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Electricity not conducted by copper wire Im am quite stunned as to what might be wrong. electronics beginner I bought today some copper wire, 20SWG /0.9MM from maplin (UK), for a project I am working on. There is no power going through the wire. I got positive and negative of 9V battery going to breadboard, then two pieces of copper wire one in + one in -. No voltage on the multimeter. Exact same setup with jumper wires work.. Copper wires are inserted fully into the breadboard... What am I missing ? p.s Label says: 250g EN COPPER 20 SWG 0.9MM <Q> That kind of wire is meant for winding transformers, inductors, and electromagnets etc. <S> The enamel coating insulates the wire and stops a coil turning into a single lump of copper. <S> You need to remove the enamel from the ends of the wire, either with a small craft knife, or burn it away using a hot soldering iron and solder. <A> I used to remove enamel off copper wire (small: #34 guage) in a production environment by dipping in a Kester solder pot. <A> As a fun fact, my grandfather used to remove the insulation of these wires using a very special technique. <S> He'd get a a pill of aspirin or penicillin (don't remember which) and glue it to a piece of wood. <S> He would then take his soldering iron and while heating the pill, the wire would be drawn through the molten medicament. <S> This horrendous hot chemical compound was apparently aggressive enough to disintegrate the coating entirely. <S> The only downside was that it produced an enormous reek. <S> The intensity of this smell isn't comparable to anything I've ever sensed.
| The molten solder burned off enamel and tinned the wire for transformer leads very nicely. My guess is the "EN" of the code means "Enamelled" - I.e., it's coated with enamel.
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Reduction of DC motor noise I am designing a circuit with a DC motor 12V DC Reversible Gearhead Motors - 70RPM and some other stuff including a MCU and a LASER, all driven from a single 12V source and am concerned about big HF noise ripple from the motor (electrical rather than radiative but no harm in reducing both). I haven't worked much with motors much before, however from reading the articles in this community and some searching elsewhere on the internet, it seems there are a few techniques for dealing with this noise and I was wondering if I could get some educated response on the validity and drawbacks of a few of the techniques that I have encountered. Small capacitors (1 or 10nF) connected across the terminals in a variety of combinations including between Vcc/Gnd, two between Vcc/Gnd with the middle connected to the case exterior, and a combination of the above two. Non-polarised if the motor needs to run both ways. Directly grounding the case of the motor. Choke inductor in series with the Vcc of the motor. Employing a more complex filtering topology close to the motor. Twisting and shielding the cables of the motor and physically isolating them from the remainder of the circuit. Keeping the ground of the motor separate from the ground of the remainder of the circuit and connecting it directly to the terminals of the power source if possible (or as close as possible if not) to avoid ground loop problems (star grounding?) Enclosing the motor physically inside a metal case (and grounding that case). Using large (1000uF+), low ESR electrolytic capacitors connected as close as possible to other sensitive equipment between their Vcc and Gnd (Anode to Vcc, Cathode to Gnd), or placing these large capacitors next to the power source itself on all of the lines leading out. Running some of the other equipment through a linear regulator (Not sure if these are particularly good at rejecting HF noise) Placing diodes next to the power source for different lines leading to different systems. Looking for a generic answer regarding the effectiveness of the above techniques and perhaps more regarding protection from DC motor noise, not something specific to that motor as that project is actually over, now I am just curious and think it would be useful to have this information available in one spot for future projects and other interested people. <Q> The usual recommendation is to install two 0.1uF ceramic capacitors, one connected from each motor terminal to the case. <S> This 'grounds' the case to rf without the danger of having an exposed DC connection. <S> Ripple can be a problem for sensitive equipment which has poor power supply rejection, but normal filter capacitors and regulators will usually eliminate it. <S> Another concern is the current spike and voltage dip that occurs when starting the motor. <S> This motor has a stall current of only 390mA, so provided your 12V supply can handle that you shouldn't have to worry about it. <S> Just make sure that the motor and its control circuit is wired directly to the power source, and run separate wires to the other devices. <A> Regarding your points to reduce the noise: Small capacitors (1 or 10nF) <S> That's correct, except of the mention of capacitor polarity: anyways the capacitors must be ceramic, designed for working under high frequency, not electrolytic or paper even if motor will work only in one direction. <S> Place these capacitors as close as possible to motor, and to the motor driver if you are using a PWM driver. <S> Using large (1000uF+), low ESR electrolytic capacitors connected as close as possible to other sensitive equipment between their Vcc and Gnd (Anode to Vcc, Cathode to Gnd), or placing these large capacitors next to the power source itself on all of the lines leading out.. <S> Most likely using large capacitors will only be partially effective, primarily during start/stop/reverse of motor. <S> Your p.9 is exactly about this. <S> Twisting and shielding the cables of the motor and physically isolating them from the remainder of the circuit. <S> The major cause of noise transmitting by motor (through cables and by air) is the ignition of brushes. <S> So if your motor is not brand new please check the brushes and connector conditions, and grind in the connector if needed. <S> Also plan your wire topology as a star(with power supply in center) with rays (parts of your scheme) and try to avoid the making of chain contained of consumers. <S> Star: consumer 2 <S> <---wires--- <S> > <S> PowerSupply <---wires-- <S> > <S> consumer 1 Chain: <S> PowerSupply <---wires-- <S> > <S> consumer 1 <S> <---wires--- <S> > <S> consumer 2 <A> Your motor is relatively low current, so unless you have a good model of your motor, the best approach is experimental. <S> Leave room on your board for choke inductor. <S> Have small capacitor directly soldered on the motor terminal. <S> Have sufficient decoupling on the power lines that supply the motor driver. <S> Then try with a few values for the capacitors and for the choke inductor and measure the noise on the supplies with a scope (or a spectrum analyser if you have one). <A> Several points that somehow were not yet mentioned. <S> The whole power path from your bridge to the motor must be carefully shielded. <S> The shield, if possible, should be connected to motor's enclosure. <S> On your board side the shield must be connected through capacitors to the ground, and somewhere in the system it should be connected to the ground directly. <S> Near the bridge you should have input capacitors to provide power during the switching. <S> I mean small caps to react quickly and several large caps that will have enough energy to hold the voltage even during ling current peaks. <S> Normally it's hundreds of microfarads. <S> Use common mode chokes of better ready made filters on your power line to virtually only allow DC current. <S> All switching effects must be limited to the bridge and cable towards the motor. <S> Of course, good layout is very important. <S> Always think about where the current goes, reduc magnetic flux, remember that in switching system it's not just the motor current that is switched, it's alse bridge transistors gates, boost capacitors, and sometimes other stuff. <S> Good luck! <A> currently the Arduninos are very sensitive to this type of noise, the lcd are much more... <S> The main solution for this type of situations, is a; snubber they are wide recomended to be used in conjunction with inductance loads as motors. <S> Some articles with subber desings and calculators <A> Same here, I found that cheap drones are really bad and it is normally one motor which is the culprit. <S> The really strange thing is that its typically no weaker than the rest but adding ceramic capacitors helps a lot. <S> I found a physically larger part seems more reliable, ended up using the ones from old flat screen LED backlight PCBs. <S> Also relevant <S> , this helps with "mad drone disease" which is normally the RF interference problem.
| You should always put a capacitor across the motor terminals even if your circuit is not affected, because brush arcing creates rf noise that can interfere with other equipment (eg. AM radios). Better noise protection is - to make separate power supplies for power circuit and for control part even if both require the same 12V. Also ferrites around the cable are a good idea.
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What happens if you put the connector of a live charger in your mouth? So what exactly would happen if you put the connector of a plugged-in charger in your mouth? I am sure it is a bad idea but can it really hurt or even kill you? Some posts on social media led me to this question. <Q> This is not the same as putting the terminals of a battery across your tongue. <S> Modern wall warts/chargers are likely to have an open circuit voltage to earth (on both output conductors) of many tens (if not hundreds) of AC volts due to the way they are constructed. <S> Normally this isn't a problem because the impedance behind this voltage is quite high. <S> As a side note, this AC voltage is due to the way EMI filter components are used - at regular AC frequencies they are fairly high impedance (tens of kohm) but they will give a bigger than expected jolt on the tongue and this is not to be recommended. <A> It causes a very unpleasant, painful even, sensation if the adapter output voltage is high enough. <S> When my son was an infant one time he was sitting happily on the floor next to my desk chair while I pounded away at the keyboard on some kind of engineering stuff and briefly started to cry a couple of times. <S> Turned out he was picking up a loose end of an AC adapter and putting it in his mouth. <S> I tried it on myself to see what he had experienced (I think it was more than 12V- possibly 24V for an EPROM eraser). <S> Quite painful. <S> 5V would likely not be too bad judging from 9V batteries. <S> Andy has a valid comment that the Y capacitor reactance on an isolated may cause some sensation (I think it wold be slight, but I'm not going to try it) <S> even touching a single side depending on how the adapter is plugged in- <S> it would be more noticeable in a 240V country. <S> I certainly don't suggest doing this yourself, however it's unlikely to be anything but unpleasant, depending on voltage, if the adapter is working properly. <S> It could, however, be fatal if the charger were to be faulty <S> and there was was a path to earth through the body. <S> This is an increasing possibility with the proliferation of criminally bad overseas-made chargers, such as the fake 'Apple' ones that sell for $1 including airmail, with fake safety agency approvals and definitely dangerous internal construction. <S> A similar device is already known to have killed a woman. <A> https://www.youtube.com/watch?v=hp97GjuULX8 <A> It is non-linear, since it depends on the applied voltage, a higher voltage might result in a lower impedance! <S> As said, it is also time variant. <S> At first, when one puts a DC source on his body, electrical resistance is relatively high (a few mega-ohms), after a few moments ions in the blood stream approach the positive and negative poles. <S> This facilitates electrical conductance in the body. <S> This makes a positive feedback and more ions populate the pole sides and hence even more current passes through the body. <S> This all was about skin stimulation, however situation in mouth is a little bit different. <S> Impedance in mouth is not too high , since the saliva makes mouth surface a good conductor, and in addition, the mouth texture is made by membrane cells, which are more near to the blood stream (which thanks to ios floating, is a very good conductor). <S> The placement of DC poles are too important, if their in a setting such that a current path is made through the chest, the heart might be in severe danger. <S> Since the poles of the charger are too close to each other, I do not think it affect the heart. <S> Remember even if the charger is protected against high current draw, it is still to dangerous to humans! <S> Wiki Refrence
| In a nut shell, if the charger voltage is low and the exposure time is also short , no critical damage might occur but if the voltage is high and/or exposure time is long, severe damage might occur . I would expect that putting one or both terminals on your tongue would be slightly more shocking that the equivalent battery. It's far more of a danger than with dry hands because your skin resistance is less inside a wet mouth. After that, the body impedance lowers and more current passes through the body. I can say this for sure because I tried it. OK this is not really an answer, but the videos by "Electroboom" are a hoot. The impedance of body is non-linear and time variant.
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How does the resistance of the load affects a voltage divider? Every explanation I've seen so far about voltage dividers, only consider the two main resistors (that do the "dividing"), and pretty much ignore what happens after Vout . The circuits in the explanations look more of less like this: In this example the voltage at Vout should be 2.5V. This really gives the impression, that whatever goes on for the rest of the circuit, Vout is at 2.5V. But what happens in practice, if I get something connected to Vout ? E.g, I connect some components, and their equivalent resistance is 100Ω: Since R2 and R4 are in parallel, an equivalent resistor would be of 50Ω. Now, if that's correct, then the voltage divider really operates between a 100Ω and a 50Ω resistor. I.e. the voltage at Vout would be 1.67V and not 2.5V, as I originally wanted it to be. Is this really how it works? How should one design the part that connect to Vout to avoid messing up the divider? <Q> Yes this is how it works. <S> The load needs to have very high impedance so that it doesn't affect the voltage division. <S> An OP buffer could be used between R2 and R4. <S> Perhaps using a zener diode as a voltage divider could be of use here. <S> This picture is from the site http://www.electronics-tutorials.ws/diode/diode_7.html <S> which explains this further. <S> Depending on which zener diode you're using the voltage, Vz, will (under the correct circumstances) be the same independent on the load, Rl, connected to the circuit. <A> Voltage dividers should only be used to provide a reference voltage to a high-impedance load. <S> They are very inefficient when used to provide a reduced voltage to a low-impedance load since the currrent through the divider must be ten times the current delivered to the load to hold the load voltage within reasonable limits. <A> You are quite correct and, when taking your load into account, the top resistor R1 would need to reduce to 50 ohms to accommodate the extra load you have or, use a voltage reference chip that provides a stable 2.5 volts up to several milli amps being drawn from its output. <S> Alternatively use an op-amp to buffer the output of the divider - this is basically how a voltage reference works. <S> The same applies to capacitive dividers when trying to create an AC voltage ratio that is precise. <S> It gets more complex when resistor loads are added to capacitive dividers though. <A> How should one design the part that connect to Vout to avoid messing up the divider? <S> A simple voltage follower: simulate this circuit – Schematic created using CircuitLab Q1 conducts Vin to load while Vload < (Vdivide - VbeQ1) Edit: <S> As Andy_aka pointed out, Vload <= <S> Vdivide - VbeQ1 (~0.7V)
| Yes, that is really how it works - the load does appear as a resistor in parallel with the bottom voltage divider resistor.
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Forward biasing a diode past built-in voltage Why can't we forward biase a diode past its built-in voltage/potential? What's the problem in doing so? <Q> Eventually this exceeds the thermal capacity of the diode. <S> This kills the diode. <S> $$I = <S> I_s\Big(e^{\frac{V_D}{nV_T}}-1\Big)$$ <S> where I is the diode current,IS is the reverse bias saturation current (or scale current),VD <S> is the voltage across the diode,VT is the thermal voltage, andn is the ideality factor <A> I like to liken a diode to the wall of a dam. <S> You can fill up the water behind the wall up to the height of the dam, but once you reach that level (the forward voltage) <S> the water flows over the top <S> so it stays at that level. <S> Current literally flows :) <S> The only way to get the water higher than the level of the dam is to flood the land beyond the dam also to the same level, in which case there won't actually be any current flow since there's no difference in levels. <S> This site has a good description of it: http://www.talkingelectronics.com/projects/Diode%20-%20How%20A%20Diode%20Works/How%20A%20Diode%20Works.html <A> Why can't we forward bias a diode past its built-in voltage/potential? <S> What's the problem in doing so? <S> No problems doing so at all - precision op-amp rectifiers do this to obtain what is virtually the rectification properties of an ideal diode: - Here the op-amp's output finds whatever level is necessary to forward bias the diode sufficiently to obtain signal rectification of Vi. <A> Take a look at a typical rectifier - the 1N4007 ( datasheet ). <S> Look at figure 2, which shows voltage vs. current. <S> Consider how the current increases with voltage. <S> And let's also calculate the power which the diode dissipates for each case (which is just the product of voltage times current). <S> $$\begin{array}{c|c|c}\text{volts} & \text{amps} & \text{watts <S> } \\ \hline0.7 & 0.01 & 0.007 \\0.8 & 0.1 & 0.08 \\0.9 & 0.7 & 0.63 \\1.0 & 2.0 & 2.0 \\1.1 & 3.5 & 3.85 \\1.2 & 5.5 & 6.6 \\1.3 & 7.5 & 9.75 \\1.4 & 10 & 14 <S> \\\end{array}$$ <S> You'll notice that you can indeed get more voltage than the diode voltage ( <S> ~0.7 volts) <S> but it takes more and more current to get even fairly small increases in voltage. <S> Plus, of course, the diode can only handle a certain amount of power before it bursts into flames (or better yet, explodes). <S> Since the 1N4007 used for this example is rated for 1 watt continuous, trying to get a continuous voltage of more than about 0.9 volts will cause it to fail.
| As you increase the voltage across the forward-biased diode, the current increases exponentially , as does power dissipation in the diode.
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How to inverse behaviour of a switch? I have a sustain pedal (which is just a momentary switch!) with the wrong polarity (according to the device I'm using it with) : circuit is CLOSED when pedal is pressed circuit is OPEN when pedal is released I would like the opposite. Is there an electronical solution to invert the behaviour of such a switch, with only passive components (without having to use a battery-powered component, etc.) ? <Q> simulate this circuit – Schematic created using CircuitLab <S> This might work. <S> Before trying this, check the equipment it needs to work with and measure the voltage across the open switch. <S> If you see 3 to 5V, then the chance of this working increases. <S> Otherwise, it may not be worth the trouble to try. <S> Connect the plus side to the plus of the diagram. <S> When SW1 is open, R1 conducts, Q1 turns on, voltage between <S> + and - becomes around 0.7V (if current from equipment not too high). <S> There is a reasonable chance that the equipment would interpret 0.7V as switch closed. <S> When SW1 is closed, base of Q1 is grounded, some small amount of current drains through R1. <S> If the source resistance is small compare to R1, the voltage is high. <S> May need to experiment with different value of R1 but most likely something around 10K to 100K. <A> That type of switch is normally open only. <S> It works by pressing a conductive pad across the exposed PCB traces. <S> There is no passive way that I know of to allow current when the controlling device blocks it and then block it when the controlling device allows it. <S> At best, you'd end up allowing current all the time or not at all. <S> If you're okay with potentially breaking the one you have, you might try a mechanical solution. <S> Somehow, press the pad against the PCB using something other than the pedal, and use the pedal to release it. <S> Good luck! <A> The simplest method is to use a relay!Connect your switch to the relays coil pins and circuit that you want to control to the pins that are normally open. <S> Thats it! <A> Almost all keyboards have this behaviour. <A> I was thinking a bit more about this problem. <S> There is no need for arduino/uC/external supplies and so on. <S> The solution is as simple as it can be: <S> You need to get SPST switch which is NC (normally closed). <S> You will need to resolder it in the place of the original one. <S> Here are some SPST NC which I found from my quick search. <S> Make sure it can get back to its original posistion <S> and you shall have no other problems. <S> Some images: <S> You have this kind of switch (call it button). <S> It is normally OPEN. <S> When the pedal is pressed it gets CLOSED. <S> Instead, for your needs, you need button which is normally CLOSED, then when switched it gets normally OPEN. <S> I hope this is the simplest solution available. <S> Get the proper button and there you have it.
| If you hold the pedal down while Turning the device on it will switch the polarity.
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Square wave into emitter follower I put a square wave of relatively low rise and fall time (of about 10ns ), and wanted to see the response from an emitter follower output. I put together the circuit and here's what I see on scope. I could not explain why the fall times differ so much, and the fall time of the wave at the output increases with the emitter register. In spice the waveforms look entirely different. The spice model is quite simply showing the discharging of the \$C_{BE}\$ capacitor, with the downward spike across R1 as the direction of current flows reverses while discharging. So what is actually going on ? The \$C_{BE}\$ capacitor should have discharged the way shown in the spice, but it does not in reality. PS: I have done the circuit in Veroboard. PS: Here are some Ft values with different resisters. R1 Rise time Fall Time220 ~10ns ~20ns1k ~10ns ~57ns10k ~10ns ~522ns <Q> This is as expected. <S> On the rising edge, the transistor is actively providing current to charge up the inevitable parasitic capacitance across the emitter resistor. <S> On the falling edge, there are two effects that slow down the edge. <S> First, the transistor is just off. <S> It's not actively removing charge from the emitter node, as it was doing in the inverse on the rising edge. <S> The voltage on the parasitic capacitor only discharges thru the resistor exponentially. <S> Second, it takes a little bit of time for the charge carriers to be swept out of the base region of the transistor, so for a short time it is still conducting. <S> Actually this effect is minimized by the transistor not being saturated. <S> To put some numbers on this, let's see how things work out with 3 pF of parasitic capacitance across the resistor. <S> (1 kΩ)(3 pF)= 3 <S> ns. <S> The actual parasitic capacitance can vary significantly by type of resistor and build technique, so we don't really know what it is. <S> Still, this shows at least that observing this effect on your time scale is plausible. <A> Looks like you've got about 20-25pF from the scope probe. <S> Try a 10:1 setting. <A> I suppose that line (1) is the input and line (2) is the output of your follower. <S> What happens is that in the first case (rise), the transistor allows for a large current to flow through C-E and therefore you have a rapid rise. <S> In the second case (fall), the current is limited by the resistor and you have a simple exponential decay (and the B-E coupling).
| Unless you have a slow transistor, I'd say the dominating effect is the exponential nature of the voltage decay.
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Input and output resistance of an opamp circuit Recently we have been working with opamps in the lab and I never can understand how to determin the output and input impedance of a circuit with op-amps (not the input and output inpedance of the circuit itself) We have been measuring it by placing a voltage signal in the output of an inverting amplifier with its imputs grounded. Measuring the voltage over an impedance we have found the current flowing in simulate this circuit – Schematic created using CircuitLab Then the output impedance is found to be the voltage at the output (Vout) divided by the current. I would like to know what the input and output resistances in opamp circuits represent and if possible how to obtain them from equations <Q> I would like to know what the input and output resistances in opamp circuits represent and if possible how to obtain them from equations <S> I assume you are asking for "theortical" formulas, right? <S> OK - from system theory <S> we derive the following expressions for the whole circuit; all opamp resistances without feedback will be drastically altered due to feedback (Loop Gain LG): 1.) <S> Non-inv. <S> input: r,p=rp,o*(1+LG) <S> 2.) <S> inv. <S> input: r <S> ,i=rn,o/(1+LG) <S> 3.) <S> input R2: <S> r,2 <S> =R2+r, <S> i (very close to R2) 4.) <S> output: r,out= <S> r,o/(1+LG) <S> rp,o and rn,o: <S> dynamic input resistances without feedback; r, <S> o: dyn. <S> output resistance without feedback; Loop Gain: <S> LG=Aol*feedback factor=AoR2/(R1+R2). <S> PS: <S> Your measurement of r,out is OK (in principle). <S> However, I suggest (a) to use a smaller external resitor (R3=1...5 Ohms) to realize a suitable voltage divider (together with r,out) or (b) to use a much larger resistor for realizing a "good" current source (current practically determined by R3 only). <A> As you may have covered in class (or maybe not yet), the three rules for (ideal) opamps are: <S> The inputs take no current. <S> The output voltage travels in the direction of (+in minus -in). <S> If +in is bigger, the output increases; if -in is bigger, the output decreases. <S> The output will go as far as it needs to to make the two inputs equal. <S> (Note: for real opamps, it cannot exceed the power supply and so will stop there) <S> Now, you've tied +in to ground, so it's going to do whatever is necessary to hold -in also at ground. <S> The two resistors R1 and R2 make a sort of lever, if you will, because their center tap takes no current. <S> By tying one end of that to ground, the opamp trying to keep the center tap also at ground must therefore keep its output at ground. <S> So now you can replace the entire opamp circuit with ground and see what you get. <S> If you're studying non-idealities, like the finite output impedance of a real opamp, then R3 is physically inside of the opamp chip. <S> Not completely literally, but it behaves that way, because of actual resistance in the silicon die and the wires that connect it to the pins, and because of the overload protection that's built into even the dirt cheap ones. <S> However, by including that non-ideality into the feedback (consider what would happen if you moved your right-most vertical wire to the right of R3), you can all but eliminate that effect at the expense of some headroom. <S> (the output will clip easier because it's swinging farther to compensate for the weirdness) <A> In your case, assuming the input is on the negative side of the opamp, the input current is \$V_{in}/R_2\$ (because of the virtual ground on the negative terminal). <S> So the input resistance would be \$R_2\$. If \$V_{in}\$ is applied to the positive side, the input resistance will be close to infinity, since there is no input current. <S> As for the output resistance, it can be obtained by connecting a known load \$R_L\$, measuring the voltage on it \$V_L\$, and then calculate the simple voltage divider problem:\$V_L= <S> V_{out}R_L/(R_L+R_o)\$, where \$R_o\$ is the output resistance, and \$V_{out}\$ is calculated as for ideal opamp. <A> For the model of an ideal opamp with negative feedback, the output impedance of the circuit is zero. <S> The next more realistic model (for the inverting amplifier) would be$$R_{out\_of\_circuit} = <S> R_{o\_of\_opamp <S> } \frac{A_{CL\_CloseLoopGain}}{A_{O\_OpAmpOpenLoopGain}}$$ <S> Which in the ideal case, \$A_O=\infty\$, it reverts back to the model on top. <S> I hope this helps.
| The input resistance is the ratio between the change in the input voltage and the input current (or just between the input voltage and current in case of linear systems).
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Why can't I get a output lower than 0V in this diode and capacitor circuit? In the image above green is my input (node left side capacitor) and blue is my output (node right side of the capacitor).I would expect that the blue signal would follow the input, but it does not do this. Why is this the case? Can you explain this with open circuit and short circuit equivalents for the diode? <Q> Here's a step by step list of what happens: <S> The input starts positive. <S> The diode is reverse biased, so no current flows and the capacitor doesn't charge. <S> Around \$t = 0.5s\$, the input becomes negative. <S> The diode is forward biased, so the capacitor charges to \$1 <S> V\$ <S> (the negative of the minimum voltage applied). <S> Note that there is no resistor, so there is no time constant - the capactor acts like an open circuit (as if the source is DC). <S> After the capacitor finishes charging, there is an offset of \$+1 V\$, and the diode never becomes forward biased again. <S> The output is \$V_{in} + 1V\$. <A> Kynit's answer is correct, but I think what makes this circuit confusing is the location of the ground node. <S> Let's pick a different node and see what happens. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Here's a simulation. <S> Download the image or open it in a separate tab to see the full-size version. <S> V(Old_right_node) <S> - V(Old_ground) is your output voltage. <S> V(Old_ground) is the negative of your input voltage. <S> Now you can see what's going on. <S> For the first half-cycle, the diode is off. <S> The capacitor is uncharged, so the voltage across it is zero. <S> This causes your input to follow your output. <S> During the second half-cycle, the diode conducts, charging the capacitor to 1V. <S> There's no voltage across the diode, so your output is clamped to your ground. <S> Once the capacitor is charged, the diode never turns on again. <S> V1 can't produce more than 1V, so V(Old_ground) can never be greater than V(Old_right_node). <S> The voltage between your output and input is equal to the capacitor voltage, which is a constant 1V. <S> In your circuit, V1 varies your input voltage. <S> In my circuit, V1 varies your ground voltage. <S> Either way, the result is the same. <S> It's just a question of which reference point you use. <A> That diode model you're using is totally unrealistic. <S> You should get a shifted sine wave that goes negative by somewhat less than a volt and <S> (very slowly) rises as time goes on. <S> Something like this (1N4004 model): <S> Eventually (after a very long time) it will resemble your curve, but it will take a very long time because the diode does not conduct much current at lower voltages and it has to charge a 1 Farad capacitor. <A> The circuit given in the question is a positive clamper. <S> Figure below shows a diode clamper that inserts a positive dc level in the output waveform. <S> The operation of this circuit can be seen by considering the first negative half-cycle of the input voltage. <S> When the input voltage initially goes negative, the diode is forward biased, allowingthe capacitor to charge to near peak of the input(\$V_p(in) - 0.7 V\$), as shown in figure. <S> (It can take many negative cycles to get the capacitor charged to the peak. <S> It depends on the value of capacitor, forward resistance of diode and amplitude of voltage source.) <S> This is because the cathode is held near \$V_p(in) <S> - 0.7 V\$ by the charge on the capacitor. <S> Assume it is \$R_L = <S> \infty\$ and hence capacitor does not have a discharge path. <S> So once the capacitor is charged to peak, the diode is reverse biased and The capacitor voltage acts essentially as a battery in series with the input voltage <S> The dc voltage of the capacitor adds to the input voltage by superposition as in figure (b). <S> So the output voltage will be $$V_o = V_{in} + V_c = V_{in} + V_p(in) - 0.7 $$ <S> That is why you are not getting negative voltages.
| Just after the capacitor getting charged to negative peak, the diode is reverse biased.
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Why must resistors be on the respective anode terminals instead of the common cathode terminal of a RGB LED? Different from the best answer of this one: Why does a resistor need to be on the anode of an LED? But well, I'm also an electronics noob ^.^ Recently I bought the official Arduino starter kit and played around with the common cathode RGB LED that came with this kit. In the project book, the instructions for the colour mixing lamp is to use separate 220ohm resistor for each of the RGB legs. But I thought, why can't I put the resistor on just the cathode of the LED? I tried that by plugging the LED into the breadboard and connecting the cathode through a 220ohm resistor to the ground of Arduino UNO. When I touched any one of the R, G or B anode legs with a jumper wire connected to +5V, the respective colour lights up alright. However, when I bridge the G and B pins together, only the green LED lights up. When I bridge all three anodes together, only the red LED lights up. But if I connect the anodes to +5V through their separate 220ohm resistors(according to the instruction), the colours will combine. Why is it so? <Q> The red led will hog all the current because it might only need 2 volts across it to begin conducting. <S> The green and blue LEDs need a higher voltage but because they are all in parallel the red led dominates. <S> Try measuring their respective volt drops when each operates. <S> It's like putting a 5 volt zener in parallel with a 10 volt zener. <S> The 10 volt zener will never conduct. <A> However, an RGB LED is a rather different animal since one side of all of the elements are tied together. <S> This presents a bit of a problem. <S> If you wire the common terminal directly to a power rail and put 3 resistors on the other side, it will work as expected. <S> However, if you a stick one resistor pin on the common terminal, you will have some issues. <S> If you only try to turn on one element at a time, it will work correctly. <S> However, if you try to turn on more than one, then the LEDs will be in parallel and they will behave in an unexpected manner. <S> If the forward voltages are the same, then the LEDs will split the current and light up at half brightness, approximately (not exactly because the current will never quite split exactly). <S> If the forward voltages are different, then only the LED with the lowest forward voltage will light up as it will turn on and steal all of the current before the other LEDs will exit cutoff. <S> Bottom line: don't put LEDs in parallel as they will not evenly share current. <A> The answer is simple: Its because if you have only one resitor and three leds, you have a "variable" load for a fixed value resistor. <S> If red and blue leds are turned on, this means a load of a value. <S> If green and blue are lit, this means a load of another value, and so on. <S> You cannot match all possible states with a single resistor, so you place three on the other side, allowing each resistor to deal with a single led. <S> If, for any porpuse, you would use the leds in a fixed load state, ie, keep the same leds lits forever, them a single resistor would be enough. <A> To put everything in the answer and comments together: First, RGB LEDs are just three separate LEDs packaged into the same package, with their anode or cathode tied together. <S> The other end is also tied together if only a single resistor is used. <S> So that is three diodes in parallel. <S> Second, the voltage each LED need to begin conducting is called the forward voltage, usually noted as Vf. <S> In the ideal diode model, a diode does not conduct when the voltage applied to its terminals <S> is lower than Vf or when it is reverse biased (negative to anode and positive to cathode), and conducts like a short circuit when the voltage is higher than that, while keeping the voltage between its anode and cathode constant. <S> so it's no longer a short circuit. <S> Else the LED will burn out. <S> Or to think about it in another way, it doesn't add up for a 5V source and a 1.8V voltage source to be shorted together. <S> Either the LED or the source will burn out so the voltage settles. <S> The LED will usually be the one to burn out first as it usually tolerates a lower current. <S> In the example mentioned in the question, taking the following Vf for different LEDs: Vfred = 1.8VVfgreen = 2.1VVfblue = 3.0V When applying 5V to the anode of all three, the red LED will conduct first, so the cathode would be fixed at 1.8V below 5V. <S> The voltage difference between the other LEDs' anode and cathode will also be 1.8V since they don't get to conduct. <S> As a result, they are as good as not in the circuit at all.
| For a single LED you are correct - which side you put it on does not matter. The resistor is added to limit the current through the LED
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Device which will turn off my water motor automatically after the tanks are full I have to build a system which automatically turns off my water motor after the tanks are full. I already have build up a system which lights up a bulb and an electric bell and tells me that the tanks are full. Now I want the water motor to be switched off automatically after the tanks are full. Is there some a device I can make which will turn off the water motor automatically after filling the water tanks. <Q> There is no need for fancy electronics here. <S> Look for something called a float switch . <S> Various types are readily available in hardware stores. <S> I have one of these turning on a sump pump in a bucket that my basement sink drains into. <S> In this case I have a float (a small re-purposed pickle jar <S> if I remember right) connected to a threaded vertical shaft. <S> That shaft has two pair of opposing nuts on it, which are the stops to activate the float switch to click between on and off. <S> I can adjust the water level for on and off by adjusting where the nuts are on the threaded shaft. <S> In my case, the shaft going higher turns on the switch since I'm trying to empty the tank, not fill it. <S> However, I could have mounted the switch upside down to get the effect you want. <S> Most switches are either mechanically symmetric so that you can mount them either way, or are available in normally open and normally closed options. <A> If the tank is a sealed system use a pump/pressure switch in the pipe line, these are common and commercially made for household water systems. <S> They include a mechanical relay which is controlled by the water pressure high & low points. <S> Example: <S> http://www.zoro.com/i/G2713557/?utm_source=google_shopping&utm_medium=cpc&utm_campaign=Google_Shopping_Feed&gclid=COaXi-_c38MCFVFp7AodnlgAgw <S> If the system is open then use the float device switch as described by the other replies. <S> Example: http://www.zoro.com/g/Float%20Switch/00086553/ <A> Find this controller on Amazon: <S> RMG Automatic Water Level Controller for Motor Pump Operated by Switch/MCB upto 1.5 HP - Tank only Also for water motor comparison check the Industrial matrix website: Industrial Matrix: <S> Electric Motor Catalog . <A> I have also used the RMg Automation water level controllers and such devices work perfectly and are reliable for long operation times: I have used it, say for 3 years without no issues.
| There are many water level controller devices available in the market that can automatically turn off motor when the tank is fully filled with water. I have used Float switches but I found them to be a bit unreliable.
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How to generate Assembly drawing in Altium How to generate Assembly drawing with designators inside the boxes like this in Altium? Is it by placing designators (silkscreens) inside manually? I guess not. <Q> It can be somewhat automated, but it is tedious to do the first time, because you have to modify all your footprints <S> What I do is Rename two mechanical layers as "Top Assembly Dwg" and "Bottom Assembly Dwg". <S> Desginate these two layers as a mechanical layer pair . <S> Now anything on "Top Assembly Dwg" will get flipped to "Bottom Assembly Dwg" when you move the part to the bottom side of the board. <S> On each footprint, add the text ".Designator" on the Top Assembly Dwg layer. <S> Place this text centered on the part, and sized so that your maximum length designator (maybe 4 characters) will fit within the boundaries of the part. <S> You might also want to copy some silkscreen features onto this layer because you won't want to make prints that contain both the Assembly Dwg layer and the silkscreen layer. <S> Make a PDF output that prints each of the Assemlby Dwg layers, along with whatever other layers are appropriate to make your assembly drawings. <A> There are some helpful Altium scripts for centering and sizing ref designators that can be used for your purposes. <S> Lots of useful scripts live in here <S> (this used to be a project on google code): <S> https://github.com/Altium-Designer-addons/scripts-libraries <S> These two can center and resize ref designators according to the component outline size: <S> AdjustDesignators AdjustDesignators2 <S> And this one can move ref designators onto designated mechanical layer pairs: CopyDesignatorsToMechLayerPair <S> These are also a good starting point to write your own script if you want to customize the behavior further. <A> Altium 16.1 includes "Draftsman" documents, and there is a view that does this automatically. <S> Before this version, I used Photon's answer to create a similar drawing, and while it did take a while, the results were very useful.
| Altium will automatically replace the ".Designator" text with the actual designator when you generate output.
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Why are watts used as a measurement for some appliances? Why do some appliances tout watts and some others advertise volts. I know watts is a measurement of volts*current but looking at just watts alone, what does that tell us about the appliance? Can't volts be low and current be high to get the same amount of power in watts? Why is the watts measurement necessary is what I'm really trying to ask. Why not just provide volts and current? Here's a drill advertising volts . Here's a juicer advertising watts . <Q> There are two principal AC mains household voltages in use in the world today, (120 and 240 volts) and the appliance's voltage rating is given in order to indicate what input voltage the appliance is rated to support. <S> Watts is specified since that's the work that's being done by the appliance when it's ON, and that's what your power company is charging you for. <S> EDIT/ <S> UPDATE: <S> In the examples you've shown, the juicer ad is non-deceptive in that it discloses how much power it uses and, knowing that and what your power company charges for energy, you can determine its run cost. <S> For example, if your power company charges 10 cents per kilowatt hour of energy delivered to your home and your juicer uses 1 kilowatt, (1000 watts) then for every hour <S> it's ON <S> it'll use one kilowatt-hour of energy <S> and you'll be charged 10 cents for that. <S> Not a bad deal, and to look at it another way, if it takes you six minutes to juice some fruit and your juicer is running balls-to-the-wall <S> , that'll cost you a penny. <S> On the other hand, there's nothing to be gleaned from the Sears ad since all that's mentioned <S> is that the battery is an 18 volt NiCd pack, and that it has a capacity of 1.1 ampere-hours, which means nothing if the running current, under load, isn't specified. <S> It isn't, and even though the magical power number could be backed into knowing torque and RPM, the torque is specified, at 1700 RPM, as zero inch-pounds, shutting all the detective work down. <S> There may be more exhaustive data at DeWalt's site <S> re. <S> the performance of their drill. <A> For example, two 2000 W hair dryer will produce the same heat, even though one might be American and run from 120 V and the other European and run from 240 V. <S> The same is true of a microwave oven, a light bulb, a toaster, and a motor. <S> Motor are often rated in horsepower, but that just a different unit of power. <S> 1 horsepower = 746 Watts. <S> Some appliances nowadays have a universal power input. <S> That means they can run from around 90 to 260 V and 50 to 60 Hz, which makes them compatible with normal line power anywhere in the world. <S> The amps varies inversely with the volts, but the watts remain mostly constant. <A> Watts*hour is what you are paying for to the electricity supplier. <S> It is a power, or energy per unit of time consumed by the appliance. <S> Given your device is plugged into 110VAC or 220VAC wall outlet, you can derive the current consumed and estimate whether the outlet/entire home wiring is capable of powering it. <S> If you see the voltage rating on the appliance, it is likely to indicate whether or not it will work with 110V or 220V. <S> If some other voltages are stated, it is most likely DC voltages, for operating with some kind of AC/DC adapter or even battery, and the purpose of it to prevent you from plugging the wrong adapter in. <A> It's true that you can calculate wattage from voltage and current - the wattage given is redundant. <S> For example, if I see a 20 V, 4.5 A laptop charger, I know that it's a 90 W charger, no matter whether it says that on the box. <S> The main reason why manufacturers give wattages is because not everybody knows this. <S> Some people have a concept of how much power 90 W is and <S> how much current 4.5 A is, but don't know how to calculate one from the other. <S> Labeling products with both makes it easier for the average joe to figure out how much power this thing needs. <A> A very practical explanation for the outlier (the drill) is because the drill uses a battery pack (see the black handle in the bottom of the picture in your original link). <S> The juicer runs off of mains (AC), so (for the USA) <S> the voltage is assumed to be 120 VAC, so wattage is specified such that current can be determined if you will overload the breaker for that line. <A> Voltage and Amperage markings are required per safety labeling (on the appliance). <S> However for advertising proposes just about anything can be used to attract the customer. <S> For eg: You may want to conserve power so you buy the lowest wattage device, or you want the strongest car battery <S> so you look for the highest cold cranking Amp rating. <A> There are a lot of devices, where watts is the output, and voltage and amps is the input. <S> For example my microwave states it is a 900W one, but it says that it draws a max of 6A. <S> This can be especially true for all kinds of power supplies where the DC wattage is specified, but the AC power draw is higher, due to losses. <S> Also on quite some devices, wattage is a pure marketing number, giving some kind of overall power, whereas the momentary amps it draws can lead to much higher values for short times. <S> This can be good to know when you don't want to trip your breaker...
| Watts are specified when the device's power usage or power output is important to you.
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Digital circuit for adding multiple frequencies The goal is to implement a kind of a digital piano on an FPGA. Until now, I managed to produce individual notes using counters. Basically, I generate a square wave of a certain frequency, that corresponds to a musical note (of course, in order to do that, I rely on the frequency of the clock, but that's not an issue). How can I design a circuit that can produce the overlapping of multiple notes? Basically, I want to be able to produce 2 sounds of different frequencies at the same time. <Q> When I was a teenager in the 1980's, I had an Apple II+ computer, which had a speaker output similar to the IBM PC speaker. <S> Accessing a memory-mapped I/ <S> O port location had the effect of toggling an output connected to the speaker. <S> I wrote a machine-language program which played triads: three simultaneous notes. <S> A look-up table converted note values (indexed by semitone) into counter values. <S> The approach was simple: initialize three counters and cycle them independently in a big loop. <S> On any iteration when at least one of the counters has rolled around, toggle the speaker. <S> This routine, coupled with a BASIC program to drive it with data, played a very good sounding rendition of the synthesizer chord progression from Van Halen's Jump . <S> It's a unique sound. <S> There is distortion, but the note separation is quite clear. <S> Variations in timbre can be produced, by the way, with duty cycle variation. <S> I seem to recall that this translates to perceived volume. <A> It depends a lot on the type of hardware you have. <S> If you only have a 1 bit output on the FPGA, you can oversample the audio signal (1MHz or more) and use a PWM or delta-sigma modulator, effectively generating a variable amplitude signal (you may add a low pass filter, a simple RC circuit is sufficient). <S> You can then add your synthesized signals (sine, square, triangle...) and generate polyphonic sounds. <S> For example, with a 8bits (signed) adder, set the amplitude of the two square wave generators to +50/-50. <A> One simple way is to take each output of the counters, pass it through a resistor, then connect the other end of the resistors together. <S> Now the individual counter outputs would be attenuated while the final output would be the sum of the counter outputs. <A> Two ways (that I can think of): <S> With a DAC Multiply signals individually to get different individual amplitudes. <S> Sum the signals you want to combine Output n bits to the n-bit DAC <S> You can make a simple R2R DAC yourself using only resistors. <S> Without a DAC, using a single pin and a low-pass filter Add the two signals Modulate <S> the output using delta-sigma modulation Output through a lowpass filter (RC is simplest). <A> This produces an equivalent result to Kaz's toggling approach.
| If you're only dealing with a single, digital output, and generating square waves, the simplest way to do this is to have two (or n) counters at each desired frequency, and XOR their outputs together before outputting them to the pin.
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Connecting two power supplies with same voltage rating to increase current output? My h bridge requires 5v 1 amp. I have a 5v 500mA and another 5v 850mA power supply lying around. Can I join the +ve of both and send that signal to the h brdige so that it can have enough current of 1 A ? <Q> Possibly. <S> Possibly not. <S> It depends on whether your power supplies have linear or foldback current limit. <S> The problem is that the power supplies don't have exactly the same output voltage. <S> The power supply with the highest output voltage will provide all the current that the circuit is requiring until it reaches <S> it's current limit. <S> What happens then depends on the current limit circuit inside the power supply. <S> If it has linear current limit, its' voltage collapses down to the point where the other power supply begins to supply current. <S> All is good. <S> However, if that first power supply has foldback current limit, it will effectively drop out of circuit and the other power supply will attempt to supply all of the current. <S> It, too, will go into current limit. <S> And the output voltage will fall. <S> You have to test your particular power supplies and see how they behave while in current limit. <A> Looking at this from a mathematical standpoint we can explore this concept by looking at the equivalent circuit, specifically the Thevenin equivalent to start. <S> Below is the circuit you are suggesting to build (including the output impedance of each source): Rearranging things a bit you will agree that the following the same circuit: <S> We we have is a simple potential divider between Vy and Vx . <S> The process of finding the Thevenin equivalent is simple. <S> First, find the open-circuit voltage and then the short-circuit current . <S> The former can be found by disconnecting the load and calculating the voltage. <S> Since we know this is a potential divider circuit we have: $$V_{open-circuit} = <S> V_x <S> + (V_y <S> - V_x)\frac{R_x}{R_x + R_y}$$ <S> Then we find the short-circuit current . <S> To do this, simply short the output node to ground and determine the drawn current. <S> In this case, the current is simply that which is drawn from each source independently. <S> $$I_{short-circuit} = I_{V_y} + I_{V_x} = \frac{V_y}{R_y} + \frac{V_x}{R_x}$$ <S> And then, by the Thevenin Theorem, the equivalent circuit has voltage source with magnitude equivalent to the open-circuit voltage and an output impedance equal to the open-circuit voltage over the short-circuit current. <S> $$V_{open-circuit} = V_{TH} <S> = <S> V_x <S> + (V_y <S> - V_x)\frac{R_x}{R_x + R_y}$$$$R_{TH} = \frac{V_{open-circuit}}{I_{short-circuit}} = <S> \frac{V_x + ( <S> V_y - V_x)\frac{R_x}{R_x + R_y}}{\frac{V_y}{R_y} + \frac{V_x}{R_x}}$$ Therefore, mathematically this is acceptable, but will it yield the results you intend for this combination to provide? <A> Generally speaking this won't work, unless the power supplies are specifically designed for such operation. <S> Some bench or stackable system power supplies have a tracking/cascading mode for this purpose (Share a single regulation loop). <S> a) <S> If the power supplies are able to sink current, the following scenario may occur: The internal voltage regulator always has a certain regulating margin/accuracy. <S> Imagine for a moment that this is +/- <S> 0.2%. <S> For a 5V (4.99 to 5.01V) supply this would mean a potential voltage difference of 0.02V over a very low wiring resistance. <S> At for example 0.001Ω wire resistance this would result in a 'circular' current of 20A!!! <S> ! <S> between both units. <S> b) <S> If the power supplies don't sink current <S> the excessive "circular' current flow won't occur, however mutual influencing of internal voltage regulation loops may still result in unexpected behaviour. <S> Work arounds: 1) <S> However this solution has disadvantages: A voltage drop (typ. <S> 0,7- 1V) <S> Output is actually no longer regulated since the voltage drop varies with the drawn current. <S> The biggest problem however is that due to differences in the dynamic resistance of the PS and wiring and diode resistance the current distribution between both can vary widely (Not 50/50 as one may expect). <S> 2) <S> You could make use of what is called ideal diodes, which are basically MOSFETs with the accompanying circuitry. <S> These components exist in a variety of current ratings have a very low voltage drop and therefore perform much better with respect to the above. <S> Google "Ideal diode". <S> You may want to read this . <S> 3) <S> Best solution is generally a single PS or PS with parallel tracking mode.
| It depends on the power supplies being connected. You could tie both output together via conventional diodes which solves the problem.
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Diode homework: How is Vo calculated? For this circuit we're asked to calculate V o using the 0.7V diode model. simulate this circuit – Schematic created using CircuitLab I understand that when the potential is higher at the anode than the cathode then the diode is in the forward biased direction, current will flow through the diode and a (in this model) a 0.7V drop across the diode will occur. My guess was that D1 is in reversed bias mode b/c the V2 is at a lower potential than ground, so D1 can be modeled as an open circuit. I don't know where to go from there, my gut says nodal analysis but I'm not quite sure how to set it up. Plugging this into circuit simulators I get ~525mV when using 1N4004 diodes, but I'm not sure how to calculate this by hand. <Q> From the sounds of it, the diode model you are using is the simple "ideal diode" with a fixed forward voltage. <S> This model is an open circuit when \$V_{\textrm{Anode}} - V_{\textrm{Cathode}} <S> < V_D\$ (reverse biased), and a fixed \$V_D\$ voltage supply otherwise (forward biased). <S> Start by making assumptions about the state of D1 and D2 (for example, D1 is forward biased, and D2 is reverse biased). <S> Your circuit would then look like this: simulate this circuit – <S> Schematic created using CircuitLab <S> What is \$V_o\$ here? <S> The last step is to check your assumptions on each diode. <S> If the assumptions are correct, the model is applicable. <S> If not, permute your assumptions (ex. <S> : what if D1 is reverse biased, and D2 is forward biased? <S> or D1 and D2 are reverse biased, etc.) <S> Only one of the 4 possible permutations will have a consistent answer. <S> As a side note, the answer you get will not match what the circuit simulator gives you (though it will be close). <S> This is because the circuit simulator uses a more advanced diode model . <A> Start by assuming D1 is open, and D2 is shorted. <S> What is the output? <S> So which way are the diodes biased? <A> It looks to me as though, it is a trick question. <S> The V2 supply has the + positive lead hooked to R2. <S> If their is no current flow <S> Vo should be +6 volts. <S> Even if V2 had the negative lead hooked to R2 the diode would be reverse bias and not conduct. <S> Still +6 volts. <A> If you run with your guess and either work out the voltage divider or simply apply symmetry (can be done in your head) <S> you get a number for Vo. <S> Since that number is less than 0.7V your guess checks out and D2 is forward biased and D1 is off. <A> In above circuit D1 is forward biased for v1 but D1 is reversed biased for V2 <S> so I think V2 (-6volt) has no use or contribution because it is blocked by Diode D1, so if used 0.7volt diode model the Vo will be <S> 0.7volt. <S> (In my opinion)
| If you use 0.7V diode model then I think the Vo will be 0.7Volt.
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What's the best way to add USB support to a project It just so happened that I need to add a usb communication for my next project that I am working on.What's the best/cheapest way to do this. Should I be using a microcontroller with USB support like maybe the ATMEGA32U2 or use a ftdi chip,or maybe something else. I have never really worked with USB before,so any help would be nice. <Q> It depends on your volume. <S> If you're making thousands of your board, then it's probably best to find a chip you like, lay out your PCB for it, and use that. <S> My personal favourite is NXP's LPC11U-series. <S> They're nice because they're modern ARM Cortex cores, have a built-in bootloader that emulates USB storage. <S> So when you enable the bootloader and plug in your device, the computer it's plugged into will show it as a USB drive. <S> You can then copy/paste in a new firmware binary there and upgrade it like that. <S> It makes it super easy for you and your end-users to upgrade firmware. <S> Writing firmware for them is also very nice and easy thanks to the mbed project which supports those chips. <S> If you're doing something low-volume, then use modules, the cheapest you can find. <S> Go to Aliexpress.com or Alibaba and look for Arduino clones. <S> The clones of the Arduino Nano are great and they have ATmega32u4-based ones too. <S> These boards are ridiculously cheap, going from $2.5 USD to $5.50. <S> In low volumes, a USB chip, plus passives, plus connector, will run you more than that, so it's a crazy good deal. <S> Avoid FTDI at all costs. <S> They're a rotten company and have screwed over people pretty bad by bricking clones of their chips on purpose. <S> You can read about that here: http://arstechnica.com/information-technology/2014/10/ftdis-anti-counterfeiting-efforts-sit-between-a-rock-and-a-hard-place/ <S> While it may seem like breaking cloned devices is OK, a lot of people paid FTDI-prices thinking they got genuine chips but were actually counterfeit. <S> The supply chains for these are complex enough, and the clones good enough, that it's very difficult for manufacturers to tell them apart. <A> I am going to assume "best" way is easiest. <S> You may want to check first if you can find a library with proper USB APIs for a specific MCU or for a chip. <S> Many vendors provide software support. <S> Simply put, do not go for one unless they offer API support. <S> Its a painfull process to build the protocol from scratch and it shouldn't be. <S> Easiest would be if you have USB built into chip. <S> Its cheap and comparatively easy to program than having to accommodate two ICS, increasing cost, effort in soldering etc. <A> To a large extent, it depends on if your concern is development ease or unit cost. <S> For example, you can implement serial port, keyboard or mouse, audio device, storage device, or entirely custom types of communication schemes. <S> However, there is a downside in that servicing the USB port regularly adds complexity to your firmware. <S> Further, if you are depending on USB to obtain status or debug message from your firmware, various common types of misoperation (leaving interrupts disabled, code runaways, hitting a breakpoint, etc) can break the USB interface in situations where a simpler UART output channel would probably keep working. <S> Resetting your microcontroller can also be problematic, as it may require re-enumeration by the host. <S> Using a distinct USB-serial converter, be it a fixed function chip (FTDI, SiLabs, CH340, etc), or a separate micro providing only bridge functionality (ATmega16u2 as on an Arduino Uno, etc) <S> has a key advantage in keeping the moderately time-critical servicing of the USB bus distinct from your firmware. <S> Normally, your USB interface chip will keep happily running and proxying serial data, even as your main microcontroller firmware may crash, be restarted, reprogrammed, paused with a breakpoint debugger, etc - all operations which tend to break a USB interface and require re-enumeration to the host. <S> That can be a huge timesaver during development, but is less important as a project matures. <S> FWIW, when I build something intended to utilize an on-chip USB interface, I habitually break out the signals of a UART to a header or solderable vias, so that I can use an external logic-level USB-serial for debug output early in the development cycle.
| Using a USB-enabled microcontroller as the heart of your project tends to be the cheapest (or at least lowest parts count) solution, and has the greatest flexibility in how the USB interface is used.
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How is this simple counter implemented on an FPGA without a clock? As part of an assignment, I must create these blocks that tie in to a larger top level module. (there are more blocks not pictured). I have everything working fine, except this UP/DOWN counter because I can't really work out how this can possibly be implemented without a CLK . The EN_UP and EN_DOWN signals are simple pulses that should increment or decrement an internal 16-bit value, which gets split into nibbles and put on the output. I feel as if this should be fairly simple, but I can't work this out. I've tried multiple approaches. 1) - Inside a single process count : PROCESS (RESET, EN_UP, EN_DOWN)BEGIN if(RESET = '1') then countSignal <= x"0000"; elsif(rising_edge(EN_UP) and EN_DOWN = '0') then countSignal <= countSignal + 1; elsif(rising_edge(EN_DOWN) and EN_UP = '0') then countSignal <= countSignal - 1; end if;END PROCESS; This ultimately compiles with no errors or warnings, however the compiler ends up creating the wrong circuit, tying the EN_UP to the CLK of the flip-flip, and the EN_DOWN to the CE (clock enable) . While yes, that is part of the equation, it doesn't mirror that for the opposite case. 2) - Separate processes countUP : PROCESS (RESET, EN_UP)BEGIN if(RESET = '1') then countSignal <= x"0000"; elsif(rising_edge(EN_UP) and EN_DOWN = '0') then countSignal <= countSignal + 1; end if;END PROCESS;countDOWN : PROCESS (RESET, EN_DOWN)BEGIN if(RESET = '1') then countSignal <= x"0000"; elsif(rising_edge(EN_DOWN) and EN_UP = '0') then countSignal <= countSignal - 1; end if;END PROCESS; This results in: Signal countSignal[15] in unit UD_COUNTER is connected to following multiple drivers: 3) Multiple processes with Hi-Z states I tried some attempt with Hi-Z which also failed. <Q> Otherwise there is nothing to tell it when to count. <S> Either the clock is missing from the diagram, or you will have to edge-trigger on the enables, in which case the circuit you think it should have synthesized to is impossible, as you can't drive the same register with two different clocks. <S> You might try something like: en_either <= en_up or en_down;process <S> (en_either, rst)begin <S> if rst = '1' then count <= <S> (others => '0'); elsif rising_edge(en_either) <S> then if en_up = '1' then count <= count + 1; else count <= <S> count - 1; <S> end if; <S> end if;end process; There may be better ways, but if you really can't have a clock, I suppose that will work. <S> If the enables overlap in any way, of course, this may not work as intended. <S> As the error message indicates, your two-process version will not work because you're driving the same signal from two different processes. <S> There may be ways to work around this, but it would be much more trouble than the 1-process version. <S> As David points out in the comments, this will not work properly unless your enables are bounce-free. <S> The source of the enables isn't clear from your post. <A> You don't need a clock for this, but you need to better understand what you should do for every possible input given to your network. <S> The first approach is the best, so let's analyze it. <S> You start by checking RESET and that's great, elsiffing whatever comes after is just how I would do it. <S> Now to the tricky part. <S> You check if either EN_DOWN or <S> EN_UP had a rising edge, check that the other signal is zero and act accordingly. <S> That makes no sense to me, at least given the (little) specification you provide. <S> Since you are in a process you are sitting on edges already, i would just check what happened with the event thingy. <S> Something like count : PROCESS (RESET, EN_UP, EN_DOWN)BEGIN <S> if(RESET = '1') <S> then countSignal <= x"0000"; else if(EN_UP'EVENT and EN_UP = '1') <S> then countSignal <= <S> countSignal + 1 <S> ; end if; if(EN_DOWN'EVENT and EN_DOWN = '1') <S> then countSignal <= <S> countSignal - 1; <S> end if; <S> end if;END PROCESS; <S> This seems better to me and more adherent to what you ask. <S> For the record, an heuristic way to do this <S> (so you understand you don't need a clock): <S> you have your 16 bit outpt register made with DFF, and two combinational networks that compute output - 1 and output + 1. <S> these two networks are connected to a dual input mux, and the output register input is connected to the mux. <S> the DFFs clocks are the or between count_up and count_down, while the mux control bit is one of them. <S> When a pulse arrives the register is then refreshed with the correct value. <S> If you are not satisfied with the implementation you get after synthesys you can try to implement my latter idea and see if that works for you. <A> It is possible to design a counter from purely combinatorial logic, if it is given two inputs rather than one, and the inputs change state in a well-defined pattern. <S> For example, suppose one has two inputs P and Q, two eight-bit values X and Y, and computes them combinatorially as follows: <S> When P is high <S> X=Y+1Else <S> X= <S> XWhen Q is high: Y= <S> X+1Else <S> Y= <S> Y <S> If P and Q alternately go high, then X and Y will count the number of times they have done so, provided that P and Q are never high simultaneously; if they do ever overlap, then the values of X and Y will be undefined. <S> Although flip flops have largely replaced the above style of circuitry, the two-phase clocking approach has historically been quite popular, especially since in NMOS technologies, given a minimum clock speed, one could use latching circuits that were much cheaper than flip flops. <S> Consider: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Each relay should simply be a single NFET, but I can't get those to simulate correctly, so I substituted a relay instead. <S> This circuit implements a divide-by-two counter using five active transistors and three passive pull-ups. <S> Implementing an edge-triggered flip flop would require more circuitry, so even though it's necessary to generate and distribute two clocks rather than one, the overall effect is a "win". <A> FPGAs really, really don't like to run without a clock. <S> You get wildly unexpected behaviors. <S> So you're much better off using a synchronous up/down counter with the up/down inputs sampled by a faster clock. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The clock period should be no more than 1/3 the up or down pulse width.
| You're right in thinking that you will have to use something for a clock for the counter.
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Lower Voltage With A Zener Diode I am trying to lower the voltage from a 9V battery to 5V for use on an Arduino pin. I bought a 5.1V zener diode from RadioShack. Will this do what I want? Do I have to lower the amps as well? It looks like the battery delivers 9V and 1A. Should I first use a resistor to lower the amps, then use the zener diode to lower the voltage? I tried using a 330 ohm resister directly with the battery and a multimeter just to see how much it lowered the amps, and it got very hot so I am not sure if I should be using it. Zener Diode: IN4733A Voltage: 5.1V Current: 49mA Maximum power dissipation: 1.0W I looked at some example equations here: http://www.electronics-tutorials.ws/diode/diode_7.html but am still not comprehending it. I am not even sure which side the black band on the diode should point. So my question is: Will the diode work to drop from 9V to 5V and do I need to do anything extra? I wanted to use this: http://www.amazon.com/Voltage-Sensor-Detector-Divider-Arduino/dp/B00S4PCCG8/ref=sr_1_1?ie=UTF8&qid=1424025730&sr=8-1&keywords=voltage+divider+arduino , but don't want to wait so thought I would try something else until I can get this ordered. <Q> A much better way to measure/detect a signal from a battery powered device would be to use a high value resistor divider. <S> In the case of a battery powered fire alarm you do not want to alter the function in any way or run down the battery prematurely. <S> Using a series resistor and a Zener to ground could result in a moderate amount of current being drained from a constant 9v signal. <S> When set as a digital input without the internal pull-up resistor an I/O pin on the Arduino has a very high input resistance, (near 100M). <S> For example two resistors of 470k each could be used as a simple voltage divider for detecting a 9v signal. <S> So long as the internal pull-up is not enabled the resulting voltage on the Arduino pin will be 4.5v when the input signal is 9v. <S> The current draw from the 9v signal will only be about 0.0096ma. <S> With a voltage divider using equal value resistors <S> the signal voltge could be as high as 10v and a 5v Arduino system would still see a safe 5v signal at the I/O pin. <S> The digitalRead() instruction would be used in this case. <S> If you want to detect and actually measure the 9v signal with the Arduino ADC system the voltage divider could be made using resistor values a bit lower to give resonable operation. <S> (Very high impedance signals can cause errors related to the internal capacitance of the analog input). <S> Using a voltage divider with two 220k resistors and a small capacitor (about 0.1uF across the grounded resistor) could make a reliable voltage divider to measure a 9v DC signal and still limit the current draw to a low level, (in this case about 0.020ma). <S> The ADC can now safely measure a DC voltage from 0v to 10v and output an actual value related to the input voltage, (multiple this by 2 to see the original signal value). <S> The analogRead() instruction is used here. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> The zener diode you reference has a zener-voltage of 5.1V and a power rating of 1W. <S> You want to place a resistor in series with the 9V source and <S> the zener diode to limit current draw. <S> The datasheet for the IN4733 states that at a zener current of 49mA <S> the zener voltage is at the rated 5.1V. <S> The only calculation you need to do is to size the resistor such that you have at least 49mA, although lower current would also likely give you around the 5.1V zener rating. <S> Normally you would look at the datasheet for a chart to give you this information, but the one I found left this information out. <S> $$ R = <S> \frac{9-5.1}{10mA} = 390 <S> \Omega$$ <S> But, again, larger sized resistors are likely to give you the voltage you want with lower power dissipation. <S> Test to verify. <A> If no other circuitry is using 9V, I might suggest a 6V configuration with some "AA" batteries. <S> Instead of a Zener, you could use an LM7805 (orderable from Digikey). <S> There will be less IR drop across that regulator (less waste heat).
| So if only a high/low detection is needed from a 9v signal a simple high value resistor divider is a good method that also limits the current draw from the device.
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ATtiny85 to control a relay, which will turn on and off a motor I am trying trying to control a 5V relay (SRD-05VDC-SL-C) with the ATtiny85. I have a simple sketch uploaded (Blink without Delay example): const int ledPin = 0; // the number of the LED pinint ledState = LOW; // ledState used to set the LEDunsigned long previousMillis = 0; // will store last time LED was updatedconst long interval = 1000; // interval at which to blink (milliseconds)void setup() { pinMode(ledPin, OUTPUT); }void loop(){ unsigned long currentMillis = millis(); if(currentMillis - previousMillis >= interval) { previousMillis = currentMillis; if (ledState == LOW) ledState = HIGH; else ledState = LOW; digitalWrite(ledPin, ledState); }} The whole thing is being powered by an 18650 Li-Ion battery (3.7V-4.2V) and the measured input is averaging 3.88V. Here is a quick sketch of how the circuit looks like: Some issues comes up WITHOUT the motor connected: If I connect the battery and relay directly, you can hear it click, despite it being rated at 5V input. If I try to trigger the relay from Pin0, the oscilloscope will show the voltage of Pin0 to be about 2.72V, when the relay looks to be switching. If I replaced the relay with an LED, the voltage of Pin0 is about 3.88V. What causes the drop in voltage? I believe the current should be enough since it's an Li-Ion battery. What sort of protection do I need to implement in this circuit? Will the Li-Ion battery be enough to power this circuit? If I connect just battery, relay, and motor, it runs fine. The issue comes with the addition of the ATtiny85. <Q> You should not really drive the relay coil directly from the microcontroller pin. <S> Connect a diode backwards in parallel to the coil (i.e. diode cathode to positive supply, diode anode to transistor collector). <S> This is a "fly-back" diode to protect against the back-EMF you get with the coil switches. <S> Put something like a 330 ohm resistor in series between the transistor base and the microcontroller pin. <S> This limits the current into the base of the transistor to a level that is within the capabilities of the microcontroller. <S> That's the most common way I know of to control a relay with a GPIO pin of a micrcontroller. <A> Recommend adding some serial debug output to that sketch. <S> Motors + micros without some type of supply isolation normal equals a microcontroller that is constantly reseting. <S> Serial outputting the loop count might show it just keeps reseting. <S> You'll likely need a few capacitors to filter it. <S> Maybe even an LC filter. <A> The output circuit of the processor is not a zero-resistance switch. <S> It is a couple of MOSFETs that have a finite resistance when turned on, and the voltage differences you are seeing are due to the voltage drop across the MOSFET due to the current you are attempting to draw from the output pin. <S> The datasheet for the ATTiny should give the maximum recommended output current, and the voltage drop at that current. <S> You should have <S> the output pin drive a transistor or FET that, in turn, controls the relay. <S> This should allow nearly the full Vcc to be available to drive the relay.
| Instead, I would recommend using an NPN transistor between GND (emitter) and the coil (collector), then connect the other side of the coil to the positive supply directly.
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Series vs parallel for microphones (electric guitars) I want to create a passive "mixer", with which I can connect two of my guitars to one amplifier, so basically one with two 6,3" input jacks and one for output. For passives without a transformer as far as I know I have two options:Connecting them parallel or series. simulate this circuit – Schematic created using CircuitLab Which one should I use and why? If there is no specific, what are the advantages of each?Won't parallel produce too much voltage and damage my pre-amp or amp? (PS: I know that the volume knobs will affect both guitars, but that doesn't matter for me.) <Q> If this is two separate guitars, pretty much your only option is a parallel connection <S> That's because these are unbalanced, hi-impedance signals. <S> If you were in fact to wire these in series, touching any of the metal associated with the top guitar's wiring(metal connector shell, etc) would cause hum and other noises. <S> Pretty much any of the passive audio mixers that I've seen just connect the hot signal from each guitar to the top end of a potentiometer with all of the ground connections connected together and to the bottom end of each pot. <S> The wiper connections of each pot go to a summing resistor (about 100k for guitar signals) and then to the output. <S> Pot values for a passive guitar mixer would be somewhere between 100k to 1 Megohm, audio taper. <S> A passive mixer such as I have described has a lot of flaws but they do work. <S> And they will allow you to adjust the level from each guitar individually. <A> That's wrong, the question was about microphones on Guitar cabs and not guitar signal. <S> The Microphones outputs are balanced. <S> I've seen people doing it with a paralel connection, but Series sounds better to me, and it seems there's less impedance mismatch. <S> Check this thread on the subject: <S> https://groupdiy.com/index.php?topic=60308.msg765021#msg765021 <S> Most people do it in paralel because they dont even know you can connect/sum two mics in series. <S> Try both and check wich you prefer <A> If you want to connect two or more guitars in parallel, I would suggest adding a 100K resistor in series with each output before joining them. <S> The output impedance of a typical guitar will vary depending upon its volume-pot position, but will drop to almost nothing if the volume pot is set to 0. <S> Adding a resistor in series with a guitar's output will reduce the signal level from that guitar when its pot is turned up, but will reduce the amount of signal the guitar steals from the other when its pot is turned down. <S> If you don't have a resistor, turning either guitar's volume to 0 will kill the output from both guitars.
| "If this is two separate guitars, pretty much your only option is a parallel connection That's because these are unbalanced, hi-impedance signals."
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Relay not switching correctly, controlled from ATtiny85, output to motor I have an ATtiny85 switching an NPN transistor, which will in turn switch on a relay that will run a motor. Everything is run off a 3.7V 18650 Li-Ion battery, but I have since moved over to a variable power supply with same problem: Without the motor, the relay switches fine, ATtiny85 will switch the relay on for one second, and then off for one second, and repeat, and the oscilloscope (connected to relay output and GND) and clicks are showing correct behaviour for the relay. When I connect the motor, the relay still clicks, but the oscilloscope is showing zero voltage. Seems like relay tries to switch, but doesn't. I connected an LED to Pin0 of the ATtiny85, and it flashes instead of staying on for one second and going off for 2 second. Seems to me that the problem could be: The ATtiny85 is resetting because not enough current going through. This doesn't make sense because I've switched over to a bench power supply and same thing happens, voltage is still the same. Some sort of feedback from the motor? The motor runs fine if I connect it directly to battery so I don't think power is an issue. Some additional information: Relay switches on for 1 second then off for 1 second. I am not doing PWM, but just have this on/off behaviour for testing. This is the motor: 3V-12V water pump motor Power supply is set at 3.9V with 2A current available. Increasing it to something like 5V still doesn't work. EDIT: Although the answers all helped me figure out my problem, I chose the one that helped me the most. Found that it wasn't back-emf or your standard type of decoupling required. The motor was dropping the voltage down hard and no capacitor was going to fix it (already had these in place), although these are common issues. More findings here <Q> I have two suggest i ons for you: <S> 1) check to make sure that the relay is able to handle the current the DC motor is pulling. <S> 2) use the variable power supply to run the control circuit, and the battery to run the motor with relay. <S> if this proves to be working, you can then work your way back to re-integrate the motor back to your system. <A> This is why you see sparks at the brushes of a spinning motor. <S> This kick could glitch the chip and cause exactly the kind of stuff you are seeing (relaTo remedy this, you can try putting another flyback diode across the motor leads just like the one across the relay coil. <S> It is also good practice to use a decoupling capacitor across the leads of the chips. <S> This will filter any sudden power spikes or dips that could otherwise cause the chip to glitch. <S> Try putting a ~1uF capacitor physically as close to the leads of the chip as possible. <A> Something like this: simulate this circuit – Schematic created using CircuitLab <S> Are you actually PWM'ing? <S> Why relay? <S> Why not FET?
| Depending on how the motor is wound, it could be giving an inductive kick onto the power supply rails as the brushes disconnect from the coils.
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Non tinted text on pcb altium I added a string in PCB Editor and it is on Top Layer, which is good. But when viewing in 3D, its tinted. Its behind the color layer and not appearing clearly as it should like pads(the carbon part). I couldn't find any setting to remove its region from color layer. How can I make it appear clear ? Edit: added screenshot <Q> You placed the text on Top Layer, which is the top copper layer. <S> Your screenshot shows solder mask covering your text. <S> You need to place a feature on the solder mask layer to remove solder mask over your text if you want it to be visible. <S> Solder mask is a negative layer, so features placed on this layer in your design are areas where solder mask will not be present. <S> Once you do that, at least for the default 3D view settings (which it doesn't look like you're using), Top Layer copper will render as yellow or gold. <S> If you want your text to show like the "P1" in your screenshot, you need to put it on the Top Overlay layer, not Top Layer. <S> This will put it in the silkscreen, which is an ink layer on top of the solder mask, instead of a copper layer beneath the solder mask. <A> If you're placing text on the top copper layer, there will be soldermask over the copper, which I think is what you're seeing. <S> If you want to leave the soldermask off the copper, you actually place a pour/polygon on the soldermask layer. <S> The soldermask is a negative layer, so where you draw things on the soldermaks layer, it will be removed . <S> You may need to enable the soldermask in the 2D view before you can draw on the layer, o , <S> m in the PCB view will open the relevant dialog, and then check the entry under "Mask Layers" for "Top Solder". <A> Mate, you've got the highlighting thing turned on. <S> Down the bottom right is a "clear" button which clears all the selected/highlighted layers and returns everything to proper colour. <S> EDIT: <S> The Photon is right <S> , it's the Top Layer being copper, you wanted Top Overlay layer for the string object.
| If you want it to render in a different color you probably need to adjust settings for the 3D viewer, not for the text itself.
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Why do wifi channels overlap? In the United States and Canada there are 11 channels available for use in the 802.11b 2.4GHz WiFi Frequency range defined by the IEEE. But there are only 3 non-overlapping channels available in the 802.11b standard.These are Channels 1,6, and 11. Why is it that we couldn't define a large band width so that all channels will have no overlapping? If it were done, would we have better performance when using Wifi today? <Q> Bandwidth is money. <S> The more they can fit into a frequency space, the less it's going to cost. <S> If it were done non-overlapping you might get marginal gains, but it was likely an engineering/cost tradeoff. <S> At the fringes of the channel the amplitude of the signal will be lower due to imperfect filtering resulting in less noise travelling into the next channel anyways. <S> so the other channels would just be there to try and avoid the other 1 or 2 in a given location. <S> If you find a location with 4+ wifi signals, you'd likely make some gains by giving the channels more bandwidth, otherwise, you're just burning money by requiring more bandwidth. <A> If there's space for only three non-overlapping bands, there will only be space for only three non-overlapping bands, regardless of how the channels are numbered. <S> A more interesting question is why there are more than three channel numbers; the answer to that comes from the fact that WiFi is the 2.4GHz band is used for many kinds of devices besides WiFi; some of those devices, such as analog audio/video transmitters, may require a "full-time" channel allocation. <S> Such devices are no longer terribly common, but their existence would have influenced the creation of WiFi standards. <S> If a few such devices which each uses 5% of the available bandwidth have four-way channel selector switches, with different devices offering different but overlapping sets of frequencies, it may be that the only combination of available frequencies which would allow everything to work without interference would have something in each of the three "main" WiFi bands, but that a shifted range of frequencies would be available for the WiFi. <S> That having been said, I think things would have been far less confusing if the channels had been labeled "1-5", "6-10", and "11-15", (along with "2-6", "3-7", etc.) <A> The channel assignment was done for 802.11 and 802.11b, where we get 11 non-overlapping channels. <S> 802.11g with its increased bandwidth demand came later, but the channel numbering was kept to avoid confusing the users, as most "g" devices also support a silent fallback to "b".
| Another reason this is done is because it's not likely to have more than 2-3 802.11b signals in any given locations
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How long will this appliance run from a battery? How long can I run a water pump which requires 12V and 50W from a 12V, 40Ah, 400A car battery? Can you explain how this was calculated? <Q> Short answer: 9 hours, 36 minutes <S> Long answer: <S> Assuming the system is 100% efficient, (i.e. no resistive losses in cables, or internal resistance from battery) Power (Watts) <S> = I(Amps) <S> * V(Voltage)Therefore, <S> Amps = Power/Voltage50/12 = <S> 4.16666A <S> Assuming that the battery is fully charged, then it will have the full 40AH capacity. <S> This battery is not linked to a boost converter or anything like that, so it would be 4.16 amps at 12 volts, so we don't need to do any more calculating here. <S> If the power output was not at 12v, then we would reverse the power equation to do 12*40 = 480WH. <S> This means the battery can provide 480 Watts for 1 hour. <S> 40AH <S> (Amp Hours)/4.16Amps = 9.6 hours, or 9 hours and 36 minutes.or480WH <S> (Watt Hours)/50 (Watts) = <S> Longer answer: <S> Sadly, in the real world, nothing is as easy as this. <S> Batteries are not perfect. <S> They have some internal resistance, which we have to factor in. <S> We can also calculate the resistive losses from the internal resistance of the battery from the 400A rating R = <S> V/IR = <S> 12/400R = 0.03 <S> ohmsP = <S> I^2 <S> * RP = <S> 4.16 <S> ^2 <S> * 0.03 <S> = 0.52083W <S> lost <S> In 1 hour = <S> 0.52083Wh <S> lost, as you multiply by one. <S> Therefore, in 9.6 hours, it means 5 watts is lost. <S> 480Wh - 0.52083Wh <S> = <S> 479.49716 <S> Wh479.49716/50 = 9.589583 hours. <S> This is 9 hours, 35 minutes and 22.5 seconds <A> Your battery energy capacity is 12 volts x 40 amp-hours, for a total of 480 watt-hours. <S> Divide this by 50 watts, and you get 9.6 hours. <S> BUT. <S> This is a car battery, not a deep-discharge marine battery. <S> If you discharge it by more than 50%, you will shorten its life. <S> If you drain it completely you will shorten its life a lot. <S> So, assuming you want the battery to last, don't discharge it more than about 50%, which will take about 5 hours. <S> Limiting run time to 4 hours will do even better. <A> 9.6 Hours First figure out how many amps your device consumes, in this case ~4.16 (50W/12V).Amps = Power/Voltage. <S> Then divide your amp hours by the amps <S> , 40Ah/4.16A gives you ~9.6Hours. <S> This doesn't take into account specific battery chemistry, temperature, etc.
| 9.6 hours, or 9 hours and 36 minutes. Practically speaking, about 4 - 5 hours.
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How can I view, debug, or analyze data being input to my FPGA? I'm working with a Xilinx Spartan6 on Digilent's Nexys3 board. I've also purchased their PmodMIC so I can try to get some audio data onto my board to perform some signal processing. The PmodMIC has a simple SPI connection, and is supposed to serially output 12 samples per "run" that get shifted into a 12-bit vector inside the PmodMIC VHDL module. I have created some top-level code and supporting code that I think might be working, but I find myself completely lost when I ask myself - "How do I verify I'm getting correct data, or any good data at all for that matter?" I decided to map 8 of the 12 bits in the output vector to the LEDs on the board so I could see something at all. When I press my button mapped to start the conversion process, I can even see which LEDs are lit occasionally change and move around, but I know this is essentially useless for anything meaningful. At this point, I'm entirely clueless. If only there was a way to serially output data back over USB to the PC, or freeze the FPGA's state and examine the contents of its memory after storing some samples there. If methods like this do exist, I certainly don't know about any of them. Logic Analyzer I bought the Analog Discovery (another Digilent product) and tried to view the serial data coming from the PmodMIC to the FPGA. I simply routed the serial data to another unused output on the JA bank and hooked up the analyzer, but my results have been pretty poor. I am sure much of this has to do with my inexperience with the device (I can't even figure out how to control its sample rate). In any case though, my ultimate question is a bit more broad, which is: What tools, methods, or processes exist to view, debug, or analyze data on an FPGA? <Q> Xilinx has an in-system debug tool called " ChipScope ". <S> It uses spare FPGA resources (mainly block RAM and counters) to create a logic analyzer that can connect to any node in your design, capture waveforms and display them on your host PC, using the same JTAG interface through which you program the chip. <A> I have an Atlys board and have a couple of different solutions that I use to validate data, depending on the application and datarate: <S> I use a UART for low-speed transfer and connect it to my pc using a USB cable. <S> I think the nexys has an onboard FTDI chip converter <S> so you just need to use the USB cable. <S> There are a number of uart opencores , or it's relatively simple to write your own. <S> It's an easy interface, protocol-wise. <S> I suspect that this is what the Digilent USB debugging interface uses, however it has only got windows support. <S> Since I mostly use linux <S> I use my own implementation. <S> UDP isn't too difficult to code/understand and a lot of the header data can be hard coded. <S> I made my own custom packet header and used MAC addressing to route the packets the correct way. <S> On the PC side, I usually write a python utility to parse in UART/UDP packets and display the data in a meaningful way. <S> It's also possible to use something like hyperterminal in windows. <S> I also use wireshark to check that my UDP packets meet the ethernet spec. <S> Chipscope is useful for seeing what's going on inside the FPGA (for debugging purposes) but you can't capture data for long periods of time. <S> As far as I remember, it uses JTAG <S> so it isn't that fast either. <S> For applications where you're only interested in the a specific data output, it's easier just to continually feed that data stream to a PC over some interface. <A> Most of Digilent's FPGA boards, including the Nexys3, implement a simple parallel interface which can be accessed over USB. <S> The interface is documented at: http://www.digilentinc.com/Data/Products/ADEPT/DpimRef%20programmers%20manual.pdf <S> See also: Implementing the Digilent EPP <S> You can interact with this interface using code libraries provided by Digilent, or using the I/O sections of Digilent Adept.
| For high-speed interfaces, I use a modified version of the opencores UDP stack over Ethernet.
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CAN Bus Physical Layer Twisted pair or not CAN bus is very common in automotive or vehicular applications. In our case, we are using aerial platforms with have CAN bus links between operator remote and ground control box. As is very typical of such products, the cables carrying the CAN bus around are neither shielded nor twisted pair. The typical run of cable between devices is something along 20 meters with a few Deutsch automotive connectors along the way. The original products from the manufacturers work flawlessly, as is expected. However, if we are to change the cabling/connectors configuration some glitches start to appear, although the terminating devices at both ends are identical: shortening the overall cable length to a total run of 10 meters going through Entrelec terminals, then on individual wires for about 2 meters, then connecting to an Harting industrial connector, which is finally attached to a 3 meters spring cable, Every once in a while (say 3-4 times a week of regular usage), communication glitches occurs between the devices. The on-board diagnostics signals an error and just halts everything. It is as though the physical link "is now"/"was originally" just within margins of proper operation and those issues start to appear. Thus, the question is, how can we diagnose/improve this cabling? Does using twisted pair/shielded cables runs going to help or harm? Is there a specific impedance of cables we should be aiming for (like 50Ohm or 75Ohm BNC cables for Network of cable applications)? Other thoughts on the subject, as we are ill-equipped to diagnose the waveform or the protocol on the line with a few occurrences every week. It also makes it very hard to confirm the validity/improvement of any changes we are introducing? <Q> Yes, use twisted pair. <S> Failing to do so for a differential signal is self-defeating. <S> Don't forget to include 120-ohm termination resistors at each end of the transmission line. <S> The less connectors in any transmission system the better. <S> Connectors are almost guaranteed to present an impedance discontinuity, and hence will cause reflections. <S> Transmission line stubs of any length are also a source of reflections and (if I'm reading your description correctly) you have transmission line stubs of several metres . <S> The longer the stub <S> , the worse the impact of the reflections. <S> Reflections are bad because they can cause destructive interference, i.e. they can corrupt any transmitted data. <S> I'd use a longer primary (twisted pair) cable, less connectors and shorter stubs. <S> Run the primary cable as close to each node as you can, even if it means that the cable has to be longer. <A> •Does using twisted pair/shielded cables runs going to help or harm? <S> •Is there a specific impedance of cables we should be aiming for (like 50Ohm or 75Ohm BNC cables for Network of cable applications)? <S> OMG, <S> yes. <S> Standard impedance for twisted pair is typically 100 - 120 ohms, but the exact impedance is unimportant. <S> What is important is that the cable HAS an impedance, and the terminating resistors at each end of the cable match the cable impedance. <S> Pick a cable, find the impedance from the data sheet, and go with that. <S> Or, since you seem to be using terminators, find a cable with impedance within +/- <S> 10% of that. <S> Also, check the CANBus spec to be sure that your cable length is appropriate for the bit rate you're using. <A> At the higher bitrates that CAN is capable of (125kbits/s+), the cabling is important The differential pair is important for keeping the error rate low, so using twisted pair cabling is also important. <S> 110 ohms is the nominal impedance. <S> don't create stubs (up to 30cm can be permissible, but it depends on lots of other factors). <S> Take the bus to each device - many automotive systems now have a pair of pins for each of the CAN wires, an "in" and an "out" to avoid having any significant stubs <S> In terms of fixes: How loaded is your bus, and can you clarify what kind of comms glitches you get? <S> Are you logging your CAN error counters regularly to see if there are lesser glitches on a more regular basis - that will allow you to have a "figure of merit" for your current system which you can attempt to improve upon. <S> Also, a single corrupted message is going to happen every so often, so your system needs to handle that. <S> If you are seeing long error bursts, I would look for large external sources of noise which push the system over the edge. <S> (As an extreme data point: I have seen CAN buses with error rates of several error-frames per second operate "just fine" from a system perspective!) <A> Regarding the wiring, in ISO 11898-2 (CAN high-speed standard) is specified: "pair of parallel wires, shielded or unshielded, dependent on electromagnetic compatibility (EMC) requirements". <A> Twisted pair + shielded cable, yes, but to get a longer CANbus <S> you can lower the baud rate. <S> The book "Embedded Networking with CAN and CANopen ISBN 0-929392-78-7 Page 515" gives a max length of 25m for 1Mbps and 1km for 50kbps.
| You must use twisted pair, and shielded twisted pair will help with reliability.
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With today's technology, is it possible to charge a battery over 100% without exploding? We all know who Iron Man is. Hollywood makes a nice job to make him look like a genious in electronic. However, there is a scene where his suit charges it's batteries to 475%. There is another question, on another website, adressing one part of this issue. You can read it on Can Iron Man recharge his suit from lightning? . What I'm asking here is if it is possible to charge a battery (no sci-fi stuff, please), with current technoligy, to over 100% of it's charge without exploding? <Q> Well if you say 100% is the capacity of the battery specified by the manufacturer, then yes nearly all batteries are charged beyond 100%, as the manufacturer will indicate the minimum capacity, so they are not made liable for any failures concerning too little capacity. <S> But that's in the range of maybe a few percent and not a massive 375% or something like that. <S> The reaction to overcharging the battery is very dependent on the chemistry involved. <S> Some will tend to explode (lithium ion and lithium polymer), others will just waste the additional charging current to heat. <S> The point made by PlasmaHH in the comment is actually a quite good one. <S> If a battery could hold significantly more charge as advertised, the manufacturer would be quite dumb. <S> Now you could argue, that a lot of ICs will also run at much higher speeds than advertised. <S> But on closer inspection, not all ICs are able to, and that's why manufacturers turn the speed down to a level where every chip runs - it increases the yield and thus you can make more money. <S> Batteries seem to behave more predictable for manufacturing. <S> Maybe a capacitor can handle more voltage and <S> so store more energy than advertised. <S> As the energy stored on a capacitor increases with the square of the voltage, you would need to double the voltage to store four times the energy. <S> That doesn't sound too bad actually - you might find some capacitors which are able to handle double the voltage specified as they are tested quite hard. <S> But then the energy stored on capacitors is quite low compared to batteries. <S> (There are massive capacitors, but you wouldn't want to carry one around) <A> Fundamentally a battery works by adding and removing electrons from atoms in solution. <S> This sets an absolute physical upper limit on how much energy it can hold, in the same way that you can't put more than 1L of fluid in a 1L bottle: once you've run out of atoms to change the charge state of, there is nowhere to put extra charge. <A> NiMH batteries will take the excess charge current and liberate flammable hydrogen gas (which is vented to prevent the cell from exploding) plus heat, permanently reducing the capacity of the cell (damaging it). <S> In extreme cases (like the example), thermal runaway takes place which doesn't just damage the battery, it destroys it on the spot in a geyser of electrolyte steam and liberated hydrogen. <S> Lithium cells can turn into pyrotechnics, presumably incinerating the unfortunate Mr. Stark. <S> While it's possible that you might be able to get a small amount of extra capacity at the risk of damaging a battery in general, if 475% was possible they'd be using it. <A> As others have explained, what you call "100% charge" is definitely not fully charged. <S> Battery people don't necessarily figure out what 100% charge really means because the battery usually explodes when you get there. <S> I'm most familiar with Li-ion batteries, so here's a quick "oh my god" situation: <S> The Li-ion batteries I work with have two flat electrodes rolled up with a separator in between so that the electrodes don't short together. <S> If you hold a battery at high voltages and temperatures for a while, something similar to tin whiskers can form on the electrodes, causing a short and generating a lot of heat. <S> The battery will probably catch on fire (and lithium fires are particularly nasty). <S> Note that charging batteries to higher voltages will also cause them to die in fewer cycles without a whole lot in return. <S> Most of the capacity of a battery is in the lower (regular) voltage ranges, so adding an extra 100 mV of charge won't make your battery last as long on one charge as you think.
| There are plenty of battery chemistry options which can be left on a charger without damage - but that doesn't lead to higher charge levels!
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Using a 1.5V battery to increase the voltage from a headphone jack so it can switch a transistor at lower volumes I recently built a circuit using a 2N3904 BJT transistor with the hopes that when connected to a headphone jack, it would light a LED whenever an audio signal was passed through. However, I found that this only worked when I put my devices to full volume (or at least > 80%). Trying a Darlington transistor setup didn't help. I need this LED to activate at much lower volumes (this is for a research project in which I need a visual indication that a sound is playing, often at 50% volume). My proposed solution: Add a 1.5V battery and resistor to the Audio line in before the transistor base so the battery will amplify the baseline voltage enough that it will trigger at lower volumes. Then I could just use the appropriate resistor to adjust the voltage amplification of the signal for different sounds I'm using. Before I try this, I thought I'd post on here and get some second opinions. My two questions are: -Will this work to amplify the voltage of the audio signal as I'm intending? -If so, will I run into any risk of damaging any headphone jacks which are plugged into this? On the image, the ground and empty negative terminal of the battery at the top are connected to the ground and Left audio channel of the headphone cable, respectively. Any advice is greatly appreciated! This is similar to the question posted here: How can I effectively reduce the voltage needed to activate a transistor? But since I'm a beginner, I'm looking for easier ways to amplify the signal. This "just has to work" without being fancy. <Q> Here is a solution that minimizes the input threshold voltage as battery voltage changes: simulate this circuit – <S> Schematic created using CircuitLab Adjust R2 to control sensitivity to low-level signals. <S> Adjust R5 to control sensitivity to high-level signals. <A> Edit: <S> Actually, I found out the answer I posted is already given in the question you rejected as "too complicated". <S> I ask you to reconsider that judgement, as anything involving operational amplifiers (suggestions also found in that question) will be even more complicated. <S> While the commentors are right that your circuit is a really bad idea in practice, your thoughts that lead you to the that circuit are actually quite good. <S> You notice correctly that the voltage you obtain at typical volumes is too low to turn on the transistor (you need around 0.6V to get a significant effect). <S> You problem is that your voltage is too low, not that your supply can't deliver enough current, so forget about darlingtons. <S> Electronics people often call this process "adjusting the DC bias" of a transistor amplifier. <S> It can be done quite easily on your circuit without adding a dangerous low-impedance power source like a battery. <S> Let's work on your circuit. <S> First, add a current limiting resistor into the LED circuit and put the LED on the other side of the transistor. <S> It makes no difference in your circuit as it is now, but it will make soon: simulate this circuit – Schematic created using CircuitLab <S> You can fix your problem, as you correctly identified, by providing DC bias (a baseline voltage) to the transistor base. <S> But you should not force a DC voltage into the headphone jack, so you need a capacitor for DC decoupling. <S> This capacitor should pass the AC music signal, but block the DC required to bias the transistor. <S> With the circuit as drawn, you can get the required DC bias (around 0.45V) from the battery using a voltage divider like this: simulate this circuit <A> Few simple options. <S> Germanium Transistors would be better because of the lower VBE voltage drop of ~0.2 instead of <S> Silicon Transistor drop of ~0.6. <S> Resistor values would vary, you gotta play with them. <S> Two NPN transistors, one as a simple switch, the other providing the current gain. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Non inverted, with 1 NPN and 1 PNP. <S> I am not very sure if this is accurate or not (never very good with PNP logic). <S> simulate this circuit Simple Opamp circuit. <S> Look for any common battery 3V powered op amp. <S> Nothing special. <S> It provides plenty of gain. <S> Finally, look for dedicated ICs like the LM3914/15/16, which provide led bar graphs (VU meter), essentially designed to light leds to sound. <S> (It's basically a bunch of the opamp circuit above, in parallel).
| Using a darlington transistor doesn't help, as you experienced, because you need even more voltage to turn on a darlington transistor, but you can get away with a lot lower current. Inverted, so led turns off as audio gets louder/sound plays. Increasing the "baseline voltage" as you call it, is exactly the way to go.
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What is the meaning of negative value of alternating waves? What does the negative values in AC mean? If we have a symmetrical sine wave with positive and negative amplitudes, then what is the physical significance of negative values? If the answer is related to direction, then the current should travel forwards and backwards, then how does the current complete its path in the circuit? <Q> It is used to indicate a phase relationship. <S> A negative amplitude on an AC value indicates that it is 180 degrees out of phase with whatever convention we have picked for positive values. <S> To explain better, let's look at the simple circuit below. <S> The relationship here is: \$v(t) = <S> i(t) <S> \cdot <S> R\$ <S> Note that to be careful, you have to use time dependent voltage and current values. <S> Assuming that v(t) is a negative amplitude AC sine wave, you will get something like this. <S> \$-A_Vsin(\omega t) = <S> i(t) <S> \cdot <S> R\$ <S> But from trigonometry, this is equivalent to this, which is just 180 degrees out of phase with the equivalent positive amplitude sine wave. <S> \$A_Vsin(-\omega t) = i(t) <S> \cdot <S> R\$ <S> So, physically, a negative amplitude AC wave is just another way of describing a 180 degree phase <S> shifted AC wave. <S> Note also what the sign used on the schematic actually means for AC. <S> In the schematic, current is marked as going clockwise. <S> This does not mean that the current is always flowing in that direction. <S> Let's say we evaluate \$i(t)\$ at a particular time and get a positive result. <S> This arrow just says that positive results mean clockwise current flow. <S> If we get a negative result at a particular time, then current is flowing opposite the arrow, or counter-clockwise. <A> Is a negative amplitude eqivalent to a phase oft 180deg? <S> I don`t think so. <S> To speak about "phase shift", I think, we need to compare two different continuos waveforms - and this is not the case in the example under discussion. <S> If we add a certain positive dc voltage two the symmetrical sinusoidal signal (symmetrical to 0 Volts) <S> the phase of the signal is not touched at all - and the negative half waves are shifted to positive values. <S> And - yes - in case of positive and negative half waves <S> the direction of the current changes periodically. <S> Why should this be a problem? <S> The formulation "to complete the path" is relevant for dc currents only. <S> What matters is simply the fact if the circuit allows a movement of charges. <S> And this is ensured also in case of an ac signal. <A> Negative voltage means that the node marked ' <S> +' is actually at a lower potential than the node marked '-'. <S> Consider the figure: <S> The value of V is negative if top node in the figure is at a lower potential when compared with the bottom terminal. <S> (V is measured by connecting the red probe (+ve) of voltmeter to the top node.) <S> The assumed direction of I is in clock wise. <S> The value of I is negative if the actual current flow is in anti-clock wise. <S> In the negative half cycle of an alternating source, the voltage at +ve terminal will be less than that at -ve terminal and current will flow from the -ve terminal to +ve terminal through the circuit connected to it. <S> The sign actually depends on how you measure it. <S> You measure the value of a 5V DC voltage source by connecting red probe of multimeter to + terminal and black probe to the - terminal multimeter will show 5V. Instead <S> if you exchange the connection and measure it again, it will show -5V. <S> (If you are planning to test this, use a multimeter or a voltmeter that can show a -ve reading)
| Negative value of current, whether it is alternating or direct, means that the actual current flow is in the opposite to the direction marked.
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What is the typical pin designator convention for footprints? Altium (and I'm sure other PCB Design software) supports both numeric and alpha for pin designator. This is not the pin display name, like you would see on a schematic. This is how the footprint will align with the schematic symbol. I've used pin numbers to align symbols to footprints in the past, but I'm curious what the general practice is - pin numbers vs pin names ? Are there any benefits (short term or long term) to choose one or the other ? <Q> If in doubt, use the same designators that the datasheet for the part is using. <S> Usually, datasheets use simple numbers for pins. <S> Some parts require alpha-numeric string for pin "number". <S> Both drawings from datasheet for MSP430F5500 . <A> This pin number may be a number, letter, or any combination (the pin numbers on diodes are usually "A" and "K"). <S> The pin name shown on (usually inside) the schematic symbol is not used to link between schematic symbol and PCB footprint. <A> I would say convention is on a square or rectangular parts or a part where all the pins are along the outside they are numbered. <S> Anything that is in an array with pads underneath the part is typically in an alphanumeric grid. <S> For example bga, PGA, lga, etc. <S> Benefits? <S> Well most people do it that way <S> :). <S> This will most likely march the datasheet. <S> For a rectangular part I expect to be able to count around the outside of the part to find my place when debugging. <S> For bga where you could have 2000 pins it would get a little messy with the large numbers. <S> Plus it's easy to find your way around in review or debug (or when designing your pin map if you've made the asic). <S> A2 is right above b2, <S> but if it was numerical you'd have to count or do some math in your head to find the next pin below. <S> Again this will likely match most datasheets keeping in mind <S> they often skip ambiguous letters like O. Just some thoughts hope it's helpful. <A> For simple components (diodes, transistors) it's convenient to use alpha names in footprint - it allows to use one graphical component with multiple footprints. <S> For ICs is more convenient to use names as is in datasheet.
| In my experience (Protel/Altium and KiCad), the pin number on the schematic symbol must match the pin number on the PCB footprint.
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Are solder bridges on chips always bad? I just bought my Arduino Ethernet shield – a knock-off one for $10 instead of $60. I connected it up, but it doesn’t work – LEDs light up but do not blink, and no ethernet link is established (ethernet switch doesn't light up, can’t ping, etc.). It’s as if the Arduino cannot communicate with the board. Trying to work out why, looking closely it appears as if the soldering on the Wiznet chip is messy, and a few sets of pins are bridged. Is this potentially the source of my problem, or could this be done on purpose? <Q> In the old days of DIPs, sometimes entire sides of chips were designed as heatsinks, so it made sense to connect them with a big blob of solder too. <S> But this heat-sinking technique is extremely uncommon for SOPs, which usually use pads underneath the package when moderate heat-sinking is needed. <S> (Before someone says that using pins as heat-sinks is nonexistent for SOPs, the SO8 version of OPA551 uses a couple of [non-adjacent] pins as heatsinks). <S> As for your chip, Wiznet W5100 (which alas I first read as WS100 making me waste some time) there's a datasheet for it; it seems the guys who put together that PCB soldered together most if not all the address pins, which couldn't possibly work. <A> I think that you may have found your problem. <S> I usually clean the board afterwards - if you use Rosin flux, a toothbrush and IPA works well. <S> Use enough IPA to wash all the flux off the edge of the board. <A> It has already been said that these bridges should not be there. <S> This can also be seen easily on the left side (lower picture), as the short but thick trace connects to only three pins. <S> If all pins should be connected together, there should be tracks to each pin. <S> About general design of PCBs, a colleague of mine used to connect several pads in a row by a single track through all pads. <S> As the result looks like solder bridges, it was indistinguishable from them. <S> I would never do so, but always connect the pads at the side of the pad row. <S> Then, any solder bridge found is clearly an error and should be removed. <A> That is surely the problem. <S> They may have reflowed the solder joints, and not fixed the pins later. <S> However, it may well work if you fix it; try Dwayne Reid's solution
| Cure is fairly easy - all lots of liquid flux on top of all those solder bridges and either drag the excess solder off with a hot, clean soldering tip or use solder-wick to suck up the excess.
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Consume 12v to 5v converter I am not a genius in eletronics but I know some information about converting 12v to 5v. There are different devices that can do these stuff like linear or switching converter, and I read that switching converter is the best one. Now, my problem is to understand what is my relation between device using 5V@300ma from a 12v battery that has the capacity of 100Ah.How do I calculate the life time of my battery? I am very confused even because I got a very big battery (100Ah) that should run my Raspberry Pi for a while ( hopefully over a month ) but not sure. I got a HKU5 5V/5A UBEC from hobbyking that converts from 12v to 5v. <Q> A switching converter is not the best one for every application, for yours it probably is though. <S> A switchmode converter will transfer the power with a certain efficiency, which is sadly not given for that device, so I'll assume some bad numbers. <S> $$P_{in} = <S> \frac{P_{out}}{\eta} = \frac{5V*300mA}{0.7} <S> \approx 2.2W $$ <S> Even a bad switchmode power supply should be able to get 70% efficiency. <S> So the power supplied by the battery is 2.2W, but you have the capacity in amperehours, so you'll need the current. <S> The power supplied is constant, so the worst case current draw (highest) occurs when battery voltage is lowest. <S> For a standard 12V battery it's nearly empty at 11V, so I'll use that. <S> $$I_{bat} = \frac{2.2W}{11V} = 0.2A$$ <S> With that you can get the hours the battery will be able to supply the current <S> : <S> $$\frac{100Ah}{0.2A} = 500 h \approx <S> 21 days$$ <S> It will probably run longer than that, as this is basically a worst case view on this, but there is some data missing to give a better estimate. <S> You should also be aware, that the output voltage ripple might be too high for your system to tolerate depending on how good that switch mode supply is. <A> Normally one would just divide the Ah capacity of the battery by the current requirement, and get a figure in hours. <S> However here it is a little more complicated because you have two different voltages. <S> So we'll use watt hours instead. <S> $$100\ <S> Ah <S> * 12v = <S> 1200\ <S> Wh$$ <S> The requirements of the Raspberry Pi are: $$5v <S> * 0.3\ mA = <S> 1.5\ <S> W$$ <S> So the amount of time the Raspberry Pi can run is: $$1200\ Wh / <S> 1.5 W = 800\ hours$$ <S> You can't really drain the battery all the way to 0 <S> so let's say 750 hours. <S> Furthermore, there will be some loss in the switching regulator, let's assume it has an efficiency of 90%. <S> So $$750 <S> * 90\% = 675\ <S> hours$$ <S> $$675 / 24 = 28\ days$$ <S> almost what you were hoping for. <A> The battery rating is 12V 100Ah. <S> So total energy the battery can deliver at 12V:$$E_{tot} = <S> 12 <S> \times <S> 100 \times 60 \times <S> 60 \ <S> J$$ <S> Now the 5V, 300mA device will be consuming \$5\times .3\ <S> Joules\$ per second. <S> So If we are connecting this battery and device with an ideal 12V-to-5V converter in between then the battery backup will be = total energy/energy consumption per seconds. <S> $$t = \frac{12 \times 100 \times <S> 60 \times 60}{5\times 0.3} = <S> 800\ hours <S> \approx 33\ days$$ <S> But practically the values can change because The converter will have losses inside it <S> The converter may give 5V output even if the terminal voltage of battery falls below 12V also.
| To calculate the battery lifetime, you need to calculate the current which is drawn from the battery.
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Connections for Panel Mount 12VDC Power Supply I'm using a 12V Panel Mount Power Supply , taking a 110/220VAC input and giving a 12V/5V DC output. What is the best practice for connecting this power supply to the AC mains? And how should this unit be mounted for use with a PCB, ie: do you mount this directly to the PCB, or is there a special mounting platform that we can mount both the PCB and this power supply to? <Q> This power supply is intended to be mounted somewhere inside your enclosure. <S> The screw terminals are intended to be used with either stripped wire ends or spade crimp connectors. <S> We normally just use stripped wire ends. <S> Stranded wire <S> but not tinned. <S> The pressure plates are designed to spread the wire strands out over a large surface area and provide long-term clamping pressure. <S> If you tin the wires, all of the clamping force is concentrated over a very small area and the solder between the strands will cold-flow. <S> This results in the connection becoming loose over a long period of time. <A> If you look at the left side of the terminal strip, there are 3 connection points labeled L, N, and FG. <S> L should be connected to line, N to neutral, and FG to Frame Ground. <S> For this type of supply, ordinarily it is mounted to the chassis. <S> The PC board is mounted separately, and wires run from the power supply outputs to the PC board power connector. <A> In addition to what other have answered, typically you would connect the input connecter such as Digikey or Digikey . <S> The second would be preferred as it includes a fuse so that if your wiring shorts out, you'll be protected from fire. <S> You would run wires from the appropriate terminals of the module to the L, N and FG terminals of the power supply.
| In other words, just use wires to connect the power supply to both the incoming power as well as to your circuit.
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Why does this kit only support 24V However this question may seem naive, it is still something that worries me. I bought this twilight relay kit (to turn on/off my home entrance light obviously). Thing is: The guide shows that it can only be used with 24Volt/5A. But I (and maybe others alike) want to use it for the home power line aka 230V (110V). As seent on the picture, on relay is written 240V/10A. So what's the problem then ? I can guess this: 1) Manufacturer safeguard themselves. 2) the cobberlines on the print is too small to support the voltage. 3) Some other security thing is missing on this kit (like fuse or...). Anyaway - I dont like to use this with 230V before I are certain about what could be the risk. note: I dont have a pic of the underside with the lines, but there is two straight 1.5mm lines from the relay to the connetion socket <Q> I'm pretty sure that this board is intended to operate from a 12 Vdc power supply. <S> This is based both on the relay coil voltage as well as the LED series resistor. <S> So long as you have sufficient creepage distance from the relay terminals to any other nodes within the circuit board, you should have no problem using the relay to switch AC Mains voltages. <A> The writing you're looking at on the relay (the 240v/10A) is the contact rating. <S> The coil rating is 12V, and the silkscreen on the power input is also "12V". <S> You have a 12 volt board, that can switch up to 240V, <S> 10A. <S> But that's the relay rating! <S> To be safe, find out the specs from the manufacturer or the source of your board. <S> If you can't (or don't understand Chinese), then look at the distance between the traces from the relay to the connector. <S> From having seen other similar boards, I'm thinking you're safe switching a couple of amps, but with 1.5mm traces I wouldn't go over that. <A> I post this as an answer (with the risk of getting a mod after me.. <S> ;) <S> From other answers/comments allready posted - the problem, if such - seems to be about the isolation of the high voltage from the relay. <S> I had an idea about this, but i needed some comments and confirmations..and surely got that I recall something about the authorities and the laws about high voltage devices. <S> There is a ocean of rules and regulations. <S> There is ISO, EU regulations etc.. <S> If such a kit was sold assembled to users as an 240V device, it should go through regulated tests, which include (as far i know), temperature (cold/hot), overvoltage, moisture, HF-noise and....and a fuse here and there.. <S> An ironic fact about this is, that You surely could start a fire with their rated 24V/5A (120W) on the kit (by shotcut). <S> So the simple answer must be:This kit is not designet for high-voltage (even though it may handle this). <S> And If You want to use it for High-V, then be d... sure to make some secure modifications. <S> And maybe I could add: This goes for all such home-assembly kits. <S> sometimes theres no answers, only comments and assumptions: <S> just ask about the univers and the black holes.
| Please note that relays have 2 ratings: First is the coil (the power necessary to activate the relay) and second the rating of the contacts (the power that can be safely switched on & off). The board may not be designed to switch that much power, and I'm betting if you tried, you'd burn up the traces. So the producer safeguard themselves (even it is a kit) and go below the regulations.
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What is a suitable candela for a status led? What is a suitable candela value for a LED to have that doesn't blind you if you stare at it, and will still allow you to view silkscreen markings on the PCB but not to dim that you can see the internal workings of the LED ? I suppose this question is more of an experience question. This question originally had "lumens" in the text but what I was really asking was for candela. The question has been edited with candela. <Q> For a status indicator, I'd look for an LED with an angle of 130 degrees or so. <S> If you aren't sure how bright you want it to be, buy one that is brighter than you think you need. <S> You can always use a larger resistor to make it dimmer. <A> Digikey P/N 732-4981-1-ND will work. <S> No relation to Digikey or the manufacturer, just something that happened to lay in front of me <S> so I just read the label. <S> The brightness depends on the current setting resistor and voltage, I use 10K@3.3V. Ambient light level matters too. <A> Without knowing the number, I recommend looking at some examples. <S> I find the raspberry pi LEDs to be Just Right, and the Arduino nano LEDs to be much too bright. <S> Looking at the resistors (and LED models) should give you an idea about the current you want. <S> After that, as Erik says, you'll need to experiment. <S> Different colours are perceived differently, the color of your silkscreen matters to the eye, etc.
| It depends on the brightness of the LED and the viewing angle.
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Calculate resistor for red Led in series Hello guys i am really confused about the whole resistor issue. I have visited so many sites that tells you what resistor you need but i don't know the volt drop that asks to complete or the current. I want to power 3 RED LEDs with a 9 Volt battery <Q> If the LEDs are wired in series, they will drop about 5.1V. 9V - 5.1V leaves about 3.9V across the current limit resistor. <S> I'm going to further assume that you want to have the LEDs run at 20 mA max. <S> So: 3.9V / 0.02 Amps = 195 Ohms. <S> The closest standard (E12) resistors are 180 or 220 Ohms. <S> I'd choose 220 Ohms. <S> Now let's see what happens as the battery dies. <S> A standard Alkaline battery is considered to be dead when its' terminal voltage drops to about 1V under load. <S> A 9V battery contains 6 cells. <S> 6 <S> * 1V = <S> 6V. <S> (6V - 5.1V) <S> / 220 <S> Ohms ~= 4 <S> mA. <S> The LEDs will be lit but dim. <A> The red LEDs I've used have had a forward voltage of about 1.8 volts (but there may be some newer technologies with higher voltages). <S> Typical <S> 5 mm LEDs usually have a recommended maximum current of 20 - 30 mA, but do produce ample light at lower currents. <S> So, three LEDs will drop 3 x 1.8 volts = 5.4 volts, which leaves 3.6 volts across the series resistor. <S> I usually aim for 10 mA current, so the resistor is R = <S> E/ <S> I = <S> 3.6/.010 = <S> 360 ohms. <S> The resistor value is not critical - a higher value will reduce the current, and make the LED dimmer. <A> Lots of "ballpark" figures and guessing here. <S> Red LED's are typically ABOUT 1.8V [as others have mentioned] but the actual voltage varies for each device. <S> The current you are pumping through it will also effect the voltage slightly. <S> A quick Google search for "red LED datasheet" got me devices with voltages ranging from 1.65V to 2.25V <S> so 1.8 is about average. <S> USUALLY, the limiting resistor will average out these variations. <S> If you put them in series, however, you COULD get one LED running at 1.9V and another at 1.6V... <S> which impacts the voltage across the resistor... and your overall brightness is lower. <S> It might not be a big change, but it CAN happen. <S> If this isn't an issue, put the LEDs in series: you use less current, and batteries last longer. <S> For greater control <S> [at the expense of using more current ] you can have the LEDs in parallel, with one limiting resistor on each LED. <S> 9v - 1.8V = 7.2V, which at 10mA is 720 ohms [or 360 ohms for 20mA]. <S> If you find one LED is brighter/dimmer, you can change its resistor: especially useful if you ever decide to mix LEDs with different colors.
| Red LEDs usually have a 1.7V drop. 390, 470, and 560 ohm resistors are common values which should work.
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Is it possible to solder this non-standard package (sim900) using hot-air station? I don't have any experience with hot-air soldering. Would it be possible for a complete newbie to solder this kind of package using a hot-air station: (source: soselectronic.pl ) I only need to solder pads closer to the edges (the round pads closer to the middle should not be soldered). To give you an idea about how small are the pads the whole package is 24x24mm. <Q> Without the BGA pads, that's just a QFN package with 1 mm pitch. <S> Yes, you can solder that with a hot air station. <S> The way I usually do this is to use a soldering iron to put a bead of solder on all the pads. <S> Then smear paste flux over everything and position the chip over the pads as accurately as you can. <S> A mag light is handy for this. <S> The paste flux acts like goo to hold the chip in place. <S> Now heat with hot air at around 700°F. <S> Make sure you have reasonable air flow, but no so much that the force of the air can move things around. <S> For a package this large, you need to have one of those nozzles that has a long and thin vent for each side of the package. <S> Be careful to hold the hot air wand steady and centered over the package. <S> It may take 5-10 seconds, but you should see the solder melt. <S> Make sure it is melted all around on all four sides, give it maybe another 2 seconds like that to be sure, and remove the air. <S> It is important that the solder on all pads be molten at the same time. <S> Molten solder has significant surface tension. <S> In the beginning, some of the pads will be molten, which pulls the chip down against the remaining pads harder. <S> This is how the process deals with the inevitable mismatch of a higher solder beads than others. <S> Once the solder on all pads is melted, they all pull together to line up the chip nicely with all its pads. <A> We used the SIM900 in a project last year, we hand soldered 25~ of them, It worked perfectly fine. <S> Both with soldering wire and soldering paste. <S> I tried to "answer" this as a comment <S> but I dont have enough rep. <A> It can be done with the use of solder paste and hot air method. <S> I have soldered a number of these in the past i usually perform a preheat method as well as <S> the hot air to ensure solder paste will flow evenly.let <S> me know if you need help <S> a little bout my soldering skills <S> http://www.turbotronicstech.net/work-history.html
| It is actually possible to solder it with a standard soldering station.
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Can a 6.3v capacitor be replaced with a 16v capacitor? I'm repairing my old 478 socket computer motherboard and two 6.3v 2200uF capacitors near the CPU are bust. The computer still powers on, but it doesn't pass the BIOS screen without the ps/2 mouse plugged in, and the keyboard stops working after about 20 minutes. I'm wondering if I could replace the broken 6.3v 2200uF caps with 16v 2200uF ones instead? <Q> Yes. <S> The voltage rating of a capacitor is the maximum it can take. <A> Yes, but ... . <S> A 16V 2200 uF capacitor will provide the same capacitance as a 6.3V 2200 uF one. <S> It should work properly - initially. <S> Added: [ <S> [ Spehro notes that the capacitors should be low ESR (equivalent series resistance) parts. <S> It's likely that caps with an adequate ripple current rating will also have low ESR as the two are related. ]] <S> A 6.3 V capacitor of the same capacitance <S> MAY have a higher ripple current rating - the amount of current at operating frequency that it will tolerate long term. <S> In a power supply situation this is likely to be the case and is reasonably likely to matter. <S> (ie ripple current rating is a factor in cost and designers tend not to over-design $ wise if they can help it.) <S> A capacitor run above its ripple current rating will run hot, dry out quicker and die sooner. <S> If the capacitor dis it MAY do more damage than when it first fails - but probably not. <S> Aluminum electrolytic capacitors should be run at 80%+ of their rated voltage for longest life. <S> Not liable to matter too much here. <A> If they will physically fit, probably not. <S> You must use low impedance (low ESR) capacitors on a motherboard. <S> Ordinary 2200uF 16V caps will not work properly and could cause something to be damaged if it isn't already.
| In theory, a 16 V cap is a superset of a 6.3 V cap if the other specs are the same.
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Why does a closed circuit with a pull-up resistor output "0"? I have a switch setup with a pull-up resistor. With the switch open I get a "1" or HIGH output and with the switch closed I get a "0" or LOW output. I understand why I am getting the "1" with it open (because it is connected to 5v) but I don't understand why I get the "0" when it's closed. Aren't 5v's still passing through the switch? <Q> The switch is essentially an ideal conductor, with a negligible resistance that can be ignored. <S> The resistor, as ohm laws would dictate, carries the 5v across it, "using" the voltage. <S> Since the path of least resistance is a direct connecting to ground, the node is pulled to ground. <S> When the switch is open, the node is open, and no current flows, bringing the node to 5V through the weak resistor. <S> Think of the switch as a direct wire. <A> The other side of the switch is connected to Ground (0V rail). <S> The other side of the switch is grounded. <S> When the switch is open, 5V from the pull-up resistor is fed to the controller input pin. <S> When the switch is closed, the input pin of the controller is connected to ground. <A> With the switch open,measuring across the switch will give you five volts because no current can flow through the resistor,hence you will find no voltage drop across it. <S> With the switch closed, assuming an ideal switch with zero resistance, according to our man Kirchov, all the voltage must drop across your resistor. <S> So, you measure 0V in this case.
| When closed, the node between the pull up resistor and the switch is pulled to ground or 0 potential. In other words, you have a pull-up resistor connected from the 5V rail to the top side of the switch and also to your controller input pin.
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Open circuit volts and short circuit amps I am looking to buy a few solar cells to mess around and learn with and I don't understand what exactly this part in the description is telling me: "...offering 4.5V (open circuit) and 18mA (short circuit)" Can someone please explain? <Q> It means what it says: if you have the cell unconnected driving nothing, there will be up to 4.5V across the terminals. <S> If you connect the terminals together through an ammeter, it will read 18ma. <S> There will be some sort of standard assumption about the light on the cell during these measurements. <S> This does not mean you can get 18ma at 4.5V; <S> drawing current reduces the voltage at the terminals. <S> There will be an optimal level that maximises the power delivered by the cell, and this is what a Maximum Power Point Tracker (MPPT) is for in large PV installations. <S> The actual power to be expected should be quoted separately in watts. <A> If you were to connect the two leads to your multimeter in a current mode, essentially shorting them, it would read 18 mA. <S> So this is the theoretical maximum current the solar cell can output (in practice, it will be lower). <A> Open-circuit voltage is the maximum output voltage with no load applied. <S> short-circuit current is the maximum output current with 0 ohm load. <S> Assuming the V-I curve (graph of voltage output vs current output) is linear, then the source impedance is 4.5V / 18mA = 250ohms. <S> In other words, this solar cell behaves like a 4.5V battery with 250 ohms in series. <S> This source impedance is not a real resistor component, but a characteristic of the physical voltage source. <S> When you model its behavior, you model it as though it were a Thevenin voltage source of 4.5V in series with 250 ohms. <S> At the extreme of open-circuit, the load current is 0mA and therefore the power transfer is 0 watts. <S> At the other extreme of short-circuit, the voltage provided to the load is 0V and therefore the power transfer is 0 watts. <S> To get maximum power transfer, the load impedance needs to be matched to the source impedance. <S> (If you could choose the source impedance you'd obviously choose 0 ohms, but that's not possible.) <S> If the load is 250 ohms and the source impedance is 250 ohms, then the power transfer will be 2.25V <S> x 9mA = 20mW (assuming linear V-I curve). <S> In reality, the V-I curve is probably not linear, but without more data it's probably good enough for initial estimates. <S> If you're trying to get 10-15mW that should be possible, but you won't get more than 20mW out of this solar panel.
| It means if you were to measure the voltage across the leads coming from the solar cell with a multimeter with no load on them the voltage would be 4.5v.
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FTDI chip based custom board. Driver not working I designed a board containing a USB to UART bridge in order to program a ATmega microcontroller. I copied the schematic of this breakout board from Sparkfun https://www.sparkfun.com/products/9716 This is my schematic: I install FTDI drivers from this link: http://www.ftdichip.com/Drivers/VCP.htm in "windows - setup executable". It doesn't work. My computer doesn't recognize it. I have windows 8.1 and I think it could be related. I disabled device driver signatures. I don't know what to do now. I don't think it's related with my schematic, but I can't solve this problem. Any idea? Thank you. <Q> If by "my computer doesnt recognize it" you mean you can't run the exe to install the drivers, you probably have a corrupt download. <S> Download the file again and see if it works. <S> If you mean the installer worked just fine, but the drivers aren't working properly, then you either hooked it up wrong, have a defective FTDI chip, or have a counterfit chip. <A> The problem is solved. <S> As Eugene Sh suggested, it was a problem with USB lines. <S> I didn't assemble the board <S> so I supossed it was properly soldered, but it wasn't. <A> Were the D+/D- differential pair laid out on the PCB appropriately? <S> Have you attempted to look at the USB enumeration with a protocol analyzer? <S> I believe for a device to be recognized, all that needs to happen is <S> the 1.5K pull-up is present, so I would lean towards incorrect signaling environment as the culprit. <S> Also, your schematics don't show an oscillator, but there are pins for an oscillator. <S> While I'm unfamiliar with this exact part, most of the USB device ICs I have used have required a 12/24/48MHz crystal.
| I fixed one resistor (R4 in my schematic) and now it works fine.
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MOSFET not fully OFF when Gate Voltage is 0V A N-channel MOSFET STP16NF06L is used to drive a 12V load. Pin 1 is connected to Arduino Uno digital pin 4. Pin 2 is connected to the negative terminal of the 12V load. Pin 3 is connected to the GND terminal of a 12VDC power supply. The positive terminal of the 12V load is connected to the +12V terminal of the 12VDC power supply. Problem: When the Arduino output pin is at LOW state and measured to be at 0V, the MOSFET still turns on and passes 6V to the 12V load. When the output pin is in the HIGH state, it's measured to be at 4.9V and 7.5V across the 12V load. Shouldnt the MOSFET provide 0V across the load when pin 1 is 0V, and just above 10V when pin 1 is at 5V? Pin 1: Green clip to Arduino pin Pin 2: White clip to negative terminal of load Pin 3: Black clip to GND of 12V power supply <Q> For your connection to work, you must have a common ground for the MOSFET and the Arduino. <S> The MOSFET's Source must go to the common ground. <S> Also your (12V) power supply's ground must be the same ground. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> If the Arduino pin swings from 0V to 5V when disconnected from the MOSFET, replace the MOSFET. <S> And use proper anti-static working procedures this time. <S> Also note that most Peltier coolers take 6-10A, so with an ON resistance of 0.1 ohms the MOSFET will dissipate 3.6 to 10 watts, and die rather quickly unless you mount it to an appropriate heatsink (say 5C to 10C per watt) <A> I think you missing a 10k resistor. <S> Just run a 10k resistor from ground to mosfet <S> pin no. <S> 1 (digital pin 4) <S> and then it will work fine! <S> here is the working example and circuit diagram. <S> http://www.circuitmagic.com/arduino/run-small-brushed-motor-for-mini-quadcopter/
| If you have the device connected correctly, and the Arduino ground is commoned with the 12V supply ground, the next suspect is electrostatic damage to either the MOSFET or the Arduino.
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why does the model s use multiple batteries? The Tesla Model S and NASA's Robonaut both use a battery pack that consists of an array of hundreds of battery cells. Why don't the engineers just create one giant cell? <Q> Your question is self nullifying: <S> Why don't the engineers just create one giant battery? <S> Google: "define: battery" <S> Battery <S> /noun/ <S> a container consisting of one or more cells, in which chemical energy is converted into electricity and used as a source of power. <S> A battery is a collection of cells. <S> So that is exactly what they have done. <S> They have created a giant battery. <S> A battery is made of cells. <S> A cell gives a small amount of power. <S> Combine the cells together and you have a battery that gives a large amount of power. <S> It's like building a house. <S> Would you rather: a) build it from bricks you can buy from the brickyard, or b) <S> build it out of one gigantic brick you then have to hollow out by hand to make your house? <A> The voltage out of a single battery cell is a function of the battery chemistry. <S> For example, carbon-zinc cells generate about 1.5 volts as does alkaline. <S> Lead acid cells are about 2.2 volts and so on. <S> For an electric car, the power required would necessitate having currents of hundreds and even thousands of amperes if only a single cell was used. <S> This would require very heavy cables and would be very inefficient as a significant amount of the battery energy would be wasted in the cables. <S> For this reason, these cars use a large number of cells in series to produce several hundred volts. <S> This reduces the magnitude of the current and hence the cabling to manageable levels. <S> Even a simple 9 volt transistor battery uses 6 cells in series because there is no simple battery chemistry that will directly generate 9 volts. <A> A battery consists of a number of individual cells, usually connected in series, although sometimes several groups of series-connected cells may be connected in parallel, depending on the required battery capacity, and practical sizes of the cells. <S> The individual cells may be more easily arranged to fit in awkwardly-shaped spaces, and if one cell fails, you only have to replace that one cell, rather than the whole battery.
| It may be more practical to use individual cells, rather than use a pre-assembled battery containing several cells.
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How to deal with low space and high current? In my circuit I am using a stepper motor driver, providing me up to 3 A current per trace. Thus I calculated the required trace width with a trace calculator , which told me to use a trace width of 0.07'' (or 1.78mm). Unfortunately, the chip itself does not allow me to connect them directly onto the IC legs. These are only 0.012'' (or 0.3mm) traces. Therefore I created this routing: Pin 1, 21, 24 and 26 are responsible for the high current. I used 0.012'' traces right after the pin connections, then switched to 0.016 and 0.024 and as soon as possible to 0.07''-traces (the big vias). Is that enough, or will the possible heat damage my board? How can I improve it (instead of soldering copper bars on it)? Driver Datasheet <Q> Here's a little tip for you: <S> The thickness of trace needed can be seen as a function of the length of the trace. <S> The longer the trace the greater the voltage drop, and thus the thicker the trace needs to be. <S> Online calculators all assume the trace will be a constant width the entire length. <S> This is not the case in your (and many other) situation. <S> So treat your layout as two traces . <S> One trace that is the length from the chip to the start of the wide portion, and a second that is the length of just the thicker portion of the trace. <S> Perform two calculations - one for each portion. <S> You'll find the shorter portion of the trace can generally be thinner than the rest. <A> At this stage, fine modifications to placements can make a difference. <S> It can be tedious, but it's worth knowing <S> how. <S> Start with the unnamed passive component, 2nd in from the bottom left. <S> The upper track to it takes a tortuous route occupying critical space. <S> Delete it, move the via above it down as far as you dare, then squeeze that track back in. <S> Now you can massively straighten out the track from Pin 1. <S> That should let you move the capacitor from Pin 28 left a bit, maybe half its length, without changing the overall length of its routing. <S> Which opens up quite a lot more space above the chip to improve the routing there. <S> And so on. <S> You won't be able to get full width tracks all the way in, but you'll be able to get intermediate width tracks within a pad-length or so of the pad. <S> Takes practice and time. <S> At some point, you decide it's done... <A> The A4983 data sheet states that peak current is 2A <S> so this means you are going to get into some trouble when driving 3A loads. <S> Additionally it is the RMS current that is important for calculating track widths - it is the RMS current that heats the tracks. <S> I'd estimate, that if you were using the device at the full 2A peak current, you should be plugging in something like 1.5 amps into your trace calculator.
| Of course, if you go too thin then the heat dissipated could get too high, so it's a bit of a balancing act - try and keep the thin trace as short as possible.
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Appropriate fuse for 12V battery - solar power controller connection I'm a biologist tracking migratory bats and need to build a bunch of antenna towers to listen for pulses from tagged bats flying nearby. These are off-grid, so we'll have 55W solar panels charging 12V 100Ah deep cycle batteries during the day, and the sensor gnomes use that battery power to listen for bats at night. I need to put a fuse between the battery and the solar controller. The first version of this system used 20A class CC fuses and a bulky fuse box meant for home solar use. Is there any reason I can't use a smaller and cheaper fuse and box, perhaps an in-line model made for car batteries? The controller is the Morningstar Sunsaver SS-6L-12V if that matters. Let me know if you need more information. Thanks! <Q> Just to follow up on this: the project is over now without any fuse-related mishap. <S> I deployed about 40 systems and for most of them I just used in-line auto fuse holders with 20A fuses, purchased from radio shack or auto parts stores, on the positive line between the battery and the solar controller. <S> Occasionally fuses were blown, but rarely. <A> For a 12V 55W solar charger, you would need approx. <S> 10A fuse. <S> But please remember, my guess is based on the fact you give me and better safe than sorry. <S> But if you find a 10A breaks too often, you can safely insert 15 or 20 A for cars also. <S> Jst ensure you use appropriate wire gauge. <S> 10A car fuses are fine for that. <S> With 10A fuse you could deliver 120W before it breaks. <S> I guess the fuse it ment to safeguard the Solar Charger if you by accident should short the wires or swap the wires on the battery. <S> Sine <S> this is outdoor equipment, this could be the reason why the expensive fuse holder is used. <S> Bad wheather and electricity are not good friends. <S> You can get cheap electricity boxes with rubber covered holes for sticking wires through, thus ensuring dry connections. <S> Good luck with your project <S> , it sounds interesting. <A> Normally the fuse is sized based on the wire diameter. <S> The idea is that the fuse will blow before the wire gets hot enough to start a fire. <S> But there is another concern with large 12V batteries. <S> The fault current may be extremely high. <S> A large 12V battery can deliver maybe 500 Amps or more during a fault. <S> A small 12V fuse may essentially explode with such high fault currents. <S> Maybe it could even form an arc and continue to conduct current into the fault. <S> I am not sure. <S> " The most conservative thing you could do would be to use a physically large fuse with a high interrupt rating. <S> AFTER this fuse, you can use a normal automotive fuse with no problem, if necessary. <S> For boats, where fire is a major concern, the types of fuses you can use are ANL, Type T, and MRBF. <S> These have interrupt ratings high enough (thousands of Amps) to operate (blow) without drama when you short out a large battery. <S> Probably the easiest solution is to install an MRBF fuse holder directly on the positive battery terminal. <S> There are 30A fuses available with the MRBF rating <S> Then use 12 AWG wire after the fuse. <S> Probably no additional fuses will be needed in your application, but you could put a smaller fuse after the 30A fuse, and then use smaller wire, if the smaller wire was adequate for your needs. <S> Here is a picture of an MRBF fuse holder and fuse. <S> This type mounts directly on the battery terminal. <S> This is not an endorsement of the specific product pictured. <S> I just thought a picture would be helpful.
| But the bottom line is that fuses typically have a maximum interrupt current rating, in addition to the normal rating for where they will "blow.
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Replacing 18650 cells in laptop battery with greater capacity ones My laptop battery died and I can't find a replacement (old laptop) so I'm going to need to replace myself the 18650 cells it has. It is an 8-cell battery (a series of 4 banks of 2), making the whole package 14.8V and 4Ah. So each individual 18650 cell is the typical 3.7V and 2Ah. As a possible replacement I found these: http://www.ebay.com/itm/2-x-3-7V-4000-mAH-18650-Protected-Lithium-Based-Rechargeable-Battery-Flashlight-/131347594474 Which are 4Ah (twice the capacity of my current ones) and also they are protected. My current battery cells don't have any marking that indicates whether they are protected or not but I have the impression they are not since the battery charging circuit should provide such protection. Also, the physical dimensions of the cells are exactly the same as my current ones. My question is: Would there be any problem that the new cells have twice as much capacity? Also, does it affect in any way that the new cells are protected and the old ones are not? <Q> Having protection circuits does not guarantee it will not happen. <S> Cells without protection are intended for use by either manufacturers or experts or enthusiasts who add their own or for suckers. <S> Whole device protection and cell protection are complementary and serve overlapping but different roles. <S> 4,000 mAh 18650 LiIon batteries are ~~~= 99.9% +0.1% - 0.0% sure to be rubbish. <S> ie <S> not just < 4000 mAh but << 4000 mAh and low quality. <S> The people who bogus label cells almost never feel an obligation to use the best cells they can and almost always decide to add injury to insult by using junk as well as lying. <S> Real world experience shows that the value of 'almost' is very high in both cases. <S> Higher capacity cells can usually be fitted OK. <S> MUCH higher capacity will lead to long CC tails and overcharge but not an issue here as mAh_new is < to << 4000 mAh. <S> ADDED: <S> Notice that in this ad and all their other ads they ALWAYS show non-brand-label views of the battery. <S> However, you may find that the racing stripes and general colour scheme a good match for the well known "Ultrafire" brand batteries. <S> This may in fact be a real brand and these may be real examples of it <S> BUT you can buy empty shrink wrappers to apply to the battery of your choice with this (or other) branding on it , so caveat emptor. <S> Better nullius emptor <S> I'd hazard. <S> These ones are a stunning 6800 mAh - a steal at the price. <S> Available here <S> You'll find others similarly arrayed here and here - 3000 <S> mAh and 4000 mAh and 6800 mAh !!!!!!!!!!!!!!!! {again} and unspecified but with GENUINE CREE 2000 lumen {so 20+ Watt} flashlight for $9.27 and only 4000 mAh and <S> that's better - 4200 mAh and ............. <S> Flee! <A> A few comments. <S> 1) Usually no problem replacing old 18650 cells with new. <S> 2) <S> You really should take your new cells to a professional battery build company and have them do the welding for you. <S> Under no circumstances should you attempt to solder directly to the end terminals of a Li-Ion battery. <S> 3) <S> There are two types of "protected" cells. <S> Those having only over-current protection and those having low-voltage cutoff boards built in. <S> The cells having only over-current protection are a tiny bit longer than cells without; cells having low-voltage cutoff can be as much as 1.5mm longer than cells without. <S> In general, you are better off purchasing cells that don't have any protection added. <S> The battery management circuit within your battery pack will take care of that for you. <S> 4) <S> It's probably wise to actually test the cells you purchase before you spend the time and money to rebuild your battery pack. <S> Charge them fully, then put a load on each cell and monitor the cell's performance as it discharges. <S> This can be as simple as a resistor on each cell. <S> Bonus points if you are able to actually match the capacity of all the cells going into your rebuilt pack. <S> You should immediately recharge your cells after discharge-testing them. <S> You don't need to take them all the way to full - around 30% net charge <S> is good. <S> Testing your cells first is a good way to weed out any duds. <S> 5) <S> Your battery pack may not work after you have replaced the cells. <S> The cure is usually simple: apply a charge voltage to the end terminals on the pack connector. <S> This voltage needs to be current-limited to a low value (50 - 100 mA) with the voltage high enough that to ensure that current flows into the battery. <S> That usually resets the shutdown circuit on the protection board. <S> Note that when I say "end terminals" on the pack connector, what I mean is the terminals that correspond to the most positive and most negative pins in the connector. <S> These are not necessarily the actual end pins on the connector. <A> Usually LiPo battery chargers implement a timeout feature. <S> Since you are doubling the battery capacity, there is some remote chance that your charger will timeout before they completely recharge. <S> Even if this does happen, it is probably not a big deal. <S> You will still get a lot of life from the pack. <S> Apart from that, I suspect it will be OK. <S> I recommend that you buy protected Panasonic cells. <S> I bought mine from Orbtronic. <S> Personally, it makes me nervous to be at the absolute cutting edge of high capacity LiPo battery technology, and also to use unknown vendors for the cells. <S> So I would stay away from anything rated at 4 Ah, and I would stay away from any cell vendor I have never heard of.
| Cells in devices that you do not wish to be a flaming ruin MUST have protection. You can purchase cells with tabs already welded on but those may take up too much room in your existing battery case.
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Voltage drop when measuring battery capacity I'm trying to build a battery capacity tester for Li-ion batteries. I made this schematic using Arduino: While bush button is not triggered arduino only shows battery voltage on Serial & when button is triggered it turns on the MOSFET and starts the test. My problem is while MOSFET is off the battery voltage that arduino reads is battery's 100% voltage (e.g. 3.97v) But when MOSFET turns on the battery voltage drops to a lower value (e.g. 3.21v for that initial 3.97v) ! And when the test is finished (arduino reads 2.7v) and it turns off the MOSFET that dropped voltage comes back! and 2.7v becomes 3.5 !! for different amount of load resistor this drop is different (larger resistor ,less drop!) Why this happens? am I doing something wrong? <Q> A battery has some internal resistance that for Li-ion batteries is usually in the 0.15 \$\Omega\$ region when fresh. <S> As you should know from Ohm's Law, \$V = <S> I <S> \times <S> R\$ <S> so if you put a load on the battery this will cause a current to drain from the battery and this will cause, depending on the current, a small voltage drop over the battery's internal resistance as it would on any normal resistor in a circuit. <S> This can be explained visually like so: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> As the battery gets used and 'ages' this internal resistance increases <S> so the voltage drop over it will also increase for a constant current. <S> The 'open-circuit' voltage of the battery will always be more or less the initial voltage as the 'load' <S> will be near infinity so all the voltage drop will be over this air-gap between the batteries two terminals. <S> As you mentioned the drop in voltage is less with a larger load, this is because the current draw will be less and hence the drop over the internal resistance will also be less. <S> You should look into a constant current source using a low voltage op-amp (I did something similar to this a couple of months a go and ran into the same issue - hence that question) and you'll be able to work out the mAh easily enough as the current should be more or less constant and you'll know the time taken for the voltage to fall. <S> Hopefully this helps and answers your question <S> - it is expected that a voltage drop will be seen when applying a load to a battery. <A> The constant current and endpoint voltage should be specified in the battery datasheet. <S> It is normal for the battery voltage to recover when you remove the load. <S> That is, the open circuit voltage will be higher than the voltage under load. <S> But the voltage under load is what is used for capacity measurement. <S> Your circuit is not a constant current circuit, so you may get slightly different results compared to the datashseet. <S> Your circuit (if you fix the mosfet problem Spehro mentioned) could still be useful for comparing cells to each other, but not for comparing to battery specifications. <S> Make sure that the discharge current does not exceed recommended maximum for the given cell. <S> For example, if you are testing small cells (less than 2 Ah), your 2 Ohm load could discharge the cell very fast, resulting in abusive conditions for the cell. <A> A simple workaround is to measure voltage under load and without load. <S> So you can calculate the change in internal resistance. <S> Regarding the capacity of the battery, it is important to know under what discharge conditions this capacity is calculated. <S> Very often this is a 20 hour discharge with low current - conditions too favorable for good results.
| The normal procedure for measuring battery capacity is to discharge at a constant current until the battery voltage drops to some endpoint voltage.
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Dupont connector for 5A I have to do a connector adapter, that has to handle 5A current, from a pwm motor driving circuit(power mosfet), to the motor.I have not found the current rating for dupont connectors It is possibile to use them for a prototype connection or it is dangerous?Thanks <Q> I think those cables are 26AWG (American wire gauge). <S> You can use an online calculator to get the AWG you need, depending the Voltage, Amperage and length. <S> http://www.solar-wind.co.uk/cable-sizing-DC-cables.html <A> As a rule of thumb you can pull 3A through such a connector. <S> It is very hard to find current ratings for 0.1"/2.54mm dupont connectors in most datasheets, some do specify one though, eg: <S> http://www.molex.com/webdocs/datasheets/pdf/en-us/0901200121_PCB_HEADERS.pdf <S> However I have certainly seen poor examples of these connectors, <S> usually the pins, which I would not rely on for even 3A. To address your question "is it dangerous", you could certainly try this as a prototype, but you should never leave the device unattended. <A> Buy your connectors from a supplier which provides the specs for it, then you'll know the current rating. <S> Example: Dupont 2.54 mm pitch connector , rated for 3A AC or DC. <S> Using no-name parts without any specs is asking for all kinds of surprises. <S> I know it's impossible to prove a negative, but it's very unlikely <S> the pins you've got can handle 5A. <A> Most of the connector I havefound are only 2.5Amp, but for safety it should be derated a bit depending how much overall power is going through it.
| You may also see lower current consumption in such a prototype compared to the final design due to the extra resistance in the connection. BTW, I have never seen Dupont connectors rated for more than 3A.
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Are normally open contactors virtually safe from failing closed? Quick question: are normally-open contactors (which I assume are simply high breaking capacity relays) virtually safe from failing closed? I want to cut off a 400V 3-phase supply using an emergency stop, and I have 2 options: Use a 3-pole emergency stop, and wire the three phases to that button. I don't like the idea of 400V being so close to the user's hands. Wire the contacts of a 3-pole normally-open contactor on the supply line, and wire the emergency stop on the 230V coil. Three phase is used for heating, not for a giant chainsaw, but the emergency button should still do its job should it be pressed. So is option 2 commonly accepted, and safe enough for an end product? Any norms stating this on which I could rely? Edit implementing the answer I am quite keen on the idea of forcing the user to fix a failure in the redundant system, but I'm trying to limit the number of components as well. Is this a good comromise? I found this contactor , which has 1NC and 1NO auxiliary contacts that we can use. Since the current in the coils is already very small compared to the switches capability, I made them single pole, if that is acceptable. In particular, I'm not sure whether there is a possibility that the auxiliary contacts do not follow the main contacts (meaning the main contacts may fail closed or welded without the auxiliary contacts staying open when normally closed), could you confirm? <Q> EDIT - the original question referred to relays and not contactors. <S> It might be confusing to see the question now referring to contactors with my answer suggesting the use of a contactor! <S> Here's what wiki says about contactors and relays: - <S> Unlike general-purpose relays, contactors are designed to be directly connected to high-current load devices. <S> Relays tend to be of lower capacity and are usually designed for both normally closed and normally open applications. <S> Devices switching more than 15 amperes or in circuits rated more than a few kilowatts are usually called contactors. <S> Apart from optional auxiliary low current contacts, contactors are almost exclusively fitted with normally open ("form A") contacts. <S> Unlike relays, contactors are designed with features to control and suppress the arc produced when interrupting heavy motor currents. <S> Here is what Rockwell say about safety contactors (you may decide on this as a preferred product): - <S> Safety Contactors and 700S Safety Control Relays provide mechanically linked, positively-guided contacts up to 97 A which are required in feedback circuits for modern safety applications. <S> 100S-D Safety Contactors provide safe isolation of hazardous motion loads, using mirror contact performance. <S> And finally, the EU machinery directive implies that hazards be assessed and appropriate safety measures taken. <S> The impact of this is that equipment/installations can be categorized with a letter/number and appropriate equipment purchased to maintain that level of safety. <S> I would recommend that you look into this. <S> Here is a website related to safety relays that discusses the relevant safety categories. <S> Incidentally you may need to use a category 3 contactor. <A> No, they are not safe from failing closed. <S> No switch is, mechanical or solid-state. <S> For your option <S> #1, you would wire these in series, but it's not recommended because of the reason you stated and because there's no way to tell that one has failed until they both go. <S> For your option #2, you would wire these as part of two identical circuits, with all relevant safety switches wired in series, and the two channels wired in parallel and kept separate electrically. <S> (no crossover wiring) <S> Then each channel drives the coil of its own contactor, and the contactors are wired in series to control the load. <S> Also with option #2, you can now create a latching/lockout circuit using the two safety channels and the contactors' auxiliary contacts so that a separate button is required to turn it back on once the safeties are satisfied, and only if both have dropped out. <S> This forces you to fix a stuck contactor before they both become stuck. <S> Per a comment, here's one possible version of option #2: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Or, if you have sufficiently deep pockets, you could buy a safety rated PLC and do all of this in software with even more fault-checking and detailed diagnostics. <S> Please note that I am in a different industry now, and so there may have been some legal changes since I left. <S> Check the latest electric codes, OSHA regulations, etc., before trusting this (or anything else really) to an operator. <A> Even if the relay is 100.000% reliable from that kind of failure, the driving circuitry could fail or something could jam the contactor test button or whatever. <S> Some products have three safety devices- <S> one electrical, one hydraulic and one purely mechanical, because a failure could result in severe injury to an operator.
| Therefore, an E-STOP or other safety switch requires at least two overrated contacts that operate independently, so that one getting welded does not prevent the other from operating. People normally use a contactor for this sort of job: - Found here . Whatever the manufacturer may or may not say, I would strongly urge you to limit the consequences of such a failure by adding redundant safety features such as a thermal cutoff or mechanical stop so that injuries and intolerable property damage are prevented. Mirror contacts provide reliable indication about the open or closed status of the main power poles.
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Working with a tiny IC I have recently purchased several Atmel's touch sensor IC's AT42QT1010 which are extremely tiny (1.6mm x 2.9mm) and I would like to be able to mount them on a breadboard. I have tried soldering them to a PCB but I just can't solder that precisely so I always end up soldering the pins to each other. Is there any IC socket (or anything else) that would increase its size or at least the space between the pins? <Q> I presume you got the SOT23-6. <S> Could be worse, they do make chips in even smaller leaded packages, and extremely small leadless packages. <S> They don't make sockets for chips this small, possibly aside from expensive and very bulky ZIF sockets. <S> These chips are just meant to be soldered down onto a board. <S> I would recommend getting a breakout board that will convert this chip to a DIP form factor for breadboarding. <S> Just search on ebay for sot23 adapter <S> and you'll find a bunch of breakout boards, make sure you get one with the correct pin pitch. <S> You'll have to solder the chip to the breakout board yourself, but once it's on there you should be good to go. <A> I've had success soldering these onto a small piece of veroboard using thin copper wire to contact the actual leads, and some .1" headers to make legs. <S> It helps somewhat to bend the four outer leads to the side (be careful! <S> only bend them once or you stand a good chance of breaking them off) so that it's easier to reach the middle legs. <S> Pictures is of a SOT-23 part, but it's the same deal. <S> That said, if you end up soldering the legs together then your problem may be that your soldering iron is too large. <S> Also: Solder wick. <S> Don't attempt this without it. <A> Proto-advantage. <S> Com will sell you a breakout adapter pcb, with legs, and will even order your chip and solder it down for you at very reasonable cost.
| If you intend using parts like these - which you often have to because all the good stuff is in tiny packages - then it's a good idea to invest in a decent soldering iron with a fine tip and plenty of power (so it can heat up solder with just the tip).
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Fuse: why does it connect in parrallel with resistor? I have a wire in normal condition runs 4A. The device I am using is auto-transformer, and its data sheet recommend to use fuse of 10A. So I went to buy the 10A fuse, and mount it onto a fuse holder. However, my friend recommend to insert a 150k ohm resistor connected in parrallel with the 10A fuse. I don't know what he is talking about, is he doing this correctly? Why do we need a resistor? In normal condition, can we still use the fuse without the resistor? Because he doesn't seem to know what he is talking about. Please refer to the figure below. The fuse holder can be found here <Q> This fuse holder has a small neon bulb in series with the resistor. <S> The neon bulb acts as an indicator that the fuse has blown. <S> When the fuse is intact, the voltage across the fuse is small. <S> So is the voltage across the neon light with resistor, because it's in parallel with the fuse. <S> When the fuse blows it becomes open circuit, and the supply voltage appears across the neon bulb with resistor. <S> The neon bulb lights up. <S> The purpose of the resistor is to limit the current through the neon bulb. <S> [from Amazon product customer review] <A> Your fuse holder already has a resistor and neon bulb in parallel with the fuse. <S> As the fuse is basically a near perfect short in parallel with the resistor and bulb, the resistor and bulb will have little to no current flow. <S> It looks like it is a Brown Black Yellow resistor with Gold or Silver tolerance, which is 100K 5~10%. <S> When the fuse blows, the resistor allows a little current through the neon bulb, lighting it to let you know the fuse blew. <A> The FUSE is a short circuit PAST the resistor and bulb for the supply voltage. <S> If/When the fuse blows the voltage will take the parallel path and light the Neon Bulb (probably an NE2) indicating that the fuse is blown. <S> Same circuit is good for DC using an LED and a current limiting resistor.
| Wired across the fuse terminals and contained in the plastic cover is a resistor and neon indicator that would presumably light up if the fuse was ruptured.
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I want to replace a car dash cam battery with a capacitor My car dash cam (G1W) has a weak battery. When it dies the camera will not function at all. I have already replaced it once. The original battery was 200ma, but I found a 320ma one that would fit. It lasted about the same as the original - a year. The newer models (G1W-C) of the camera have a capacitor as original equipment, but I can't get any information from the Chinese manufacturer(s) except that it's a super capacitor. How do I calculate the necessary capacity? It only powers the camera for about 10 seconds so that it can save the recording before the camera shuts down. <Q> There are two electrical relationships that are useful for approaching this problem. <S> First, current is the amount of charge per time (one Amp is one Coulomb per second):$$I=\frac{Q}{t}$$Charge is measured in Coulombs and literally represents a quantity of electrons. <S> \$6.241x10^{18}\$ electrons, to be exact. <S> So if you were controlling current flow through a wire and could magically see the electrons flowing by, you would achieve exactly 1 Amp when exactly \$6.241x10^{18}\$ electrons were flowing through the wire every second. <S> Second, capacitance is the amount of charge per volt (one Farad is one Coulomb per Volt):$$C=\frac{Q}{V}$$The capacity of a capacitor is measured by the amount of charge (Coulombs <S> ) it takes to change the voltage across the capacitor by 1V. <S> It's easy to see the relationship between the two equations:$$Q= <S> I*t <S> = <S> C*V$$Rearranging <S> , we get:$$C=\frac{I*t}{V}$$Now we have an equation that tells us the capacitance necessary to support a given current flow for a given time, given a desired change in voltage across that capacitor. <S> For example, let's say the capacitor was initially charged to 5V while the camera had power. <S> Let's assume the camera can operate down to 3V, so we can afford to lose 2V during those 10 seconds. <S> We'll also assume the camera draws a constant 100mA during that time.$$C=\frac{I*\Delta t}{\Delta V}=\frac{100mA*10s}{2V}=0.5F$$So <S> you'd need a 0.5F capacitor in this example. <S> Do not assume any of these numbers actually apply to your specific camera. <S> An important thing to note. <S> In this example, I assumed a constant current flow which makes everything nice and linear. <S> In reality, as the voltage across the capacitor drops, the camera will draw less current, which will reduce the rate at which the voltage drops, etc. <S> To properly solve the equation, you need to form it into a differential equation. <S> The end result is actually an exponential decay curve. <S> That said, if the change in voltage is relatively small, performing a linear approximation as above is "good enough" for a rough estimate. <A> So, let's say your battery has 3V, and the cam would just switch off if the voltage drops below 2.5V. <S> Also, 3.5V would be an over voltage and could damage your device (These are just random numbers, to find out the upper limit, you would need a few cams...). <S> Further more, let's assume your cam sinks 100mA for 10s. <S> You have to charge your capacitor to 3.5V, and after 10s delivering 100mA <S> (Charge: Q=100mA*10s=1As), it still needs to have 2.5V. <S> It is <S> $$ C=\frac{\Delta Q}{\Delta <S> U}=\frac{1As}{1V}=1F$$ <S> So you need a 1F capacitor, which is quite huge. <S> Even if you find one, it has to be charged to 3.5V. <S> You did not say if your cam recharges the battery, but even if , it will for sure not try to charge it to 3.5V <S> And there is still plenty of charge in the capacitor at 2.5V. <S> To use a capacitor of reasonable capacitance, you would need a circuit which charges the cap to a high voltage like 12V from the car and then a voltage converter to 3V. <S> There are voltage converters on the marked which give you 3V from lets say 0.8V to 12V with >80% efficiency. <S> (Yes, they also boost a lower voltage to a higher.) <S> As you can see, it's not just plug-and-play. <S> It now depends on how much effort you want to put into it. <S> Another way is to use the un-switched (i.e. always-on) <S> 12V from your car, regulate it to 3V, and put this into your cam. <S> But make sure you do not drain the battery and also use fuses, as there's a lot of power behind the 12V in a car. <S> (Again: The numbers were just exemplary, I have no idea what the real numbers are. <S> I also doubt that 100mA could be a far to low guess.) <A> What kind of measuring equipment do you have ? <S> Have you tried to create a charging circuit of your own , supplying enough power (amps !) bypassing the original circuit ? <S> Not being specific, but it seems to me, that the charging circuit of the battery (capacitor ?) was not well dimensioned from the start. <S> Or is this a "new / modern" way of emptying your pockets ? <S> Any short circuits ? <S> ( I don't think so, but I just have to ask ). <S> Kris Norway <A> This is a 2.5F Supercapacitor assembly, that is good for up to 5.4V, which is typically used as a direct replacement for internal prismatic lithium ion polymer cells: http://www.amazon.com/Mobius-Action-Camera-Super-Capacitor/dp/B00JG7CRYU <S> The unit is quite small, and will fit into a variety of dash cam units (not just the Mobius), and can be parallelled with additional units to increase the overall C. <S> The major advantage gained, here, is that the Supercapacitor arrays will better withstand the typical high temps developed in a vehicle, during the Summertime months, than do the lithium ion polymer cells that you're having short service life issues with (LiPo cells hate heat). <S> Now, to give you an idea of the successful implementation of this particular supercapacitor, the Mobius utilizes a LiPo cell of 820mA/h capacity, with a nominal terminal voltage of 3.7V (full charge terminal voltage is 4.2V). <S> The above indicated supercapacitor assembly, having a capacitance of 2.5F, is sufficient to soft power down the Mobius unit, allowing it to close the video file current at the time of power down, preserving the recording successfully, while maintaining sufficient charge to maintain the internal clock function and parameter settings for several days. <S> I believe that you will find this to be a satisfactory solution for your needs, in this instance.
| The main difference between a battery and a capacitor is that the battery maintains the voltage while it sources current, while the voltage of a capacitor will decrease linearly over time.
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Is current flowing backwards a bad thing? In the circuit below: I find it awkward that the direction of the current on the 2V power supply is running opposite to the direction of the current on the 5V power supply. In the 5V power supply the current flows from - to + but in the 2V power power supply the current flows from + to - . Mathematically, it all works out and the numbers add up but I am having a little bit of a hard time gaining some intuition regarding what is going on with this "backwards" current. Here are a couple of questions: If the 2V power supply was a battery, would this mean that the battery would be charging? If that was the case, would the battery eventually explode due to overcharging? If the 2V power supply was a regular power supply (connected to the wall) why is the power supply not breaking? The power supply is not a battery, right? So there should be no such thing as "charging" a power supply so why do I not see smoke coming out of the power supply given that I am going against the natural flow of electrons? <Q> As the other answer says, from a purely theoretical point of view, there's nothing wrong with current flowing either way through a voltage source. <S> That's exactly what an ideal voltage source is: a circuit element that maintains the same potential on its pins, no matter what the current through it is. <S> In the real world, our voltage sources are not ideal. <S> A linear regulator will not usually maintain regulation when current is reversed. <S> A battery will charge (if its chemistry permits it) when current is reversed. <S> But we also use voltage sources to model things besides power supplies. <S> For example, a voltage source can model the output of an op-amp. <S> And an op-amp output typically can maintain its output voltage whether sourcing or sinking current (within limits). <S> Or a voltage source could model a shunt regulator or zener diode. <S> These devices will only maintain regulation when they are sinking current. <S> Or a voltage source could model a forward-biased diode. <S> Diodes also will only work as voltage sources when current flows in to the more positive terminal. <S> When thinking about real circuits, you need to consider what actual device the voltage source is modeling and then whether the model is still valid with whichever current direction the model predicts. <A> In pure circuit theory, power supplies can absorb or deliver power (they don't need to deliver power). <S> If your circuit has only one source, <S> then it <S> will only deliver power. <S> But if your circuit has two or more independent sources, it's possible for one/more of them to absorb power (as long as at least one more is delivering power). <S> Keep in mind that these circuits are purely mathematical constructs. <S> An independent voltage source is not a battery, but can be part of a model for a battery. <S> For intuition's sake, though: think of a rechargeable battery. <S> When you recharge it... <S> which way do you suspect current flows? <S> If you guessed out of the battery... think again =) <A> Addressing the bullet points: 1) <S> Yes, that's what charging a battery looks like: pushing current back through it by connecting it to a larger voltage. <S> What happens depends on the chemistry and size of the battery relative to the current. <S> Some types (NiFe, larger lead-acid) can be kept on a float charge of a few miliamps forever. <S> Other rechargeables will be slowly reduced in charge capacity by plating at the electrodes until they're useless. <S> 2) <S> It's a simulation, not a wall power supply. <S> Again, what happens when you try to drive a voltage source with another voltage source depends on how it's implemented; it may turn itself off, lose regulation, sink current (as in your simulation), or something else. <A> Q1: <S> this would be a charging current, but it depends on the type of battery if it actually stores the charging energy or will explode. <S> Q2: <S> a regular power supply is not designed for negative loads and may have a diode to prevent a negative output current. <S> But power supplies may work in 2 or 4 quadrants , to be able to sink a negative output current.
| So from a pure theory point of view, you should be prepared to accept current flowing either way through a voltage source.
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Efficiency of Antennas vs operating frequency I have refereed a document from TI which mentions that antenna efficiency decreases as frequency is further decreased. Although 136 - 240 MHz has highest range for equal power, the efficiency of antennae at that frequency is not acceptable. What causes this effect? <Q> An antenna needs to be of a "certain size" to be most effective at turning EM waves into electrical signals and vice versa. <S> The size for best effectivity is related to the wavelength of the transmission/reception frequency. <S> For instance, a simple quarter wave monopole at 433MHz needs to be about 17 cm in length whereas at 2.45GHz this drops to about 3cm. <S> Wavelength of an EM wave is: - \$\lambda = <S> \dfrac{c}{f}\$ <S> And an effective monopole is one-quarter this length <S> So, if you have a bunch of antennas of approximately the same physical dimensions I would expect the higher frequency antennas to be more superior because the lower frequency ones just cannot be as effective given the limited space they occupy. <A> It's actually a bit more complicated, it's the gain bandwidth product that is limited. <S> There is a theory called the Chu-Harrington limit which tells us the maximum gain and bandwidth possible for a particular size antenna. <S> Lower frequencies do give better range for the same power and antenna gain. <S> Higher frequencies give better range for the same power and antenna size. <S> It depends what is important to you. <S> A 136 MHz antenna can be made with excellent efficiency and gain, if you can make it full-sized for that frequency, perhaps 1 m long. <A> The antennas in the paper are clearly intended for confined spaces, consequently there is a trade-off between shape, overall size, matching (to the receiver and/or transmitter), bandwidth, gain, radiation resistance, lobe pattern, etc. <S> Generally to make an antenna shorter than an optimum length at a given frequency, inductance can be added, thereby providing an approximate 50 ohm load (or whatever) but sacrificing some aspect(s) of performance. <S> Alternatively, to fit into a given space, the antenna can be formed into a zig-zag, but this also introduces inductance and some capacitance between adjacent zigs and zags, once again reducing performance.
| Making an antenna much smaller than a wavelength generally results in it becoming less efficient.
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How to order a custom PCB I need a tiny PCB for a tiny IC. Is this something I have to design in autocad? Do I have to go to a shop to print it for me? Ultimately I want to be able to flux solder the IC on the tiny board so that I can plug it into my prototype where I need it. I would use a breakout accelerometer on Amazon but they are too big for what I need them for. Ideas? <Q> If the IC is a standard size, then you can probably buy a pre-made breakout board for it. <S> This is much faster, easier, and cheaper than making one yourself. <S> Check out these (511!) <S> breakout boards of various sizes and shapes at Digikey... <S> http://www.digikey.com/product-search/en/prototyping-products/adapter-breakout-boards/2360393 <A> The company that makes the cheapest boards for one size is not going to make the cheapest boards for all sizes. <S> There is no such thing as "the cheapest" PCB manufacturer. <S> Once you've designed your board, you should go to PCBShopper.com. <S> It's a price-comparison site for PCB manufacturing. <S> You enter your board's size, number of layers, soldermask color, silkscreen options, quantity, the country to ship to, and how quickly you want them to arrive, and PCBShopper gives you prices from over 20 different manufacturers. <A> To get a PCB made, use any of the many CAD programs for that purpose. <S> I use Eagle, for example. <S> There is a free version of Eagle that is limited in board size, but it sounds like it will be fine in your case since your board is so small. <S> There are also many other programs out there, all the way from free to quite pricy. <S> However, AutoCad isn't one of them. <S> You want something that inherently understands PCBs and can write out Gerber files for the PCB layers, and Excellon drill files for the holes. <S> That's what board houses will want. <S> With the right package of files, you can send it to just about any board house in the world and get the right PCBs back. <A> There are a lot of ways to build and design a board yourself. <S> Certainly any of the free tools <S> , Eagle being one of them but <S> the are many others that can get you where you need to go. <S> There's a bit of a learning curve as you will have to define the symbols for the part(s) <S> you are using, then create a schematic. <S> From there you need to create the footprints, or the physical drawing of what the part will look like on the board. <S> Place those on the board and then route your signals and gnd between them. <S> I'd suggest looking through a tutorial on how to make a simple board with whatever tool you pick. <S> If you're going to send these out to a shop it pay attention to the design rules the shop has for the cheap price you likely want to pay. <S> You can't just make lines and spacing anything you want. <S> Often the cheaper the price the larger the spacing, the larger the holes etc. <S> To actually make the PCB there are many options. <S> You could print and etch it yourself, heck there have been times when I've carved one out of bare copper board with an exacto-knife. <S> If you're going to have it made at a shop then start looking around at prices. <S> I've seen prices as low as $10 for tiny protos of a few 2 layer boards (with a little waiting time), but keep in mind what I said about verifying their capabilities. <S> It's probably a worthwhile learning experience to make your own board <S> but it is more work than you maybe suspecting. <S> One last thing the last time I saw an accelerator breakout board <S> it was a little smaller than a quarter. <S> If you're going to make something smaller than that you might want to add some break out tabs and maybe put a bunch of them on a single pcb. <A> here is a list of companies that produce PCBs, mostly in connection to eagle.lot of them will let you order also small amounts, just check it out: http://www.cadsoftusa.com/partners-references/board-houses/ <S> did you even find what you were searching for?
| There are many places to order boards like that on the internet.
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What is the easiest way to stagger a signal chain on a stack of identical boards? I am designing a stack of identical boards that each generate an analog signal. What I am trying to do is take the signal from the board above, mix it with the signal from this board, and then send it down to the next board. The problem is that the board-to-board connectors need to be through-hole for mechanical support, which means that the output from one amplifier would be connected to the output of the next amplifier if they are all identical designs. I need a way to alternate the connector pinout from one board to the next, but I would prefer not to design two versions of the board. Is there an elegant way to do that? <Q> What kind of analog signal? <S> One technique that works well is to convert the analog signal to current. <S> You receive the sum of all those currents at one location. <S> The receiver can be a proper current-to-voltage converter or a simple resistor. <S> For what it's worth, most professional audio intercom systems work this way. <S> You can have a hundred or more users on a single intercom line and adding or removing users does not affect the audio level significantly. <S> The current source that is used for audio intercom systems is usually some variation of the Howland Current Pump (also known as a Howland Current Source). <S> There are many resources available on the web to show you what it is and how to use it. <S> I first saw this circuit described many years ago in a National Semiconductor app note where it was called a "bilateral current source". <S> I promptly adopted it for the intercom system that I was building for myself at the time. <S> It was quite some time later that I learned of the proper name for the circuit. <S> The advantage of this system is that the input pin is electrically connected to the output pin. <S> That means that you can stack boards on top of each other and use the same pin to collect the signals from all the boards. <A> The use of a pair of smd headers would easily solve your issue, as the would be no connection between the layers of the board or connectors. <S> These can be used in conjunction with through hole headers used for the rest of the board. <S> In either case a single board is designed, the change is done in assembly. <A> All of your suggestions helped me come up with the solution I will use. <S> Since it is highly preferable that the boards are not physically different from each other, and because I want to avoid stress on surface mount connectors (there will potentially be a lot of plugging and unplugging), here is my hybrid solution:
| You use current sources on each of your boards (a proper current source has very high output impedance) and simply sum all of your currents together onto a common bus. Alternatively, a 4 pin block of jumpers to allow rerouting of two pins, and alternating a blank pin (breaking or cutting short) on the connector would work as well.
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Use transistors to amplify a very small voltage I'm new to electronics and I have a problem. I want to use a LED to detect light but the voltage it provides is very small (30 mV) and I don't think it can open a transistor. How can I amplify this voltage? (I don't want to use op amp) And, I have only NPN transistors. Can you help me? <Q> Yes, you can do this with just NPN transistors. <S> 60mV is small, but probably enough. <S> Resistor values will depend on your voltage source, but try this: simulate this circuit – <S> Schematic created using CircuitLab <S> In normal conditions, R1 and R2 set Base voltage about 0.57V - not quite enough to conduct. <S> Adding 30mV gets you 0.6V. <S> With no Emitter resistor, you should get a decent current through R3. <S> One example is here EDIT : <S> there should be another resistor (RD) in series with D1. <S> Just as R1/R2 bias the transistor in cutoff, RD would limit the voltage across, and current through, D1. <S> LDR's have a resistance of Mega ohms in dark, dropping to about 10K in bright light, so voltage across them is not as big an issue. <S> The same can't be said for diodes. <A> You can do what you want if you add some amplification to the output signal from the LED. <S> A single op-amp can do this or you can use a pair of transistors configured as a differential amplifier (the "long-tailed pair" that Brian Drummond mentions above. <S> I'm going to show you an op-amp version because it is easy. <S> That makes them easy to use with a single supply voltage. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Upon rereading your question, I see that you asked if you could do this using only NPN transistors. <S> Probably <S> yes <S> but it's not either easy or obvious. <S> The problem is that the signal provided by the LED is both low voltage and very high impedance - some of the references that I checked show the LED being loaded by a very high value resistor (30M or 100M). <A> If you just want to detect light then use this circuit. <S> Reverse bias the LED into the emitter of an NPN transistor. <S> When light hits the LED then reverse current will flow proportional to the incident light. <S> The transistor provides current gain, boosting the LED current from say (10uA) to around 1mA. <S> The behavior of the circuit will be this. <S> In the dark the output is high, in the light the output is low. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> If you really want to use an LED (not good but cheap) you should probably use it as a current device and make a one op-amp current-to-voltage converter. <S> Very simple and the output would be linear. <S> Use a TL061 for its low cost and its low bias current which might be important depending on the amount of current the LED puts out. <S> Do not use a TI brand TLO61 which has given me problems in years past. <A> I've come up with something that might work. <S> You will need to breadboard it and see. <S> simulate this circuit – <S> Schematic created using CircuitLab Q1 & Q2 need to be identical transistors, preferably from the same batch. <S> They should also be touching each other if possible so that they are at exactly the same temperature. <S> Cover D1 and adjust R1 so that the output LED (D2) goes out. <S> Then shine light at D1 and see if D2 lights up. <S> No guarantees that this works but it should. <S> I'm not anywhere near my lab bench, so <S> I can't try it myself.
| You can also look at "super alpha" designs for very high gain. The last time I built this circuit, many years ago, I used a Light Dependenet Resistor (LDR) instead of a diode. I like using the LM358 or LM324 op-amps for these simple circuits - they are readily available and the input common-mode voltage range includes ground.
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How do holes conduct current? I have some questions about holes in a semiconductor. When I checked the net I found holes are said to be equivalent positive charge and they say because the hole moves from one place to another when it is occupied by an electron, and electrons leave holes behind, etc... But I couldn't find a clear answer for P-type semiconductor. It is written the holes in the valence band increase, but my problem is I see that lags and reduce the current not increase it because the holes somehow will attract electrons and get them from conduction band to valence band, and since the conduction band is the band that allows current to flow, won't that reduce the current instead of increasing it? <Q> It is not true that the valence band cannot contribute to conduction. <S> That's just what happens in P-type semiconductors. <S> The doping alters the band structure of the semiconductor so that there are "missing" electrons (holes) in the valence band. <S> This allows other electrons to "move" from an atom to a nearby one without jumping into the conduction band: they fill a hole "near to them", leaving a hole "behind them". <S> This mechanism is modeled by virtual charges (the holes) moving in the opposite direction. <S> All this happens in the valence band, and this is (intuitively) <S> the reason why the mobility of holes is less than that of electrons. <S> Actual conduction is always due to moving electrons, but when conduction happens in the valence band all is more "difficult" ( <S> the energy of the moving electrons filling hole after hole is less than the energy they would have if they were in the conduction band). <S> Bear in mind that this is only a qualitative explanation. <S> This is something in the realm of quantum mechanics and solid-state physics, and the equations involved are rather nasty. <S> BTW, what you mention by "holes attract electrons from conduction band" is called recombination . <S> In a P-type semiconductor there are very few electrons in conduction band, and they are due to thermal generation (the higher the temperature the higher <S> the probability that a free electron-hole pair will be generated). <S> So it is true that very few electrons in conduction band will contribute to current in P-type semiconductor (they are the thermally-generated minority carriers). <S> But the bulk of the current is supported by holes "moving" in valence band, as I explained above. <A> I remember when I was following an introduction course about semiconductors, from which I can remember something that I never tought before. <S> The very basic meaning of how holes are conducting and have less mobility, and before understand how a PN junction conducts, it's all about bubbles: <S> an air bubble in the ocean does not move toward the surface, it is the water that slides down around the air due to gravity. <S> It is intuitive, then, that moving everything around is more difficult, from which you have less mobility. <S> This is the virtual hole which seems to move, like a bubble that is pushed up by the gravity (which pushes down, though). <S> How a PN junction conducts is a bit more technical, but if you understand the band diagrams of a junction, it is pretty easy. <S> Good luck. :) <A> Since you asked an identical question on physics. <S> SE, I'll give you an identical answer: because the holes somehow will attract electrons and get them from conduction band to valence band <S> The reason that a p-type semiconductor is p-type is that it contains acceptor impurities. <S> These are atoms that tend to capture electrons in localized states around their nucleus. <S> For example, group III boron is a typical acceptor impurity in silicon. <S> Because the captured electrons are in localized states, they aren't free to contribute to conduction. <S> But, consider if we start with intrinsic material and start to increase the density of acceptor impurities. <S> In intrinsic material, the conduction band has very low occupancy, so the electrons can't be captured from there. <S> Instead, they're captured from the valence band, leaving holes behind. <S> And indeed, these holes do attract electrons from the conduction band, but to make p-type material you typically add many more (orders of magnitude more) impurities than the intrinsic carrier density, so there simply aren't enough conduction band electrons to fill the acceptor states or to fill the holes resulting from the acceptors attracting valence electrons. <S> Another way to look at this is to look at Fermi level. <S> As the acceptors capture electrons and create holes, then we know the occupancy of the valence states has decreased. <S> Since the valence state occupancy is reduced we realize the Fermi level must be closer to the valence band edge than in the intrinsic material. <S> And since the Fermi level is closer to the valence band edge, it must be farther from the conduction band edge, resulting in even fewer conduction band electrons than in intrinsic material. <S> In (quasi-)equilibrium, we find a Fermi level that gives a statistical balance between density of valence band holes, occupancy of acceptor states, and density of conduction band electrons. <S> And this gives tells us in what ratio holes and electrons are available to carry current.
| Short version: The holes can try to attract electrons from the conduction band all they want, there simply aren't enough electrons there in p-type material to fill all the holes.
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100W Soldering Iron Risks And Arduino So I bought a 20W soldering iron a while ago and I found the experience very frustrating as the tip does not get hot enough quickly and when it does melt the solder it gets cold so quick I cant get the solder to flow long enough i.e. I end up with blob all over the place. A friend of mine heard all my cussing and praying for a more powerful iron when I was putting a circuit together and out of the goodness of her heart she bought me 100W soldering iron from Hong Kong. I was praying for a 60W model. This new soldering iron gets hot, about 420 C and I an easily get the solder to flow but I am worried about this high temperature killing my pcb or components. So main question to ask is this? Is a 100W iron safe to use on Arduino type projects? What are the risks I run with using this type of iron? Just a safety check before I destroy something valuable. Update: Been soldering for a while now with both the 20W and 100W and they both have very different uses. The 100W is NOT recommended for smaller job and finer components it is way to powerful. I replaced the tip of the 20W iron and tinned it properly and been keeping it tinned. It is actually not that bad so I suspect my blues started with the tip not being maintained properly. Lesson learned experienced gained. Update 2: Finally invested in a proper soldering station(Hakko-FX888) and I am in soldering heaven. I felt like I have been trying to paint the Mona Lisa with rock with my other irons. Life is beautiful again. <Q> Using a 100W soldering iron will obviously get the job done. <S> It will melt nearly any width electronic solder, and maybe do some moderate plumbing jobs too. <S> With an iron like this you will likely never learn the important aspects of soldering sensitive electronics. <S> A 20w soldering iron should be well enough for most electronic soldering applications, (except for very thick wires or heavy duty solder lugs). <S> This helps transfer the heat to the item being soldered. <S> Tinning a solder tip involves cleaning the tip and getting a thin coating of melted solder on it. <S> This is easy to recognize as the tip will stay shinny with the liquid solder. <S> When using the soldering iron it is also important to frequently clean the tip using a damp sponge (often sold with a soldering set). <S> See some additional tips here: <S> http://www.wikihow.com/Solder https://learn.adafruit.com/adafruit-guide-excellent-soldering/tools <A> With practice yes. <S> Don't stay on the part too long, and everything should be OK. <S> I have a 100W that I use when not at work <S> , never had a problem. <A> A good compromise is 40W iron for larger components and 20-25W for small, sensitive ones. <S> 100W is way too much. <A> 100W solder guns (not as much always-on irons) are somewhat more tenable for unsoldering large components in old TV sets and household appliances, the stuff that has really fat solder points and partly metal cases soldered to a robust PCB. <S> Neither are suitable for soldering integrated circuits and small components. <S> I've worked for most of my electronics experimental time with a 16W solder iron without active temperature control (basically, the resistance of the heating coil went up when it got too hot). <S> The main thing is making sure that the heat actually flows well from heating element to solder tip: the tip needs good contact and heat transfer from the shaft it sits on (if necessary, taking it off and bending it slightly narrower if it sits too loosely) and needs to be clean and tinny at its tip. <S> While more (and actively controlled) power makes this less of an issue, good heat transfer remains something that makes things work better.
| 100W always-on irons are more for soldering plumbing and sheet metal. You will possibly destroy many parts as you learn. One of the most important things is to learn how to keep the tip "tinned".
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Is this a reasonable way to switch a circuit on and off? I'm using optical slotted switches to detect the position of a wheel. I may have a dozen or so, and at any given time I only need to have one switched on. The device might be battery powered, so I want to conserve power where I can (the opto switch draws about 25mA and I guess the comparator a little bit too.). Also, I want a way to quickly see that a) the opto switch is powered on and b) if the beam is broken. This allows me to simply connect a 5V source to the circuit and fine tune the positioning of the plastic tab that breaks the beam without having to connect and run the whole system. Here is my stab at a diagram of what I've got so far on the breadboard: simulate this circuit – Schematic created using CircuitLab Is there some glaring deficiency in this circuit? I suppose there must be, because although there are plenty of examples of common emitter configurations, there don't seem to be many examples of sticking a NPN transistor below a whole circuit to power it on and off. Edit: The Slotted Optical Switch is part no HY505. It specifies an IR diode forward current of 50mA, which I took to be the max. Experimentation revealed that 20mA seems to provide adequate distinction between a blocked slot and not. The op amp was just one corner of a quad comparator I had lying around. <Q> You get the local ground above the universal ground. <S> Which generally speaking is bad. <S> The main reason is it can cause signals generated by the circuit to interact unpredictably with circuits connected to the "true" ground. <S> You also need to remember that ground is not universal when expanding the circuit. <S> Generally it's commonly accepted to disconnect the positive (and/or negative) rail and not ground. <A> Your LED resistors are vastly too small for their purpose. <S> A modern indicator LED should give an adequate result at 1 mA. (2 to 5 mA <S> if desired but 30 mA is (O.O.)OTT.) <S> If D4 is a red LED it draws about 30 mA (which is above what many LEDS are rated at) as does D3. <S> You should say what OA1 is as it may matter. <S> The optos should not need to draw 30 mA. 5 mA should do and maybe less. <S> Again, part number needed. <S> An opamp should draw perhaps 1 mA quiescent. <S> I've saved you about 20 x that - let the poor creature run all the time. <S> IF you must have a low side switch <S> so be it, but high side is about as easy and <S> the "ground" of the main circuit can then be at true ground potential. <S> User61143 will tell you why that is a good idea. <S> You may wish to excise mention of "Arduino". <S> This circuit is quite universal and some people have the strange idea that if it's associated with Arduino's it must be rubbish. <S> While you probably [tm] find that idea puzzling, at least, if it is essentially a general microcontroller question then omitting the unneeded reference to A's helps <S> keep such people perhaps a tiny bit happier. <S> Added: <S> User61143 used the term "local ground" <S> What he calls local ground is Q1 collector, the Cathode of D2 and -ve power supply of the op-amp. <S> This is the point to which all current in the circuit flows when Q1 is on - and it is then at true ground potential <S> + the Vce drop of Q1. <S> When Q1 is off this point floats to near V+ <S> so the whole circuit is at V+. <S> But if anything is electrically connected to any point on this circuit it is liable to be driven in an ill defined way. <A> Just a few ideas that may help you out... <S> You mentioned you wanted to be able to tell if the LED is on or not. <S> One thing you have to tell is "is the LED on but blocked by something in the slot?"To do this, you could measure (with another comparator) <S> the voltage drop across the series resistor feeding the LED (a sense resistor). <S> Your other question was a way to make the whole circuit more power efficient. <S> Depending on your application of what is breaking the beam, you could PWM (pulse width modulate) <S> the signal driving the LED <S> so it would be some fraction of the 20mA. <S> This comes at the expense of sensitivity of the phototransistor side of the "slotted switch", but you can give it a shot.
| As long as the circuit remains totally isolated it's probably OK.
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Am I reading my multimeter correctly? I'm trying to wrap my head around the 20m and the 2000µ setting. On 20m I see 0.01 but on the 2000µ setting I see 005. Does this mean my circuit is using 5µA of current? Why the discrepancy? <Q> This is normal for the accuracy of your meter, at 20m you read 10µA at the more accurate for your range 2000µ <S> you read 5µA. <A> This multimeter looks suspiciously similar, look at its spec sheet . <S> It is important to realize that every meter has an accuracy and it varies per range. <S> At 2000µA the meter's specification is +/-1.0 <S> % + <S> /-5 counts. <S> (Check the manual that came with your meter for its own specifications). <S> This means your reading 005 can vary: <S> min: <S> 5µA - 1% - 5 counts = <S> 5µA <S> - 0.05µA - 5µA = -0.05µA <S> max: <S> 5µA + 1% + 5 counts = <S> 5µA <S> + <S> 0.05µA + <S> 5µA = 10.05µA <S> The same accuracy applies for the 20mA range. <S> This means your reading 0.01 can vary: <S> min: <S> 0.01mA - 1% - 5 counts = <S> 0.01mA - 0.0001mA - 0.05mA = -0.0401mA = <S> -40.1µA <S> max: <S> 0.01mA <S> + 1% + 5 counts = 0.01mA <S> + <S> 0.0001mA + <S> 0.05mA = 0.0601mA = 60.1µA <S> This clearly shows that you always want to use the range that is as close as possible to the measured value. <S> It also shows that the last digit can often only be used to view the trend of a measurement rather than measuring its absolute value. <S> This specification does not account for the burden voltage as mentioned in other answers. <S> Unfortunately the higher the resolution of the range, the higer the burden voltage becomes. <S> So if you switch your range to an accurate range, the meter influences the circuit with a relatively large internal resistance. <A> With a 4 digit DVM the reading of 00.01 is the averaged up reading, equivalent to 5uA. (0.005ma). <S> The leading 0's being shown are misleading though. <A> Depending on the circuit you measure, the internal burden resistance may influence the current you are trying to measure. <S> Especially on low current settings the burden resistance gets larger and has a bigger influence. <S> But as others have noted in 20m you are hitting the accuracy limit of your DMM.
| The internal resistance of your meter influences the circuit and with that may influence the measurement too.
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What is the name of this circuit? simulate this circuit – Schematic created using CircuitLab I've tried to Google it but my keywords don't really yeild anything. I'd like to read more about this type circuit ; what applications its useful in ? How to select C1 ? From what I recall, it provides a DC gain of 1, but provides an AC gain set by the feedback resistors. Does it have a name ? Added I should have emphasized that the focus for this question is C1 and its location in the circuit. <Q> Yes - it has a name. <S> In control theory this circuit is known as a PD-T1 unit . <S> It has a proportional-derivative behaviour with a certain delay term T1. <S> In filter terms, it works like a first-order high-pass with a superimposed constant gain. <S> The transfer function is \$H(s)= 1 <S> + sR1 <S> \cdot <S> \dfrac{C}{1+sR2C}\$ <S> This device is used to enhance the phase (for stabilizing purposes) in a certain frequency range. <S> Please note that application as a PD-T1 element requires \$R1 <S> >R2\$. <S> More than that, the shown circuit is used as a simple non-inverting amplifier ( <S> gain: \$1+R1 <S> /R2\$) for single-supply operation. <S> For this purpose, the non-inv. <S> input is dc biased with 50% of the supply voltage - with the consequence that the input signal must be coupled via an input capacitor. <S> Because the dc gain remains unity, the bias voltage is transferred to the output also with the gain of "1". <S> BODE diagram : <S> The magnitude starts at unity and begins to rise at \$wz=\dfrac{1}{(R1+R2)C}\$, then it stops rising at \$wp=1/R2C\$ at a gain value of \$1+(R1/R2)\$. <S> The rising of the gain is connected with a corresponding phase enhancement. <A> I'd just call that a non inverting amplifier. <S> Calculating the transfer function is quite easy if we can consider the op amp ideal. <S> In DC C1 is open, so you don't have current in R1 nor R2, so the op amp is in the buffer configuration and the gain is 1. <S> When f gets very big C1 is closed, the gain of the circuit is the usual 1+R1/R2 leading to a 2 gain for your values. <S> You then expect a finite pole and a finite zero, the zero comes first than the pole kicks in. <S> You can calculate the pole with the "seen resistance" method: C1 sees R1+R2 <S> so \$\omega_p=\frac{1}{(R_1+R_2)C_1}\$. <S> You can now calculate the zero pulsation as \$\omega_z=\omega_p\frac{A_0}{A_\infty}=\omega_p\frac{1}{2}\$ <A> There is no generic name for it I believe. <S> It has unity gain at DC and, at some point in the spectrum the gain will have risen to 2. <S> This is dictated by: - High frequency gain = <S> 1 <S> + R1/R2 <S> The frequency where the gain is nearly 2 (the 3dB point) is determined by R2 and C1. <S> When the reactance of C1 equals R2 this is the 3dB point and the reactance of the capacitor is: - Xc = <S> \$\dfrac{1}{2\pi <S> f C}\$ = <S> R1 <S> therefore f = <S> \$\dfrac{1}{2\pi <S> R_1 C}\$ <S> For your values this is 1591 Hz.
| Because of the mentioned phase enhancement properties the PD-T1 block is also known as a "lead controller".
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How much is a transformer's secondary floating? Consider an isolation transformer with its secondary referenced to ground via a resistance R as in the following picture. simulate this circuit – Schematic created using CircuitLab If R is infinite, the output voltage Output1-Output2 will float with respect to Ground. And if R is a short circuit, then that voltage is obviously referenced to ground. However, what happens in between? Can the "floating offset" be quantified or at least bounded depending on that resistance? Reason of the question: I'm wondering if a simple high resistance path to ground like this one could protect people from shocks (in series with the person's body) due to single point contact (as far as I'm aware, nothing protects against one finger on output1 and one finger on output2)... But letting it float is a danger as well. I am aware of RCDs, I'm just wondering if it's not a safe alternative, out of curiosity. <Q> Even a screened transformer (where the capacitive current is conducted out to earth) will have some voltage on each output with respect to ground. <S> If the construction is symmetrical you might see half of the secondary voltage on each output terminal. <S> Without the screen you might see that plus perhaps half the primary voltage. <S> I don't think there is a safety advantage in a resistor (it could be worse if the resistor can conduct harmful current from the secondary voltage) but it could be useful in low level circuits to keep the secondary from waggling around at 120VAC or whatever if no other ground exists in the circuit. <A> I am in Canada, and we have 600V, 3 phase delta, ungrounded and floating. <S> If a phase gets accidentally grounded, nothing happens, but if another, different phase gets grounded, fuses will blow. <S> These systems have a monitor for grounding faults. <S> Ungrounded systems can be safer, but only when they are monitored for faults. <S> In older homes receptacles didn't have a ground. <S> A ground-fault receptacle can be used in this case, and ground is not needed. <S> In solidly grounded systems, the problem is that grounding and bonding offers a good path for a fault current to harm. <S> When I touch a stove, I have completed half of the circuit. <S> If I touch a live wire, I don't get a shock, but if at the same time I touch a stove, I get harmed. <S> Another problem is arc flash. <S> It is when a live wire touches grounded surface. <S> electric arc melts materials. <S> The high temperature causes air to expand, and propel the melted materials at high speed. <S> Resistance grounding solves some of the problems, but it has to be monitored for faults. <S> Resistance grounding reduces shock to ground by limiting current. <S> It also eliminates arc faults to ground. <S> In North America resistance grounding is not common. <S> Hospitals in some instances use isolating ungrounded transformers to isolate circuits. <A> I know this is a late response <S> but I wanted to give my opinion here. <S> I look at this situation in terms of Kirchoff's laws. <S> You said you want to find a way to 'quantify' the in-between state when \$R_{isolation}\$ is neither a short nor a open circuit. <S> Consider the following circuit, which is essentially the same that you have except for a load that I added: simulate this circuit – Schematic created using CircuitLab <S> \$V_x\$ and \$V_e\$ are referenced to ground. <S> If a person touches \$\text{OUT}_1\$ with respect to ground, you can find (after applying Kirchoff's laws): $$ <S> V_e = <S> -\frac{R_{Isolation}}{R_{Isolation}+R_{Human}}V_{in}$$ <S> And $$ <S> V_x = <S> V_{in}+V_e$$ <S> As you said, if \$R_{Isolation}\$ is very large then (and with respect to \$R_Human\$): $$ V_e \approx -V_{in}$$ <S> And therefore \$V_x\approx0\$ which means almost no current or very small current run through the person. <S> If \$R_{Isolation}\$ is a short to gnd, then \$V_x\$ is the same as \$V_{in}\$. <S> So the current through the person may be substantial (think \$V_{in}\$ is at main). <S> The in-between state can be quantified by the values of \$R_{Isolation}\$ and \$R_{Human}\$ because they determine how much potential will be across the person and therefore the current through. <S> The assumption is that earth ground is at zero potential which could very well be different. <S> But at least, this gives some insight.
| There will, in general, be a voltage on the output with respect to ground with a high value of R because of capacitive coupling and leakage between the windings.
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Minimum number of turns on a transformer I am trying to wind my own transformer for a power supply. How many turns do I need on my primary winding? I know that the turn ratio determines the voltage ratio, but how do I determine the actual number of turns? I can imagine that for an air-core transformer, having many windings can help keep the flux from leaking. However, the more wire I use the more materially expensive and electrically inefficient it becomes. If I'm using a toroidal core, can I get away with just one loop for the primary? Thanks! <Q> Here (from Wikipedia ) is a fairly complete linear model of a transformer: Note the magnetizing inductance Xm across the primary. <S> If that inductance is too low, you'll get excessive current flowing even with no load on the secondary. <S> While a single-turn primary is certainly possible, with sensibly-sized cores it implies either a very low voltage (for example, a current transformer, which is typically toroidal) or a very high frequency (or both). <S> The inductance is proportional to the number of turns squared, and a small 120/240V 50/60Hz mains transformer primary might be some hundreds of turns, so you can see how far off a single turn is. <S> At a fraction of a volt, or higher frequencies at relatively low voltage, a single-turn primary might make some sense. <A> The calculations for the transformer are complex; however, since you want a toroid where windings are always on top of each other you can't just make something, measure and adjust - you want to know before laying out the coils <S> so I suggest to just bite the bullet and start understanding the formulae. <S> If your power source is 120V <S> and you want to get <S> 12V then the smallest secondary is one turn and your primary can't have less than an integer multiple of 10 turns. <S> This is only close to real life for high frequencies, for 50/60 Hz frequencies of typical household mains the number of turns in the primary will be in the thousands and the number of turns in the secondary must reflect that. <S> A workable shortcut will be to grab a ready-made toroid transformer that has its secondary on top, remove it, figure out the turns ratio by winding and measuring test coils, then wind the desired secondary - this can be done without calculations. <A> The minimum number of turns required for a 120 V, 60 Hz primary is a closely guarded secret. :-) <S> Not really, <S> but everyone always talks about turns ratios, but forgets to mention the minimum number of turns for the primary. <S> Well, the iron/silicon/metal core of the transformer can accept only so much magnetic flux before it saturates and can't take more. <S> If you go beyond this, the inductance drops a lot <S> and you end up drawing a lot more current off the powerline <S> and it will get really hot - not good. <S> You measure the height of the pile of laminations, and the width of the center leg of the E lamination. <S> This width is measured along a line that would go vertical when looking at the E as the letter E appears here. <S> In other words, imagine a single turn of wire tight on the center core, the area of the loop <S> this single turn forms is the area. <S> Do not include the outer legs of the E, or the I. A bigger area will make for a lower number of turns. <S> Once you have the number of primary turns, then you can do the turns ratio to get the number of turns for the secondary. <S> Add a few more turns to make up for resistance of the wire voltage losses. <S> If your powerline frequency is 50 Hz, you need 60/50 times the above result for your primary for 120 V, and twice that for 240 V. <A> https://en.wikipedia.org/wiki/Transformer gives the universal EMF equation as:E(rms) = <S> 4.44fNaB(peak), where:E(rms) is the rms voltage that you are applying to the primaryf = <S> the operating frequencyN = <S> the number of turnsa = <S> the core cross sectional area in square metersB(peak) = <S> the peak magnetic flux. <S> This is a property of the core material, typically 1.2 - 2 for silicon steel and 0.7 for soft steel.
| A rule of thumb, for transformer laminations you may salvage from a junked 60 Hz transformer: Number of turns needed for the 120 V, 60 Hz primary = 800/(area of the core in square inches).
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What are the main causes of Overshoot and undershoot of a signal? I'm not an EE major, but I was curious what causes overshoot and undershoot. Where can i find a detailed explanation of the process? I heard it has to do with the parasitic capacitance, but there were no details. parasitic capacitance seems to have the effect of drawing more current into or out of the line, but i don't directly see how this would make a signal overshoot. Also, how does an improperly terminated signal result in overshoot? If the signal is not properly absorbed and reflects backwards, I would expect some of the energy not to be transmitted into the receiver. Thus, if anything wouldn't this cause the voltage to undershoot because it doesn't receive as much power as it is expecting? <Q> You really only need to read about and understand the reflection coefficient. <S> Wikepedia: <S> http://en.wikipedia.org/wiki/Reflection_coefficient is a good simple starting point. <S> As you can see, it's all about impedance mismatch. <S> You do not need standing waves or any of that nonsense - just some kind of edge going from one characteristic impedance ZS to another ZL. <S> If ZL is bigger than ZS <S> you get overshoot. <S> If it's the other way around you get undershoot. <S> Positive or negative reflection coefficient. <S> Let me know if you need more to fully comprehend this. <A> If you have a series inductance and a shunt capacitance with very little resistance, you have a resonant circuit. <S> If you stimulate the resonant circuit with a fast positive or negative edge, it will resonate, which includes overshoot and undershoot. <S> Traces can act as series inductors, and digital inputs typically have some capacitance. <S> So you have all the makings right there in the typical high-speed digital circuit. <S> Also, some digital signals have controllable output strength. <S> Lower drive strength will reduce overshoot and undershoot. <A> I suppose, your question (overshoot, undershoot) not only concerns "standing waves" or similar effects but also "simple" 4-poles (like amplifiers, filters), correct?In <S> this respect, the terms "overshoot" and "undershoot" are used to describe the step response of such a device. <S> (a) Overshoot: If a system of (at least) second order has two complex poles (a pole pair) <S> the step response will exhibit overshoot. <S> It is possible to find a relation between the pole quality factor Qp and the amount of overshoot (in %). <S> For passive circuits, this is possible only for RLC topologies (resonant effect); for active RC circuits overshoot can be observed in case of feedback (wanted or unwanted). <S> (b) <S> Undershoot: This phenomenon can be observed for active circuits which have a "non-minimum phase" zero in the right <S> half of the complex s-plane (RHP). <S> A simple example for a system having a step response with "undershoot" is the second-order allpass . <A> In a nutshell: impedance mismatches can cause standing waves or "moving" waves along a conductor. <S> If they build up constructively you get overshoot; if they build up destructively you get undershoot. <S> The specifics are complex and depend on the driving source, the transmission line, the length and the losses in the line. <S> There is an absolutely fantastic video on impedance matching which helps to demonstrate this and other concepts better than any drawing or text <S> I've ever seen. <S> It uses mechanical wave generators and wave guides, but the principle is exactly the same in electric circuits. <S> It was made by AT&T in the (I think) 1950s. <S> It's called Similarities in Wave Behaviour <S> and it is worth the time to watch.
| Adding a damping resistor between source and load can dramatically reduce overshoot and undershoot.
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How to turn a switch on/off repeatedly in 1 minute intervals? I have a light bulb plugged into an outlet and I need to be able to turn it on and off again and again in 1 minute intervals for a period of 10 hours per day (a total of around 600 on/off events per day). I have found a similar question on SO here: How can I turn on and off a switch every 20 minutes? However, besides the fact that the question was for 20 minute cycles long I saw a number of answers about micro controllers which I understand will take a lot of work and know-how to achieve. Is there any pre-packaged solution or something that can be done easily ? As a side note most digital timers that can be set for 1 minute increments can only be set around 7 times per day (max I found was 50 settings per day) so this isn't a viable solution. The timer will need to withstand 500-1000 watts (if this matters). Thanks <Q> Digital timer (you could find them on Amazon for around $25-$50) <S> Microcontroller (a bit tedious) <A> You use the WatchDog timer on an AVR with the maximum prescaller to generate an interrupt once every 8 seconds and use this interrupt to wake form sleep and update a counter until it overflows past <S> 225 <S> (use an unsigned u_int8 register for efficiency) at which time you can write a 1 directly to the PIN bit to toggle an IO pin that is in turn connected to the input of an SCR that is wired in series with the lamp power input. <S> ... <S> OR... Get a 0.5 RPM motor like this... <S> https://www.amazon.com/Dayton-2L003-Gearmotor-0-5rpm-12vdc/dp/B000TKAZ4M/ref=as_sl_pc_ss_til?tag=joshcom-20&linkCode=w01&linkId=7XSZKUJRMIJYSETK&creativeASIN=B000TKAZ4 <S> M <S> This is a motor that will make one full rotation every 2 minutes. <S> Look around and you can probably find one for $20. <S> I have seen ones that plug directly into 120VAC plug, or you can get one that used 12VDC and a wall wart transformer. <S> Mount the motor onto a 2x4 with screws or duct tape. <S> You can also duct table directly to a table or wall if no 2x4 is available. <S> Mount <S> a switch that controls the light to the 2x4 - also using duct tape or screws. <S> Find another 2x4 about 1' or 2" feet long. <S> Could also be a 1x or a broken broomstick. <S> This will be the rotating arm. <S> Find 2 objects on the floor. <S> Maybe kids' blocks, or old packs of chewing gum (they must be old enough to be hard). <S> Glue or duct tape the two objects to opposite sides of the rotating arm such that one turns the switch on as it passes it. <S> The other turns it off. <S> Use duct tape or screws or partially chewed gum (from above step) to attach the rotating arm onto the motor shaft. <S> drilling or poking a hole in the middle of the shaft might help. <S> Lots of value engineering improvements are possible depending on the atual switch configuration. <S> The arm will now pull the string once per minute, toggling the lamp on and off. <A> What you are looking for is called a repeat cycle timer. <S> It's not surprising that you can't find one for 600 cycles -- that seems to be quite a lot. <S> However, it should be possible to combine a cycle timer with a regular timer to get the 600 cycles over a 10-hour period. <S> Like the other answers you found, I'll repeat that it is quite simple to do with a microcontroller, but if you have zero experience then it can be a steep learning curve. <S> It can be custom built, but that will cost a lot more than any off the shelf solution. <S> Why are you turning the light on and off so frequently? <S> Maybe there's another, easier way to accomplish the underlying thing you need to do. <A> Perhaps an intermittent switch / outlet will work for you. <S> These are used with aquariums to cycle from a pump plugged in to Circuit A to a second pump plugged in to Circuit B. <S> You can set the time for each circuit and it cycles continuously. <S> This creates a wave effect in saltwater aquariums. <S> Prices start around $8 and go up from there.
| If, for example, the lamp has an existing pull string switch, then you could substitute a 1 RPM motor and connect the string directly to the end of the rotation arm using a thumbtack or hair tie. Couple options: Repeat cycle timer (you could find them on ebay for around $20)
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Are \$\frac{1}{4}\$watt 4x\$\Omega\$ Resistor and \$\frac{1}{2}\$watt x\$\Omega\$ Resistor equivalent? We know Ohm`s law: $$\Omega = \frac{V}{I}$$ $$W=\frac{A}{t} \space\space\space\space\space\space V = \frac{A}{q} \space\space\space\space\space\space I = \frac{q}{t} \space\space\space =>\space\space\space W= VI$$ $$\Omega = \frac{W}{I^2}=\frac{V^2}{W} \space\space\space\space\space\space Right!$$ Lets say I have a \$\frac{1}{2}\$watt x \$\Omega\$ resistor in a main circuit, I need to know can I replace the soldered \$\frac{1}{2}\$watt x \$\Omega\$ resistor using another two \$\frac{1}{4}\$watt 2x \$\Omega\$ resistors in parallel or using single \$\frac{1}{4}\$watt 4x \$\Omega\$ resistor? I think that if the power supply gives the same voltage before and after replacement across the resistor, then it will flow half current through new resistor and wattage requirement might be satisfied. I mean if before replacement everything was working correctly and the original resistor was not burned, then the new \$\frac{1}{4}\$watt 4x \$\Omega\$ resistor also might work just fine! Am I right? Is it possible to do? will any problem about voltage drop? Is it possible to replace Anytime? <Q> Resistors in parallel dissipate proportional amounts of the total power from each depending on the proportion of the current that is flowing through them. <S> Two equal resistors in parallel will share the current equally. <S> Ergo, they will dissipate half the total power each. <S> To reiterate the maths differently: <S> Original = 1W, <S> 50Ω. Current would be \$\sqrt{1/50}\$ = 0.141A. <S> 100Ω resistor with half the current flowing would be \$(0.141/2)^2) <S> \times 100\$ = 0.5W. <A> I'm having a hard time understanding your notation. <S> R is usually resistance. <S> Stick to that. <S> R = <S> V <S> /I <S> Anyways, if you have a 0.5W resistor then the equivalent of that same resistor is two 0.25W resistors in parallel. <S> However, because the resistance is in parallel, that means you need to increase the value of each resistor to compensate. <S> For example 500 ohm @ 0.5W would be equivalent to two 1kohm @ <S> 0.25W <S> in in parallel. <A> Is it possible to replace Anytime? <S> No they are not equivalent - <S> the 1/4 watt resistor can dissipate power up to a nominal maximum of 1/4 watt whereas the half watt resistor can dissipate more power without burning out. <S> If the resistor is just across a fixed voltage then <S> yes you can <S> but, it could be across the output of a constant current circuit in which case the current will remain the same thru either resistor and the quarter watt 2X ohm component will burn off twice as many watts as the half watt 1X ohm resistor. <S> Without knowing the precise detail of the circuit there are no other generalizations.
| So yes, two equal resistors of twice the value and half the power in parallel will handle the same power as the original resistor.
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what is quadruple in IC gates What is quadruple in integrated circuit gates? Is it just about 4 gates inside the IC or anything else more special? <Q> Yes, it is usually an IC with 4 of the same gate inside it sharing a common \$V_{CC}\$ and GND. <S> The datasheets sum this up pretty well. <S> Here is a quad NAND gate pinout: <S> Notice that there is one set of power connections and 4 identical gates. <A> Normally, yes, it refers to the number of gates in a chip. <S> A "Quad 2-input NAND gate" (74xx00) has 4 gates in it, each with 2 inputs. <A> Quadruple, or more commonly just "quad", means 4 of something in a package. <S> For example, the 7400 is a "quad NAND gate" chip, and the LM324 a "quad opamp". <S> Yes it really is that simple.
| If there is a 4 input or 4 bit gate it is normally referred to as such - a "4-bit ripple counter" or a "4-input OR", etc.
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Why don't DC PSUs need a fuse? This feels like a silly question, but after taking apart some of our 6W 12V DC power supplies I've noticed that none of them are fused. Out of curiosity more than anything else, I'm wondering why a fuse isn't required for these power supplies - for the record these PSUs are: Input: 100-240V~50/60Hz Output: 12V DC 500 mA CE compliant (tested by a UK test house rather than just relying onmanufactures word) UPDATE: Thanks for the responses, so on closer inspection (and based off comments and answers) I think I have found the fuse like component on each board. My confusion came in that I was expecting a removable fuse as found in a standard 3-pin plug socket. If someone were to ask me what these were fused to though what would I say? Whatever the input current is stated as? 0.18A? <Q> Short answer. <S> I'm not a Transformer expert but <S> in the UK the plug is fused and rated to protect the wiring and also the transformer sometimes. <S> Also transformers can be self limiting/protecting due to the design (internal protection device or even the wire resistance <S> is high enough to limit current). <A> This was introduced in the UK at the same time as switching from a large number of low current circuits to a smaller number of 30A (later 32A) circuits. <S> In the case of a wall wart there is no flexible cable to protect. <S> Power supplies (whether wall warts or otherise and whether british or otherwise) will need to have overcurrent protection to protect against faults in the power supply itself. <S> This is generally provided by a suitable fuse soldered to the board. <S> Making the fuse user replacable would just be a liability, if it blows it means there is something seriously wrong. <S> The component you have identified in your first picture is almost certainly a fuse (the text "f1" is a dead giveaway). <S> I'm not so sure about your second picture. <S> I'm not so sure about your second picture <S> ("L" as a comonent designation would normally indicate an inductor) <S> The output cable will likely be protected by designing the PSU itself to limit the current. <S> IF a PAT test report or similar asks for a fuse rating <S> I would just put N <S> /A <S> as there is no user replacable fuse. <A> Just to clear up the fusing in UK AC: <S> The fuses present on the UK plug are part of the 5 pronged approach to safety 1) <S> the sockets are switched 2) <S> the sockets are EARTHED 3) <S> the LIVE-NEUTRAL are blocked until the EARTH is inserted <S> 4) <S> the LIVE-NEUTRAL are partly shrouded so that the exposed brass isn't "accessible" or contacted until fully inserted 5) each plug is fused due to a ring system in use & the fuse rated for the appliance NOT the wiring - why put a 13Amp fuse onto something that only needs 3Amps? <S> Under a radial system the fusing could be done at the distribution board <S> BUT it would be rated for the maximum that could be connected (+ any spurs...) <A> The input resistor (red circle) limits the input current and work as fuse to. <S> If PS draws to much current, the input resistor blows up like a fuse.
| The primary purpose of the fuse in the plug is to provide overcurrent protection for the flexible cable and to a lesser extent the appliance.
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ADC with I2C interface with multiple addresses I want to connect 31 ADC slaves to a master. The ADCs simply should measure voltage and response its values to the master after request. I thought about using I2C as bus system. The problem is that most ADC ICs with I2C interface only offer 2 or 3 bits for address (4 or 8 addresses possible) I need 5 bits for my 31 addresses. What can I do to use 31 I2C ADC slaves in one bus system? Are there such devices like address mappers, expanders, switches, multiplexers? What is common and good practice for my requirements?The reason I would like to use a bus system is to keep wiring and pin usage of the master low. To be more specific I need to measure voltage from multiple in series connected sources: simulate this circuit – Schematic created using CircuitLab I am aware of that I need to provide a galvanic isolation between I2C master and slaves. I am afraid that I cannot use multichannel ADCs in this case am I right? <Q> You didn't specify the A/D speed or accuracy, so any A/D will do. <S> One possibility is to use A/D chips with multiple input channels. <S> You can find them with 8 or 16 or even more inputs. <S> These are essentially a A/D with a analog mux front end integrated into a single package. <S> Four 8x <S> A <S> /Ds would do it, and you can probably find some that do IIC and have at least 2 address bits that can be set via the pins. <S> This is like the previous case in that there is a analog mux integrated with the A/D, but there is also a microcontroller integrated with it. <S> The integrated micro can then do the continuous sampling, perhaps some low pass filtering, and communication back to the master. <S> Now you're not limited to IIC, although many micros have IIC hardware that can act as a slave device. <S> Since the address is set by firmware, you can have as many of these on a IIC bus as it can electrically handle. <S> However, no more than 2 would be needed since a number of micros have 16 or more A/D channels. <A> There are quite a few SPI bus ADCs that can be daisy chained together. <S> This means they all receive a common clock and "start conversion" signal but the data output from the "furthest" ADC feeds the data input of the next furthest. <S> They all convert simultaneously and the last in the ADC chain clocks out all 31 slave values: - Master MISO in the above diagram <S> only connects to the lowest slave's data. <S> It in turn receives data from the middle slave as its own data is being clocked out. <S> Here's another diagram: - <A> A better way to handle your requirements would be an IC like Linear's LTC 6803-2/4 series. <S> They can hadle up to 12 cells per IC, and moreover the ICs can be stacked together with isolation. <S> You need to interface them with a micro using SPI though.
| Another possibility is to use a small micro with a large number of A/D channels.
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Why are MOSFETs better than BJTs in application of Voltage Controlled Resistor? Why are MOSFETs better than BJTs in application of Voltage Controlled Resistor ?Thanks for your responses. <Q> The following refers to NPN BJTs and N channel FETs <S> It's really hard to get good accurate representations of collector current versus collector-emitter voltage for different levels of base current. <S> It's a bit easier for JFETs because, traditionally they claim to encompass the functionality of a voltage controlled resistor. <S> Anyway, here are the best three I could find and they show a similar story: - <S> Top left is a BJT. <S> Top right is a JFET. <S> Bottom left is a MOSFET and bottom <S> right I have drawn my own picture of the saturation/ohmic region. <S> One thing that makes a JFET much more superior is that the drain current and voltage can be reversed (made negative) and the characteristic shape of the resistances are largely similar. <S> They are not hi-fi quality the same but good enough for controlling volume in a cheap transistor radio or amplitude in a sine wave oscillator. <S> BJTs wobble a tad when the voltage is reversed because you are then relying on the transistor being symmetrical and having the same gain. <S> MOSFETs usually (but not always) have a parasitic diode which makes them unsuitable for negative voltages when used as a VCR. <S> I'd go for JFETs every day. <S> As a footnote don't believe that the BJT characteristic looks like this: - It completely fails to show the ohmic/saturation region properly, Rather, it seems to indicate that there is a one value of resistance irrespective of base current. <A> Why are MOSFETs better than BJTs in application of Voltage Controlled Resistor ? <S> Thanks for your responses? <S> If such a device should be able to act as a voltage-controlled resistor it should have a voltage-current characteristic that is (at least) similar to an ohmic resistor. <S> In particular it should exhibit a symmetric behaviour in the vicinity of the origin (positive/negative current for positive/negative voltages). <S> More than that, we reqire that I=0 for V=0. <S> It is to be noted that both requirements cannot be fulfilled with a bipolar transistor (BJT). <S> This is because collector and emitter cannot be exchanged without a drastic change in voltage-current characteristic. <S> More than that, for Vce=0 the current Ic does not cross the origin for any finite control voltage Vbe. <S> Hence, the BJT cannot be used as avoltage-controlled resistor. <S> However the situation is completely different for FET´s in particular for JFET´s. <S> These devices have another working principle than BJT`s - they can be seen as voltage controlled resistors within the so-called "linear" region (ohmic region) for rather small VDC values (up to 1V). <S> And the resistance between D and S can be varied (controlled) with the voltage VGS. <A> I have 2 images above. <S> The first one is Ic vs Vce which i captured in a lab at university. <S> The second I took off the internet of a FET because i didnt have one. <S> In the FET one you can see that all the curves split off before they come flat. <S> Giving a R = V/I for those curves will give different resistances. <S> However, in my BJT one, you can see that before the curves become flat, they're all relatively close together, this means that R = V/I for all those curves will be slightly different, but more or less the same. <S> Therefore, thats why MOSFETs can be used as a variable resistor, whereas BJTs can't really. <S> I hope this helps
| Basically, what I'm trying to say is that all three devices can be used as a voltage controlled resistor (VCR) because the base/gate control can vary the slope of the graph and the slope is basically conductance and that, in turn translates to resistance.
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Is it OK for traces to come off SMD pads at the corners? I'm working on a PCB and have found it very convenient to connect adjacent passive components by running traces between them diagonally, as shown below: Is this bad for any reason? <Q> I do it all the time. <S> I like the way it forms <S> nice smooth 135 degree angles trace to pad, pretty much <S> no chance of having etchant traps. <S> On the other hand, you need to be careful. <S> Depending on the component size, having traces come off pads asymmetrically can prevent them from centering during reflow. <S> I've never had a problem with this with anything 0603 or larger. <S> 0402 <S> and smaller, you better think really hard before doing it. <S> It helps to keep your solder mask as tight as board house tolerances will allow. <A> No, it's fine. <S> Very sharp corners (internal or external) are bad. <S> Also, make sure your clearances to other pads/traces are OK. <A> You may want to consider some other changes though. <S> Here is one example: <S> moderator note: <S> I would agree that this is more of a comment than an answer. <S> But Michael needed to post images (and image is worth many words), so this had to be posted as an answer.
| Just make sure that if the pad is rectangular rather than square, then the trace still comes out if the corner.
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Could I use a surface mount thermistor to measure ambient temperature? I'm working on a project where a device I'm using has this thermistor , a Murata NCP15WF104F03RC, to measure the temperature of the device itself. This is a surface mount thermistor. The implementation of my project requires ambient temperature sensing, and comparing the two temperatures. I figured the easiest way to do this would be to use the same thermistor for ambient temperature sensing, as they would have the same thermal resistance curves and other relevant characteristics, which would probably help with precision of comparison. Are there any glaring flaws with this design decision? <Q> I see one potential flaw - if the device is warm it will thermally bias the board to which a second thermistor is attached and the second thermistor will read higher than ambient. <A> Sure; just mount the thermistor on a small flat heat sink which is exposed to ambient. <A> I thought the following by Maxim was a good read on selecting a temperature sensor and it talks about the trouble of measuring ambient down the page some(after Table 1): http://www.maximintegrated.com/en/app-notes/index.mvp/id/3229 <S> To summarize what it says:The two best ways to ensure the sensor measures ambient temperature and is not affected by other sources is to a) use a thermistor or other temperature sensor with long leads and keep it extended up away from the pc board. <S> b) use a satellite board that you can place the thermistor on that will keep it away from other electrical components that could heat up the pcb and keep it away from other parts of your system like fans that will alter ambient readings <S> Just avoid having any components on that satellite board that will dissipate power.
| So it sounds like your solution of having the thermistor on a separate board away from the fans/main pcb components will work just fine.
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Why does a metal sheet block all noise, but not when wrapped around a cable? If I place a flat sheet of aluminum foil between a near-field probe and a noisy wire, the sheet completely blocks the noise. However, if that sheet is then wrapped around the wire, it seems to have no effect on the noise. Why? <Q> Without being grounded it does what all shielding would do - it reradiates. <S> When its not wrapped around the wire, the noise is likely being distributed widely enough to reduce the power available to the probe, and hence the reading. <S> Also, as the sheet becomes better coupled to the wire, the more effectively it behaves as an antenna. <A> But.... <S> ... <S> If the near field probe were a magnetic loop, things would be different; it wouldn't make much difference where the aluminium sheet was placed because the mag field would get thru the same but, attenuated because of eddy current losses in the aluminium sheet. <S> So, back to electric fields AND importantly for this question, the aluminium sheet appears to be unearthed. <S> When I say unearthed <S> I mean galvanically unearthed. <S> However, the aluminium sheet will still have capacitance to ground and that capacitance to ground will tend to make it act like a faraday shield. <S> In other words it will block electric fileds and reduce noise pick-up to the near field probe. <S> If the aluminium sheet were close to the near field probe, then the electric field impinging on it is a bit weaker than if it were close to the noise radiating wire. <S> This is primarily due to distance. <S> This is important to remember. <S> This means you can regard it as having more capacitance to ground than capacitance to the noise source. <S> Try this for an equivalent circuit: - simulate this circuit – Schematic created using CircuitLab <S> In this scenario the noisy electric field trying to hit the probe is dramatically attenuated due to capacitance to earth dominating the capacitance between the noise source and shield. <S> It's a potential divider made from capacitors. <S> Now, consider the scenario when the shield is really up close to the radiating noise wire. <S> In effect, the shield becomes massively capacitively coupled to the noise source. <S> It "effectively" becomes part of the noise source because the capacitance between it and the noise source is far greater than the capacitance to earth. <S> There is some slight attenuation because there is still some capacitance to ground but basically the aluminium shield re-radiates <S> the noise minus a dB or two. <A> I totally agree with Sean Boddy. <S> The electric field will changed because of the sheet.attached is a example the change of electric field because of the metal ball. <S> If you wrap the sheet around the wire (not grounded), the electric field will almost keep the same shape, but an advantage in this case is that the wire is kept away from the noise outside. <S> If you wrap the sheet around the wire (grounded), the noise generated by the wire will only be inside of the sheet. <S> So the outside probe will not affected by the noise from the wire.
| It's about shield capacitance to earth versus shield capacitance to the noise source aka electric field interference.
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Why are 32.786Khz crystal cans soldered to PCB? Every development board I have ever used that includes a 32.786Khz crystal has always had its can soldered directly to board. What is the reason for this ? Is this to minimize radiated energy from the crystal during its operation (assuming that the pad its soldered to is ground) ? If the can was not soldered to the board, what would be the effect ? Added If you look at this image, the crystal has 3 solder points. Two for the legs, and one for the case/shell/can (green arrow). My question is about the reason for soldering the case/shell/can to the board. <Q> "Radiated energy" is the right idea, but your question suggests that you are thinking about electromagnetic energy here. <S> It's rather mechanical. <S> 32768Hz is ultrasound with an arial wavelength of about 1cm, and a crystal oscillator works via mechanical vibration. <S> If you want to have constant Q and constant frequency response, fixing the case mechanically makes sense. <A> 32kHz and 32.768kHz crystals are physically large. <S> The lower the resonant frequency, the larger the crystal - the higher the resonant frequency, the smaller the crystal. <S> So you might expect the case and leads to be larger than higher frequency crystals which obviously have leads that are strong enough to support the can, crystal, and all. <S> However what you find instead is that the can is just barely larger than the crystal itself, and the leads are very, very thin - not strong enough to support the weight of the can by themselves. <S> The reason these low frequency crystals have such seemingly poor packaging methods is due to the watch industry. <S> There are 1.2 billion watches sold each year . <S> The majority of them are inexpensive digital watches, requiring a small, 32kHz crystal. <S> The crystal can is securely fixed inside the watch body using pressure or glue, while the leads are soldered to the circuit board. <S> There is no need for strong, supportive leads. <S> This enables cheap, small, lightweight watches. <S> As a result, these crystals are extraordinarily inexpensive - in their existing form. <S> You can get more expensive 32kHz crystals with better mechanical fixing, but they cost 10 to 100 times more in quantity than these inexpensive watch crystals. <S> Some manufacturers have adapted them by bending the leads and putting them into reel packaging appropriate for surface mounting use, suggesting a pad and soldering layout that allows the can to be affixed using solder. <S> Since the boards are already going through a solder paste, placement, and oven process, this requires no extra steps, such as gluing, hand soldering, wave soldering, etc and reduces the labor required to use these devices. <S> So this all comes back to using a part intended for another big industry in your design, and having to deal with the tradeoffs that suit the other industry, but might have to be accounted for in your design. <A> I think you'll find more information in this old answer <S> Is case grounding compulsory in typical 32.768kHz crystal for Real Time Clock? <S> I would have said it was more for mechanical reasons and that those packages they typically use are the cheapest possible. <S> They're basically old through hole packages converted to surface mount that wouldn't have much mechanical stability without soldering the case down. <S> However another person in that answer notes that it may help with parasitic capacitance. <A> It's done because it's cheaper than providing a socket. <S> 32.768 KHz is used because it can be divided down to 1 Hz. <S> Systems which do it this way typically don't require an enormous amount of accuracy, so an external crystal is used in conjunction with an on-chip oscillator, rather than a complete oscillator which can be trimmed to final value as part of the assembly process. <S> ETA: <S> (after the new picture and a bit of clarification in the OP) <S> Ah. <S> There are at least 2 reasons. <S> The third tab is the case - this provides physical attachment for the crystal and keeps the legs from failing as a result of vibration. <S> It also provides a fixed capacitance to ground for the internal leads, which helps keep the frequency constant. <S> While it does (also) provide shielding, this is not a particular concern. <S> The crystal is working at fairly high voltages (a volt or more counts as high in this case), and has an extremely high Q, so it's hard for radiated noise to affect it significantly. <S> Plus, of course, there really isn't a lot of radiation going on at 32 kHz. <S> Everything else on the board will be running in the MHz range. <A> If you want to lay the case down to minimize height, soldering it is more stable than just leaving it loose to move around. <S> This is probably the main reason you see it done. <S> There are also electrical reasons to solder the case to the board. <S> One reason is to ground the case to prevent damage from ESD if you plan on touching the board near the crystal frequently. <S> Usually in production this is not needed, but if you're testing/programming a board and plan on handling it a lot this might be a good idea. <S> In most cases this doesn't matter because it is very small, but in the rare occasion it can be significant. <S> SHG's mention of parasitic capacitance as well is not going to be an issue most of the time. <S> The electrical reasons for soldering the can are just not very valid in most projects/situations. <S> If I had to bet I'd say the boards you've seen with the cans soldered to the board were done purely for mechanical reasons, and there wasn't an electrical reason behind the decision. <A> Gluing, which is the other common option, is used when the case is smaller and the required temperature range is very narrow. <S> It is difficult to get an adhesive which is both a "good glue" and also very temperature and water stable. <S> Even more important, it adds an extra step -- even if you are already gluing other components, but particularly if you are NOT already gluing other components.
| One reason to solder the case to the board is purely for mechanical stability. So to keep the cost of hundreds of millions of crystals low, the leads are thin, the can is small and thin, and it's produced as inexpensively as possible. There is also a potential coupling path from the case of the crystal to the crystal itself.
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3 leg potentiometer - what's the 3rd leg for? In a potentiometer, I often see the 3rd leg connected to ground. I understand exactly what a potentiometer is physically (a wiper along a log or linear resistance), but I don't understand the benefit of connecting that 3rd leg to ground. <Q> Using the third leg turns the rheostat into a potentiometer. <S> The whole thing acts as a voltage divider, with the moving wiper varying the ratio between the two resistances connected to the differing voltages. <A> Although it didn't quite make sense to me at first, i stumbled upon this page that gives information about voltage dividers. <S> Sparkfun : Voltage dividers <S> Since a potentiometer consists of two resistors, just like the most basic voltage divider you see (two or three resistors along a wire, they cut voltage each time) it can be used as a voltage divider. <S> It makes more sense if you study the basic voltage divider before the actual potentiometer. <A> simulate this circuit – Schematic created using CircuitLab <S> A 3 terminal pot used with 3 terminals, is basically just a voltage divider. <S> As you move the wiper, you increase one resistor in the voltage divider, while decreasing the resistance in the other. <S> But each section of the voltage divider can be considered to be an infinite number of resistors in series such that the total sum is the pot full scale resistance. <S> Now if we use this infinite resistor model, if you connect one leg of the pot and place in somewhere in within this infinite set, that means we have shorted out those resistors. <S> The total resistance from end to end is no longer the full scale pot resistance (because we have shorted out some of those "resistors". <S> A rheostat is a variable resistor. <S> You don't have to connect one the wiper to ground. <S> So long as the wiper is connected to one leg of the potentiometer, it will behave a variable resistor.
| So a 3 terminal pot is a variable voltage divider. Connecting a wiper to one leg of the pot, is a rheostat.
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How to best attach a wire to a switch without soldering? I've got a switch like this one: Now my (quite novice) question is: How do I best attach a wire to the connector without soldering, e.g. for quickly trying something out? Right now I've folded the wire so that I got a hook at its end, put that hook into the hole, and wrapped everything with electrical tape. Is there a better way? (I can't use alligator clamps as the two connectors are too close to each other.) <Q> Quick disconnect terminals. <S> They are good for permanent attachment too. <S> The blades under the switch were intended for this type of terminal, so they should have a correct width and thickness. <S> ( page where the picture came from ) <A> Female 6.3mm (1/4" approx) <S> They're also known as Spade Terminals. <S> Example <S> They come in different colours according to the thickness of wire you intend to connect, which in turn relates to the current which will be flowing through the wires and switch. <S> Each colour is available in different blade sizes also, but you require the most common, 6.3mm size, judging by your photo. <S> One important consideration before twisting the wires around the blade terminals on the switch: are you connecting up a safe voltage/current? <S> If you're dealing with mains electricity, fit fully insulated terminals before attempting to power anything up. <S> Any loose or frayed wires could make the whole switch body 'live'. <S> Fully insulated terminals have rubber or rigid plastic sleeves which cover the metal blade part as well as the crimp section. <A> I would go and buy a cheap automotive crimping tool. <S> They come in kits complete with push-on connectors for your switch terminals. <A> Create a pair of wires, alligator clips on one end and the flat connectors match with the switch on the other.
| So connect the new wires to the switch via the flat connectors and use the alligator clamps on the other end to connect to the wires to be tested. Blade Terminals are what you require, plus a crimping tool.
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How do I find out about the technical details of an LED if I don't have the fact sheet? I have bought a bunch of LEDs from the rummage table of my local electronics store. Now, I don't have a fact sheet, neither I do know the vendor and / or the article number. How do I find out about the desired voltage and current of the LEDs (without doing it trial-and-error)? <Q> Short answer: you can't. <S> But you can do some smart non destructive trial and error. <S> First of all you need to at least know what current/voltage you are expecting: <S> are they power leds, are they small 3mm leds, are <S> they smd or not, what color are they? <S> Let's assume that you find a somewhat related datasheet <S> so you have the forward current and voltage. <S> You will need a bench psu for this, or a power adapter and a variable resistor, and in both cases a multimeter. <S> You start by building a circuit that will let some 1/10 of the current from the datasheet at the same forward voltage, you then slowly increase the current until you are satisfied. <S> You can possibly burn some leds but you'll probably get away with it. <S> Shorter answer: no datasheet, no buying. <A> A lot of common LEDs have very similar specs. <S> For example, if I picked up a jellybean red LED, my first guess at its specs would be \$V_F = 2.0 V\$, <S> \$I_F = 20mA\$. <S> If you're just using these as a hobbyist, numbers like those should be close enough to make them light up on a breadboard. <S> You probably don't need to get close to the rated current anyways - for me, \$1\$ or \$2 mA\$ is enough to tell that the light is on. <S> See if you can find some generic LED datasheets for the colors of LED that you have, and use those sheets as a baseline. <A> Measure the forward voltage and observe the color (preferably comparing to known wavelength LEDs) in order to find the chip type. <S> You can make some guesses by looking at them- if they're of cheap construction and look like overruns from a toy maker, they're probably not a chip comparable to the best Avago or Cree ones. <S> Measure the Vf vs. I <S> curve- <S> easily done with a bench power supply and inexpensive handheld multimeter. <S> Use something like a 499 ohm 1% resistor and gradually increase the voltage, measuring the voltage across the LED and the resistor to give Vf and If. <S> Observe the size of the die and package and compare with datasheets you can find online (using all the before-mentioned information) to get an idea of the maximum ratings. <S> Take off some factor (25% 30% or whatever) for the possibility the unknown ones may be made using inferior materials or design. <S> You can easily check if the leadframe is steel or copper with a magnet. <S> Test the LEDs and discard ones that seem unusually dim compared to others- <S> they're probably damaged. <S> Or just buy known good ones like most of us...
| Try to find a datasheet of a led you think is similar.
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Using DC current to operate a fish feeder that takes 2 AA batteries I'm a complete electronics newbie. I have a auto fish feeder that takes 2 AA (1.5V) batteries. After changing the batteries quite often I got curious to find out if the feeder can be modified to run using the electricity at home. Some research led me to find these: The AC current has to be converted to DC An adapter is needed that converts 220V AC to 3V DC I need a multimeter So I looked around and found an adapter at home which says: Output:3VDC 1000mA 3VA . I actually have no clue what that means but I'm guessing that it will output the 3V required to run the auto feeder. Bought a multimeter and used it to measure the output of the adapter. I was expecting to see 3.00 . Nope. I'm getting a reading of 6.46 every time. So this is my question - is it OK to go ahead and wire the adapter to the fish feeder despite the higher reading? Will everything magically work or something is sure to blow up? Will adding some register help? I would've gone ahead to find out if the feeder was available in my Country. I imported it from the US so I don't want to mess it up. Could you guys please give me advice on how I can accomplish what I'm trying to do? I'm totally ok to start from scratch. EDIT: As for the adapter, it's from an electric shaver and this is how it looks: <Q> Judging from the picture that you posted: You have an unregulated DC adapter. <S> See this link for some more details. <S> In simple terms, the amount of voltage that your adapter puts out goes down as the amount of current draw goes up . <S> In your case, the adapter is rated for 3 VDC at 1000 mA, so it will put out: 3 V when you draw 1000 mA < 3 V when you draw more than 1000 mA 3 V when you draw less than 1000 mA <S> (as you've measured, 6.46 V at 0 mA) <S> Since I don't know anything about your fish feeder, I don't know how much current it'll take or how much voltage is safe to put on it. <S> It might work. <S> It might not. <S> If you can get your hands on a regulated 3 V adapter, you and your fish would be much happier. <A> You can add a regulator, but since good engineers not only solve problems but save money , I'll suggest a cheaper solution. <S> Put a resistor in parallel with the fish feeder. <S> You will probably find that if you give the wall wart just a little bit of load, the voltage will drop to something more reasonable. <S> It still won't be well regulated, but I doubt your fish feeder will care. <S> I'd start with a 2.2kΩ resistor, which by Ohm's law, would draw: $$ {3\:\mathrm V \over 2.2\:\mathrm{k\Omega}} <S> = 1.36\:\mathrm{mA} $$ and consume energy at a rate of: $$ 1.36\:\mathrm{mA} \cdot 3\:\mathrm V = 4\:\mathrm{mW} $$ <S> That's of course based on the assumption that with this added load, the adapter will supply its rated 3V. So try a resistor, and <S> if the voltage is still to high, try a slightly smaller resistor. <S> Be sure that the power doesn't exceed the resistor's capabilities, 1/4 W for the most common variety. <S> And make sure you aren't drawing an appreciable fraction of the supply's rated 1000 mA through the resistor, otherwise there will none done left for the fish feeder. <S> You will probably find that you are nowhere near these limits before the voltage drops. <S> You don't have to drop it all the way to 3V, either. <S> 4V is probably just fine. <S> In fact if you really wanted to save money, you might be able to put the unregulated 6.46V into the fish feeder. <S> I'd say you should read the fish feeder's datasheet to make sure you aren't exceeding its specifications, but I'm guessing you don't have that datasheet. <A> Please keep in mind, that the fish feeder may have been designed to run on batteries only as it may assume battery internal resistance. <S> I may not be the case, but if it is you will blow the shit out of it when you connect it to DC adapter, even if it's just 3V. <A> Since this is an engineering forum, I'll say that you could add an LDO regulator to your existing adapter- using a part such as the Diodes Incorporated AP1186T5-33L-U . <S> It should have a small heat sink to be safe (1 square inch of copper is probably enough). <S> That will take care of the variation in your adapter output without dropping too much voltage at full current. <S> They're $1.58 each in singles. <S> This part is on the way out, but for a one-off <S> it's not a problem. <S> It needs a few electrolytic capacitors (the can type) in addition to the chip (3 x 100uF will work). <S> That will use less vampire power so your electric bills are less (assuming 24/7 operation this can be a factor- <S> the type of adapter you show typically runs noticeably warm and you're paying for that wasted heat). <A> Your meter probably has the ability to measure resistance. <S> You can connect it across the heater and read some number of ohms. <S> Say you get 30 ohms. <S> The current at 3V is then 3/30=0.1A=100 mA. <S> As long as the resistance is greater than 3 ohms you will be within the 1000 mA current rating of your adapter. <S> The power of the heater is V^2/R=9/R Watts. <S> On the heater package you may find a heater power, which would also let you calculate R. <S> I suspect that the 6.46 Volts you measured was open circuit, without the heater connected. <S> It will be lower with the heater connected.
| Personally, I'd probably just buy a switching adapter (make sure they have genuine safety-agency approval or listing markings.. not just a generic PRC "CE" mark that has no real value)-- some of the ones on eBay etc. are criminally bad.
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Shut down Atmel µCs I was wondering if it is possible to shut an atmel microcontroller down with a software command? I want to shut my controller down either via a remote command via USART or if an internal error happens (as BOD, for example). Unfortunately the whole data sheet tells me nothing about that. <Q> They can be put into a low-power sleep mode, and additional hardware can be added to disconnect power externally, but there is no equivalent to turning power off internally. <A> Microcontrollers, and indeed CPUs in general, don't understand the concept of "shutting down". <S> They just run and do what they are told. <S> When you shut down your computer you're not shutting down the CPU, you're shutting down the operating system. <S> The last thing that shutting down of software does is to interface with a part of the hardware on the motherboard to indicate to the power supply to switch the power off. <S> The CPU is still running as per normal at that point since it is the CPU that has to perform the software instructions to send the signal to request the poweroff. <S> Microcontrollers and CPUs do have different internal power states though where they can turn different parts of themselves on and off - these are often referred to as "sleep" or "idle" modes, and different modes can cause different effects on the internals of the chip (memory loss, etc). <S> The datasheet for your specific chip will detail these modes and what their effects are - as well as how to utilise them. <S> No chip ever has an "off" mode though, only very low power modes. <S> Boards like the BeagleBone Black which you can "shut down" in software and switch themselves off have a separate power management chip which is instructed by the CPU to switch the power off to the whole board. <A> There is the 'sleep' instruction. <S> It waits for an interrupt to occur. <S> Clearing the interrupt flag (CLI) will disable them and it will never wake up. <S> However how are you going to tell it's done, without some kind of outside indicator <S> (LED?) <S> BTW, software code is usually referred to as instructions, not commands. :)
| Microcontrollers are not meant to be shut down in the same sense as a PC.
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PCB alterations post-production. Or, how do I not stab myself while scraping off the copper? I made a PCB that I am happy with (it works!). This is for a power circuit, and I have realized (way too late) that I want a switch on the board as well. I bought a switch from a hardware store that seems to be ok for the purpose. I have the location of the thick VCC trace where I want the switch identified, and I think I can drill two slots using a small drill. The only problem that remains is the copper trace between the two slots. I want to remove that copper to make sure it cannot turn on without the switch being on. I have 2oz copper traces. The trace I will need to remove is about 7 mm in width and probably 1 cm in length. The only two ways I can think of getting rid of this trace region are: Try to scrape it off with an exacto knife. I have only ever taken solder mask off with a knife, so I don't know how well this will work on copper... Try to overheat that region of the PCB and hope the copper peels off. I've done this accidentally a few times. Well, the transistor burned and took the traces with it). Is there a better/safer way? I could probably just not put the switch on the board, but put it on a separate prototype board by itself and wire that to the power board. <Q> Assuming your PCB is low-voltage, then just a wide-ish cut from a scalpel will do. <S> Personally, I've had success with a technique akin to a lumberjack cutting a tree with an axe. <S> Put one cut in as deep as possible, then shave off bits from either side, making a "V"-shaped "canyon" until you're satisfied that the cut is big enough. <S> If your circuit is high-voltage, then make two cuts with a decent separation distance, then use the scalpel to cut underneath the trace and lift it off the board. <S> If possible, hold the board tightly in a vice whilst you're doing this, to allow you to have only one hand (the one holding the scalpel) near the blade. <A> The one time I had to cut a thick power trace, I used a Dremel with a round bit. <S> It was enough to get into the copper, but leave the FR4 beneath it with only a little bit of surface scuffing. <A> Don't try to scrape it off. <S> Make sure you get all the way through. <S> That is, all the way through the copper, not the substrate. <S> Also, cut through the resist at the edges of the area. <S> Now take the point of the knife and start poking under one of the corners. <S> You'll find that the copper peels away from the board without heat. <S> When you have a large enough corner pulled up, grab it with your needle-nose pliers and simply peel the piece off. <A> Well I'd just score two ends of it with the exacto knife to make a little section. <S> Then if you want you can heat that up a little and just peal it off with the knife or a pic. <S> Do you really need to remove the trace or will cutting it suffice? <A> The easy way to remove larger portions of thick, wide trace is to scrape the solder-mask from the copper, cut through the copper trace at each end of the section that you want to remove, then flood the trace in between those cuts with a soldering iron and lots of solder. <S> You will need to keep the solder molten for anywhere from 10 to 30 seconds, sliding the soldering iron tip from one end of the trace section to the other. <S> The adhesive holding the copper to the PCB substrate will fail at some point and the trace section simply peels off after it has cooled. <S> Been there, done that, works well.
| Instead, use an Xacto knife or something similar, and make cuts at each end of the area to be removed.
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OP-AMP Flip Flop start up predictability A customer had asked for a circuit design which I felt would be fun to implement using a couple of OP amps. (Its always fun to see if you can do something with op amps instead of a small MCU ). Anyway my circuit worked fine, but of course the customer then requested "just one" additional function, and it warranted an S/R flip flop. So, wanting to stay with the original design philosophy, I found the Op-Amp based SR flip flop circuit below. I implemented it and again everything worked fine. The circuit needs a "high" on either the SET or RESET inputs to toggle its state and conveniently (for me anyway), the circuit always powers up in the reset state. I've tried this with an LF353 I had handy, as well as a spare op amp in a TL084 in the customer's circuit, at a variety of voltages. For me this works out well, but I'm usually paranoid about start up "maybe" states. I suppose I could add a little RC network to the reset input (to force a little pulse on start-up). But if there is a good reason why its likely to be unnecessary, it saves me a couple of parts. :-) Can anyone see a good reason why this circuit reliably powers up reset? <Q> Sure can. <S> It's the 10k to V+. <S> As the circuit powers up, assuming the set and reset lines are held at the same value, the - input to the op amp will reliably be held higher than the + input. <S> During the early stages of the power-up cycle, while V+ and V- are low, the output transistors in the op amp output stage are not able to produce much current. <S> As a result, the output is not able to override the pullup resistor before the output state is established, and the reset state is established. <S> And your paranoia is justified. <S> If you cannot guarantee the condition of the set and reset inputs during power-up, you cannot guarantee that they will not override the pullup resistor. <S> So a POR (power on reset) pulse or level is ALWAYS a good idea. <A> This does not answer the original question exactly but demonstrates a means by which the startup-reset state can be assured and shows another low cost and effective "Heath Robinson" means of achieving the desired aim. <S> I have built many circuits using this approach with excellent results. <S> Schmitt trigger input inverters are an immensely flexible and powerful building block for small tasks such as this. <S> A 3rd inverter can be used as an optional reset with a startup delay. <S> With care the reset function could be implemented with an RC arrangement without the 3rd inverter. <S> Below: IC1 + IC2 form a no inverting buffer with Schmitt triggered input. <S> R1 feeds buffer output to input forming a self driving latch. <S> D1. <S> D2 provide Reset & Set signals which must be able to override the feedback latching signal via R1. <S> IC3 is an optional reset circuit. <S> C1 is discharged at startup so IC3 input is HIGH on power-up <S> so IC3 output is low and provides a reset signal to IC1. <S> The R2 C1 time constant governs the length of reset signal. <S> R3 ensures that the reset signal level provided by IC3 dominates the startup conditions regardless of the state of the set or reset inputs. <S> The three "spare gates" are available for other uses - oscillators, delays or Diode-Resistor-Inverter logic gates. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Without any inputs applied, the 10k pull down on the non inverting input will cause the opamp to always power up with the output low but, on a different opamp there could be reasons for suspecting a different result especially on an opamp that requires a minimum power supply voltage of several volts. <S> My reason is because if the output is closer to the positive rail whilst the power rails are below the value guaranteed for operation, there could be enough output voltage to forward bias the feedback diodes making the non inverting input higher than the inverting input when the rails finally get to the minimum level required by the chip. <S> I note that you have two diodes in series and presumably one diode isn't sufficient to guarantee the output being low at power up. <S> Else why use two diodes?
| As an option to using an op-amp you could consider the use of 2 x Schmitt trigger input inverters (74C14 etc) as an RS latch.
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How do I detect human touch on a point on the PCB? I want to print buttons in the silk screen of one side of the PCB. And, I want to detect if a human finger touches any of these areas. What kind of sensor should I use? Can I do this without using a sensor product? For example, by placing traces behind the button shapes in some certain topology, and detect change in the capacitance between two points on the traces? <Q> Any capacitive touch sensor will work for this. <S> Most vendors have dedicated ICs for exactly this application. <S> You could even get away without a dedicated sensor if you used a MCU with built-in touch capability and an appropriate library, such as pairing Atmel's AVR chips with their QTouch library . <A> Yes, this is commonly done with capacitive sensing, often referred to as cap touch or cap sense . <S> This is a deep subject, way too complicated and with too many details and pitfalls <S> you really need to know about to get into here. <S> Most of the major microcontroller vendors have numerous app notes, sample code, working subsystems, and evaluation boards. <S> Microchip even has some stand-alone hardware products for this purpose, and other vendors may too. <S> I'd start on the Microchip web site and look thru app notes about cap touch or cap sense. <A> You can create the capacitive sensors on the PCB itself right under the areas of interest. <S> A filled circle with surrounding ground plane will form a capacitance you can measure. <S> You can get not so bad results quite quickly, but outside of the lab, things start to get tricky quickly as well. <S> I implemented something just with a timer and the comparator of the GPIO - still works after 5 years, but never left the lab (it's still running on the same battery as well). <S> It is based on the application note from Texas Instruments. <S> The basic idea is this: simulate this circuit – Schematic created using CircuitLab <S> You set the GPIO at first as output and charge the sensor element. <S> Then you switch it's functionality to input and as a trigger for a running timer. <S> The external resistor will "slowly" (still a matter of microseconds) discharge the sensor element. <S> When the voltage falls below the trigger point of the pin, it will capture the timer value. <S> If a finger comes close, the capacitance rises and the time to discharge will increase, with this you can detect if a finger is there. <S> In the application note there is a not so dumb algorithm to track environmental changes, as the will disturb your system in the long run. <S> For a real robust implementation, you are probably better off with a "real" touch sense controller and a study of how to design the touch sensing elements. <A> Very simple to implement as you can use individual copper planes beneath each buttuon and route them off board to a capactive touch sensor like Atmel AT42QT1070. <S> For reference look a how these TTP223 sensors do it: https://hobbycomponents.com/sensors/901-ttp223-capacitive-touch-sensor
| A lot of MCUs have a capacitive touch function built in, but you can also get away with just a timer and a GPIO pin. This is something you need to study before implementing.
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Formula to find out the resistance per meter of a specific gauge of wire if I already know the resistance of a different gauge of wire per meter? So if I know a 30 gauge wire's resistance at 1 meter, how can I figure out the resistance of a 29 gauge wire at 1 meter long of the same material? Or 5 gauge, or any other gauge? I need a formula, not a chart. Working with multiple materials, and I know the resistance of them at different gauges, but need to find out how to get the resistance per meter of varying gauge sizes using math. <Q> Where <S> p = Resistivity of material , L = length of wire and A = area of cross section <S> Assuming DC, and that the cross section is uniform. <A> Resistance of a constant cross-section of material from end-to-end is \$\frac{ \rho <S> L}{A}\$ <S> Where: \$\rho\$ is the resistivity of the material (can be considered a scalar constant at a given temperature for a given homogenous anisotropic material) <S> L is the length of the material from end-to-end A <S> is the cross-sectional area of the piece of material. <S> The cross-sectional area of a cylindrical wire is simply \$\pi r^2\$ where r is the radius of the wire, so you can see that the resistance is inversely proportional to the radius squared. <S> The relationship between AWG (American Wire Gauge) and area can be described by a formula (from Wikipedia ) \$A_n = 0.012668~\mathrm{mm}^2 <S> \times 92 ^ \frac{36-n}{19.5}\$ <S> Where n is the wire gauge. <S> By combining those formulas you can get your requested results. <A> If A is the wire gauge (American), and the resistance of 1 meter of wire equals R, then R = .01 <S> (10 <S> ^ <S> ((A - 15)/10)) for copper. <S> In other words, the resistance of 15 ga wire is .01 ohms/ meter, and increasing the the gauge by 10 multiplies the resistance by a factor of 10. <A> Looking carefully at the AWG tables, you will find the relations between gage. <S> For example -The ratio of any diameter to the diameter of the sixth greater gage number is equal to 2.0050 -The ratio of any diameter to the next smaller is constant number -The difference between any two successive diameters is a constant ratio times the next smaller difference between two successive diameter. <S> -….and <S> much more. <S> What is related to your question is that the resistance, mass, and cross section vary with the square of the diameter, hence by the use of the square of the ratio of one diameter to the next, or 1.2610, it is possible to deduce the resistance, mass, or cross section of any size from the next. <S> It is easy to remember the 1.25. <S> The approximate factors for finding values for the next three sizes after any given size, are 1.25, 1.6, and 2.0. <S> Furthermore, every 10 gage numbers, the resistance and mass per unit length and the cross section are approximately multiplied or divided by 10. <S> For resistance a very simple approximate formula may be remembered for computing data for any size of wire. <S> ohms per 1,000 feet at 20 degr. <S> Celc. <S> Where n is the AWG gage number. <S> The same you can do for mass and cross section.
| The equation to find resistance in a wire is: R=(p*L)/A
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How to split a floating point number into individual digits? I am helping a friend with a small electronics project using a PIC microcontroller 16F877 (A) which I am programming with mikroC. I have run into a problem which is that I have a floating point number in a variable lets say for example 1234.123456 and need to split it out into variables holding each individual number so I get Char1 = 1, Char2=2 etc etc. for display on LCD. The number will always be rounded to 3 or 4 decimal places so there should be a need to track the location of the decimal point. Any advice on how to get this split would be greatly appreciated. <Q> There's numerous ways of doing it. <S> You may find your compiler has a library function to do it for you. <S> dtostrf() dtoa() <S> Alternatively, it's not too hard to write your own routine to do it. <S> So in pseudo-code it may look something like: If the value < 0.0: Insert - into string Subtract value from 0.0 to make it positive. <S> While the value <= 10.0: <S> Divide by 10 Increment decimal counterFor each digit of required precision: Take the integer portion of the value and place it in the string Subtract the integer portion from the value Decrement decimal counter <S> If decimal counter is 0: Insert decimal point <S> Multiply the value by 10. <A> Use sprintf(). <S> It works just like printf(), but "prints" to a string (also known as an array of char). <S> do: <S> char stringvar[10];sprintf(stringvar, "%9.4f", floatvar); <A> On a small MCU without hardware floating point support we should do as little floating point math as possible, and unless you really need the printf family of functions, try to avoid it because it bloats and slows down the code a bunch. <S> I suggest converting the float to an integer by first multiplying the float by 1,000.0 (assuming you want three decimal places) and then convert it to a long integer, round-off as appropriate. <S> If you will be displaying the result on a 7-segment or dot matrix LCD then I think this format is ideal. <S> Let's assume that the float can be in the range 0 to 999.999 (negate if negative, save sign for display later.) <S> The corresponding long int then has the range 0 to 999999. <S> We will convert the number starting with the most significant digit. <S> Pseudo code: <S> dig6 = <S> -1 <S> // <S> Init MSDigitwhile number <S> >= 0 <S> number = <S> number - 100,000 <S> dig6 = <S> dig6 + <S> 1number = number + 100,000 // <S> Restore number, Dig 6 is donedig5 = <S> -1while <S> number >= <S> 0 <S> number = <S> number - 10,000 <S> dig5 = <S> dig5 + <S> 1number = number + 10,000 // number can be switched to 16-bit here for speed... dig4 and 3 in similar fashiondig2 = <S> -1while <S> number <S> >= 0 <S> number = <S> number - 10 <S> dig2 = dig2 <S> + <S> 1dig1 = number + <S> 10 <S> At this point you have all six digits stored in a byte each and the minus sign saved. <S> If you are using a 7-segment LCD, pass the digits to a 7-segment encoder function before writing to the LCD. <S> If you are using dot-matrix display with serial interface, add 0x30 to each digit for ASCII encoding. <S> We also need to remember the decimal point between dig4 and dig3. <S> This algorithm is quite fast since there is no multiplication and division involved. <S> I have used it on tiny 4-bit MCUs with good results.
| It may be possible with: sprintf() / snprintf() It's just a case of first working out how many digits before the decimal point there are, dividing it by 10 that many times, then taking the integer portion repeatedly while multiplying by 10, making sure you add the decimal point in at the right place.
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Two voltage sources - selecting with digital signal I'm trying to build a fan controller based on Arduino.It will be (at least I hope so) possible to either control with pure PWM or voltage regulator based on buck converter.Now what I'm trying to do is: build a circuit that will allow me to choose, if I want to power the fan with full 12V (for PWM fans) or buck converter output (for 3-pin fans). Selection should be possible with Arduino digital output (either LOW or HIGH).I to google some solutions and tried to simulate some circuits based on diodes or transistors, but it wouldn't work as I envisioned. I lack knowledge to fully design what I want.I would like to get i either: sample circuit (if possible simple and cheap) phrases to google (I'm not native english speaker and maybe I'mnot looking for the right thing) some insight if my "vision" is achievable and makes sense (alternatives?) Thank you! EDIT.According to "pure PWM" or "12 V". 4-pin fans use 12V supply all the time with additional cable to control via PWM.3-pin fans on the other hands should be controlled by varying input voltage. This is conceptual image of what I'm trying to do: Input voltage to buck converter is 12V. Thank you for tag suggest! <Q> This is not ideal as you have a diode drop in each case but should work well enough n most cases. <S> Use Schottky diodes for minimum voltage drop. <S> Once we understand the requirement better we can show you how to use transistors to do the switching. <A> As a simple and largely unbreakable device try using a device called a relay. <S> Pick one that has a changeover contact. <S> In effect it is a mechanical switch controlled by an electromagnet. <S> The electromagnet (coil) would be activated by the microcontroller via a transistor. <A> First, to make things simple I would pick a fan for the design like the first option (12V, GND, Tach). <S> Next you will want to protect your Arduino, the fan and the power supply. <S> I would recommend looking at Texas Instruments website for a fan or motor controller demonstration board and drive the PWM signal to it from your arduino, which in turn will PWM the actual motor for the fan. <S> This has several benefits <S> The problem of control/feedback has been solved/built for you (in a demo board) <S> If you choose the right part, there is motor stall/overcurrent protection built in, so you won't need to worry about cooking your motor (or the driver circuitry) <S> It will take the control/feedback signals that the arduino can support.
| If you want either Output from buck converter or 12V applied to the fan then you can use diodes from both sources and enable one at a time. Alternately, there is a motor shield for arduino that will get you where you want to go like this one here
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How dangerous is a bare wire? My cable to charge my iPad has recently exposed some metal wire (due to the plastic insulating splitting). How dangerous is such a wire? If I was to touch the wire when plugged in to the mains, what would be the likely outcome? <Q> As pictured, the outer insulation jacket has cracked or split (because of cheapness on apple's part which they refuse to acknowledge and recall, but I digress), exposing a metal braided wire. <S> This wire is a shield cable, used to protect the signal carried by the wire from interference, and prevent it from interfering with other devices. <S> That shield braided wire surrounds another, thinner piece of insulation, which has the actual power and data wires inside. <S> The shield is often tied to the ground return wire of the cable, on one or both ends, and typically is safe to touch. <S> The typical metal cased computer would be grounded in the same way, and these can be freely touched without harm. <S> As the iPad is a low voltage DC device, and that crack appears on the lightning connector end, you would not be exposed to mains voltage. <S> That said, the cable has been compromised, and continuing to use it as such is up to you. <S> It is now a weak spot for the inner wires, a stress point where bending will have a greater impact. <A> It can be lethal if you are not using an properly isolating charger, Ken Shirriff explains why in this detailed blog posting . <S> In a standards-compliant charger, there are various design aspects which ensure your safety, which include using recognised safety components, materials and spacings between live parts and connections which lead to the device. <S> If you are using the correct charger, there should be no effect from touching any of the connection points on the USB side, but when it comes to safety, extra layers of protection are always a good idea, so get a new lead as soon as you can. <A> If you touch it, it will be harmless. <S> It may however cause damage to your charger or to your iPad if you short circuit it or if you touch it to a grounded conductor. <A> Are you referring to the braided shield overlay or the positive or negative wire. <S> The shield is just for interference and is connected to the return on one end. <S> The positive is 5vdc, current in the range of milliamperes, not sure how much, and the negative is the return, so no real danger there unless shorted. <A> Mechanically compromised cables are dangerous because: 1) There is a small risk of fire, the cables could short out and heat up thus causing a fire. <S> 2) <S> It could compromise the device (brownouts are never good, especially if they are momentary. <S> There probably isn't that much risk for electrocution, because most laptop adapters are lower than 20V.
| You should insulate the crack by some means, heat shrink tubing, electrical tape, plastic molding, etc, not just to prevent accidents, but to strengthen the cable at that point to prevent further damage. So proceed at your own risk.
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What soldering kit for tiny electronics? This is a my first foray into building electronics. I am building a project that uses chip/SMD LEDs and resistors. What type of soldering kit should I be using for those devices? <Q> Are you using SMD parts because you need to keep the the size of a board to a minimum? <S> If not, there are often through-hole parts that provide the same functionality. <S> Obviously in you case, the resistors and LEDs, but there might be other ICs as well. <S> You don't have to use SMD parts for everything just because you have to for a couple, for example one or two ICs that you can't get in a through-hole package at all. <S> You can get one from SparkFun for $100. <S> I worked for a company that got these for all the engineers because they didn't want to spend four times as much on a Weller. <S> You will probably want to buy some extra tips in different sizes. <S> I suggest getting a T18-I-T18 ; it has an extremely fine tip (0.2mm). <S> The next step up for working with SMD parts is a stereo microscope ; the minimum for a decent one used to be around $700 but they are now appearing at half that price. <S> The important thing is to get one with a boom, like the one below, so it can swing out over your PCB. <S> If you are going to be doing a lot of SMD work in the future, and can justify the cost, you won't believe how much easier it is to work with. <A> I have never heard of a soldering "kit" per-se, but here are the ways that I solder SMD components. <S> My favorite for SMD is soldering paste, then heated in a toaster oven. <S> (Be sure not to put food in the toaster afterwards!) <S> http://www.instructables.com/id/Toaster-Oven-Reflow-Soldering-BGA/ <S> Followed in close second is doing it by hand with a variable temperature soldering iron. <S> (I like wellers) <S> http://www.instructables.com/id/How-to-Solder-SMD-ICs-the-easy-way/ <S> I prefer to use 60/40 solder & solder paste. <A> All you need to know is addressed in Dave Jones dev video blog: https://www.youtube.com/watch?v=J5Sb21qbpEQ <S> He also mentions Hakko and many more tools and techniques. <S> It is worth watching his tutorials, not only for beginners.
| That said, a good soldering station for SMD work that is a fraction of the cost of most others is the Hakko FX888D .
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Controlling the power from a 9V battery with a potentiometer I'm working on a project where I'm powering an LCD screen backlight via a 9V battery. As it is currently, I'm powering the LCD via a 9V battery. This works great, but I'd like to have the ability to easily change the brightness via adjusting the voltage with a potentiometer. I have a 10k Ohm one currently, but that only works when it's fully on. Anything less, and the screen is off. Would a higher Ohm potentiometer solve this issue, or is there a better solution? <Q> First of all, is 9V the supply voltage of the backlight? <S> LEDs usually have voltages of about 3V, and a higher voltage usually destroys them immediately. <S> Maybe, there are multiple LED in series and a resistor in your display. <S> Your problem is the characteristic U-I curve of LEDs. <S> A little lower voltage, and the current drops significantly, and the LED becomes much dimmer. <S> If your potentiometer is too large, a small angle will already cause a high resistance, which does not allow enough current to flow through the LEDs. <S> Hence, a smaller potentiometer may help. <S> As said, the correlation of angle and resistance is logarithmic, so the resistance will change only a little over a wide range of the angle. <S> This allows a very fine adjustment of the current. <S> However, I wonder that the LED is fully off as soon as you touch the potentiometer... <S> Also, you have to take into account the power dissipation of your backlight. <S> Potentiometers usually can not dissipate much power, and depending on your backlight, you may overload it. <S> Another solution would be a PWM, for example using a NE555. <S> Here is an example from <S> http://www.reuk.co.uk/LED-Dimmer-Circuit.htm <S> : <S> The benefit is that there is a linear potentiometer, and the output power also depends linearly on the potentiometer setting. <S> Also, the behavior is independent from the load connected. <A> No, a potentiometer is not really suitable for this job. <S> Instead you should control the back light through a transistor and drive that transistor with a PWM signal. <A> One simple solution is to use a transistor as an emitter follower: simulate this circuit – <S> Schematic created using CircuitLab <S> The benefit is the simplicity of the circuit, you only need to add a NPN transistor to your potentiometer. <S> The drawback compared to the PWM solution is that the output can only go up to about 8.3v (so you can't achieve the full brightness), and the power consumption, since the excess voltage (input-output difference) multiplied by the output current is consumed on the transistor as heat.
| You could use a much lower resistance potentiometer as a variable resistor, but the currents involved at higher brightnesses would probably burn out the carbon track. Also, have a look at logarithmic potentiometers.
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Why can't the atmega328p accept 5 and 3.3v signals at the same time? The same chip can be run at 5v or 3.3v so it's tolerant to 5v so why when I run it at 3.3v can't I send in a 5v signal on an input pin? Curious on what's in the chip that makes this a bad idea. <Q> This is to expand on Olin's comment about protection diodes . <S> The protection diode will clamp the 5V input, and it can be damaged in the process. <S> (fig. <S> 10-1 from p.60 in the datasheet for ATmega32u4 . <S> Purple scribbles mine.) <S> On a different note. <S> From the O.P.: The same chip can be run at 5v or 3.3v, so <S> it's tolerant to 5v [...] <S> [emphasis mine, N.A.] <S> There is a flaw in this reasoning. <S> 5V tolerant means that an IC can use 5V input while it's powered from +3.3V (or some other voltage lower than +5V). <S> If the IC can use a 5V input while it's powered from a +5V supply, that doesn't constitute 5V tolerance. <A> It's because of the input protection diodes. <S> Each pin has a (normally) reverse biased diode from the input to Vcc. <S> If a high voltage spike comes in the diode conducts it to the rail instead of it damaging the actual input circuitry. <S> You can actually use this to your advantage though, put a resistor in series with the input. <S> Then when you apply 5V the diode will turn on and clamp the voltage to a safe level, but the resistor will limit the current to something safe. <S> Your other option is to power the chip with 5V. <S> The datasheet says it will accept anything over 0.6*Vcc as a logic high, so anything 3.0V or high will be read correctly. <S> Which method is better depends on the output voltage you want. <A> Look at Table 28.1 Absolute Maximum Ratings - in that table, it says <S> "Voltage on any Pin except RESETwith respect to Ground ................................ <S> -0.5V to VCC+0.5V <S> " There will be protection diodes on the I <S> /O pins that will prevent the voltage on the pin from going outside those limits. <S> The "Vcc+0.5" limit applies regardless of Vcc.
| If you power the chip with 3.3V a 5V input will forward bias the diode with a lot of current and damage it, then possibly the rest of the circuitry.
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Push power assisted trolley with one finger I recently saw a demo. One man pushed a power-assisted trolley using only one finger. The trolley was loaded with ~40kg stuff and it started from standstill on a up-slope when the man pushed it using one finger and apparently without much effort and it could move up quite swiftly. I was told there was no force or proximity sensor to detect the finger, but solely depend the motor encoder and controller. I didn't understand how a trolley loaded with 40kg and pushed by a finger could cause any change in the motor's encoder. Did anyone ever worked on power assisted trolley or bike or something else before to understand how to realize it? <Q> The answer they gave was probably misleading. <S> One method, and it is very likely that the one used is generally similar, is to have a "position control" lever that is spring loaded to centre. <S> The lever is vertical and spring loaded to the vertical centre position. <S> If you push on the lever from behind it moves forwards and the system is designed to move the cart forwards until the lever is again vertical. <S> If the pushing "finger" is stationary the cart will move away until the prior condition again exists. <S> If the finger (and the attached person) moves forwards at a steady space the cart can be arranged to move at that pace. <S> The biggest challenge is probably to get a control loop such that the cart neither "shoots away" when the lever is slightly pressed or stops topo suddenly (or moves backwards once pressure is taken off the lever. <S> The user's brain is probably part of thge control loop - getting the cart to move at desired speed up a slope or down a slope probably requires experience in just how hard and fast to push. <S> Arrangements like this have been "common enough" for a very long time. <S> These may operate with a caternerary loop (cord hanging in a downwards loop) so the user exerts almost no force at all. <A> Based on the video you linked it sounds like an inverted pendulum control system, similar to a Segway. <S> When you push on the top of the inverted pendulum the wheels move to bring the CG overtop so that is balanced. <S> This is a closed-loop control system. <S> I've never ridden a Segway <S> but I tried a Chinese ATV version recently- as Russell says there is significant interaction between the machine's control system and your own <S> so there is a learning process if you're actally aboard such a beast (during which you may be somewhat of a hazard to yourself or bystanders). <A> To expand on what Spehro said. <S> The video explicitly says that the machine uses an inverted pendulum principle to control the wheels. <S> What this means is that it has a very sensitive vertical sensor in the body. <S> When you push (even very lightly) forward on the top of the machine, it senses the resulting tip, and drives the wheels forward to get the bottom of the unit directly under the top. <S> This is like a hanging body, which will move the bottom to get under the top if the top is moved. <S> Except, of course, that top and bottom are reversed, which is why it's call an inverted pendulum.
| Rather than a lever they may use a control plate or a "cord" that the user pulls on and the cart follows.
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What is the typical error of a voltage follower opamp I am trying to identify design elements that might have error from unit to unit. In the design I am looking at, there is a voltage follower configuration for an opamp (in this case, an LM2014), which has an ideal gain of 1 (by definition). Of course, in the real world nothing is ideal, everything has some amount of variability and error, and I would speculate that the gain of a unity gain opamp is no exception. I was unable to find any rule of thumb on real world gain errors for this configuration, and I'm not good enough with discrete electronics and opamp specs to narrow the answer down for myself. What should be considered a typical error in the gain of a unity gain opamp? (e.g., is it safe to say 3, 4, or 5 decimal places?) How should I go about figuring this out from the specs of a given opamp so I can do this myself next time? simulate this circuit – Schematic created using CircuitLab <Q> With two voltages (input, output) I would expect that it is likely a "voltage follower". <S> More than that, are you sure about the part number? <S> When speaking about classical voltage opamps the real closed-loop gain for the unity gain amplifier (follower) <S> is <S> \$Acl = \dfrac{Ao}{1+Ao}\$ with Ao: Open-loop gain. <S> Fort a typical value <S> \$Ao=10^5\$ <S> (100 dB) <S> we have <S> \$Acl = <S> \dfrac{10 <S> ^ <S> 5}{1 <S> +10^5}\$ which is very close to unity. <S> However, it is to be noted that the open-loop gain Ao continuously decreases with rising frequencies and causes - in addition - phase deviations. <S> Hence, we have Ao=Ao(jw). <S> However, each opamp has an input offset error between µV (very good devices) and some mV (universal types). <S> This voltage is amplified with unity and, thus, appears with the same value at the output. <S> But it is not an amplification error but a fixed dc shift of the operational point. <A> This can range from microvolts to millivolts. <S> Traditionally, op-amps with \$V_{os}\$ below 1 mV have been sold as "precision" op-amps. <S> And you'll find this as one of the main categories of op-amps on most vendors webpages (Analog, TI, Linear, ...). <S> If this parameter is critical, you should also look for a part whose datasheet specifies the temperature coefficient of the offset, and be sure that your circuit will remain in spec throughout its expected operating temperature range. <A> The error will be composed of the input offset voltage and change of input offset voltage with input common mode voltage (CMRR) plus error due to finite gain- <S> so a small offset and a small gain error. <S> The error due to finite gain with a precision op-amp is usually pretty low at DC, but increases with increasing frequency. <S> Noise will generally be higher than the number(s) <S> shown on the datasheet since it's usually specified with a high gain, which pretty much eliminates noise from the output stage. <A> Many have told you to beware input offset voltages, which will be your main error. <S> You may also see slew rate limits, which may distort your output at fast transiotions. <S> Also, many op amp outputs can not reach closer than a few volts below the power rails, so you may have saturation problems.
| The error is essentially the input voltage offset of the op-amp.
|
Determining if a diode conducts or is cutoff? Can somebody please explain to me how you exactly determine whether a diode conducts (shorted) or is cutoff (opened)? I found a couple of posts on this but I wasn't able to justify the answers myself, probably something I'm doing wrong. It was said that the anode should be greater than the cathode for the diode to conduct but it's not making any sense to me in these examples. The answers are: for (a), D1 is cutoff and D2 is conducting. V = 2V, I = 3.5mA(b) D1 is conducting, D2 is cutoff. V = 1V, I = 2mA <Q> The model you are using is: If \$V_{anode} < V_{cathode}\$, \$I_D = 0\$ <S> (cutoff) <S> If \$I_D > 0\$, \$V_{anode} = <S> V_{cathode}\$ (conducting) <S> Try to solve the circuit with those assumptions. <S> If you have a solution, check to make sure that solution is consistent, i.e. all of your original assumptions were correct. <S> If there is no valid solution, your original assumptions were probably wrong. <S> Pick a new set of assumptions and repeat. <S> For example, take circuit a . <S> Let's assume D1 and D2 are both conducting. <S> By definition then, the node labeled \$V\$ must be at 1V and at 2V at the same time, which is nonsense. <S> So we know that both diodes can't be conducting at the same time. <S> Now pick new assumptions and try again (left as an exercise for the reader). <A> From the answers, it appears the question is using ideal diodes, which have no forward voltage drop. <S> (Real silicon diodes have about a 0.7 volt forward voltage drop, which will change the output voltage, but not affect which diode is conducting.) <S> In (a), if D1 is conducting, it will pull the output up to 1 volt. <S> However, the anode of D2 is at 2 volts, so D2 will pull the output up to about 2 volt - that will make D1's cathode more positive than its anode, so D1 will not conduct. <A> If the p terminal of the diode is towards the p terminal of the battery then the diode is ON. <S> Similarly, if the n terminal of the diode is towards the n terminal of the battery then also the diode is ON. <S> But if battery and diode is facing opposite sides of each other then the diode is OFF. <S> I hope u got my point. :)
| The general procedure for solving systems with diodes is: Make an assumption about the state of every diode (either cutoff or conducting).
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Does the type of capacitors matter? The IC ICL232 needs some capacitors. Should these capacitors be ceramic or any other type? When does the capacitor type matter? <Q> Different types of capacitor have different properties. <S> Some of the properties that vary between capacitor types: <S> Polarised vs unpolarised Max voltage Equivalent Series Resistance (ESR) Lifetime <S> (electrolytics are particularly bad in this case) <S> Physical size (e.g. a 100,000 uF ceramic capacitor would be HUGE!) <S> Tolerance of capacitance (again, electrolytics are bad here, often being +/- <S> 20% <S> In your link, the diagram shows polarised capacitors, which suggests that they weren't intended to be ceramic (which are unpolarised). <S> Two types of polarised capacitors are aluminum and tantalum electrolytics. <S> In your case, I'd use tantalum. <A> From the datasheet it is clear that the capacitors have been used for a charge-pump circuit. <S> So it's always preferred to have polarized capacitors in such applications for stability purposes. <S> Since the values are comparatively smaller, you can go for tantalum capacitors . <A> Capacitor type can matter quite a bit. <S> Small, high value, type 2 Ceramic capacitors have two significant disadvantages, one of which applies to use in this application with the ICL232: <S> The capacitance varies substantially with voltage. <S> Under certain circumstances, your 4.7 uF cap may act more like an 0.33 uF cap. <S> Maxim has posted a very nice tutorial on this topic: http://www.maximintegrated.com/en/app-notes/index.mvp/id/5527 <S> The ceramic dialectric exhibits the piezoelectric effect. <S> This means that the capacitor may exhibit microphonics undesirable in low-level audio circuits. <S> (This could also cause the capacitors to emit sounds with substantial changes in applied voltage.) <S> So, for large value bypass caps, or for switched-capacitor power supplies, try to use polarized tantalum (smaller and more stable, but more expensive), electrolytic (cheaper, larger packages, less stable with age), or larger-bodied ceramic (moderate cost, more stable, require more board area). <A> The datasheet is not very specific about the capacitors. <S> X7R ceramics should work well. <S> Small, inexpensive, can't put them in backwards. <S> Class 2 ceramics are modulated by voltage so use the highest voltage part you can, 25 volt maybe.
| Yes, the type of capacitor can matter.
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What is that? chip identification What is that? No HD-SDI signal.Black magic camera motherboard <Q> It might be a custom chip built by Gennum/Semtech. <S> The numers on the chip seem to correspond to those of HD-SDI cable drivers they make (e.g. the GS1678), in particular 0002E3 should be batch number 0002 <S> (this very low batch number and the absence of an identifiable part name make me think of a custom chip), E3 on their packages is meaning RoHS compatible, and 1352 <S> should mean that it has been manufactured on week 52 (end of December) of 2013. <S> it has the same QFN16 package, and the pins seem to be used for similar/identical purposes: <S> Image from datasheet <S> In particular, you can notice that pins 9 (VCC) and 6 (DISABLE*) are connected together, which perfectly makes sense, and RSET (pin 4) is connected as required to a resistor (possibly going to VCC through the via). <S> The top four pins are not connected (except maybe pin 16? <S> not sure from the image where the track nearby is going), the DDI pins (1/2) have two ceramic capacitors nearby as expected per datasheet, and the output signal (going to the BNC through the matching network) seems to come out from SDO (pin 12) as it should. <A> Well, that looks like an impedance matching circuit coming from the right of it, going to the BNC connector, so my guess is <S> it's some kind of RF (or similar) <S> transceiver chip maybe. <A> For HD-SDI? <S> It's probably the cable driver and output filter. <S> Hard to see the numbers in that pic.
| The chip looks very similar to their GS1678 HD-SDI cable driver
|
How can I protect a DC switch from arcing and sparks? I have a DC switch that is being used as the main power switch for an audio amplifier project I am working on. The schematic is shown below. (a few things are wrong with that circuit, but have since been fixed.) Anyway, I am assuming that, since this is an amplifier, I have mainly a capacitive load? What is the best method I can employ to "protect" the switch from the effects of arcing and pitting on the contacts? <Q> You might want to add a fuse to the input (either a polyswitch resettable fuse or a one-time fuse). <A> An amplifier I took apart for repair had its power switch on the AC side, operated remotely inside the case by a long arm from the front panel button. <S> If you get the slightly more expensive sort of LDO which has an enable pin, you can use that to turn the regulators on and off. <S> Then your power switch doesn't have to carry any significant current. <S> Mind you <S> , I wouldn't expect much arcing if you're within the rating of the switch! <S> It's only 1A, and unless you're turning it off at full output volume it won't have that much through it at power-off. <A> You can put integral circuit between contact instead of putting big capacitor. <S> Circuit is as follows. <S> simulate this circuit – <S> Schematic created using CircuitLab
| You can move the input filter capacitor to the other side of the switch (leave it connected to the input power) and add a small capacitor (maybe 100nF) on the actual input of the regulator.
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High power PCB to PCB connector names? Trying something new out, I'd like to make an 'attachment' PCB that will slot into my mother-PCB, and all the questions here have said to use a kind of edge-to-edge connector or a ribbon connector. My question is what exactly do I use for high power? Something like 2-3A rectified mains 170VDC. Basically I know the 'mother' board works, but I don't know if the child board works, so I'd like to 'modularize' it. Perhaps these things; Would work? I just have no clue what they are called. EDIT: Uhp, they're called banana connectors. Still don't know if they're good for high power PCB-PCB connection. EDIT2: The boards will be side-by-side, edges touching. EDIT3: pic <Q> Use connectors <S> that are rated for the power you want . <S> They might be edge-to-edge or board-to-board, or wires. <S> At some point they might have product names that are power-edge-to-edge or something similar at some arbitrary threshold, but they're all in the same category. <S> For instance: <S> This connector is a standard card edge connector that is rated for 40A and 250V. <S> The main thing you need to figure out when choosing a different type of connector is how you want the boards to mate. <S> Do you want a wire between two loose boards? <S> Do you want them to slot in perpendicular? <S> Do you want them to connect end-to-end? <S> Etc. <A> Maybe something like this? <S> http://www.digikey.com/product-detail/en/6643220-1/A101989-ND/2056032 and http://www.digikey.com/product-detail/en/6643274-1/A114321-ND/2305428 <S> I think they are considered a "two piece power connectors" meant for hot-swap power electronics. <S> I've seen them mostly on hot-swap power supplies. <A> Something like http://uk.rs-online.com/web/p/din-41612-connectors/5424631/ <A> I work for GCT so express an interest. <S> We have some board to board connectors for co-planar PCB mating, these are aimed at LED lighting applications. <S> There are SMT and thru hole variations, they are available in 2-6 contacts and rated at 3A per contact. <S> I think you will find this an inexpensive solution for co-planar mating which is also available from stock in Newark/Farnell. <S> SMT = <S> BG300 socket mating with BG301 Header Thru Hole = <S> BG302 socket mating with BG303 Header <S> Here is the GCT link for general information: <S> GCT White Lite PCB Connectors Link to Newark SMT headers <S> BG301 Link to Newark SMT sockets <S> BG300 Link to Farnell SMT headers <S> BG301 <S> Link to Farnell SMT sockets BG300
| You could also try DIN41612 connectors - there are variants with special high-current pins.
|
What is the difference between Microchip and ICD and PICKit? I understand that Microchip has In-Circuit debuggers (ICD) which gives insight into a running PIC (or dsPIC for that matter) in real time. Then we have the ICE which goes a step ahead by replacing the microcontroller within the board and emulate it. The ICE gives full view of what would be happening inside the PIC had there been a real microcontroller on that board there. Microchip also has something called the PICKit. It can be used to program a PIC. Can it also be used to replace the ICD? I am specifically referring to PICKit 3. For Easy PIC PRO v7 here is a section for "Programming with ICD2/ICD3". Can the PICKit be used instead or shall I have to buy an ICD? <Q> There is a nice comparison between the PICKit 3, ICD 3, and Real ICE here . <S> The big difference between the PICkit 3 and ICD 3 is speed -- the PICKit run at USB 1.1/Full Speed (12 Mbs), and the ICD runs at USB 2.0/High Speed (480 Mbs) as does the Real ICE. <S> This really does make a difference. <S> Otherwise, the PICKit and ICD are similar in features, except the ICD allows for more complex breakpoints, including software breakpoints. <S> The Real ICE does not replace the microcontroller on the board with one inside the Real ICE (although the name seems to imply that). <S> The chart says the PICKit is not suitable for production programming, whereas the ICD and Real ICE are. <S> I'm not quite sure what they mean by that, except perhaps the interface to the board is more failure proof. <S> The interface for the Real ICE is on its own little daughterboard, so if you blow the output circuitry, you only have to replace that piece. <S> I strongly suggest you get an ICD 3 if you can afford it. <A> The main differences between ICD4 and PICKit 4 are as follows: o Power to Target: ICD4 can provide up to 1A, whereas the PICkit4 can only provide 50 mA. o <S> Breakpoints: ICD 4 supports complex breakpoints, while as the PICkit 4 only supports simple breakpoints. <S> o <S> Over Voltage / Current Protection: In the ICD4 the Over Voltage / Current Protection is implemented in the Hardware, while as in the PICkit 4 <S> it is implemented in the Software. <S> o <S> SDCard: PICkit4 has an SD Card slot for storing programming images for Programmer to Go. <S> It's not fully implemented yet, but will be in the near future. <S> References: <S> • <S> microchipdeveloper.com/icd4:start <S> • <S> microchip.com/icd4 <S> • <S> microchipdeveloper.com/pickit4:start <S> • <S> microchip.com/pickit4 <A> Reviving an old thread since PicKit 4 has came out. <S> PicKit 4 is much faster <S> the PicKit 3 so ICD is not needed anymore if all you wanted is speed. <S> Also PicKit 4 will soon allow you to put your HEX on a micro SD card to program on the go far from the computer.
| The main difference between the ICD 3 and the Real ICE is that the latter uses the trace capability (like JTAG) built into chips like the PIC32 series.
|
Which files to version control for an Altium PCB project? I'm trying to create the .gitignore file for an Altium project that is versioned with git and stored remotely on GitHub. I don't want my teammates to continually struggle with having to update every single time I make a small change, like re-run design rule checks or re-compile the project. What are the minimum files I should I add to my version control system? # Ignore the subdirectory where output job outputs are placedGeneratedOutput# Ignore autosave files (anything that begins with a tilde)~*# What else to ignore? <Q> Here is the ignore list I use for managing Altium files in SVN. <S> I'm sure they can be converted to a .gitignore format (if it's even needed) without too much trouble: <S> */History/**.PrjPCBStructure*.SchDocPreview*.PcbDocPreview*\__Previews*\History*.PrjPcbStructure*~$ <S> * Note that I disagree with @KyranF <S> , you do not want to archive the prjpcbstructure files. <S> They're entirely regenerated every time you compile the project, and there's not really anything in there anyways. <S> They are functionally just build-artifacts, and those should not be committed. <S> Also, I've had my ass saved a few times by the History stuff, <S> if a few hundred MB of local storage is a problem, you need 1. <S> A bigger hard drive, and 2. <S> To fix your priorities, if a few hundred MB is a serious consideration at all. <S> I also disagree strongly about committing gerbers. <S> If you're trying to canonically link a actual PCB to a set of files, having the gerbers can be essential, particularly when things go wrong in the gerber export and/or processing stage. <S> Admittedly, you shouldn't be comitting gerber files every day (you shouldn't be bothering to create them daily either), but you should ALWAYS commit (and ideally tag) <S> each set of gerbers you release to manufacturing. <S> I think <S> * <S> ~$ <S> * may be the temporary file indicator for SolidWorks , not Altium. <A> As an addition to the other answers: Do not version Obvious files: Potentially all generated output, like BOM, report files and so on. <S> .Dat <S> - Is (re-)generated as part of BOM generation. <S> Does only seem to contain redundant information. <S> .TLT <S> - GUI layout information. <S> .REP <S> - Teardrop Report file. <S> Do version / consider versioning Obvious files: <S> .PcbDoc, <S> .PrjPcb, <S> .SchDoc, <S> .OutJob <S> , .PrjPcbVariants .DsnWrk - I often find, that I open several unrelated PCB projects while working through the day, which results in changes to the workspace file. <S> Since that is not relevant for version control, I tend to leave it out. <S> PcbLib, .SchLib, .PvLib - A project may contain some project specific libraries. <S> .PrjPcbStructure <S> - Contains information about the hierarchical structure. <S> It gets (re-)generated on compilation. <S> Can be versioned, since when the structure changes, a commit is pending anyway. <S> .Harness <S> - Those files contain Harness definition and can potentially contain arbitrary definitions that can't be generated. <A> You want to version control the .prjpcb (assuming you have a PCB project!) <S> , .prjpcbstructure, .pcbdoc, and .schdoc files. <S> The rest, such as the history, previews, gerber outputs etc are sort of transient files, not important to the project. <S> edit: also be aware of the huge size of "history" files, they can become many hundreds of megabytes, so you should adjust the history size in the project/Altium preferences to keep it under control, even if that whole folder is being ignored by sub-version file control, it will still eat up HDD space on your PC.
| .Annotation - Contains Board level component annotation information.
|
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