHz and this has a wavelength of 4.3 km. <S> General rule of thumb is that a badly terminated cable will be OK if the length of the cable is below one-tenth of the highest wavelength and clearly it is. <S> Further reading: <S> Why does the CAN bus use a 120 ohm resistor as the terminating resistor and not any other value? <A> You need to have a 120Ω resistor at the beginning of the chain and at the end of the chain. <S> If one is missing, it won't work (trust me!). <S> CAN is intended to be a linear chain with termination resistors at each end. <S> Best to figure out a way to get rid of the middle termination resistor. <S> Sometimes this is just a jumper, but you might want to desolder or clip it if that is necessary. <A> If, for example, you had ten 120-ohm resistors, each transceiver will have to deliver 5x the normal current. <S> This will make them heat up, make them unable to complete proper transitions, or both. <S> Three resistors instead of two may not be the end of the world. <S> But it's still stressing the transceivers unnecessarily.
| If you have too many, it might (MIGHT) work, but you're asking for trouble. The more resistors you have, the bigger a load there is on the transceiver trying to drive the bus.
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Cheapest timer for radio trilateration I'm building a device that must find its own position on a field (400x400m). Since there will be lot of devices on the field I could not send them their position individually and I want them to be passive (regarding signal emission). My current plan is to setup 3 rf emitters on the field one master and two slaves. The master on one side emits a signal every second. On the opposite corners are the two slaves emitters, they each send an echo signal when they get master signal. In the device I need to measure the time between master pike signal and slaves echoes. This makes 2 durations. Then send these duration to a micro-controller that will compute position. I don't need a big accuracy, 1m will be enough. But this means a timing accuracy of 3ns wich is quite a lot. It must be cheap, let say less than $50 each for 20 devices (prototype), and less than $2 each for 10.000 pc (production batch). What kind of cheap component can measure a time gap with such an accuracy ? I'm also interested if you know any other solution to this cheap positioning problem. ---------- Edit I'm not asking for any design service. I explained the context to help other understand the needs. I think my question is quite simple for someone experienced. Let's ask it another way : Given two impulse signals, how to mesure the time gap between them with 2-3ns accuracy ? I'm thinking of a crystal and a counter. Does it look ok, or plain crazy ?The microcontroller could calibrate itself using the master signal, so if the crystal frequency is not precisely set it's not a problem providing it is constant over time. <Q> This IS a design service, within reason, AND people will close your answer before starting to think (under 5 ns) if you mention commercial alternatives. <S> So, maybe you can't win :-). <S> A fast enough comparator (they exist) will allow time of an "edge" relative to a local clock to be turned into a pulse. <S> A local clock and counter will allow clocks between start time and trigger time to be determined. <S> People make short range time of flight ICs made for cellphone gesture detection (believe it or not) with times approaching what you want. <S> You should conceptually look at as many alternatives as you can think of. <S> One or more may even work. <S> You may be able to do computed tomagraphic position detection by seeing which targets are blocked or affect the signal along a number of paths. <S> You say there are "many" devices but you need to be more specific about quantity and blocking characteristic. <S> If there are eg 100 each 1 metre square and 5 metres tall "there may be problems" with some systems. <S> If they are say 300mm x 300mm and 200mm tall <S> so you can "look down on" the field fromm an angle you may be able to use cameras or barcode or similar pattern reading or ... <S> You say you want them to be passive wrt emission but do not say why or how close to a "MUST" your "want" is. <S> IR LEDs each coded could be very helpful. <S> Transponding LED tx's could be VERY helpfil (Poll and respond. <S> Resonant circuits in each device that respond to whatever could be helpful but <S> does this violate your radiation objection if passive - or active? <S> A lot more flesh and blood will help with a good answer. <A> Does it look ok, or plain crazy ? <S> The microcontroller could calibrate itself using the master signal, so if the crystal frequency is not precisely set it's not a problem <S> providing it is constant over time. <S> It's plain crazy using your method <S> I'm afraid - a slave receiver would have to decode the master message with an MCU and the time taken to lock into the signal <S> could be several tens if not <S> hundreds of cycles of the transmitted carrier. <S> If the carrier is 1 GHz then somewhere between 10ns and 100ns (or longer) would be taken up just by the locking-in process. <S> That's a big gap in time to issue a slave response. <S> Also, one cycle of an MCU might be +/- <S> 20nsec error. <S> You cannot guarantee that the MCU operating frequency is somehow synchronized to the master transmission hence the uncertainty error. <S> I think you are miles off getting 3ns accuracy or even resolution. <S> BTW it's called Trilateration and not triangulation <A> Check out the amazing TDC7200 Time-to-Digital Converter... <S> http://www.ti.com/product/tdc7200 <S> It can measure time with ~55 picosecond resolution (!) and costs about $2 at 1K quantity.
| Rotary scanning of a field of sensors with a number of transmitters may be good enough.
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Why are Arduino peripherals called shields? When looking up electronics shields the only useful information I could find pertained to Electromagnetic shielding. But when inspecting Arduino shield schematics most offer no protection to the Arduino. So it begs the question, why are Arduino peripherals called shields? https://en.m.wikipedia.org/wiki/Electromagnetic_shielding http://m.instructables.com/id/ATtiny-Programming-Shield-for-Arduino-1/ <Q> I doesn't look like they've registered a trademark for it ("shield"). <S> The mostly obvious explanation is the mechanical position of the daughterboard. <S> One book says: <S> These boards are called “shields,” because they usually fit over the top of Arduino like a protecting shield. <S> The setup is by no means new. <S> I had an ancient (1990-era) video card that had a 2MB addon-VRAM card (for a total of 4MB); it looked like the image below <A> It's hard to answer questions "why somebody did something non-technical in a certain way". <S> I can think of 2 possibilities, though. <S> Primordial ardweenies haven't thought it through. <S> More likely, they knew better, but decided to go with a fun terminology like "sketch" and "shield" and <S> sacrificed technical accuracy to marketing. <S> Recently, I was reading an application note written by a large company. <S> When they had to mention Arduino extension board, they wrote "Arduino extension board". <S> [I can't seem to find that app note again. <S> It was by Texas Instruments, iirc.] <S> Finally, the question has been already answered here . <S> (trivia: Arduinos are named after Bar di Re Arduino pub, where the primordial ardweenies were meeting up. <S> Source. ) <A> Daughterboard, shield, hat, cape, ... <S> The standard name for a subsidiary printed circuit board (PCB) that plugs into a main PCB is "daughterboard" <S> However, the somewhat artsy communities that produce these small experimental microcontroller boards each tend to invent their own name for daughterboards that have the required dimensions and connector to mate with the primary form of main PCB in that community <S> Arduino: shield. <S> Raspberry pi: hat. <S> Beaglebone: cape. <S> So they tend to borrow the name of some type of accoutrement that a human might loosely attach to themselves. <S> There is a vaguely protective theme here <S> but I think that is largely irrelevant. <S> I'm not sure what the first Arduino shield was <S> but it was probably something that mediated between the Arduino and the outside world. <S> Also from its shape it isn't too dissimilar to a rectangular shield <S> , I suppose it also (uselessly) shields the Arduino from light (and fingers). <A> It's a brand name. <S> Compare it to the names that other manufacturers of other systems give to their plug-on, stackable, add-ons. <S> Why did they choose that brand name? <S> It had to be different from the others. <S> And easy to say. <S> And it helps if it conjures up something small and single-layer. <S> And has good rather than bad connotations.
| Most Arduinos have no (or little) electrical protection on their inputs and outputs, at least some shields provide a modicum of additional protection - although that is usually not their main purpose.
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Why is designed active low? I buy a 2 channel relay module for Arduino.I surprised that why this relay module is designed active low? Circuit: i means that when we connect In1 to low V (GND) relay turn on. Is there any reason for using this way instead of turn on relay with high V? sorry for bad English. <Q> One reason is historic. <S> Early bipolar logic families such as TTL, DTL or RTL used only NPN bipolar transistors, they could easily drive an output signal to be pulled to within a few millivolts of ground but not easily drive a high signal. <S> So many integrated circuits tended to use an active low signal output. <S> That also allowed WIRED-OR functionality where a virtual gate could be created from open collector signals. <S> The use of active low-signals predates integrated circuit logic and is often used in cars where a switch may just ground a signal to assert it - for example the brake switch or dome light switch often just connects the signal to ground. <A> Having the input <S> active low makes it easy to drive the module with an open-collector NPN transistor, or other open-collector-like things such as the ULN2803. <S> There may be a historical reason, as the outputs of bipolar TTL logic chips could sink (pull down) much more current than they could source (pull up). <A> I surprised that why this relay module is designed active low? <S> Well you could remove the "Jumper" and connect 5V directly to JDVcc, short In1 to ground and put your input into R1 via the Vcc labelled connection on the jumper: <S> - Now you have both options because this version is now active high. <S> I'm assuming that you have shown all the connections and that these three points are wirable independently of other channels.
| In systems where ground is available everywhere (the chassis of a car for example) it saves a wire as you just need one wire to the load rather than bringing power to the switch and also taking the signal away to the load.
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Distributing power to varying resistance Can a regulator be built to divide the output of a DC power source to supply identical amperage, in the range of 1 to 5 Amps, to 16 individual conductive parts in an electrolytic solution Said parts are connected to the system with varying contact resistance. It will be required that voltage to each different location will be controlled to assure the amperage to each part is constant The parts would be plated in an electrolytic solution. The desired end result will be the delivery of the same amount of charge (number of coulombs) to each part. <Q> This paper describes a good overview of the entire electroplating process, including a blurb about current density and plating thickness. <S> "The current divided by the apparent area yields an average figure. <S> Except for the simplest geometries of a cell, such as when the anode and cathode are concentric, the current is not uniform over the surface of an electrode." <S> So if you were hoping for a "magic current value" to make everything plate wonderfully, then it doesn't exist. <S> What will help, is placing more anodes in or near the hard-to-plate areas, such as inside circles or tubes, depressions, cavities, etc. <S> Surprisingly, there is little info to be found about power supplies specifically for electroplating. <S> There are a few big vendors, such as Caswell or Mastech , and these may be fine for a commercial or industrial application. <S> Some specialty supplies may be a "pulse" type, which either pulses the current on-of-on, or reverse the current altogether for a short time. <S> There is some debate as to whether this may improve the resulting plate, and like everything else in electroplating, depends on many factors. <A> If each part has a different resistance, then each part will need a different voltage to achieve the same current. <S> So you'll be able to do this, but not with one regulator - you'll need one per part. <A> You can use the one unregulated power supply to drive multiple (adjustable) constant current sources. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> A simple way of doing it might be to use a lamp as a 'constant current' source on each circuit to balance out the currents. <S> Not perfect, not efficient, but very simple and gives a bright indication of how each piece is doing. <S> Would a 55 W, 12 or 24 V car / truck headlight bulb do the trick?
| What you're after is a constant-current source , where the current is set, and the voltage varies according to load resistance.
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What kind of microscope is used to see the electronic flow on a microchip? I've found an image, where anybody can see the electron flux on an active microchip (especially the conducting-path's). It's pretty amazing to see such electric flow and so I would like to know what kind of microscope is used for that? Can somebody help me please. <Q> I've seen old still photos from the 1970s where a reversed diode junction is visible, and where metallization has black or silver color depending on surface charge. <S> I've always wondered if they ever used this technique to observe slow-clocked logic. <S> (The e-beam may inject negative charge and interfere with analog, or with any floating inputs.) <S> Just went looking, and immediately found a youtube vid of animated voltage signals , an ad for someone's SEM system. <S> Here's a much older one: MOS transistor sections viewed with SEM. <S> Search for keywords "voltage contrast," also VC-SEM <A> That's not the electronic flow, it's the metalization layers on the die. <S> In general, for larger feature size lithography, a normal optical microscope is completely sufficient. <S> I'm not sure what you mean by "electric flow". <S> If you're saying you can see the electrical activity of an active IC, I haven't seen anything like that. <A> I think this is indirect visualization of the current via thermal imaging; i.e. what you see is the Joule heating. <S> The thermal imaging itself could be via a AFM or something coarser depending what scale we're talking about; that's unclear from the question. <S> AFM can usually be used only for the surface temperature; for sub-surface temperature there's Scanning Thermal Wave Microscopy (STWM) ; it has been used for VLSI. <S> Again, depending on the scale, it could be something as simple as an IR macrolens. <S> The hot line in the middle below is a PCB trace. <S> Image from: http://www.photonics.com/m/Article.aspx?AID=57307 <S> As for scanning thermal microscopy, you can do much more detailed stuff with it, like visualize the pinch-off effect in silicon nanowires : <S> (For background, these nanowires are "junctionless" FETs ).This <S> visualization technique was used on "classical" MOSFETs too , a decade ago, but the B&W images in that paper are of poor quality (scanned at bad contrast by IEEE). <A> Has to be an atomic-force microscope . <S> They are the only type that could get reasonable resolution and sensitivity in such an application.
| Conventional electron microscopes, SEMs, can 'see voltage' on conducting surfaces when adjusted just right.
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PIC12F509 Switch Debouncing On Press Release Or Both? I have a question regarding switch debouncing I'm using a PIC12F509 and the internal RC oscillator operating at 4 MHz. I'm toggling an LED on and off connected to GP1 (output) via a momentary push button switch on GP3 (input) which is held high via a 10K resistor when the switch is open. I'm using the following macro and the code below to debounce the push button switch. It utilizes timer 0 set to a 1:256 prescale. It waits until the contacts have remained in the same state (open in this case) for 10 milliseconds before executing the rest of the code. If they bounce (go low) timer 0 gets reset. DbnceHi MACRO port,pinlocal start,wait,DEBOUNCEvariable DEBOUNCE=.10*.1000/.256 ; debounce count = 10ms/(256us/tick) pagesel $ ; select current page for gotos start clrf TMR0 ; button down, so reset timer (counts "up" time) wait btfss port,pin ; wait for switch to go high (=1) goto start movf TMR0,w ; has switch has been up continuously for xorlw DEBOUNCE ; debounce time? btfss STATUS,Z ; if not, keep checking that it is still up goto wait ENDMmain_loop; wait for button presswait_dn btfsc GPIO,3 ; wait until button lowgoto wait_dn; toggle LEDmovf sGPIO,wxorlw b'000010' ; toggle shadow registermovwf sGPIOmovwf GPIO ; write to port; wait for button releaseDbnceHi GPIO,3 ; wait until button high (debounced); repeat forevergoto main_loopEND Now onto my question up until now I've been debouncing both the button press and release. It's been brought to my attention in the tutorial I'm following that it's not normally necessary to debounce both the press and release of a switch. I can understand why in theory because after the switch is read as low via the btfsc instruction (even if it's momentarily because the contacts are bouncing). The LED is toggled and the debounce high code will wait until the contacts have opened and stopped bouncing before executing the goto to loop the code from the beginning. On the positive side this saves program memory. But if something seems too good to be true it usually is what are the potential pitfalls using this method? Obviously for a simple application like this the code is overkill. But for something more critical like putting the PIC in sleep mode and waking up from a pin change any bounce must be eliminated in order for the micro to enter the leave sleep mode reliably. What are your opinions on switch debouncing do you debounce the press and release or only one? My guts telling me to go belt and suspenders and just debounce both to be safe but I have no logical reason for it. <Q> Mechanical switches bounce on both transitions. <S> The bouncing on opening is usually shorter than on closing, but can certainly happen. <S> I generally use 50 ms debounce time in both directions. <S> A long time ago, I tested a bunch of pushbuttons. <S> Most stop bouncing in 10-15 ms, but some took much longer, like 40 <S> ms. <S> I settled on 50 ms since that seemed to be long enough to be safe with just about any switch unless you deliberately try to keep it at the transition. <S> I also did some testing on human delay perception, and concluded that 50 ms was about the limit you can get away with without the delay being noticed, unless perhaps someone is specifically looking for it. <S> Most microcontroller projects have a 1 ms periodic interrupt for other reasons anyway, so I usually add debouncing code to that. <S> I keep a counter and a global flag for each switch. <S> The flag indicates the current debounced state of the switch, which is what all the application code uses as the switch state. <S> The counter is private to the interrupt routine. <S> It is set to 50 whenever the actual switch state matches the debounced state, and decremented by 1 when they differ. <S> When the counter reaches 0, the switch becomes officially debounced in the new state, and of course the counter is reset to 50 since now the switch and the debounced state agree. <S> I have used this basic approach in many microcontroller projects, and haven't had any trouble with bouncing buttons or users complaining (or even noticing) <S> the 50 ms delay. <A> Although the data might be skewed one way or the other, I wouldn't trust either direction to be completely bounce-free. <S> When I first started with embedded software, I would use timers for debouncing - ignore further transitions within x time of an accepted one - but as my I/ <S> O count went up, it quickly became unwieldy. <S> Now I scan the entire user interface at a low enough rate that the switch-bounce no longer matters but still fast enough to appear responsive. <S> ** <S> This includes both buttons and LED's. <S> In other words, all of my user interface code is in one place and runs periodically at that rate. <S> Your low-level drivers (PWM, shift registers, etc.) can run faster than that if necessary or convenient, but there's no point in refreshing their values any faster. <S> ** <S> The reason this works is if I read it mid-bounce: <S> I can read the same as the previous sample, meaning no change <S> and I'll catch it next time. <S> I can read the same as the next sample, meaning that I caught it this time. <S> Either way, I get a clean transition in software simply because I sample it slower than the bounce time. <S> But it's still faster than the user's perception of "instantaneous", so one sample period worth of jitter is okay. <A> Why would you need to debounce a button that controls an LED?If the LED's state changes rapidly for a few milliseconds while your switch bounces, who cares? <S> If you are trying to learn more about debouncing, I suggest connecting your switch to the clock of a binary counter . <S> Your counter should advance only 1 count per button push. <S> If it advances 2 or more counts, you know there has been some bouncing happening. <S> Alternatively, you could program your PIC to count the button pushes. <S> In fact, it might be more insightful to have your PIC (or logic chip or whatever you are using) to count the number of edges that it sees. <S> This may indicate whether your button is bouncing when you press, when you release, or both. <S> TL;DR <S> If you want to learn about debouncing, create a circuit that captures the bounces. <S> Otherwise, if you are just trying to control an LED, debouncing probably isn't necessary. <A> I have PIC32 Ethernet Starter kit . <S> It has 3 buttons which do not have any debouncing circuitry as per the document . <S> I only use it for button press event <S> I am not good in assembly language <S> but you can use many debouncing software techniques for your problem. <S> For ex: Check for button press Add a delay of 100ms Again check for button press <S> Add your code <S> if(BUTTON <S> == PRESSED){ DelayMS(100); <S> if(BUTTON == PRESSED) { <S> LED = ON; }} <A> I can see where debouncing the "press" action only will probably work, since during button release, the routine will immediately jump to back the "press" function, but be rejected. <S> That sounds a bit unclean, but should work well enough. <S> You might consider a routine that doesn't check "press" or "release", but rather keeps a "now" state and just checks that the input has changed (for the 10 ms or whatever). <S> Then update the "now" state, and the program can just read "now" to decide where the button is at any time. <S> This has a number of advantages. <S> This method debounces the switch in either direction, since it's only looking for a change. <S> So you get "press" and you get "release" for free. <S> "Now" can be a byte, and therefore hold up to eight switches. <S> And you can debouce all of them together! <S> Once any and all of them have settled, you change "now" to reflect their states. <S> If you arrange a method of calling the checking routine periodically (commonly done with a timer interrupt, but there are other creative ways), you have the rest of the time between checks to be doing useful work (meaning, extra cpu cycles). <S> You might guess that a lot of this is written from experience; and it is, so I'll share that I check the switches every 10ms, and I wait until they have been stable 3 times in a row. <S> At 30ms, you still won't sense any delay.
| So for these buttons I use software debouncing techniques.
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Is it possible to build an H-Bridge with only N-MOSFETs (and these other components)? I am trying to design an H-bridge circuit and I am only given the following components: Opamp Zener Diode Resistor Capacitor N-MOSFET NPN BJT Diode 2-input NOR 2-input NAND Infra-red Detector Infra-red Emitter D flip-flop I know that it is easy to design an H-bridge using 2 P-MOSFETs and 2 N-MOSFETs, but I am not given P-MOSFETs. Is there an equivalent circuit for only N-MOSFETs? <Q> Without even considering the concept of bootstrapping it is totally possible to build an H-Bridge with N-Channel MOSFETs in both the upper and lower switch positions. <S> This is done all the time in high power drivers due to the much larger selection of high performance N-Channel FETs over that for P-Channel FETs. <S> The key parameter to keep in mind is that the gate drive voltage for the upper FETs needs to be driven above the output voltage positive level by a good amount to ensure that the upper FETs are turned on fully. <S> A starter way to achieve this is to provide a higher voltage supply rail for the gate drivers than what is used for output positive voltage rail. <S> Once you have learned and understood the concepts of what it takes to drive the upper N-Channel FETs with sufficient gate voltage you can then begin to investigate how to get that higher voltage without having to provide an additional higher voltage power supply. <S> One way to do that is to build a charge pump circuit, with some of your other components, to "double" the positive voltage rail up to near 2X the voltage. <S> To utilize this higher voltage will require a circuit that can swing the upper FET gate up to this 2X voltage and then low. <S> The Op-Amp would be a good candidate to do that for you. <S> In the industry there is a tendency for engineers to be tasked with designing circuits that cost less but give desired performance and function. <S> There is technique called boot strapping that uses the switching action of the H-Bridge itself to drive a capacitor in a simpler form than the completely separate charge pump circuit. <S> The bootstrap capacitor can achieve almost twice the positive voltage rail level of voltage and again provide the necessary gate drive bias for the upper N-Channel FET. <S> We never get anything for free however and in this case the simpler bootstrap circuit has the requirement that the H-Bridge needs to keep switching all the time to provide the pumping action. <S> This means that a motor drive would need to be switched PWM type drive and never get to full 100% ON drive level. <A> but you'll not be able to saturate the high side devices so, depending on what your load looks like, they (the high side transistors) may get too hot for comfort and your load probably won't perform the way you want it to. <S> However, with all the stuff you have available you could easily build a charge pump and use it generate a voltage higher than your supply voltage. <S> Once you did that you'd be able to drive either the gates of the high-side MOSFETs or the bases of the high-side BJTs at higher than your supply voltage and get the H bridge to work properly. <A> Since this is a school assignment, I suspect the solution is expected to be simple, not industry grade. <S> The simplest bootstrap that works for the high-side N-channel MOSFETs looks like this (half bridge shown): <S> This is from a textbook by Grant and Gowar . <S> Now since you're not given any PNPs, you can't have the complementary pair drive Q3 there. <S> Instead you have to use a two-NPN totem pole in their place. <S> You seem to be given all the logic gates you want, so this looks like will satisfy your assignment requirements. <S> A bit more searching even found ( <S> in an article referenced by Grant and Gowar) something more economical along the lines I was talking about: <S> I have no idea why you're given "Infra-red Detector" and "Infra-red Emitter" <S> but if by any chance that means you're given optocouplers, you could also do this: <S> You're not saying what kind of motor you need this H-bridge to drive. <S> The above circuits are suitable for BLDCs or motors where the on-time of the high-side switches cannot be arbitrarily long because the capacitor can only recharge when the high-side is off (it's shorted while high-side drive is on). <S> If you need arbitrarily long on-time, then a slightly more complex circuit called a charge pump is needed to continuously top off the boostrap/tank capacitor. <S> The textbook-level circuit is (only the high-side driver is shown): <S> Again you'll have to substitute the complementary pair to meet your design requirement of no-PNPs-allowed. <S> This circuit also requires a clock signal (square wave generator) to drive the pump, which you can make in many ways with what you've got, e.g. op amp. <S> If you don't know how to do that bit ( <S> and you need it), you should search the site and/or ask a separate question.
| If you have a single-output power supply and only NMOS and NPN transistors available, you'll certainly be able to build an "H" bridge, using either, (or both)
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What supplies ATS control circuit when the mains fail? In a church, I've seen a large generator and ATS panel (automatic transfer switch) so that the generator turns on when there is a failure in the mains.If there is no power in the mains, How does the control circuit of ATS work and send a signal to the generator to start? I know that in power distribution rooms (stations) which contains switch gear panels, There are DC batteries (UPS). So that the control circuits still works even if there is no power in the mains. Does the church have at least one battery? or there is another technique? <Q> I was right about capacitors, partly. <S> Some models of ATS have a center position, in which they wait until either supply is stable. <S> In order for this to work, the ATS is powered by a capacitor during this time. <S> For this particular model, it takes 10 minutes for the capacitor to charge. <S> Based on this long time, I suspect it's some kind of supercapacitor . <S> And it's possible to have the ATS' own battery (or supercap) send the signal and power to start the generator. <S> And the generator has its own built-in battery too in that particular setup (the charge needed to start is substantial). <S> So yes there are batteries or supercaps involved in the transition, but they may be built-in the devices involved. <S> A more serious setup puts UPS after the ATS so the consumers don't lose power while the transition happens. <S> Someone even wrote a paper on <S> The importance of the generator starting battery . <S> Re: <S> "Do you mean that ATS can be supplied from the built-in battery of the generator?" <S> Some ATS models normally do that, for example : <S> M2G units require a DC power supply which is normally taken from the generator’s starter battery. <S> If connection to this battery is not convenient then an alternative must be found, otherwise the ‘integrated DC supply’ package must be added to the ATS panel. <S> Actually the whole bit on ATS types in that brochure is worth reading: <S> There are more details on the internals of a cheap/household ATS in this paper . <S> That one needs/uses a 12V battery. <S> For some [more expensive] ATS units, no-external-DC-power-requirement is the standard option. <S> This one from ABB doesn't detail how it powers its own circuitry, but it says The new ATS family is designed to work without auxiliary power supply. <S> Auxiliary power supply is only required when Dialogue Modbus RS485 is used or in networks with rated frequency of <S> 16 2/3 Hz . <S> It probably has internal battery or supercap but <S> its charging circuitry needs at least 50Hz to operate; that ATS is not rated for operating on anything in between 16.7Hz and 50Hz mains frequency. <A> The church has at least one battery, and a contactor with its coil wired across the mains so that the normally open contacts connect the mains to the church's downstream electrical loads. <S> When the mains fail, the contactor's normally closed contacts then connect the church's downstream electrical loads to the generator, which is started with the battery. <A> It's simple, in the company where I work <S> The ATS receives DC power to turn on its display lights and to control/ operate the generator. <S> It gets its DC power from the plus and negative wires in the generator's control panel (which come from the starter)and <S> the starter gets these same wires from the battery. <S> And to ensure that the generator battery can Continually keep powering the ATS and the generator's own display screen without the battery going dead, then what is done is that 220 volts is taken from the building cables (which are connected in the ATS) and this 220 volts is given to a battery charger, which receives 220 volts on one side and on the other side it sends wires that charge the generator battery. <S> And that is how the ATS always has power because the generators battery always has power to give to the ATS because the battery charger ensures this.
| In other words, the generator's battery is giving DC power to the ATS.
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Powering NiChrome wire from battery I'm looking to power nichrome wire from battery. The problem I ran into is that desired length/width of the wire has very small resistance ~ 8 Ohm. Several kinds of batteries I've tested have internal resistance much higher then that, which leads to wire staying cold and battery getting hot. Is there any way to increase current drawn from the battery? Some kind of amplifier maybe? Edit: 1.2V(rechargable) and 1.5V one, while not getting hot itself did not provide enough power to heat the wire. 3V did not provide enough power, even stacking 3 of them in series, only provided about 1V of voltage on the wire, and not enough amperage to heat the wire. 9V battery did provide enough amperage to heat the wire, but at the same time got very hot very quickly itself. PS. I was wondering if https://en.wikipedia.org/wiki/Joule_thief might work for this purpose? <Q> The issues you are facing are pure Ohm's Law issues. <S> First, the coin cells. <S> They have a high ESR, Equivalent Series Resistance (or Internal Resistance). <S> 20 <S> ~ 40 Ω, depending on the load or battery state. <S> Stacking them increases the ESR as well. <S> So from a 3V coin cell like the cr2032 <S> (250 mAh average), and a 8 Ω nichrome wire, we solve for I. <S> I = <S> V / R where R = R-Wire + R-ESR. <S> 0.108 <S> A = 3 V / (20 Ω + 8 Ω) <S> And since we know the current through the two resistors (current is the same when in series) and the resistance, we can see how much voltage is actually on each, with more Ohm's Law. <S> V = <S> I <S> * R . <S> 2.16 V = 0.108 <S> A * 20 Ω and 0.86 V = 0.108 <S> A * <S> 8 Ω. 2.16 Volts of the 3V available, are wasted inside the coin cell. <S> Only V * <S> I = <S> P <S> .86 <S> V <S> * 0.108 <S> A = <S> 0.09 Watts or 90 MilliWatts of Power are going through the Nichrome Wire. <S> Coin Cells are useless in your application. <S> So pure Ohm's Law applies. <S> 0.1875 <S> A = 1.5 V / 8 Ω and 1.125 A = 9 <S> V / 8 <S> Ω A 1.5V AA battery, at a sixth of a 9V battery's voltage, with the same load resistance, will produce a sixth of the current. <S> But the Power difference is much greater. <S> In Power, that's 0.28 Watts and 10 Watts. <S> While the ESR of the 9V is negligible, it still exists, at likely <= <S> 1 Ω, and while reducing the power going through the wire some, the amount of Power drawn by the load is enough to heat the battery up. <S> 9V batteries are designed for low current draws. <S> Drawing 1+ Amps through it is not ideal. <S> AA, C, and D can handle this better. <S> Each has less ESR than the smaller, and a higher capacity. <S> Due to the lower voltage, you need to combine a few in series to get the desired current and power draw through the Nichrome Wire. <S> See <S> How do i find nichrome temperature for specifics on heating and Pulse-powering heavy loads with a coin cell on battery loads. <A> At a given ambient temperature, it takes a set amount of energy to heat the nichrome wire to fusion temperature, and then another amount of energy to melt it. <S> Calculate the power required across your nichrome wire depending on the power losses and the required time-to-melt. <S> Deduce what the required voltage and the corresponding current across the nichrome wire are Rule of thumb: choose a battery technology that has a capacity (in Ah) equal to or greater than the required current (in A) <S> (10x greater for sustained use) if possible (otherwise step 4 is mandatory). <S> Stack as many cells in series as required to make up the required voltage <S> AA batteries are usually quite good in terms of resistance (~0.2Ohm for alkaline) and current capability (up to 1 or 2A, not for long though). <S> They wildly vary in specs depending on the manufacturer though, therefore if yours didn't work, you should try with a better quality one before trying the C and finally D type batteries. <S> If the best stack you can afford (in terms of space, mass or whatever) can't provide the current or gets too hot, here is a potential solution: insert a Schottky diode that can handle the voltage and current in series with the stack (note that the stack voltage will be reduced by 0.2 to 0.5V), and duplicate that assembly in parallel as many times as required to reduce the current each stack has to provide. <S> The diodes will prevent batteries fighting for the net voltage. <S> Note that this technique (step 4) is normally used to switch between power supplies, not increase the current capability. <S> Only one diode will conduct at any time*, meaning that here the entire current will be supplied by one source before its voltage droops sufficiently to make another one conduct, and so on. <S> This could work okay for this application though. <S> *: considering an ideal diode <A> Your problem is defined by the amount of power <S> you can extract from the battery. <S> Once you are operating at the limit the best you can do with fancy circuitry is trade off current for voltage. <S> But the power won't increase. <S> For example, (disregarding efficiency effects), if the 9v battery gave you 100mA, you could change that to 1.5v at 600mA. <S> The resistance changed from 90 ohms to 2.5 ohms, but the power is still less than 1 watt. <S> Your best bet is to determine which cells present the lowest internal resistance, and then, if necessary, connect several in parallel to reduce the resistance further.
| Alkaline Primary Batteries, like your typical AA, C, D and 9V have much lower ESR, and can be considered negligible for this purpose.
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How much hold current is consumed when there is no load? Given a 24V 1A motor with no actual load connected (just a piece of orange tape so we can see the shaft moving). How much current will the motor use if I configure the driver board for 2A maximum hold current? This is the scenario: We have a breadboard that is configured for 2 motors at 24V 2A. However, we only have 1 motor and axis configured for now. So until we get the second motor, I asked for just any motor to be connected to the second channel so that I can see that the software is working. The project EE found a 1A motor and connected that. The EE then warned me to reduce the maximum hold current to 1A. I was told that the driver board would "push" 2A of current to the motor. This is where I get confused: I don't understand how the motor could draw 2A when there is no load, even if there was no current limit at all. I thought that the "maximum hold current" was the current limit . Not that the board would push 2A to the motor. At 24V, supposing that the motor was drawing say 200mA just sitting there idle, wouldn't the driver board have to supply 24V x 2000mA/200mA = 240V to get 2A of current into the motor? <Q> A stationary stepper motor supplied by DC is just a resistive load, so the maximum current which will be drawn is the driver's maximum output voltage divided by the resistance of the relevant coils. <S> When you're operating the motor at speed, changing coil current very quickly, you need a lot of voltage to impose the rapid changes of current. <S> For this reason, stepper motor drivers tend to have lots of voltage available, much more than you need to maintain a suitable holding current. <S> Because of this, the holding current is something which the driver has to control <S> - it's not a matter of the motor 'drawing' it, so much as the driver imposing it. <S> It's going to be the case that a 1A motor has a higher resistance than a 2A motor, so <S> if you had a constant voltage supply you'd see lower holding current when the 1A motor was connected without changing anything. <S> However, the mere fact that you've been told 'set the holding current lower' implies that actually you have a more sophisticated driver, which will have lots of excess voltage and will need to be told what a safe holding current is. <A> A stepper driver usually has lots of extra voltage in hand for slewing the current quickly against the inductance of the motor coils, most of the voltage ends up getting lost in the driver circuit. <S> This means that if you have a motor that is rated for 1A operation, and your driver is configured for a 2A hold current, there could well be enough extra voltage available at the driver to push 2A through the motor. <S> This would generate 4x the rated heat in the motor, and would probably overheat it. <S> You ought to be able to find out what the motor coil resistances are, if not from a specification then by measuring with a meter. <S> Then calculate what voltages would be needed at the terminals to make 1A or 2A flow. <S> What voltage is available at the driver? <S> Do what your EE says and reduce the driver max current to the motor's rated current. <S> I've melted a motor during a lab, it can be quite embarrassing. <A> You're right. <S> The EE is wrong. <S> When both voltage and current are specified, they are in fact limits , not forced values. <S> Whichever it hits first is where it stops at each moment in time. <S> V = IR still holds. <A> A driver has a chopper and countinously feeds the motor with set current. <S> You say 1A is needed. <S> When motor is not spinning, there is no back EMF and the voltage applied is V = <S> R_winding <S> * I_set. <S> When you drive the motor at higher speeds, the driver has to compensate for BEMF that is induced from rotating motor, but the current is still the same. <S> As for holding current, this is a feature of your driver, that detects no pulses - standstil <S> and it reduces the current to a lower value.
| It depends on the resistance of the motor, the supply voltage, and the driver limits, which one will limit first.
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When measuring a resistance using a multimeter it is a few percent out. Why? For example, if you were to put 5V across a \$150\Omega\$ resistor and measure its resistance with a multimeter, the reading would always be a few percent off the true value. What reasons might there be for this inaccuracy? <Q> The resistor may have a 1% or 2% tolerance (5% resistors used to exist, but are pretty rare these days, at least for professional suppliers). <S> Then check the specification of your meter. <S> I would be surprised to find it had a better accuracy than 2% unless it was a really expensive one. <S> These tolerances can add up worst case, to give you 3% to 4% error, with meter and resistor fully inside specification. <A> When measuring resistance, it is generally good to disconnect it entirely from any circuit. <S> In this way, the multimeter isn't measuring the resistance of the circuit, but is only measuring the resistance of your resistor . <A> The connection between probes has a finite resistance that needs to be zeroed out. <S> Touch your probes together to see if you get zero volts. <S> If you don't, some meters have a way to zero this out, or you can do it in your head. <S> At 150 ohms, 3 or 4 ohms in series in the probes/meter is a significant percentage. <S> Of course, you need to remove the resistor from the circuit for accurate measurement. <S> If you are measuring HIGH resistances, you also can't hold the ends of the resistor in your hand, as your body impedance in parallel can impact the measurement. <A> The multimeter itself is actually correct because all resistors have a tolerance (so the resistance of the resistor could be +/- <S> a certain percentage). <S> The multimeter measures much more accurately and gives you the real value of the resistor. <S> You can get more accurate resistors if you need however for simple, 'normal' uses, <S> a normal resistor with a 1%-5% tolerance should be fine. <A> Commercially available resistors will typically be 5% or 1% tolerance, which means that within their operating temperature range, they will deviate no more than that much from their nominal value. <S> When you put some voltage across a resistor, it will conduct current and start dissipating some heat. <S> This will cause a change in temperature, which will in turn change the actual resistance, but should still keep it in tolerance range. <S> On a side note, Your multimeter's manual suggests to remove components from circuit board while measuring their attributes.
| If a component is within a powered circuit and you attempt to measure resistance/capacitance/hFe etc, it may give widely incorrect readings, and may also kill your multimeter.
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Detecting start bit in software UART I'm experimenting with writing a software UART on my microcontroller using GPIO pins. This is to temporarily add a UART channel on a project until we get the new design implemented that uses a uC with more UART ports. What I'm having difficulty with is correctly detecting a start bit in a serial stream. The source of the stream is external and doesn't care when my device powers up. So it's very likely that my device will power on and start seeing data bits in the middle of a byte transmission. Undoubtedly, that will cause my software UART to read erroneous values, as it won't be able to tell the difference between a start bit and any other high-to-low transition. Is this an inevitable issue with a UART channel? Or is there some clever trick that the uC manufacturers use in their hardware UARTs? <Q> If you use a stop bit length that is easily discerned from the rest of the data stream, such as 1.5 bit time, then it should be easy to start receiving mid-transmission. <S> However, this comes at a cost of increased overhead. <S> Your total available data throughput will suffer as you increase the length of your stop bit. <S> If you're not using the bus that heavily, and frequently have gaps between frames, then it may just be a matter of waiting for one of these gaps to occur, and then picking up the first hi-lo transmission as the beginning of your next start bit. <S> Keep in mind that the number of data bits should be predictable, as should the frame size, so even if you're using 100% of the bus's capacity and your stop bit is a single bit time, you should still be able to find the start bit if you collect enough frames. <S> Every frame is guaranteed to have a hi-lo transition in it. <S> The stop bit is the one that is always high. <S> The start bit is the one that is always low. <S> Assuming your data is random (or random enough), you could do something as simple as creating a buffer the size of your frame, set every bit in it, and then keep collecting frames and ANDing them into this buffer until the buffer only has 1 bit set. <S> This bit is your stop bit. <S> The one after it is your start bit. <S> Voila! <S> You've found it. <S> The downside of this approach is that finding the start bit takes a bit more time than the alternatives, but the advantage is that you can maximize throughput by minimizing your stop bit time and your bus's idle time. <S> If it matches, then you've (probably) found the start bit. <S> If it doesn't, pick the next low bit and repeat until you get a good checksum. <S> If you can't find a bit in your two frames of data that checks out as a valid start bit, then your data was corrupted, and you'll need to grab two more frames. <A> Hardware UARTs have the same problem. <S> But it's usually one that solves itself in short order anyway. <S> At the end of each frame, check the stop bit, and if it isn't high, discard the frame and wait for the next high-to-low transition. <S> Assuming that the data from the source isn't totally pathological (e.g., long strings of "UUUU", or ASCII 0x55), the UART will eventually "walk" itself over to the real start bit. <A> Assuming 8N1 transmission. <S> You must wait for a string of 9 high or low bits in a row. <S> If high it will signify either a idle gap in the data or a 0xFF character and STOP bit or if low a START bit and a NULL 0x00 character . <S> Either one of these conditions will allow resynchronisation. <S> To speed it up:If you know certain characters that are not possible in the data you could parse the incoming data repeatedly (after the fact) for each bit and if you get a series of 7 characters that are nonsense (high bit set, lower case, control codes, punctuation or whatever) followed by a valid character you can be fairly sure you are resynchronised. <S> You will have the similar problems when you use a built-in UART peripheral and be unable to do bitwise evaluation and also have to remember to reset all framing error bits and such whenever they occur (especially on power up).
| If you're using a parity bit, another option would be to grab two frames worth of data, pick the first low bit as the start bit, and then calculate the checksum and compare to the parity bit.
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3.15A and 4A fuses - can I swap? I have an old oscilloscope which doesn't work. I tried swapping the 3.15A/250VAC fuse with a 4A/250VAC fuse. The fuse immediately blew. My questions are: Did the fuse blow because it was the wrong fuse? Or did it blow because there's likely something else wrong with the oscilloscope? Must the fuse be exactly as specified? Thank you. <Q> In this case, assuming the two fuses were the same type (fast or slow blow), then there is definitely something wrong with your scope. <S> Since the scope is not meant to exceed 3.15 A, but actually exceeded 4 A, there is clearly something wrong. <S> By making it take more current before the power was shut off, you may have even damaged it further. <S> This is all assuming the two fuses were of the same type. <S> If the original was a slow blow, and the new a fast blow, then what you observed could still happen in normal operation. <S> Look at the old and new fuses carefully. <S> This isn't universal, but usually a fast blow fuse is just the melting link suspended between the ends. <S> A slow blow usually has the melting wire wrapped around a ceramic rod. <S> If the original was fast blow and you replaced it with a slow blow, then there is really something wrong with the scope. <A> It is also possible that you have different types of fuse between the 3.15 A and the 4A. Fuse comes in two types: Slow-blow and regular. <S> Slow blow fuses are used when there can be a short-term current surge in the system, like when you power on your oscilloscope. <S> If the old 3.15A is a slow-blow and you put in a 4A regular, it will blow right when you turn it on, which is what you are seeing. <S> I'd suggest looking at the 3.15A to see if it's a slow-blow kind <S> or if it's a regular. <S> Then you need to match the 4A to it. <S> I'd bet that it's a slow-blow. <A> By changing the fuse to a higher current, you are changing the power which the oscilloscope can draw from the wall. <S> The power is$$ P = IV $$for current \$I\$, voltage \$V\$. <S> So by moving from a 3A to a 4A fuse you are increasing the power available inside the scope by about 30%. <S> You're unlikely to start a fire by switching to a 4A fuse, but I wouldn't go any higher without debugging the actual problem.
| First, replacing a fuse with one of a higher current rating is a bad idea.
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Charging a battery with float voltage Is it possible to charge a lead acid battery(12V) with float voltage(13.6V) instead of charge voltage(14.4V)?I understand it will take lot of time to charge if we charge with float voltage. <Q> Yes, it is possible but it takes a lot of time, first your battery will be charged at the constant current until the folat voltage (13.2) is reached (75%) <S> then it will take a lot of time to charge the 25% (low current). <S> The best way to charge your battery is the 3-stages charging it like the float charging and continues to charge until your BOOST voltage (14.4V), then the current is reduced to 1/4 of the maximum to the 90% then the 10% are charged slowly by the float charging. <A> If you charge at float voltage, not only will it take a long time to take as much charge as it's going to get, it will also not reach 100% charge. <S> If as a camping/caravan battery where you do want to know that you are up to 100% SOC, no, it won't give you that. <S> A word about lifetime and battery chemistry. <S> Lead acid doesn't like being discharged to below 50% SOC <S> , it shortens its life significantly if you do this repeatedly. <S> It therefore makes a good alarm standby battery, but not such a good camping battery, where you might be tempted to run your peltier fridge up to 100% of the Ah claimed on the battery. <S> There are better battery types for deep discharge, NimH and LiPo to name the obvious ones. <S> Neither are as cheap as lead acid (if you have a vehicle to transport the weight). <S> If you want a long lifetime from your lead acid battery, then double the <S> Ah you buy initially, so you don't need to fully discharge it. <A> I believe it will work fine. <S> The battery should last for years if you don't over-discharge it. <S> An AGM might be the best choice due to low maintenance requirements. <S> The exact float voltage depends on battery type and temperature. <S> You need to reduce the float voltage at higher ambient temperatures. <S> For me, it would be worth the money to buy a multi-stage charger intended for RV's or boats. <S> Then it will re-charge quickly when needed, but also float at the correct voltage. <S> Some chargers also have temperatures sensing. <S> If the battery will be exposed to a wide temperature range, I think a temperature sensor should be considered a requirement.
| If as a mains failure backup battery for something like a burglar alarm or safety lighting, then floating it is a sensible thing to do, as it maximises life of the battery under those conditions, and makes for a very simple charging circuit, one voltage regulator and you're done. It depends what it's going to be used for.
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How to manufacture cheap custom flexible circuit boards? (strips) Let's suppose I need two long series of IR emitters and receivers to detect position of people along a hallway (corridor). I would use surface mount components to make the sensors less noticeable, and one series per side of the hallway. This is only the framework, no need to suggest improvements here :) I could order custom PCBs online, but the thickness would be significant and I would have to run wires all the way to each PCB. Another solution is to produce some custom plastic strips with metallic traces, on which I would solder directly the components. Only one emitter and only at most three receivers would be active at the same time, the expected current consumption is 500 mA, at most 1 A (this if the IR emitters are pushed a lot), therefore metallic traces on plastic could work. I have experience with the plastic circuit of the dashboard of my old ('90s) car and it was able to carry enough current to light the dashboard bulbs. I also successfully soldered stuff on it, without the circuit melting or deforming. See attached image (found online) as example of what plastic strips I mean. However, I was not able to find any means to produce said plastic strips. How could I get them? I would need a periodic design, repeated over and over along the strip. I would say 10 mm width by 20 metres length (2x 10 m). <Q> There are PCB manufacturers that offer flexible PCBs. <S> However, PCBs which are some meters long will cost a significant amount of money. <S> You may look into a conductive silver paint (e.g. http://www.maplin.co.uk/p/electrically-conductive-silver-paint-n36ba ). <S> With that you can "paint" your electrical traces onto a thin substrate. <A> You might be able to make something yourself. <S> I'm assuming the components are several cm apart, and the components are hence a small fraction of the overall length. <S> Use small, simple, PCB's for the components. <S> Their would likely be an emitter PCB and a receiver PCB. <S> Then use adhesive copper tape for conductors between component PCBs. <S> AFAIK, that adhesive wouldn't conduct well enough for your application, so it would need to be removed with some kind of solvent. <S> I believe I've seen narrower copper strip with non-conductive adhesive. <S> EDIT: I believe companies on, e.g. Alibaba, have 2mm copper tape on e.g. adhesive-backed PET. <S> However I don't know if they sell in quantities small enough to be useful. <S> I hadn't thought of stacking layers, mainly because I thought connecting it to the components might be awkward. <A> The problem with PCB prototyping services is that "unusual" tends to imply "expensive". <S> The economics are very different for mass production. <S> The cost per unit of PCBs drops massively as volume increases.
| There are plenty of services out that that will make flexible PCBs on kapton substrates but they aren't cheap. The copper strip might be soldered onto the PCBs, though I'd expect that to be quit hard unless the strip is quite short, so it may need a mechanical fixing or maybe silver-loaded epoxy. Thick aluminium foil might work too.
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Good way to affordably manufacture low-quantity PCB? I am thinking about making an open-source development board for a specific field in power conversion. If there is interest from local universities, I will manufacture a few hundreds of these boards. Is it better to engage a manufacturer or buy a small pick-n-place machine? EDIT: It is assumed that the PCBs will be sourced out. The number of components should fall in the <20, 50> range. The DSC/MCU could have ~200 pins, TQFP package. EDIT2: The answers so far indicate that it is very costly and time demanding to make PCBs at home. But what is the approximate cost to get the minimum viable production line for a few hundred boards? To get done ~20 per hour (e.g. 2" by 4"). <Q> This is a no-brainer. <S> There is absolutely no way that the cost of a pick and place machine spread over a few 100 boards <S> will be cheaper than getting the boards assembled by a contract manufacturer. <S> And, a pick and place machine is only one part of the assembly process. <S> Are you going to get a reflow oven, inspection station, etc, etc, too? <S> Added <S> As Scott Seidman mentioned in a comment, you have to think about testing whenever you have someone else build your boards. <S> If you don't give the assembler a way of testing the finished board, then they could ship you 100 units that all don't work <S> and you'd have no recourse. <S> The assembler won't like that either since they have no way to find problems with their process and fix them, hopefully without anyone outside ever finding out. <S> What I usually do is build three test jigs. <S> One goes to the contract manufacturer, one is for our own use, and the third is a spare. <S> Fancy high end ones intended for high volumes usually contact the board with pogo pins and pads designed onto the bottom of the board for that purpose. <S> Then there is a mechanism that closes down on the board. <S> That holds the board in place and provides the force to compress the pogo pins a little, and usually has a switch on it that the test program can read to start the test. <S> You don't need to have something that automated for a small run like 100 units. <S> You can require the operator to connect a programmer, then run a program on the host to dump the code into the microcontroller, connect cables, push buttons, look at lights, and the like. <S> Consider whether it makes sense to add test functions to the production code, or have special code loaded into the micro just for testing <S> , then the production code loaded after the test passes. <S> I generally figure the test jig is about of the same complexity as the product being tested, or a bit more. <S> Designing the test jig is something that you need to plan on right from the beginning. <S> In a larger organization, it can be useful to have separate engineers designing the product and the test jig, each working from the same specification of the product. <A> Technique is not necessarily straight forward. <S> For optimal quality it is not always as easy as raising the package to 260C and cooling it down, for instance large QFPN packages are now famous with lead free solder for cracking and premature failure due to the technique used when soldering. <S> Manufactures are familiar with some of these details and what extra steps may be needed to ensure quality such as under-fill or footprint modifications. <S> All machinery has quirks and learning curves, you will spend a lot of time and material figuring out optimal settings for re-flow ovens, pick and places etc. <S> As long as you're not doing any fine pitch BGA components, everything can be cheaply inspected by just about any fabricator or yourself with a microscope, this is usually a cheep but somewhat reliable testing method before you move to a fixture with electronic testing or X-Ray inspection. <S> That being said again even basic Chinese manufactures will often have these sorts of tools available. <S> Finally, if you do end up with quality problems, having someone to talk to can be very helpful (again using experience that is hard to find). <S> To you the problem could be a needle in a haystack, to a fabrication house, they can take one look at your board and identify the common components that have problems. <S> If you think about some of what I mentioned, it becomes clear doing it yourself <S> has a lot of hidden costs, time commitments and risks. <S> Unless you plan to start doing a lot of fabrications of different boards yourself in low quantities, I can't come up with a single reason to ever do it yourself. <A> I agree 100% with Olin that you should contract with an assembly company. <S> Make sure you get pricing from a couple of sources that are happy with making your quantities before <S> you quote your customers (sounds obvious, but if you make assumptions <S> and they turn out to be wrong <S> it could be expensive). <S> Different PCBA companies may or may not be happy with very small orders. <S> Naturally the ones that are happy with small orders will tend to charge significantly more per unit. <S> There will be setup costs and per-unit costs. <S> Usually any kind of change incurs additional setup charges, so be sure to allow for iterations as may be required. <S> It's possible to buy the equipment to make such small quantities <S> but it's not economical, and there are many processes to learn (any one of which can seriously compromise the quality). <S> If your board is open-source and your competitors have access to significantly lower cost manufacturers that are local to them , your run as a profitable manufacturer may be truncated.
| As someone that has looked into this in the past even if you break even buying your own equipment, it would still be much better to have the boards professionally fabricated for a few reasons.
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Need some help in understanding this circuit This circuit is taken from this website. http://www.circuit-finder.com/categories/sensor/temperature-sensor/905/temperature-controlled-switch What is the purpose of R2 and R3 in this circuit? I use voltage divider rule to calculate R2 which is R2/(R2+P1+R1) * 5V = 4.6987VConverting it to degree Celsius , it is 46.987 degree Celsius. Does it mean that the range is from 0 ~ 46.987 degree Celsius? Only when the sensor voltage rises above the ref voltage level for P1 (which is around 0.06 V = 30 degree Celsius since P1 is set to 10k ohm) , the output of the comparator toggles to the full supply voltage level The purpose of R3 is it to limit current flow? <Q> Note that the temperature signal from the LM35 is compared to the output of the voltage divider formed by R2, P1, and R1. <S> You could have asked similarly what the point of R1 is. <S> In theory, you could connect the ends of P1 to power and ground directly. <S> It could still be adjusted to the desired temperature. <S> However, the adjustment would be more coarse. <S> Both R1 and R2 limit the adjustment range, which gives you finer control over the range you do get. <S> R3 together with R4 controls the amount of positive feedback, or hysteresis in this case. <S> The bigger R3 is relative to R4, the larger the hysteresis "dead zone". <S> The purpose of hysteresis is to crisply switch the output on or off, not have it being partially driven when the temperature is near the threshold. <S> Put another way, hysteresis provides the "snap action" so that the load is always driven full on or full off. <A> Regarding R2: You're using the formula to calculate the voltage across R2 !?!? <S> More sense makes: (R1 + P1)/(R1+P1+R2 <S> ) <S> * 5V = 300 mVAnd and the lowest side of P1: 180 mVSo that would make the range 180 - 300 Kelvin or - 93 to 27 degrees celcius. <S> This results in that for example the on voltage would become 41 mV while the off voltage is 39 mV. <S> Without hysteresis the trip voltage could be 40 mV with the chance that the circuit quickly starts switching on and off around 40 degrees. <S> You want to avoid that ! <A> Pad Pad Pad Brute force verify. <S> :)
| R3 works in combination with R4 introducing a little bit of positive feedback , this adds some hysteresis to make the opamp switch more abruptly.
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MOSFET: Very high Rdson Iam trying to test my mosfet because my project which involved connecting a load to the drain did not work. After some tinkering Iam down to suspecting the mosfet (n-channel, IRF630B). I have connected it this way to test, +12v -> gate,ground -> source Then I switched my multimeter to measure resistance (200 ohm option) and connected the red lead to the drain and black lead to source. Q1. Am I measuring the Rdson correctly? (it does not need to be accurate, good approx is enough) Now, with power switched off, I see that the circuit is open. After switching on, the Rdson value is in the range of 10-11 ohms. I have tried 10 different mosfets and all show the same range. According to specs, anything above 10V should bias this mosfet correctly and bring down the Rdson drastically (ideal value is 0.4ohms). I also changed the power source to +19.5v to see if a harder drive makes a difference. I still got similar results. Q2. What am I doing wrong? <Q> I can't answer questio 2n since my computer got a problem and I can't see your schematic <S> but I can answer question 1: Q1. <S> Am I measuring the Rdson correctly? <S> (it does not need to be accurate, good approx is enough) <S> Answer: <S> No you don't. <S> The resistance measurement need to be done without power on the board. <S> If you want to measure Rdson you can do the same operation but putting your multimeter in voltage mode. <S> You measure the voltage between the drain and the source. <S> With another multimeter you measure the current flowing through the transistor. <S> then You do R = <S> U/ <S> I and <S> you have your Rdson. <A> Direct in-circuit resistance measurement with a regular 2 wire multi-meter is not possible. <S> You need to measure voltage and current separately and do the calculation. <S> because this is a very simple circuit, I suggest you to first make sure 12V is really 12V <S> (measure the source voltage). <S> then make sure the R1 is really 5 Ohm (measure the resistance). <S> then run the circuit and measure the Drain Voltage. <S> it should be around 0.9V (assuming the RDS=0.4 Ohm as Datasheet says). <S> 1) Measure the Drain Voltage: V(Drain) 2) <S> Calculate the current: <S> I(R1) = I(Drain) = <S> [12V - V(Drain)]/R1 3) <S> Calculate the channel resistance: <S> Rds = V(Drain)/I(R1) <A> The circuit you have will not allow you to measure \$\text{Rds(on)}\$. <S> The MOSFET goes into saturation when two condition are true:$$V_{GS} > V_{Th <S> } \\\text{and} \\ V_{DS} \ge \left(V_{GS} - V_{Th}\right)$$The <S> first condition is true for your multimeter circuit. <S> However, since the drain is floating the second condition is not satisfied. <S> Connecting the multimeter in resistance mode between the drain and source will bring the drain to a voltage close to the source. <S> The second condition is definitely not satisfied. <S> The MOSFET is in the linear region, and one would expect the \$\text{Rds(lin)}\$ to the bigger than \$\text{Rds(on)}\$. <S> In order to measure \$\text{Rds(on)}\$, you would need to use the first circuit you posted, where the drain is connected to the power supply through the resistor. <S> This will keep the drain positive enough to allow the MOSFET to go into saturation. <S> However, you can't measure the resistance of the MOSFET directly while it's in the circuit. <S> The best way to measure \$\text{Rds(on)}\$ is to use the multimeter to measure the voltage and current and use Ohm's law to determine \$\text{Rds(on)}\$. <S> Edit: You can calculate the current by measuring the voltage across R1 <A> It could be oscillating. <S> Have you looked at Gate and Drain with a scope? <S> Looking at the breadboard photo, there is a huge loop for the Gate circuit. <S> Wire loop inductance is roughly: \$L\$ <S> ~ <S> \$\mu_o\$ \$\pi d\$ <S> If \$d\$ is about 0.2m, then \$L\$ is about 800nH. <S> Put that with ~800pF for \$C_{\text{iss}}\$, gives \$Z_o\$~32\$\Omega\$, and \$f_o\$~8MHz. <S> This would be a pretty high Gate Q. Coupled with \$R_1\$ of 5\$\Omega\$ and \$g_{\text{fs}}\$ of 7S, \$C_{\text{gd}}\$ would act like ~1000pF. Also, \$L\$ could be higher, it's hard to tell how far the wire extend out of the photo. <S> It would seem a good case for oscillation. <S> You might try putting about 100\$\Omega\$ in series with the Gate to dampen things out.
| A multimeter is not meant to measure a resistance if there is current flowing through it.
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Is it ok to clip component leads after soldering? When working with through the hole components, I have read that you should clip leads before soldering so the shock from clipping doesn't break the joint. On the other hand I've seen a ton of instructions for (more or less professional) DIY kits, where the picture depicted all components of a type being installed, then soldered, then clipped. The latter method is much faster and easier of course. <Q> I was told to cut the leads to length before soldering. <S> There was an IPC standard around on which basis we were taught that. <S> I don't remember which standard it was exactly. <S> A reference I found is this document by IPC . <S> [...] <S> The next part of the through-hole solder joint is the component lead. <S> [...] <S> Either way the leads are formed – usually with a 90 degree bend – and trimmed to the correct length. <S> After the lead is trimmed, it may be clinched or bent over to hold the component in place during the rest of the assembly process. <S> [...] <S> The last part of the connection is the solder joint. <S> [...] <S> So to prevent parts from falling off, you bent the already trimmed leads over (in a direction where it doesn't cause problems). <S> And after that you solder them. <S> It is probably taught that way just as @GummiV mentioned in the comment, that the force and shock of the trimming might cause the joint or the PCB to be damaged. <S> I actually managed to break of a fragile ring with the track once because I was trimming after soldering, so I think the advice of the IPC is quite sound. <S> There are also frames around which help you in the assembly of through hole boards. <S> You can put the board in, put all your components in from the top, then put down a lid and turn the whole thing around to solder on the bottom side. <S> Looks something like this: <A> If you are interested in doing very high quality soldering <S> I recommend you search for military soldering specifications. <S> You will find instructions for bending leads with stress reliefs and to the proper radius. <S> The leads should be inserted into the board and bent slightly, just enough to keep them in place during soldering. <S> After soldering, you cut them off with cutters that have blades where the side toward the component is perpendicular to the lead. <S> In other words, the angle part of the blade faces away from the board. <S> Ordinary toe-nail clippers are the cheapest adequate illustrative cutter for this, but tend to wear more quickly than purpose-built cutters. <S> If you examine toe-nail clippers you will see they are flat on the side that faces your toe and angled on the side that faces the cutoff nail. <S> This is why your toenails don't shatter, crack, etc. <S> when you cut them off, instead the cutoff nail gets enough energy to fly across the room into your coffee cup. <S> If you use a cutter with blades like this, the shock will be transmitted to the cut-off lead and not to the board/lead/component. <A> As well, how do you intend to manually cut and bend the component legs before inserting so perfectly that they fit and can be soldered? <S> I can bend them fine, but getting the length right is just too fiddly. <S> Bend, insert, solder, clip. <S> If you have a machine doing the work, then the question is moot. <S> The machine will bend, cut and insert however it was built. <A> One common process for industrial assembly of through-hole boards is: <S> Insert all the parts <S> Crop them all with a cropping machine (e.g. 'Cropmatic', though that might be a UK thing) <S> Wave solder them. <S> You couldn't wave <S> -solder the board sensibly with all the leads still on anyway. <S> For a bit of hand assembly, if you're damaging things cropping the leads you just need to go and get a better set of side-cutters. <A> 40 years in the industry.. <S> so, 1) clinch, clip and solder, or 2) clinch, solder, clip, and then RESOLDER. <S> do not leave exposed base metal. <S> corrosion can and will creep up through the metal until it causes the connection to fail. <A> J-STD-001 and IPC 7711/7721, both instruct you to cut to length and solder to prevent shock to the solder joint. <A> As a master instructor for IPC and previously the Navy it is always best to trim the leads before soldering. <S> If trimmed after soldering per IPC J-STD-001 specs they must be either reflowed or inspected at 10 power magnification. <S> Any time a solder connection is reflowed you stand a chance of reducing the reliability of the solder connection by increasing the thickness of the intermetalic layer which reduces strength. <A> Another option is to solder components, clip them, and then reflow the solder. <S> This has the advantage of easy initial soldering, easy clipping, a clean looking finish, and a solid bond. <S> That being said for the strongest bond I would follow the other answers and clip first.
| I've always been told to clip the leads after soldering to reduce the shock that gets transmitted to the part - this reduces the chance that the part will be damaged.
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Routing power into power plane (4 layer PCB) In a four layer PCB scenario with standard layer stack-up (signal, GND, power, and signal), how do you get power into the power plane? In my case, there is a voltage regulator sitting on the top layer (SMD) and I need to get power from it into the third layer which is my power plane. Is it enough to put some vias near the output pin of the regulator (or output cap)? It is a mixed digital-analog design. (STM32F4 + some DACs). <Q> That's the general idea the number of the vias depends in how much current you want to carry and how low of an impedance you are looking for. <S> I suggest you look at the datasheet for your regulator. <S> No doubt there is a recommended layout at the end and <S> if not I'm sure there is an eval board layout you could examine as well. <A> Generally you want the vias to be as large as possible. <S> The trace width, via diameter, and via count will depend on the current <S> you expect they will carry. <S> You can use online calculators like this one based on IPC-2221 to determine how wide you need your traces to be. <S> For the vias, I generally try to make the via diameter match the trace width. <S> I also put multiple vias in to ensure good connection and good current carrying capability, because the surface area of the trace could be reduced in the via barrel. <S> A good rule of thumb is one via per half amp. <S> Put these vias as close to the regulator output capacitor as you can. <S> These are just some basic rules of thumb to get you going. <A> As mentioned, use a sea of vias to create a low resistance path between layers. <S> There are online calculators for trace resistance, and these can be used to calculate the resistance of a via based on its diameter, length, and copper thickness. <S> Plus, there probably are specialized calculators specifically for via performance. <S> The design becomes a tradeoff of the number of vias versus the diameters. <S> You can optimize this for the space you have to work with. <S> http://www.ultracad.com/articles/viacurrents.pdf <S> http://www.pighixxx.com/test/tools/pcb-via-calculator/ <S> Using this technique, I did a 21 slot VME backplane with 50 A of +5 V per slot on internal layers. <S> First article had less than 15 mV drop between any two points.
| If you have a voltage regulator on the top layer and want to route it to another layer, then yes, you use vias.
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Best approach for designing 40A laser diode driver? I need to design 40A 1.8V laser diode driver and need some word of advice as in the past I did from scratch only much smaller DCDCs. As it's a laser diode there will be no fast current changes. My idea is: do synchronous buck DCDC, use PWM from STM32 microcontroller, then MOSFET driver then MOSFETS. 40A for mosfets does not looks like a problem. Sense current using high-side (on 1.8V rail) 0.001 Ohm shunt with opamp amplifier. Few questions though: 1) Vin: I am looking more at 12V instead of 24/48V, to avoid too short transistor on times. Reasonable? Input current 8A vs 2A does not looks like a deal breaker. 2) Inductor. 40A inductor looks scary, and I probably won't find anything readily available. I was thinking about taking standard 13x13mm SMD inductors (which usually 3A), and parallel lots of them. But this will reduce their inductance a lot, probably below acceptable level. Any suggestion? 3) Is there a benefit of implementing multiphase DCDC for this application? I have bunch of PC motherboard inductors, they are like 1uH and too low for any paralleling, but might be ok for multiphase. For the craziest approach I might just cut piece of PC motherboard with transistors, inductors and capacitors and redo gate drivers. Will look weird, but beating 8-layer board design is something out of reach for me at the moment. Does that look too crazy? <Q> This is no easy task to do by simply re-using bits of old motherboards. <S> It's a problem that I feel requires a properly designed solution. <S> I'm initially thinking of maybe the LTC3882: <S> - The output parameters can be pretty much controlled from an MCU (Digitally Programmable Voltage, Current Limit, Soft-Start/Stop, Sequencing, Margining, AVP and UV/OV Thresholds) <S> so I thinks it's a solution with high probability of working. <S> Another chip that looks promising is this: - There are plenty of options from LT - if you use their search engine it narrows down the requirements of 11 to 13 v operation with 1.2 volts and 40 amps output to this web page here . <S> As for using a micor to control this read what folk are saying about this here: Digital SMPS <S> Vs. <S> Analog SMPS <A> Multiphase is the way to go. <S> Ideally, build a single, low-power, multiphase controller. <S> Then build multiple, small, power switcher sections, more than you need. <S> This way, if one fails, you can substitute. <S> When debugging, you will have good ones and bad ones to compare, there's nothing worse than seeing a waveform and not knowing whether it's meant to be like that or not. <S> If you can use a PC motherboard, do, saves so much building. <S> But don't cut it, use it whole. <S> There's no knowing how far vital buried tracks go a-wandering. <S> By all means remove components, but don't take a saw to it. <A> One thing to be VERY careful of is the output capacitance you typically need with buck converters of this type, it is a massive danger to your expensive pump diode. <S> If the output wiring becomes open circuit even for a moment the caps will charge to whatever you have set the voltage limit to be, when the diode gets reconnected the resulting current pulse will kill the diode. <S> When I did this for a one off <S> (And I have, slightly different parameters but a large low voltage diode bar) <S> I punted on the efficiency slightly and brought a couple of off the shelf isolated DC/DC module (GE Critical power as I recall), which brought my input down to a few hundred mV above the diodes operating voltage, then regulated the current with a passbank made of butch bipolar transistors on a heatsink <S> (The isolated converters meant that I could simply parallel the current ouputs from the two halves of the passbank). <S> It was a while ago, and I dont have my BOM to hand, but I am thinking it was something like one of these <S> http://uk.mouser.com/ProductDetail/GE-Critical-Power/ESTW025A0F41Z/?qs=sGAEpiMZZMvGsmoEFRKS8A9yhiuKixFL2sWuTv9sg4E%3d <S> There are 1.8 & 2.5V versions of the same idea available which should get you something you can trim to give the required headroom for the passbank. <S> The advantage was that the tricky to make efficient DC/DC was a solved problem, and I had isolation so accidents around the diode bar became less destructive, and I was not trying to regulate current in the presence of hundreds of microfarads of output cap. <S> Modulation was brought in via a linear optocoupler and optos were used for the status, interlock loop and fault reporting output. <S> The isolated module was important because big laser diode bars have an annoying tendency to be positive on the mounting flange, the isolation meant that an accidental short could not bypass the current regulator. <S> For one off things you do NOT <S> in general want to be designing your own heavy current switchers <S> , apart from anything else the appropriate magnetics are a problem in small quantities. <S> Be careful out there, a laser diode bar like that is not a good toy, you only come with two retinas. <S> 73 Dan. <A> Computer existing technology is valid to do your job .So <S> is the LT chip .In <S> other words a polyphased buck current mode hard switched running off a not too high input rail would work and would be sensible if your rail is there .If <S> you want to go offline <S> then you can make a 10: 1 transformer possibly using a planar <S> E I core and prototype your winding with teflon wire .A <S> 60 amp <S> but your job is different. <S> Maybe you could use lots of 10microfarad ceremics .Now that the output has been dealt with your problem becomes a 4 amp problem . <S> I ran a ZCS invertor in a half bridge on the 10 turn side .I used 2 cheap BJTs and <S> so could you .Remember the power is less than 100watt. <S> I fed my simple two transistor ZCS half bridge invertor with a "s trap buck convertor "This was prototyped using a mains isolation transformer for safe debug and now can go off rectified mains .You could use a buck convertor thats more familiar to you .I suggest that you control the current .My <S> little job was to put currents through wires which wasnt a laser but showed some similarity .For <S> current measurement <S> I used a cheap CT on the primary AC side of the planar transformer where the currents were very managable. <S> This meant no shunt .The error due to magnetising current was insignificant because the planar transformer had no gap and was running at low flux density because of 1 turn making such a low output voltage .
| I can't imagine that the laser diodes are cheap and because of this I would recommend a properly designed circuit board with a design solution that is tailored for this application. TO247 shottky on a 1plus 1 ct winding will do 40Amp easy .Place lots and lots of output caps across your Low volt DC .I used 680 microfarad sprauge for a 1 volt job
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PCB tracing - Ground plane and Power plane? Is it generally a bad idea, in simple or moderately heavy circuits, having two layers, to have a power plane in the top and a ground plane on the bottom? The tracing would become much easier, there wont be virtually any noise, better heat dissipation and aesthetically the circuit would be layed out neatly. Also, is there any handy guide to refer, when to use a thicker trace for different values of current? Thanks. <Q> Well, I suppose this is one of those topics where opinions may vary. <S> Hower <S> it's somewhat useful to hear opinions backed up by some kind of logic/argument. <S> So here's one from http://www.ti.com/lit/ml/sloa089/sloa089.pdf <S> There has been a lot of confusion in the past over what is the optimum order for PCB layers. <S> Take, for example, a 4-layer board consisting of two signal layers, a power plane, and a ground plane. <S> Is it better to route the signal traces between the layers, thus providing shielding for the signal traces – or is it better to make the ground and power planes the two inner planes? <S> In considering this question, it is important to remember that no matter what is decided, there will still be signals exposed on one or both of the top and bottom planes. <S> The leads of the op amp PCB package, and the traces on the board leading to nearby passive components and feed-throughs will be exposed. <S> Therefore, any shielding effects are compromised. <S> It is far better to take advantage of the distributed capacitance between the power and ground plane by making them internal. <S> Anyone who has had to change connections on buried traces will appreciate this feature. <S> For more than four layers, it is a general rule to shield higher speed signals between the ground and power planes, and route slower signals on the outer layers. <S> Hope <S> this helps. <A> I do this all the time, ground planes and power planes are very useful in a lot of boards. <S> It not only makes routing easier, but also helps with signal quality if you're using high frequency signals. <S> You'll want to pick your planes wisely though, I would recommend a ground plane to be on the same side as high frequency signals to ensure there is a nice, short return path. <S> What you're looking for when you mention trace width/current is in the IPC-2221 standards. <S> There are lots of calculators out on the web, like this one: http://www.4pcb.com/trace-width-calculator.html <S> You enter your current and it tells you the minimum trace width for that current. <A> This is probably the norm instead of a bad idea. <S> At least for a 4-layer board you effectively have a board-sized decoupling capacitor, and it makes signal routing all easier.
| Another advantage of placing the planes internally is that the signal traces are available for probing and modification on the top and bottom layers.
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Is it a good idea to connect the Neutral point to your common ground in your PCB? I am making a PCB that uses AC Mains power where 120VAC is converted to 5VDC with the help of AC/DC converter. The 5V output is then supplied to my digital circuitry. The main confusion I have is to whether to connect the neutral point of the Mains power to my common ground of my PCB or not. In theory, it shows no harm. I have shown the example circuit below where I=0 between mains's circuit and my digital circuit but I am still not sure if its still a safe thing to do. If I did not connect my common ground to the neutral point, would there be any EMI/crosstalk issue on my digital circuitry? simulate this circuit – Schematic created using CircuitLab <Q> Let me give you the simple answer. <S> No, do not connect any current carrying conductor from AC mains to your low-voltage circuit. <S> Regardless of whether it is "hot" or "neutral". <S> Don't connect it. <S> Some people are suggesting that in some situations you might have to connect it for the power supply to function. <S> If that is the case, do not use that power supply. <S> Also, you would be better off whether you are a professional or an amateur, if you used external AC-DC converters (agency approved) for all your projects. <A> When thinking about if this is a good idea (it isn't) you MUST keep in mind that for plug-in kit in vast chunks of the world, you can never know which conductor is 'live' and which is 'neutral'. <S> For this reason safety standards make little distinction between the two wires, both are treated much the same. <S> So if you've somehow convinced youself that something about 'neutral' isn't very dangerous, then ask yourself if your reasoning still applies when the wires are swapped. <S> There are supply techniques which legitimately play non-isolated games, and contrary to much of what you read they are not dodgy or dangerous at all, in their proper place. <S> However, they ARE completely unsuitable for hobbyists or test equipment or anything like it. <S> If you have to ask, then the answer is definitely not. <S> In summary: Both mains conductors are live <S> Don't connect either of them to anything other than your psu primary. <A> As Will Dean noted, it is quite common for neutral and hot to be reversed, but even if they are not, neutral usually does carry some voltage on it relative to earth ground, and this tends to cause problems like ground loops when you connect to some other device using a proper ground, and also can cause GFCI circuit breakers to trip. <S> To be UL listed, your device must not bleed current onto its ground pin or its chassis even if the neutral is cut. <A> First, for safety reasons, you'd have to make sure that the chassis of your device is connected to earth ground so it doesn't become live if there is a ground fault. <S> Second, if your circuit is to be connected to other devices, you must ensure that they have the same ground reference. <S> At best it would create a ground loop and at worst it could blow something up. <S> The usual way of doing this is to use an isolation transformer and make your circuit floating ground. <A> If it's transformerless, your circuit must share a common ground (AC neutral) in order to function. <S> If you're using a transformer (followed by diode rectifier, capacitors, and regulator), your low voltage side is galvanically isolated from mains and is thus floating. <S> You can connect DC ground to neutral, but absent a concrete reason to do so I'd avoid it - having a floating ground can be useful if you want to interface with other systems that might have different ground references.
| Connecting your circuit ground to mains neutral is probably a bad idea. This depends on the architecture of your AC-DC converter.
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How to measure the component value on a PCB? I have faced a few instances in which I need to measure resistor or capacitor value which is soldered to the PCB board (SMD component). These are very small and difficult to look the value written on it. When I measure using a multimeter I am not getting the correct value (maybe due to other components connected to it). Is there a way by which I can measure the component's value without desoldering it? <Q> No, this is generally not possible because of the rest of the circuit. <S> By ideal, I mean that you have a perfect understanding of the circuit, how you measuring instruments work (for example the waveforms used to measure capacitance including the voltages on each range) and how parts such as the chips <S> Nick mentions behave when not powered and subjected to those waveforms). <S> If a much smaller and/or less accurate part is effectively in parallel it may be impossible to get an accurate number. <S> The most reliable way is to remove the parts with a tweezer type desoldering tool and measure it, then replace. <S> Some larger (0603 and up) resistors are still marked but there are many smaller resistors and most SMT capacitors that are not marked at all (photo from here ). <A> What are you asking called "In-circuit resistor measurement" and is discussed in different ways. <S> Regard copy right low, I attach the exact file of a known method in this field. <S> 'in-circuit' resistance measurement <S> The original link is : ioScience- "in-Circuit" resistance measurement <S> Consider you have a equivalent circuit as below : that R1e and R2e are input and output impedance respectly. <S> Now assume a circuit as below : <S> That node 1 and node 2 are your probe signal. <S> In this Signal VD is equal to : <S> Which delta-V is op-amp DC offset voltage and K is inverter circuit gain. <S> When we adjust circuit so that VD=0 then we have : It will be satisfied by using an variable resistor for R0 or Rs. <S> Suggested values for R0, Rs and op-amp model is described in original document that is attached. <A> In general, no. Consider, for example, a component that's shorted out by PCB traces. <S> Obviously it's going to be impossible to measure it while on the PCB. <S> Conversely, a component that's not connected to anything isn't going to pose any problems at all. <S> If you're trying to make an in-circuit measurement like this, you have to look at the schematic and consider what other paths could be formed between your probes besides the one you want. <S> Consider not just the marked parts, but also ESD protection diodes in any ICs. <S> If the only current paths other than the one you want <S> go via semiconductors, you may be able to measure your component accurately if your meter uses a test voltage lower than one diode drop. <S> Otherwise, you're stuck with unsoldering the component in order to measure it. <A> This is a difficult problem. <S> Components tend to be connected on a board, and these connections corrupt the readings. <S> Many manufacturers, starting in the 1960s when automatic test equipment was just starting to be computer controlled, like Teradyne and Marconi Instruments to mention a couple, came up with a way of making some of these measurements. <S> In the best case, these connections can be 'guarded out'. <S> This is essentially an extension of a 4 wire measurement. <S> Let's say you have a pi attenuator , and you want to measure the series resistor R2 (from that link). <S> You would apply a known voltage to one port, so the current down R1 is irrelevant, and measure the current through R2 using a virtual ground amplifier across R3. <S> To do this, you also need a connection to the ground, this is called 'guarding out' the ground connetion. <S> As there is no voltage on R3, no current flows in it, so all the flowing current comes through R2. <S> This allows calculation of R2. <S> Unfortunately, some components are just not feasible to guard out. <S> The guard node might not be accessible. <S> You might not have enough dynamic range to control the guarded components.
| If you have a schematic you may be able to analyze the circuit and figure out what the value is from the 'incorrect' reading, but that is not always possible, even under ideal conditions.
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Does this schematic with an old doorbell electromagnet look ok? And could it be improved? I have a circuit which seems to work and I would just like someone to confirm that it looks ok and isn't going to damage a 3.3/5V system (Raspberry Pi). Also, if there are improvements / simplifications I could make, that would also be great. Basically, the circuit is used to get an old doorbell electromagnet to work to make it chime. When I set pin 24 high on my Pi, which is connected to a transistor and in turn a relay, it dings and when I set pin 24 low, it dongs. You should know that initially I was just connecting this all straight to pin 24 on the pi which I was advised here that wasn't a good idea, so thankyou for that ;). Below is what I put together in Fritzing to show what this version looks like. Initially, I was just trying to use the transistor on it's own, which would be simpler, but for whatever reason, that doesn't give me enough voltage to get the magnet to work -- I will understand why one day... So any thoughts / feedback / criticism greatly received. Many thanks,D. <Q> It probably didn't work with the transistor because the transistor didn't have enough base current to fully turn on and therefore the voltage developed acorss the electromagnet <S> would have been too weak. <A> The circuit looks feasible, but you don't give a transistor type or an indication of what current the relay coil wants, so we can't be sure. <S> It's not surprising that a doorbell doesn't work very reliably at 3.3 volts, especially with a series transistor. <S> In my experience, they might use a 24 volt AC transformer, expecting lots more juice than 3.3 VDC. <S> But maybe your doorbell has a different voltage rating? <S> Also I would be cautious about tying an inductive load to your 3.3 volt line, even with the relay. <S> An inductive kick or sag could give your Pi some trouble. <S> Better to use an isolated supply, probably with higher voltage. <A> E-Magnet is driven by Current not by Voltage. <S> So I think if you reduce the resistor R1 value, you may get enough current to drive. <S> Secondly, I may understand the Ding but when switch is OFF, how will door bell Dong? <S> I am just curios, really did not understand. <S> I hope this works.
| You should probably put a diode also across the doorbell electromagnet to prevent sparking acrss the relay contact when the circuit opens for the "dong".
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What is the problem with charging a (lead-acid) car battery with AC power? The closest thread I've found (which didn't show up in the automated suggestions) was What happens if an AC voltage is applied to a battery? . Unfortunately, I felt like the answers were more aimed at the device and schematic drawing instead of a directed explanation of why AC cannot be used to recharge a DC battery. I have a vague understanding of why AC cannot be used in electronics (something about the cycling between voltages would not fit the either "0" or "1" needs of a dc circuit) and I wouldn't expect you to try to teach me a bunch of theory that would require long study and research on my part to go down that road, but I was hoping for a clear explanation on specifically why AC power doesn't work for just charging a battery. My educated guess is that the cycle from negative to positive voltage would some how cancel out the charge. <Q> To charge a battry, you must essentially force the positive and negative ions each to their electrode. <S> When you apply AC you periodically swap the direction in which you force the ions, hence they shuffle back and forth, achieving nothing (except causing heat, and probably damage to the battery). <A> If you put a short across the terminals it's the same as applying 0V to the terminals. <S> You'll get a thousand+ amps and melted wires and possibly a fire. <S> Considering that an AC voltage produces zero volts 100 to 120 times per second (50 Hz or 60 Hz) as the waveform passes from positive to negative or vice versa, you surely must be able to start realizing the problem. <S> It's a lot worse than this because when the waveform is negative it's like having two car batteries connected to each other with the wires crossed (smoking big time). <A> An AC current keeps reversing direction; it spends half the time flowing one way, then the other half flowing the other way. <S> Another way to look at it is that the positive and negative terminals swap on each half-cycle. <S> A battery has distinct positive and negative terminals which don't change. <S> Connecting up a charger the wrong way discharges the battery, rather than charging it. <S> If you were to connect an AC supply to a rechargeable battery, it would be charged up for half the cycle, but then discharged for the other half. <S> The diode pack in a car alternator stops the reversal of the voltage. <S> A single diode could be used to block the current from flowing the wrong way. <S> But a bridge rectifier made from four diodes can re-direct <S> the flow of current so that the alternator then has positive and negative terminals which always have the correct polarity. <A> To charge a battery the charging voltage needs to be higher than the battery terminal voltage. <S> Putting one or more diodes into the circuit to "rectify" the AC is how we solve the problem. <S> The diode only allows current through in the right direction. <S> Your vague understanding regarding the "0" or "1" needs of an electronic circuit is, as you say, vague and <S> I would recommend that you don't spread it around. <A> AC is short of "Alternative Current". <S> What does alternative current means?It means that the positive terminal of an AC supply is not positive all the time. <S> Sometimes it becomes negative and sometimes it becomes positive. <S> The negative terminal changes its polarity as well. <S> As a result, We can NOT say that a terminal is positive (or negative) in AC. <S> For more information about AC: https://en.wikipedia.org/wiki/Alternating_current <S> After understanding what AC is, let's imagine what will happen if you connect a battery with AC supply to charge it: <S> The supply has two terminals. <S> When the first terminal is positive, the battery will start charging. <S> I thing you agree with me. <S> After some time, the first terminal will not become positive anymore. <S> It will become a negative. <S> Then, The battery will Discharge electricity again because the positive terminal of the battery will be connected to the negative terminal of the supply. <S> After some time, the first terminal will become positive again and the battery will charge electricity and so on ...
| As you have realised, alternating voltage will not only go below the battery terminal voltage but will go inverse polarity too.
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BPM Generator with display? Trying to figure out how to design this with decent precision but having trouble getting this started. I've worked with 555's before but not for precise counting that can be modified in steps. I'm trying to come up with a 5 volt system that accurately pulses an output of +5v that is measured and displayed in pulses per minute, or beats per minute, between 60 and 240. The BPM should be adjustable by 2 momentary buttons which will step the timer +1 and -1. It would also be great if there could also be a second set of buttons for +10 and -10 steps. Is using a 555 timer the wrong way about this? Is there something more precise I can use to reference the gating? Is there already an IC out there that can do this on its own and do a compare to calculate the maths? Apologies for not posting what I have so far. I'm on version 80-something on my breadboard and nothing has been working accurately so far. I also want to stay away from having to program roms and stick with discrete IC's. Again, I DO NOT want to be programming chips. I think what I'm after might be a function generator with a frequency counter, but I don't know if that's the best way to go. <Q> This could easily be done with a microcontroller. <S> 240 BPM is 4Hz. <S> By using a microcontroller with timer peripherals it would be easy to get accuracy in the +/-0.01% range. <S> Suggest a microcontroller such as PIC or AVR or MSP430 with a small display, buttons. <S> An LCD display could be used to save power, and there are micros with LCD controllers on board. <S> Cost and circuit complexity would be very low. <S> You can find open-source LCD module display frequency counter designs based on the PIC16F628, for example, but they're probably not directly usable for such low frequencies. <S> To get 1 BPM resolution with a simple frequency counter requires a 1 minute gate time, so period counting and math would be a better approach- actually more difficult than generating a set frequency. <A> If you want to implement something without using any programmable parts beyond a custom-frequency oscillator, it should be possible to build a circuit that takes a 3-digit BCD frequency and outputs a signal with that many beats per minute using an oscillator plus five off-the-shelf chips. <S> Feed a 1,092,267Hz oscillator into a CD4060 to scale it down by a factorof 16 (DIP-packaged oscillators at Digi-Key have a 1Mhz minimum speed).Feed that 68,266.7Hz signal into a cascaded sequence of three CD4527 chipsset up for the "ADD" mode to yield an output of (1-999)/4,096bpm, andfeed the output of that into a CD4040 to get the desired output rate aswell as various power-of-two multiples and submultiples of that. <S> Highertaps of the first CD4060 may be used to provide various power-of-twomultiples of 66.7Hz [perhaps usable as "beep" tones]. <S> If you have three BCD thumbwheels, you could would only need six mainelectronic parts, all DIP; quantity-one prices at Digikey would be: 3x CD4527BE -- $0.80ea ($2.40 total)2x CD4060BE -- $0.56ea ($1.12 total)oscillator -- $3.02ea ($3.02 total) <S> $6.54 total Assembly should be fairly straightforward on 0.1" perfboard sincethe <S> only interconnections other than power and ground would bethe oscillator output feeding the first CD4060, the output of thatfeeding all three CD4527, each of the first two CD4527 feeding twosignals to the next, and <S> the last CD4527 feeding one signal to thelast CD4060. <A> As the others have said, a PIC or Arduino is the way to go <S> but if you're determined to avoid programming ... <S> You could consider using CMOS chips with a a high frequency oscillator with a crystal for stability. <S> You would then use a counter chip to count pulses and give the present count on its output pins. <S> These would be fed to some logic to give an output pulse at a certain count and reset the counter. <S> Alongside this you would need some other counters to set the reset point for comparison. <S> Unfortunately, here is where things get messy. <S> If you were happy to count 100, 200, 300, 400, etc. <S> , pulses it might be doable but <S> the problem is you want to specify beats per minute <S> so you'll need to calculate 1/BPM to give you counts for even BPM steps. <S> Think again about the programming. <S> Micro's have been doing well for the last while. <S> I think they're going to catch on!
| Whilst you could use a 555 and attach a frequency counter to it (and adjust the knob as it drifts off frequency), the frequency counter would best be implemented with a micro and thus it's easier to simply synthesize the frequency you want correctly in the first place. You can prototype this with something like an Arduino.
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Can I use a resistor before a bridge rectifier to lower inrush current? I have been looking for ways to limit inrush current in a power supply circuit I am designing. In LTspice, with no current limiting, I see a spike of around 24A before my caps are charged. Looking around, I see NTC inrush current limiters as a common solution as well as some ideas with inductors and resistors inline with the capacitor, each with its own advantages and disadvantages. My idea though was to put a resistor before the bridge rectifier in my circuit, so the resistor would always provide some resistance before the capacitors got charged. Playing around again in LTspice, this drops my inrush current to about 4A (and changes with the resistor value obviously). Is this ever done? Are there any major good or bad reasons to do this? I suspect no one does this due to decreased power efficiency, but is that the only reason? <Q> If you get an inrush, does it matter? <S> Well, only if it breaks something. <S> So what could break? <S> Supply fuse, transformer, or diode rectifiers. <S> Fuses have 'T' rated versions that take a long time to blow, for just this application. <S> The transformer is a heavy lump of copper, that's not going to fail. <S> Read the specification of your diodes carefully. <S> You may be surprised at how much the 'single cycle surge current' is. <S> In a 1N40xx (cheapo workhorse mains diode), the continuous current is 1A, the surge is 30A. For 1N54xx series, the figures are 3A and 200A. <S> This is specifically to allow them to survive inrush. <S> You may find you already have enough stray resistance in your circuit to limit the current to the safe surge value. <S> If not, you maybe won't need much more, and it still may give you acceptable efficiency. <S> If it doesn't, then try smarter solutions. <A> Using a resistive element to limit inrush current is done. <S> However, placing a permanent resistor in the AC (or DC ) lines is very inefficient as it is always in-circuit. <S> The use of a Negative temperature coefficient (NTC) thermistors is often done instead as it provides a level of resistance during the precharge event that is then reduced due to heating. <S> This however does impose a "cooldown" period to facilitate the NTC returning to room temperature resistance in the event of a power loss or power cycle. <S> An established method is to use a form of soft-start http://www.ti.com/lit/an/slva156/slva156.pdf simulate this circuit – <S> Schematic created using CircuitLab <S> There is equally monolithic chips that provide the same functionality. <S> http://www.onsemi.com/pub_link/Collateral/NCP330-D.PDF <A> Adding resistors will create a permanent power dissipation. <S> This is usually undesired. <S> It will also create fluctuating input voltages caused by changes in current draw (voltage drop at resistor). <A> There are fancier ways of limiting inrush current than the simple resistor ,Some of which JonRB has stated. <S> If you are using a resistor there are a few more things to watch out for .Firstly <S> the surge power dissipation of the resistor can be incredible .I have seen Christchurch design engineers blow up such resistors .Its really important to get specifications for peak power dissipation .Melf <S> are good that way but some manufacturers are vauge. <S> Secondly the average power dissipation is dependant on the RMS current not the average current .RMS <S> current can be and often is significantly more than the average .I <S> have seen people unwittingly undersize these resistors . <S> The horrible part about this is that the resistors take some time to fail open circuit .The way the resistor fails is also important .Some <S> resistors will go on fire which is bad .Some <S> are flame retardant and go out ,some dont <S> .If <S> you still want to use a resistor test it under fault conditions . <A> A quite simple solution (I'm surprised it's not mentioned already) is to add a small coil after the rectifier: simulate this circuit – Schematic created using CircuitLab <S> A coil doesn't add much active resistance and doesn't practically affect the efficiency under "steady state" operation. <S> The current spike is suppressed by the inductance which limits the current rising slope by $$L=\frac{dI}{dt}$$
| There are good and complicated solutions, relay, FET, and simpler, thermistor, but it still may be possible to use the simplest, fixed resistor, without too much dissipation for your application.
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Eclipse + GNU ARM + STM32 - HAL or SPL I am going to start with ARM development (after 2 years of AVRs) and have picked up the STM DISCOVERY board with the stm32f4 microprocessor on it. I have decided to go with eclipse + ARM gcc since I don't like the code limit on Keil and I don't have the money to get a paid version. Following the tutorials I have installed eclipse along with gcc ARM tools + openocd + make utils etc. My question is about the 'packages' plugin. Like every beginner, I am confused as to whether to use new STM HAL or the older SPL. My understanding is that HAL has implemented abstraction to a level where it can be referred to as Arduino equivalent for arm. SPL on the other hand provides just enough abstraction to make coding faster but you still need to deal on chip level. With this understanding I would like to stick with SPL to understand things better rather than using HAL. What I would like to know is, does using packages for STM implicitly force me to use HAL? If so, can someone point me on how to use SPL with my setup? <Q> The SPL, as I see, has nothing to do with what IDE you are using. <S> You can simply include the relevant modules (e.g. stmf4xx_dma.c and stmf4xx_dma.h) in your project and use the functions exposed (and described very well) in the .c <S> and .h files. <S> In fact I've been learning on the stmf411 nucleo with gcc, openocd and SPL using just the windows command prompt; no IDE. <S> Packages in eclipse probably would force you to use the HAL (since inside the downloaded 'Packages' folder for eclipse, I only see HAL modules). <S> The HAL itself <S> IMO seems far much layered than necessary. <S> Whereas accessing the registers directly gets tiresome and is hardly readable. <S> The SPL seems just right. <S> clive1, the guru on the st.com forum, also prefers the SPL over HAL. <S> Here's my question on that forum... <S> might be helpful. <S> Need help with USART on Nucleo stmf411 <A> I don't have any experiences with HAL,but used SPL many times for saving my times. <S> In my believes Target community of this Embedded processors are 2 groups: <S> First group which are not interested to engaged with hardware layers. <S> Software programmers,usual hobbyists and Arduino ,Raspberry worshipers.if <S> you are in this group <S> seems HAL is good choice for you. <S> Seconds which comes from electronic and hardware community, whom prefer GPIO_A->PIN & <S> = ~(1 << 15); to LED_On(1) for turning on the LED and want to know what they are doing basically.then if you have in this group and have enough time to reading reference manual and programming manual of your MCU maybe register level programming is another choice.but if you want to decide between only above 2 option:HAL has a better future because of ST' supporting but SPL is a easier way to understanding for a new starter. <S> Maybe this can help http://www.eevblog.com/forum/microcontrollers/stm32-and-their-hal-library/ <A> - it's free, based on Eclipse and have both arm-gcc and openocd in one package. <S> And about libraries: besides SPL and HAL <S> now exist LL.Eache one have some advatages and disadvantages, and you must choose what you need. <S> And as I understand , all of them have experimental status for ST. <S> Below my grades to each of them: SPL: old, cumbersome, no extra ram usage, flexible HAL: actual, cumbersome, extra ram usage, not flexible <S> LL: actual, ligtweigth, no extra ram usage, <S> flexible Short description for my grades: cumbersome - big flash usage, "super" universal functions for work with periphery extra ram usage - it's about HAL, it have copy of peripheral state in ram located structs and use it everywhere and everytime <S> not flexible - and again about HAL, it have a lot of functions for differents cases, but! <S> most of them are not usable for real devices (peoples try to reinitialize HAL for receive byte by byte from usart <S> > <S> _< , all functions for TIM+DMA are implemented for rewrite TIM register and no any other...) <S> For a little rehabilitate HAL: it have one big advantage for newbie - it suppoted by STMCubeMX. <S> EDIT: <S> I forget about libopencm3 - it's alternate library. <S> I haven't used it.
| Get this IDE: System Workbench for STM32
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ALT key in Eagle (Linux) I am facing an annoying problem while using Eagle in Linux.It is not possible to set the desired position of a component while ALT key is pressed. I hit the ALT key to active the alternate grid and then specify more precisely the correct position.I also tried to move the component while ALT is pressed, then release the key and left-click to set the position, but then the component moves to the main grid automatically. I know there is other ways to solve this tricky bug during component placement, but it would be easier using this shortcut. Does someone have already faced this bug? Thanks, (I am using Linux Mint with Cinnamon.) <Q> EAGLE interferes with your operating system settings. <S> I'm using OpenSuse with KDE5 Plasma, where ALT + mouseclick resizes or moves the window. <S> EAGLE may notice the ALT key is pressed, but a click is intercepted by Linux. <S> Just try to remove the ALT + mouseclick key binding. <S> As every linux differs a little, I can't tell you how exactly to do it. <S> For me, it's in workspace settings / window behaviour . <A> In KDE Plasma 5 it can be disabled by: System settings -> Window management - <S> > <S> Window behavior -> <S> Windows Actions <A> The solution that worked for me in order to solve this bug in Linux Mint with Cinnamon: System Settings> <S> Windows <S> >Moving and Resizing Windows <S> > "Special keys to move and resize windows: Disabled"
| You can change the modifier key or disable the left button action.
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heat sink calculation for parallel bjt I'm working on a project and I need to use 3 parallel bjt to give me 20 A current.Total power dissipation is 356 W.I know how to calculate a sink for one transistor,but in this case I'm not sure what is Pdiss in Rja=(Tj-Ta)/Pdiss, is it 356 W or ,as I'm using 3 2SA1302 bjt,356/3 W.Also is Tj (junction temperature) 150 C (for one 2SA1302) or is it 3*150 C ? <Q> The situation is the same as if you had a single transistor with a Tjmax of 150C dissipating 360 watts. <A> If you want to make a single sink for the three transistors <S> Each BJT is dissipating 356W/3 but the sink has to dissipate the total temperature. <S> With this amount of power <S> I think the sink will be quite huge. <S> Another thing you can do is designing one sink by BJT, then each sink will have to dissipate 356/3 and then Pdiss will be 356/3. <S> For Tj it is a caracteristic of the BJT <S> you can not change it. <S> It means at 150°C your BJT will crack. <S> Then even if you have 3 of them, at 150°C they will crack. <S> Which means your sink should be designed to have Tj<150 and not Tj<3 <S> *150 <S> And if you want your circuit to live long enough you need to design your sink to have a Tj around at most 125°C cause the hotter it is the earlier it will break. <A> You are expecting to get rid of 118 watts in each transistor assuming that your proposed 3 transistors share perfectly .Look <S> at the total thermal impedance from the junction to the heatsink because this is the important parameter .Now <S> if I ballpark 1 degree per watt which is reasonable for normal packages and normal metalwork <S> then things start to get bad. <S> So the junction will be 118 degrees hotter than the heatsink <S> .This means that the heatsink will have to be in Ice for good reliability .This <S> means 118 degrees for Tj .So to make mil461 it would have to be on the north or south pole .This <S> is the hot junction ,cold heatsink syndrome that I have seen so often <S> .The sensible thing to do these days is to use more and more transistors until the heatsink becomes reasonable .I <S> use the term " Silicon Aluminium balance " when I have to teach people this stuff <S> .The <S> heatsink allows you to get more out of the transistor but over the decades the heatsink has got dearer and the transistor has got cheaper. <A> This must be split into two parts: junction to case (package) and case to ambient (heat sink). <S> Let's assume (since you haven't stated it) that you will use the same heatsink in both cases, but in one case you will use a single transistor, and in the other you will use 3. <S> In both cases, the heatsink temperature will be the same, since in both cases the total power being dissipated is the same - 356 watts. <S> Just as an arguing number, let's say you have a big heat sink with a thermal resistance of about 0.1 deg/watt. <S> Then the case temperature of the transistors (assuming a real good thermal mounting) will be about 25 + (0.1 x 356), or 61 C. <S> Now look at the junction to case temperature drop. <S> The data sheet says that the power must be derated from 150 watts by 1.2 watts per deg C at the case. <S> 356/3 is 118 watts, so the maximum case temperature is about 150 - (118/1.2) or about 52 C. <S> So your heatsink will not work - <S> the transistors will get too hot. <S> You can do the math to find the required thermal impedance of the heat sink required to keep the transistors cool, and it will obviously be less than 0.1 deg/watt. <S> Just off the top of my head, that's a pretty big heat sink, and may well require forced air cooling.
| Pdiss will be 356W because it is the power you need to dissipate in the sink.
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Is there any benefit from installing 2 resistors in parallel instead of 1 resistor? I'm restoring an old tube radio. When looking into the schematics I noticed that they drew 2 resistors in parallel instead of 1 resistor, between the power transformer and the rectifier tube. Is there any benefit from installing 2 parallel resistors instead one resistor? <Q> Two resistors have an increased ability to dissipate power/heat, up to twice the time one can, when mounted with enough space between them. <S> Also for a few applications it is easier to find two standard value resistors that together form one resistor value that is not among the standards you already have in your BOM (or are available). <S> Lastly for the gray beard hand chosen value resistor, it might have a higher yield to use two that are slightly off than to find the one that matches perfectly. <A> The benefit is the power dissipation. <S> If you have a voltage V and a current <S> I the power dissipation across a resistor is <S> $$ P= <S> V <S> \cdot <S> I$$ <S> Now if you put two resistors in parallel you divide the current in each resistor by 2 (if they have the same value) and then the power dissipated by 2. <A> As others pointed out, the main reason increased is heat dissipation. <S> I had a sound amplifier that had a huge power resistor which every week was burning out literally. <S> One day I replaced it with 2 resistors and since then never failed. <S> Another experience of mine is regarding PCB design. <S> I wanted to order minimum variety of components. <S> So the pick and place machine won't charge me for extra rails. <S> I had many 5k resistors in the design. <S> Anywhere that I had 10k or 20k resistors <S> I simply used 5k in series. <S> A bit stupid but saved me few cents!
| Sometimes it is cheaper to put 2 resistors in parallel to dissipate more power than only one bigger resistor.
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Move silkscreen text from component on EAGLE CAD How can I move a silk text from a component? I cant use the button move to the text and when I click smash nothing happend. I need to move the number 8 inside the board Edit1 I try to smash but when I press it, a new text appear and this I can delete it, but I cant move the numbers or delete them, how can I modify the footprint? or erase the numbers and add it manually?(you see the opcion unsmash because I already smash it) Edit2 I did it! Thanks all! I could modify the footprint going to library, then selecting the library, then package :) <Q> While @crasic's answer is correct for attribute labels, it does not work with regular text. <S> Basically when the component is designed, the name and value are added as a text label set to a special placeholder - >NAME and > <S> VALUE to be precise. <S> Other attributes can have similar placeholders. <S> When you smash a component, it only allows free movement of any of these attribute labels (anything beginning with a > ). <S> When you add other text like the number 8 <S> you have there, they are not attributes, so you cannot move them after smashing. <S> This is actually quite sensible as you could move that number 8 label to the other end of the connector which would mean as a reference it becomes useless. <S> The only way to move that label is to edit the footprint and either (a) delete it, (b) move it where you need it, or (c) move it to the References layer. <S> To elaborate on (c), you would normally not include the References layer when generating Gerber files, so if the label is on that layer it won't appear on your final PCB. <S> What you can then do is in your board layout manually add a number 8 using the TEXT tool and place it where you want. <S> By doing it this way it means that if you want to use the same footprint on another board you don't need to modify the footprint again for that new board. <S> Furthermore, by leaving the label on the reference layer, if you move the component around you will be able to still see the reference as a reminder where pin 8 is. <S> If the component is moved, you will have to move your added text label and place it in the correct place. <A> Smash is a command, you need to select Smash tool (or type SMASH into the command interpreter) and then click on the part. <S> There are some caveats - the part origin is either top or bottom, while the pads are on both sides, if you place the part on the bottom but hide the bOrigins layer you will not be able to select the part to smash but this is not obvious because you can still see the pads of the part <A> I think, the 8 is the number of a pin, and its location is fixed in the footprint (unlike a component designator). <S> Two things could be done. <S> Change the footprint. <S> Move the 8 to the desired position. <S> [This is what I would do.] <S> Add another 8 as free text to the PCB. <S> Since it's not a part of a footprint, you can position it where you want. <S> You will end up with two 8s. <S> One inside of the board, another cut off by the board edge. <S> By the dame token, you could also add a 1 next to pin 1. <S> as an aside: Smash un-docks the designator from the footprint. <S> Select the smash too on the toolbar. <S> Click on the component you want to smash. <S> (You can also do it from a command line, like @crasic had described above. ) <S> The designator's own drag handle appears.
| You can explicitly invoke smash on the part by typing SMASH JP1 in the command interpreter (replacing JP1 with whatever the reference designator for the part is)
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RF FETs - how exactly are they different from regular FETs? I've browsed through DigiKey and there are these RF FETs . Are they for final stage amplification? Say you have an opamp that doesn't go for the final desired voltage, so one uses these FETs as Common amplifiers? Is that why they are rated with gain and frequency in the page, rather than Transconductance and rise/fall times? <Q> There are hundreds of types of RF FETs all with their specific (sometimes niche) application. <S> There are RF FETs for high, medium and low power, switching operation or linear. <S> High, medium or low transconductance. <S> Many frequency ranges. <S> That a certain RF FET is made for a certain application <S> does not mean it cannot be used in a different application. <S> You will have to look at the datesheet to find out if it suits your needs. <S> But some specialist RF FETs can be very expensive so in general you would only use it for it's intended purpose. <S> "Are they for final stage amplification?" <S> Some are, some are not. <S> "Say you have an opamp that doesn't go for the final desired voltage, so one uses these FETs as Common amplifiers? <S> " <S> You could <S> but as opamps in general aren't considered "RF" using an RF FET with an opamp would be silly. <S> General purpose FETs are good enough for such a task and probably a lot cheaper as well. <S> "Is that why they are rated with gain and frequency in the page, rather than Transconductance and rise/fall times?" <S> Yes, in RF the "analog" parameters are important like gain (which is actually the transconductance under specified load conditions), cutoff frequency, input impedance, biasing requirements, noise, etc. <A> If you look at the frequency column, many go up to many GHz. <S> They are rated with gain and frequency because that's the best model to use when designing the circuit RF function. <S> They are still FETs, they still have a DC transconductance, which may be handy when checking the DC stability of the bias circuit. <S> Once you look at the price, power handling and frequency, you'll see why there are different classes of FET for different jobs. <A> RF fets are specified for RF .Power mosfets are not <S> so you are on your own when you use them .Power <S> mosfets are optimised for low on resistance at the expense of much greater input capacitance .This <S> can make gate drive more difficult .In <S> fact <S> the product of gate resistance and Cgs makes many of the newer low on resistance cheap power devices useless at RF .The newer low RDS powermos devices are not suitable for analog operation which could be a showstopper in some RF applications. <S> The RF fet has higher RDs on and much lower capacitances and is totaly suitable for analog mode. <S> Lateral construction is a popular way of achieving this. <S> The RF fet costs more to produce and is made by few manufacturers in relatively small numbers these days. <S> The characteristics are more linear which allows good performance with little or no feedback. <S> Decades ago the differences between power and RF fets were not so pronounced but cost and on resistance meant that the BJT was dominant in power conversion. <S> An old school example of such a FET is the VN88AF which I have seen being used for RF and for Audio and for power . <S> The complimentry audio fets that have been available for decades at a price are more like RF fets than power fets.
| They are for use as RF amplifiers.
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How to improve this amplifier so I can get up to 5 Watt output in 8ohm load This is Class A amplifier I trying to make up to 5 Watt output for 8 \$ \Omega \$ load. Is this amplifier circuit OK?t gives about 800 mW out on 8 \$ \Omega \$ load. <Q> Just looking at this simply, you have a "Darlington" output stage and this inevitably "loses" about 1.4 volts when powering a load - basically, from a 12 V supply, you can't deliver a "high" voltage greater than maybe 10.6 volts. <S> To avoid distortion maybe 10 V is the maximum. <S> So, 10 V is the range of the signal that can be produced. <S> In other words the output maximum level is ~10 Vp-p. <S> This is the same as an RMS voltage of about 3.54 volts. <S> If this feeds an 8 ohm load then the power delivered is about 1.56 watts. <S> That's the limit from 12 V <S> I'm sorry to say. <S> If you want to know why your amplifier only gives 0.8 watts you have to tell me more about the bias points in the circuit and how much distortion you are tolerating when 0.8 watts is being produced. <S> Clearly you could produce an output that is a square wave of 10.6 Vp-p and, into 8 ohms this is a power of about 3.5 watts <S> but I suspect you are doing a sinewave measurement. <A> You can't, and you shouldn't. <S> You can't because the theoretical maximum you can get from a 12V powered single-ended amplifier is a 12V peak-peak sine, which is 6V/sqrt(2) <S> = <S> 4.25V RMS, which delivers (4.25)^2 / 8 = 2.25W into an 8 Ohm load. <S> (This is a theoretical maximum, in practice it is considerably lower, because you will always have some voltage drop in your output stage.) <S> You shouldn't because the simple emitter-follower with a resistor load circuit you use is very ineffective, it will waste much more power in the load resistor than it delivers to the real load (the speaker). <S> That is the reason you will only see it in low-power settings. <S> Next you'd have to worry about cross-over distortion, DC bias, feedback and stability. <S> But unless you are in it for the learning process, you'd better buy one of the numerous chips that are used in car radio output stages. <S> They cost next to nothing and are very easy to use. <S> A more modern approach would be to use a class-D amplifier, which is essentially a switching power supply acting as audio amplifier. <S> You'd still need a bridge topology. <A> You wont get 5w on 12V like Andy <S> says.but you can do better by replacing R10 with a current sink maybe frying an egg at 1 amp. <S> Your amp will now sound lovely .If <S> you know what you are doing you can modulate the current source with the speaker return current which will allow you to halve the Idle current with no sonic penalty. <S> Consider bootstraping your discrete darlington,or feed the driver from a higher rail like 15V if possible .Now you can expect a totaly normal output power considering 12V and 8 ohm.
| A realistic way to get 5W from 12V is to use a bridge amplifier (two output stages, one fed with the inverted signal), with push-pull (complimetary emitter follower) output stages.
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why is the voltage zero across the resistors in this circuit? simulate this circuit – Schematic created using CircuitLab <Q> The answer is, it is, and it isn't. <S> You have missed one critical piece of information in your question, when. <S> At steady-state, or during the transient period when the voltage is first applied. <S> It seems you added the information in the comments. <S> In the transient the capacitor will charge up through the resistors until it reaches \$1\mathrm{V}\$. <S> Once the capacitor has reached this voltage (i.e. it is fully charged), assuming it is ideal and the voltage source remains constant, then you will have: $$ <S> V_s= <S> V_c$$ <S> Clearly that means all the voltage is dropped across the capacitor, so there cannot be any voltage across the resistors. <S> For completeness, we can look at the steady state condition in another way. <S> The reactance of a capacitor (similar to resistance, but frequency dependent), is given by: $$X_c = <S> \frac{1}{2\pi fC}$$ <S> Where \$f\$ is the frequency, and \$C\$ is the capacitance. <S> At DC, the frequency is \$0\mathrm{Hz}\$, so the reactance is: $$X_c = <S> \frac{1}{2\pi C\times 0} = <S> \frac{1}{0} = <S> ∞$$ <S> So what will the current be if the reactance is infinite? <S> \$I=\frac{V}{X_c}=0\$. <S> If there is no current flowing, there can be no voltage across the resistors \$V= <S> IR=0\times R=0\$. <A> Falstad's simulator might be helpful for your understanding. <S> Here's your circuit with the voltage shown on a green/gray scale: <S> Notice how the voltage is constant (+V) across the entire top of the circuit. <S> The entire bottom of the circuit is at zero volts. <S> You can't tell from this picture, but the dots representing current are not moving. <S> What's going on here is that at DC, a capacitor acts like an open circuit. <S> (It's a pair of conductors with a gap between them.) <S> Since the circuit isn't complete, there's no current, and thus no voltage drop across the resistors. <S> That's just Ohm's law. <S> Physically, all of the charges in the top half of the circuit have the same electric potential energy relative to the bottom half. <S> Since they're not moving, there's no change from one side of the resistor to the other. <A> It's supposed to be fully charged <S> If the capacitor is fully charged it is not taking any more charge i.e. no more charge is being passed through it. <S> With no current flowing through the resistors, there can be no voltage across them (apart from self-generated thermal noise <S> but that's a different story).
| Given that Q=CV in a capacitor and also that the rate of change of charge is current, there can be no current flowing through the circuit. It is fully charged, i.e. steady-state condition.
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How to solve a resistor nest efficiently Of course, one way to do it is to split it into series and parallel circuits and apply Kirchhoff's law, but that seems inefficient. Is there a method by which a circuit nest can be solved (finding the voltage and current at each resistor) efficiently (suitable for a computer simulation, perhaps)? (Of course, any approach would be based on Kirchoff's laws, but they can be rearranged to be more efficient for this task.) I am thinking it would involve finding relative voltages at every node, instead of voltage differences at each resistor. What should I do? <Q> Well in the particular diagram you have shown there appears only one wire coming in. <S> So there will be no current flow in any of the resistors and all the nodes will be at the same voltage. <S> Edit: <S> looking at the full nonsense circuit on XKCD there is a wire coming into the top of the subcircuit you pictured which you cropped off. <S> If you connected a voltage source to the top and bottom of the circuit you could solve it by the means below. <S> But in general you can solve a circuit with lots of resistors by the following procedure. <S> Choose <S> one node to be your 0V reference. <S> Assign a variable for the voltage of each node other than the reference node. <S> Assign a variable and direction for the current through each component (including your voltage source). <S> Don't worry if the direction you assign is backwards, that just means you will get a negative answer for it's value. <S> Write and equation for each node setting the current into the node equal to the current out of the node. <S> The voltage source equation will relate the voltages on two nodes directly. <S> The resistor equations will relate the voltages on two nodes to the current through the resistor. <S> Solve (or get a computer to solve for you) <S> the resulting simultanious equations. <A> Well you didn't give any other information, so from that perspective I can't help. <S> What I can tell you though is that those resistors seem to all be in parallel, and if they are, voltage would be constant throughout the whole circuit. <S> My advice would be to clean it up from that circular/triangular mesh mess, simplify to an equivalent resistance to find the initial current into the circuit, then keep doing calculations as you re-expand backwards. <A> Here's an example circuit. <S> If R1=R2 and R4= <S> R5, you have the same voltage at each side of R3, so VR3=0, therefore current ( <S> I2-I3)=0 <S> and you can ignore R3. <S> This simplifies the calculation. <S> I got nerd sniped and distracted here <S> (see another XKCD on that!). <S> Mathics is opensource code that let me solve the Kirkoff law's linear equations symbolically. <S> Maybe the closed form equation will entertain someone. <S> It's been a while since I did circuit analysis, and a lot of the professional tools are expensive. <S> There's a free opensource tool called Qucs, which lets you draw your resistor nest circuit diagram, and you can calculate the total resistance by adding a voltage source to the circuit and finding the current, R=V/I. Last but not least, the XKCD nest is small enough to fit on any breadboard with a dollar's worth of 1K resistors, and you can measure it.
| Write an equation for each component (including your voltage source). There may be symmetries in resistor nests that you can use to simplify things considerably.
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Is it OK if a newly purchased electrolytic capacitor is supplying higher capacitance? I purchased new electrolytic capacitors in order to restore an old tube radio, and when I tested them with my capacitance meter I noticed that one them is having higher capacitance than it should. It's labeled as 22 mf / 450 V. but it's actual capacitance is 27.5 mf, would it cause any problem if the needed capacitance is only 20 mf ? <Q> Power supply electrolytic capacitors often have a tolerance of +/-20% or even -20/+50% (I've seen -20/+80%). <S> Usually a larger capacitance does no harm (as you might guess from the way the tolerances are specified). <S> Your capacitor measures +25% (assuming the meter is measuring it accurately), which may or may not be within tolerance, but normally a larger capacitance does no harm, within reason. <S> A much smaller capacitance than nominal would be a much larger cause for concern. <A> Electrolytic caps I commonly work with are labeled with +40%/-10% tolerance on the capacitance. <S> Since electrolytic caps decay over time, making them far above the spec to start with allows the unit to stay in tolerance for longer. <S> In general, if the capacitance value was critical, nobody would have used an electrolytic to begin with. <A> It's difficult to say without knowing the application, but electrolytics are not the highest quality capacitors available. <S> They have the advantage of high capacitance in a given volume, and are typically used where bulk capacitance is needed and precision is not. <S> Most likely it would be okay to use. <A> It depends where you put it. <S> If it's for the PSU, don't put it in the first position after the rectifier: put it after the first resistor or choke. <S> The rectifier will have specified limits on the capacitance that it sees. <S> If it's for a cathode resistor bypass, don't worry about it.
| It's imaginable that having too much capacitance could damage a precharge circuit, but at that capacitance level I'd be very surprised.
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Measuring multiple analogue signals on 1 analogue Input? Is it possible to detect multiple analogue signals from different analogue sources onto 1 analogue input pin on a micro controller. Another way of asking this question is that we can detect multiple digital signals from a single analogue input by attaching different values of resistors to each digital input source and then when a digital source is activated the specific voltage can be detected using an analogue input, so similarly would it be possible to detect multiple analogue voltages using a single analogue input pin on a micro controller? <Q> Is it possible to detect multiple analogue signals from different analogue sources onto 1 analogue input pin on a micro controller. <S> If they are for example sinusoids (of different frequencies), then you can do a [fast] Fourier transform and find them. <S> For arbitrary signals, I think the answer is no. <S> You need to clarify what you mean by signals. <S> Another way of asking this question is that we can detect multiple digital signals from a single analogue input by attaching different values of resistors to each digital input source and then when a digital source is activated the specific voltage can be detected using an analogue input <S> This is not the same problem as in your previous question. <S> Here you have multiple input pins via the resistors and you differentiate the signals (presumably of the same amplitude) by giving them different amplitudes using your resistors. <S> would it be possible to detect multiple analogue voltages using a single analogue input pin on a micro controller? <S> This is more or less a repeat of your 1st question. <S> If you ask however whether an arbitrary DC sum can be detected/split: the answer is obviously no. <S> How could you tell the difference between 1V+1V and 0.5V+1.5V by just measuring the total/sum? <A> No. <S> There would be no way to know which analog signals were contributing to the sum. <S> The problem is not unusual, however, and the standard solution is time-division-multiplexing. <S> Your micro probably has one ADC (analog-digital-converter) internally and several input pins can connect to it in quick sequence, the only requirement being that the switching and conversion start are correctly synchronised. <S> In your case, if you really have only one analog in then you will need to do the analog switching externally. <S> Something like CMOS 4016 chips might do the trick. <S> The analog switches will require control by output pins from the micro <S> so it won't help your pin count. <A> No. <S> Your digital analogy works only in digital domain since the inputs have just two possible voltage values (ideally). <S> What you described is basically a D/A followed by your A/D. <S> If you make a weighted sum of two analog signals, one with weight 1 and other with weight 10, how could you tell the difference from a 10 mV variation in the first from a 1 mV variation in the second? <S> But if you elaborate more on your signals (amplitude, required measurement precision, bandwidth) and your A/D capabilities, there might be a way to mix them (no just adding). <S> But it is very likely that this solution would not have good performance and cost more than replacing your micro-controler.
| It depends on the signals.
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Filter of DC does not work? I have the following simple circuit: I thought that the capacitor under R1 can be used to filter off the offset of the DC, but instead I am still measuring (steady) voltage on the output. Is this the correct way to filter the DC? What am I doing wrong? P.S. Capsule=Microphone <Q> You would need a resistor from the output to Ground to remove the offset. <S> Without a resistor to ground, leakage in the capacitor will hold the output voltage near the voltage at the bottom of R1. <A> Depends on the value of the capacitor and resistor, and the load applied to the other side of the capacitor. <S> The "charge" and "discharge" of a capacitor promotes a current into and out of the capacitor plates. <S> If there is enough current to load the capacitor plates, at a time this current will reduce to virtually zero (never zero), and you will measure almost zero volts. <S> Your voltmeter has a resistance, normally in MegOhms range. <S> When measuring voltage, the current that flows through the voltmeter is the responsible to develops a voltage that the voltmeter will shows. <S> If the capacitor is fully loaded, no more current will flow through it, so you will not measure any voltage at its output. <S> But if the capacitor is not fully charged yet, what must be the case in your example, you will measure voltage. <S> If you multiply the value of the resistor by the capacitor (ohms and farads), you will have a value in seconds that means how long the capacitor will take to charge 66% of its total charge. <S> If you don't have a load at the right side of the capacitor, the circuit is incomplete, so the capacitor never charge. <S> However, if you add a small resistive load from that capacitor lid to ground, lets say, 10k to 100k ohms, the capacitor will charge fast and you will measure no more voltage. <A> Nearly all capacitors will have some small amount of DC current leakage. <S> So with a high impedance voltmeter at the output the small leakage current will register as a voltage. <S> Large value polarized capacitors can have fairly large amounts of leakage, especially electrolytic types. <S> The manufacturer's specifications will give the value of DC current leakage at various operating conditions. <S> If you were to use a high quality low value ceramic, disc, or mica capacitor you would most likely measure 0 VDC with most standard voltmeters.
| As already mentioned, adding a pull down resistor after the capacitor can reduce the measured voltage to near zero (as the voltage divides passing through the high DC resistance of the capacitor).
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How does a non contact voltage tester work? How does a non contact voltage tester pen detect voltages and/or currents? Are they limited to voltages of a certain range or type (AC or DC)? Here's some experiments I have conducted that lead to this question: Using a cheap pen I bought for a couple bucks, I can detect the usual 120V AC in an American outlet, but I also have been able to detect voltage in a USB cable connecting a switching power supply to a smartphone (this setup is usually referred to as charging one's phone). Here, of course, the voltage in question is DC with a negligible ripple. I also noticed that while the detector can detect the 5V of a phone charger cable, it cannot detect the voltage in a USB keyboard cable. The only difference between these two scenarios is the current level and perhaps some minor signaling differences. A final question: In what voltage/current scenario would you have to use a current clamp sensor and not a non contact voltage detector for merely detecting the presence of power non-intrusively? <Q> they work by capacitive sensing of the AC voltage on the live conductor. <S> they only work with AC. <S> It's obviously responding to some varying signal on the USB cable, (possibly "ground bounce" due to the varying current draw of the switching regulator in the phone. <S> (USB cables are usually well shielded so that's about all that's likely to be there. <S> As the pickup is capacitive, the higher the frequency <S> the more sensitive it is (up to the frequency limit if the amplifier it uses), so 100V at 60Hz produces the same output signal amplitiude as 10mV at 600Khz <S> if the charger has a captive cable (doesn't use a detachable USB cable)then <S> it's probably not shielded and the signal may be from the powersupply itself. <S> Current clamps are for measuring how much electric current is flowing through a single conductor (eg to get an an indication that the circuit is not only live but is also in-use), as you need to separate out the individual conductors of the cable, you typically use them at a junction box or some other place where the cable is opened up. <S> using a current clamp on a bundled cable will usually give you a zero reading (unless there's some sort of electrical fault) <A> You need to realize that there are two different things: voltage and current. <S> Voltage is present in an outlet even if nothing is connected to it (therefore you can have a voltage and no current). <S> Alternatively the neutral wire is at 0V earth potential, but you can have a big current flowing through it. <S> A voltage creates an electric field, while a current creates a magnetic field. <S> Therefore depending on its construction you can have devices sensitive to static electric fields (therefore a DC detector), and devices that are only sensitive to alternate electric fields (an AC detector). <S> The same applies for magnetic field detectors, such as the ones in a current clamp sensor. <S> Actually, if you were to cut a wire, for example, you should first measure both: that it has no harmful voltage in respect to earth, and that there is no current flowing through it. <A> How does a non contact voltage tester pen detect voltages and/or currents? <S> Are they limited to voltages of a certain range or type (AC or DC)? <S> They rely on capacitive coupling which limits them to AC and they are generally designed for mains voltages. <S> Your body being a large object has some capacitance to ground. <S> This makes a (very weak) circuit from the item with AC voltage on it, through the tester, through your body and through the capacitance to ground. <S> I also noticed that while the detector can detect the 5V of a phone charger cable, it cannot detect the voltage in a USB keyboard cable. <S> The only difference between these two scenarios is the current level and perhaps some minor signaling differences. <S> Wrong! <S> In order to supppress electromagnetic interference from switched mode power supplies capacitance must be placed between the input and output sides. <S> In a class 1 <S> (Earthed) power supply the earth is used as a barrier between input and output either by connecting the output to mains earth or by splitting the capacitance into two parts in series, a part between output and mains earth and a part between mains earth and mains live/neutral. <S> In a class 2 (non-earthed) power supply the mains earth is not available and so can't be used as a barrier. <S> The result is that the output is often at a significant voltage relative to earth (half the mains voltage is common). <S> This should not be a safety hazard if the power supply is properly designed as the capacitors have a high impedance (low capacitance) and hence the "touch current" is low despite the high open circuit voltage. <S> The capacitors will be special safety types so that short circuit failure of the capacitors is extremely unlikely. <S> As a general rule PC power supplies are class 1 while smartphone power bricks are class 2. <S> This is why your tester lit up on the cable for charging your phone but not on the cable for your keyboard. <S> A final question: In what voltage/current scenario would you have to use a current clamp sensor and not a non contact voltage detector for merely detecting the presence of power non-intrusively? <S> There is no surefire way to detect electricty non-intrusively. <S> Especially when dealing with multicore cables rather than individual wires. <A> In regards to the final question for reasons when to use current clamp sensors: Current clamps allow to measure the current in a conductor minimally invasive. <S> You don't use them to check whether a cable is live, but rather to see the current consumption of a load for example. <S> If a clamp uses the transformer principle then you can only measure AC. <S> A clamp based on a hall effect allows to measure AC and DC. <S> There are clamps, with a usual maximum frequency around 500 kHz, which are designed to connect to an oscilloscope. <S> This allows to analyze the behavior of a load or source in detail. <S> Be aware that a non-contact voltage detector is an unreliable way of looking for live wires. <S> It does not measure the actual voltage nor will it allow you to distinguish different phases.
| Therefore if you only need to measure the presence of a harmful voltage you would use a voltage detector, and not a current clamp (that will indicate 0 A if no equipment is drawing current from the mains at this moment).
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How to check if a DC power supply is a floating source? Imagine a laptop power supply with 2 prongs and I'm using this as a power supply for a circuit, and the output of this circuit is an input signal Vin for a data acquisition hardware. How can I verify if this signal is floating? My idea is to apply continuity test between gnd of the Vin and neutral of AC in of the laptop power supply. Could it be said that then the power supply is a floating supply if continuity test doesn't beep? <Q> There are a couple of ways that you can verify if the power supply is floating. <S> 1) <S> This is a device that provides a current-limited high-voltage (either AC or DC) across the device under test. <S> You simply short both of the output DC terminals together and connect to the Hi Pot Tester Ground lead, then short both input AC terminals together and connect to the Tester HOT lead. <S> Apply voltage and ensure that the Tester does not show excessive current. <S> You should follow the test procedure that your country's Electrical Authority provides for the proper usage of the Hi Pot Tester. <S> In Canada, our test voltage is (Two times AC Line Voltage) + 1000V for one minute or about 1250 Vac for one minute. <S> 2) Simple method but not as thorough as (1) above <S> : Connect a pair of identical resistors in series and connect the free leads of the resistors to the power supply output terminals. <S> Connect the middle connection of the series resistors to earth ground. <S> Measure the voltage across each of the series resistors. <S> The voltage should be identical across each resistor. <S> Pick the resistor value such that you get about 100mW dissipation in each of the resistors. <A> By UL standard 840 or IEC 61010-1 (I can't recall which one exactly) (if it is sold in America) it must be isolated from line voltage. <S> The European union has basically the same rules for anything CE marked. <A> I would check on the case if it is Class II (with Roman numerals), as this refers to power supplies with either a double or reinforced insulation barrier between the input and the output. <S> Class II supplies do not rely on an earth connection to protect against shock hazard. <S> Many laptop power supplies are Class II. <S> The symbol is a double square (seen below). <S> You should be able to do your continuity test to make double sure. <S> = <S> floating powersupply
| The easiest method involves using a piece of commercial test equipment called a "Hi Pot Tester".
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8x1 header in OrCAD I'm trying to create a LCD Arduino Shield for the Arduino Uno using OrCAD. I want to add 8 pin through hole to solder headers into them. However, the 8 pin header in OrCAD library under Connector library is 4x2 header in the PCB editor. But I want a 8x1 header for Arduino and two 8x1 header for my 16pin 16x2 LCD as well. Or is there a way I can modify the footprint of the jumper header in Connector library? <Q> Make your own footprint. <S> If you do any amount of PC design, you will have to make your own schematic symbols and PCB footprints - you can't expect the CAD developers to supply symbols and footprints for every conceivable part. <S> (and you should save any parts you create in your own library, so you won't loose them when you upgrade the CAD package.) <A> Which version do you use? <S> i use layout plus 10.5 if it is 10.5 you can use below 8x1 pcb footprint: BLKCON.100 <S> /VH/TM1SQ/W.100/8; 16x1 footprint: BLKCON.100/VH/TM1SQ/W.100/16; <S> Also you can create how you want it <A> @PeterBennett is correct. <S> Of course, making a part in OrCAD <S> kinda sucks :) <S> I use Library Expert whenever I can. <S> The "Lite" version is free, and creates IPC-7351-compliant footprints from the part's dimensions. <A> As has been said, create it yourself. <S> This is an easy one compared to some of the complex footprints you'll see, such as USB connectors and the like, but will allow you to begin to explore the package creator. <S> Try the "package symbol (wizard)" tool under File > New. <S> This makes it super simple, you basically just have to specify the pitch (likely 0.100" = 100 mil), package length and width, and ref des (J). <S> The length and width you should get from the datasheet, but are most likely approximately 0.800" (800 mil) and 0.100" (100 mil) respectively.
| You need to be able to make your own.
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Multiple Ground Referenced Sensors on Single Floating Supply Some background: I'm designing an industrial control board with isolated analog and digital IO. The hope is for it to be pretty foolproof, such that you could connect any 4-20 mA or 0-5V sensor with little configuration (i.e. a DIP to switch between current and voltage measurements) and have it work without having to worry too much about grounding specifics. For example, I'd want to a technician to be able to connect both ground-referenced and non ground-referenced sensors without having to think about how the grounding is configured. Some of these sensors may be powered by a separate mains supply, while others may be powered off the board's (isolated) 0-5V supply. To me, it seems like the way to do this is to have a floating ADC that digitizes the analog signals then sends them to the earth-referenced board through opto-isolators, rather than deal with analog isolation. My question is, to safely do this, will each analog input require a separate supply? Thinking about it, I think if I could guarantee that all sensor signals would be floating and not referenced to their case ground it would be fine, but I can't. As in the schematic below (sensors are modeled as voltage sources), if two sensors on the same floating supply have their analog ground connected to the case ground, you create a ground loop, conceivably the two sensors at different locations could have different ground potentials/ An example of this would be two pressure transducers with case grounds in different water tanks. simulate this circuit – Schematic created using CircuitLab If so, what's a good cheap way to get several floating supplies? Or is there a better way to do what I'm thinking? Thanks! <Q> I've had the same constraints and had to find the solution that others are talking about. <S> I don't know what capture speed you want or <S> what sensors you have <S> but there is almost an integrated solution using TI's chipsets like this: <S> - This is a generalistic circuit suitable for multiple options but you can get more bespoke variants of the same theme suitable for the individual sensor types. <S> The solution that I ended up using was with ADI because it was self-powered thru the isolation device: - <S> (source: analog.com ) <S> The <S> ADuM5401 has a small DC-DC converter internal to the device that can provide enough power for the front end AD7793 chip. <S> You might also consider the good-old isolating amplifier: - <S> For less problematic ground potentials you can even use the LTC1043 isolating capacitor thus: - <A> If you want each of your inputs to allow for separate ground references, then yes, you will need either a separate isolation amplifier on each, or a separate isolated ADC on each. <S> For analog isolation, isolation amplifiers such as AMC1100 are an option, but will require a separate isolated power supply for each channel. <S> Isolation amplifiers that include an isolated power supply such as AD202 exist, but tend to be both bulky and very expensive. <S> There are also digital isolators that have their own isolated power supplies built in, such as ADUM6403 , which you could use in conjunction with a regular ADC with an SPI interface. <S> You can even craft your own linear isolator by PWMing your input signal and filtering the output, as documented in this SILabs appnote . <S> Finally, if you can put some reasonable constraints on the difference in ground levels, you might consider a regular differential amplifier with a wide common-mode range. <A> Yes, each input requires its own, isolated supply. <S> Transformer's will get you true isolation. <S> Since the transformer is the most expensive component and your question says you want to do it cheaply, and if you don't have too many channels, you could look into a transformer with multiple, identical secondary windings. <S> One winding is needed for each channel (to get the isolation). <S> If you used 4-20MADC transmitters, you wouldn't need much, or possibly any, power supply regulation as long as the power supplies provided clean DC within the transmitters range, taking into account voltage drop across whatever else the current loop is going through (recorders, meters, computer points, etc.) <A> Usually the PLC ADC is isolated from bacpklane bus (not the cheaper ones) and sensors 4-20mA <S> they don't have a minus connected to the casing or ground. <S> What you need is connect the ADC's minus to the ground in single point only, then connect sensors. <S> If they are incapsulated in metal case with shielded cable, then proabably the shield is connected to case and both to the ground trough construction, in this case you don't connect the shield in the cabinet. <S> I don't know what type of sensors you use, but it would be to expensive to have multiple isolated ADCs and sources, not the case in normal PLC engineering.
| For digital isolation, you can get ADCs with integrated isolation barriers, such as AMC1203 , but again, they typically require separate isolated power supplies.
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Sallen-Key filter with polarized capacitors Maybe someone can help with this. I need an active filter with a very low cutoff frequency (~2 Hz). I think to make it with aluminium electrolitic capacitors. Is there any draw back? The signal to be filtered varies from 0 V to 3 V. Here is the circuit: Is there any problem with this conection? The simulation runs OK, but I want to know if there is any problem in using electrolytic capacitors as shown in the diagram. <Q> It's crazy - look at the resistor values - 2k2 - why not use 22k and reduce all your capacitor values by 10 times. <S> Why not use 220k and reduce your cap values by 100 times. <S> Worried if it will work with 220k? <S> You might need an op-amp with lower input bias/offset currents. <S> The LM358 offset current is about 5nA and this creates an error of 1.1 mV thru a 220kohm resistor BUT given the offset voltage error for the LM358's inputs <S> is about 2mV <S> it's not going to make thing much worse than they are. <S> Personally I'd go for a better op-amp and get rid of these errors by about a 10:1 factor (off the top of mu head) <S> but it might come down to cost. <S> Get rid of the electrolytics - you don't need them!! <A> In order for electrolytic capacitors to perform correctly, they need to be forward-biased. <S> You cannot reverse-bias them and have them perform properly. <S> Even though your input signal has a DC component, your feedback capacitor is still going to spend some of its time with its polarity reversed as the op-amp attempts to follow the filtered input signal, which means the feedback capacitor won't perform as a capacitor. <S> Instead, it will function as a capacitor half of the time, and an ever-hotter short-circuit the other half of the time. <S> It won't blow up as quickly as if you attached it in reverse to, say, a 12V power supply, but it won't last very long, and your circuit won't perform as expected. <S> What will probably happen is that your voltage regulator feeding the op-amp will get very hot and burn out before the capacitor boils. <S> So, you cannot use an electrolytic for the feedback capacitor. <S> But, if you modify this circuit and remove the DC component of the input signal, and feed the op-amp with two supply rails, one positive and one negative, then you can't use an electrolytic capacitor in either position. <S> It's probably safest just to go with non-polar capacitors, rather than mix-and-match. <A> Electrolytic capacitors used in an AC filter circuits are bad for all the reasons listed so far. <S> (Use electrolytic parts for filtering unwanted AC from a DC line.) <S> Even using non-polarized foil capacitors of the values shown would be awkward as they would be physically very large. <S> The suggestion of increasing the resistor values and decreasing the capacitor values is the best way to go. <S> To achieve an accurate filter response consider using mylar type capacitors with values at about 1/50 to 1/100 the original values (and recalculate the necessary resistor values). <S> Mylar types will more often include the actual tolerance so that you can better determine the potential error of the circuit's frequency response. <S> Film capacitors at these lower values would also be reasonable with perhaps a somewhat wider tolerance range.
| An electrolytic may work for your capacitor connecting the positive differential input to ground, assuming your input signal is centered at some DC component.
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power absorbed in an inductor in an RL circuit during step response strong text In the following circuit , what I found was that , the current through the circuit reaches a steady state after some time . but , then when I plotted the power through the inductor VS time in pspice , I found that the power increases , and after some time , starts to decrease . My question is , if the current through the inductor reaches a steady value , then how come power through the inductor first increases , and then decreases ? <Q> The inductor stores energy when a current passes through it. <S> I don't want to carry around the fact that both current and energy are functions of time, so let the instantaneous energy stored be \$\small E(t)=E\$, when the instantaneous current is \$\small i(t)=i\$. <S> Then we may write \$\small E=\frac{1}{2}Li^2\$. <S> Now, the current in your circuit is exponentially increasing, and given by: $$\small i= <S> I(1-e^{-t/\tau})$$ <S> where \$\small I\$ is the maximum current, and \$\small\tau\$ is the time constant. <S> Hence, the energy at time, \$\small t\$, may be written:$$\small E=\frac{1}{2}Li^2=\frac{1}{2}LI^2(1-e^{-t/\tau})^2=\frac{1}{2}LI^2(1-2e^{-t/\tau}+e^{-2t/\tau})$$Power, \$\small P=P(t)\$, is the derivative of energy, thus: <S> $$\small <S> P=\frac{dE}{dt}= <S> \frac{LI^2}{\tau}(e^{-t/\tau}-e^{-2t/\tau})$$ <S> This is the equation of the graph of power vs time shown in your question. <S> Let's check a few important features of the graph against those predicted by the equation: The equation gives \$\small P(0)=0\$, and \$\small <S> P(\infty)=0\$ <S> Maximum power occurs when: <S> \$\small \frac{dP}{dt}=0\$. <S> Performing this differentiation and ignoring the constant term gives: <S> \$\small (-e^{-t/\tau}+2e^{-2t/\tau})=0\$, <S> hence \$\small \frac{t}{\tau <S> } = <S> 0.693\$. <S> From the circuit, \$\small \tau=\frac{L}{R}=1ms\$, so the peak power is at \$\small t=0.693ms\$ <S> Given the maximum current, \$\small I=10mA\$ and inductance, \$\small L=1H\$, the value of peak power given by the equation at \$\small \frac{t}{\tau}=0.693\$ is \$\small P=25mW\$ <S> When the current has reached its maximum value, the energy stored in the inductance will also have reached its maximum value (\$\small \frac{1}{2}LI^2\$), and the power transferred to the inductance will, henceforth, be zero. <S> Power and energy transfer to the inductor only takes place when the current through the inductor is changing (i.e. work needs to be done, against the back-emf, to establish a current through the inductor) <A> Power has been taken from the circuit to store energy in the inductor. <S> There is no getting away from the fact that the instantaneous voltage x instantaneous current = the instantaneous power (at that instant). <S> This is an absolute irrefutable truth. <S> So, the inductor stores (or maybe accumulates if you prefer that word) an amount of energy and, this energy is proportional to the current-squared at any instant in time. <S> The power of the signals across and entering the inductor (i.e. volts and amps) totally cause this energy to be stored. <S> If you differentiated the value of energy stored with respect to time you would get the power curve shown in the question. <S> For your RL circuit, after a while, the current eventually (and exponentially) rises to a constant value and, at this point, the energy stored becomes a fixed numerical value. <S> Therefore, no more power is taken by the inductor - the voltage across the inductor is now at zero and the current is at some fixed value hence, power = 0. <S> The power curve began at zero watts with 10V and zero current and ended at at zero watts with zero volts and some current. <S> It's all over now! <S> Power will still be taken by the resistor but this is not stored; it is emitted as heat (or maybe light if it gets really warm). <A> For a simple visual answer just plot both the current and the voltage for the inductor. <S> At some interim point there will be a maximum power flow. <S> (If you can plot the product of voltage and current values you can also see the power curve, P=VxI). <S> Also note that in the real world an inductor has some DC resistance (in the wire) that would truly dissipates some of the input power. <S> If you assume a perfect inductor then all of the input power is recovered when the source is disconnected.
| You will see the current increasing (and leveling off) as the voltage decreases.
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Starting motor for 1 minute after every 2 hours My requirement for hydroponic fodder trays, I wants to spray the water in every 2 hours for 1 minute. Can anybody tell me how I can achieve this in lowest budget as wanted to distribute this technic on large scale to farmers. With the information I have, I need to purchase Option 1. 555 timer IC and Relay? And please suggest how? Option 2. Some sort of ready made controller like adriuno? And please suggest how?Option 3. ???? ? And please suggest how? <Q> Here's a complete circuit using a dual 556 timer chip to obtain a long duration delay with a secondary short delay. <S> (Note the circuit drawing shows 565, but its a 556). <S> To get the short delay time correct refer to the 555 or 556 data sheet (link at the bottom) and adjust the RC values as needed. <S> http://www.electroschematics.com/9082/long-duration-timer/ <A> You haven't defined large scale. <S> Let's assume you produce 1 million devices. <S> You can get 8051 processors featured for motor control applications and 32kB flash for 25cents, so I assume the MCU you need is sub 0.10 USD. <S> NE556 is probably around 0.05USD. <S> I would go for MCU, as 0.05 USD is irrelevant against the price of the relay, and MCU is a much more flexible solution. <A> If you need to use this system for agricultural applications you can go for the 89C51 micro controller you can use one of the port of microcontroller to control the motor.
| By inserting delay in the microcontroller assembly language program you can controll the motor as per your any timing requirements
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How to avoid Attiny84 being powered on by a pulled up data line? I've developed a simple sensor board based on the Attiny84a microcontroller which is able to communicate over a serial connection. I'm using an arduino bootloader and the SoftwareSerial library on pins 5 (RX) and 3 (TX). I would like to be able to keep the serial line attached while power cycling the sensor. But it seems that the Attiny84 is powered through the RX line of the Serial connection, which is pulled high by the other communication partner. It seems the RX line is pulled to 3.4 volts when nothing is attached and I can measure 1.4V across the Attiny84's Vcc and GND pins when the serial line is attached. Apparently that is sufficient to turn it on, since my debug LED starts flashing dimly. Is there a simple way to keep the Attiny84 off even when the serial cable is attached? <Q> This happens because there are internal protection diodes on the GPIO pins so any voltage that comes in is shunted to the internal power rail... <S> Even though these are little diodes, this chip uses so little power that it is often enough to run it. <S> This is the source of one of the top-10 hardest bugs <S> I've ever hard to find! <S> Method #1 <S> - Pull ~RESET <S> Low As long as the voltage on the ~RESET pin is less than 0.2*Vcc, then the chip will stay in a reset condition and not running. <S> Note that you need the pull-down because there is an internal pull-up on the ~RESET pin. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This method wastes a little current, but will keep the chip in a reset state when it is not actively powered. <S> Keep in mind that there will be a voltage drop across the diode. <S> Method <S> #2 - Isolate Input Pins simulate this circuit <S> Here we use the power supply to enable the input pin. <S> If there is no voltage coming from the power supply, no significant current to flow into the input pin. <A> If your comms speed is low-enough and your power-budget will allow it, you can often avoid this simply by adding a diode between the RX pin and the outside world (anode to the RX pin, cathode to the outside world), and then having a pull-up resistor on the RX pin (your micro might be able to do this internally). <S> Then the remote device can only pull the RX pin down, but the pull-up resistor will pull it back up. <S> You will probably want to use a schottky diode (e.g. BAT54), and you do need to think about the input-low (Vil) levels you'll get at the micro pin - check that they're still OK. <S> If you can't do it like this, you could use a buffer in that line - selecting a buffer which didn't have this protection-diode-to-VCC behaviour (e.g. 74LVC or one of the others like it - Texas Instruments <S> call this 'Ioff' support - see this for example) <A> Depending on the margins you have available, this might work for you. <S> It allows the Rx signal to draw the ATTiny Rx input down to 0.6-0.7V, but does not allow current to flow back into the processor. <S> The node "ATTiny VCC" is switched on and off with the processor. <S> If this is 5V (rather than 3.3V) you need to size the resistor so that it doesn't cause backflow problems in the Rx circuit. <S> If 0.6V is too high to trigger a good LOW in the ATTiny, you might use a low-drop diode like a Schottky or a germanium. <S> simulate this circuit – <S> Schematic created using CircuitLab
| You can connect the ~RESET pin to the power supply pin and use a pull-down resistor to keep the ~RESET pin below the threshold when there is no active power supply.
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Is a ground/common needed for proper CAN bus communication? Since CAN Bus is a differential signalling system, what is the significance of GND/COM signal? Can I get by without it? I've got a very simple 2-node system that has some weird happenings going on. (see here for that issue, if you're interested) The two nodes are my device and a USB-CAN monitor; both have 120ohm termination resistors, and the wire run between them is <1m. I just remembered that in my earlier tests I only had the CAN_H and CAN_L signals connected between the nodes; I did not have anything connecting the CAN_COM of my device to the ground pin on the USB-CAN monitor. A (the?) point of differential signaling is to be much less (not?) effected by common mode noise. So by not having the ground connected, I would guess we lose the absolute reference for CAN_H and CAN_L... but does that make a difference? <Q> A perfect, theoretical, differential system can run with no ground, because the input can always do (A-B) without worrying about the absolute values of A or B. <S> Unfortunately, CAN is not such a system, and the receiver will have limits to its 'common mode' capability, beyond which it be damaged. <S> So you have to have a ground to keep the common mode voltages within limits. <S> 'Common mode' just refers to stuff (voltage, in this case) which is common to both wires. <S> There are other ways of building receivers which are far freer of this sort of constraint - for example an opto isolator can do the (A-B) detection with maybe 1000s of volts of common mode offset. <S> Transformers are often used too (e.g. in twisted-pair Ethernet, which has no ground). <A> Yes, a common ground is need. <S> A can transceiver has a maximum common mode voltage. <S> If the common mode voltage of the differential CAN signals exceeds the maximum, then the transceiver will not be able to recognize the bits. <S> In practice, this means that the grounds of CAN nodes have to be connected (although a small voltage differences between grounds can be tolerated). <S> Special case. <S> Some CAN devices are galvanically isolated. <S> Then there is not common ground throughout. <A> It depends what system you are building. <S> If you are building in your house or in your car it is better to use common ground. <S> Actually on some cases you can blow everything by using common ground - when the nodes are powered from different power sources with different ground references. <S> I am always building large industrial CAN networks without common ground and it is working with no issues.
| But for normal CANbus, you need a ground. Because it is differential pair type communication the common ground is not required as what is matter is the potential voltage difference between H and L.
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Watts used by chargers eg laptop, wall warts etc This is my first question on here, its quite easy and I am sure I have half the answer, maybe. I was thinking about how much power was being used by these wall chargers etc, you read the label it may say 1A 240V that equates to 240 Watts? Can it be that high or is there another equation to use? Don't be shy to show me math. Thanks <Q> More likely, you're reading "Input 240V; Output 5V 1A" (or similar). <S> This means that it's outputting 5 watts and (unless it's a linear regulator, which is unlikely) it's drawing just a little bit more than 5W. <S> Even then, the 1A is the maximum that can be supplied by it. <S> If the load is only drawing 1/10 of a watt, the power being supplied is 1/2 watt and the input draw is a little bit more. <A> Quite often there is a huge discrepancy between the label rating on a power supply and the power it actually uses, on the input side, while its output ratings are likely to be reasonably accurate. <S> Picking one power supply at random <S> I see: <S> INPUT : 100-240V(AC) 50-60HZ <S> 1.5AOUTPUT : 19V(DC) <S> 3.95A <S> It's quite small and runs warm (but not hot) possibly dissipating 5-10W as heat. <S> So it supplies just under 20*4 <S> = 80W and probably never consumes more than 90W sustained. <S> That would mean 0.9A at 100V or 0.375A at 240V. <S> The input ratings are deliberately conservative, because their purpose is not to estimate your electricity consumption, but to ensure safety by encouraging you to budget your current to choose the correct fuse, breaker, and not overload the mains circuit. <S> The fuse rating has to cope with the inrush current, which may exceed 1.5A but for such a short time that the fuse will not blow. <S> This explains why one current rating is given despite the huge range of input voltages. <S> If you have fused cables, a 1A fuse will probably not last, but a 3A fuse (the next standard value in the UK) will. <S> If you add all the rated loads on an extension cable on a 13A plug, they should come to less than 13A. <S> If you add all the loads on 13A plugs on a 30A circuit, they should come to less than 30A. Given the expected current for a circuit, you can choose the appropriate wire gauge, and so on. <A> That is usually the maximum input current under the maximum load. <S> $$Input <S> Power = Output Power + Efficiency <S> Loss$$ <S> The 1A 240V should encompass both the power passed through to the load and the losses involved in converting the voltage. <A> Here are two additional factors to consider: For AC power, Power = <S> Voltage X Current X Power Factor. <S> The concept of power factor was developed to account for the fact that the phase relationship between the voltage waveform and the current waveform effects power. <S> Power factor is a number between zero and one. <S> Power factor is also used to account for the effect of waveform distortion. <S> All electronic power conversion devices tend to have a distorted input current waveform. <S> Devices designed to prevent that are more expensive. <S> I measured the power factor for a small power supply using a Kill A Watt (TM) meter and found it to be 0.48. <S> That means that the actual power used by the power supply is less that half of the power calculated as V X A. <S> The various electrical safety standards dictate how the information marked on products is determined. <S> It is likely that the output current of a power supply is the current that can be drawn without causing the voltage to decrease too much or without the device getting so warm that it's useful life <S> is shorter than expected. <S> The input current is likely the maximum current that the will be drawn if the output current is just below the point that the power supply fails or shuts itself off. <S> The power supply will likely put out at least a little more than the rated output current before failing or shutting off.
| If you read "Input 240V 1A" then yes, it's supplying 240 watts (maximum).
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How do I know if my microcontroller clock frequency is fast enough? I've been using ARM microcontrollers from NXP in commercial products with great success for the last 2 years. In most of my projects I don't use the PLL to increase the oscillator frequency because I never found it was needed (never felt that clock frequency was an issue), and because the PLL slightly increases power consumption. However I don't know how can I measure what clock frequency is enough to power my code. I'm stuck with this question in my head because as you may know the faster the clock you use, the more power you end up consuming. Since these days portable equipment with limited batteries are common this is an important topic. The main question here is how can I know if the current clock frequency is enough for my code? In this question I state "stopping or delays" because I assume that if my clock frequency isn't enough the first symptom that will happen is a slow response time from time consuming routines. I use in almost my projects a state machine system that has an OnIdle() function that is called when the system doesn't have any events to process, so it can put the MCU in a low power mode. I was thinking about measuring how much time the system stays in this function and record min and max watermarks of this measure to have real numbers about if the current system clock is or isn't enough. Does anyone have suggestions for this? <Q> A common technique to write an embedded real time application is by having the main loop to be fixed time long. <S> Than your main loop will look something like that: while (true){ doStuff(); waitSync();} waitSync <S> here is driven by some kind of timer interrupt. <S> By adding an indicator like GPIO pin output like this: while (true){ gpioOn(); doStuff(); gpioOff(); waitSync();} and placing a scope probe on that GPIO you will get a nice pulse train, whose duty cycle will indicate the percentage of the CPU time, when it is actually "doing stuff". <A> You're thinking about this backward. <S> The question isn't "how can I know if my code running with its current clock is enough? <S> " , it's "how to I design my code such <S> that I know if it's running correctly? <S> Microcontrollers are somewhat opaque. <S> They're a hard platform to debug on. <S> You need to develop the habit of splitting the code up into testable chunks, using techniques such as bit toggling to test each chunk, and then test the entire system. <S> Building in the tests should become part of your though process, and starts before you write any code <A> There are different ways to run out of time with microcontroller code. <S> You may run out of total bandwidth, in which case you've definitely got to change something (better algorithms, higher clock frequency, maybe a better architecture). <S> Before that happens, you may get more subtle effects such as excessive jitter in interrupt service routines or servicing of low-priority events such as the user interface. <S> It's usually desirable to have a fair bit of margin in total bandwidth- running out of speed (or memory) is not much fun unless you have a clear upgrade path (such as crank <S> the clock speed up or drop in a more expensive micro that is upward compatible). <S> You can profile code in simulation or by toggling a GPIO pin as Eugene suggests, however note that in micros such as the ARM <S> the GPIO is decoupled from the micro core by a relatively slow bus, so the profiling should be done on relatively chunky time slices. <S> You should also look at the specifications (and perhaps test) to see what the trade-offs are with clock speed and the typically numerous operating modes of the micro. <S> If you're just idling in wait loops and not doing any power control, part of the supply current will be proportional to frequency (to a first order). <S> However, if you're waking up, doing some stuff, then going to sleep, the effect of higher clock frequency may not be so significant <S> (could even be an improvement), since the duty cycle will be less and perhaps some peripherals can be put into a low-power mode for longer at the higher clock frequency.
| In case it is difficult or impossible to estimate the required frequency theoretically (based on calculations and the knowledge of the microarchitecture) or based on experience (similar code on similar machine), the empirical approach can be taken.
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Help with understanding op amp circuit using diodes at the output I'm trying to understand this circuit and having trouble figuring out how the diodes will affect it: I'm trying analyze this and understand how \$V_{\text{out}+}\$ and \$V_{\text{out}-}\$ will be affected by different values of \$V_i\$ but I am unsure. I believe \$V_{\text{out}+}\$ will see a higher voltage with lower \$V_i\$ values and once \$V_{\text{out}+}\$ is past 0.6V the current will go through both of the diodes and \$V_{\text{out}-}\$ will begin to see an increase. I'm unsure of what to think about the op amp in the center though. <Q> Vo will be negative. <S> So, the current flows through R3 and the upper path (R1, D1), while current through the lower path is blocked by D2. <S> So, for a positive input, Vo+ will be negative and Vo- will be zero. <S> The exact value on Vo+ is determined by the ratio of R3 and R1, as the output terminal <S> Vo will be a value which cancels out the voltage at the negative input terminal. <S> To examine this a bit more, you can do a simulation, even here. <S> I've drawn your circuit using the circuit designer of EE. <S> Click on simulate this circuit and do a DC sweep for Vi. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The following picture shows the result, blue is Vo+, orange <S> Vo-. <S> I've also added Vo (brown) just to see the effect of the diodes. <S> (Click the image to enlarge) <A> Andy's answer is spot-on if this circuit operates in isolation. <S> But if there happens to be a positive bias current source connected to Vo+, and a negative current source connected to Vo-, then it is something different. <S> (For simplicity, these bias current sources can be resistors connected to +V and -V respectively.) <S> Then the outputs form three parallel lines, about 0.6V apart, with the central one (Vo = -Vi). <S> Vo+ and Vo- may be used to drive the bases of two complementary emitter followers, providing a crude power amplifier. <S> Their emitters are connected together, and each base-emitter voltage is cancelled by the voltage across each diode. <S> Therefore the final output (at the emitters) is approximately Vo <S> = -Vi. <S> But only approximately : errors here lead to crossover distortion. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> In its current form, it's not very efficient : if you think about what happens when VO+ = <S> V+/2 you'll see that it runs out of bias current and starts clipping (with the current resistor values), but this illustrates the basic principle, and you'll often see something like this in an audio power amplifier. <S> Further circuitry is usually added to fix the defects in this basic configuration. <A> Compare this to the standard "ideal diode" op amp design. <S> If Vi is higher than A1+ (GND), the feedback loop will try to null the input voltage, driving the V+out line through the diode to "pull" A1- down. <S> If Vi is lower, the V-out line wil similarily be pulled up. <S> This circuit then behaves as a standard inverting amplifier, but with any negative output being delivered to the V+out line (for a positive input), and any positive to the V-out(for a negative input)
| The OpAmp is used in an inverting configuration, i.e. when you apply a positive Vi, the output
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Rpms required to generate electricity by means of cycling I'm making a bicycle generator, and am trying to search for the right motor. I need to know at how many rpms a motor would need to be rated to generate a significant amount of electricity (like, to charge a laptop). Also, is there any type of motor from an old device that would work well for this (like an electric weed wacker motor).Edit: I'm planning on using a 2750RPM scooter motor (24V, 250W) Thanks <Q> This got a bit long for a comment: <S> All motors have a different base speed for the same voltage because of their geometry. <S> The formal specification for this is Kv, or RPM/volt. <S> Look for something that has a permanent magnet in it, like Brush[ed/less] DC, and then look for a Kv rating that is roughly close to what you can achieve. <S> (Note: it may be difficult to find in some cases and must be calculated from other specs) <S> The permanent magnet ensures the motor does not require external excitation to generate electricity. <S> Two types of motors that can't do this are Universal and Induction, both of which are very common in AC-powered equipment. <S> A Universal motor may also appear in DC-powered equipment because it works there too and may have some advantages over a permanent magnet in its intended application. <A> The problem you will face is that there are few common motors that will generate high enough voltages for your application while spinning only at 100 RPM or so. <S> For instance, your 2750RPM scooter motor (24V, 250W) will be outputting less than 1 volt at the 100 RPM you are likely to pedal. <S> So you will either need a gearing system or a very small RPM/volt constant. <S> You could also use a 3 phase brushless dc motor and run the leads through a three phase rectifier to get DC. <S> This will open up a lot of motors that are used for hobby boats/cars/helicopters. <S> If you are really gung ho, you could even wind your own to get a voltage constant in your wheelhouse. <S> Probably 300 Watts for around a minute before you give/throw up. <A> It will work as a generator without needing a battery to initiate power generation. <S> The output voltage will be proportional to RPM. <S> At the motor's rated speed for motor operation it will generate something close to its rated voltage. <S> If you turn it at half of that speed you will get about half of rated voltage, but you can load it with its rated current and get about half the rated power. <A> You could use a rear wheel hub motor. <S> Pick a non-geared one, without freewheel (or jam the freewheel). <S> These usually reach peak efficiency at 20 km/h. <S> Just add a bicycle and a solid stand to keep the rear tyre away from the ground.
| A permanent-magnet, brushed DC motor would be best. Unless you are in really good shape, you can probably only get around 100 Watts average on a good bike with gears.
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Knife Blade On/Off Switch for an A/C circuit I'd like to build a fun audio amp. Its a blue tooth module, so there are no controls. I just want a simple on/off switch. This amp is to be used in a workshop, and because I live in Las Vegas (where it gets quite hot in the summer) I don't want to leave the power supply engaged all the time. I'm not all that worried about speaker pop on startup. I'd like to shut off power upstream of the switching power supply. I was considering using this switch, just because it looks cool. I intend to use it in this circuit What I'm not clear about is safety using this switch. I'm guessing the power supply output is approx 2 or 3 amps at 14v. I don't know the input power ratings. I'm thinking worst case is that somebody might bridge the open knife blade to blade seat with a sweaty hand? The parts aren't all on hand yet, so I don't know the actual operating current in a single leg of that A/C circuit. The other option is to use a double pole, single throw switch of the same design, to shut off both legs of the 120v wiring, but that seems much more dangerous to me. Somebody could bridge the 120v terminals on the hot side of the switch. (In fact, why do they even make a double pole single throw switch like that? That just seems unsafe.) Note: the intent here is that this amp assembly hangs on the wall, about ten feet from the door of the shop. I am not worried about tossing a tool on the workbench and having that short out the switch. I am not worried about hitting the switch with my hands when I enter the dark shop (while trying to find the light switch) Your thoughts here? Is this crazy unsafe? <Q> Hmmm. <S> Sounds like this is safe. <S> But that doesn't mean you can't have fun with it... <S> Reference, lots more here. <S> Hmmm. <S> I'm not ready to give up on the knife switch yet <S> , Maybe I'll put it inside the box behind glass with a wooden actuator, or figure out a way to use a cool mechanism with an old skool traditional light switch. <S> I'm not exactly a rookie when it comes to mechanism design. <S> update: I also ran into this implementation of a knife switch. <A> Exposed 110V is a horrible idea and very unsafe. <S> Never do this. <S> 14V is going to be generally safe in the dry conditions: for example, UL 1310 "Safety Requirements for Class 2 Power Units" requires "A maximum potential of 42.4 Vac peak / 60 Vdc for exposed wires / terminals. <S> " I personally would not go above 24V DC. <S> A separate issue is temperature: an incompletely closed switch may heat up so much as to cause a burn. <S> This depends on the design of the switch and on maximum current consumption -- you mention "2-3A" which is not much; moreover, this is likely a peak current, not continuous. <S> Also, a lot of bluetooth receivers reset if the power is too low, and this causes their current consumption to be reduced. <S> Yet another thing is a damage to amplifier. <S> Someone walking on a dry carpet can have few thousands volts of static electricity and all of this will go to your amplifier's power input if they touch the metal part. <S> The biggest practical danger with that switch is the accidental short circuit. <S> You really want to make sure that the switch is the only exposed metal part of this circuit. <A> To safely use that switch to connect/disconnect AC to your amplifier I would consider having it control a relay. <S> The relay handles the high voltage AC supply thus isolating you from danger. <S> However, you'll still need a permanently connected low voltage supply to be able to activate the relay's coil when you wish to use the amplifier. <S> So scout around for some relays that have the required AC voltage and current ratings for the contact and try and find one that doesn't need much power to energize the coil. <S> Then look for maybe a wall-wart to supply the low power and safe voltage to activate the relay coil. <S> Here's a CEC (California Energy Commision) compliant power supply that says it uses less than 0.3 watts in standy: - <S> It's got a European plug arrangement <S> but I'm sure you can find one that is compatible with the US.
| Otherwise, if someone shorts the switch directly to the other transformer leg, the transformer will be overloaded and may get damaged; also the object that shorted the circuit may get very hot thus presenting the burn hazard. You will likely be fine if you use automotive grade parts -- they are usually designed to withstand this.
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How to start 220/380V 3 phases induction motor by star-delta method with 3 phases 380V power source? I would like to ask a question about how to start 220/380V 3 phases induction motor by star-delta method with 3phases 380V power source. Practically I think that it is impossible to be able to do it without step-down transformer by convert the power source from 380V to 220V. But, my boss tell me that I need to be able to makes it happen without using transformer. So, is there any suggestion to solve this problem? I just start working so any help will be appreciated. Thanks <Q> That's simple. <S> First the motor is wired in star (Y) and the voltage is distributed over two windings. <S> Each winding gets Vline/sqrt(3) if you calculate from 380V <S> you get 220V. <S> The current is also reduced by factor 1/sqrt(3), while the power is reduced by factor 1/sqrt(3)*sqrt(3) <S> = 1/3. <S> Once the motor reaches the rpms, it is switched to delta connection, now you get full voltage on windings, current and nominal power. <A> If the motor is designed to run in star a 380V 3-phase power supply, then it cannot be connected in delta on the 'same' supply . <S> This is similar to applying 380 volt to 220 v windings so clearly the motor would fail. <S> The solution is either to get a 3 phase step down transformer to get 220 3 phase voltage and you need to calculate the ratings if the KVA of the transformer based on the load. <S> OR get an inverter <S> , simply provide a 220V single phase (Line and neutral of the 380V supply) to it and get a 220V 3 phase. <S> Hope <S> the answer is useful and clear <A> There are better options than using a step-down transformer. <S> and then switches to full voltage. <S> You could also use an electronic starter that does the same thing. <S> Search "reduced-voltage motor starters."
| You can use the 380 V motor connection with an autotransformer starter that reduces the voltage during starting
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Why do PCBs have big interfaces? I am not sure if this is a good question but I am curious. Consider the following PCB: I have realized that although the lines leaving the IC in the middle start fairly close to each other, they are significantly separated through the edge of the PCB. Now I am guessing that the width of and spacing between the leaving lines are determined according to a standard but I am not sure why they are decided to be this wide. Could you explain the reasoning behind? <Q> What you're looking at is a PC expansion board using the ISA expansion bus. <S> This uses the then-standard .100 inch contact spacing which was widely available. <S> There are a number of reasons for this. <S> First, the connectors were cheap and fairly reliable. <S> Second, when the PC was introduced, DIP package ICs were the norm, rather than surface mount. <S> This means that the large size of the connectors was not a problem, since any reasonable circuit board was also large by today's standards. <S> EDIT <S> - It's worth keeping in mind that the ISA bus was introduced in the original IBM PC - in 1981. <S> You can read about it here . <A> Mechanical tolerances, mostly. <S> Fine pitch (close spacing) on edge pins is highly prone to shorting out or even being misaligned by an entire pin space due to tolerances required for insertion, and in some cases even thermal expansion could come into play. <S> Essentially, robust and reliable (also, the bus spec tends to be in use for a long time, and cause great gnashing of teeth when altered, while the pitch for the on-board parts is more free to alter as manufacturing tolerances improve over time.) <A> Supplementing WhatRoughBeast's answer, it doesn't look so unreasonable if you look at a contemporary card from when the connector was designed. <S> This is apparently a CGA graphics card. <S> Note the connector pitch is the same as that of the IC pins. <S> The connector takes up a small part of the huge card. <A> 0.100 was ubiquitous in through-hole, DIL-era electronics, and using a .100 edge connector also meant you could use a socket that connected to the mainboard with pins at a .100 spacing - that's what everyone had experience using, and it probably also was easiest to develop (and manufacture) PCBs with .100 parts using the prototyping and design tools available then. <S> Also, digital electronics back then tended to take a more naive approach to the problem of having to deal with relatively high frequency signals: one tried to just not get lines too close to each other to control capacitive coupling and lowpass filtering...
| In addition, there are practical problems with making the other half of the connector (spring-pin contacts) at a very small size. Finally, the contact fingers had to be fairly large in order to make the contacts physically strong enough (by having enough glue area) to hold up to repeated insertions.
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Why are components hard to insert into a breadboard? I have just bought a breadboard, but I have notinced that the components are hard to insert into the holes (and sometimes only one pin go in and the another one go out, like a LED). In videos I see people insert componenets into breadboard very easily. Is it normal because it is a new breadboard? Will it be fixed soon ? Any way to make inserting compomemts easier? I tried to use a jumber wire to help making holes bigger, but it didn't helped. <Q> I have had this happen to me <S> and it was essentially a faulty breadboard. <S> I have managed to loosen up my bad breadboards by forcing large tough component legs(like pin header) into and out of the stiff holes repeatedly, they eventually become more accepting of parts. <S> If it is really bad you may just need another breadboard. <A> problem solved by using a male jumber wire, I put then remove it in every hole (and reapeted it 3 times). <S> I will also try using a pin header as HighInBC mentiond <A> Depends if you are using the correct AWG(gauge/thickness) of wire. <S> Normal jumper cables are 22AWG, and they fit fine. <S> If you are using 22AWG wire(it should be written on the wire), then it's a problem with your board, and you should return it and get a new one. <S> You can buy one here for about $2.50 and there's 820 pins.
| The components should easily go in and out if the breadboard is good.
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Breadboard Resistor/Motor Not working I'm pretty new to 'breadboard-ing' and I'm beginning to work with and need help understanding some basic concepts that I can't grasp. Basically I've made a basic circuit with a push button for an LED light. There circuit is a basic circuit, powered by two double A batteries and there is a 220\$\Omega\$ resistor on the cathode side of the LED (from what I understand it doesn't matter which side it's on but that's the way I have it set): The LED lights up when I press the push button. Victory. The next set up is with a 3VDC Motor (350mA) and when I plug that into the batteries directly it spins. When I plug it into the circuit where the LED is (with the resistor) and press the push button, it doesn't spin: BUT, when I remove the resistor and put a straight wire into the circuit it works great when I press the push button: So obviously the resistor causes enough resistance to not spin the motor in the circuit BUT 220\$\Omega\$ seems like a very small resistor and would seem as though that should resist the circuit so much that the motor doesn't work at all. I don't have a resistor smaller than 220\$\Omega\$ to try out but obviously anything more than that won't work if this doesn't work. I'm just trying to understand the concept of why this wouldn't work because I'm trying to chain some LEDs and motors together and the resistance obviously causes an issue. I tried a capacitor after the resistor too but that didn't work (but maybe my concept is off so far; as I said I'm pretty new to this and am having trouble even knowing what to search for to get answers). Thanks for your help! <Q> You say that the motor draws 350 mA. <S> If you pass 350 mA through your 220 ohm resistor, Ohm's Law says there will be a 77 volt drop across the resistor - since you only have a 3 volt battery, this obviously won't work! <S> Ohm's Law indicates that 3 volts across 220 ohms will only result in a current of 13.6 mA. Common LEDs typically have a maximum current of 20 - 30 mA, so they cannot be connected in series with your 350 mA motor - <S> the LED would vanish in a puff of smoke! <S> If the motor is designed to operate on 3 volts, it should be connected directly to the 3 volt battery, with no series resistor. <A> 220 ohms is enough for the motor not to spin. <S> A lot of motors need a specific minimum input current(you <S> can always go higher). <S> You should look at your motor datasheet for this. <S> There is an important thing that you should know about motors: When connecting a motor to a circuit, you should always add a diode to either the ground or the positive voltage input of the motor. <S> This prevents the inductive kickback of the motor from potentially damaging other parts of the circuit. <S> According to what I think you want to do(which is connecting motors and led's to the same circuit) you should connect the motors and led's in parallel. <S> Go to this link for more info about series and parallel circuits: <S> https://www.swtc.edu/ag_power/electrical/lecture/parallel_circuits.htm <S> A parallel has multiple paths which current can flow through and because of this connecting a motor and led to the same circuit in a parallel circuit won't cause any resistance problems. <S> On the other hand, a series circuit only has one path for current to flow through, which causes the resistance problems that you talked about. <S> Just a note, when asking a question like this always mention how you are powering the circuit and at what voltage and current. <S> Also, try making a circuit diagram with the schematic maker. <A> This here is a simple diagram of what your circuit should look like: simulate this circuit – <S> Schematic created using CircuitLab ] <S> [1]][1] Basically, make sure the resistor, LED, and motor aren't on the same wire, but are in parallel. <S> I didn't put this in the diagram, but put a regular diode before the motor to prevent hazardous backflow
| Your question has basically been answered here: How to wire circuit DC Motor with LED If you connect the led and motor in series, then there is a resistance issue, but if you connect it in parallel, you should be fine.
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Why led is on when there is no input at terminal 1 &2 for 7432 IC? I have two questions under same heading:First why the led connected to terminal 3 and ground is always on when terminal 14 is connected to 5V and 7 to ground when there is no any input at 1 and 2? Second As we give inputs to 1 and 2 and have output at 3 why we need to connect 14 to power supply and 7 to ground? Thanks! <Q> 1st question: First of all, all unconnected pins are like little antennas, and maybe can have a floating voltage on it, possible from every radiation caught around it from wifi, tv, radio and so on. <S> So to assure the LED will go off you have to short pins 1 and 2 to the GROUND. <S> Only connecting they to zero voltage will guarantee the led to turn off. <S> 2nd question: <S> The OR gate actually look like this . <S> You can see this circuit has 4 terminals besides the output: two digital inputs A and B (1 and 2 inputs at 7432) and two feeding pins (Vcc and Gnd). <S> The OR gate does not copy the input as you were taught. <S> Actualy <S> it sends the same voltage expected by the OR table but from the Vcc(from the pin 14) <A> The 7400 series (including the 7432) are TTL (Transistor-Transistor Logic) devices. <S> So by not connecting anything to the inputs you have both inputs HIGH and so the output will be HIGH. <S> In order for the transistor circuit to work (the gate) we have to supply external power. <S> This is done through pin 14 (+5V) and pin 7 (0V). <S> The chip does not take power from the inputs. <A> Old-style TTL gates interpret an open input as a logic '1' so the output of an OR or AND gate will be '1' with the inputs open. <S> The inputs are always high but somewhat susceptible to noise so a 1K pull-up resistor is recommended, or they may be connected directly to GND if a logic '0' is desired. <S> The 1K is recommended for TTL rather than a direct connection to Vcc for TTL only. <S> As you can see from the above schematic, the 4K resistors source current out the inputs, so about 1.1mA flows when an input is grounded.
| The inputs are multiple emitter transistors which assume a logical HIGH if not connected.
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What is the low limit of today's power supply voltage generation and voltage amplitude of sinuosid generation? Voltage is not really generated, current is, but let me use this colloquial term. What is limitation for generating electrical voltage very low (DC power supply)? What voltage is the real low limit just above 0? Also if one wishes to generate a sinusoid function of very low amplitude above 0, what is the limitation of that amplitude? <Q> Also if one wishes to generate a sinusoid function of very low amplitude above 0, what is the limitation of that amplitude? <S> You can generate a sinusoid with as low an amplitude as you like. <S> Simply generate a 1 V sine wave, then attenuate it to the level you like. <S> You could easily apply 100 dB attenuation, for example, giving 10 uV output; Then apply another 100 dB attenuation giving 100 pV amplitude; and so on. <S> The practical limitation is that there is inherent noise in any circuit, and a small enough signal will eventually be too small to detect relative to the noise. <S> In a 50-ohm rf system with 1 Hz detection bandwidth, for example, you'd have about 0.9 nV rms noise inherent in the receiver, so any signal much below that would be very difficult to detect. <A> 20mV with this any lower voltage <S> and you'll have to roll your own <A> As commented by 'PlasmaHH' that it depend on whether you are asking about a power supply or any other voltage generation method in lab. <S> Consider this one: In BJT Ciruits currnets in the Base region(Ib) is nearly in uA range. <S> You can easily generates a voltage in few mV/uV by using a small registor connected with Base and taking output across it. <A> You mention DC power supply, but you don't specify the output current capability, which maybe a limiting factor. <S> This may count as a "power supply". <S> This in theory, neglecting the major limitation in low voltage tracking: noise. <S> If the generated voltage is too low it will be swamped by electrical noise, which depends on many factors: noise of the reference voltage, noise introduced by the opamp, environmental interferences, etc.
| Anyway, from a theoretical point of view, there is no limit, if you don't want high power: a simple opamp powered with dual supplies and connected as a voltage follower whose input is tied to the output of a voltage divider allows you to maintain a stable voltage equal to the voltage at the output of the divider.
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Create binary sequence to use less inputs Let's say I have 63 buttons, but there are only a few inputs on Arduino. I want to know which one was pressed. To minimize the number of inputs used, I could tell from a binary sequence. If I press button 61, I need to have six inputs returning 1,0,1,1,1,1. How would I convert the button number to binary? If you know any easier way tell me, it's the only that came to my mind. <Q> 6 pulldown resistors, one to each data line. 1 to 5 diodes from each button, the button connected to vcc. <S> Button 1 has one diode to pull up the d0 line when pressed. <S> Button 61 has 5 diodes to the d5, d3, d2, d1, d0 lines as for your example above. <S> Button 0 not used. <S> When any button is pressed, at least one of the data lines will go up. <S> Can be used for interrupt on button press. <S> Suitable pulldowns, almost any value, 10k will do. <S> Suitable diodes, almost any diode, 1N4148 will do, quite cheap by the 100s. <S> This is a static solution, simple, but uses shed-loads of diodes. <S> There are others. <S> You could put them into a shift register. <S> As Arduino has ADC inputs, you could use each button to switch various value resistors, to change the voltage into the ADCs. <S> Whether you could get enough resolution to resolve all 61 buttons without calibration or adjustment is another matter, you might want to split it into 4 channels of 16 buttons. <A> One possible way is to use input shift registers to perform a parallel to serial conversion. <S> Shift registers are covered in great detail on the net (try reading this description ), so I won't go too far into it here. <S> You would need several (9 in your case) of them, as typically each one has eight inputs. <S> They will be wired together in a chain to create a single 64-bit number, or several smaller numbers: the choice is yours. <S> I do this kind of thing all the time, and it is pretty easy. <S> You could create three 24-bit values and read them periodically, then perform an XOR with the previous values in order to detect changes. <S> This will require 3 inputs and one output (for a clock) from your Arduino. <S> If you are limited on inputs or simply prefer it, chain the registers together to make a single 64-bit number and then you'll only need 1 input pin, although you may not be able to store all of those bit in one value on Arduino, so it would take more than one read. <S> In any scenario above, it is stuff a beginner can do. <A> You could follow the ideas from the decade counter reading question. <S> There are many ways to multiplex inputs in digital or analogue ways. <A> what you need is a keypad/keyboard matrix.here's a complete guide of using keypad matrix,easy to understand. <S> http://pcbheaven.com/wikipages/How_Key_Matrices_Works/
| You could put them into a matrix, scan and decode.
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Is Saving Through-Hole Components Viable? I recently got my hands on some old electronics. After getting some to work and failing to get others to work, I decided to simply desolder some of the through-hole components to possibly use in future project. After desoldering a few, I realized that resoldering them on to a new board would be difficult considering the tiny amount of remaining wire and that the quality was quite poor. My Question: Is desoldering and salvaging through-hole and surface-mount PCB components viable, or should I just buy the components online? <Q> I routinely salvage through-hole and even SMD parts, with the following considerations: 1) <S> If the device is broken, the components from it are suspect; check everything. <S> I have one of these which is invaluable (It tests more than transistors). <S> 2) <S> If the leads are not long enough to go into a breadboard, it's probably not worth it. <S> I find that most transistors are OK, but resistors often are not. <S> Capacitors and LEDs are a mixed bag; your mileage may vary. <S> Leads tend to be longer on older devices. <S> 3) Be careful of static and heat limitations <S> so you don't destroy the part you're trying to salvage. <S> 4) <S> If you get even a few components out of a piece of garbage, it is probably worth trying. <A> The answer is a resounding yes or maybe. <S> Yes salvage <S> what you are likely to use in some way for play, study or repairs and maybe don't keep stuff that you will never use. <S> I find that keeping a couple of old and a couple of contemporary PCBs of the sort that I am likely to work on or experiment with saves a LOT of time compared to ordering or shopping for the odd part that I want to try out. <S> I have very rarely made use of salvaged parts for any kind of production run of more than 2 as it it <S> plain too slow collecting multiples of anything on scrap boards or in the junk boxes. <S> Also remember to sort the pulls so you can find them speedily when you are in need. <S> Exotic parts are much more useful to pull as generics are easier to buy in assortment kits. <S> Having 1200 resistors (US$10) on tap is much more fun that hunting for a desired value and not finding it. <S> However of much more value than the components when starting out with repair , design or hobby is the amount of learning you can gain from figuring out what the boards generally (and later more specifically) did and why designers did what they did which will teach you good skills. <S> You will learn to tell which boards were well designed with a glance and pay attention to new tricks they have used and which boards are commodity junk that is a second (or more) generation copy and contains defects or errors that have been copied over badly without understanding what they are doing. <S> Generally industry gets better electronics because they pay more because their equipment makes money, consumers shop by price <S> so they get junk. <S> It is always fun to see huge mains isolation areas on part of a PCB and then have the mains switch traces routed around to the other side of the PCB alongside all the low voltage circuits with 1mm track spacing. <A> For me, I would only salvage parts you'd need for a project you're working on now. <S> Otherwise, you'll keep piling up components you'll never use, and space is precious. <A> You can salvage THT components rather quickly if you fix the board in a vice on an edge and then pull off components with pliers while heating up the other side with a heat gun. <S> Take it outside, though, you will create very smelly smoke if not well practiced in that technique.
| If it's an IC or other non-obvious part, take a moment to look up the data sheet before removing it; if you can't find the data sheet, there's no point in having the part, unless you will repair more of the device you're tossing. SMD salvage is difficult to justify and though through hole is becoming scarcer it still has a place in robust electronics.
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Capacitors burning through in PSU (AC-DC) of 3W LED Basically my problem is probably simple, but I could not find anything searching google and here - maybe I was looking for the wrong things, but here is the problem: I work in a Hostel and we bought five 3W LEDs with each having their own PSU. (see Photo 1) After installing them just three weeks ago, one of the LEDs stopped working and we didn't think much of it and my boss replaced it incl. the PSU (because they come hooked up to each other and he didn't know that they are easily disconnected). Then the next another one stopped working today, not even 4 days later so I decided to take a closer look at it. Took it all out, connected a new LED to the old PSU and it didn't work. Must be the PSU, opened that up and it was all burnt and black on the inside, quite obviously at the capacitor. So now I am wondering what can cause this? Kinda important to know as there are 5 of those in a wooden wall in a wooden building and the last thing I would like to do is burn in this place :) The input of the PSU is (supposedly) rated at 100V - 265V (Photo 2) Thanks in advance! Misha Photo 1: Photo 2: <Q> It is not a case of overpower. <S> Most probably you have a not so stable AC grid in Patagonia and the drivers are being destroyed by Surge or Burst events on the AC side. <S> Normally in the no-name drivers the protection against such events is not really effective since some money can be saved when not implementing. <S> Question: <S> Is there any construction work, with drill machines for example, around? <S> Is there a motor connected on the same circuit? <S> Such loads are well know to cause surges on the mains damaging sensitive devices. <A> 2 out of 5 burned out in just a few weeks <S> sounds too much like really bad hardware. <S> I'd replace them with name brand equipment before the next one burns out and takes the Hostel with it, killing people along the way. <S> With the schematic, someone here could probably recognise the weak spots in the design. <S> From the photo, someone might be able to recognize an incorrect part (one that isn't up to the job.) <A> The power supply is overloaded or at least out of its specified operating conditions. <S> The specification says Power: <S> 3x 1W Output: DC 9-12V, 300mA <S> A 3W LED has a forward voltage of roughly 3-4 volts and draws between 700 and 1000 mA. <S> The problem here is that the power supply is made for three 1W LEDs in series (thus the 9-12V output voltage): simulate this circuit – <S> Schematic created using CircuitLab <S> Consequently, one single 3W LED overloads the power supply, as it is not specified to operate on an output voltage lower than 9 Volts. <S> The solution is to buy another power supply or other LEDs.
| There really isn't anyway for any one to tell you what is going wrong without seeing a schematic of the power supply, and maybe a picture of the circuit board before it burns out.
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Can thin sections of copper traces be used as fuses? Is it safe to use thin sections of copper traces as one-shot fuses when cost is important, but when it is also vital to protect the rest of the circuitry? Should the solder mask be removed at that location? What about using 0R resistances in small packages as resettable fuses? That's for applications when time to fuse is not critical compared to the location of the failure. For more demanding applications, are there I-t graphs of tracks of various widths available? I have not found any. <Q> Source <S> It's a bit more of a crapshoot than a traditional fuse, like a printed spark gap, but can be done. <S> The trace should not be exposed. <S> An exposed trace will be subject to contamination, possibly conductive contamination, which changes the amount of conductor you designed for a certain current. <S> Admittedly a minor concern, <S> but I don't see any pros for exposing the trace. <S> The above board was designed with pads to replace the fuse, it's from a car stereo system so they're likely expecting shorts downstream rather than surges from upstream. <S> Having exposed pads for the latter case would be less desirable as a surge might burn the fuse and leave a conductive creepage path between the pads. <S> This paper explores and provides calculations for determining trace size for a printed fuse with a variety of copper weights. <S> A salient point from the paper, in case it isn't always available is the approximation for the time in seconds (\$ t \$) before a trace reaches melting temperature given current passing through it (\$I\$) and its cross sectional area in mils squared (\$A\$): $$t <S> = 0.0346 <S> \times {{A}\over{I}}^2$$ Note that this is the approximate time to melt <S> the copper given a 20°C ambient temperature. <S> It might fail long before or a short time after this value. <A> I've done this in a real production design, and regretted it. <S> "It seemed like a good idea at the time." <S> I wanted to protect a backplane against a card being plugged in with a power-to-ground short in the card connector. <S> I put a necked track on the power line to each socket on the backplane. <S> Sure, it worked, and never blew by mistake. <S> However, when the trace did blow, it meant the poor field engineers had to replace the whole backplane. <S> I should have used thicker traces and just let the power supply shut down. <S> Perhaps another consideration is the possibility of a fire in a PCB. <S> A low current in a PCB trace will not raise temperature. <S> A high current will blow the trace. <S> An intermediate current, limited by some other effect? <S> Maybe that will raise the trace temperature enough to burn the PCB but not melt the copper. <S> I have seen that happen with very, very fine traces, and it can set fire to a fibreglass PCB. <S> The epoxy changes to just carbon, then the carbon starts to carry more current which heats up more, then... the results are not pretty. <S> There must be ways of designing the trace so this cannot happen, but I looked for design rules and could not find them. <S> So, yes, you can do it. <S> But I wouldn't! <A> I have got PCB layout people to do this for me for over 20 years now. <S> The applications were mainly on Automotive stuff where the 12Volt battery has a huge fault current. <S> I would call this a necked track. <S> I have never used this for any precision application. <S> The necked track is good for protecting the cars wiring loom in the event of a fault when the inline fuse has been bypassed. <S> We did tests by shorting a 12Volt marine battery into wire of various lengths to emulate the cars wiring loom. <S> Our setup would have had microhenries of inductance not millihenries <S> so it would be referred to as resistive. <S> SO at low volts resistive and high fault currents the necked track is safe. <S> We used 10 thou track. <S> At the time we were much more concerned with safe blowing than precision. <S> The thinking at the time was that if you wanted precision you would use a MCB. <A> Why use a trace for a fuse? <S> It means the board will be unusable if there's an overload until someone repairs it. <S> And how do you repair such a fault? <S> Throw away the board and get a new one, right? <S> If you're just messing around, go ahead and use a trace. <S> If you have a serious application, consider a resettable fuse, for $1 or so, which will automatically reset itself after an overload. <S> Your repair techs will write blogs about how much they like it. <S> An ordinary PC board mounted fuseholder with a fuse is a cheap solution, but service is more costly, in manpower, inconvenience and downtime. <S> A circuit breaker costs more and still requires human intervention. <S> If it's panel mounted, it's easier to replace, but that adds cost, labor and extra design effort. <S> It's hard to envision a commercial application where a circuit board trace used as a fuse is a good idea. <S> It's more likely a very bad idea. <S> And in many cases, a circuit designed with overload protection is better all around. <S> Sometimes it's as simple as choosing the right voltage regulator chip with built in overcurrent protection. <A> As a reminder, arcs due to DC voltage will not go away (unless the voltage is removed).
| One issue I see - and please let me know if I am mistaken - is that once the trace blows up, how can you reliably be sure there will be no arcing between the traces (in the case high DC voltage develops across the blown fuse). It's certainly been done.
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Two parallel motor under PWM now have slightly different speeds I have a couple of peristaltic pumps to wash my cells with different conditions for live imaging with a fluorescence microscope. I now have the two pumps with a PWM pre-assembled module. It has a timer and 3 mostfets and a poti obviously. I need to put media from a tube to the plate with a pump AND take the media from the plate to a waste container with the other pump. However when I was using the peristaltic pumps prior to the PWM the liquid flow was exactly the same with both pumps. (the level of water in 50ml tubes was constant in both tubes when crossing the in and out from the pumps).However, now with the PWM signals there is a bit of a difference and one of the test tubes goes empty while the other gains volume. I still don't know if it is mechanical or an electrical thing. I changed the peristaltic heads between motors and the same happens. Could it be the stall start or the back EMF doing this?The pumps are connected in parallel. Any way to solve this? Diode to reduce the EMF or a capacitor to equalize the start of the pump? This was my previous setup in which the liquid pump rate was exactly the same (with the same motors and peristaltic heads) Drop the current of a couple of peristaltic pumps for a microscope use Thanks for any help! <Q> You cannot expect two open-loop motors to have the same performance - even if matched initially they may age differently. <S> I'd especially expect that with brushed DC motors where brush wear can be a part of aging, or any system with sleeve-type bearings. <S> Instead, Lab-grade peristaltic pumps typically allow stacking multiple pump heads on a single motor, which will get you identical rotation rates and with the same size tubing fairly close flow rates. <S> If that is not close enough (some fluid property making pumping different? <S> one line pulling air?) <S> , you can either go back to closed loop and have a manual differential trim control to run one faster, or you can build a closed loop controller that watches some liquid level sensor, or perhaps a scale under the experiment. <A> If you want to verify that a small difference between the motors is the problem, swap the motors and see if the flow discrepancy reverses. <S> The simple way to equalize the speed would be to put a rheostat in series with the faster one. <S> Something like one ohm is probably what you want. <S> That won't work as well as the more sophisticated methods, but it might be good enough for your purposes. <S> Brian Drummond, <S> Once I setup the same speed with a rheo, won't it be problem when I change the speed (PWM pulses to a higher frequency?) <S> – João Mamede <S> That could be a problem. <S> Differences between the motors would likely have less effect at higher speeds, but the adjustment will really only be exactly right for one speed and temperature. <S> Using a rheostat is simple, but won't provide the best performance possible. <A> The 2 motors in general wont lock which is what Chris Stratton said and Charles Cowie is probably right about the rheostats. <S> If you make a simple electronic Variable resistor possibly with a mosfet with say 1 ohm fixed resistance across the drain and the source. <S> Use say a 10K pot on a 12V supply to manualy get the motors to track .The <S> mosfet will be easier to find than the rheostat .Once <S> you have manual tracking <S> you are half way there. <S> Now measure the motor current on an oscilliscope <S> and you should see a waveform that sorresponds to the commutator segments going around. <S> I tested a 12V Automotive windscreen wiper motor on a 12V lab supply and got a recognisable waveform. <S> I filtered out some sparking transients with simple RC filtering .You will have to filter some PWM junk also. <S> Your motor is probably higher quality than the ones I tested and if your PWM is worth its salt it should be much faster than the commutator grind that we are looking for .Once <S> you get a reasonable waveform then just square it up to make a 5V logic signal. <S> You now have motor speed without an expensive shaft encoder ,The rest is easy just use a micro to lock up the speeds by driving the mosfet gates with filtered PWM or analog outputs.
| You can used closed loop control monitoring a rotary encoder to slave both motors to a desired rotation rate, or one to the other other, or switch to stepper motors, however electronics may not be the best solution.
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What's a freewheeling diode I was studying single phase rectifier circuit with RL load. There I found that they are using a freewheeling diode in parallel to the RL load. So is a freewheeling diode different from a power diode ? Or it is just the same with a little higher rating? <Q> The freewheeling diode, for power rectifier applications, usually is a diode of the same type as used in the rectifier. <S> This improves manufacturability. <S> The function of this diode is to provide a path for the current generated by the back emf when the rectifier is loaded with an RL circuit. <S> Thus, damage to components of recitificador are avoided. <A> An inductor can't change current immediately. <S> Attempting to change the current rapidly such as when a switch opens after an inductor has built up energy, will cause the inductor to generate immense back EMF's. <S> If a flyback diode isn't present, the immense voltage will build across the switch potentially damaging it. <S> The flyback diode gives the current through the inductor a path to continue until the energy in the inductor has dissipated (in the inductors internal resistance). <A> Whenever there is an RL load, it's always good to have a diode connected in reverse biased across the inductor. <S> This diode is called freewheeling diode. <S> It's purpose is to block the back emf generated by a coil whenever power is disconnected from this. <S> This back emf could damage other components in the circuit especially solid state components.
| A free wheeling or flyback diode is used to prevent damage to circuits typically with any load that has an inductor and any switching potential.
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Why is this LC filter where it is, and what function does it perform? My customer has a motor controller that apparently only works with an LC filter in front of it. This configuration trips out on undervoltage: simulate this circuit – Schematic created using CircuitLab This works fine: simulate this circuit (I'm leaving out details like a line impedance and precharge, but that's the general idea. I can add them back if anyone thinks they're relevant.) I've never seen a motor controller require an LC filter like that. It is of interest to me that this filter has a pole at 49 Hz, and this equipment is designed to operate at both 50 and 60 Hz. Tests are being run at 60 Hz. If I've computed the frequency response right, that LC should have a gain magnitude of roughly 2x at 60 Hz. Is this thing really just a tuned voltage doubler? What other function could this possibly perform? <Q> Without the inductor the ripple is greater and the under voltage trip is likely due to the bottom of the ripple tripping some comparator in the controller and it decides to shut down. <S> The inductor will reduce the tops of the ripple AND significantly increase the bottom of the ripple hence the instantaneous voltage seen by the controller is never as low as the case without the inductor. <S> Speculation based on experience! <S> You could probably simulate it easily. <A> The reason to create an L-C Filter with regards to an inverter is for power quality reasons on the AC side. <S> A 6 pulse rectifier, in theory, will draw current harmonics: fund, 5,7, 11,13 .... <S> This is however dependent on continuous DC current. <S> Depending on the output power draw and the size of the DClink capacitor, the operating power point that continuous DC current occurs (and thus harmonic content of the AC current) may be too high with regards to requirements. <S> This inductor lowers the power at which continuous DC current occurs <A> I think I get it now. <S> The gain of the LC shouldn't be computed at 60 Hz. <S> The signal the LC is seeing is the rectified three-phase, which I shall henceforth call the humpty-hump wave. <S> This signal has a fundamental of 360 Hz, with some energy content at the harmonics of 360 Hz, since it's not a perfect sine wave. <S> At 360 Hz, with the components we're using, the gain of the LC filter is \$\frac{1}{1-\omega^2LC} = <S> \frac{1}{-52.7}=-.019\$. Negative gain means the output is 180 degrees out of phase with the input, but at this juncture we don't care about that. <S> At higher frequencies the gain is even lower, so less than 1.9% of the total energy content gets through the filter. <S> Instead of having a peak-to-peak ripple \$1-sin120^o = <S> 13.4 \% \$ of our peak voltage <S> , we get \$13.4\%x1.9\%=.25\%\$. <S> So the DC content of the wave (.955 Vpk per this answer ) is unaffected, remaining 324VDC. <S> The AC content was 45.5V peak-to-peak, and is now .86V peak-to-peak. <S> So our troughs, instead of being ~300 volts, are now ~323 volts. <S> Adding a resistive load reduces the AC gain further, and reduces the phase angle, but the effects are relatively small at the loads of interest. <S> I've run simulations that match up with the math quite well. <A> The principle for using the LC is the same as the one when it is used at the output of the switching regulators or buck regulators. <S> You have noisy current riding on the output voltage that has a frequency. <S> This is filtered using the inductor. <S> There are many hidden parameters like dcr,frequency ...etc of the inductor and the ESR of the capacitor that basically governs the selection of these specific component.
| The capacitor filters out voltage ripples with the capacitor acting as a small battery that kicks in during ripples in voltage or DC rails.
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What is the simplest way to turn off some circuit when supply voltage reaches a minimum level? Knowing I have only a 12V battery to supply my circuit, how I can turn off the circuit if the voltage reaches 11V? I cannot use simple comparator CI since I don't have another supply voltage. I could I do that? <Q> Is a supply-voltage supervisory IC along with a FET switch not an option for you? <S> One example is the TL7712A , but there are many others -- 12V versions aren't as common as 5/3.3/... <S> V counterparts, but they still exist. <S> An example circuit can be found below. <S> U1 is a generic 3-pin voltage monitor/reset supervisor, designed for 12V service -- more sophisticated parts can be used, of course! <S> M1 can be any N-ch depletion-mode power FET with suitably high ratings <S> (Supertex/Microchip and IXYS make 'em, the DN1509 shown being the former's work), while R1 is not needed if U1 has a push-pull output. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Yes, you can do that. <S> You should create a reference voltage for one leg of the comparator with a zener diode or similar voltage reference device. <S> The other leg of the comparator has a resistor divider that divides 11V to be equal to your reference voltage. <S> When these two values meet, your comparator will change states. <S> Just make sure your reference is less than 9 or 10 volts...perhaps 5v. <A> The LDO regulator is 10V and this can be potted down to (say) 5V (easy). <S> This is your measurement reference input. <S> This reference feeds one input of your comparator. <S> The other input is fed from your battery via another potential divider. <S> Set this 2nd potential divider to produce 5V when the battery is 11 volts. <S> The output of the comparator will switch as the battery falls below 11 volts and this switching output can drive a relay to disconnect your load. <S> You could also use a mosfet instead of the relay. <A> I really didn't understand why you can't use the same supply for comparator? <S> Correct me if i am wrong, but you can use a LDO regulator to regulate the voltage like 6 ou 7V and search for an Ampop that run on this voltage and generate a vref with a voltage divider and another voltage divider to compare with main voltage source 12V.
| Get a comparator that runs from (say) 10 volts and, feed power to it via a low drop-out regulator.
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15v to 3v LED - resistor vs regulator I've been reading about resistors vs. voltage regulators, but can't settle on which would be better in my case. I have a power supply for a mic preamp that has 15V and 48V. I want to add an LED to the power switch. Would it be best/most efficient to use an appropriate 1/2 Watt resistor, or a voltage regulator? I have both so I'm not worried about cost, but which is the "right" way to do something simple like this? The regulator I have on hand is the LM317T. I also have a 7805CV, and I think any of the 3 options will work fine with the 15V, but if someone could briefly explain which works best here or point me to some literature on the topic, I would appreciate it! <Q> I want to add an LED to the power switch. <S> Any linear regulator or any resistor will "lose" the same amount of power when dropping a higher voltage to a lower voltage. <S> If the input voltage is 15V and the output voltage is 2V <S> then the power dissipated as heat is 13V x current. <S> If the current is 20mA (standard red LED) then the power given off as heat is 260 mW. <A> It depends. <S> What are the characteristics of the 15V supply? <S> is it a nice stable 15V or does it vary? <S> If the supply voltage is unstable is corresponding instability in the brightness of the LED a good thing (because it indicates a problem to the operator)? <S> a bad thing (because it's a distraction to the operator)? <S> or something you don't care about? <S> The resistor solution is simpler. <S> The linear regulator solution isolates the LED from variations in the supply voltage which may or may not be a good thing. <S> The linear regulator solution and the resistor solution are both very inefficient (approximately equally so). <S> The efficient solution would be a switched mode LED driver but that comes at a significant cost in terms of extra complexity and is probablly not worth it for a low current indicator LED. <A> If you are just driving an LED, you want to control the current, as the LED will set the voltage, so a resistor is most suitable. <S> For almost anything else, where you want to set a fixed voltage, then a voltage regulator is required. <A> Resistors Vs Regulators ... <S> They waste the same amount of power as does an analog constant current source .The <S> expected efficiency is 3/15 or 20% of the options so far. <S> In fact this will make your LED less efficient than the incandescant lamp of last century. <S> If efficiency doesnt bother you then stick to the resistor .If <S> you think a SMPS is an overkill you are probably right. <S> Well you have volts to burn in your circuit so place the LED in series with something that doesnt mind losing 3V and already has a sensible LED current already flowing <S> .Double check that there wont be big current surges and now you have actually powered the LED for free .I <S> have used the input of a lightly loaded 7805 <S> so the idle current + the load current runs the LED <S> .Remember <S> that the human eye responds in a logarithmic way to brightness just like the ear registers sound .This <S> means that the current doesnt have to be precise.
| If you are happy to dissipate that power to activate a LED then a resistor is by far the superior choice because it current limits the LED naturally.
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Cannot understand the attenuation in signal from PC to microcontroller(NI myRIO) I am working on Active Noise Cancellation Project. I came across a problem, where the signal is getting attenuated as the audio bit depth changes from PC data to Microcontroller and I cannot understand it! My circutit :I connected Audio Output(Head phones) of PC[which is supposed to be analog signal] to Audio IN of NI myRIO 1900 device using AUX cable (two way 3.5mm jack cable). I have a pre-recorded signal(.wav) of sinusoidal waveform with Amplitude 1, Frequency 2kHz, 40000 Sample Rate, 80 Number of Samples, Bit depth( bits per sample) = 16. Now, I just played .wav file using windows media player and tried to record the waveform on NI myRIO module. Surprisingly, I am getting an attenuation of the signal corresponding to the volume reduction in PC(not linear!!). Only I can see the waveform in microcontroller close to amplitude 1, when I increased the volume of the system to 100%. Please explain this !! The ADC of NI myRIO has its Resolution as 12 bits but the audio Output(.wav file) is 16 bit data. I am expecting it does not have to matter except in quantization errors, as at the PC Audio Out, the audio data should be converted to analog and and again at the 'NI myRIO Audio In', this analog signal will be sampled! (please correct me if I am wrong !) Another question is, I have tried the same with sinusoidal sound of Amplitude 2.5, now the audio signal is clipped off at 1V above and below -1V. ADC(NI myRIO) nominal range is 2.499V to -2.5V! Is it bacause of the bit depth conversion? Please explain these things !TQ EDIT: I am not converting either 16 bit digital data to 12 bit digital data neither 12 to 16 !! I happen to have a 16 bit audio data, when I played this through windows media player will get converted to analog signal at head phones terminal, and I am feeding this ANALOG Signal to an ADC of NI myRIO, which has 12 bit resolution. <Q> Surprisingly, I am getting an attenuation of the signal corresponding to the volume reduction in PC(not linear!!). <S> Only I can see the waveform in microcontroller close to amplitude 1, when I increased the volume of the system to 100%. <S> Please explain this !! <S> Humans don't perceive sound intensity linearly, but rather logarithmically. <S> For a series of doublings in sound power is perceived as a linear volume series. <S> This has been well known for over a century, so the people that designed the volume control in your PC software know this too. <S> To give you what you will perceive as a linear volume control means actually varying the voltage exponentially. <S> This is why the voltage drops quickly from maximum as the volume is turned down only a little. <A> I came across this strange problem, where the signal is getting attenuated as the audio bit depth changes <S> and I cannot understand it! <S> You have a 12 bit resolution signal and you convert it to 16 bits. <S> One of two things happen. <S> Either: - The original 12 bits are mapped to the top 12 bits of the 16 bit number and the lower 4 bits of the new 16 bit number are set to 0 or, <S> The original 12 bits are mapped to the lower 12 bits of the 16 bit number and the upper 4 bits set to zero. <S> Number 1 will make no difference in the loudness of your signal and #2 will significantly reduce the amplitude by 16:1 whilst retaining the same signal to noise ratio (not a bad thing). <S> So why is #2 preferred? <S> An example would be when mixing sounds digitally. <S> You can now add 16 such converted waveforms and avoid clipping. <S> You can't really do anything with #1 that you couldn't already do when it was 12 bit format. <A> After you've added more info to your question, it would seem that the cause is less to do with the sampling (bit) depth and more to do with the analog connection between two systems. <S> PCs use line level output, which ranges from +1V to -1Volt. <S> That lines up with the level you are seeing on the myRUI 1900 <S> - you've got a chart there showing the signal in ranging from 1V to -1V which is to be expected. <S> The signal is clipped because your signal is being clipped by the PC. <S> If you try to send a signal out of the PC with an amplitude that is too high, it will be clipped. <S> It might be clipping in software before it hits the hardware, or it might be clipping in the hardware. <S> Some parts of the volume controls are in the hardware, some parts are in the software. <S> The parts of the volume controlling in software can amplify a signal to the point that it can't be represented in 16Bits anymore and gets clipped. <S> The parts of the volume control in hardware can amplify the signal to the point that the output amplifier itself clips. <S> In summary: You cannot generate a signal with a PC that will cause a full scale input on the myRUI 1900. <S> The PC simply cannot generate a signal with the needed voltage range. <S> Cranking up the volume on the PC to try to force a higher level will only cause clipping, which results in the signal you have captured with the myRUI 1900.
| It looks like you are reaching the limits of the PC's audio output.
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Bandwidth of inverting and non-inverting op-amps If the same op-amp is used in both inverting and non-inverting modes (with same closed loop gain using appropriate resistors), will the closed loop bandwidth of the op-amp in both cases be the same? For example, Now, if I assume unity gain frequency = 10 MHz, is the bandwidth for both 5 MHz? If I am correct, then why is the GBWP of the closed loop inverting op-amp less than that of the non-inverting counterpart? <Q> In your example circuits the loop gain is not the same - hence, the bandwidth will not be the same. <S> The circuit with the largest loop gain (non-inverter) has the largest bandwidth. <S> Explanation why the Loop Gain (LG) determines bandwidth: <S> The denominator of the closed-loop gain formula is \$ D(s) = 1 - LG \$ <S> From this, we can derive that "something" happens when \$LG=1\$ (0 dB).At the corresponding frequency \$ \omega_{o} \$ <S> we have a real pole (think of the behaviour of a first-order lowpass). <S> And this pole gives the frequency where the 3dB-bandwidth is defined. <S> I should add that this is a simplified explanation; a detailed explanation involves the open-loop gain Aol and its frequency response: \$ A_{CL} = <S> \dfrac{H_{FW <S> } \cdot <S> A_{OL}}{1 <S> - Hr \cdot A_{OL} <S> } \$ with \$LG= <S> Hr <S> * A_{OL}\$ and forward factor <S> \$H_{FW}\$. <S> We can see that for low frequencies (large \$LG\$) and negative feedback factor (\$Hr\$ negative) the "1" can be neglected and the gain is <S> \$A_{CL}= \dfrac{H_{FW}}{Hr} \$ <S> = constant. <S> However, for large frequencies (\$A_{OL}\$ and \$LG\$ smaller) we cannot neglect the "1". <S> When we reach the frequency \$ \omega_{o} \$ <S> where \$ |LG|=1\$ the "1" starts to dominate for larger frequencies and we can neglect the loop gain LG. <S> In this case the numerator \$H_{FW} <S> \cdot A_{OL}\$ determines mostly the frequency response (\$ <S> A_{CL}= H_{FW} <S> \cdot A_{OL}\$, approximately a first order lowpass). <S> Hence, the transition from the first region to the second region is at the cut-off frequency wo. <S> For inverter <S> : \$H_{FW}=\dfrac{-R2}{R1+R2}\$ For non-inverter: \$H_{FW}=1\$. <A> The derivation of the formulas can be found in multiple places <S> e.g. 1 or 2 and the books cited therein, so I'm not going to do it here; you've also done it correctly. <S> In a nutshell, if \$f_T\$ denotes the Unity Gain Frequency for the open loop opamp and \$f_B\$ denotes the same for a given circuit with circuit gain \$G_0\$, then for the non-inverting opamp circuit, the equation is simply \$G_0 f_B <S> = <S> f_T\$. for the inverting circuit, the equation is however <S> \$G_0 <S> f_B = - f_T (1-\beta)\$, where \$\beta = \frac{R_1}{R_1 + R_2} \$ using your notations. <S> What this means is that for the inverting opamp circuit the worst case is going to be \$ \beta=1/2\$, \$G_0=-1\$, when you'll only get half the bandwidth of the non-inverting circuit! <S> And to actually apply these equations to your example[s] <S> : for the non-inverting one: \$f_B <S> = f_T / 2 = 5\text{Mhz}\$. <S> for the inverting one: \$ \beta = R_1/(R_1+R_2) <S> = <S> 10/30 <S> = 1/3\$, so $$ <S> f_B = - \frac{f_T}{G_0}({1-\beta}) = <S> -\frac{10}{-2}\frac{2}{3 <S> } = 3.33\text{MHz}$$ <S> Here's an even quicker way to remember/solve this right, based on J.H. Krenz's textbook . <S> The equality \$f_B = <S> \beta <S> f_T\$ is valid for both inverting and non-inverting opamp circuits, and \$\beta \$ (which is called the feedback fraction) has the same formula as above for both circuits, i.e. \$\beta <S> = \frac{R_1}{R_1 + R_2} \$ where \$R_2\$ is the resistor in the feedback loop. <S> However, to get a gain of 2 for the inverting amp you need a beta of 1/3 as above, while for the non-inverting circuit (of gain 2) beta will be 1/2. <A> Okay, so I am pretty late to the party, but I just thought I would provide future ponderers with a single equation to solve this for either inverting or non-inverting amplifiers of this nature: \$f_B = <S> \frac{f_t}{1+\frac{R_2}{R_1}}\$ <S> To show this with the above circuits <S> : Non-inverting circuit: \$f_b = \frac{10\text{MHz}}{1+\frac{10}{10}} = <S> 5\text{MHz}\$ <S> Inverting circuit: <S> \$f_b = \frac{10\text{MHz}}{1+\frac{20}{10}} = <S> 3.33\text{MHz}\$ <S> EDIT: <S> The above example assumes an ideal op-amp. <S> If you wish to find the true bandwidth of the circuit taking into consideration the effect of finite open loop gain and frequency dependence, you must consider the gain of the circuit at the -3dB point. <S> \$G_{\text{dB}}-3\text{dB}=20\text{log}(A)\$ then: \$A=10^{\frac{G_{\text{dB}}-3\text{dB}}{20}}\$ <S> Thus: <S> \$f_b=\frac{f_t}{A}\$ <S> So for the non-inverting circuit: \$G_{\text{dB}}=20\text{log}(2)=6\text{dB}\$Then \$A=10^{\frac{3}{20}}=1.41\$ <S> Finally: \$f_b=\frac{10\text{MHz}}{1.41}=7.08\text{MHz}\$ <S> Then the same is also true for the inverting circuit. <S> Therefore, to answer your original question... <S> yes, the two circuits will indeed have the same bandwidth. <S> However, the bandwidth is not 5MHz, it is 7.08MHz. <S> I hope this helps. <A> Yes, barring limitations in the external components. <S> To get a rough idea of minimum bandwidth, divide the opamp's gain-bandwidth-product by the absolute value of the closed loop gain. <S> That is the same whether inverting or non-inverting. <S> Therefore in your example, assuming the opamp has a minimum GBP of 10 MHz, then both the circuits have a minimum bandwidth of 5 MHz. <S> However, you also have to think about the external components. <S> There will always be some parasitic capacitance. <S> To get the value computed above, the R-C low pass filters formed by any resistance and some parasitic capacitance must have a rolloff comfortably above the bandwidth you want. <S> To be pessimistic, assume 20 pF caps are added to ground and maybe 10 pF across components wherever they would reduce the bandwidth. <S> For example, assume 10 pF across R2 in the second example. <S> 10 pF <S> and 20 kΩ have a 800 kHz roloff, so 5 MHz is way past reasonable expectations. <S> We can work this backward and find the resistance that has 5 MHz rolloff with 10 pF, which is 3.2 kΩ. <S> Since you'd actually be 3 dB additional down for every filter at the rolloff frequency, you want that to be at least a octave, preferably 2-3 octaves, past the frequency of interest. <S> In this case, 1 kΩ would be a good choice for R2, with the other resistance scaled accordingly. <S> High bandwidth requires low impedances and costs current.
| There is a simple answer: The bandwidth for the closed-loop gain is determined by the frequency where the LOOP GAIN is 0 dB. You're are basically correct.
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Why is parasitic capacitance much more mentioned than parasitic inductance in MOSFET? It seems to me that in MOSFETs, parasitic inductance would be equally as important as parasitic capacitance. However, I have never seen anyone discussing parasitic inductance seriously. Is parasitic inductance's influence on MOSFET operations not really important, or is of much smaller influence compared to parasitic capacitance? <Q> For typical uses (very generally 10 - 100 V; 1-10 A, < 10 MHz), the physics of silicon mean that the capacitance of the FET structures (and parasitics associated with them) have values which have a more significant circuit effect than the inductance (generally associated with bonding wires and the package structure). <S> However, at high frequencies (certainly > 100 MHz); with certain DC/DC converters (low V, and high currents), the inductive parasitics can become significant and are critical. <S> In these operating ranges, inductance in the gate lead can significantly affect the rate at which the transistor can be switched; inductance in the source can also affect this. <S> inductance in the drain can cause large damaging voltages to appear between the internal transistor source and drain nodes, potentially damaging the device. <A> The structure of the MOSFET accounts for variety of capacitances which include junction capacitances,sidewall capacitances which come into picture at high frequencies to limit frequency response. <S> Talking about the inductances,to operate, the MOSFET must be connected to the external circuit, most of the time using wire bonding These connections exhibit a parasitic inductances. <S> However, when a MOSFET is used for high frequency switching applications, they become equally important as the parasitic capacitances. <S> As an example,the gate inductance and the input capacitance of the transistor can constitute an oscillator. <S> This must be avoided as it results in very high switching losses. <A> The parasitic inductance is the sum of the bonding wire and the PCB traces .So <S> its really a function of package type and board layout. <S> Most of the time the parasitic capacitance is more significant .When <S> you compare the energy stored in C to L you find that C is much higher <S> .This is why for powermos its generally more rewarding to implement ZVS rather than ZCS despite the fact that they are both valid <S> .At really high power and high currents and low voltages the parasitic inductance does get much more significant .Remember that the voltage across a inductive trace = <S> L times the rate of change of current .The parasitic inductance can and does resonate with the voltage variant mosfet capacitance causing parasitic oscillations generally at VHF .If <S> these are not dealt with you <S> can fail radiated EMC.The induced voltages due to rapidly changing currents can muck up your circuit or sensitive circuitry elsewhere in the product .Over <S> the decades the parasitic capacitance of powermos has got worse as on resistance has come down <S> but inductance has got a little better with SMD.
| These parasitic inductances are neglected when we typically use the MOSFET as an amplifier because they operate in saturation where current is almost a constant.
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Resistance of a diode Resistance of a semiconductor diode I was reading this question about the resistance in a diode. I have a question regarding the resistance of the depletion region. I know that the conductivity of depletion region is infinite but due to presence of electric field there's a dynamic resistivity thus a diode is represented by an EMF with a equivalent resistor. Does this resistor dissipate heat? My second question, the recombination of electrons and holes does dissipate heat? <Q> As I said in my answer to that question, a forward-biased diode (or an IGBT or BJT for that matter) can be approximated as a fixed voltage drop in series with a resistor. <S> This "resistor" does dissipate heat; a current flowing through a voltage drop always means energy is either being stored or dissipated, and diodes don't store substantial energies. <A> Recombination of electrons and holes doesn't really dissipate heat. <S> Keep in mind that "holes" are really a handy abstraction for thinking about semiconductors. <S> They don't really exist as physical entities other than a place where a electron could be but isn't. <S> Electrons aren't completely free to move around. <S> Their movement requires some power to sustain inside a conductor. <S> A inaccurate but still useful mental picture is that there is "friction" between the electrons and the molecules they move between. <A> Like all semiconductors, they have a dynamic conductance which is dependent on the applied voltage (electric field) and load applied. <S> All diode datasheets have V-I curve which is basically giving you its "resistance". <S> Since R = V/I. You can draw a load line on this curve. <S> The greater the current the larger the voltage that develops across it. <S> Thus P = VI, and P = <S> I^2 R or V^2/R <S> so yea your dam <S> right it dissipates heat. <S> Also its dangerous to think of quantities as infinite. <S> Nothing is infinite. <A> The dynamic resistance does dissipate heat, as does the current flowing through the 'fixed' V in series with it. <S> Imagine a diode with 1 mA, and 0.726 V. <S> this is 726 uW total power. <S> The small signal dynamic equivalent model is 700 mV in series with a 26 ohm resistor. <S> The power is 1 <S> mA * 700 mV + 1mA^2 <S> * 26 ohm = 700uW + 26 uW = 726 uW -- <S> same as before. <S> This model remains valid for small excursions <S> around 1 mA (e.g. 0.1 mA). <S> Changing the current to 1.1 mA increases power to 1.1 mA <S> * 700 mV <S> + 1.1mA^2 <S> *26 = <S> 770u+31.5u = <S> 801.5 <S> uW. <S> Now, I know that the voltage on a diode increases by about 2.6 mV when the current increases by 10 % (from the exponential behavior). <S> Therefore, the 726 mV increases to 728.6 mV. 1.1 mA in 728.6 mV <S> is also 801.5 uW (basically because of the same reason that 10 % gives 2.6 mV). <S> While separated, the energy stored in them is similar in nature to the energy stored in a capacitor (although there are complications analogous to the small-signal resistance with the capacitor also; not least of which is that the capacitance is not constant with voltage).
| Recombination of electrons and holes does dissipate heat, just as separating them across a potential barrier does.
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PWM voltage boost for mosfet I need to run 12v motor with using Raspberry Pi. I made a circuit which works but then I noticed a "little" problem. Pi's software PWM signal is only 3 volts so BUZ11 doesn't open enough. I found this schematic : Is it possible to somehow crank my PWM signal up? Ex. Using that schematic given with inverted pwm? Or can I just use transistor before FET to raise pulse voltage? <Q> What you have there with that NPN is a workable albeit basic MOSFET driver. <S> Which of these two approaches is better depends on various factors... <S> with the former, you need new MOSFET[s], with the latter you can use your existing ones, but the circuit becomes more complicated and driver isn't free either. <S> There is also catch with lower Vgs MOSFETs: <S> those that are both low Vgs and high current tend to come in packages that are nasty to solder by hand, e.g. NVTFS4823N is similar enough to BUZ11 in Rds and max <S> Id terms, but will give you more <S> Id current at lower Vgs. <S> The catch? <S> Difficult to solder by hand. <S> They aren't all that bad though; BUK9219-55A comes in DPAK and does 50A with 3V gate drive; also has good enough Rds on (compared to BUZ11): <S> Mosfet drivers can be realized in many, many ways. <S> A bipolar transistor is used often enough because it is cheap. <S> Sometimes a pair of push-pull transistors are used; these allow both transitions (on and off) to happen fast. <S> You can find such transistor pairs as a single package e.g. MCH6541 . <S> Additionally, if you need level shifting (and here you do) add another transistor in front) as in this answer : simulate this circuit – Schematic created using CircuitLab <S> (The gate resistor depends whether you have any ringing problems with a given MOSFET.) <S> However, for slow switching requirements, you'd be hard pressed to tell the difference between the 1-transistor (just level shiftier) and faster driver (i.e. with extra push-pull transistors): <S> Actually, if you also do level shifting, you can optimize away one of the transistors by using a twin-NPN totem pole instead of a push pull (but you need to add a diode): simulate this circuit <S> There are also more sophisticated IC drivers <S> that use custom-designed gate driving curves. <S> These will usually reduce switching losses and prolong the life of the MOSFET. <S> Some don't look much more than a transistor in size and connectivity requirements (4-pin devices) but basically have something like the above inside: <S> The catch? <S> They cost more. <A> Use a better MOSFET. <S> The BUZ11 isn't specified to work as low as 3 volts: - If you had 4.0 volts on the gate, a 2 amp motor (for instance) would probably drop about 0.3 volts and dissipate 1.2 watts. <S> Not too bad but, to my mind the simplest solution is to get a better MOSFET like the SOT23 package <S> AO3416 : - Volt drop looks about 0.1 volts to me and power dissipated would be about 0.4 watts. <S> If this device is mounted on a small PCB with some PCB copper intentionally acting as a heatsink it should be good for a couple of amps. <S> But, as always, read the data sheets and think about stall currents and don't forget about the reverse diode across the motor to prevent back emfs destroying the MOSFET when the motor current is interrupted when the FET switches off. <S> If not this device then there will be others more than likely that will suit your application. <A> If you change R1 to say, 2K2~1K, you will notice a considerable improvement on turn on time (note that more current will flow through T1 also).
| If the MOSFET doesn't open at logic level of your MCU you either need: a MOSFET that opens at lower VG (and gives you enough drain amps at that VG), or you need to use a MOSFET gate driver that does the logic-level translation. While the other answers are far better technically and more recommended on the long term, a cheap fast way to improve your circuit (so you can test it and decide what you need to do) is to use a smaller pullup resistor.
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Why could a UPS device need of a microprocessor? I have read about some UPS devices that are microprocessor-controlled , and I would like to understand the fundamentals of needing such a microelectronical device like a CPU for what it just seem a current management machine. Why do some (all?) UPS devices incorporate a microprocessor? What do my UPS loose if it has not one inside? How complex/advanced is this microprocessor and what are the advantages of having a better microprocessor for a UPS? What happens if this microprocessor is not good enough, and how may detect such flaw ? <Q> There are many advantages. <S> First, an MCU can replace quite a bit of discrete logic and monitoring circuitry for less than the cost of discrete parts or an ASIC. <S> UPS units need to monitor several different parameters (line voltage, battery charge, load current, etc). <S> Many MCUs have built-in ADCs, so using a microcontroller means that most of the necessary logic can be implemented in software on the MCU instead of discrete components. <S> This logic would include things like undervoltage and overvoltage monitoring, battery level monitoring, hysteresis, timeouts, etc. <S> In general, the MCU will replace a number of discrete logic chips and simplify the control circuitry. <S> It also becomes trivial to add external status reporting over a serial port or even over Ethernet. <S> Using an MCU also allows the UPS to be remotely managed - thresholds and timeouts adjusted, output enabled/disabled, etc. <S> Microcontrollers capable of this would be rather simple. <S> The atmega328p used in the popular Arduino boards is pretty representative of what you might find in a UPS - 8 bit processor, a few KB of SRAM, a few KB of flash, clock of a few MHz, ADC of a few ksa/sec. <S> For one that supports Ethernet, it would probably use an ARM-based MCU of some sort, or at least something 32-bit, to deal with the network interface. <S> Possibly even two MCUs, one for interfacing and one for control and monitoring. <S> Personally, I would be more surprised to see a UPS that didn't use a microcontroller of some sort. <A> Why do some (all?) <S> UPS devices incorporate a microprocessor? <S> What <S> do my UPS loose <S> if it has not one inside? <S> The microcontroller inside provides an easy way to control the UPS, including reporting status information, managing charging, responding to events, etc. <S> You lose a lot of flexibility without one. <S> How complex/advanced is this microprocessor and <S> what are the advantages of having a better microprocessor for a UPS? <S> It depends on the UPS. <S> Most likely, it exactly as advanced as the lowest cost option capable of performing the specified needs. <S> If you have more needs, then you may need a more advanced device. <S> What happens if this microprocessor is not good enough, and how may detect such flaw? <S> If it's not good enough then it's not used in the UPS . <S> You're not going to detect this perceived flaw. <A> Ask yourself: what does a microcontroller do ? <S> What would it add to a UPS ? <S> If you just want a box that powers a load from a battery when the mains voltage is interrupted, you could build that without a microcontroller. <S> For the basic UPS functionality, a microcontroller is NOT NEEDED. <S> An indicator that tells you to test the battery every 6 months. <S> A beep when the UPS kicks in. <S> You could also implement that without a micro controller but it would be more easy and flexible with it. <S> And since microcontrollers are so cheap nowadays almost any device has one.
| But if you do incorporate a microcontroller you can more easily add extra functionality like monitoring from a PC. The UPS is an engineered product, they don't use parts that can't perform the tasks required.
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Relation between Voltage & frequency in DVS As per Wiki - "In DVS, Overvolting is done in order to increase computer performance" How Overvolting will increase performance?? Also, it says " By applying a higher voltage to the devices in a circuit, the capacitances are charged and discharged more quickly, resulting in faster operation of the circuit and allowing for higher frequency operation." Aren't the above 2 statements contradiction with for eg DDR2/3/4 where we are decreasing voltage to decrease rise time & further increase performance?? What am I missing!!! <Q> DVS is dynamic voltage scaling. <S> However, dynamic power consumption in CMOS is proportional <S> to frequency times voltage squared, so you get a 'triple whammy' increase in power consumption. <S> In DDR2/3/4, the bus is a different animal than the components on the chip. <S> The data bus is subject to different limitations outside the chip because it is physically much larger. <S> The spec is designed to use lower voltages for lower power consumption as well as faster transitions. <S> This is done at design time and the supply voltage and threshold voltages are changed. <S> In DVS, the threshold voltage is NOT changed, as it cannot be changed without changing the circuit, transistor doping, transistor body biasing, etc. <S> Since the PCB traces between the RAM chips and the CPU act like transmission lines as opposed to RC limited lines, the performance characteristics are completely different. <S> The RC time constant does not apply to these traces, so raising the supply voltage does not have the same effect. <S> Instead, output driver slew rate and supply current are more important. <S> With a smaller swing, the line can be driven faster and with less power. <S> On the receive side, amplifiers are required to properly receive the data. <S> A DDR2/3/4 RAM array can certainly be overclocked with DVS just like any other chip, so long as enough cooling is provided to deal with the extra power dissipation. <S> However, the interface circuitry may or may not work properly at the higher frequency, even if the supply voltage is increased. <S> TL;DR: <S> in DVS, the voltage is changed on a fixed CMOS chip. <S> In DDR2/3/4, the bus is designed to operate at a lower voltage and it is not RC limited like on-chip components. <S> They cannot really be directly compared. <A> What's DVS, kind of overclocking? <S> Elements built for lower voltage operation are faster, because the transition time from low to high state level is shorter compared to the same element with higher nominal voltage. <S> But if you apply higher than nominal voltage to the element this transition time is even shorter, because the threshold level 0,1 <S> remains the same, but the slope is now steeper due to higher supply voltage. <S> Well, this fact is known and used for gamers, while it is useless for normal operation. <A> For what I know, when speaking of DVS, Overvolting is done in order to achieve Overclocking. <S> First thing, its not always necessary everywhere <S> that Overvolting results in Overclocking. <S> Secondly, don't confuse the case of DDR2 memories with just the overclocking method. <S> DDR2 are dual-pumped memories, they are not faster just because of Overclocking, principally they are faster because they transfer data on the rising and falling edges of the bus clock signal.
| Increasing the voltage allows the internal capacitances of the wires and transistors to be charged up above the transistor threshold voltages more quickly, allowing the device to operate at a higher frequency.
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Using more than one voltage regulator in series Is there any problem (in addition to a likely loss of current) in using more than one voltage regulator (L7812CV) in series? And is there a real gain in stability? I ask this because I 'm going to use some high precision components in a car, using it's battery. Edit.: My bad, I'm new to electrical engineering. I was thinking that voltage regulators are more like a stabilizer. For ex: inputing either 14v or 9v, it will force 12v in output by varying its current. So i guess that this question doesn't make sense. There's any other way then to stabilize the voltage of a car battery to guarantee 12v? <Q> To add to the other answers, here is some general information about getting a low noise power supply from a car: <S> The voltage from your car will normally be between 12V (engine off) and 14.5V <S> (engine on) but may have drop-outs, glitches and spikes up to about 30V. <S> So you are correct to think about smoothing it for precision use. <S> The simplest starting point is a DC-DC converter, such as one described in the other answers. <S> On its own, it'll give you the set voltage with probably 100mV or ripple, depending on how much you paid for it. <S> If you need a cleaner power supply, you can pay for a better DC-DC converter, or you can add a filtering stage. <S> A low-ESR ceramic capacitor connected across the output is a good start, and a LC pi filter will do better, but be careful to select low resistance parts to make the filter. <S> If that's still not good enough, you can set your DC-DC converter to 14V and use a linear regulator to drop the voltage to 12V. <S> That will usually produce much lower noise on the supply, but make sure the linear regulator has a good PSRR at the switching frequency of the regulator. <S> For absolute best results, use a DC-DC converter, then an LC filter, then a linear regulator, and finally a large bypass capacitor near each IC you are powering. <S> Put the DC-DC converter in a metal box, the regulator in another box and your actual circuit in a third box, so they are all shielded from each other, and use shielded cable to connect the 2nd and 3rd box. <S> There are very few applications where you'll need to go beyond a decent DC-DC converter and maybe a capacitor. <S> Taking all of the steps above would only be necessary if you were doing something like powering an ultra-low-noise instrumentation amplifier to make sub-nV measurements. <A> The L78* series of parts are linear voltage regulators. <S> What that means is they essentially work like this: simulate this circuit – <S> Schematic created using CircuitLab <S> The potentiometer R1 can be adjusted to achieve some target output voltage. <S> A voltage regulator isn't exactly an automated means of adjusting a potentiometer, but by considering it as such you can have a good intuitive understanding of the basic physical limitations of any linear voltage regulator: <S> the output voltage can not be higher than the input voltage, and the drop in voltage (\$V_\text{in} - V_\text{out}\$) is achieved by converting some of the input electrical power into heat. <S> A higher voltage drop or more current though the regulator will mean a hotter regulator. <S> By "in series", I assume you mean to connect the input of one voltage regulator into the output of another. <S> I don't think this will have the advantages you hope to get. <S> Since a linear voltage regulator's output must be a lower voltage than its input, you can't take a 9V input and make it 12V, no matter how many regulators you have. <S> I also suspect just one voltage regulator, properly used, will do a quite sufficient job of giving you a steady output voltage. <S> If you must be able to boost the input voltage, then you need some kind of nonlinear voltage regulator, like a SEPIC . <S> Furthermore, since this is for an automotive application, you should be aware that a car's electrical system is pretty hostile. <S> You'll want to consider conditions like load dumps and include appropriate protection. <S> I'm hardly an automotive engineer <S> so I couldn't even begin to give you advice, but you might read some application reports for ideas. <A> So you want something which will take an input of 9 - 14 V and will output 12 V regardless. <S> Luckily such modules exist. <S> Here's an example: Boost Buck DC adjustable step up down <S> Converter XL6009 Module Voltage <S> There's a trimmer on it to adjust the output voltage to your desired <S> 12 V. Optionally you could also set it to 15 V and add the L7812 regulator to have some extra stable voltage. <A> You can use a SEPIC DC DC Converter to guarantee your DC output output even if car battery goes above or below your operating voltage. <S> Give yourself a little margin on the top end <S> and then you can use a low-noise high PSRR ratio linear regulator to give you a nice clean 12V. <S> Linear Tech DC-DC SEPIC calculator TI LDO article
| I suggest you read the datasheet if you need to know "how good" the voltage regulation is, but unless you are making some precision lab equipment, you can probably bet it's good enough.
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Converting a 12V pulse to a shorter 24V pulse? I have a bunch of Point-of-Sale printers which can emit a ca. 1 second long 12V pulse in order to open the attached cash-drawer. Because of a procurement SNAFU, I now also have a bunch of cash drawers which would like a 24V pulse to activate the solenoid and open the drawer. In spite of this, the 12V pulse USUALLY works. But not always. So, my question is: is there a way to convert the 12V, 1 second pulse to e.g. a 24V, 0.5 second pulse, using only basic electronics which would NOT require an additional power source? <Q> buy a bunch of $1 boost converters from your favourite online <S> marketplace.the drawer will not be adversely effected by the longer pulse. <S> buy some heatshrink too to house it in. <A> You can use the following idea (charge-pump-like): simulate this circuit – Schematic created using CircuitLab <S> In this configuration the two capacitors are charged in parallel to 12V each. <S> Switching to this configuration: simulate this circuit makes the caps to be connected in series, thus producing 24V on the designated point. <S> Of course this circuit should be augmented with some automatic switching logic (transistors/comparators), and the pulse might be not very "square" one, but it is a general idea. <S> Update: <S> Actually there is no need in the C1 <S> capacitor, if the input is remaining 12V there.. <A> another way to double voltage, but this is fairly hard on the relay simulate this circuit – Schematic created using CircuitLab rly2 represents the cash drawer solenoid 1M resistor is jut to help the simulation resolve. <A> Consider the situation of charging a capacitor up to 12V via an inductor. <S> Due to losses, eventually the capacitor attains a steady state voltage of 12V <S> but, between 12V first being applied and the steady state situation is a whole different story. <S> So the capacitor starts charging up and the current through the inductor attains a peak value as the capacitor capacitor voltage reaches 12V. <S> At this point the energy stored in the inductor continues to raise the capacitor voltage higher and eventually there is no current left in the inductor i.e. all the energy has gone. <S> The voltage on the capacitor is 24 volts and this voltage is about to fall and repeat like this: - <S> (source: swarthmore.edu ) <S> For a very low damping ratio i.e. very low dc resistance in the coil, the first peak will hit 24V then subside down to 0V then rise up to 24V continuing basically forever if the damping ratio is zero. <S> Of course it won't be zero <S> but you might hit a 1st peak of 22V for instance. <S> Now here's the trick the inductor is placed in series with a diode so that the capacitor charges up to about 21 volts and (wait for it...) <S> stops there. <S> There's no more energy left in the inductor and a stable situation has arisen. <S> The capacitor remains charged at over 20 volts. <S> You could then use that capacitor to discharge energy into the solenoid that opens the cash draw on your POS system. <S> Here is a simulation: - The blue trace rises to about 22+ volts in about 3 milli seconds. <S> The red trace is the current through the inductor. <S> About 24 mJ are stored in the capacitor. <S> If this isn't enough to activate the solenoid then a bigger cap and inductor are required and a longer time delay to charge can be expected. <A> Connect the other end of the solenoid to the ground of the 12 V source. <S> In this case, the polarity of the transformer windings matters -- and they are unlikely to be labelled. <S> If it doesn't work, then exchange the terminals of either the primary or secondary. <S> However, I don't know where to get a suitable transformer :-(
| You might be able to use a 1:1 transformer -- just connect the primary to the 12 V source, and connect the secondary between the 12 V pin and the drawer solenoid.
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How to choose which usb connector type is best for Processor working as a device mode? Based on my processor specification, USB device mode is supported.USB has different kind of connector based on its host/device/ OTG modes. IF USB-OTG mode supported processor used Micro AB usb connector, Then if my processor only support USB device mode, which is the exact connector should i use? <Q> The connector is really specified by whether a device can supply power (Type A), but as the vast majority of devices consume power, they use Type B. <A> In USB device mode, your device must use one of the B receptables . <S> Type B is comparatively large, Mini-B was deprecated , so you should use Micro-B. <A> USB is designed to be a connection between a host (PC) and a client (device). <S> For the host, type A connectors are foreseen, and for clients it's type B. <S> USB OTG means that a device can also act as host, but it does not specify any connectors. <S> As the device is used as client more often than as host, it should have a type B connector. <S> OTG connections are established via special OTG cables which have type B connectors on both ends.(By the way, a USB OTG device must also be able to deliver power via USB, as most USB devices are USB powered) About the question what connector to use:The standard type B is common, but a bit bulky. <S> Micro USB is THE standard, as it is used by most mobile phones and power supplies with Micro-B are widely available. <S> This is your best choice if you also plan to power your device by a simple wall wart. <S> If you think Micro-B is a bit too fragile for your use case, have a look at Mini-USB, which is more rigid than the Micro type. <S> However, this connector isn't used that often any more. <S> Further more, there is UBS 3.0 and Type C as replacement of Micro-B, but as those are not so common yet, I'd avoid them.
| Generally, USB devices use one of the various Type B connectors.
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Is it sensible to always use larger diameter conductors for carrying smaller signals? This question as originally written sounds a little bit insane: it was originally asked to me by a colleague as a joke. I am an experimental NMR physicist. I frequently want to perform physical experiments which ultimately boil down to measuring small AC voltages (~µV) at about 100-300 MHz, and draw the smallest current possible. We do this with resonant cavities and impedance-matched (50 Ω) coaxial conductors. Because we sometimes want to blast our samples with a kW of RF, these conductors are often quite "beefy" -- 10 mm diameter coax with high quality N-type connectors and a low low insertion loss at the frequency of interest. However, I think this question is of interest, for the reasons I'll outline below. The DC resistance of modern coax conductor assemblies is frequently measured in ~1 Ω/km, and can be neglected for the 2 m of cable I typically use. At 300 MHz, however, the cable has a skin depth given by $$\delta=\sqrt{{2\rho }\over{\omega\mu}}$$ of about four microns. If one assumes that the centre of my coax a solid wire (and therefore neglects proximity effects), the total AC resistance is effectively $$R_\text{AC}\approx\frac{L\rho}{\pi D\delta},$$ where D is the total diameter of the cable. For my system, this is about 0.2 Ω. However, holding everything else constant, this naïve approximation implies that your AC losses scale as 1/D, which would tend to imply that one would want conductors as large as possible. However, the above discussion completely neglects noise. I understand that there are at least three main sources of noise I should consider: (1) thermal (Johnson-Nyquist) noise, induced in the conductor itself and in the matching capacitors in my network, (2) induced noise arising from RF radiation elsewhere in the universe, and (3) shot noise and 1/f noise arising from fundamental sources. I am not sure how the interaction of these three sources (and any I may have missed!) will change the conclusion reached above. In particular, the expression for the expected Johnson noise voltage, $$v_n=\sqrt{4 k_B T R \Delta f},$$ is essentially independent of the mass of the conductor, which I naïvely find rather odd -- one may expect that the larger thermal mass of a real material would provide more opportunity for (at least transiently) induced noise currents. Additionally, everything I work with is RF shielded, but I can't help but think that the shielding (and the rest of the room) will radiate as a black body at 300 K...and therefore emit some RF that it is otherwise designed to stop. At some point , my gut feeling is that these noise processes would conspire to make any increase in the diameter of the conductor used pointless, or down right deleterious. Naïvely, I think that this has clearly got to be true, or labs would be filled with absolutely huge cables to be used with sensitive experiments. Am I right? What is the optimum coaxial conductor diameter to use when carrying information consisting of a potential difference of some small magnitude v at an AC frequency f? Is everything so dominated by the limitations of the (GaAs FET) preamplifier that this question is entirely pointless? <Q> You're substantially correct on everything you've mentioned. <S> Bigger cable has lower losses. <S> Low loss is important in two areas 1) <S> Noise <S> The attenuation of a feeder is what adds Johnson noise corresponding to its temperature onto the signal. <S> A feeder of near zero length has near zero attenuation and so near zero noise figure. <S> Up to a meter or several (depending on frequency), the noise figure of a typical cable tends to be dominated by the noise figure of the input amplifier you are using, even cables of pencil diameter (you can get really <S> thin cables, sub-mm even, and in these you do have to worry about meter lengths). <S> To get signals down off your roof into the lab, any feasible cable will be so lossy, even unusually thick ones, that the solution is almost always an LNA on the roof, straight after the antenna. <S> That's why do tend not to see really fat cables in labs, they're not needed for short hops, they're not sufficient for long drags. <S> b) <S> High power handling In a transmitter station <S> , you tend to have the amplifier in the building, and the antenna 'out there' somewhere. <S> Putting the amplifier 'out there' as well is usually not an option, so here you do have fat cables, as fat as possible given that they have to remain TEM, without moding. <S> That means <3.5mm for 26GHz, <350mm for 260MHz etc. <S> Have a look at this cable manufacturer's tutorial on why we have different cable impedances, so 75\$\Omega\$ for lowest loss, and 50\$\Omega\$ as a compromise that has settled itself in as a standard. <A> For most folks posting answers on this particular stack, the answer to optimum cable size generally has a lot to do with economics, service life, ease of use and such. <S> Each individual problem has its own set of defining parameters, which in turn will be used to create a specification that will be met or exceeded. <S> This is an important step to take, because premature optimization is a real problem. <S> I can absolutely guarantee several things about electronic design that are always true. <S> Larger diameter cables experience less heat waste due to improved conductivity <S> , higher voltages allow more power to be transmitted per unit current, and larger batteries have more capacity. <S> But the solution must actually fit the problem, so frequently you'll find yourself using the specification to choose just exactly what is acceptable for the particular problem you are having at the moment. <S> You have demonstrated a more than adequate understanding of the issues at hand, and I humbly submit that you are likely better suited to the details than I am at the moment. <S> You also seem to be engaged in research, rather than design. <S> That being the case, I would offer this advice - having a firm understanding of the noise terms and how they are affected by increasing temperatures over time, decide on a firm, non-zero value of Johnson noise that is currently acceptable for your work, and design around that as a specification. <S> Set conductor sizes and types, and if necessary consider active cooling (provided, of course, that it doesn't interfere with or invalidate your research). <A> While you're correct in your details, I think you've missed the forest for the trees. <S> With 50 ohm loads, you don't need to worry about losses in the cable due to resistive effects. <S> at least not for RF measurements. <S> Consider your N connector example. <S> An effective conductor resistance will give a voltage drop of approximately $$\Delta v = \frac{0.2}{50} = <S> 0.4\text{%}$$which as about 48 dB down. <S> To put it another way, a 10 uV signal would give a nominal -100 dBV, but a 0.2 ohm conductor will produce a signal at the load of 9.96 uV, or -100.035 dBV, and somehow I have a hard time believing that will be a problem.
| The impedance of the cable also matters, as well as the size.
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Is that single triangle a tri-state logic symbol or an AND gate? Currently I'm a bit confused about that single triangle from that schematics used by an AVR-Microcontroller. Currently I've figured out, that the: DDRx is on the AVR an configuration-Register (What is IN, what is OUT) PORTx is on the AVR a register for writing or setting the Pull-Up-Register. PINx is on the AVR a register for read-only So the next step for me is to understand that triangle. I've searched for that and found similar symbols. One was an AND-Gate and the otherone a tri-state-logic (buffer). Has somebody an idea what kind of job the triangle implements? <Q> A triangle is basically an amplifier. <S> In digital circuits it signifies a buffer (or inverter, if there's a small blob on its input or output). <S> In this case it's controlled (turned on or off) by DDRx. <S> N. <S> You would turn it OFF if you wanted to use the pin as an input. <S> As the other answers say, this structure is called a tri-state buffer, where the three output states are '1', '0' and 'Z' (undriven, or off). <A> A triangle is just a (non-inverting) buffer. <S> If it has a connection to the 'side' of the triangle, it is tri-state and the 'side' connection is the enable input. <S> There may be a circle on the output or on the enable input to indicate inversion. <S> For example (from here <S> ): <S> The above represents four tri-state non-inverting buffers with common (inverting) enable inputs. <S> When the EN4 input is high the outputs are high impedance. <S> When EN4 is low the outputs will be driven to follow the inputs. <A> When enabled by the Data Direction Register, it conneccts the PORT register to the I/ <S> O pin, so the pin will act as an output. <S> When the buffer is disabled, it allows the PIN register to read the level of the I/O pin. <A> It is a tri-state buffer. <S> If DDRx.n is set, PORTx. <S> N register value is transferred to the output pin and the PINx.n <S> register, thus making this port output. <S> If it is unset, the value from the physical pin is transferred to the PINx.n <S> register only, thus making the pin an input.
| I would consider the triangle to be a tri-state buffer.
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Will mains kill me if I'm using an adapter? I'm sorry this is an ignorant question. I am working on hooking up leds for my raspberry pi. I keep hearing about mains power and how dangerous it is. I'm using a 5v adapter that is plugged into an outlet. If something goes wrong and I get shocked will I be okay? This might be a really stupid question, but I figured I would be better safe than sorry. <Q> As long as you are using an approved consumer power supply (such UL, TUV etc depending on your locale in the world) it is designed such you are not normally exposed to any hazards. <S> This assumes you don't do anything foolish such open <S> it's sealed enclosure or dunk it in water. <S> So - no the 5V cord coming out of it should pose no risk to you. <S> But don't stick any wires into an AC wall outlet. <A> Short answer is: its possible. <S> There is a famous case about a woman who died after answering her phone. <S> Reportedly it was a fake chargers fault. <S> Take a look at this article reviewing different chargers. <S> Note that many cheap chargers do not respect regulations and recommendations needed for safety certification. <S> If you take the right precautions and use decent gear you will be much safer though and shouldnt worry. <A> The adaptor is supposed to be insulated to prevent this, even in the case of internal failure. <S> The only circumstance it could go wrong is if it's an illegally sold one with false CE/UL certification.
| Voltages below 48V are not considered hazardous to humans.
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Why would a computer power supply output 5.2V on the 5V rail? I am using a computer PSU for lab power supply. In the current setup, I am drawing about 1ma on the 5VDC rail. And still the multimeter indicates 5.18VDC. I can upload a scope show if needed, but this seems too large of an error for a multimeter. Is this normal? Why would a regulated power supply output a very incorrect voltage value? <Q> If the power supply is rated at several amps and you have it pretty well unloaded it may have difficulty regulating accurately. <A> Low cost power supplies only regulate the voltage at the (internal) terminals of the supply. <S> The drop across the resistance of the leads between the internal terminals and your load is not controlled by the supply. <S> in order to ensure that there is 5 V at the load at the max. <S> current, a supply generally has a regulation set point at a slightly higher output voltage. <S> If you draw maximum load current, you will find the voltage will be much closer to 5 V. <A> As others have mentioned, 1) you are drawing only a tiny fraction of the current usually drawn by the computer; and 2) the 5.18V is probably within spec for a 5V rail (although they are usually closer than that). <S> If you need the voltage to be nearer to 5v, you could try putting a 10Ω 5W resistor across the output terminals, which will put a load of 500 mA across them, consuming 2.5W. <S> See if that helps. <S> RadioShack sells a 10Ω 10W resistor, 2 for $2.49, which will run cooler than a 5W one anyway. <A> 5.2V is well within USB spec. <S> Especially knowing that this is a no load voltage. <S> The USB standard voltage is 5V, but can range from 5.25 to 4.45 while remaining in spec. <S> If you needed tighter regulation, you could use a boost switching regulator to get the voltage up past 6V, then use a 1% 5V LDO. <S> However, this is not necessary if you are using the 5V rail as a USB charger, or for CMOS logic such as the 74AHC family or related.
| You might also find that having a fraction of a volt extra will guarantee 5 V on the computer board despite voltage drop along the wires.
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Could a 220 V wire shock me if I hold a 4 band Orange Red Orange Violet resistor in series? A noob level question and I am afraid of electric current since childhood shock. So a wire goes into wall socket, 200v input. If I hold a resistor of 32 Kilo-Ohm and touch wire with it's other hand and at another place with my other hand, something like: Wire end 1 ---- resistor --- my hand 1 -- body -- my hand 2 --- wire end 2. Could I die? (I meant I calculated current, I = V/R, it's low. But am I making newb-mistake?) What about parallel? (I do not know concept here.) Edit Imaginary scenario: Guys, I am afraid of electric current. I have now started to look at electrical and electronics concepts to get over this fear. Relax, all I am trying to do is to find definitive answers which books do not provide (and not electrocute myself or someone else). Regarding color combination. It's not a real resistor. I picked up high enough value 32 K-Ohm, because 220/R is 0.006875A, read that it's low to kill a human and I wanted to keep the tolerance low thus 1%; did that in gResistor in ubuntu. I will read up on what you guys have mentioned in the answers and get back with on topic but better questions. Problem I find with books is that they tell how to calculate and what to calculate, but do not relate to real life scenarios. <Q> Anyone that has to describe a resistor by reciting its color bands definitely shouldn't be sticking it into a wall outlet, especially when holding on to the other end. <S> Put another way, if you have to ask, the answer is NO, don't do that. <S> Just for interest, each color is just a digit. <S> ORN RED ORN VIO is the same as writing 3 2 3 7. <S> However, that doesn't mean 3237 <S> Ohms. <S> Usually such things are written using a floating point scheme, so that might mean something like 232 x 10 7 , but that's a rather unlikely value. <S> Four bands would usually end in gold, meaning 5%. <S> 1% resistors have 5 bands, so it's not clear what you actually have. <S> Perhaps the first orange is really gold, so the colors are actually VIO ORN RED GOL, which would be 73 x 10 2 , 5 %, another unlikely value. <S> Again, don't stick it in a wall outlet. <S> Added <S> Now that you've provided a diagram, it's clear that this is even a worse idea than it might have been. <S> Not only are you directly connecting yourself to the power line, but you are doing it between the arms, which means the current will flow close to your heart. <S> Unless you're going for a Darwin Award, this is a really dumb idea . <S> Ever. <S> No exceptions. <A> Yes I think you could get shocked or even die. <S> Please don't try this :) <S> According to V=IR, if you have 200Vac (with a peak of 1.414 * 200 = 282v) then divide this over the resistance of 32k, you end up with around 9ma of current. <S> This is enough to shock or possibly kill you. <S> I certainly would not try this regardless of the resistor's value because only some kinds of resistors are rated to be safe with mains voltage. <S> Often those "touch" lamps will have a big blue resistor going to their sensor, not because it needs to dissipate a lot of heat, but because it is the only thing separating the operator's hand from mains voltages so needs to be specially rated. <S> I am also concerned that your "Wire-resistor-hand-body-hand-wire" model doesn't include your legs that are going to earth. <A> Short version: you clearly don't have the experiance to be designing circuits involving mains, let alone circuits that involve both mains and the human body. <S> Messing around like this could get you killed. <S> Long version: <S> orange-red-orange-violet seems an unlikely code. <S> If we interpret it as digit, digit, multiplier, tolerance <S> then we get 32K but with 0.1% tolerance. <S> It seems unlikely to me that a 0.1% resistor would use a coding scheme with only two significant digits and while 32K is a standard value it's only in E192, again I find it unlikely that an E192 resistor would use a code with only two significant digits. <S> Other possible readings of the code don't seem to lead to anything sensible for me either. <S> Colour codes are easy to misread, if in doubt measure. <S> Anyway lets say you actually do have a 32K ohm resistor. <S> That's about 7ma into a short circuit load, a bit less (but same order of magnitude) into a good connection to the human body. <S> This is definately enough to give you a shock and quite possiblly enough to kill in the right circumstances. <S> Then there is the risk of failure, cheap resistors are often not designed for mains and also not designed for the power levels that will end up being dissipated in the resistor <S> (1.5W if driving into a short, a bit less if driving into the human body).
| If the resistor is on the neutral wire, you're basically grabbing the active wire directly and you could die very easily. Let me make it really clear for people with your level of understanding: Don't poke anything into a electrical output other than a commercially made plug for that purpose.
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Compute the minimum number of 120Ω resistors to get 80Ω of resistance? I recently had to take test in basic electronics. I didn't get one question right, but I don't quite understand why. How many 120Ω resistors are at minimum required to get a resistance of 80Ω? The possible answers to this question are 2, 3, 4 and 6 . The only answer I can come up with is 6 , with the resistors arranged as seen bellow. But 6 isn't the correct answer. Question: How many resistors are required and to arrange them? simulate this circuit – Schematic created using CircuitLab I only know the very basics of electronics, so I hope my thoughts are correct. <Q> A direct solution can be found through the application of continued fractions . <S> If what you have is 120Ω and what you want is 80Ω, write down the fraction: $$\frac{80\Omega}{120\Omega} = <S> 0.6667$$ <S> Since the integer part is zero, you'll be starting by putting resistors in parallel. <S> Invert the fractional part: $$\frac{1}{0.6667} = <S> 1.5$$ <S> This tells you that you'll have 1 resistor in parallel with some number of resistors in series. <S> Invert the fractional part again: <S> $$\frac{1}{0.5} = 2.0$$ <S> This tells you that you need 2 resistors in series. <S> Since there is no fractional part at this point, you're done. <A> 120 || (120 + 120) <S> If two 120 in parallel give 60, you want one of the branches to be a little higher, so... <S> that's the next thing to try. <S> And the method is true in general for getting a 2/3-valued resistor using just a bin of the same kind. <S> And in general for solving problems like this, it's worth <S> remember that the equivalent resistance of two parallel resistors is less than that of either of the branches. <S> You can also get 3/4 (so 90) for example by adding one more to a branch. <S> N.B.: <S> Thanks to Massimo Ortolano 's paper <S> , now I know that what I've done above following just intuition is that I basically followed the search path indicated below in the Stern <S> –Brocot tree : <A> You can change your solution by swapping serial and parallel: simulate this circuit – Schematic created using CircuitLab <S> You can then group R2, R3, R5 and R6 into a single 2x2 group: <S> simulate this circuit And <S> those 4 \$120\Omega\$ resistors make a single \$120\Omega\$ resistor: <S> simulate this circuit <A> Take your solution but without a central point in the middle: you can rearrange this as three parallel sections of 120+120 Ohm each (connecting the middle points does not make a difference since they all are at the same voltage). <S> Now two of the three parallel 120+120 Ohm sections combine into 120 Ohm again, so you can replace those 4 resistors from the two parallel groupings with a single one, leaving just one 120 Ohm resistor parallel to 120+120 Ohm. <S> There is a plethora of solutions proving the correctness of this solution once you have it. <S> But this rearrangement shows how to find it without reverting to mathematical trial and error. <A> Elaborating on @RespawnedFluff's answer, one way to find this is to think in the following way: What resistors do I have, ok 120. <S> What do I need to make, 80 <S> What equations do we know? <S> Well the two resistors in series or parallel are the simplest starting points. <S> Clearly series doesn't help immediately - that would increase the resistance, not reduce it. <S> So we will need to try parallel. <S> We know the equations: $$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{R_1 + R_2}{R_1 R_2}$$ <S> So maybe lets start with that: $$\begin{align}\frac{R_1 R_2}{R_1 + R_2} &= 80\\\\80R_1 + 80R_2 &= <S> R_1 R_2\\\\R_2 &= <S> \frac{80R_1}{R_1 <S> -80}\end{align}$$ <S> So can you find any combination that suits? <S> Well, start with \$R_1=120\$ and then see what value \$R_2\$ needs to be. <S> Can you make that value easily? <S> In this case yes, so great. <S> For other values, if you can't get a value immediately, you may need to try either the same approach as above iteratively to find the value for \$R_2\$. <S> If that fails to work, you could also try changing \$R_1\$ - maybe two in series or parallel, and then try again for \$R_2\$. <S> This approach is quite iterative, but in this case it would have quickly found both the answer you got (using 6 resistors), and also the answer @RespawnedFluff got (using 3 resistors). <S> If you were trying to grow the resistance (i.e. the required resistance is larger than your available value), you basically do the same thing, but start with a larger available resistance, or split the larger resistance up in series chunks and solve for them (e.g. if you wanted \$180\Omega\$, you could pick a chunk of \$120\Omega\$ and \$60\Omega\$). <S> You may wonder how the method would have gotten to your answer - given that yours has 3 parallel branches, whereas this approach uses two. <S> Well, in calculating \$R_2\$ above, iteratively, you would introduce \$R_2\$ being a parallel branch, which topologically is the same as if there were 3 branches to begin with. <A> Basic resistance in series and resistance in parallel logic. <S> Very simple.. <S> We know the output is 80Ω, so solving the problmm is easy. <S> Try the formula for parallel resistance:$$\frac{1}{Rp} = <S> \frac{R1+R2}{R1\cdot R2}$$Therefore.. <S> $$Rp = <S> \frac{R1\cdot R2}{R1+R2}$$Here put \$Rp = <S> 80\Omega\$. <S> Now as we know we have resistors of only 120Ω. Put \$R1 = 120\Omega\$ <S> Solving ... you will get \$R2 = <S> 240\Omega\$. <S> But we cannot use a 240Ω resistor here as it's said that we have only 120Ω resistors. <S> So instead of 240Ω, we will use 120Ω + 120Ω (in series) in parallel with a single 120Ω resistor.
| The answer is a total of 3 resistors.
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Where/How should I fuse this transformer? Background: I have a step-down transformer P241-5-36 , which is a single mains primary coil to 36V Center Tapped (18-0-18) secondary. I would like to use this transformer to build a linear regulated dual rail (+15V/-15V) DC power supply for DIY synth work. I'd like for both me and the power supply to not be destroyed if I happen to accidentally produce a short circuit on the secondary side. The data sheet says for my particular transformer, "36V C.T. @ 0.35A" under the "Secondary RMS Rating" column. Question: Do I place a fuse on the primary or secondary coil, or both and should I use a 0.35A value fuse or something else? I think the fuse needs to be on the primary coil (though I'm not sure the value), but when the transformer spec identifies 0.35A at secondary, it is confusing me. Research: I researched linear regulated power supply circuits and found the following schematics: Example 1 Example 2 Example 3 While these schematics all to very different things, they all share a commonality in that they show a fuse on the primary side of the transformer. I read through this post asking general safety questions about a PSU schematic. That individual's schematic showed fuses on both the primary coil AND secondary sides. Proposed Schematic: Here is the schematic I plan to use for my power supply (up to the linear voltage regulators). It is dependent upon the correct location and value of a fuse - I took a guess here: simulate this circuit – Schematic created using CircuitLab Please note: I included a switch to float ground because I may want to probe with my oscilloscope and connect the probe's ground clip to something other than ground on my circuits. Thanks, EE Community. <Q> Your transformer has 0.35A rating for its secondaries (each). <S> You can fuse those with a [say] 0.3A fuse each. <S> (The center tap doesn't nee a fuse.) <S> The fuse in the primary [if you want it] needs to be much smaller. <S> For 115V primary, you have a turns ratio of about 3.20 (to both secondaries in series), so the max current in the primary is about 0.11A. <S> A 100mA fuse should work there. <S> Beware that if put the secondary side fuses after the rectifier they need to be DC fuses. <S> AC and DC fuses are not the same because the latter need to cut the arc while in the former it gets cut by the AC sinewave itself. <A> Use a fuse for the primaries rated for 1.5 to twice the current. <S> Reason is, that the inrush current is much higher than the average current. <S> Fuses on the secondary are optional but if you have the space for them I suggest to add them. <S> It's much easier to drop in a new fuse than to change burned out voltage regulators. <S> Speaking of voltage regulators, the LM317/LM337 can be used with a overcurrent protection. <S> There is an example for this in the datasheet. <A> Fuses won't necessarily protect against a brief overcurrent thru silicon (that may be enough to destroy it before the fuse melts). <A> Fuse always connect primary side of distribution transformer the reason <S> is that when any surge come and also overloading this fuse protect the transformer from damange...... <S> Transformer is costly <S> so we protect transformer from damnage by putting fuse at its primary side.
| The fuse on the primary is there to prevent a short on the transformer melting wires in your wall and starting a fire in your house. You just have to add a few resistors and a transistor.
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Occupancy Sensor for Arduino I working on a home (rather room) automation system using a Raspberry Pi as 'brain' and an Arduino as sensor board. I find it quite difficult to come up with a solution to reliably track if a person is currently inside the room or not (to trigger the lights). Here are my current ideas and what I think about them: PIR: only for movement detection, probably can't see when I'm sitting at my desk: would need to wave to the sensor every once in a while. Door counter: not reliable if someone decides to turn on his heels. camera: too difficult to program reliably. IR thermo array: probably the most reliable, but expensive for high resolution, not sure if something about 4x16 would be enough. [Edit] I have found that the Panasonic Grid Eye 8x8 thermopile sensor (~30$) is advertised by Panasonic themselves as that it could span a whole room and indicate the position of a person in the room. Even though I think that this won't work for the entire room, it will surely be enough to get a nice movement vector in the area of the door if someone enters or leaves the room (if the sensor is faced more to the door instead into the whole room). microwave sensor: haven't really found information about sensitivity in comparison to PIR. Does anybody have some more ideas or thoughts on the ones listed above? <Q> You could set a different delay time for each and OR the results. <S> So if you haven't moved, burped, or opened a door in a long while your either sleeping, not home, or just plain dead. <S> (Its lights out in all cases.) <A> This subject is called presence detection and is very tricky. <S> As stated in other answer, you should combine different inputs to try guessing if someone is in the room or not. <S> I don't think a simple OR would do the trick. <S> Is it acceptable for people accesing the room to have an RFID tag or other wireless form of identification on them? <S> Have you consider using ambient noise as an input too? <A> One way would be using a person's smartphone's Wi-Fi to track where he/she is. <S> It is an existing technology called Wi-Fi positioning system (WPS). <S> The way it works is that you connect to the Wi-Fi hub (use the R-Pi for this, loads of material online on how to convert it into a Wi-Fi hub) in the house/room <S> and then the Wi-Fi hub keeps measuring the intensity of the receiving signal from your smartphone. <S> Then based on the intensity of the signal it approximately detects your position in a house/room. <S> One more way is a through a similar thing called Bluetooth positioning system (BPS). <S> I believe Apple had used this technology in manufacturing a product called iBeacon. <S> It works on the same principle as the WPS. <S> I think bluetooth is better suited to your application since it uses less power and detects proximity instead of location. <S> I guess if you only want to switch on the lights in a room, detecting proximity would be enough. <S> Both of these methods would be implementable with the help of an Arduino and a Raspberry Pi. <S> The only drawback of these methods are that if you go for a large scale implementation .i.e for the whole house, then you end up using a large amount of Wi-Fi/Bluetooth nodes for better accuracy. <S> But for a room it's a good enough solution.
| Combining the outputs from a few simple sensors such as a PIR, sound sensor, door switch, etc., might work well. I recommend you go for higher level of programming (meaning interfase the arduino with a PC or rpi), and use some more calculations like probabilities and such, to enrich the output.
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Gate with 2 inputs, one being "select". If select is high, output is always low regardless of other input I am tring to figure out how I can have some sort of module that can create the following scenario: The module has two inputs and one output. One of the inputs is "select". If select is high, then the output of the module is always low. If select is low, then the output of the module will equal the state of the other input. Here is the truth table for my desired module: Here is the circuit I am trying to achieve this outcome with: Subtract is the "select" input and each of the "overflow" inputs are the 2nd inputs. Therefore I need 3 of my desired modules where each of the XOR gates are. The XOR gates achieve what I want when "overflow" is high, but not when it is low. I'm sure this is relatively easy to achieve but I just can't seem to think straight at the moment. I think that some type of flip-flop would be my best bet. <Q> Your truth table is implemented by ~SELECT & INPUT . <S> Assuming you're re-using the same select line for all three instances, you need one inverter and three AND gates. <S> I think that some type of flip-flop would be my best bet. <S> A flip-flop has memory. <S> Its output depends on both the inputs and the past state of the output. <S> Since your truth table doesn't depend on the past state of the output, there's no reason to think a flip-flop would be useful here. <A> An alternative would be to use a simple non inverting tri-state buffer with active low select line and a pull down resistor on the output. <S> A detailed description of tri-state buffers is available here: <S> http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/CompOrg/tristate.html <S> In simple terms, when the select line is not active the output is not driven at all, so will take the logic state of the pull down resistor. <S> A real world example of this device would be the CD74HCT125 see: <S> http://www.ti.com/product/CD74HCT125/technicaldocuments <S> The advantage of this device is that you can 'wire or' the outputs of multiple gates if only one select line is driven at any time. <A>
| As The Photon pointed out in a comment above - the answer is simply ~SELECT AND INPUT as shown here:
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Current Limiter -- How to limit 16v 3.36a to 16v 1a? I am trying to equalize charge a 150AH Lead-Acid 12v battery. To do this, 16v at 1a needs to be delivered to the battery for a period of +-10 hours. If a 16v 3.36amps power brick (laptop charger) is directly connected to the battery then, amps drawn reaches 4.3amps and the power brick becomes extremely hot -- in danger of frying the power brick. <Q> Applying 16V to a lead acid 12V battery is a bit steep so try using a 2 ohm resistor in series to drop the voltage on the lead acid to <S> about 14V. Voltage dropped is 2V and with the resistor being 2 ohm the current taken should be about 1 amp. <S> Power dissipation of the resistor will be 2 watts but the resistor should be rated more like 10 watts for when a more discharged battery is applied. <S> This is probably the simplest method and you could look to using some silicon to provide better regulation. <S> You could even use three silicon diodes in series to drop maybe 2 to 2.5 volts. <S> The continuous current ratings og the diode need to be round <S> about 5 A just to be on the safe side. <A> There are two kinds of supplies, constant voltage and constant current. <S> The supply you have is constant voltage. <S> The supply adjusts the current output to make sure that the supply voltage output does not droop below 16V. <S> The other type of supply adjusts the voltage to make sure that the current stays the same. <S> What you want is not possible because to reduce the current you have to reduce the voltage. <S> The best way is to use a constant current supply set to 1A and then cut the charging off when the battery reaches the fully charged voltage of about 14.75V. <A> The traditional current limiter for crude chargers was a suitably rated filament light bulb. <S> When it is glowing orange to yellow hot it works as a current controlled resistor that will reduce the voltage that reaches the battery if the current is too high. <S> It is a very good robust low tech solution for float charging (if you top up cell with water occasionally as required). <S> You do want to regulate somewhat as a 16V supply <S> will eventually dry out a 12V lead acid battery cells by boiling (electrolysis) of the water when full charge at about 14.2V has been reached.
| You can switch in a beefier lamp for a period (of time or until the voltage reaches some set point) if you want it to start with fast charging if your charger can cope.
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Methods for removing solder from inside a header socket While soldering a header socket I sometimes apply too much solder and it flows through, partially filling in one or two of the female sockets holes. This prevents the male headers from being inserted all the way. What are some methods for removing the solder from within? I've tried to insert a jumper wire and then heat the wire with the soldering iron, but that did not seem to help. I do not own a solder sucker (yet), so I have not had a chance to try that yet. Edit: I am attempting to remove the solder from within the header socket itself, not from the via/holes on the PCB. <Q> Some headers use a forked contact, others use a hollow square. <S> In both cases, the solder gets wicked up with Capillary Action . <S> Once inside, the heat and time needed to reflow the solder, and wick it up with any typical desoldering wick or desoldering sucker tool will melt the plastic of the header . <S> You won't be able to get close enough with most soldering irons to do it. <S> Desolder the entire header, and replace it with a new one, using less solder of course. <S> It's the only proper and practical way to do it. <A> Besides solder sucker which you already know (the mechanical and motorized ones) <S> the only other method that comes to mind is a solder wick.. <S> p.s. <S> : I have in extreme cases removed solder by melting the solder and bumping the pcb on the table, but I'm not sure if this qualifies as a real method.. <A> If solder is in side the contact area, then sorry, game over. <S> Remove the header, fit a new one, solder more carefully this time.
| If I understand you correctly, the solder is inside the contact, not the solder hole, then there is no practical way to fix it.
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Why is home voltage value higher during the night? I have been monitoring the voltage in my own home (Mexico 120V 60Hz). I have been graphing the values every minute and I'm starting to notice that during the morning voltage gets up to 127, and during the day (4pm) it can go as low as 118V — why does this happen? The effect is quite similar to when I turn on the microwave for short periods, however these variations from 118V to 127V happen slowly during the 24 hrs of the day. <Q> TL,DR; What you are seeing is primarily the result of reactive load droop sharing, with secondary loading voltage drops in the transformers and lines. <S> When generators run in parallel like they do to supply national grids, there are two mechanisms at play that largely go unnoticed by the local population. <S> Perhaps the most critical one is speed droop load sharing. <S> It's actually pretty clever and fairly simple - every generator in parallel is programmed to have the same, predictable slow-down from no load to full load. <S> This way if something large starts on the grid, all generators respond roughly equally, sharing the load in proportion to their full load ratings. <S> There is a bit more to this, of course. <S> On a big grid, some generators like nuclear plants will be base load stations, and will hardly respond at all to these changes. <S> On the other extreme are peaking stations like natural gas turbines, which can easily react very quickly to truly large shifts in grid loading. <S> This kind of a set up makes the best use of both resources. <S> But this speed regulation trick doesn't help with reactive loading. <S> In brief for those uninitiated, reactive load consists essentially of those back-and-forth exchanges of current between capacitive and inductive loads that do not result in the delivery of real power. <S> But we can use a similar trick by using a voltage droop regulator on the output of every generator on the grid. <S> From no-load to full-load, every generator will have roughly the same drop in output voltage. <S> People live and work in daily cycles, and as a whole bunch of inductive, motor operated loads start up, the grid voltage comes down a bit to keep things balanced. <S> This usually hits hardest in the early evening, with a noticeable bump around breakfast time (usually). <S> These two systems work in concert to ensure that no single unit on the system ever absorbs an amount of real or reactive load that it can't handle (at least until something really unexpected happened). <S> Of course, additional losses are incurred in steps between you and the grid, but I cannot give you exact proportions without detailed equipment information. <A> Your power utility will generally try to generate at constant voltage. <S> The power distribution network of power lines and transformers will have some resistance and this will cause voltage drop along the network. <S> The voltage drop will be proportional to the current drawn and this varies through the day. <S> From your voltage monitoring you should be able to determine periods of high use in your locality. <S> (This my be not be so easy if the utility company switches on and off peak-time generators.) <S> The other thing that may happen is that the grid frequency may drop slightly from 60 Hz (in your area) during peak loading. <S> If this occurs then the utility company has to run slightly faster than 60 Hz during periods of light load to make up for this otherwise any mains-powered clocks will lose time. <S> That is, they will track the mains frequency and lose and gain time during the day but on average they maintain their accuracy. <A> Most likely your utility does not perfectly regulate voltage. <S> The voltage is 120 V nominal ; in practice devices would be designed to tolerate a range of 110-130 V or more. <S> If you happen to be near a large industrial load (factory) that could explain some of the variation. <S> Some power plants are easier to regulate than others. <S> There are effects that will cause a local voltage drop within your own home, but the variations you are seeing are greater than that would produce unless you have serious wiring problems. <A> Less power demand means less current demand. <S> Less current demand causes lower transmission loses, this allows the voltage to go higher (towards its "no load" value).
| The simplest answer is that the power demand is less when people are asleep.
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AC and DC signals If a AC+DC signal passes through a junction from where a capacitor is connected to ground and resistor is also connected from the same junction to the ground ,will the voltage measured across resistor and capacitor be same? IF yes then how?AC signal will pass through capacitor while dc signal through resistor...wont it make a difference? <Q> An AC signal simply means that your signal level is changing sinusoidaly. <S> Due to AC signal in both componants current in them will vary in them such that the voltage across them remain constant. <S> Consider input voltage as, $$ <S> V_i=a+bsinwt$$$$ <S> I_r =\frac{a+bsinwt}{R} $$$$ V_c=C\frac{dV_i}{dt}=Cbcoswt$$ <S> Therefore as these equation explained that current in both componants is different and hold it's value <S> such that voltage across them remain input voltage. <A> From your description it sounds like you are looking at a resistor and capacitor in parallel with respect to a voltage source and that this voltage source can either be AC or DC. <S> Correct this if my interpretation is wrong. <S> Maybe something like this: simulate this circuit – Schematic created using CircuitLab <S> It is important <S> ot remember that anytime you hook anything up directly across a volage source, it must have the same electric potential across its pins. <S> You can view this as you are effectively measuring the voltage across the terminals of the voltage source. <S> Similarly, two devices that are connected in parallel both must be at the same potential (or have the same voltage) across their terminals. <S> You can again view this as you are effectively measuring the voltage of the source. <S> My other interpretation of your question is that you are asking if both types of current flow are going through these two devices in parallel, will the voltage be the same? <S> If this was your question then you need to realize that there are only two types of current flow and voltage sources <S> cannot produce both (at least not at the same time). <S> If a signal contains postive and negative voltages, then it is definitionally AC. <S> If a signal contains only positive voltages, then it is definitionally DC. <S> (These are slightly oversimplified but represent the effective definitions). <S> Following this, either source signal (AC or DC) will have a voltage across it so any components must also have that potential. <S> For the future you should clarify your questions a little more. <A> Well it seems that you decribed a parallel RC circuit. <S> The voltage will be the same. <S> How? <S> The components are in parallel. <S> Please read about the superposition theorem <S> : Link for the superposition theorem <S> Basically, the current over the resistor will have an Ac+DC component, while the capacitor's current will have only the AC component, value is depending on the frequency of the AC signal and capacitance. <S> Again, check for the superposition theorem and the calculation of impedance in AC circuits. <A> Yes. <S> Whenever two circuit elements are connected in parallel, the potential difference across them will be the same. <S> how? <S> AC signal will pass through capacitor while dc signal through resistor... <S> wont it make a difference? <S> Because the capacitor can sustain a DC voltage across it without passing any current. <S> And a resistor can sustain an AC voltage across it while passing much less current than a capacitor (when \$C \gt\gt 1/\omega{}R\$).
| In your situation Both AC and DC signal will flow through resistor and Only AC through capacitor.
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It is better to fix x's in the simulation or in the design? I have a question about how to deal with x's in Verilog netlist simulations. I have a disagreement with another engineer (who is a bit more senior than I am) about what the right approach is. Although this question might seem opinion based, I do think there is a right answer even if the views are divided. I do development for mixed-signal ASICs. I'm doing netlist simulations of a microprocessor which doesn't have it's program registers reset (it's an ARM Cortex-M3, there is an option to not have the registers reset during synthesis). We have a ROM which the processor starts executing from after reset. During that program execution, one program register (R6) has x's in it because it wasn't reset. At a point in the simulation, x's from that register spread like wildfire through the rest of the design, and breaks the simulation. We don't see this issue in RTL simulations. I would prefer to make a design change to cause the registers to be reset or to have the ROM program write zeros to those registers first thing. My college is very resistant to making any design change to clear out these x's, and he would prefer to mask them in the simulation somehow. His contention is that the "x's aren't real because x's don't exist in real hardware". He therefore concludes that the x's in the simulation aren't real, and that any design changes based on these is not a good thing, or at lest, way too extreme of a response. My contention is that although it is undoubtedly true that x's don't exist in hardware, they represent unknown or unpredictable values. I believe the gate library models x's to propagate pessimistically. Therefore, if x's are propagating to kill the simulation, it suggests there could be a combination of bits that would cause a problem. Since that is a possibility, I don't see making a design change to clear the x's as being too extreme, even if I can't prove they are absolutely real. (I suppose I could try a search for the bad combination of bits, but that would be a lot of work.) Now, I can imagine an answer that the right approach depends on the type of quality that is being developed here. But, I think that the design changes I've suggested (adding in the resets or clearing them in the program) cost very little (especially modifying the ROM program). I think that going through the process of masking the bits in the simulation would be much more labor intensive. What could I be misunderstanding from his point-of-view? What is the best approach? Is it really so bad to make a design change for a problem you can't prove is absolutely real? <Q> I suppose it depends on where the x 's are in the design. <S> Take an example communication scheme within the chip. <S> You may want to pass data around between two components, but lets say not on every clock cycle. <S> You might decide then to have a data bus and a valid signal. <S> The valid signal says when the data is valid. <S> Because of this, whenever the valid signal is low, the value of the data signal doesn't matter (because it is ignored). <S> In this example as long as the valid signal is never a don't-care, the design will never do anything unexpected. <S> The data signal can be don't-care, it doesn't really matter, because you always know that there will be valid data (whatever it is) when the valid signal is high. <S> If the valid signal was ever don't-care, then you do have a problem. <S> Why? <S> because it means you have some scenario where you have no idea what will happen. <S> This may or may not cause an issue, but boy is it a nightmare to track down. <S> So, with that in mind, it is my opinion that: If you come across <S> x 's in a data bus, it isn't the end of the world <S> , in reality you don't know what the data will be at any given time anyway. <S> If they appear in control signals it may or may not be bad, you don't know. <S> In this situation you should run your simulation twice, once making sure that the don't care is forced to be 0, and a second time making sure it is a 1. <S> This way you know what will happen in both cases. <S> If you cannot be certain (i.e. from the code you have written) that a don't care wont cause issues, then you should not ignore it, <S> x is an unknown potential disaster. <S> Also, all control registers, whatever they are for, should be initialised to a value at reset. <S> An uninitialised control register at power on could be disastrous. <S> Imagine you make a control system for launching nuclear weapons and didn't initialise the 'launch' register to 0. <S> For all you know when you turn the power on you could start WW3. <A> There are many different causes for the X state in a simulation but unfortunately they are all represented by one state in Verilog. <S> VHDL has the U state to represent an uninitialized value. <S> That may not have a physical representation in hardware, but is certainly a real property of a hardware design in addition to the voltage level of signal. <S> Accurate propagation of the X state is a difficult in dynamic simulation for any RTL or netlist. <S> You may want to look at formal tools for this purpose - especially for reset checking. <S> They are well suited for this application for exhaustively proving all combinations. <S> As far as the cost associated with adding reset logic, designs are always at some sort of limit in their specification; power, area, features. <S> So you certainly don't want to add unnecessary logic to to make your net list simulation work. <S> Some simulators allow you to randomly initialize logic, so you can at least get some level of confidence before you try to mask out the X's. <A> If you're running code generated by a 'C' compiler (rather than raw assembler) on an ARM core, you're likely to have problems with netlist simulations unless you start by initialising the register bank. <S> Depending on how aggressive the compiler has been, you might need to initialise all state (and this then starts getting risky). <S> The most obvious problem is that flag-setting operations on in-initialised registers will move a fairly harmless data-path X into the control path in a way that isn't always harmless. <S> This isn't something that should happen in sensibly hand-written code, but the 'C' compiler is (I think) at liberty to re-order operations and sometimes do things where it will ignore the result. <S> The RTL will behave consistently if it's given arbitrary numeric input, but sometimes reacts badly if it's given X as a data value. <S> As soon as you have prefetch logic which is exposed to X data, it's not hard to imagine that more control logic can generate an X result instead of a fast/slow type of performance behaviour. <S> RTL is frequently written to be X-pessimistic, so any X values (that represent don't know values) will be passed through a design. <S> If your tests run without initialising the register bank to 0 in your init code, keep it like that. <S> You're more likely to detect a problem this way (like running off the code image into executing data). <S> Don't despair if you have to initialise some registers in software just to run on a netlist though. <S> If you find your netlist needs all flops initialised at reset, you need to check the core documentation (Integration manual) and follow up with support if necessary. <S> To support this view, look in the boot code provided with the execution testbench. <S> If there are some zero-initialisations there, you'll need them for netlist simulation (but probably not in real silicon).
| If you know that the value at any given point doesn't matter, then x is perfectly valid.
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analogue and digital ground connection I have been reading up on grounding and what is AGND and DGND i.e / analogue ground and digital ground respectively. Suppose , I have an ADC chip and a MCU in a PCB. Now, previously I would have a ground plane beneath them and split them into Analogue ground and digital ground. I have now realised that splitting the ground planes will be detrimental to my return path at higher frequencies of the signal. So, the best way is to have a single ground plane or have 2 different ground plane which is connected by a ferrite bead. Then during layout remove this bead to have a thin connection between the 2 grounds. But, this approach has an issue according to me, which might be wrong. Will this thin line not act as a sort of high impedance that blocks currents thereby making the 2 grounds at 2 different potentials ? <Q> There is no law about that. <S> If you had ten ADCs, under which would you connect the grounds? <S> So see how you can do the same with a single plane. <S> After all, current doesn't just flow anywhere, it needs a mesh. <A> The point of having separate grounds is to have the ground currents for each section confined to that section. <S> Most importantly, to keep high-speed digital transients out of sensitive analog circuits. <S> If you have done this successfully, the two grounds can be connected by a relatively high-impedance connection. <S> Since the current between the two sections will be very low, the voltage across the connection will also be low, per Ohm's Law. <S> Then the two sections will be at nearly the same voltage. <S> If you're not trying for extreme performance, it's often possible to simply place the digital circuits on one side of the board and analog on the other. <S> If the power connector is properly placed (more or less between the two sections) this will often be adequate. <S> Current is like water, and physically tends to take the shortest path. <A> Don't split your grounds. <S> Drop the word split altogether. <S> Instead, use partition. <S> Partition your grounds. <S> As the previous answers have stated, the purpose of this is to keep noisy digital currents away from the analog. <S> So you partition your board, and keep all digital traces in one area, and keep analog in another. <S> The use of ferrite beads on PCB to seperate grounds is a hack. <S> Dan Beeker (Freescale) and Rick Hartley have both both expressed that ferrites are for people who do not know what they are doing. <S> Be aware of where your currents are, and you wouldn't need to use any ferrites at all. <S> So keep a single plane, but partition your board such that analog signals are kept away from digital. <S> It helps alot if the IC has pinouts which make this easy. <S> There some good information from Henry Ott <S> http://www.hottconsultants.com/techtips/split-gnd-plane.html <S> If you look at the following image You can see that the board has been partitioned. <S> To minimize any potential leakage, or stray feilds, there are cuts in the ground plane. <S> This creates bridges that go from analog to digital grounds. <S> This is ok so long as the routing ensures that no trace passes over such gaps. <A> The whole AGND/DGND thing on ADC chips is a bit weird as it actually refers to the ON CHIP division and is done because the bond wire inductance is sufficient to cause crosstalk in modern converters if they are not separated. <S> Personally, I favour a single ground plane, no splits unless proven to be needed (they seldom are) with careful layout and especially component placement, for certain you must never pass a fast signal net across a split in a plane, that is just asking for trouble. <S> High frequency current is seldom an issue as the return current at high frequency will flow so as to minimise loop area (And thus stored energy in the magnetic field), this usually means that it will flow in the plane right underneath the relevant signal track (For this reason fast digital should be kept away from the analogue stuff). <S> This is a good reason to fit low value source termination resistors on any digital that has to approach the analogue bits, it helps with ground bounce apart from anything else (ADC output pins should always be source terminated (and ideally locally buffered)). <S> Now there is something to be said for being careful about where you take things down to the plane in the analogue section, in particular with single ended stages, it pays to consider where the current loop is (Hint, it involves the driving opamps decoupling caps), and return the reference of the next stage to that point. <S> 73 Dan.
| The reason to separate the ground planes is that you want to keep digital currents out of analog signals' way.
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