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SMD Crystal solder(?) problem and any recommended test procedure I use a 32.768 kHz SMD crystal ( datasheet ) for a MCU. Here is the layout part of the crystal. Here is a view from an actual PCB I mounted the components by hand-soldering, and produced about 30 modules. Most modules' crystal didn't start working at the beginning, but I solved it by re-soldering the crystal by applying a little bit more solder under it. The interesting thing is that when I touch a pin of the crystal by a tweezers, the crystal starts working. Here is an an oscilloscope view of the crystal's signal. The glitch is where I touched the path by a tweezers, and then the oscillation starts. Another fact is that some of board's crystal work if I touch to the C451 cap's path, some of them start working if I touch to C441 cap's path (e.g., if it works by touching the C451 cap's path, it does not work if I touch to C441 cap's path). This made me doubt if it is related with the solder under the crystal (maybe uneven contact surface or another reason that I can't think of). Or if it is not a purely solder related problem, as I sometimes needed to perform re-soldering process several times till the crystal problem is solved. On the above PCB view, the extra solder sticks out from side of the crystal (no short circuit to another pad or case of the crystal), there should be for sure a connection between the crystal's pin and the PCB pad, but the problem still remains. Another issue is that I experienced on 4 boards that they work after I re-apply solder under them, but when I test them the next day, the crystal has the same issue. Question.1 Has anyone experienced similar problem or could think of what could be the actual problem? Question.2 The boards will be on the field, I have a concern that they will work here but have problem when the customer needs to use them. How I test them is to start the modules couples times at a day and observe if there is any failure and spread this test to a week long. Is there any method/technique to test (or get indication) if the crystal will probably work fine in the near feature I inspected the PCBs by a microscope that there is no short circuit between any trace or there is no any solder connection from the case of the crystal to any path On the problematic boards, I re-placed the crystal with the ones that I removed from the board that works OK, therefore it should not be a component problem I cleaned the flux residues on PCB but it does not change the result I did search if there is a special procedure/technique to solder such type of SMD crystals but couldn't find any related information. EDIT I tried placing a different capacitor values but didn't help. EDIT2 Here is the gerber view of the reference design of TI, as it is an optional crystal, it is connected through 0 ohm resistors(R451, R441) <Q> Your layout looks a lot worse in the screen shot than it actually is with the ground pour, however I'd still try to isolate the ground return from the load caps and have it go right back to the ground on the chip. <S> I'd also like to see a ground pour under the caps and crystal (maybe you have one) and connected to the same ground point as the load caps. <S> That's not likely <S> your problem- there seems like there are several possibilities. <S> First, the metal case of the crystal may be shorting or there is some PCB problem. <S> Your description tends to indicate not. <S> This is an extremely low power-level crystal (100nW drive 500nW maximum drive). <S> Make sure it's well matched to your chip. <S> You can't change the load capacitances willy-nilly if you want accurate timing (+/- <S> 40ppm is required) but make sure they're appropriate <S> (if there are no caps inside the chip they should generally be a few pF less than double the specified capacitance for <S> the crystal- that datasheet shows several possible values for that crystal). <S> Finally ( and perhaps the best news! ) <S> , if your assembler is using no-clean flux, and especially if the chip is extreme low power, I'll bet that is exactly your problem. <S> Removing the residue takes a strong solvent and scrubbing. <S> Your re-soldering the parts may affecting the residue and allowing them to start working again. <S> Edit: I would suspect the no-clean flux residue- <S> it is extremely difficult to remove. <S> You can also compare the load cap values and specs of the Epson crystal you are using to the crystal used in TI's reference design to see if anything jumps out at you. <S> But also scrub the top of that board in the crystal area with some nasty solvent PCB cleaner and a toothbrush. <S> Solvent alone (even something as nasty as lacquer thinner) is not enough. <A> You mention that the oscillator starts when you touch it with tweezers. <S> That indicates to me that maybe you need a bit more capacitance on one or maybe both (depends on how this particular oscillator works) of the pins of the crystal. <S> I'm talking about "(3) load capacitance" in the datasheet. <S> Are there any load capacitors connected to this crystal ? <S> If not I'd start with 10 pF, if there is, increase <S> it's value with 10 pF. <S> I find it hard to believe that it has anythind to do with PCB, in my experience soldering a crystal on a PCB <S> should not be that troublesome. <S> I mostly use hot-air for such small crystals and that almost always works fine. <A> Which is the clock this board is supposed to manage? <S> It seems it's problem with capacitors. <S> When you touch a terminal things work due to your capacitance. <S> Are the values of crystal's capacitor right? <S> Have you tried to re-solder (only re-heat the terminals) <S> the capacitors? <S> 3 terminals crystals may behave more properly, specially for high frequency circuits. <S> You also should want (depending of clock's speed) to match (and reduce to minimum as possible) <S> the lenght of crystal's traces. <S> Always start to route from GND, Vcc and clock.
| You should test the boards for start-up at temperature extremes- marginal startup at room temperature indicates a probable issue related to gm of the chip, and that changes with temperature.
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Is there a standard for maximum torque that should be used to fasten PCBs to an enclosure? I am looking for a standard or guideline which describes how much torque should be used to fasten a PCB to an enclosure using different types of fasteners. For example, we use M3 hex nuts in a system. There are ~25 people doing assembly. We do not want everyone to guess how tight the nut should be turned. We would rather give an appropriate tool to everyone and make sure each PCB is fastened the same way. <Q> It depends most on what and how you are fastening. <S> Many different materials will have different properties. <S> For example, the compressive strength of FR4 material is at least 460MPa, or 460N/mm^2. <S> That number is one you want to stay well clear of, especially if you have multi-layer boards (since you don't know the exact specs of the prepreg your fab uses). <S> So let's say "definitely don't go over 100N/mm^2". <S> Now here's the rub. <S> That's the compression strength, but it does mean you have to have proper rings to prevent extra warping forces when the nut is tightened and even then some warping may still happen if tightened too far. <S> Then how that translates to a Nm reading depends on: The surface area of the rings you use: Smallest surface determines the maximum number of downward Newtons. <S> The threading on the bolt; It's neatness and the number of threads per unit length. <S> Doubtless <S> there's conversion charts for that on the internet for certain types of screws, but I couldn't find them just now <S> (though I did only spend about 1 minute searching, since that's also what it took to get the compression strength of FR4). <S> All that said, <S> I do think the tiny surface area of the threads on your screws, unless you use 13.9 grade metal, maybe, will be the weakest link in your system. <S> The cheap DIY store screws will definitely break before up to 8 layers of PCB. <S> So will, as I know from experience, RVS A2 screws. <S> That is, I did once split a 4 layer board with an M4 screw and two washers, but the RVS A2 one broke on hand-tightening, then I took out a military grade 13.9 one and needed a power wrench to tighten it enough to crack open. <S> But if you want to stay super safe, get the data of a brass spacer in the thread size you use and then apply the maximum allowable torque for that, even if you use harder types of fasteners, a metalised PCB hole will certainly survive quite well. <S> Brass spacers are relatively weak compared to FR4 compression strengths and usually with brass spacers you can still tighten stuff well enough to mount PCBs and such. <A> I use a TackLife 1-4 Nm adjustable electric torque screwdriver. <S> 1 Nm is too much torque for standard PCB material to handle before cracking if your screws don't strip. <S> I'm assuming most PCB material is similar in composition. <S> 0.6 <S> Nm torque adjuster is required at work when assembling PCB components, <S> 1 Nm for body/frame connections <S> (metal on metal). <S> Our electronics are exposed to vibration. <S> Lock-Tite (blue) is used in critical components subjected to excessive vibration. <S> Most of our fans are mounted using rubber screws as they are optical instruments sensitive to vibration and even BLDC fans are a problem. <S> These are imperical statistics, not out of a datasheet. <S> Most people don't have a < 1 Nm torquer so the "hobby rule" is 1/4 turn past "hand tight." <A> If your fasteners are crappy soft metal that might be too tight. <S> You can get a skilled assembler to do a few (as Andy has suggested!) <S> and then compare with a torque-controlled electric driver if nothing else- <S> then they'll be consistent. <S> Or deliberately break a few and use a fraction of that torque. <S> Using a nut with integral lockwasher is not a bad idea if there is vibration, or you can use Locktite type thread locking (or buy fasteners with it already applied). <A> Nothing beats experience. <S> Try a few other guys too to see what they say. <S> Lock-nuts will need a bit more torque of course.
| You can find maximum torque values for various strengths of M3 fasteners, for example Holo-Krome recommends 1-3 N-m depending on the type of fastener. It will depend on PCB and fastener materials so my advice is to get the best guy in the assembly shop to decide how tight to turn the nuts and measure that torque.
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Battery Wiring - 3 Independent systems with common ground Background: Building a radio control wheel bot. Using a motor controller that recommends independent power supply for 1.) power to the controller and 2.) power to the motors. Also want a 3rd power supply for accessories (lights, horns, etc.). I am using 8 identical lead acid batteries. 6 in parallel for the motors, 1 for the controller power, and 1 for the accessories. All 3 systems are 12v. I want to hard wire a solar charger (with controller) to charge the systems, but I will put a switch on all the positive connections to the charger, so I can choose when each system is being charged. I do not want to put switches on the negatives though. So now the question: Is it ok to connect all the grounds from all the batteries together? Then, if I want to charge all 3 systems at the same time, I will turn all the charging switches on, which will effectively put all 3 systems in parallel. Is this ok? Is it ok to charge one system while all the grounds are connected? Most likely i will want to charge the controller power and accessory at all times, but only charge the motor batteries when I'm not driving it (to prevent voltage/current spikes from going to the controller power). Any explanation is greatly appreciated. Thank you! <Q> It's not only okay to connect the grounds together, it's mandatory if you want them to communicate or interact with each other in any way. <S> Without a common ground reference, what you have is several entirely separate systems, and any communications between them would have to be isolated using something such as an optoisolator or a transformer. <S> Charging one set of batteries when they all share a common ground is also no problem. <S> That said, you really don't need separate batteries for different parts of your system, whatever the motor controller manual says. <S> At worst, you will need separate regulation for high-noise parts of the system, and input filtering (such as LC circuits) to reduce noise from the motors. <S> Just look at something like the Tesla- <S> you don't see that having several different battery banks for different parts of the cirtuit. <A> Your charging circuit doesn't sound like a good idea to me, because you will interconnect the batteries with different charge level. <S> The moment you will start charging, you will short a fully charged battery to 6 discharged batteries, causing overcurrent. <A> I would likely stop charging the the controller power and accessory batteries, and only charge the motor batteries. <S> But that does leave room for operator error, so I understand your point. <S> The controller I'm using is a Roboteq AX2550. <S> https://www.roboteq.com/index.php/docman/ax-documents-files/ax-documents-1/ax-userman-1/ax25xx-user-manual/88-ax2550-user-manual/file <S> Page 32 is where they discuss powering the controller.
| Imagine your motors were used a lot and depleted their batteries, while the accessories were not used at all.
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Impedance matching routes in Eagle board I am making my second board using Eagle CAD. I have different components with a set of different sizes. I set the minimum width for lines to be x. That way, the autorouter creates EVERY line in that width. The problem is that other components are wider, so I think I'd need to somehow widen the lines on the connections to that component (or probably not, the highest frequency in my circuit is of 16 MHz). How may I do that? <Q> Though the other answers already said that you don't have to care about impedance matching in your case, the time will come where you may need it. <S> More often, you may want to specify wider tracks for higher currents or more clearance for high voltage. <S> Eagle supports this, and it's called net class . <S> In the schematic or layout editor, goto menu <S> Edit <S> > <S> Net classes... and define sets of wire width, via drill size and clearance. <S> By Rightclick > Properties... on a wire / net you can assign one of the defined net classes. <S> This way, the auto router automatically uses these different settings for different nets. <S> Unfortunately, there seems to be no difference when routing the board by hand. <S> At least, the DRC throws an error when you used the wrong size somewhere. <S> When you pour a ground plane, the clearance is taken into account. <S> Here is an example. <S> The first wire is of the standard net class, the second from a net class with 36mil trace width and the third from a net class with 36mil clearance: <A> Your wavelength at 16 MHz is 18.75 meters. <S> Unless the traces are very long, you shouldn't have impedance matching problems (rule of thumb is wavelength/10) <A> If you are doing a digital circuit, the frequency of the (clock) signal is irrelevant for impedance matching. <S> Rise/fall time is the interesting parameter. <S> If electrical lenght of traces are more than 1/5-1/3 of the rise/fall time, you will see reflections. <S> No matter the frequency. <S> And this can cause issues. <S> That would normally not be required for the type of digital stuff you do today.
| If I read your question right, you wonder if you should taper the trace width out to match wider component pins.
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Does a coiled wire have a higher resistance than straight wire? If you take a straight piece of wire that is insulated and coil it around a non ferrous object like piece of wood would the resistance change? <Q> No the resistance only cares about the electron path through the material. <S> Regardless of the shape or if it is wrapped around something it will not change. <S> On the other hand the inductance, which is related to the energy stored in the magnetic field created around the wire as current passes through it, will change with geometry and the material it is wrapped around. <A> Everyone so far seems to have forgotten (or not heard of) proximity effect . <S> Ignoring the inductance of a coiled wire and only concentrating on losses i.e. the resistive part of the wire then, at DC there will be no change. <S> However, as frequency increases proximity effect between the coils will certainly increase the resistance of a conductor. <S> This is very much akin to skin effect where the rapidly changing magnetic field of the wire (due to the alternating current it carries) causes the electrons to flow in the outer circumference of the wire <S> thus all that nice copper in the centre carries very little current. <S> Current not using the full cross sectional area of the conductor will have a higher resistance to flow. <S> Remember <S> I'm not talking about inductance. <S> So, proximity effect is where two coils are close by and the magnetic field of one wire causes the current flow in the next wire to be even more constricted than by skin-effect alone: <S> - The above is with current in opposite directions. <S> With currents in the same direction the effects are opposite to that shown above. <A> The simple answer is no. <S> Winding wire on a non-ferrous form will not change its resistance. <S> Of course, it all depends on the details. <S> If the wire is uninsulated, and the form is conductive (copper, silver, gold, platinum, etc) the form will short out the wire turns and reduce the total resistance. <S> Or were you thinking of insulated wire? <S> You didn't say. <S> Ah, you say, but it is insulated. <S> To be precise, it's insulated with enamel, and called magnet wire. <S> Well, now it gets tricky. <S> Let's assume your reference point is a straight wire suspended between two contacts in still air. <S> As you run current through the wire, it will heat up, even if only slightly, and for most wire materials the resistance will change. <S> If the wire is now wound around a form of some material, the form will change the rate at which power is dissipated to the environment, and thus the resistance. <S> A thermal insulator, such as foam, will increase the temperature of the wire. <S> A good thermal conductor, such as silver, will tend to reduce the temperature changes, and even this effect depends on the physical details of the form to determine the long-term temperature effects. <S> Size and shape will matter, and the result can, in principle, go either way. <A> Contrary to popular belief, the resistance will change - if only by a miniscule amount - because of the wire being work-hardened by winding it around the form. <S> From: http://www.copper.org/publications/pub_list/pdf/a1360.pdf <A> This is one of those questions that starts the real meat of EE. <S> A components impedance consists of the DC resistance, inductive reactance and capacitive reactance, all of which get "sumed" up to form a frequency dependent electrical impedance. <S> I would rather not get into the down and dirty details, but this wiki article sums it up pretty nicely! <S> Electrical impedance <A> Andy aka gave an awesome answer <S> the question was so vague. <S> Sounds like you are trying to make an inductor.. <S> If so, I suggest you read up on inductors. <S> If you are asking about your ac extension cord and whether or not it matters to leave the cord coiled up while trying to use it, you can test that as well. <S> You will discover you are turning your extension cord into an inductor and depending on how many feet you have coiled up, you will notice a significant drop in the voltage, increase in impedance, and an increase in power consumption. <S> Even with household appliances, you should never coil up your excess cord. <S> If you do, straighten them out and enjoy the money you will be saving on your electric bill.
| The DC resistance will not change if you coil the wire, but the overall impedance will change.
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Capacitor connected directly with battery This may be a dumb/begginer question, but I'm having trouble to understand what exactly happens when we connect a real capacitor directly with a battery. In my understanding, theoretically, when an uncharged capacitor is connected directly to a battery of, let's say, 9 volts, instantly the capacitor will be charged and its voltage will also become 9V. This will happen because there is no resistance between the capacitor and the battery, so the variation of current by time will be infinite. Obviously, this is true when talking about ideal components and non-realistic circuits. I thought that doing it in real life would cause sparks, damaged components, explosions, or whatever. However, I saw some videos and people usually do connect batteries directly with capacitors. Also, the current that flows from the battery to the capacitor is somehow of low magnitude, since it takes some considerable time to make the capacitor have the same voltage as the battery. I would like to know why this happens, thanks. This is an example of the circuit I talked about: <Q> Both the battery and the capacitor have an internal resistance. <S> Your capacitor looks a bit like this on the inside: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Of course, I don't know your capacitor, so I don't know the exact internal resistance, but 3Ohm will be a close enough approximation. <S> The same happens in your battery, so in fact you are doing this: simulate this circuit <S> So now for a tiny amount of time the current will be maximum, but it is only about 0.9A <S> Of course when you put a capacitor onto a battery like that, you will not make great contact, so there will be some extra resistance there as well, so it might even be 0.7A. <S> The reason it now takes time, is that when the capacitor charges, the voltage across the resistors decreases, so the current decreases as well, so the voltage on the capacitor will increase more slowly, and so on and so on, so it will actually approach the battery voltage slower and slower. <S> The larger the resistors or the capacitors the more time it will take. <S> The moment it is at 67% can be calculated by R * C. <S> So in the example that is: t(67%) = <S> R <S> * C = 10 <S> * 220u = 2.2ms. <S> But if the capacitor is 22000uF (= 22mF) <S> then the RC-time, as it is called, will be 220ms, or 0.22s for it to charge with a total resistance of 10Ohm. <S> But with a capacitor that size it might also have a slightly higher resistance, so that'll make it even slower. <S> And then it's only at 67%. <S> The next 30% will take much more time. <S> EDIT: <S> Note; increased the 9V-bat resistance as per Nick's comment. <A> Real batteries and capacitors have an internal resistance which will act to reduce the current charging the capacitor. <S> This will prevent the death and destruction you were expecting. <S> :-) <S> In any case, it is hard to see a spark produced with 9 volts... <A> In addition to Asmyldofs helpful answer, it is worth noting that even if all the conductors were superconducting with zero resistance, the initial current would not be infinite and the current would decay to zero . <S> Why not infinite current? <S> As there is a loop of current the circuit will have some inductance. <S> So the current will initially rise at a rate of Vbatt/L. <S> The voltage across the capacitor will shoot past Vbatt to nearly twice that value and then reverse, giving a damped sinusoid centered at Vbatt. <S> Why damped? <S> We are generating a time-varying magnetic field. <S> That is how to make an electromagnetic (radio) wave. <S> The power in the radiated field will cause the oscillation in current to die out. <A> As you say, only in "theory," can we obtain "ideal" results. <S> Using realistic power sources and capacitors, one obtains non-ideal results. <S> This is because real components have "additional" resistance, inductance, and capacitance. <S> Although one can never obtain ideal results, by keeping the "additional" components as small as possible, we can obtain results "close" to ideal. <S> In your particular case, the reason there were no "dramatic effects," is that the battery and the capacitor have internal resistance. <S> Therefore, the capacitor will not instantly charge up to the battery voltage. <S> It will "slowly" charge up at the "normal" rate specified by the product of Rint and the capacitance C.
| To sum up, the reason why capacitors take time to charge up is - internal resistance .
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Thevenin equivalent in a nonlinear circuit Is it possible to use the Thevenin equivalent of a voltage divider driving an emitter follower? simulate this circuit – Schematic created using CircuitLab In general, when is it safe to use the Thevenin equivalent of a source if it is connected to nonlinear components? Common exercises always imply a resistor as the load. <Q> Yes, it is possible to use the Thevenin equivalent of a voltage divider driving an emitter follower. <S> The sub circuit you are replacing by a Thevenin Equivalent is a linear circuit (just the part consisting of 15V constant voltage source and the voltage divider; you are not replacing any of the non-linear part). <S> Therfore using the Thevenin Equivalent for the subcircuit is not only an approximation, it is completely equivalent and perfectly ok. <S> It doesn't matter that that subcircuit is connected to a non-linear subcircuit. <S> If you would put both implentations of the linear subcircuit each in a black box, there'd be no way to distingiush them from outside; not even by connecting an external non-linear circuit like your emitter follower. <S> That's why it is called Thevenin Equivalent and not Thevenin Approximation. <S> What you can not do (in general) is to replace the non-linear subcircuit (right part with transistor) by a Thevenin Equivalent. <A> To answer this question you need to consider two things: will the transistor have the same biasing ? <S> will the small-signal transfer change ? <S> Let's make things easy on ourselves and assume the beta of the transistor is very high so we can ignore the base current. <S> Then the base voltage would be the same in both circuits. <S> The rest of the circuit (V2,Q1, R2) is identical <S> so YES, the transistor will have the same biasing. <S> This means that the small signal parameters of the transistors will also be the same ! <S> But will the overall small signal equivalent change ? <S> We only need to consider how the signal comes from V1 into the base of Q1 compared to V3 into the base of Q2. <S> Now I do see a difference ! <S> Do you see it ? <S> R1 // <S> R3 is indeed 333.3 ohms (= R4). <S> BUT while the signal from V1 is attenuated by R1 and R3, the signal from V3 is NOT ! <S> So the circuit with Q2 will have a higher voltage gain ! <S> Even though from a DC point of view, the transistor cannot tell the difference if it was in either circuit. <S> This is assuming the input signal has the same amplitude on V1 and V3. <S> But V3 is 10 V DC instead of 15 V DC. <S> If you would also lower the signal amplitude of V3 with the same ratio then the signal at the output would be the same ! <S> Then the circuits would indeed be equivalent !!! <S> Do you also see that it is not so much the non-linear component that makes the difference ? <S> I mean, the transistor is biased in the same way <S> so it behaves identical in both circuits. <S> But the transfer to the Thevenin equivalent did change the way the signal is attenuated through the circuit. <S> I do not think you can state that in general it is safe to use the Thevenin equivalent. <S> You have to look at it case by case. <S> And use common sense ! <A> Generally - no. <S> It's because the Thevenin's theorem is defined for linear circuits. <S> For other (nonlinear) circuits one can't say about Thevenin equivalent. <S> However, you want something that is similar in its concept but for non-linear circuits . <S> Normally, there are two ways to approach this: you solve circuit equations and check if for some values it is linear or similar to linear, and then you try to approximate, you transform the set of equations of circuits (eg. <S> with Laplace's theorem ) and find the transfer function for this circuit. <S> It's generally solving algebra equations.
| There isn't a definition of a Thevenin Equivalent for a non-linear circuit.
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Best solution for compact wiring on stripboard When transferring a circuit from breadboard to stripboard, I often struggle to manage the wiring; often it starts looking like an overgrown garden; the components inevitably start disappearing under the wires. Not to mention the wires disappearing under each other, making mistakes near impossible to trace. But I want to keep the board as small as possible. What kind of wires should be (or could be, or are most often) used for low voltage, low current circuits? Up til now I've used the same wire as with the breadboard, (single strand) as I find it easy to work with since it retains its shape and I can route it round components. However, it's quite thick so doesn't lend itself to compact boards. See pic below: I searched on Google and decided that this looked like the neatest (or at least most manageable) stripboard circuit out of the results: What kind of wiring is that? I can't find any cable like it for sale. (At least not single core, but I don't fancy spending hours taking the outer sheath off ethernet cables or the like.) (I assume they're multi-strand.) The closest I can find is 30awg hook-up wire, but the outer diameter of that doesn't seem much less than what I use for breadboards. Advice would be great. Thanks <Q> Wiring by Markus Gritsch <S> Fine (34 AWG) <S> solderable enamel wire <S> (aka solderable magnet wire), isn't enamelled but is coated with polyurethane, etc. <S> No stripping required, soldering turns the polyurethane into a flux and exposes the copper. <A> It's easier if you use sockets: http://www.jameco.com/1/3/dip-wire-wrap-ic-sockets although it adds to the cost. <S> It can be less time-consuming than soldering thousands of joints by hand, too. <S> http://makezine.com/2009/07/27/lost-knowledge-wire-wrapping/ <A> Back in the day people used some sort of 30awg wire for this job. <S> But instead of a proper insulation they had something like a thin layer of colour. <S> The problem was that after some time using the pcb this insulation would break and you end up with errors again. <S> This is maybe why no one uses it these days. <S> The closest you can get might be transformer or coil wiring. <S> https://en.wikipedia.org/wiki/Magnet_wire <S> Or you try something like this <S> http://www.adafruit.com/product/1446 <S> They even speak about the old day methods there. <S> Or you wait until the next 5 people suggest to make a pcb. <S> ;) <A> I think PCB making is not a expensive thing. <S> for your circuit it will not increase a meaningful cost.you can design a PCB using software FREE. <S> and you also Eatch your PCB using IRON and FeCl3
| Wire wrapping with a proper tool is actually a great technique for building one-off circuits with lots of DIL ICs on.
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Long 24V DC cable not working We have a really weird issue which we just cannot get our head around. In high level: We have a unit which contains some electronics, we can access it with 4 cables 2 for power and 2 for communications. The power is 24V DC and communications is RS485.We could not establish communication using a 10m test lead, it has been noticed that when we moved the cable the device functioned sometimes. The conclusion we drew was that the cable was broken somewhere. The cable was cut shorter to around 1m long and worked like a charm. Fixing: We tried to use two other long cables as we still needed to be sure that the system works with the required cable length. It did not work. This is when we started to do some more testing and try to rule out the possible faults: Correct length power and comms cable works fine with test apparatus Short power and comms cable works fine for our actual apparatus Short power and long com works fine for our actual apparatus Long power short comms does not work with out actual apparatus Long power and long comms does not work with our actual apparatus We have concluded that the issue is with the power cable. The only variable that seems to have an effect is length of the power cable. We have experimented with three different long cables of different qualities, none of which have any effect. We have one of the electronic components(comms and power as well) as a spare unit which we have tested with the same long cable and it does work. In the unit it self which we are trying to fix we have a power converter (in 18-75V out 24V) which we have no spares for. We have called the manufacturer and asked if they have encountered anything similar but they said they have not. We have hooked up a multimeter to check the current draw, at short cable we get 0.6A while at long cable we pretty much have nothing. There is a really small voltage drop at the end of the cable. The power supply we use is 24-28V 15A we have set the power supply to around 25.6V at the end of the cable we had 25.4V. Currently we cannot access to any electronic as they are in an enclosure, we could open it up but we would like to avoid that if we can solve the issue without it. In short , how could the length of the power cable effect the current draw from the system, why it would not draw anything when the cable hits a certain length? Thank you in advance for any ideas you may share to help solve this problem. <Q> Essentially the IR losses in the supply negative connection (actually the cable screen) were sufficient that the control signals from the head were outside the RS485 common mode range of +12,-7V. Going to an isolated RS485 port at one end <S> fixed it. <S> One other gotcha with DC/DC bricks like that buck <S> /boost you have is that they sometimes misbehave if the supply is insufficiently stiff on startup, try adding a fairly large electrolytic across the DC/DCs input to stiffen the supply. <A> No we do not use any data termination resistors <S> Use data termination resistors. <S> Don't think about it, just do it and come back and let me know if this solved the data issues. <S> Put this right first. <S> Try 120 ohms across the ends of the data cable. <A> I had a similar problem once - a problem that appeared with long cables only (and symptoms changed depending on which long cable was used...) <S> It turned out that the power supply would oscillate (at frequencies in the 10's of MHz) with the long cable which apparently was acting as a large inductive load. <S> These oscillations were picked up by coax cables that ran in the same cable bundle - so we were seeing "noise" in long cables but not always on the same channel! <S> It was initially resolved with an extra 100 nF ceramic across the output of the power supply to get the customer up and running; subsequently the PSU was replaced... <S> Decoupling was the key.
| One issue I have had in a similar situation (An underwater electronics pod on a few tens of meters of cable with power and RS485 control wiring), the fault was due to the voltage drop in the ground connection....
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Current flow through resistor and insufficient source In a simple circuit consists of a source and resistor, according to Ohm's law, if 10V is applied to a 10 ohm resistor, the current flow through the circuit will be 1A. But what if the source in this circuit has a max output of 0.5A? What happens in that case? <Q> If the source acts like a bench power supply and actively limits the current to 0.5A you will have 5V across the resistor. <S> If the source acts like an ideal voltage source with a 20\$\Omega\$ resistor in series (10V open circuit, 0.5A into a short), the voltage across the external 10 ohm resistor will be less and thus the current will be (1/3) less than the above case. <S> There are other possibilities- <S> batteries would behave somewhat differently. <S> If the limit is because it's an auto retry switching supply with current limiting the current will be short pulses and probably average fairly low. <A> Then you won't get 10V across the resistor, you would get 0.5 * 10 = 5V across this resistor, by V=IR. <S> The resistance is fixed but you have varied the current so the across voltage changes. <S> The other 5V will be across the current limiting element. <A> Then you need to add this output resistance model to your circuit. <S> 0.5A current limit indicates that when the terminals of the voltage source is shorted, the current through output resistance of the voltage source is 0.5A. <S> This shows us that the value of output resistance of 10V source is 20Ω. <S> So, you should modify your circuit model including voltage source output resistance and analyze it as usual in case any current limit is of interest. <S> The answer to your question is that the current through 10Ω resistor is 0.33A. simulate this circuit – Schematic created using CircuitLab
| If the 0.5A limit is because of a fuse the current will be zero. A current limit for a voltage source implies that the source is not ideal and have an output resistance.
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Can I use a voltage divider and a multiplier to power a 12V component from a 5V supply? My idea was to use a the two points in a voltage divider circuit (5V and let's say 2.4V), feed those through a multiplier to power a 12V component. Is this possible? Any problems that may be encountered? EDIT: I am limited to a 5V supply. <Q> Any circuit that will perform any actual operations on a voltage must have a supply that is at least as great as the smallest result. <S> So in order to mulitply 5V and 2.4V you'd need a 12V supply to begin with. <S> Use a boost regulator instead. <A> From your reasoning it seems that you see electronics in a very mathematical way: a division operation followed by a multiplication. <S> Although there are parts of electronics that deals with such things (signal processing, for example), here we are talking about power supplies and the things are not so "simple". <S> A voltage divider is extremely simple (just two resistors), but is very power inefficient. <S> I'll add that you will be better off, probably, by using an already built DC-DC <S> boost converter module, instead of designing one yourself. <S> They are cheap, reliable and very efficient. <S> Nowadays they are also very tiny, so they don't take much board space. <S> For example one like <S> this or others like the ones in this list from the same manufacturer . <A> Seems like I misunderstood the question when I posted the original comment. <S> No you can't multiply the voltage between two different points on a circuit to give you a bigger voltage. <S> However, if your project already has a micro-controller capable of driving a PWM signal (which I'm guessing is a possibility since you're working with 5V), what you can do is to create a Dickson multiplier that will multiply your 5V supply to a 15V supply (minus diode losses). <S> You take that output and put it through a 12V linear voltage regulator to give you 12V output. <S> Here's a schematic of a very basic one that could work. <S> You could improve efficiency and the current delivering capabilities by using Schottky diodes and a LDO regulator. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> For an idea on how this circuit works, see Dave Jone's excellent video explaining it. <S> If you don't have a way of generating a PWM signal or if the voltage multiplier isn't able to source the amount of current you need in your 12V peripheral, you could of course, use a boost regulator as suggested in the other answers.
| This can be a simple way of boosting your 5V to 12V if you already have a means of getting PWM output. On the other hand voltage "multiplication" is not easy at all when the voltage you want to multiply is DC (for AC you just use a transformer). Anyway, as also @Ignacio said, you could use a boost regulator chip.
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PCB design few questions Quick couple of questions, I am designing my first PCB and want to have a few boards prototyped with OSH Park. For a board I will be manually soldering components to, will I need to include a tCream layer or is this for automated soldering only? And, do hobbyists generally make the surface mount pads a little larger than the components data sheet specifies, to help with soldering? Are there any other tips/tricks a first timer might like to know to make life easier? And, for milled slots, i.e. two holes for a USB receptacle to mount to the board, do I need to include a note for these or is the data on the 'holes' layer sufficient? Thanks in advance <Q> The tCream layer contains the top side solder paste data for SMD, normally used to make stencils for printing the paste to the board before assembly. <S> So I do not think that it will be necessary for a few prototype panels. <S> We was told back in class to add a couple tenth of a millimeter in both direction to the pads (red area below). <S> At thorugh-hole components, hole galvanization should be considered as follows: <S> drill_size= d + <S> 0.3 mm where d is pin diameter. <S> The copper ring diameter around the hole should be drill_size + <S> 0.5 mm . <S> These were advices from our teachers when I was desiging my first PCB. <A> OSH Park have some guidelines which answer your 'cream layer' question. <S> The answer is no, you only include layers that they will use for manufacture. <S> I have always found Laen at OSH Park is extremely helpful and supportive, so <S> I recommend you email if you have any concerns. <S> As for cutouts, it depends a little on how you are making the slot. <S> I have used overlapping drill holes (not allowed at many PCB manufacturers, but OSH Park did it). <S> Otherwise I have put slots on an outline layer. <S> OSH Park have extensive help under Support , for example creating slots is explained here . <S> For any devices which have pads under the package, I do extend them beyond the package boundary. <S> Otherwise it is difficult to solder. <S> In general I don't make pads larger, but I often use Sparkfun libraries (at least for packages and footprints) which are pretty good. <A> Make the paste layer. <S> Sometimes it's good to order the mask, put paste on the board and solder components with hot air. <S> Also it's good for QFN. <S> For manual soldering you don't need wider pads, you rather have to make wider spacing between the components, so the soldering iron will not touch what it shouldn't. <A> tCream layer or is this for automated soldering only? <S> This is the Solder paste layer, which tells you where to apply solder paste. <S> It can (and often is) used for soldering manually too. <S> The paste/cream layer can be used to get a solder paste stencil from OSHStencil (I believe they are related to OSHPark), then use a squeegee to apply paste to the board, then reflow or use a hot air gun to solder. <S> make the surface mount pads a little larger... <S> In addition to Bence's answer above, make through hole pads larger as well. <S> This helps greatly in the event some component fails and you need to remove/resolder the component. <S> Having larger pads reduced the possibility of delamination. <S> milled slots <S> This is highly manufacturer specific. <S> Some manufacturers do not support slots at all, some require it to be in a specific format. <S> Talk to them. <S> Are there any other tips/tricks a first timer might like to know to make life easier? <S> Once all your gerbers are ready, print them all out at 1:1 using your desktop printer. <S> Lay out your critical components (SMD ICs, connectors) and make sure the pins all line up. <S> This is an easy way to catch the most common footprint size issues. <S> Use a vernier to actually measure the pin dia for all your through hole components, and check the drill holes dia in the drill file.
| As for pads, it is recommended to use larger pads when hand-soldering because we need more space for the solder-iron.
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Concerns in H-bridge switching times with common (one) gate control line I have a FET based H-bridge being controlled by a micro. I was worried about software errors being able to turn on the top and bottom FETs at the same time, so I opted for the control scheme below for each half of the bridge. While this works fine, difference in the switching times of the two FETs means a partial short occurs - I see the voltage on Vcc is pulled down (even though it has quite a lot of capacitance) during switching. The duration of the 'short' is very brief (looks like <10us on my scope). Is this inevitable with this control scheme, or can I get some improvement? I tried making R1 and R2 bigger in simulation and then strapping diodes across them (pretty sure I had seen this done before somewhere), so that the FETs turn off faster than they turn on. This seems susceptible to variations in threshold voltage, temperature changes etc. Is there any better solution? I'm aware that there are ICs dedicated to this function, but I'd like to try and keep this simple. If the only way to do it is with more transistors, then so be it, but I wanted to make sure there isn't some simple solution involving R/Cs, diodes etc that I haven't thought of. <Q> Add a NPN/PNP emitter follower stage before the MOSFETs (just two cheap transistors) <S> then do the resistor/diode to the gate thing. <S> Resistors R2/R3 will probably be in the hundreds of ohms to get the shoot-through to a negligible value. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Make sure to pole M1 the other way around (D to VCC and S to load)! <S> In your circuit the body diode of the p-channel MOSFET is forward biased all the time M2 is turned on. <A> Having started with a high speed H-bridge design as my first proper project, it really sounds like not having to use an IC driver is simpler; it's not. <S> At all. <S> If you want a simple, quick way to make an H-bridge, without having to worry about signal inversion and dead-time, use a driver IC . <S> IR2183 works just fine and covers everything up to the beefy 7000pF <S> drive gates at like 100kHz. <S> (Then it breaks down a bit, drive speed wise). <S> If you really don't want to, you must use another transistor set to drive each gate of your H-Bridge transistor. <S> This is a lot harder, but a lot cooler.
| You could improve the circuit by reducing the switching time and then adding the diodes.
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RAM and stack memory I'd like to know what is the difference between RAM ( data memory ) and the stack memory in PIC micro controllers if both of them is volatile( not sure about stack ) and the stack is temporary memory? <Q> All RAM memory requires a constant current to hold its state. <S> That is, as soon as you turn the power off, all of your RAM memory vanishes. <S> Thus, RAM is volatile. <S> Each function call "pushes" data to the stack. <S> The data is "popped" when the function returns. <S> Because the stack is stored in RAM, it too is volatile. <A> In 8-bit-data PIC MCUs, "the" stack is a small area of RAM dedicated to the storage of return addresses only. <S> You cannot access it other than implicitly through CALL and RET instructions. <A> When a subroutine is called, the processor pushes the return address on the stack, and may make space for local variables (this can very with processor and compiler). <S> A processor register called the stack pointer contains the address of the next free location on the stack, and is decremented when something is pushed onto the stack, and incremented when something is removed from the stack. <S> The stack is usually placed at the top of RAM, and extends downward as items are pushed on it. <S> The stack is usually part of the volatile RAM memory space, but actual implementation may vary between microcontroller families.
| The stack is a specific area of RAM memory used to store temporary variables during program execution.
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How to power off a dual supply op amp? What is the best way to power off a dual supply op amp (LM741)? Should I cut the positive, negative, or both? Thanks! I’m building an audio oscillator using four op amps, and I’d like to be able to turn the oscillator on/off by cutting power to the entire circuit. The four op amps are configured in a feedback loop, so I believe that when I cut power there is no input signal to any of the op amps. <Q> Cutting one would be open circuiting <S> the internal circuit therefore powering off the op-amp. <S> As for the "best way": the rest of the circuit is of importance here, since probably powering the op-amp off change other stuff. <A> You DO need to disconnect both batteries when powering the circuit down. <S> There are a couple of options to consider. <S> 1) Modify the circuit to run from a single-ended power supply. <S> This is easy to do in your circuit because the only nodes that connect to ground are the 4- (+ <S> ) inputs on the op-amps and the load. <S> The changes needed are simple: add a voltage divider reference from V+ to V- with the mid-point going to the 4- <S> (+) inputs on the op-amps. <S> Add a bypass capacitor across the lower resistor for stability. <S> You will also need to add an output coupling capacitor. <S> The (+) side of the cap goes to the op-amp output pin, the (-) side goes to the load. <S> I normally also add a bleeder resistor from the (-) side of the capacitor to ground (V-) so that you don't get a thump when connecting the load. <S> 2) Use a DPST or DPDT switch to disconnect both batteries. <S> 3) <S> There is a simple trick that can work well under certain circumstances: connect the batteries in series with blocking diodes. <S> You then use a SPST switch to short the diodes. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The downside of this approach is that the ground terminal will bounce around by one diode drop as current consumption changes from (+) to (-). <S> Depending on the PSRR rating of the op-amps, this can lead to distortion at the output. <S> However, I have successfully used this technique in the past. <A> As an alternative: There also are op amps with a built in shutdown function, have you considered using those? <S> This example <S> consumes <1uA when you pull the dedicated pin low. <S> Perhaps you can find one that suits your needs.
| What you need is to is to either make the voltage difference between two nodes zero or make open circuiting so that they can not reference each other in order to actually power off.
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Why do RC applications use such a small PWM duty cycle? I know that RC applications, such as a drone, use PWM signals to drive the motors. This PWM signal is mostly 50 Hz (0.02 s). The pulse itself varies from 1 ms to 2 ms. So a 1 ms pulse corresponds with a minimum motor rotation and a 2 ms pulse with a maximum motor rotation. So basically the other 18 ms of the period the signal stays idle. Why does the PWM signal have such a format? Why is the active part of the signal not spread over 1 ms and 10 ms? What is the advantage of using such small pulses? <Q> The reason for the long gap is so that the transmitter can send all of the other servo positions. <S> In the days of clothes-pegs and crashed aircraft from frequency collisions, radio control was done with AM at 27 MHz. <S> The transmitter would send a sync pulse, and then a series of 1-2 ms pulses, one for each servo. <S> The earlier ones delayed the later ones, didn't matter much. <S> These are just RF pulses, no special modulation. <S> The receiver would receive the pulse stream, synchronise on the first one and then direct each successive pulse in turn to a different servo socket. <S> So to allow for maybe 8 channels set to 2 ms, and have some gaps, you need about 20 ms. <S> With an 8 channel transmitter, the duty cycle on the combined RF channel would have been over 50%. <S> This servo protocol, of 1-2 ms every 20 ms, has just stuck around from then. <S> This site about making a PC digitiser for your remote control has some oscilloscope graphs showing four or five channels. <A> The point there isn't really Duty Cycle. <S> The 1ms to 2ms pulse is one that is easy enough to "decode" both in analogue and digital circuitry, so it is adopted as a standard. <S> You need standards to be able to mix and match stuff, and in RC systems there's many different applications and sub-devices, so the standard is strictly adhered to, to keep the market alive for all hobbyists alike. <S> No translation requirement = <S> more sales, because easier. <S> Hobbyists like easier. <S> But many devices that need higher response rates perfectly support a pulse repetition of 1s to 5ms, allowing update rates of once per second to 200 times per second. <S> Some normal response types don't even "fail to default" with many seconds between pulses, but the most used standard says "be at least compatible with 50Hz update rate" and most seem to interpret that as "be 50Hz". <S> But it's technically not a hard requirement. <S> I have certainly had 200Hz polled systems pulled from high-end flying equipment, but I have also seen sensory-systems in olden days that only sent a pulse 10 times per second. <S> (Probably since the analogue needles weren't quick enough to fall back fast, even if they got 5 pulses of info per second) <A> An typical RC signal contains one pulse for each servo being controlled; a typical six-channel receiver (at least historically) wouldn't capture the signal from any of the input channels, but would instead include a counter circuit that would reset itself after a sufficiently-long gap and would advance a little bit after the falling edge of each pulse; each servo output signal would be high only when the input was high and the count held the right value for that servo. <S> If a servo wishes to be usable in a system with eight or more channels, it must be able to accept a signal with a very low duty cycle.
| Having encoders respond to pulse lengths in the 1-2ms range independent of how often they receive pulses makes it possible to have servos that can accept a large number of servo readings at a relatively low update rate, or that transmits a smaller number of servo readings at a faster update rate, without needing any sort of "configuration".
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Why are vias bad? I am designing a PCB with EAGLE and saw that it was trying to limit the amount of vias through the PCB. Why do you want less vias? Why are they bad? Do they bring extra manufacturing cost or is it OK for low-frequencyand low-power solutions? <Q> On the first picture a TH vias allow us only four pads to be placed. <S> But with a blind via or without a via we have place for six <S> (or more if we have more rows) pads. <S> A larger BGA component could be placed here this way. <S> source <S> And at the end reduced size means reduced cost. <S> For example at high power dissipation componenets <S> thermal vias could be used to help dissipate the heat by leading it to large copper-pours. <S> All in all it is very application-specific and could have both advantages and disadvantages as well. <S> It is up to you to find the balance. <A> I wouldn't say that vias are bad. <S> They are not! <S> One useful way to use vias is to shield RF energy in a RF board, a technique called via stiching: <A> It is just one of the parameters you can use to tweak the autorouter. <S> Via's add a little cost in drilling (even though this might not be explicitly shown on the bill), they take up space, and other things being equal it is better for a route to stay on the same layer. <S> I can imagine (but I am not sure) that a via is just a little bit less reliable than a simple copper trace. <A> For High Speed Buses, vias will lead to impedance mismatch and cause reflections. <S> Vias also cannot tolerate high current. <S> Multiple vias are needed for high current planes. <S> This is obviously going to increase the spacing.
| I think the main problem is: vias could occupy significant space from other components, thus a larger board is necessary. But to defend the vias a little: There are cases when they are useful.
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What is the total resistance of a LED array? First time poster and a hobby EE enthusiast here. I want to turn on a LED array with a transistor switch and I'm having trouble with determining what will the total resistance on my transistor collector be. And without that I can't determine what should my base resistor be. This is part of my schema, the switch will be a pin on an Arduino that can supply 5V and max 40 mA: I got the LED array scheme from an online LED array calculator and it gave me the info that the array draws 60 mA current - from which I figured I could get the total resistance of the array with Ohms Law: 12V/0.06A = 200 Ohm. If my collector resistor is 200 Ohm that would mean that my base resistor should be 2k7 Ohm. Is that correct? Can a treat the array as 200 Ohm resistor on the collector? Or is the total resistance the parallel of 3 resistors - 110 Ohm? Any advice or a hint in the correct way is appreciated. Thanks,Tadija Edit: LED Red 2.0V, 20mA <Q> Since LEDs drop voltage when there's current through them, and since Ohm's law states that: $$ R = \frac{E}{I} <S> $$ <S> They can certainly be considered to be resistances of a particular value with a known current through them and a known voltage across them. <S> For instance, one of your your LEDs, with 20mA through it and 2 volts across it, will look like: $$ R = \frac{E}{I} = \frac{2V}{0.02A} = <S> 100 <S> \text{ ohms}$$ <S> Usually, though, the resistance of the LED is ignored because it's not needed to calculate the values of the ballast resistors or the transistor's base resistor. <S> The value of the ballast resistor is determined by: <S> $$ Rs = <S> \frac{Vs - <S> (n <S> \ <S> \ <S> Vf) <S> + Vce(sat)}{If}\text{ ohms} $$ where n is the number of LEDs <S> , Vf is the forward voltage of one LED, Vce(sat) <S> is the transistor's collector to emitter saturation voltage,\$If\$ <S> is the LED forward current, and Vs is the supply voltage. <S> In your case that works out to: $$ Rs = \frac{12V - ( <S> 3 \times 2V) + 0.5V}{0.02A}\text{ 275 ohms} <S> $$ <S> The 330's you have in there will work, no problem, with the LEDs losing a little brightness. <S> Since there will be three series strings in parallel, the current into the transistor's collector will be 60 milliamperes. <S> Switching transistors doing this kind of work are usually run at a forced beta of ten, which means that for 60 mA into the collector <S> 6 mA is forced into the base. <S> The base-to-emitter junction of a transistor is basically a diode, so in this case it'll drop about 0.7 volts with 6 mA through it. <S> That means that with the Arduino supplying 5V to drive the base, about 4.3 volts of that must be dropped across a resistor with 6 mA through it so, from Ohm's law, R = E <S> /I = <S> 4.3V/6mA <S> = 717 ohms. <S> 750 ohms is a standard E24 value and will work well. <A> No, you cannot treat the LEDs + resistor as a 200 ohms resistor. <S> would use 12 <S> V x 60 mA = 0.72 <S> Watt <S> But there is no need to treat it as a resistor ! <S> It works like this:LED current = 20 mA per LEDso <S> total LED current = 3 x 20 <S> mA = 60 mAThis <S> 60 mA will also flow through the Collector of the transistor. <S> The transistor has a current amplification (from Base to Collector)of 100 - 500 times, let's assume it's 100 times, then the Base current willneed to be: <S> 60 mA / 100 = 0.6 <S> mA <S> Now it depends on what voltage you will apply at the switch SW1if <S> that is 5 V <S> then you need to subtract the Vbe voltage of Q1, <S> this isroughly 0.7 V <S> so 5 V - 0.7 <S> V = 4.3 V, this will be the voltage across R4.Now 4.3 V / 0.6 <S> mA = 7.17 kohms ! <S> A larger base current is OK as long as it does not get too large, like 10 mA <S> or so. <S> Also in practice the current gain is usually a lot more than 100.So in practice <S> I just use R4 = <S> 10 kohms and this circuit will work. <S> Only for very high values of R4, like 500 kohms you might start to notice that the LEDs will become less bright. <A> The transistor will behave the same as if you had a 200\$\Omega\$ resistor when it is on, so the calculations below apply for your LED array or for a simple resistor. <S> A forced beta (Ic/Ib) of 20 or so is reasonable for a transistor like that. <S> The BC547 is only good for absolute maximum 100mA <S> Ic @ <S> 25 <S> °C (and you're running it at 60mA) <S> so you should drive the base reasonably hard for the best reliability. <S> Personally, I'd use something like the 2N4401 that is good for 500mA+ Ic, but the BC547 should work okay. <S> The hFE is already tailing off above 50mA even at Tj = 25°C. <S> So if you have Ic <S> /Ib = 20 <S> , then you want 3mA base current. <S> Rb = <S> (5V - 0.7V)/3mA = <S> 1.4K. <S> Since the output will be a bit less than 5V and the supply voltage might be a bit low, I'd use 1K for best reliability. <S> If you use too large a base resistor (especially at low ambient temperature) <S> the transistor will have higher than a few hundred mV across it and will run warm to hot. <S> Usually hFE will increase with temperature and the Vce will drop saving the transistor <S> but it is best to keep closer to the datasheet <S> specified Ic/Ib = 10 for Vce(sat) than hFE. <S> Keep in mind that Vce(sat) is specified at Ic/Ib = 10 and hFE <S> (min 110 at 25°C) is specified at Vce = 5V, which would make the transistor very, very hot. <S> (5mW/K) <A> See the data sheet of this led and depending on the Q point you can find the gradient of I/V graph at that Q point. <S> You can then simply find the equivalent resistance by applying series and parallel equations for resistors <S> Hope this Helps !
| The only valid thing you could say is that the LEDs + resistors draw60 mA from 12 V and that a 200 ohm resistor would also draw 60 mA from 12 V.Both
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Appropriate Method of Counting counts from a Quadrature Encoder I'm using an Mbed to drive a DC motor which will follow a trapezoidal trajectory.What is the most appropriate way to determine position from the quadrature encoder? Should I use interrupts, or is there a risk of the trajectory routine (which calculates theoretical position) falling behind? Alternatively, I could poll/sample the encoder in fixed time steps, but I'd imagine this would miss out many encoder counts. How is this problem usually dealt with? <Q> It's best if you have quadrature encoder hardware to handle the encoder pulses, unless the speed is very low. <S> Many microcontrollers have at least one such peripheral on the chip. <A> You have to poll the state of the two Q input often enough not to miss a state transition. <S> How quickly this is depends on how quickly the Q inputs can change, and only you can tell that. <S> How do you check often enough? <S> as Sphero stated, using Q-decoder hardware is a perfect way, and it will liklely be fast enough for any purpose. <S> you can use a timed or on-change interrupt to tickle your FSM. <S> How fast you can do this depends on your system, and how much of its resources you want to spend on this task. <S> for in-frequently changing Q-inputs you could just poll the inputs. <A> A very great deal is going to depend on your motor/encoder setup, which you have not specified. <S> Let's say (just as a starting point) <S> the your motor shaft can run at a maximum of 600 rpm (10 rps) and your encoder produces 64 pulses per channel. <S> That is, if you run the shaft at 10 rps, you get a 640 Hz square wave on each channel. <S> When using a quadrature encoder, you can derive 4 times as many position events, corresponding to both the rising and falling edges of the two channels, for a shaft resolution of 256 points per revolution, or 2560 points per second. <S> You have not provided a processor board specification, either, but let's say that it allows 4 channels of digital inputs. <S> Then you can look at both your encoder channels, and also a pair of channels which have inverted versions of your encoder lines. <S> Now you can write your software to generate an interrupt on each of the rising edges of the 4 inputs. <S> At each interrupt you examine the two encoder lines and determine the local position. <S> At the same time, you read your real-time clock to determine the time since the last encoder tick, and from that determine the shaft velocity. <S> Can you do that in <S> (let's say) 1/10 of the time between ticks? <S> That's 40 usec. <S> Also, you'll need to find some way to prevent any other interrupts from preempting your encoder routines for more than 400 usec. <S> If you don't, you lose position information, and any process which prevents immediate time interval measurement will screw up your shaft rate calculation. <S> With only a small about of hardware (a couple of exclusive-OR gates and an RC delay) you can produce a single input which will generate interrupts for all four encoder transitions, so you could get away with 3 digital inputs. <S> If the transition frequency is low, as Wouter van Ooijen says, you can try simply polling the encoder outputs, but be aware that this will introduce uncertainties in your shaft velocity calculations, especially if your processor is doing anything else which may have higher priority, and any such uncertainty will degrade your ability to hold to a desired velocity profile, trapezoidal or otherwise. <S> Overall, I'd advise you listen to Spehro.
| The correct way to do the counting is to use a 4-state FSM, which handles bounding and direction changes correctly.
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method of locating schematic part on board layout? In a design with a large number of parts it can be difficult to immediately pin-point a part's location on the board. Is there an easy method of finding a schematic part's location on the board layout? <Q> It's called 'cross-probing'. <S> In Eagle open the schematic and the corresponding brd , on both interfaces press the eye-shaped icon ("Show Objects"). <S> Now by clicking a specific object in one of the windows it will be highlighted in the second. <S> If you are wondering how to locate a part on the schematic, just type show <part_reference> on the command entry field just above the schematic. <S> The same will work on the PCB view. <A> In the SCH, try typing: sh @ part which highlights the part with a box, both in SCH and BRD files simultaneously <A> Use the show tool. <S> This will highlight the corresponding element in both the layout and the schematic. <A> A good thing to do on dense boards, is renumber the components after the layout is complete. <S> This will make the reference designators read left-right, or top-bottom, as you specify. <S> After this step, you back annotate the renumbering to the schematic. <S> It's a little harder if it's not your board because not every designer does this. <S> If there is no rhyme or reason to the reference designators, it's easiest to try to visually break the circuit up into blocks, and go from there. <A> If Eugene's answer isn't the one you're looking for; <S> I.e. <S> you mean a finished PCB of which you have only some PDF because you do not have source files <S> , then the answer is nope. <S> There are many ways to plan components in designs, but there are no guidelines for "you should do X or Y". <S> So some designs will number components from left to right, top to bottom, on the sheet and then they will show up on the board where they make sense, which may seem random to you. <S> Others renumber on final placement. <S> Yet others don't even renumber at all (bad form in my opinion to choose neither). <S> If you understand the schematic, you can make good educated guesses with some experience: The decoupling cap will be close to the chip, the filtering L will often be at a right angle to a filtering capacitor, which is often across the pins that need the filtered voltage. <S> Etc, etc.
| When someone wants to find a part on the board, they can just follow the reference designators until they find it.
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Can a magnetic wire be used to extend/strengthen the signal coverage of a cellphone? When i'm inside of our apartment, getting a cellphone signal is very hard, often times I wouldn't get any bar in my iPhone. My unit is in the ground floor and is surrounded by tall walls. My apartment has a small balcony ( which is still surrounded by tall walls ) where I can at least get a single or two bar signal from my iPhone. This question was inspired by this video ( https://youtu.be/pFx8tuSlCcQ ) where a guy used some copper or magnetic wire to extend the range of it's mobile phone antenna. I'm aware that there are some repeater that are being sold but they're really expensive. I'm just looking for a cheap (below $10) solution that I can do or buy. I'm planning to do a similar solution where I'll be extending some magnetic wire from the inside of my apartment to my balcony. But before I do that, I wanted to confirm if this idea would really work. I also don't know where I will be connecting the end of the magnetic wire as iPhone doesn't have any external antenna port. <Q> This will do absolutely nothing. <S> What you could consider doing is picking up a pair of antennas and a length of coax cable for the correct cell band. <S> I think that would be around 800 MHz. <S> It should be possible to get a couple of cheap 800 MHz antennas and maybe 20 feet of coax with the proper connectors for around $20 or $30. <S> Note that this may not help very much; it's going to be more useful for, say, a basement that has thick concrete walls where reception inside is zilch and reception outside is very good. <S> If reception outside is only mediocre, then you may have to get a repeater or a femtocell. <A> No wire, magnetic or not, is going to help. <S> Magnetic has nothing to do with it. <S> You need either a proper repeater (two antennas and a high gain amplifier), or an external antenna connected directly to your phone with a cable. <S> See <S> this answer for a detailed explanation with some calculations. <S> It's not a $10 problem, unfortunately, it's going to take $100-200 to make a difference. <A> Two ways to do what you're trying to achieve: Video solution: <S> Assuming some basic understanding of electromagnetic waves, the phenomenon being used here is essentially constructive interference: Any metal pin will be reflecting received signals, thus acting like a tiny emitter itself. <S> If you position the pins in such a way that the reflected waves add up in phase (=distances from the pins to your desired receiver position only differ in integer multiples of the wave length - as an exercise to you.) <S> you can definitely expect an improvement in signal reception. <S> Using active components: <S> With the solution above it is clear, the you don't actually gain any signal strength: You simply redistribute it smartly. <S> An alternative is to build a circuit that actually amplifies the received signal - As we are dealing with RF design here, this is not an easy task at all. <S> (Issues with gain, linearity, e.t.c. <S> so unless this is your area of expertise you won't get around buying an off-the-shelf solution....)
| If you know where the local cell tower is, you could consider getting a log-periodic or other directional antenna for the external antenna and mount it so that it points at the cell tower.
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can dropped voltage be greater than forward voltage? What I know is: forward voltage, Vf = The minimum voltage that is required to start current flow across the LED/Diode dropped voltage, Vd = The voltage that is used/lost across the LED/Diode. My question is, initially the equation should be Vf>Vd but can Vd>Vf be true at any moment? Besides, please correct me if I have any misconception. <Q> In general, for semiconductors diodes which are forward biased, $$I_f = I_0 e^{kV_f}$$where <S> $I_0 <S> $ is a scaling factor, and k depends on a bunch of things like material and temperature. <S> For "regular" diodes (silicon signal diodes) <S> Vf is usually about .6 to .7 volts at a handy current level like 1 mA. <S> For LEDs, Vf depends on things like the LED material, and is usually specified at a nominal operating current like 10 or 20 mA. <S> It is usually in the range of 1.5 volts (early red LEDs) to 3 to 3.6 volts (for 1 to 10 mA). <S> However, for greater currents the voltage drop does get greater. <S> So, your terminology is non-standard. <S> Vf, in your terms, is not actually a threshold voltage, but rather a convenient way to characterize normal operation. <S> The actual voltage during actual operation (Vd in your terms) can be either greater or less than the reference <S> Vf, depending on what current level is used. <A> What you're referring to is the piecewise linear model . <S> I think the Vf you refer to is the threshold voltage, or <S> Vt, while the Vd is the voltage across the diode. <S> Vd can surely be greater than Vf, unless you're using the ideal diode model (no series resistance) in which case Vd= <S> Vf for any possible forward current. <A> Sure, whenever the diode is conducting. <S> With greater \$V_d\$ <S> the current across the diode increases exponentially, and if it exceeds the maximum the diode can handle, the diode is destroyed. <S> When the supply cannot sustain the current (e.g. because it is regulated somehow), the system will reach an equilibrium where a slight increase in \$V_d\$ would reduce the resistance of the diode, which in turn reduces \$V_d\$ because the ratio \$\frac{R_{diode}}{R_{supply}}\$ drops.
| Actually, Vf is rather arbitrary, and can be as low as you like - as long as you're talking about arbitrarily low currents.
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Is it possible to create a double layer PCB by gluing together two single layer PCBs? I'm really in a hurry with a university project. I need to finish the first prototype of an electronics board by the end of the week. It's a through hole board, no SMD. Having just single sided copper boards and press-n-peel paper, I first tried to go by the single layer way. I can't get to the point of a decent routing. Even using jumper wires seems unpractical. I ended having more than 20 jumper wires connecting different parts of the board. Then I tried the double layer routing. Everything fell into place in a matter of minutes. Everything looks clean and manageable. No jumper wires, no hard-to-fabricate vias in the middle of connectors, etc... Could it be possible to develop the two sides of the two layers board onto different copper boards and the align and glue everything together? It looks promising to me. The hardest part would be the alignment, but to me it looks quite easier that aligning the press-n-peel sheets. I could make alignment holes and then do the rest of the milling. Has any of you tried this method? What are the pros and the cons? Thanks and sorry for my bad English <Q> Sometimes you need a lot of wirebridges, but you can do that pretty neat, as in the pcb I made some years ago: <S> If you really want to go for a double sided PCB, what you say seems possible to me. <S> Onther option is to drill the mounting holes on both pcbs and use some nuts and bolts to hold them together while you solder. <S> Afterwards you can drill all the holes for the component leads, so to match exactly. <S> If available you might want to use 0.8 or 1.0mm material instead of the standard 1.6mm so to prevent problems with e.g. DIP IC's that have pretty short legs. <A> PCB's are made out of fiberglass laminate sheets. <S> Could try this, should work: <S> Buy a small hobbyist's fiberglass repair kit. <S> Very thin sheet of woven glass, for repairing small nicks and cuts in things like boats and motorcycles. <S> Rough up both "top sides" of the PCB's with sandpaper. <S> (The rougher, the better.) <S> Mix up some epoxy, wet one top side, wet other top side, press and clamp together very tightly, <S> making sure there are no air bubbles. <S> It may warm slightly as it cures. <S> In 12 hours it should be rock-solid. <S> Trim off any excess with metal shears (don't grind it.) <S> I'd do this to bare boards, then do the layout, etching, and drilling after. <A> This can work. <S> Make sure that you make the bottom board a mirror image of your design. <S> That way you can glue the non-copper sides together, and have traces exposed on the outsides. <S> You don't want traces between layers where you can't get to them. <S> The only major problem that I see is that the plated through holes (and vias) won't be connected through the board. <S> You can solder a small piece of wire though the vias. <S> And you'll need to solder your through-hole components on both the top and bottom sides. <S> This is fairly easy with most components. <S> However, some parts cover up the holes. <S> For example: most connectors, and some electrolytic capacitors. <S> In this case, you can't solder the top side once the part is in position. <S> Try to route all of your connections to these parts on the bottom side of the board. <S> Your English is great! <S> Good luck.
| In my opinion it is always possible to have a single layer PCB with trough hole componentens, if the size does not matter. If you drill some pads that are both on the top and bottom PCB before sticking them together, you might even be able to fix the two pcb's just by soldering the wires on the top and bottom pads.
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Will connecting two batteries in parallel increase current drawn in simple led circuit? I have a resistor diode circuit that I'm powering with a 9V battery. The circuit diagram below is a simplified version of it. The real one has 4 LEDs, 8 resistors. I want to deliver 30mA to the diode. I had the idea to place two batteries in parallel to increase running time. Will placing two 9V batteries deliver 60mA to the LED? or does the current supplied remain constant and the batteries just drain less? simulate this circuit – Schematic created using CircuitLab <Q> This one: "the current supplied remain constant and the batteries just drain less" <S> The LED current will be unaffected by the addition of the second identical parallel battery. <S> V = <S> I <S> x R <S> In this circuit you are doubling the battery, but not changing the output voltage (two identical 9V batteries in parallel is still a 9V output). <S> On the load side, the resistor and LED have not changed (that's the R in Ohm's law). <S> Please note an LED is not accurately modeled as a pure resistance, but a complete explanation of that is not necessary to understand the answer to your question. <S> No change in V; No change in R; ...therefore NO CHANGE in I <S> (current) <S> E = <S> V <S> x I <S> x <S> t <S> What does change is the total potential energy in this circuit. <S> If you double the battery count, the total current sourced to the LED will be unchanged, but the current supplied by each battery will be 1/2 of the total. <S> Because the batteries are supplying half the current as before, they will last twice as long. <S> Energy is voltage times current times the time the current is supplied at that voltage. <S> A 1000mAh <S> Alkaline battery means that it can supply 1A at ~1.4V for ~1 hour. <S> So... <S> No change in E <S> ; No change in V; ...therefore battery life (time) is INVERSELY proportional to current <A> In your case, referring the circuit you have shared, there is no change in resistance. <S> So, V = IR remains pretty much the constant through the time. <S> So, your intended drop of 30 mA across the diode is achieved such that, 9 = I*(300) , so I = <S> 30 <S> mA. <S> So, the voltage shall remain 9V and only the runtime is increased by two times (since there are two batteries). <S> Threats in wiring the Batteries in parallel are subject the matter that whether the cell chemistry is identical or not. <S> As long as the two batteries are of the same state of charge, chemistry, ah rating and voltage, they should do just fine. <A> If higher currents are needed and larger cells are not available or do not fit the design constraint, one or more cells can be connected in parallel. <S> Most chemistry allows parallel configurations with little side effect. <S> Figure illustrates four cells connected in parallel. <S> The voltage of the illustrated pack remains at 1.2V, but the current handling and runtime are increased fourfold. <S> you can study more about the batteries connected in parallel & series here. <S> http://batteryuniversity.com/learn/article/serial_and_parallel_battery_configurations <S> In your design, since two cells are connected in parallel voltage will remain 9V only however the current capacity will be doubled but it doesn't mean it will supply double current(however <S> it can depending upon the load). <S> From the datasheet, maximum forward voltage you can apply is 2.6V (2V typical) <S> Since you are applying 9V you need to put resistor in series in order to reduce the voltage drop across LED. <S> if you see the datasheet of LED, it specifies that maximum continuous forward current <S> it can handle is 30mA. <S> it doesn't mean that it will always consume 30mA <S> it simply means it can handle up to 30mA forward <S> current(again forward current will depend upon the current capacity, supply voltage & load ) & brightness of the LED will depend upon the current flowing through it. <S> parallel connection doubles the current capacity but the current in the circuit remains the same but if you change the load, current will change & if this current exceeds 30mA, LED will get damage it won't operate more than 30mA(30mA+30mA=60mA will not be the case)
| In general when Batteries are connected in parallel, the voltage remains the same while the current gets divided between the two batteries and so the runtime will increase. Since load is unchanged current will be the same.
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How do I know current across unknown load if I know voltage If I know voltage across a load, but don't know load resistance, how do I know current going through load? Do I put a resistor of 1K and calculate? My voltage is square wave so multimeter only get average current. I want current square wave <Q> One approach is to use a hall effect sensor to measure instantaneous current and a low end microcontroller to drive an output led or lcd panel. <S> The Allegro ACS 712 is fairly simple to use, and you can find breakout boards for it at low cost at sparkfun. <S> datasheet here: http://www.allegromicro.com/~/media/Files/Datasheets/ACS712-Datasheet.ashx?la=en <A> Serial resistor is OK, but it should be several orders of magnitude smaller than the load, so 1k is a little surprising. <S> Also i would measure with oscilloscope instead relying on average and recovering peak out of it. <S> And the best part: you cand use DC instead your square wave, just put external voltage source. <S> More generally there is a four - point measuremet method. <S> You take two DVMs: one for current and one for voltage. <S> It's normally used for power metering, but also for accurate resistance metering too. <A> *the resistor has to be small enough <S> so the voltage dropped across it is small compared to the voltage on the load, however, not so small so as to be impossible to measure.
| You have to put a small* resistor in series with the load and then measure the voltage across it.
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Are ten 2W speakers as loud as two 10W speakers? If all things are the same (e.g. total magnet weight, manufacturer, cone material) and I have two 10W speakers as compared to ten 2W speakers would the decibel level be the same? <Q> The other things you list (magnet weight, cone material, etc.) are only indirect indicators of efficiency, and there are many other parameters that would apply as well. <A> If all else is the same then, yes, the power output and decibel level would be the same. <S> However, it's more likely that the physical arrangement of the speakers would not be the same, leading to either a perceptual difference or an actual measured difference due to differences in distance from source to sensor. <A> Sound is always measured at a distance from the speakers. <S> Why? <S> the inverse square law. <S> The intensity of observed radiation is inversely proportional to the square of the distance between the source and the observer. <S> The practical result of this is that the best way to configure the speakers depends on desired coverage and placement as well.
| The key parameter that needs to be the same is the efficiency of the speakers, in terms of electrical power in to acoustic power out.
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How to power a 12V fan simply using a PV panel? I am an electronics noob and I'd like to create some air circulation through my friends cellar using a salvaged 12V 0,8A brushless DC computer fan and a suitable PV panel. I have searched Ebay for 12V solar panels but their V-maximum is mostly in the 17-18V range and that worries me. What PV rated wattage would be ideal, how would I calculate? Will 17-18Vmax kill my 12V fan? If yes, then do I need a step-down buck converter or should continue looking for a 12Vmax panel? Does a step-down converter help? Does it help or hinder in cloudy weather? Id like to keep it cheap and simple but any suggestions to make it better are very very welome. Thank you in advance! This is the fan: Delta Electronics AFB1212SH-F00 These are two PV panels that I am considering, a 10W Eco-worthy and a 20W GWL/Sunny mono (not enough reputation for link). <Q> The Fan is rated at 0.53A at 12V. Voltage range of 7-13.8V. <S> If you select a nominal 12V solar panel with an output short-circuit current of around 0.5A (and certainly no more than 0.53) you will be OK. <S> Yes, the maximum panel Voltage may be 20V or more, but the panel cannot achieve that voltage when a load is attached. <S> The solar panel has a V-I curve. <S> In other words, the current and voltage of the panel are always on a well-defined curve (under constant lighting conditions). <S> Likewise, the motor will have a V-I curve. <S> When you connect them together, they will find the single point of intersection of their V-I curves. <S> As long as the voltage at that single point of intersection does not exceed the maximum motor rated voltage (13.8), then you are OK. <S> Unfortunately, we don't have the V-I curve for the motor or the solar panel. <S> But, we know that the motor will consume 0.53 Amps at 12V. <S> And, every solar panel publishes its short-circuit current (aka, Isc). <S> So if you use a solar panel which can only supply 0.53A <S> (Isc <= 0.53A), you don't need to worry. <S> Because at any lower current, the voltage will be below 12V. <S> The panel you linked to has an Isc of 0.69A. <S> So if you use that panel, there is some potential for problems. <S> There is still a chance that it will be OK, but you would have to try it to find out. <S> If you can find a panel with a lower Isc, that will be safer. <S> If you want to use the panel you linked to, you should be prepared to burn up some extra power somehow (if needed), using one or more power diodes in series or resistors in series, or using an LDO regulator. <S> Just enough to keep the voltage under 13.8 in full sun. <S> Personally, I would not use a DC-DC converter in this case just because of the complexity and cost. <S> I am also not totally sure how the DC-DC will behave when the lighting is low. <S> It may due some weird thing where it cycles on and off in an irritating way. <A> Supplying a 12V fan with 17-18V is a bad idea. <S> Using a linear regulator to convert the solar panel voltage to 12V is also a bad idea, since you will just waste energy ( <S> 18V-12V)*0.8 = 4.8 Watt. <S> The step-down converter you mention will convert the solar panel voltage to 12V with minimal loss. <S> To answer your question 4, the load (converter with fan) won't affect how the solar panel handles cloudy weather. <S> The clouding will however affect how much power the panel can put out. <S> At some cloudy point the step-down regulator will require too much power to keep the 12V regulation, which will lead to the 12V output decreasing = fan stops. <S> Step-down converter boards are possible to buy from most shops supplying electronic components or kits. <S> Search for instance for a LM2596 board like this: Step-Down <A> If you add a load or resistor in series that takes on 5V from the PV panel, the fan will operate without issue. <S> Since you are using free energy from the sun, the power loss introduced by the resistive load is irrelevant. <S> Your fan will work. <A> Wouldn't it be simpler to hook up a Charge controller to the PV panel which will accept a large range of PV panel voltages (enusre you get one for your PV panel voltage range), add a 12V battery (around 10 Ah), and hook up the motor to the 12V output from the charge controller. <S> This will allow the fan to be run when the sun is not shining. <S> The battery will be the largest cost, probably around $30, and the charge controller will cost around $15 (see ebay for both). <S> Finally, most charge controllers today also have a 5V USB output so you can charge your phone if desired.
| If you can find a panel with a published V-I curve, you can check the current at 12V to see if it is OK. If you know the impedance of the fan during its operation, you can connect it to the PV panel as long as you add an additional load to the series loop.
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PCB Holes Too Small to Fit Component Based on the VSK-S25 datasheet which states the holes to have a diameter of 1.5 mm, I used the following values in EAGLE When the PCB is made, it turns out that the pins of VSK-S25 are too large for the pads. Should we have used 0.059 inches for Drill rather than Diameter ? <Q> Drill is the size of the clearance hole, diameter is the outer diameter of the copper pad. <S> The difference between these numbers determines the width of the annular ring (donut). <A> Yes, with most (not all) <S> In your case, you specified 0.8mm, which is obviously way too small. <S> The recommended hole diameter of 1.5mm is a bit loose on the actual 1.0mm nominal pin diameter. <S> You should make the pad quite a bit bigger than the hole size for a relatively heavy component and large pin diameter. <S> Maybe 2.1mm or 2.2mm, which will give more than 10 mils annular ring. <S> If the board was one-sided you'd use a bigger pad size again for the pads that are far enough apart to allow it. <A> Sort of if you know the pin size is 1.5mm <S> do you think you can count on drilling a slightly less than 1.5mm (1.4986mm) hole and have it work. <S> Granted your fab house will just pick a drill size close to that one. <S> That 1.5mm hole is going to be plated, the thickness of which will bring down the hole size even further. <S> Then there's tolerances, <S> first off the drill size itself will have a tolerance, and the bit may wander a little from absolute center. <S> So now you find yourself needing some slack <S> so all 6 of those holes always line up. <S> I'm sure the pins themselves and their spacing also have a tolerance in the datasheet. <S> All that adds up to needing hole sizes that are slightly larger than your ideal pin sizes. <S> I've gotten lazy and like to specify to my fab house that all my holes sizes are finished sizes after plating which makes it a little easier if your fab house accepts that. <S> You can come up with your own actual drill size by considering all these requirements, and you can also consult the IPC standards for more advice.
| You could also use a slightly small hole size- perhaps 1.3mm. PCB manufacturers you specify the "drill" diameter and they will use a slightly different size of actual tool to ensure the finished, plated hole diameter is close to what you specified.
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How can voltage burn out an LED? I understand how a current higher than what the LED is rated for can burn out an LED, but how does something equivalent happen with voltage? If the correct current is on an LED but the voltage is too high, what causes it to burn out? I just don't see what effect voltage has on the LED. <Q> Voltage and current are intimately related. <S> If you attempt to increase the voltage across an LED, the current will increase. <S> Likewise, to increase the current through an LED, you must increase the voltage across it. <A> As you can gather from the other answers, voltage (U) and current <S> (I) are linked. <S> In the case of a simple resistor: U = <S> R <S> * <S> I where R is the constant resistance of the resistor. <S> A diode is only slightly more complicated. <S> Here we can use a graph to show the relationship. <S> The graph uses i for current and V for voltage: <S> The image is taken from Using larger resistor values . <A> First a diode(LEDs are diods) above a certain voltage is like a closed circuit. <S> The problem is that like every wire, the diode has a critical point after which it will "burn", basically some irreversible transformations occur. <S> So you can say the diode can sustain a certain power. <S> Now, power is related to voltage like this: P = <S> I*V <S> where I is the current, and V the voltage. <S> Since it's a closed circuit, the current is ∞ over it. <S> The source can't give ∞ current, and is limited to a maximum amount. <S> So over a diode, it will use that maximum amount and thus, I becomes a constant. <S> Since I is constant, this means power increases proportional with the variable left, which in our case is V(voltage). <A> I understand why you are having a tough time here. <S> An LED is not like a resistor/heat lamp per se. <S> An LED is like any other diode, except in the forward conducting mode as the electrons flow through the junction they cause the atoms to shake at a specific frequency, and <S> not just randomly like a normal conductor. <S> This shaking causes light. <S> Think of them as a whistle. <S> One note, one amplitude. <S> (Much like a blade of grass held between your thumbs.) --- and that takes energy. <S> If you force too much air, because of a higher pressure (voltage), you will blow out the reed that is making the vibrations. <A> As mentioned in one of the comments, this: If the correct current is on an LED but the voltage is too high ... is not possible. <S> If the current is "correct", then the voltage will be equal to the characteristic voltage of the diode. <S> For example: simulate this circuit – Schematic created using CircuitLab <S> In the above schematic, Vdiode will be about 1.9V, because 10kV/1MΩ is about 10mA, and that's the voltage this particular LED arrives at if biased on by 10mA ( datasheet PDF ). <S> Were you to change the value of R1 to 1 ohm, then approximately 10kA <S> would briefly flow through the LED, resulting it a burned-out LED. <S> A key concept to grok is the difference between Constant Current and Constant Voltage regulators. <S> A typical "bench" power supply is Constant Voltage, meaning it puts out X volts at some current, and will regulate its output to remain at X volts whatever its load. <S> Diodes approximate Constant Current regulators to a degree, because you can think of the voltage being dependent upon the current. <A> Diodes inhibit Avalanche in Reverse Bias. <S> LED are not exception they being diodes. <S> LEDs have a tolerance of a specific amount of reverse voltage that it can withstand but once this is exceeded, the LED can be damaged. <S> So any excessive reverse voltage ( V R ) applied can cause avalanche breakdown. <S> The easiest you can do is adding a simple PN-junction diode in series with the LED (if you wish to save your LED from getting damaged). <A> The above response is correct in that voltage and current are intimately related. <S> If you think of a regular resistor that follows Ohm's law, then you can see the relationship V = <S> I*R. <S> With a diode, this relationship still exists, but it is not linear, which is why in data sheets of LED's you will see plots of the Voltage and Current. <S> So if you increase the voltage across the LED above a certain threshold, the current will also increase, burning out the LED. <S> The reason why a power-line has high voltage and low current is that power-lines are super long, which increases its resistance. <S> Big Voltage = <S> Big Resistance <S> * smallerCurrent. <S> It still indicates that voltage and current are immediately related.
| It is not possilbe to have the correct current through an LED, but too high a voltage across it.
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Can this ATMega328-PU ATMEL microchip be used instead of MCP3008 I wanted to follow this tutorial as a first project. I've got all the parts I need, but I could not purchase the chip that was linked in the article. So I found another chip that sound similar (To a complete beginner) . My question is, will I be able to use the chip I brought, in place of the one suggested in the article or am I trying something thats impossible? <Q> Yes you can, but your tutorial breaks down because you are not following it. <S> The ATMEGA is actually a microcontroller. <S> It requires more attention to setup the hardware <S> as well it requires programming. <S> If you are a complete bigger and want to follow the tutorial, then this might not be the approach you want to take. <S> That's not to say you can't do it, it's just slightly more complicated to get working. <S> The MCP3008 is an external 10bit ADC. <S> It is not as complicated to get working than a microcontroller. <S> If you can find a 10bit external ADC, you should be able to follow your tutorial pretty well. <A> The Microchip IC you refer to is a ADC, the Atmel device is an MCU which has integrated ADC in it. <S> In theory you can use the ATMega's ADC + MUX to get the readings you need and send them over simulating the original ADC. <S> That is of course creating more work for yourself for no good reason. <S> You can buy the Microchip here . <A> No. <S> Indeed they're quite different: <S> MCP3008 <S> is <S> an A/D converter and ATMega328-PU is a microcontroller (which has a built-in A/D converter). <A> These are completely different kinds of devices, and they can't be swapped between If you want to learn and understand mechanics behind A/D converters, you can try and make your own in FPGA, <S> or you could use a microcontroller ( <S> like atmega328 you just linked to) to make use of its built-in ADC and communicate with your master processor with serial interface .That can be done, but requires lot of coding and attention to wiring your device. <S> If there's anything that needs to be cleared or explained I'd be more than glad to help you
| Well, atmel chip which you link to is a microcontroller, which roughly means that it is a microprocessor with peripherial circuits etched in the same piece of silicon, however that MCP you link is a A/D converter which simply means that it converts analog signal to discrete, digital interpretation of it.
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Welding Thermocouple to Test Sample I have a 3mm copper test sample that is going to get heated up, and I would like to know how hot it is actually getting. Previously, we have just held the thermocouple close to the sample, but we are considering mounting it to the sample to get a more accurate reading. I was told that it is possible to weld an exposed thermocouple to the sample, but I am concerned about making a ground loop. I guess there are two questions here: Will a ground loop be created and do I need to be concerned aboutmetal-on-metal contact with the thermocouple and copper for anyother reasons? Does anyone have advice on the best way to actually weld thethermocouple to the test sample? <Q> I normally do this by twisting the thermocouple wires together, then crimping into a standard crimp-type terminal lug. <S> That is then either bolted or soldered to the test item. <S> So long as the two thermocouple wires are in direct contact with each other, you won't have any appreciatiable accuracy problems. <S> Note that your temperature measuring device must be able to accept a grounded thermocouple. <S> This isn't a problem with a stand-alone / hand-held meter but does have to be taken into account if connecting to a multi-channel and/or computer-based data acquisition system. <A> Welding to copper is not easy because copper has very high thermal conductivity. <S> Aside from that there is no real issue. <S> You could braze it, depending on what temperature range you need to withstand. <S> This is also called " Silver Soldering " - melting point is in excess of 600 <S> °C <S> so pretty high. <S> Anyone can do this with a propane or MAPP torch and the appropriate flux and solder. <S> Soft soldering (what we do in electronics) doesn't work well with most thermocouples, but it does for coppper-constantan, so that's a possibility if you don't need to go too high in temperature <S> (the thermocouple itself is usable to 400 <S> °C, solder melting point is typically lower). <S> I'm not suggesting this, but you could even connect just a constantan wire to the copper and use a type T signal conditioner (not suggesting it because the copper you have may not be sufficiently pure to not require calibration, not because it won't work as a thermocouple- <S> sometimes they add tellurium or other impurities to aid production procceses) <S> To avoid problems, you should use a measurement device that works with a " grounded junction " thermocouple. <S> Any battery-powered device will be fine as will most industrial type thermocouple transmitters and signal conditioners <S> (they should say they have " galvanic isolation "). <S> It's also possible to use differential (instrumentation) amplifiers, but I would suggest avoiding it outside a very controlled environment. <S> Remember that thermocouple signals are relatively low level (tens of uV per K), and if you have a lot of EMI floating around you could have issues. <S> Also the temperature of the junctions inside the measuring instrument must be measured to greater precision than the desired system accuracy because (to a first order) thermocouples measure temperature differences. <S> That means that high absolute accuracy may be hard to achieve, but seeing tenths of degree C change in a lab environment is easy. <S> The best accuracy will be achieve by using a fine wire gauge since heat necessarily flows in or out of the wires. <S> For most purposes you can say as fine wires as practical is best. <A> How hot is your sample going to be? <S> For how long? <S> When testing mechanical seals, I have used a high temp epoxy to attach thermocouples to silicon carbide seal faces and test in applications over 550F. Temperatures during a seal failure would easily exceed 700F. <S> Typically, a failure to get a temperature reading would be the result of insulation breakdown of the thermocouple wire and not the epoxy. <S> I built my own thermocouple junctions and the epoxy also served to electrically isolate the thermocouple as well. <S> The epoxy was not designed specifically for this kind of application <S> and we were often operating beyond its published capability, however, in years of testing, I never observed a failure of the epoxy due to thermal breakdown or other heat effects. <A> What you might considers is attaching thermocouple to the sample using <S> thermoconductive glue/paste (preferably one which is electrically neutral) <A> I got an idea which probably might not work use a heat conducting rod solder to the sample and caliculate the heat near the sample and at end of the rod which might have cooled by the environment and near rod .then caliculate it on the rod to get the geat approximate by comparison.using weilding might cause potential differences in thermocouple as the electronic density of the sample might interfere
| Soldering the thermocouple might not be the best way to go as the joint created by soldering process will affect heating, and therefore your results.
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Basic power: drive 12v LED from 5v Raspberry Pi output — no 12v supply available There are a lot of questions on here and elsewhere about this. The consensus seems to be to use a transistor to switch a power supply — implying that I have a 12v supply available. Only 5V is available, either coming out of the GPIO pins or directly from the power supply (USB battery in this case). This is a single LED: seems like it should be possible to directly wire it to the GPIO pins with a component in the middle that raises voltage. (GPIO pins can "safely" provide 16 mA) It appears that I need a stepup/boost converter, but nobody seems to be discussing that. Am I thinking about this correctly? <Q> You can typically see 85% efficiency. <S> You still want to use the transistor on the RPI output, since it is 3.3v and low current. <S> Alternatively you can find a boost converter with a 3.3v enable pin, and control it directly from the rpi. <S> Edit <S> The button that OP linked to can be disassembled by design, to replace parts. <S> An easier solution is to replace the led resistor with one suitable for 5V operation. <S> I'd still advise a transistor setup. <S> If you feel lucky punk, well do you? <S> , then a resistor suitable for a red led at 3.3V 16 mA or less would work directly, with minimal brightness change, but a ten cent transistor to protect the RPI pin is recommended. <A> If you have a 12 volt LED and only a 5 volt supply you need a boost regulator and a means of activating the LED from a gpio pin. <S> So, look for a suitable boost converter (TI or LT) and then choose a BJT that can be activated by a gpio pin via a resistor. <S> Emitter connects to 0 volts and collector to the LED cathode. <S> Anode to 12 volts and MAKE <S> SURE that the LED will work without a current limiting resistor. <A> Although the Pi can supply 16mA (ish - drive strength is a poorly controlled parameter) at 5V, energy is conserved. <S> Since Power P = I*V, where I is current and V is voltage, stepping up the voltage means the current supplied to the LED will be lower than the current from the GPIO <S> - even with 100% converter efficiency , you'll get 4.4mA max from a 3.3v GPIO. <S> When we also take into account that The converter efficiency will be much less than 100%, especially with such a large step from 3.3V to 12V 16mA is the current with the GPIO shorted - at this point, the output voltage is actually zero , <S> not 3.3V. <S> Any significant current draw will cause some amount of voltage drop <S> We can see that driving a high power LED from a Pi's GPIO is essentially not a good idea. <S> BUT there is a better way! <S> A DC-DC boost converter (search ebay) can provide 12V from a 5V supply with a reasonable efficiency (typically ~85%). <S> A dedicated 12V supply is better. <S> The current through the LED is switched on and off by a transistor - a MOSFET <S> is ideal because no current flows into its gate while it is on. <S> The MOSFET suggested in the diagram will happily pass an amp or two with a 3.3V gate voltage from the GPIO pin, making the LEDs much brighter and avoiding any damage to the Pi. <S> This type of transistor circuit (just the lower part) is very useful for driving higher voltage loads from a low voltage microcontroller or processor.
| In general, you need a boost converter on the 5V supply input, to 12V. Any typical step up or boost converter will work, considering you keep in mind the minimum regulating load.
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What are the differences between an AC and DC coil relay? If a DC coil was just a coil, it would saturate. If an AC coil was just a coil, the magnetic field would drop to zero at 120 Hz and the relay would chatter. Presumably the designs in some way prevent these problems. How? And does it prevent me from putting DC on an AC-rated coil? <Q> The current is limited by that resistance. <S> AC coil relays have inductance as well (the DC relays have inductance as well, of course, but it does not affect the 'on' current). <S> They also typically have a shading ring that acts as a shorted turn in order to cause a magnetic field 90° out of phase with that from the coil, so that the total magnitude of flux does not drop to zero at the zero crossings. <S> Edit: As Andy says, an AC relay will work on (greatly reduced) DC. <S> Of course you can also make a DC relay work on AC by adding stuff to it (if you use a capacitor filter you might need only 18VAC (RMS) to operate a 24V relay, and 24VAC would cause it to overheat and fail early). <A> A Dc relay coil has resistance that limits the dc current. <S> An AC coil relies on its impedance for governing the current. <S> An AC relay will remain contact closed due to mechanical inertia and a little mechanical hysteresis and, the fact that an alternating north and south pole both attract the relay armature. <S> Putting dc on an ac coil should work fine but be prepared for the resistance to be low. <S> In other words, If relay is rated at 24 volts ac don't use 24 volts dc. <A> In theory AC coils can be driven with DC as long as you limit the DC coil current to the level of the AC holding current (to keep the coil from overheating). <S> For larger relays i.e. contactors, a DC current limited to the equivalent AC holding current is often insufficient to actuate the contactor effectively. <S> Since the coil inductance of the contactor increases once the contactor is closed <S> the AC holding current is less than the AC pull in current. <S> DC contactors usually have some mechanism to shift between a coil pull in current and the holding current. <S> This is the difference between an AC coil and a DC coil in large relays. <S> The flux shunt (shader ring) is external to the coil and only shorts a portion of the core. <A> Recently, I had a problem with a batch of four 2-pole plug-in relays that had coils marked as 240V AC. <S> When connected to the supply, three of the four merely 'chattered' and failed to 'pull-in' the contacts; the fourth operated correctly. <S> Believing the problem to be one of poor manufacturing quality control, I received some replacements from the same supplier which subsequently worked OK. <S> I assumed that the three faulty relays had been fitted in error with DC coils. <S> Consequently, rather than binning them, I overcame the problem by internally fitting 1000 PIV DIL bridge rectifiers allowing the relay coils to operate normally on raw DC (i.e. no filtering). <S> Perhaps the use of 1000PIV devices was overkill, but I wanted the rectifiers to have a sufficiently high withstand voltage to the back emf <S> generated when the relay released. <S> Additionally, as the relay coils were now operating on DC (albeit raw DC), I could have fitted a single rectifier wired in inverse parallel to the polarity of the raw DC to snub the back emf, thus permitting use of a bridge rectifier with a lower PIV, but a lack of sufficient space within the relay precluded this solution. <A> If you use a full wave bridge, the diodes in the bridge should act as free-wheeling diodes, and should snub the back EMF.
| The DC coil relay has a resistance from the copper wire typically used.
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How to apply HT gradually in a valve amplifier? For valve amplifiers, it is said, that applying anode voltage before the valve is heated reduces lifetime and can have other negative effects on circuit lifetime. To avoid this, I want to add a thermistor to the HT voltage line, so the valve has time to heat, and the HT is applied in a more controlled manner. Based on what parameters of my circuit should I choose a thermistor?Or what else should I use to "delay" the HT voltage? Currently I am using a manual switch, which is good for standby mode, but I'd prefer to have something, that doesn't rely on the human factor. Edit:In the past this was solved by using tube rectifiers, so they would need time to heat up too, but I want to use solid-state rectification, so that is not an option. <Q> Since it takes the tubes some time to warm up, what you need is a time delay relay that starts to time out when mains power is applied to the amplifier and then closes a set of contacts between the B+ and its load(s) <S> when the proper time has elapsed, perhaps 30 seconds to a minute, depending on which of the tubes takes the longest time to warm up. <S> Then, when mains power is cut off some time later, the relay opens immediately, allowing the tubes to cool down with no stress from the B+. <S> The relays are available commercially - quite pricey - and are called ON-DELAY relays, or you could easily put one together if you're so inclined. <A> I don't think your thermistor idea is going to work very well. <S> The tubes will represent little load before the filaments are warm so a resistance in series won't drop much voltage. <S> The current through the NTC has to be high enough to make it/keep it hot or it won't be low resistance. <S> You could automate the switch by replacing it with a time-delay relay- that would probably be the simplest solution. <S> To avoid the drop, create another supply about 10V higher than the B+ supply. <S> You'll need a few parts to protect the MOSFET gate and to discharge the capacitor at a reasonable rate when power is removed. <S> Or something like this 20 second Amperite delay relay might be more in the spirit of your 'bear skins and stone knives' technology: <A> The thermistor idea works well <S> , I use it on all my amplifiers <S> applied to the primary of the main power supply transformer. <S> Connected this way it also limits the current inrush: of the filaments when they are cold. <S> in the (big) power supply capacitors. <S> in the coupling capacitors which often discharge very slowly through high resistors. <S> This said, it will only limit the current for 1 or 2 seconds, you may wish to rise the voltage slowly on tubes anodes. <S> This is a bad idea to wait for the filaments to be hot and apply the B+ suddenly because: during the time the HT is not applied, the power supply has no load and then presents its maximum voltage because it will create a current inrush specially in the power tubes that can -- for a short moment -- drive the tubes far above their max dissipation. <S> The idea is to apply the HT gradually. <S> Using a vacuum rectifier is a way to achieve this. <S> You can also set up a mosfet regulated power supply: Copyleft Yves Monmagnon. <S> August 2009 from <S> this link (in french) <S> Q1 and R1 create a CCS. <S> This current will charge C1 through R2 and polarize Q2. <S> When C1 is charged the voltage on the source of Q2 is roughly equivalent to Iccs <S> * R <S> I use DN3545 for Q1 and IRF820 (up to 500V) for Q2. <S> The zener is here to discharge C1 when the PSU is switched off and avoid Vgs > 20V <S> (maximum specified by the DS). <A> "For valve amplifiers, it is said, that applying anode voltage before the valve is heated reduces lifetime and can have other negative effects on circuit lifetime." <S> This is not, actually the case. <S> Standby switches were the invention of Fender. <S> I would highly recommend you read the work of Merlin Blencowe as well as this piece from Peavey's site . <S> Keep in mind that tube TVs, tube radios, tube PAs, and other tube devices never had standby switches. <S> Only guitar amps have them. <S> With the exception of gain structure and features, there's little difference between a guitar amplifier and a PA. <A> Use a sensitive bi-metallic switch whose contact closes when warmed by the tube. <S> Old tech solution to match the amplifier technology. <S> I guess a related question might be, do you leave anode voltage on until cooled down OR, should anode voltage be collapsed before deactivating the heater supply. <S> You need to think about this. <A> As others have pointed out and have also provided links to other sites to add citation, a simple standby switch can actually shorten the life of valves and can seriously damage a valve rectifier if it is placed before the reservoir capacitor. <S> And as has also been pointed out, <S> theoretically at least, there is no evidence that the phenomenon of cathode stripping due to HT being applied before the valves are up to operating temperature actually occurs at least not to any significant amount that needs to be worried about. <S> There is, however subjective evidence that when both a mains soft start circuit (to manage the inrush for the whole thing lasting only for a few mains cycles) and a means of slowly building up the HT as the valves are warming (either using a valve rectifier or a pair of mosfets after solid state rectification), rather than suddenly applying full HT after a delay with a human operated switch or a timed relay, that amplifiers with a reputation for repeatedly blowing valves for no apparent reason, do then settle and stop blowing valves. <A> The time delay timer seems like a good idea, I found a delay on timer on Ebay for 99p up to 10 sec from a 12V supply, I supply my valves with a volt reg (7806), so have 12V on power up, hope this gives others some design ideas
| If you really want to apply voltage gradually (rather than with a delay), and were willing to live with a few volts of drop, perhaps you could use a MOSFET voltage follower and an RC delay.
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What is the small barrel type connector called? I just ordered a barrel connector but it's bigger then the one i need, What is the small barrel type connector called. My original one has a yellow ring around it, Is that an indicator of what type it is? <Q> What is the small barrel type connector called. <S> It's called a barrel connector with a different diameter. <S> Typically I see these connectors specified in 0.1 mm increments. <S> My original one has a yellow ring around it, Is that an indicator of what type it is? <S> Within one particular vendor's product line, it may be. <S> But you might find a compatible connector from another vendor with a different color. <A> It's still a barrel plug. <S> Whip out your calipers to verify, but I'd say that's a 4.0mm/0.7mm barrel plug. <A> Sadly you hit one of those nasty secrets of electronics: lack of standardization! <S> Power jacks are built in a wild, crazy (and admittedly quite stupid) variety of sizes. <S> Behold the horror of any designer (and any user) in this Wikipedia page on coaxial power connectors ! <S> Excerpts (emphasis mine): <S> A coaxial power connector is an electrical power connector used for attaching extra-low voltage devices such as consumer electronics to external electricity. <S> Also known as barrel connectors, concentric barrel connectors or tip connectors, these small cylindrical connectors come in an enormous variety of sizes . <S> Contact ratings commonly vary from unspecified up to 5 amp (11 amps for special hi-power versions). <S> Voltage is again often unspecified , up to 48 V with 12 V typical. <S> It is quite possible that new sizes will continue to appear and disappear . <S> One possible reason that a particular manufacturer may use a new size is to discourage use of third-party power supplies, either for technical reasons or to promote use of their own products, or both. <S> The sizes and shapes of connectors do not consistently correspond to the same power specifications across manufacturers and models . <S> Two connectors from different manufacturers with different sizes could potentially be attached to power supplies with the same voltage and current. <S> Alternatively, connectors of the same size can be part of power supplies with different voltages and currents. <S> Use of the wrong power supply may cause severe equipment damage, or even fire. <S> I would describe the situation as "wild monkeys gone berserk". <S> Good Luck!
| The smaller types usually have lower ratings, both for current and voltage.
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Effect of 2-terminal and 3-terminal Voltmeters and Arduino Voltage Sensing on Circuit I am trying to build a board to sense circuit voltage relative to ground at different points on a breadboard. I would want the board to work regardless of configuration of the circuit. The only exception would be the location of GND, which would be the same always. There are two ways I was thinking about doing this: Voltmeter Modules: My first idea was to get either a 2 terminal or 3 terminal voltmeter. I believe 3 terminal is less invasive to the circuit, but please correct me if I am wrong. My thoughts were that this would need to be powered externally so that it wouldn't effect the voltage within the circuit, whereas the 2 terminal voltmeter is actually using some of the voltage to power itself? Also I was assuming that the 3 terminal voltmeter would not have an effect on the circuit even if it was attached to a common ground within the circuit. Anyways I was wondering if you could verify these assumptions and offer some advice on the best one to use. Or let me know if this is a terrible idea (aside from the fact that is would be space intensive with all the voltmeters) :). Arduino Voltage Reading My second idea was to use an Arduino and a multiplexer to sense the voltage within the system. Again I wasn't sure what effect the Arduino would have on the system if any if it is constantly attached and sometimes reading? The voltage I could sense would be based on the resistors I use. I then could use the Arduino to do something with the signals to make them visible to the user. If you have any feedback on my questions, the solution you think would be the best, or alternative solutions, I would love to hear it! <Q> Sounds like you are concerned about something called a loading effect. <S> This occurs when you add another component (Arduino or volt meter) to a circuit. <S> Essentially, you are causing some current to be drawn from the circuit. <S> High end meters have something called a high input impedance. <S> These meters draw negligible current from your circuit. <S> The bigger your load is, the more current a meter will draw. <S> If you have a very large resistor and read the voltage using a volt meter with a relatively small input resistance all of the current will want to flow through the small resistor, aka volt meter. <S> You can learn more about current distribution by looking up "Loading effect". <S> I have more experience using the arduino for measuring voltage, so I will talk a bit about my knowledge with that. <S> First off, it isn't too hard to program an arduino to read a voltage. <S> But it does have some limitations. <S> It can only read voltages up to 5V <S> (There are ways around this) and I found that it wasn't too accurate (8-10 bit resolution). <S> Then again it depends how accurate you want the readings to be. <S> The Arduino didn't seem to have much of a loading effect (I was comparing results from a volt meter). <S> The MOST challenging part is a means of displaying your voltage reading on a screen. <S> You can buy little screens with Arduino libraries, but they can tend to be troublesome. <S> Most of my time was spent making the screen look presentable. <S> The Voltmeter Module seems to be definitely the simplest route to take. <S> I am not sure what loading effects they will have. <S> I am sure that differs between models. <S> I would personally recommend buying a volt meter at your local hardware/electronics store. <S> They have good input impedance and can be cheap. <S> Hope I was of some help! <S> Goodluck,Josh <A> Not sure about the loading effect, but the way I've used the Arduino as a voltmeter is by connecting an analog pin (say, A0) to a wire, and you can touch the wire to whatever spot on your breadboard you need to test. <S> I am not sure what output is your preference, but you can download Firmata for your Arduino. <S> When you open it up and select the correct port, you'll see a list of all the pins on your Arduino and their current readings (between 0 and 1023). <S> This will display on your computer that the Arduino is tethered to. <S> When you touch the wire to a spot on the breadboard, you'll see a number between 0 and 1023 on the corresponding pin in Firmata. <S> Since USB is 5V max, you just do (reading/1023) <S> * 5 <S> to get the voltage. <S> Since some of the analog pins are floating, for whatever reason there may be some interference, so you can ground the next or preceding analog pins to hold them at 0. <S> Not sure how accurate this is, but it seemed to work. <A> Good voltage measurement requires three things: A high input impedance, to reduce or eliminate loading effects. <S> A circuit which scales the input to the voltage range of the ADC input to minimize quantization error. <S> An accurate voltage reference, typically a semiconductor bandgap reference. <S> In the case of the Arduino, the maximum recommended source impedance is 10k ohms, so if you need to measure voltages at source impedances greater than that, you need to buffer the input. <S> Both this and the scaling can be accomplished with a simple opamp circuit. <S> You may or may not need an external reference depending how accurate you want your measurement to be. <S> However, you can get a pretty stable one for about a dollar on DigiKey. <S> The Arduino docs will tell you how to use an external reference . <S> If you do this, remember to scale your input voltage range to the external reference voltage, not the 5V internal reference. <S> EDIT: <S> Here's a page which covers using an external reference with the Arduino for voltage measurement.
| If you are looking for a quick way about this a voltage module might do the trick.
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1.5V-rated MOSFET doesn't react to a gate input of 1.8V I am not really an electronics specialist, but a software engineer (so excuses if I 'm asking stupid questions). I am trying to use a microcontroller GPIO output rated at 1.8V. When this pin becomes high, I want to enable a 12V relay. I am using a N-channel MOSFET from freetronics The specs for the MOSFET can be found here . For some reason the 1.8V seem insufficient to drive the MOSFET although it is specified for 1.5V min. I've tried a standalone setup using a 1.5V AA battery and that doesn't work either. But if I apply 3.3V with the same setup, it works (just so you know my wiring is OK). Unfortunately my microcontroller (Intel Edison) only has 1.8V GPIOs. Am I missing something? How can I make this work? Should I use a different MOSFET? And if so, which one? Your help is much appreciated. <Q> Sadly this setup won't work. <S> If you examine the datasheet carefully it states that the MOSFET has a threshold voltage which is guaranteed to be between 1.5V and 2.5V, with 1.8V typical. <S> Even assuming you are lucky and you've got a specimen whose threshold is at 1.5V (best case for you), that doesn't mean that the MOSFET magically turns ON when its Vgs voltage reaches that value. <S> That's the minimum voltage needed to make the MOSFET just barely conduct: in that line of the datasheet you can notice that the threshold voltage is specified at scant 250μA of Id. <S> That level of current is insufficient to operate a common relay reliably. <S> Note: <S> (as pointed out by @SpehroPefhany in a comment) these are the values at 25°C. <S> If the ambient temperature is lower (e.g. winter, cold climate, circuit placed in cold rooms) the current at that level of Vgs will be even smaller until the MOSFET warms up! <S> To use a MOSFET as a closed switch you must drive it into the ON region, and specifically in the ohmic region, i.e. that part of the output characteristics where it behaves as a (small value) resistance: <S> As you can see, the curves shown correspond to higher values of Vgs (~2.8V or higher). <S> You can better appreciate the problem looking at the Rds(on) graph, i.e. "the resistance of the switch": From the graph on the right you may see that Rds(on) doesn't vary much with current, but the graph on the left tells another story: if you lower Vgs under ~4V you get a steep increase in resistance. <S> To summarize: this MOSFET cannot be turned on with a mere 1.8V. <S> At least you should provide enough Vgs to make it conduct in the worst case , i.e. Vgs(TH)=2.5V. <S> And this is confirmed by your experiment at 3.3V. <A> @Lorenzo has explained why this is not working for his, and if it did work it would be marginal, which might considered worse. <S> Here is what a spec for a suitable MOSFET (AO3416) looks like: <S> The Rds(on) is guaranteed at 1.8V Vgs, and at 34m\$\Omega\$ even if it's a bit higher because of tolerance on the 1.8V supply or temperature, still plenty of drive for a 12V relay. <S> In general you should use Vgs(th) to determine when the MOSFET is mostly off, and the voltage(s) <S> at which Rds(on) is specified to determine when it is mostly on. <A> Figures 2 and 3 from the data sheet are shown below. <S> Note, in figure 2, that for a Vgs of less than about 2 volts, the drain current will be close to zero, while with with a Vgs of 3 volts the channel is nicely enhanced. <S> That's in agreement with your experiment, and shows that you need more voltage on the gate to make your circuit work, Figure 3 shows how the Rds(on) rises very quickly to a high value as Vgs falls, and even though it's given for an Id of 20 amperes, the slope of the curve will be similar in your circuit, with the ultimate effect being that when Vgs gets low enough, Rds(on) - <S> which is in series with the relay coil and the DC supply - will rise to a value high enough to limit the current through the relay coil to the point where it'll be impossible to actuate. <S> Since you don't have the gate drive required to assure that Rds(on) will be low enough to allow the relay to work, arguably the easiest way out would be to substitute a jellybean bipolar transistor for the MOSFET and drive the base of the transistor through a resistor with your 1.8 volt signal. <A> Other answers have finely explained why the FET in the question does not work. <S> I'll focus on solutions. <S> One is to use a FET designed for the purpose; e.g. FDN327N . <S> To determine the appropriate resistor, find the minimum resistance Rlmin of the relay and the maximum of the 12V supply (say V12max=13.6V), giving you the maximum current in collector <S> Ic= <S> V12max/ <S> Rlmin (keeping the saturation voltage as engineering margin). <S> Find the minimum gain of the NPN transistor at saturation for this current (be reasonably rather than overly conservative on that one; strictly speaking the BC848C data sheet only guarantees a minimum gain Gmin of 20 at saturation, but the 420 min for Vce of 5V for the C grade might give us enough confidence to use G=50). <S> The minimum current we should target in the base is Ib = <S> Ic/Gmin. <S> Then we must account for the minimum supply voltage V1_8min of the device driving the DATA port, subtract the maximum rated dropout Vdrop on the high-side FET of that DATA port under load Ib, another 0.75V or so for V BE(ON) at saturation at Ic, and the maximum resistor comes out as Rmax=(V1_8min-Vdrop-V BE(ON) ) <S> /Ib. <S> If V1_8min-Vdrop-V BE(ON) gets negative, we need a less conservative estimates of the three values in the sum, which might be helped by a less conservative (increased) <S> Gmin, which decreases Ib. <S> We must also insure that the current in the DATA port does not exceed its maximum rating (for this we must consider the maximum V1_8, the minimum high-side dropout and V BE ). <S> If that's exceeded, we must increase the resistor and justify less conservative estimates (of Gmin in particular).
| Another cheap, easily sourced and reliable solution is using a plain NPN bipolar junction transistor.
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Why is the capacitor considered lossless, even if it can store/give out energy? They say capacitor is lossless because average power is 0.But contradicting this is the fact that the energy stored/disipated by a capacitor is (1/2)CV^2.How and why is this? <Q> Dissipation of energy means transforming it to a form less usable and not easily recoverable to the original one (usually heat). <S> With capacitor, it is obviously not the case - <S> an ideal capacitor maintains constant voltage on its ends until you discharge it by connecting external load. <S> It provides you exactly with the amount of energy you have given it, equal to the expression you wrote. <S> Resistor, for example, "consumes" power as well, but when you disconnect it from the source, it has zero voltage across. <S> Here, the energy has been dissipated into heat and you cannot recover it by any means. <S> Real capacitor, though, has some parallel leakage conductance preventing it to stay charged forever. <S> When using it in an AC circuit, dynamic losses in the dielectric dissipate some power as well, this is quantified by Dissipation factor, \$\tan \ <S> > \delta\$. <A> A capacitor stores the charge Q= <S> CU. <S> Because the voltage over the cap increases linearly in an RC circuit when it's charged, it will finally hold an energy of E=1/2CU^2. <S> However, the power supply provided the entire charge with its full voltage, so it delivered an energy of E=CU^2. <S> Whenever you charge a capacitor and the resistance of the system is not zero, you waste 50% of the primary energy. <S> Using a cap as energy storage is everything but lossless. <A> Any energy put in is stored, and possibly later put out again. <S> A real physical cap can lose energy by heating up (due to internal resistance) and then radiating the energy away, so it is not lossless. <A> Average power of a capacitor is zero for perfect symmetrical ac input voltages only, e.g: sine/cos signal. <S> Here, Average power is defined as the net power the capacitor has stored/dissipated over a cycle of the input sinusoidal voltage. <S> An ideal capacitor, in the first half cycle, it stores some energy, P. <S> And it dissipates the same energy P in the next half cycle. <S> so the net energy = <S> P + (-P <S> ) = 0 <S> ; hence power = rate of change of energy is also zero. <S> This is the same with inductor too. <S> Thus, ideal C and L don't consume any energy in an ac circuit. <S> So we say they are lossless. <S> The equation E = 0.5 CV^2 is applicable in this case too; but v is a time varying signal, so we have to integrate and find the total power. <S> In fact, if you calculate the power in both cycles separately, you will arrive at the same result, when you add them finally. <S> Considering a DC circuit, if an ideal capacitor is charged with a DC voltage and allowed to discharge at power-off, the average power maybe expressed in sum of stored and dissipated energies, which is equal to 0 too. <A> No real capacitor is loss less but an ideal one is and some real ones almost are. <S> With an ideal capacitor you can store energy in it by charging it. <S> $$E = \dfrac{1}{2 <S> } <S> \cdot C \cdot <S> V^2$$ <S> But you can get that energy back by discharging the capacitor. <S> The energy you put in is equal to the energy you get back. <S> Similarly, with an ideal inductor you can store energy in it by putting current through it. <S> $$E = \dfrac{1}{2 <S> } \cdot L <S> \cdot I^2$$ <S> But you can get that energy back by reducing the current to zero. <S> The energy you put in is equal to the energy you get back. <S> There is no loss of energy. <S> For a lossy component you get less energy back than you put into it. <S> An ideal resistor for example will take energy in. <S> $ <S> $E = \int <S> V <S> \cdot <S> I \text{d}t$$ <S> But you can't get any of it back <S> so all that energy is lost to heat. <S> Some components will return some energy but less than you put in <A> The energy \$E = <S> \frac{1}{2}CV^2\$ is not dissipated by the capacitor but by the resistance of the wire connecting the battery and the capacitor. <S> The battery transfers a charge of \$Q = CV\$ across its own potential difference V. <S> So the work done by the battery is <S> \$W = <S> CV^2\$. <S> But since the capacitor stored only half of this power, we conclude that half of it is dissipated by the wire resistance and not the capacitor.
| An ideal cap is lossless since there is no mechanism by which it can dissipate energy. There is no loss of energy.
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Is there a technical reason not to use RCA jacks for speaker connections? I'm designing a home-made amplifier and am trying to choose connection options for the speaker outputs; is there any reason that RCA plugs are inappropriate, assuming I control the full system, including making the correct cables? Can they cause interference of some sort? They would be more compact than banana sockets or terminal connectors, but seem rare - is this purely convention or is there a motivating technical factor? <Q> It avoids accidentally connecting a speaker output (100W at 8Ω = <S> > 28V) to a line input (which may not be 28V tolerant). <A> RCA connectors typically have a maximum current rating of 2 amps. <S> For an 8-ohm speaker, that's 32 watts. <S> EDIT - Per a comment by Dwayne Reid, I have found that Switchcraft makes RCA connectors which are rated for 6 amps (I'm impressed). <S> Those puppies will handle nearly 300 watts at 8 ohms (200 watts RMS to be conservative). <S> On the other hand, I've found RCAs that are only rated for 0.5 amps, which amounts to 2 watts. <S> So, you can use RCAs for your speakers as long as you are absolutely sure of your source. <A> There is no reason not to use RCA connectors for (small power) amplified outputs. <S> Plenty of professional equipment uses RCA for line level signals <S> , that is more liable to pick up noise, with no problem whatsoever. <S> Maybe they use some sort of fancy connectors like gold plated <S> but you know... <S> Doesn't really make a difference in your project probably. <S> The maximum power is somewhat limited but probably the cable <S> you are going to use is the biggest problem. <A> I would never use anything other than XLR for a live show. <S> RCA has exposed grounds, which can cause a 60hz hummm if your power to everything isn't filtered and perfect. <S> For any other reason like making a home amplifier... just make sure your ohms rating can be handled by the line, as @WhatRoughBeast suggested.
| RCA plugs can carry some 2A, that is plenty of current for a satellite of an home made amp, just be sure to use appropriately sized cables and you will experience no problem.
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Moving two elements from series to parallel I got a small pizza oven with two elements (top and bottom), they are rated 115V 600W and they are connected in series, as we have 230V in Europe. Everything works fine, but I would like to control those two elements independently (so I will need to connect them in parallel), unfortunately that is not going to work with 230V, what is the best solution, in your opinion, to connect them in parallel and still have them operate within their rated conditions? <Q> You can't just connect them in parallel. <S> If you're fluent with this stuff, you can drive each one directly from 230 V, but with PWM so that the average power doesn't exceed 600 W. <S> There are various ways to do this, including triacs, but there are issues beyond what we can reasonably get into here. <S> You have to consider what happens when the system screws up and applies the full 230 V. <S> In that case the power dissipation will be 4 times intended, or 2.4 kW, for a short while. <S> That will damage the element quickly, and possibly other things around it. <S> Another option is to use a transformer to make 115 V, then drive each element independently from that. <S> That will work fine with less chance of catastrophic failure, but will require a big, klunky, and expensive 50 Hz power transformer. <A> The crude but secure and working solution <S> Since you're working on 110V, now, you can easily connect them in parallel. <S> They are not that cheap though. <S> Any other solution involves complicated electronics which might fail after 5-10 years, damaging your heating elements in the process. <S> A transformer will just work for centuries. :D <S> *EDIT: <S> Why 2kVA? <S> Because you have to account for losses in the transformer <S> and it's better to have a safety margin in there. <A> An even cruder solution would be to find a second oven, and use those elements as ballast resistors. <S> You could leave them inside their oven and just let them get hot. <S> Not the best solution, but lighter and cheaper than a step-down transformer. <A> You can use a rectifier diode in series with each heating element - it should half the average power, so if your heating element draws 1200W @ 230 V, then with diode in series it would draw only 600W. <S> However I'm not sure about safety of this solution.
| :Get a 230V primary / 110V secondary isolating transformer used for US equipment in europe with about 2kVA and use that for your oven.
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Why do we use 32.768 kHz crystals in most circuits? Why do we use 32.768 kHz crystals in most circuits, for example in RTC circuits? What will happen if I use a 35 or 25 kHz crystal? I assume because the IC internal Xin, Xout pin circuitry should be in CMOS/TTL/NMOS technology. Is it that true? <Q> The frequency of a real time clock varies with the application. <S> The frequency <S> 32768 Hz (32.768 KHz) is commonly used, because it is a power of 2 (2 15 ) value. <S> And, you can get a precise 1 second period (1 Hz frequency) by using a 15 stage binary counter. <S> Practically, in majority of the applications, particularly digital, the current consumption has to be as low as possible to preserve battery life. <S> So, this frequency is selected as a best compromise between low frequency and convenient manufacture with market availability and real estate in term of physical dimensions while designing board, where low frequency generally means the quartz is physically bigger. <A> The number 32768 is a power of 2, i.e. it is 2^15. <S> If you have a 32.768kHz clock frequency it is easy to divide it to an 1Hz frequency using binary frequency dividers, a.k.a. binary counters, i.e. chains of flip-flops. <S> Having a 1Hz frequency means you have a clock signal which provides 1s time resolution: count the seconds with a counter, do the math and you have a Real-Time Clock (RTC). <A> It's primarily due to cost. <S> These particular crystals are dirt cheap due to the watch industry. <S> This answer provides more detail, here's an excerpt: <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> As a result, these crystals are extraordinarily inexpensive... <S> [Other crystals] cost 10 to 100 times more in quantity than these inexpensive watch crystals. <S> Real time clocks are expected to run such an oscillator for 10 years on a CR2032 type cell. <S> To get low frequency, low power, small crystals in other frequencies, you're looking at a substantial increase in cost. <S> In low volumes these crystals are still less expensive than even the normal or high power the 25kHz or 56kHz crystals, but cost the difference isn't large until you get into high volume manufacturing. <S> Choose what you need, but if you are going to produce a high volume product and can adjust your design to work with a 32kHz crystal, then there's a substantial financial incentive to do so. <A> With CMOS chips, frequency is related to power consumption. <S> So a 25KHz clock would consume less power than a 32.768 KHz clock. <S> You should do the math to determine your proper minimum/ maximum clocking, coordinated with the actual chips you select. <S> There is a tradeoff between clock speed, power consumption, and the amount of work you can get done per clock cycle. <S> This varies from circuit to circuit. <S> RTCs as a class, are most concerned with power consumption when the main power is off - and you are running on the back-up coin-cell battery, but also still need to be reasonably accurate clocks as well - within a few seconds per month typically.
| Further, these crystals are particularly well optimized for low power. You can use any frequency you desire - provided your circuit is designed for it. 35 KHz clocking would comsume slightly more power.
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What sensors shall be used to detect if a dustbin is full (light weight material like paper,hair, etc)? Considerations:I am trying to detect if the dustbin is full. The dustbin is a rectangular shoe-box like structure without any lid. The kind of material that will be collected will be mostly hair, paper, carpet etc. I am thinking to use optical sensors. But I do not want to go through the pain of cleaning sensors again and again. Are there sensors that can be put in a glass case and they can penetrate through the glass? This will protect the sensor from dust/collected material. Is there a better alternative then optical sensors (IR, depth, laser,weight, acoustic, capacitive, touch, limit switches ). The system needs to be robust. Weight sensors: seems to be unreliable because the weight is not consistent. Touch sensors: The material, say paper, if it touches the sensor then we can think the dustbin is full. But because dustbin is long (not cylindrical), the indication can be false. Also, in case of hair, the weight or touch, neither can help in detection. Multiple IR sensors: These can be mounted and based on blockage, we can detect if the system is full but again I want to eliminate the need of cleaning sensors again and again. <Q> You can always use IR sensor. <S> Rather if you want you can use ultrasonic distance measurement sensors. <S> This way you can calculate the height of the trash in the bin when placed under the sensor. <S> And you don't need to clean the sensor. <S> This saves your thought on mounting the sensor inside or along with the bin and the bin remains replaceable. <S> Programming and getting results from the sensor however depend on the thoughtfulness of the programmer. <A> If the type of rubbish that the the dustbin is expected to hold is rather undiverse, you can find a statistically correct weight that will report full within a reasonable-enough degree of certainty. <S> The dustbin can also cross-reference the time of day to add an additional angle of classification. <S> This, of course, assumes that the dustbin is used in an ecosystem with low rubbish diversity, and periodic fill levels. <S> A machine-learning rubbish bin, this is interesting. <A> We can use Ultrasonic Level Measurement sensor to measure the height of the Garbage. <S> We can place the sensor at the top of the dustbin and we can measure the height. <A>
| Multiple use of the ultrasonic sensors placed opposite each other in the angled manner at all the four sides of a bin can result in analyzing the distance and decisive to account for a bin full status. Another advantage will be that no matter what container you use, the system can be used. But they need to be cleaned.
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Replacing wires with threads? Is it a good idea to replace wires that get bent a lot (e.g., around a hinge) with conductive threads? It's a 5V device using up to 1A. Thanks <Q> Conductive thread is simply thread with wire woven through it, so it won't necessarily be any more robust. <S> The critical thing for reducing fatigue in wires that are flexed a lot is to maximize the radius of curvature. <S> Instead of bending it directly around the hinge, use a generous loop of wire, so that the maximum curvature is very gentle. <A> I wouldn't necessarily use conductive thread for this, because the resistance is actually pretty high -- 10's of ohms per foot. <S> Ohms law says that if you sent 1 amp through 1 ohm (an inch or so of conductive thread) <S> the voltage drop across the thread will be 1 volt. <S> The power through the thread will be 1 Watt. <S> This may be fine for you, but you will have to pay attention to the voltage your load actually sees. <S> I recommend considering Flat Flexible Cables -- but you'd need to pay attention to how many cycles of motion will cause failure <A> Microflex Wire and Cable, among others, makes extra flexible wire for these applications.
| You may also consider looking for cable that is specifically rated to withstand a lot of flexing.
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Explantation for fried charger with Multimeter I have a 9v AC to DC wall adapter that I use to power my Arduino boards. The label says it gives 9v at 1000ma but I wanted to test the validity of the label to troubleshoot. When I connected my probes to the terminals however it said 0 amps and then the green power light on the adapter turned off. After that the adapter would not power anything. I opened it and could not identify any blown or otherwise damaged pieces. So, before I fry something else like the USB port on my computer :(, does anyone know what I was doing wrong? The multimeter still works and gives valid readings on other things and I have measured chargers before without issues. I have this multimeter from radioshack. <Q> The short-circuit current you measure this way will probably be far above the supply's rated output, unless the supply has overcurrent protection. <S> I don't think it is practical to try to verify the rated output current of a power supply by a direct measurement - you just have to trust the maker's claim. <S> You could, however, connect a load of the correct resistance to draw the rated current to the supply, and monitor the output voltage and power supply temperature over an hour or two to see if the supply survives, and continues to output the advertised voltage, and doesn't get too hot. <A> How did you test the current draw? <S> Like you would voltage or by placing the meter inline with the power? <S> Ideally in voltage measurements the meter represents infinite resistance, and in current measurements represents zero resistance. <S> If you have the meter in current mode and short the leads across the output, you're effectively shorting the circuit with zero resistance, which can damage the meter and/or the circuit being tested. <S> In current measurements you place the meter inline with the circuit so that current flows through the meter to measure it. <A> you made a short, the internal resistance of the multimeter is under/ <S> around 1 ohms in ampmeter mode ( in most case). <S> So if you have a 1 ohm res with 9 v apply to it, V = R*I , that is around 9 amps that you are requesting to your adapter. <S> Either, you burn a fuse or you broke your device. <S> Normally, you should put a nominal charge connected to your adapter than you can measure the current. <S> Regards,MathieuL. <A> Further to the other answers, if you want to avoid shorting out supplies it's better if you get a resistor, which you calculate the resistance for by using V=IR, which rearranges to get R=V <S> / <S> I <S> You say the voltage is 9v and the current is 1000mA. <S> There is 1000mA in 1 Amp. <S> Therefore the current is 1A. <S> The resistance is therefore 9/1 = 9 ohms. <S> Therefore, If you connect a 9 Ohm resistor in series with the ammeter, you can measure the maximum theoretical current. <S> That means you won't burn out another supply. <S> Of course, you have internal resistance to worry about, but you can always measure this by putting the meter to resistance mode, and then probing another power supply.
| As others have said, connecting a multimeter set to measure Amps directly across a power supply effectively puts a short circuit on the supply, due to the very low resistance of the multimeter when set to measure current.
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Checking when a car horn is pressed? I want to detect when a car horn is pressed. The only way I can detect it currently is using the voltage on its line.When the car is turned on, there is always 12V being supplied to the horn through the battery i.e. When the horn is NOT being used the voltage to it is 12V. When the horn is being used the line voltage drops between 2V-1V due to the consumption by the horn.I was thinking of using the High/Low voltage detect module of the pic 18f4520 to detect this voltage drop when the horn is pressed. But the module seems to be more for detecting a drop in operational voltage. Is there a downside to using the HLDV in such a manner, also is there a better/other way to do this? Any code snippets to HLVD or other methods are greatly appreciated. Thank you for your time and have a great day. <Q> I'm assuming that you're measuring the voltage with respect to ground, not across the horn terminals. <S> If so, what you are describing sounds like this: Except that the switch is likely a relay, MOSFET, SSR, or some other device that has a non-zero resistance even when "closed". <S> In this case, both wires across the horn will be at the battery voltage when the horn is off. <S> When the switch closes the bottom wire will drop towards ground and current will start to flow. <S> The reason it doesn't actually reach zero volts is because of the resistance of the switch. <S> It is important to realize that the top wire won't vary (much) from the battery voltage whether the horn is active or not. <S> You'll need to monitor the line between the horn and the switch. <S> To interface this to a PIC (or other microcontroller), you will need to scale the voltage so that it doesn't exceed the PIC's Vdd (probably 3.3V or 5V in your circuit). <S> You would do this by using a voltage divider : <S> Say that your Vdd is 5V. Assume that the horn voltage swings from 15V to 2V. <S> If you divide the voltage by 3, the output swings from 5V to 0.66V. <S> This is within spec of the PIC. <S> If Vdd = 3.3V, you will need to scale appropriately. <S> As far as the interface method, you have a few options: <S> The HLVD module. <S> This would work just fine, and has the benefit of "set-and-forget". <S> Once it is configured, you just wait for an interrupt. <S> An analog input. <S> This also works, but you'll need to regularly sample the input. <S> If the scaling works out, you may be able to simply use the horn signal as a digital input! <S> Make sure that when the signal goes low that it is below the \$V_{IL}\$ of the PIC, found on page 335 of the datasheet. <S> However , you'll need to add some extra protections. <S> The power systems of automobiles are very noisy and hard to deal with. <S> You can get false triggers, and large voltage spikes when starting the engine that can damage your microcontroller. <S> By the time you account for these things the circuit can become considerably more complicated. <S> See some of the other answers. <S> @tcrosley even shows how to create a safe, filtered Vdd for the PIC! <S> Good luck :) <A> Here is a circuit which will provide a digital input to your microcontroller, and also provides a filtered power supply for your microcontroller. <S> Transients can be nasty on a vehicle's 12V system, with voltages rising as high as 125v for 10 ms during a load dump . <S> This circuit provides protection against negative voltages in addition to the positive spikes due to load dumps, noise, and jump starting. <S> I have used an LDO regulator for simplicity, but you can substitute a switcher if you want for that block. <S> The Zener diode protects the voltage going into the regulator from going over the 26V maximum. <S> The 1000 µF capacitor will smooth out any noise from getting into the microcontroller and make handling of the signal simpler. <S> So 14V (typical car system voltage) will appears as 4V, and 12V as 3.43V. <S> The other side of the comparator is connected to another voltage divider which provides a 2V reference. <S> A Zener is provided to limit the voltage going into the comparator to 4.7V. <A> For example say it's a mechical relay and fails one day you might find when replaced the internal and external contact resistance is much lower <S> so you start getting a much lower voltage drop. <S> To handle that and some of the other nasty characteristics automotive systems can have I'd consider using an optocoupler instead. <S> In this circuit I've calculated the resistor so that when subjected to a 125 V load dump it will fall within the 50 mA limit of the LED in the optocoupler and the 1N4148 provides some reverse voltage protection for the LED: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The above assumes you can access the positive side that runs to the horn, but should be more reliable than trying to detect the voltage difference to ground and will keep your microcontroller input line isolated from voltage spikes on the other side. <A> Arranged as per below these will serve 2 purposes. <S> D1 will provide a constant voltage drop, this will eliminate the effect of the "On" resistance of the switch. <S> D2 <S> In combination with the R1 limits the voltage to 3.3v and deals with any nasty automotive spikes. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> As a bit of a side-note the disadvantage to this circuit is a constant draw when the horn is off, this can be minimized by increasing R1, however the datasheet will need to be checked as zenner diodes are specified at a certain current.
| The horn circuit uses a comparator to generate a digital signal which indicates if the horn is operated. Bitsmack has provided an excellent description of why you're seeing that voltage drop, but it's also one reason I'd avoid doing what you're proposing of relying on that voltage drop because it might not be reliable in the longer term. The circuit uses a voltage divider to cut the nominal 12v horn voltage to 4/14 or 29% of its value. I think more robust alternative on bitsmacks answers is to use zenner diodes. Also if you move your circuit to other vehicles they will have different characteristics as well.
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Why is it so hard to find component footprints? When designing PCB's, I find myself very often having to make footprints for a significant portion of the components on my board. This tends to be very time-consuming, as (in Altium at least), dimensioning out land patterns for strange connectors or chips (those that can't be created from a wizard) isn't very easy. It seems like anyone that uses these chips or connectors would need a footprint, so I can't understand why these aren't more commonly provided. For example, right now I'm trying to put a USB 3.0 Micro-B connector on a board, but the top 5 connectors on Digikey don't seem to provide footprints. I have access to the Altium Live design content, but even that seems often pretty out-of-date. I feel like there's something obvious that I'm missing - or else this system seems very inefficient (which usually isn't the case). Can someone enlighten me? <Q> When I was working as an engineer, I wondered the same thing, which is why I decided to create SnapEDA. <S> SnapEDA <S> is a CAD library of 25 million electronic components, for which we provide PCB footprints and schematic symbols. <S> Our PCB footprints convert to Altium, OrCad/Allegro, Eagle, KiCAD, & Pulsonix. <S> We run a diagnostic test on each CAD file, which we'll be making public in the next month or so. <S> This checks for different aspects that might go wrong with mappings, silkscreen overlaps, etc. <S> In the future we plan to expand to other forms of design data as well. <S> Would love to know what you think if you have some time to check it out. <S> We love getting feedback, and everyday we are continuing to refine the product to make it even better! <A> You've discovered the dirty little secret of the EDA industry: Thousands of engineers everywhere reinvent the wheel every day - they all create many of their sch symbols & pcb footprints from scratch. <S> It is quite ridiculous. <S> However there are reasons for it, in particular no universal (or even common) file format (nor for the schematics & PCB designs either), and that's in large part the fault of the various EDA software developers, who rely on this lack of file format compatibility to keep customers locked in. <S> Until recently there's also never been a way to have confidence in 'random' other people's sch/PCB-footprint designs, so <S> E.E.s err on the side of caution and make most of them themselves. <S> But now there are some options, like snapeda.com and circuithub.com. <A> In the mechanical world a 4-40 screw, is a 4-40 screw. <S> In electronics, we don't have that luxury. <S> Every PCB is different. <S> The footprint for even an 0805 resistor will be different for wave vs. IR reflow vs. hand soldering. <S> Some boards are small and dense, other larger and sparse. <S> It's just easier to assess the design requirements, and tailor footprints to fit them. <S> Never mind the issue of having to go through and verify that Joe Blow's footprint he put in some library is correct, and then making it fit the design requirements. <A> That tool is really easy to use. <S> Literally, you just copy/paste the measure of the mechanical data furnish by your supplier and the IPC wizard take care of drawing you a sweet footprint. <S> I made a 62 pins MCU footprint in less than 3 min with that tool. <S> Here a little demo from youtube: <S> https://www.youtube.com/watch?v=8V0ZfLsp8gY <S> If you don't want to pay for the IPC wizard, AD include a component wizard in the footprint creator, however it is more hard to use it and often you end up with a footprint that need manual rework. <S> So you can understand that the main reason why footprint aren't online is because most people are using footprint generator therefore, it take less time to literally make a footprint than search for it on the internet.
| If you are willing to spend money, you can acquire the IPC footprint wizard tool: http://landpatterns.ipc.org/default.asp
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Voltage and current measurement of battery Single ended or differential? I am trying to measure battery's current and voltage for a battery charging/monitoring project. I have read all about current sensing (including high side & low side sensing). And i have decided to use Shunt resistors for current measurement as they are accurate as compared to other current measuring devices. My battery would be a Li-Ion Battery, and the maximum rating of this battery stand would be (4.3V , 40A). However i am confused on how to measure the voltage and current using an ADC, i-e whether it should be measured single endedly or differentially. A very rough sketch of my circuit is given below.(This ADC would be interfaced with a microcontroller) The battery can be seen connected to a buck converter for the charging purpose. And and ADC can be seen as well. ( Please note that my sketches might not be accurate, but i mean all what i have written here and in the diagrams ) What i think is that, if i try to measure my Battery's voltage and current in this way (shown in the image below), my voltage would be differential(since battery's negative terminal is not directly grounded, theres a shunt in between) so i have to feed it to a differential input ADC, whereas the current is to be measured singe endedly, as one leg of the shunt is grounded. And if i try to measure my Battery's voltage and current in this way (shown in the image below), my voltage would be singe ended (since the battery's negative terminal is directly grounded), and my current measurement would have to be done differentially (as my shunt is placed in between my supply and the battery). Now, i am no expert in ADCs, but as far as i have read about them (also their data sheets), if an ADC has both single and differential ended inputs, we can use it as a single ended input ADC OR we can use it as a differential ended input ADC. Which means we cannot use it as both single and differential input at the same time. Which brings me to my question. What could be a solution to it ? Shall i use 2 different ADCs, one for single ended input and the other for differential ended input ? Or can i measure both the current and voltage differentially and feeding them both to a single ADC configured as a differential ended input ADC ? P.S i am not looking forward to use a single to differential ended AMP , as i am supposed to measure these quantities with as high accuracy as possible, and introducing such AMP would decrease my system's measurement accuracy. So it leaves the question, whether i can measure both the quantities differentially ? like given in the picture below, which is just feeding the voltage measurement connections to the '+'and '-' input of a differential ended input ADC. As the Battery's negative terminal in this case would be at ground potential, so can it be fed to '-' terminal of a differential input ADC ? (Since i dont have much knowledge in field of electronics, i dont know if it would be possible or not, or what i am asking here is totally stupid) You helful comments would be really appreciated, Thankyou. Thankyou. <Q> I think all your proposed solutions are possible good solutions. <S> If implemented properly I think it does not matter that much if you go for a full differential or partly single-ended solution. <S> But in general differential circuits are less sensitive to external disturbances. <S> Do make sure that the (differential) amplifiers have the right voltage gain such that you will be using the ADC's full range. <S> Using a low-pass filter between the shunt resistor/ battery and amplifier input might be sufficient to suppress this noise enough. <S> Some averaging of the measured values from the ADC might also help and improve accuracy. <S> And kudo's for doing your homework <S> and you already know a lot more than many answer-seekers on this forum ! :-) <A> ...as far as i have read about them (also their data sheets), if an ADC has both single and differential ended inputs, we can use it as a single ended input ADC <S> OR we can use it as a differential ended input ADC. <S> Which means we cannot use it as both single and differential input at the same time. <S> This is not always true. <S> For example, I have used ADS1015 on a couple of projects recently. <S> On this chip, whenever you switch the channel being read you also have the option to switch between single-ended and differential measurement. <S> (This is not an endorsement of this chip for your project. <S> Just an example of a chip that doesn't have the limitation you thought was universal) <S> Furthermore, even if you had a device that had to be configured as single-ended or differential for all channels at the same time, nothing prevents you from using ground as one of the inputs to a differential channel. <S> So you could just configure it as differential and move on with your design. <S> The only thing you'd lose is the opportunity to use the 4th input pin for some other purpose. <S> Another option, if you plan to use external signal conditioning, you can do differential to single-ended conversion in the signal conditioning circuit, and your ADC will never know the signals are anything but single-ended. <S> This essentially makes the amplifiers shown in your diagrams to be external devices rather than internal to the ADC chip (and add some filtering in their feedback networks to reduce noise). <A> What value are you using for a sense resistor? <S> A lot of 4.3V LiIon battery chargers/monitors use 10mOhm; at 40A, you would generate 400mV and burn 10mOhm <S> * (40A)^2 <S> = 16W. Seems pretty wasteful, not to mention it will be expensive to pay for a precision resistor at that rating. <S> Get a resistor small enough that the voltage across it is negligible; then you can measure both the battery voltage and current single ended. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You can also calibrate out the drop on the sense-resistor with VBAT' = <S> VBAT - IBAT <S> * RSNS.
| Another potential issue, since you will be using a switching converter, there will be switching noise on the measured current and voltage.
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Can I use a line multiplexer as a level shifter as well? I have 40 x 5V inputs that I want to read using 3V GPIO pins on the beaglebone black. I am planning on using 5 multiplexers (such as the 74HC151 or 74HC251) that will be sharing the 3 select lines. I'm not sure what the best strategy for level shifting would be: Power the multiplexer with 3V VCC and hope it'll work with 5V inputs Throw in a level shifter / buffer before each multiplexer, something like 74LV245. Use a series resistor to limit current on each of the inputs since I just need unidirectional 5V->3V shifting (Yes, I know this won't change the voltage but perhaps it'll protect the multiplexer from the 5V signal by reducing current?) Is there a different multiplexer I should be looking at that will be tolerant of the different voltages? I'm really hoping I can do (1). The datasheet says input voltage should be between GND and VCC. How bad will it be to power it with 3V and leave the inputs 5V? Thanks! <Q> Why not just use an open collector multiplexer powered by 5V and pulled up to 3, like the 74LS156? <A> You can use a voltage divider on each signal. <S> For example 18K + 33K will give you about 3.2V out for 5V in. <S> That will slow the signals somewhat, but if you are not concerned with MHz frequencies it won't make much difference. <A> Another option is to use CMOS buffers/inverters of the LVC family. <S> They can be powered at 3.3V, but their input are 5V tolerant, i.e. they will handle and recognize TTL logic levels on their inputs. <S> Sadly they are only provided in SMT packages, so no through hole parts in that family. <S> For example see 74LVC04 inverter datasheet (NXP) or 74LVC244 line buffer datasheet (NXP) . <S> An excerpt from the front page (yellow emphasis mine): <A> Thanks for the answers! <S> I am going to end up using a single octal buffer (74LV245) to shift the 3 shared select lines and the 5 individual data lines.
| A single additional IC is easier and faster than 40 voltage dividers, and it'll also be better to fan out the 3 shared select lines to the 5 multiplexers when going through a buffer instead of directly from the GPIO pins. If you have high frequencies, by all means use a level shifter chip.
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How to solder micro stepper motor pins? I have a hobby project where I would like to use micro stepper motors as shown in the picture below. However, as you might be able to see, the spacing between the pins is incredibly tiny and each pin is much narrower than the 24 AWG wire I was planning on using. In addition, each pin seems to have a single strand of copper wire attached to it presumably attached to the stepper windings. So my question to us all ... what technique might one use to attach wires to these pins? Assume poor/average soldering skills and "normal" mans hands with normal steadyness (and by that I mean (what I hope are) the normal amount of shakes). Here is a photo I took myself which shows the scale against a rule. <Q> I would probably use a small amount of tacky flux and pre-tin both the wire and the pins. <S> If you can secure the motor in a clamp with pins facing horizontally, that would be ideal for me (an average solderer). <S> I would feed the wire with my non dominant hand while resting my palm on the bench top and soldering with my dominant hand. <S> Hope that makes sense. <S> Pre-tinning will make it relatively easy. <A> The pins on each motor seem to be all in a single plane; perhaps you could make a custom circuit board that fits those pins and functions as a breakout board. <A> Some guesswork: <S> What is the length of those pins? <S> you wrote about alterating pin pattern (two rows), so maybe they are designed to be soldered dircetly to a PCB? <S> They seem short, so not full through-hole, but rather push-me-somewhat-into-hole style. <S> With tiny PCB you could presolder the points on PCB then press the PCB to motor (or use soldering paste), then heat the PCB from the other side. <S> Hm.. or a plastic connector with four slots.. <S> but my, they are short. <S> No, I'd bet on tiny PCB. <A> Support the soldering hand to reduce shaking. <S> Preferable wrap them a couple of rounds. <S> Cut the excess wire. <S> Use finest tip possible and a fine soldering wire. <S> Add a tiny drop of soldering wire to the tip and apply to wire pin (some flux might help). <S> Hard to see exact size, <S> but I guess the motors are about 6 mm i diameter, and the terminal less than 0.3 mm. <S> AWG 34 <S> (0.19 mm) or AWG 36 <S> (0.15 mm) wire is probably needed.
| If the motor and pins are fixed in horizontal plane with a clamp (as suggested by Stihl Alighve), it should not be that hard to hang the wires around the pin and then solder (make a J in the wire end).
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True GPS Location - At the Antenna or Receiver Chip? Suppose I have a GPS unit attached to an antenna through a 50 meter coax cable. How would the location as calculated by the GPS unit be affected by the cable length? As a bonus question, how would the time accuracy of the GPS be affected by the cable? <Q> The exact position is the phase center of the antenna independent of the length of the cable and location of the chip. <S> The time delay has to be calibrated by measuring the delay of the cable for the band. <S> (L1 band). <S> Many GPS receivers provide option to key in the delay parameter. <A> Cable delay adds an equal offset to the pseudoranges for all satellites. <S> Since GPS uses the difference in the pseudoranges to each satellite to calculate the position, positioning isn't affected by cable delay. <S> The time calculated by the GPS receiver will have an error equal to the cable delay, which is the length of the cable divided by the propagation velocity of the cable. <S> The RG174 commonly used on "puck" antennas has a velocity of 0.66 c , which is about 5 nanoseconds per meter. <A> As already mentioned, the position is determined by differences in signals received by the antenna. <S> So the cable and chip will be irrelevant for that. <S> However, when it comes to timing things get tricky. <S> As mentioned, you can calculate how long it takes for your signal to travel through the cable and correct for that, but when I did an experience a few years ago, we actually found the variability in timing to be in the order of microseconds. <S> So you can correct a few nanoseconds for your theoretical cable delay, but in practice the uncertainty in timing may be much larger.
| The position calculated will be at the antenna, not at the receiver, which you can see by realizing that moving the antenna has a different effect on the pseudoranges to different satellites due to geometry, but moving the receiver has no effect at all (the cable length stays the same and so does the cable delay).
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Debugging when oscilloscope grounds cures the problem In a nutshell : My circuit behaves erratically I connect the 'scope to find out why The problem disappears In particular, this concerns the circuit in my last question although this is not the first time it's happened to me and I'd like to know what's the correct way of finding the source of this general problem. In software, we'd call these Heisenbugs but I don't know if the same pun is used for EE problems. In this particular instance, I have a test program running on the PIC that goes from 0% to 100% duty cycle on the PWM output in 8 steps, pausing for 10 seconds at each step. Then it goes back down again from 100% to 0%. The problem is that it goes up ok, but gets stuck coming down - i.e. the fan doesn't fall in speed as it should. Just connecting the ground of the probe to my circuit ground fixes the problem, even without the scope switched on. When the scope is connected and running, all probed signals look clean and tidy and everything works perfectly. I'm guessing that I'm picking up some interference from mains hum or from the power supply, but without being able to observe it when it's misbehaving, I don't know what it is I should be fixing. What do I do next? Schematic: Board: The 2x5 header at the bottom is there just to expose all of my unused PIC pins in case I want to extend this in future (it's a hobby project). Fan connector is at the top. <Q> There are three main possibilities -- there is some external system you're connecting to, and you haven'tpassed a reference voltage between your circuit and this system. <S> Ifthe external system is connected to earth ground, when you attachthe ground of your scope cable to your circuit, you are now passinga reference and fixing the problem. <S> (my first bet if I were a betting man). <S> Something wants a little more capacitance and the scope probe is providing it <S> Some spurious currents are looking for a better path to ground, and the earth ground is providing it. <S> The first option is probably debuggable by extension, the second by adding caps as others have described. <S> The third might be caused by currents associated with voltage spikes associated with PWM control of a motor, and missing the diode you need to deal with this. <S> You might consider just replacing the fan with a few resistors (i.e., a non-inductive load) in parallel (sufficient number to deal with the power generated) and see if the problems go away. <S> One more <S> issue-- <S> If GND in your circuit is chassis ground (i.e., the "Ground" jack on your power supply), and there is no internal connection between V- and ground on the supply, you need to make a connection between chassis ground and V- or there is no reference to V+, or, instead of hooking your circuit ground to chassis ground, you use V- for GND in your circuit. <A> Unlikely, but easiest: Try a 10 µF capacitor acorss the input of you 78L05. <S> Chances arethat your supply (5 V) becomes unstable, causing the microcontrollerto... um... go all Heisenberg? <S> Very (!) <S> likely: Also, a somewhat bigger capacitor acorss your 12 V supply inparallel with a low-impedance capacitor (e.g. <S> 100 µF <S> electrolytic,100 <S> nF ceramic) <S> across your 12 V input is a really good idea. <S> Yourfan motor looks much like an inductance, and the spikes created whenQ2 turns off might disturb your 5 V regulator and/or yourmicrocontroller. <S> Ideally, connect them such that the loop from the positive end of the cap to the output of the fan, and from Q2.source to the negative end of the cap becomes as small as possible. <S> Very likely and really a thing you should try, too: <S> Put a (fast!/Schottky!) <S> diode from Q2.Drain to +12V,with the anode connected to the drain and the cathode to +12V, <S> right where your capacitors are. <S> Thisdiode will catch the spikes and clamp them onto the capacitors youjust added in (2). <S> You can actually probe from Q2.drain to ground and check if the drain spikes go way above 12 V, or maybe even above Q2's max. <S> allowed drain voltage. <A> With due respect, regardless of what the scope ground does, your layout is abominable. <S> Particularly when PWMing, you MUST maintain better ground routing. <S> As it stands, current from the sources of your FETs runs on thin little traces through the PIC ground, then to the regulator ground and finally to your input pin and decoupling. <S> I suspect you're getting ground noise like crazy. <S> Why the scope lead fixes this <S> I have no idea. <S> I would suggest placing JP2 just above your FET, with at least a 0.1 wide trace from the ground pin to the sources of Q1 and Q2, and C3. <S> Then a separate trace, at least .05 wide to your PIC, regulator, C1 and C2. <S> For now, run a short jumper, say 20 ga, between JP2 GND and Q2 pin 3, and a 24 ga jumper from JP2 GND to Q1 source (pin 1, I think). <S> In the future, always run power and ground (especially ground) first. <S> Use wide traces and run as directly as possible. <S> Only then can you consider routing problems and strategies for the other traces.
| You may just have your power wired up incorrectly.
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What can the ohms of a DC motor tell me about itself? I have a lot of motors and all type of size but mainly they are all small to mid size adult hand size or so , I took out my multimeter and switch to ohms settings and grabbed the ground and power wires of each motor to see what type of reading I would get, what confused me is that, I found that most bigger ones had high ohms of 1.5 to 2.5 ohms while the smaller ones had higher ohms of 4.5 to 14 ohms. Can someone break it down for me? <Q> <A> For a given motor, torque is proportional to current, back emf to rpm. <S> This means that at some rpm, the back emf is sufficiently large that the difference between supply voltage and back emf is so small it cannot generate a current larger than the friction. <S> This is the maximum no-load rpm (for DC motors, but AC work similar). <A> I have a brushed DC motor with 13.5 0hms resistance. <S> I use that to calculate the STALL current at maximum torque <S> (More current = More torque). <S> That is pretty much accurate with the datasheet spec. <S> At 3V, when motor is stalled current shoots to about 3/13.5 ~=170 mA
| In general terms, the motors with the lower resistance will draw more current (ohm's law), and therefore use more power, and (hopefully), generate more mechanical power.
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Numbering format in code syntax For the code, which I am referring following is syntax: Write_Parameter(0x00); In above line, what does "x" stands for and what is data size for x? My assumption is its dec format. So, What other format, we can use? Thank you. <Q> The number of digits after the x represents the number of bits, in multiples of 4: 0x0 - 4 bits (or one "nibble")0x00 - 8 bits (or one <S> byte)0x0000 - 16 bits0x000000 <S> - 24 bits0x00000000 - 32 bits <S> But as Nick Johnson points out, regardless of the number of digits in the constant, in C a numeric constant is treated as an int unless it has an l or L suffix, or preceded by a cast. <S> On 8 and 16-bit machines, an int is usually 16 bits, and on 32-bit machines it is 32-bits. <S> Because the number is hexadecimal, each digit can represent one of 16 values, 0-9 and A-F (A=10, B=11, C=12, D=13, E=14 and F=15). <S> Each digit position, going from right to left, represents a hexadecimal "nibble" or four bits, with a placeholder value of 1, 16, 256, 4096, 65536, etc. <S> So 0x0ABC, for example would equal decimal 2748: <S> 4096 <S> 256 <S> 16 <S> 1 <S> 0 <S> A B C = <S> > <S> 0 <S> *4096 + 10 <S> *256 <S> * 11 <S> *16 + 12 <S> *1 = 2748 <S> The 0 after the x could be omitted and you'd end up with the same thing. <S> The largest unsigned value in each of the fields above is: 0xf - 15 (2⁴-1)0xff - 255 (2⁸-1)0xffff - 65535 (2¹⁶-1)0xffffff - 1677215 (2²⁴-1)0xffffffff - 4294967295 (2³²-1) <A> If you are using C (you did not specify), data can be represented in the following ways: <S> Numbers starting with 0 are octal (eg 0123 is binary 001010011) <S> Numbers starting with 0x are hexadecimal (eg 0x1A is binary 00011010) <S> Numbers starting with 0b are binary <S> Numbers starting with digits 1-9 are decimal 'a' is a character which represent the decimal value 97 (binary 01100001) <S> Some compilers (not many as this is not part of standard C) accept 0b as binary. <S> Gcc does not (Thanks @Rev1.0 for the correction!). <A> In C (and many other programming languages) "0x" indicates a hexadecimal number. <S> Since your example shows two digits after the "0x", that parameter must be 8 bits. <S> Each hex digit represents four bits. <S> "0b" would indicate binary. <S> If no prefix, the number is decimal.
| 0x followed by a series of digits means a hexadecimal number in C and many other languages (other common formats are decimal, octal, and binary).
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How to tin a soldering iron after unboxing? I have bought cheap soldering iron before but when I turn it on waited to heat up, it turned black and wont take a tin. Am I doing it wrong? I wasted 2 soldering iron like this. It turned all black and not tinning.Please can you suggest what to do after plugging it for the first time .I don't know how to tin a soldering iron thats why I wasted two soldering iron. <Q> Assuming you mean really, really cheap cheap like the one below, then there is a way out of the mess. <S> The points on soldering irons like that are simply pieces of copper. <S> You can file the black stuff off and make the surface smooth again. <S> Tin the clean, shiny copper surface with solder. <S> Clean the tip often using a damp sponge. <S> This means while it is hot and you are using it. <S> Unplug it if you won't be using it for a while. <S> That kind of soldering iron is completely unregulated and will get very hot. <S> Th black gunk will accumulate faster when you just let it sit and heat up. <S> My father used to use a bit of electronic solder that had a high silver content to tin the soldering iron, then used normal solder for soldering. <S> The idea was that the silver solder would prolong the life of the tip. <S> Don't know if did any good. <S> The cheap things are also handy if you live in the boondocks and can't just buy a replacement tip at need. <S> They are easy enough to make out of heavy copper wire with a file and a threading die. <S> For the one in the picture, you wouldn't even need the die - it isn't threaded. <A> When you buy a new iron there are several things to remember. <S> The first is that cheap is rarely good enough. <S> Then again, I bought a $7 pencil type iron from RadioShack once and it's still going strong. <S> It's all about care. <S> The trick is to apply the solder AS it's heating up. <S> Do not wait for it to get hot before touching the solder to the tip. <S> That would give the tip time to burn (which, from the sounds of it, is exactly what happened). <S> Hold the solder to the tip while it's heating up to make sure you tin the tip before it has a chance to turn black. <S> Filing off the black really isn't the best thing to do--It leaves microscopic holes and fissures in the tip <S> which A) Reduces the effective surface area, and B) promotes burning inside the fissures where the tin doesn't go. <S> However, since you say you have a cheap iron, it may not make a huge difference. <S> I really recommend springing for a Haako or a Weller as soon as you can afford it. <S> Even used ones are generally very good. <S> I have a Weller that's about 25 years old <S> and I don't have a problem with it at all. <A> I had a similar problem with an inexpensive model/tip. <S> I picked up some 'tip tinner / cleaner' from radio shack for a few dollars and have been tinning along just fine ever since. <A> A sponge should work in general, but you can also dip the hot but not too hot tip into some flux. <S> That should improve the tip, but be ready to tin it immediately.
| When I used the old cheapy soldering irons, I would just tin them with regular solder and file smooth and re-tin as needed.
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How to work with the MCP1702? I'm reading the MCP1702 datasheet and there's the typical circuit on page 2 depicting a 9V source and a 3.3V 50mA output. My question is: How come the current output is 50mA when the spec says that when Vr < 2.5V, Vin > 3.45V the min. output current is 200mA? How is the output current determined? Why is it 50mA? Thanks for the help <Q> It's an example circuit. <S> They could have written 35mA. Power dissipation is (Vin - Vout <S> ) * IL, so 285mW in this example. <S> So in the SOT-23 case we'd have a junction temperature rise of about96°C. <S> If we allow a maximum junction temperature of 125 <S> °C (the highest steady state temperature for which the characteristics are guaranteed- see table below), thecircuit would be good for an ambient up to 29°C, which isn't veryimpressive. <S> Maybe they should have written 35mA. <S> It would be better in the other packages. <S> SOT-89 <S> would only rise 44 <S> °C <S> so would be good to ambient 81 <S> °C (or higher current at a lower ambient). <S> These numbers are calculated from this table: <S> They may be optimistic because they assume a 4-layer board, probably with ground and power planes, so caveat emptor. <S> Edit: <S> Yes, it does assume 1-oz power and ground planes, so if your design is 1 or 2 layer it will not perform as well: If you only require brief pulses of current from the source, such that the junction does not heat excessively, the higher current rating may indeed apply, but you would have to pay attention to the transient thermal characteristics to determine if those operational conditions were acceptable. <S> In design it's important to make sure that all the constraints are satisfied. <S> It's no good saying that the regulator is supplying current as specified if it shuts down a second later (and eventually fries) due to overheating. <A> Look at the datasheet carefully. <S> It says that for flavors with output voltages (Vr) of 2.5 V or more, the output current can be up to 250 mA. <S> The lower output voltage variants have less output current capability. <S> The datasheet then lists the maximum current capability for different Vin levels. <S> This is all quite clear and well specified. <S> I don't understand what it confusing about this. <S> As for the circuit on page 2, it is completely consistant with the specifications on the next page. <S> In this case, Vin = 9V and Vr = 3.3. <S> According to the specs, the part in that case is capable of 250 mA output minimum. <S> They are running it at 50 mA. Since 50 mA ≤250 mA, I again don't see a problem. <S> The circuit they show uses the device within spec. <S> Note that this is just a example circuit, and you are free to make your own circuit and use the device within any of the conditions specified on page 3. <S> Note also that in this example circuit <S> the chip is dropping 5.7 V. <S> That times 50 mA means it is dissipating 285 mW. With such a large voltage drop, the power dissipation becomes the limiting factor on output current, not the maximum output current spec. <S> Again, it's a example . <S> (Of course one is left to wonder why use a LDO when there is 5.7 V of headroom available, but that's a different issue). <A> With 9V in, 3.3V out and 50mA, thats 285mW dissipation for the device. <S> The SOT-23 package has a Junction-Ambient thermal resistance of 336C/W, so that means at 285mW <S> the device's internals will be almost 96C above ambient. <S> The max junction temperature is 150C, so under those conditions, the highest ambient temperature at which it can operate is 56C - <S> which is not unexpectedly high if its in a sealed enclosure with other power-dissipating stuff.
| Its probably because of power dissipation, thermal resistance of the package & max junction temperature.
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Can one instantiate RAM with 2 read ports and 1 write port as IP in Quartus? As part of MIPS design we have something called as register file. It only has 32 registers each 32 bit which only makes 1024 bits or 128 bytes. I am not sure how to tell Quartus to instantiate this as a memory block with 2 read ports and 1 write port. The Dual Port RAM in Quartus seems to be having 2 read/write ports. It is not possible to have 2 read and 1 write port. How do I create a 2 read port + 1 write port memory using the memory blocks in Altera Quartus II? <Q> There's a trick for doing this in Xilinx technologies : instantiate two memory blocks, with their Write ports tied together. <S> (Each block is configured to have 1 read and 1 write port, independently addressable). <S> I can see no reason why the same pattern wouldn't work in Altera FPGAs. <S> Last time I looked, their memory blocks had similar dual-port (just not 3-port!) capability. <A> Given from the comments your design writes and reads at different times, you can actually use a True Dual Port RAM for this. <S> For one of the read ports, you would wire up port A so that it is say read-only (tie the write enable permanently to 0). <S> For the other read, and the write, these would both share the same port on the RAM. <S> You would use a multiplexer on the address input which is controlled by the write enable signal - so basically when you perform a write, the multiplexer will select your write address and connect it to the address pins of the memory. <S> When you are not doing a write, the read address will be connected to the memory allowing you to perform read operations. <S> This should work because as you say you never need to write and read on that port in the same clock cycle. <S> If you need more clarification, I draw up a diagram to explain further. <A> Your options are: delay loading the second operand until the next pipeline stage delay writing the result <S> if the currently decoded instruction has two register operands <S> implement the register file in LUTs.
| You can't, as far as I know, because there is only the logic for two ports in the hardwired parts.
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Is a buck regulator the best option for my needs? I'm looking at using an MC33063 buck/boost inverter in a QFN package to step down 5-30V to 4.2V. This will be used to power an MCU, some modules, LEDs cosuming about 200mA max, optionally to charge a small LiPo, so the 1.5A is more than enough.What I would like to know is if this is the correct way to go considering space is of the essence and I need to use as little extra parts as possible, ideally low part heights.The additional diode, resistors and caps I saw in example schematics could probably be made to fit.Everything is SMD.Can I expect to be able to go as low as 5Vin or more likely around 6V? 1.5A isn't mandatory, but I would rather not go down to 0.5A, just in case, so 1A is a good compromise. <Q> This will be used to power an MCU, some modules, LEDs cosuming about 200mA max <S> Linear technology parts are not cheap (I haven't checked on this part) but it has a low parts count. <S> The picture below directly converts to 3V3 but, by using resistors connected to Vfb, higher regulation voltages can be obtained: - GOALPOST CHANGE: <S> 1.5A isn't mandatory, but I would rather not go down to 0.5A, just in case, so 1A is a good compromise. <S> Or, plug your numbers into the LT parametric engine here . <S> I chose to tick the radio box marked "synchronous" because then you don't need to worry about the flyback diode. <A> Rathyer than use a buck-boost converter, like the MC33063 you referenced, I suggest using a buck converter like the <S> L5973AD . <S> It is available for $3.08 from Digi-Key. <S> The L5973AD has a switch current limit of 2A, so it can deliver over 1.5A to the load. <S> With the input and output voltages specified (Vin=5V, Vout=4.2V), the efficiency is approximately 87%. <S> The package can dissipate 2.25W, with the temperature below 60°C. <A> A buck regulator suits your need BUT <S> the 33063 may not. <S> Input voltage range is OK (they work down to about 3V in) but power dissipation may be a problem at higher voltages. <S> 1.5A is the switch voltage so will be 2x to 4x Ioutmax at minimum Vin. <S> Assume 300 mA load at 4.2V = <S> 1.26W. <S> Regulator dissipation will be ~= 1.26W <S> x Z/(1-Z) where z = efficincy & 0 < Efficiency < 1 At low Vin & Vin > Vout 33063 efficiency may be poor due to drop across non saturating darlingtpon switch. <S> 33063 data sheet here <S> The example in 9.2.1 on pages 9 on has a similar Vin and power rating and only 62% efficiency. <S> While that's an inverting converter the reasons for low Z largely still apply. <S> At say 66% efficiency and 1.2 W out power in regulator = <S> 1.2 x <S> (1-0.67/)/.67 <S> ~+ 600 mW. <S> Whether that's acceptable in a QFN pkg needs checking - but it's annoyingly high. <S> Also, the 33063 has a nominal fmax of 100 kHz (page 7) and 7.9 fig1 suggests you MAY get 200 kHz+ in a gopod day downhill with the wind behind you. <S> But at say 100 kHz the inductor size is large by modern standards. <S> The 33063/34063 is often still an excellent choice in many applications, but this is not one of them. <S> There are many MUCH higher efficiency ICs, usually with synchronous rectification, and MUCH higher switching frequencies - 500 kHz to maybe 2 MHz + - allowing much smaller inductors. <S> In trhis case something else will be a better choice. <A> As already said above a synchronous step-down converter would probably be a good choice. <S> Have a look at the MP9942 datasheet from Monolithic Power Systems. <S> It's available at mouser, quite cheap and comes in a small package.
| Unlike LDO regulators, in general with a buck regulator you should be able to operate it with the input voltage as low as the desired output voltage, in your case 4.2V.
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How could I read a digital input (an open/closed switch) from a smartphone without a digitizing board? I have to read a single digital input signal (a trivial open/closed-circuit push button) and use this signal in an Android application on a smartphone. Of course, I could connect a tiny Arduino-like board to the USB port of the smartphone, use the digital input pins of this board to read the electrical signal and pass a message to the Android app but this task is so simple that I would like to avoid the use of any addictional board. So, I wonder if it would be possible to (ab)use a regular headset wire for this task. I would like to connect a regular 3.5mm headset to the smartphone, cut away the actual speakers and microphone, connect two of the resulting free wires to my push button and read the open/closed-status change of the switch from my app. As long as I can see, Android has a ACTION_HEADSET_PLUG intent and a AudioManager.isWiredHeadsetOn() function so this is probably possible. Does anybody have any idea about the correct wiring I should use? In particular, should I insert a resistor/capacitor to protect the inner circuitry of the audio board? Any suggestion about the software side? Please note that I'm only interested in reading the change of the switch status from open to closed or viceversa. I do not need to know the current status of the switch. PS: No, I'm not willing to destroy the smartphone and connect the external push button to the existing keyboard. <Q> If you have the kind of headset with built in volume controls, you could wire your switch to the mute or volume buttons and detect that being pressed from your app. <S> In fact...you could probably just use the existing button. <S> That would be the easiest and most reliable to detect. <A> There is no external access to the wires for this switch. <S> I would use a 555 to create a tone, say 1 kHz. <S> Connect the output to the MIC input of the headset jack. <S> Then use the switch to turn the tone on or off. <S> Note this would require you to poll the input on a regular basis though. <A> I'm answering myself to add some more information. <S> Android headset control system can read up to four different buttons on the standard microphone line. <S> They are normally used for answer/refuse, play/pause, scan and other commands. <S> Each button is assigned a different resistor and is recognised thanks to the resulting resistance on the line when the circuit is closed. <S> Mic is assigned the higher resistance (5000 ohms), the play/pause button is assigned the lowest one (0 hms) and the other buttons are in the middle. <S> Android platform software supplies a MEDIA_BUTTON intent to intercept and handle these "events". <S> You just have to create a single broadcast receiver and an a handling function for each event. <S> Using this HW/SW stuff it is possible to read up to four digital (1/0, open/closed) input lines and even an analogic one <S> (the Mic, that can be used a an oscilloscope). <S> Of course, you have to use a standard headset 3,5mm jack to create your own "digitizing board". <S> Here some reference and some useful link. <S> Hardware side Android specifications supply all the required information here: <S> Wired audio headset specification (v1.1) <S> (I do not know if all existing devices comply with these specs.) <S> Software side <S> There are a few guides/tutorials that explains how to detect and handle events coming form the headset buttons (both wired and Bluetooth): <S> Allowing applications to play nice(r) with each other: <S> Handling remote control buttons Add Headset button support to your Android application <S> And there is a complete sample application on Github: gauntface/Android-Headset-Example <S> Here is a commercial app that uses all of the above to implement a better control for the headset buttons: Headset Button Controller (Trial) <S> Funny stuff, really. <A> No wires, and easily supported by any app you create. <S> They are very common at the moment, part of the selfie stick craze. <S> Another wireless version is nfc buttons. <S> Same concept.
| You cannot make use of the headphone switch built into the phone, since this is a mechanical switch that opens or closes depending on whether a jack is plugged in. Current answers about the headset buttons aside, a simple way would be blue tooth button as well. There should be a way to determine if there is audio or no audio going into the MIC input.
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8 bit MCU with 12 bit ADC, possible? I am trying to measure battery's voltage and current but with a very high accuracy. So i've selected a 12 bit ADC to be interfaced (via I2C) with my micro-controller. My micro-controller will receive these values form the ADC and send them to the PC via the USB. Now my question is, can we interface a 12bit ADC with an 8bit MCU like this one ? Being new to microcontrollers i did a research over the internet and got the concept of 8bit Micro-controllers. But i couldn't find any where if i am able to interface it with a 12 bit ADC. Also IF i am able to interface it, would it be more easy to use for eg a 32 bit MCU like this one instead of using the 8bit micro-controller, and interfacing it with the 12 bit ADC ? P.S by 'more easy' i mean simplicity of coding. Looking forward for your suggestions. Thankyou. <Q> There is absolutely no problem in interfacing a 12-bit ADC with an 8-bit micro. <S> The one you show has I2C interface, so it's a bit easier if you pick a micro with I2C bus support. <S> Atmel calls theirs "Byte-oriented 2-wire Serial Interface" because of Philips' IP rights, and the Atmega processors have such an interface in hardware. <S> It is common to use more than one byte to represent data. <S> If you were programming in C you might use an int, long int or even a float. <S> You can write code in assembler to deal with 8, 16, 32 or 256 bit numbers if you really want (limited by memory). <S> Simplicity of coding is highly dependent on what exactly you are trying to accomplish, and what your tools and experience are. <S> If you are using a compiler, the details of the hardware are mostly hidden from you <S> so it doesn't much matter <S> (speed could be a bigger factor). <A> Typically you when interfacing devices through I2C or other IC level buses, you will transfer a word of information out of your device, like 8 bits at a time but the payload may vary between protocols (usually a defined protocol word or word range). <S> So to see how your 8bit MCU will work with your 12 bit DAC, first look at the internal data structure of the ADC Output. <S> You can see that this is a padded 16 bit number and can neatly be stored in two bytes/words using your MCU. <S> In c, this number would transparently be read as a u_int16 or equivalent definition for your platform. <S> Moreover, if you program in C, multi-word data structures are handled transparently, processing and mathematical operations would be slower, but you would be able to safely store a 24bit ADC in a u_int32 and the compiler will handle the word level details for you. <S> Now when it comes to transfer protocol , I2C is a serial transfer protocol and you can chomp the data on MCU side however you wish (even with a 7 bit mcu if you had such a beast) , but the bus neatly divides the data for you by insisting on an ACK after every 8 bits. <S> Let's look at the I2C signaling for your ADC. <S> You can see that all communications are neatly split into 8 bit sections, writing command words, writing registers, reading data, etc. <S> If you were bitbanging this protocol on your MCU in assembly, you would be able to easily work with your word level operations without too much headache. <S> In C this would be done transparently by the compiler. <S> Your MCU is smart and will handle a lot of bus level details for for you. <S> You can perform I2C transactions using a few control registers on your MCU without dealing with timing details like shown above. <S> For the datasheet you linked (Atmega 48/88/168) the relevant chapter is Chapter 21 (Two-Wire Interface). <S> A high level diagram of I2C communications using your MCU is shown in the following figure. <S> The benefit of using the integrated I2C is that you save a lot of program space and clock cycles that would be spent bit banging I2C in your code. <A> The result of the ADC are typically returned in two 8-bit registers labeled high and low (ADRESH and ADRESL in the diagram below). <S> The result many be returned either left-justified, meaning the high bit of the result corresponds to the high bit (bit 7) of the high byte of two 8-bit registers. <S> The other option is for the result to be right-justified, meaning the low bit of the result corresponds to the low bit (bit 0) of the low byte of two 8-bit registers. <S> Left justified gives you effectively an 8 bit conversion by just reading ADRESH. <S> The two least significant bits are in ADRESL and are thrown away. <S> This is often useful when you want a quick result and don't want to deal with the extra code and time dealing with a 16-bit value. <S> Of course the left-justified format can still be used the to access the full 10 or 12 bit result, but the right-justified format described next is actually more useful. <S> Right justified gives a full 10 or 12-bit result (depending on the size of the ADC), in a convenient format. <S> When read into a 16-bit variable, there is enough room to add readings together for averaging without an overflow. <S> However you then have to deal with 16-bit arithmetic.
| It is very common to have ADCs integrated with 8-bit microcontrollers that are either 10-bit or 12-bit.
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How to see electronically if a cup is empty or not My requirement is just to know when a cup/mug is empty. I searched for weight sensors but they are bulky and expensive for this simple task. Is there any cheap solution for this task? Using weight sensor is not necessary as long as requirement is met. The circuit will be under base of the cup. The liquid can be anything from water to black coffee. update: I asked i wanted to detect whether a cup is empty or not. The bottom surface of cup is flat. One cheap idea is to have a little transparent surface at bottom of mug like 1 cm by 1 cm. Put a photo transistor there to see what's in the mug. If the mug is empty, photo transistor will give a unique value as compared to any other liquid present regardless of its amount in mug. Any other ideas are welcome. <Q> I experimented with this round force sensitive resistor, and it can probably be used to determine whether the cup is empty or not. <S> It is the opposite of bulky, so if it works you can get a very small form factor. <S> The one caveat is that I had to place a small metal piece on the force resistor and balance the mug on it, to direct all force towards the material. <S> It turns out that mugs are often curved at the bottom so that they don't touch the ground at all in the middle where the resistor would be placed. <S> You might be able to work around this by getting a larger piece of force sensitive material since they are sold in all kinds of sizes (and can be much cheaper than the one I linked to). <S> These are the measurements I got when testing this with an Arduino (the analog signal on an Arduino varies between 0 and 5 volts): <S> As you can see from the red lines which indicate the points at which I did not pour water into the mug the response is not linear. <S> However, the data sheet includes a resistance versus force curve, and you can do your own calibration. <A> Other suggestions: conductivity (two rods in the liquid) capacity (two rods, or one is a foil outside the mug) <S> US distance (from somweher above the cup to the liquid level or the bottom) IR distance (idem) somehow measure the natural frequency (speaker + microphone?) <A> You don't provide much details on the specific application (i.e. the context where this would be applied and the specifications you want to meet). <S> Then use some simple logic to detect when the cup is full, for example if 2 out of 3 switches are on then declare the mug full. <S> I suggest an array of microswitches because with just one microswitch you could end up requiring a more sophisticated balancing mechanism in the base to prevent a non-centered mug not to trigger the only switch. <S> With 3-6 microswitch spread evenly in the base you don't need a very precise mechanics setup and you could implement all the detection logic in software with a microcontroller, for example. <S> Of course my suggestion is viable if the weight of the mug is something known in advance, i.e. the mugs are all the same weight and the base can be made not to press on the switches if an empty mug is placed on it. <A> Calle has a great answer for a weight sensor. <S> The other main way would be to sense the liquid using a level sensor . <S> Wikipedia has some useful details on level sensors here . <S> There are three main ways to do this (beyond weight) that I know about: <S> Float sensors <S> Conductive sensors <S> Capacitive sensors <S> Optical sensors <S> Float sensors rely on a float rather like in a toilet cistern moving, and switching something (either mechanically, magnetically - i.e. a magnet moves - or occasionally optically). <S> These would generally be too bulky. <S> Conductive sensors attempt to pass some current (often low voltage AC to prevent you electrolysing your coffee too much) through the liquid if it reaches a pair of electrodes. <S> Provided your cup doesn't contain the purest distilled water, this should work. <S> Capacitative sensors are immersed in the liquid and measure the change in dielectric constant. <S> These would work here, but something needs to be dipped in your coffee cup. <S> Optical level sensors (including laser optical level sensors) would remain above the liquid, but more often protrude into the liquid with a dome that is covered by the liquid, which changes the output of a phototransistor. <S> These are often used in vending machines (though not normally in the coffee cup - <S> that's normally done by weight or timing the amount of liquid put in). <S> From Farnell (randomly selected UK supplier) these seem to start at about 18 UK pounds (approx $25). <A> I would use a low level microwave transmitter (below the cup), and a receiver (at top of cup).
| Calibrate it with an empty cup (strongest signal), and any reading less that this (liquid absorbs some radiation), would be the not empty cup signal. These have the advantage that they don't give misleading results if the weight varies between cups, but would normally need to be put above the cup. I would think that a simple weight switch is the ideal solution. I repeatedly poured approximately four centiliters of water into the mug. You can build one yourself (it's a couple of wires), or search for 'immersion electrode'. Anyway, assuming you don't need much sophistication, a base with an array of microswitches can be made cheaply.
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Why do electronics components have maximum voltage and not just maximum current? I am trying to figure out how electricity works and I wonder why do electronics components have the maximum voltage limit? Why does the voltage matter? Isn't it the current that can 'kill' the component? For example if I have an LED with a maximum voltage of 2V powered by a 5V source.If I put enough resistance in the circuit - it will effectively lower the current and the LED will be working fine, the voltage would stay the same though.Which means the LED is working fine beyond the voltage limit. I would really appreciate if someone could clarify this. <Q> You most probably don't have a LED 'with a maximum voltage of 2V when powered from a 5GV source', but a LED that is rated for a maximum of 20 mA. Components that have a low resistance, and particularly ones that approximately drop the same voltage over a wide current range, are rated for a maximum current, not a maximum voltage. <S> In most cases it is their product (power P = V <S> * I) <S> through power dissipation that causes a temperature rise above what the integrated circuit material is rated and causes it to fail. <S> The maximum could be expressed in either power or voltage. <S> But when component and temperature variations can cause the same voltage to cause widely different currents, it makes more sense to express the maximum as a current. <S> OTOH, if above a certain voltage breakdown will occur, but the breakdown current is not easy to specify, it makes more sense to express the maximum as a voltage. <S> Summary: the maximum is generally expressed in the unit that is (for the components at hand) most stable. <A> From what it sounds like you are saying, You are putting resistance in series to lower the current to the LED. <S> The LED and resistors are seeing 5V but the LED is not seeing the 5V now. <S> At the appropriate current it will only be seeing 2V. <S> The resistor will be seeing 3V and the circuit will be balanced 2+3=5. <S> When devices are in series they will always see the same current. <A> While current represents the amount of units of charge that pass through a point in a given amount of time, voltage represents the amount of energy that each unit of charge possesses. <S> And energy causes things to happen. <S> Such as heating. <S> And insulation breaking down. <S> So too much voltage and bad things will happen.
| For most components, voltage and current have a fixed relation that can be found in their datasheet.
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Reducing parasitic capacitance in momentary latching circuit I'm using this circuit ( http://www.edn.com/design/power-management/4427218/Latching-power-switch-uses-momentary-pushbutton ) to get latching on/off behavior from a standard momentary switch (an Omron B3S-1000P). We have the momentary switch on a separate PCB and wired to the switching circuitry on the main PCB. The whole system is a microprocessor running off of regulated 3.3V from a li-ion battery. We've been turning the circuit on and off over and over to see if the RC constant is going to work for our debouncing needs. In doing that, we run in to a problem where we can turn it on fine, but it's really spotty in turning off. Recently, I think I found the solution: if I monitor the voltage on the 330nF cap it will reach a threshold voltage below which it won't turn the circuit off, and above which it will work fine. If I then place my hand on or near the PCB containing the momentary switch, the voltage at the positive terminal of the capacitor will start to drop below the threshold and then the circuit won't turn off. It seems like there's some capacitance from my hand/body that's getting in to the circuit and causing the 330nF to discharge a little bit. It makes sense now that holding my hand on the button PCB, turning it on and off over and over would cause the circuit to work starting working poorly. Wrapping my hand in something insulating (like a plastic bag or piece of rubber) and then turning the button on/off works flawlessly. My question is: how can I counter my body's capacitance? My first thought was to add a ground plane to the PCB with the switch, but initial tests of holding a grounded lead while pressing the button don't seem to improve anything. Adding another capacitor seems like it would mess up the RC constant of the circuit and change the debounce behavior. <Q> There is nothing discharging the 330n capacitor except your voltmeter and leakage (internal leakage should be minimal unless the capacitor is defective, as should be switch leakage). <S> Perhaps your SMT board was assembled with high leakage 'no clean' flux and you are experiencing the ill effects of that, from breathing on the board etc. <S> This may appear to be a digital circuit, but leakage resistance of a few M ohms or less could cause it to stop working. <A> I'm on my phone and see both your question and the schematic you linked to at the same time. <S> Try adding a small capacitor from Q2 G-S. <S> I'd try a value of about 1/10 the value of the other cap in the circuit. <S> The node that goes to Gate of Q2 is fairly high-impedance and susceptible to noise being coupled by your hand. <S> The added bypass cap should reduce or eliminate the problem. <A> I actually got an answer from the designer of this circuit! <S> I tracked him down, and he gave some helpful tips: <S> Does the button work if it's soldered directly to the board? <S> (Yes.) <S> For the button on the separate PCB, does twisting the leads reduce the noise? <S> (Yes -- I could produce a 0.2V drop by touching the button without twisted leads, and 0.04V drop with twisted leads). <S> So, overall, it was probably just a pretty standard noise issue.
| You could stop using that **& & $# (clean the boards you have with strong solvent and a brush) or you could try reducing the resistor to (say) 100K and increase the capacitor to 3.3uF.
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What is the maximum number of bytes that can be sent through USB-to-serial COM port at one go? Suppose someone uses hyper-terminal to send a very long ASCII string at a COM port. This COM port is created by FTDI USB-to-serial port cable. The cable used is http://www.ftdichip.com/Support/Documents/DataSheets/Cables/DS_TTL-232RG_CABLES.pdf . Is there a limitation imposed by the UART driver for PC? For example, Arduino UART tx buffer is 64 bytes only. What is the maximum number of bytes that can be sent through a PC serial port at one go? <Q> In high-speed USB, the maximum packet size is 512 bytes. <S> Most USB/serial converters, including yours, use full speed. <S> However, if you are looking at the serial output, the USB packet size does not matter, because USB packets can be sent faster than the speed of the serial line, and are buffered. <S> For example, if the PC sends 100 bytes, it will use two packets, but what you are seeing at the other end, on the serial line, is again a continuous stream of 100 bytes. <S> Similarly, if you are sending data from an Arduino, you can buffer a new TX byte as soon as some previous byte is being sent, so the size of the TX buffer does not really matter. <S> (However, a larger buffer size allows the pre-buffer more data, which allows continuous transmission even if the microcontroller must do something else for a longer time.) <S> PCs, most microcontrollers, and USB/serial converters are fast enough so that the only bottleneck is the speed of the serial line, so you will never see a gap in the data, regardless of how many bytes are transmitted. <A> Hyperterminal is a Windows-specific software program, so I don't think this is really on-topic, also it has been deprecated- <S> it is no longer bundled with any supported version of Microsoft Windows. <S> Anyway, there is no practical limitation. <S> Set it up to talk to a COM port and you can send a text file of just about any size (or a binary file via the Xmodem and other ancient protocols of the saintly days of yore). <S> If you sent a 2G file at 9600 baud <S> it would take a month or something like that. <S> Nevermore. <A> As many as the involved chips can actually store. <S> Go back 20 years: Back then, modems were attached to a PC via a COM port. <S> The "Download" was basically a stream of data which was only limited by phone fees :-)
| In full-speed USB, the maximum packet size is 64 bytes.
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Question about a phone wall adapter and it's current output My phone charger (Male wall adapter to female USB) says it outputs 5V @ 2A on the back. I want to use it to plug a Male USB into. I guess my question is; is that 2A always being output? Or is it a maximum of 2A? I have a homemade project, that is simply 10 RGB LEDs with 5V going to the common anodes (parallel), and only the R and B are going to 200 ohm resistors going to ground to show purple light. I tested this with the USB cable in my computer, but I want to make sure it's safe to use with this phone adapter before I plug it in. Thanks! <Q> Current is dependant on voltage. <S> The 5V @ <S> The other day I took an old phone wall charger (similar idea), made sure that the RGB LED's didn't go over their rated voltage and plugged it <S> in!If <S> you think 5V is too much you can always add a resistor in series with your LED's. <S> Hope that helps!Josh <A> The current rating on a power supply is the maximum current that it is designed to supply. <S> Any load you connect to the supply will only draw the current that it (the load) requires. <S> The supply does not force its full rated current through the load. <A> Josh is right and to expand on his answer. <S> The 2A is the maximum current draw allowed by your wall wart. <S> Your PC will allow probably 0.5A or 0.9A depending on the USB version. <S> So this implies your RGB project is working fine with under 1A therefore it will be ok with the wall wart. <S> To work it out properly you should use Ohms law <S> I= <S> V/R Where I is the current in amps, V is the voltage and R is the resistance in Ohms. <S> Not exactly sure how your circuit is laid out so if you want to supply a Schematic diagram we can help you calculate the exact current required. <A> 5 volts constant at 2 amperes is maximum rating. <S> You may have a load that consumes less current. <S> 5 <S> x 2 = 10 watts = rate of change of Energy <S> (the more is it, the more powerful is the device)
| If you have a small load it will draw a lot of current, if you have a large load it will draw small current. 2A indicates that it's rated to handle up to 2A at 5V.
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What happens to unused power in generators I have a 7KVA generator which i use for multi-purpose. Some times i run AC on it and some times not. Will the generator consume less fuel when i utilize less power generated from it? Thanks <Q> There should be some mechanism/electronics in the generator which will keep the rpm (revolutions per minute) of the engine constant or to keep the voltage constant. <S> If you load the generator more the rmp and or voltage will drop so the engine will be pushed harder to compensate. <S> If this was not so, where would the excess energy go ? <S> It would be turned into heat which is very wasteful. <A> There are two aspects to control of the generator- <S> the first is the motor RPM and that will determine the AC output frequency. <S> That is controlled (typically mechanically with a governor ) by adjusting the throttle automatically (you may be able to see the throttle linkage). <S> That maintains fairly constant RPM as the engine loading changes. <S> The second is voltage regulation and that is sometimes achieved by changing the field coil current (inexpensive or small ones may have fixed excitation so the voltage will drop noticeably as the load increases). <S> At a very light (electrical) load, the generator is very easy to spin so the motor is essentially idling. <S> As you add electrical load the field caused by the current flowing through the load (and the coils in the generator) starts to oppose the rotation of the rotor ( Lenz's law ) and it becomes much harder to spin the generator. <S> (The voltage will also start to drop so field current also increases to maintain the voltage if the generator has an AVR- which will also increase the required torque to compensate for the internal losses). <S> As the load increases the motor RPM will tend to drop, so the governor opens the throttle wider, admitting more fuel and air to the engine, in order to try to maintain a constant RPM. <S> It's probably a reasonable rough approximation to say that the fuel consumption increases linearly with kW loading above the baseline idle consumption. <S> TL;DR: <S> Yes, it will suck a fair bit more fuel if you put a heavy load on the generator. <A> What happens to unused power in generators <S> Power is volts x amps and if the load is not demanding any current then there is no demand for power and no power is generated. <S> Of course there will always be some power needed to combat the friction in the bearings of the generator and wind resistance in the spinning parts but this will be constant. <S> Fuel is needed for this. <S> There will also be a bit of extra power wasted proportional to the current demanded by the load due to copper losses in the windings of the generator supplying load current.
| Yes and it will consume more fuel when you load it.
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Driving DC motor using a single MOSFET, why does the motor spin without applying a gate voltage? The arduino drives the gate of the mosfet (irf540n) just fine (I've tested it with an led + 100 ohm resistor) and now I want to drive a small dc brushless motor. I've connected everything properly but it works not quite the way I want it to. The drain and source are supposedly not connected if there is no gate voltage applied and thus the motor should not spin but it does spin (I've connecting it correctly, just the standard way that is found all over the internet). When the gate voltage is made high by the arduino the motor spins a little faster. My question is thus: why does the motor spin when there is no gate voltage applied? Difference with the above schematic is:- No diode- Mosfet is irf540n- Motor is driven by 9V Other than that the configuration is the same (connected ground of arduino to the - side of the 9v battery, the - side of the battery is connected to the source and the + side of the battery is connected to one terminal of the dc motor while the other terminal of the dc motor is connected to the drain) EDIT: I got it working when I tried it again (using another irf540n MOSFET). Diode in parallel with the dc motor is indeed a good thing. Placing a resistor as you guys said is also needed indeed so that the capacitance of the MOSFET is able to discharge I think. Btw drain & source mixing up isn't that big of a deal is it? Thanks everyone. <Q> By "no gate voltage applied", I assume you're not driving it - <S> it's unconnected, or connected to high impedance such as an input pin. <S> A MOSFET's gate has extremely high impedance - from hundreds of megaohms to gigaohms - and thus it takes very little current to change the voltage on it. <S> Ambient EM fields can easily affect it, and with nothing connected the voltage can fluctuate and take on just about any value. <S> In your case it's likely above the Vgs threshold voltage, turning the MOSFET on. <S> This is why you should always have a pullup or pulldown resistor on your MOSFET's gate if there's any chance it might be otherwise undriven. <A> You need a pull down resistor on the pin that drives the gate. <S> Look at the 10K resistor in the pictures below: I quote from KyranF: <S> The resistor holds the gate low when the arduino does not send a high signal. <S> This is here incase the arduino comes loose, or the wiring is bad it will default to off. <S> You don’t want this pin to ever be floating as it will trigger on and off. <S> Floating MOSFETs are bad, and because the "on" control of the FET is essentially just a capacitor with very low capacitance, it is quite easy for it to float up and turn itself on. <S> This situation will only really happen in your arduino program if you make the output pin an Input by mistake, or during power off/on/restart states. <S> The ATMEGA328P on the Uno makes all it's pins go into high impedance state during power cycle, which is a prime opportunity for the gate of that FET to float high. <S> For your third question - MOSFET gates only "use" current for a brief time during the ON period, to charge the gate capacitor. <S> The Arduino's 40mA output maximum per pin is not going to be an issue. <S> It WOULD be an issue if the FET was a BJT Transistor instead, as those will constantly draw current into the base in order to operate. <S> MOSFETs work differently, and do not consume current to have them be "on" constantly. <S> Putting a 10K ohm resistor is also too high in general, it will slow down the ON/OFF time of the FET considerably, and cause major switching losses if you are doing any reasonable frequency PWM. <S> Use something like 100 Ohms if you want to put a resistor there. <S> Putting a resistor there may not be needed for a MOSFET, but it IS recommended to reduce the possibility of inductive feedback into the microcontroller and other forms of dirty business related to switching an inductive load like a motor. <S> Add the diode as a flyback protection or you are going to have to add a new MOSFET + diode very soon!!!! <A> My friend, you are 100% facing a MOSFET failure. <S> I don't think it is a wiring problem. <S> you can easily check your MOSFET by the DMM. <S> Just put it on the Diode mode and see if it beebs when placing its pins on the drain & source. <S> So they are shortened due to the back current coming from the motor when turned off and yet turning... and this is exactly the function of the Diode you underestimated...
| The resistor ensures there is always a known state, and only an active output HIGH from the Arduino will cause it to actually turn on.
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SPI Clock and CS signals I have one doubt regarding SPI. I am very new to this module. I know that communication is initiated by asserting the CS(high to low) in SPI. And then in every clock based on polarity and phase the data is transmitted/sampled. My doubt is once data has been transferred and CS is asserted again(low to high), does the clock pulse still continues or clock pulse is only there till the communication is maintained. The image shown is the waveform while trying to test SPI communication. The Master SPI shift register is of 16bits. I am transferring four 16 bits of data during the CS low period(i.e. when CS is enable). But during each clock I am getting 16 small incomplete waveform. I know this is wrong. Could someone give me the resolution or suggestion to rectify this. I am expecting something like this, although it transfers two data of 8 bits each at 8 clock pulse, I want to transfer 4 data each at 16 bit clock pulse. When I increase my data buffer from 4 to 5, I get 5 square pulse, and each pulse having 16 jitters. Please reply. <Q> It ends whenever the master says it ends. <S> If the master is a MCU, see the MCU's datasheet for its behavior. <A> An SPI Slave device is selected by its very own CS* (active-low) signal. <S> If the Slave's CS* pin is high, it is required to ignore any clock pulses that go past. <S> If the CS* pin is low, it must clock data in and out as the clock pulses dictate. <S> This allows you to have multiple SPI Slave devices connected to an SPI Master device. <S> For that matter, you are allowed to have multiple SPI Master devices sharing one or more Slave devices. <S> The Masters are required to coordinate their usage of the SPI clock and chip select lines, so that they don't both try to use the lines at the same time. <S> SPI does not discuss or mandate the synchronization mechanism. <A> The SPI Master will hold the clock high/low until the communication is initiated by the master, at which point the master pulses low/high to initiate and pulses for the duration (low to high, low to high, etc.) <S> of the communication. <S> Unfortunately, I do not have the waveforms to confirm this, however I was working with SPI a few months ago and confirmed that this was the case in my scenario. <S> EDIT: <S> After seeing your image <S> I realized I was confusing SCLK with CS/SS. <S> So what I said holds true for the clock in my situation. <S> Now the CS/SS is necessary when using more than one slave device. <S> The master will hold the line low for the selected slave while it communicates and then when done pull it high again. <S> Each slave needs its own CS/SS line in order to select them individually or one line to select all slaves at the same time. <S> This might help: https://learn.sparkfun.com/tutorials/serial-peripheral-interface-spi
| Once communication is done, the master pulls the clock high/low and it remains high/low until the master initiates communication again.
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Making LM78xx fail safe against short, open, reverese bias, flyback, etc I want to use LM78xx for various applications. What I am concerned about is its failure modes. 1. Failure to Short Not shorting Vcc to ground!! I mean shorting V_input to V_output. Assume that you have a delicate and expensive piece of electronic equipment that works with 5V. You are using LM7805 to reduce voltage from 12~13VDC to 5VDC. For some reasons your LM7805 fails to short and your delicate electronic device sees the 12VDC instead of 5VDC and fries. I have seen this all around the web. How to protect against this? 2. Failure to Open It is not as critical as the failure to short. It can happen as a result of over temperature (temporary) or permanent failure. What are the other things that can trigger this? 3. Reverse Bias & Reverse Discharge I have seen that too big an output cap can cause reverse current going back to input. It is recommended to add reverse discharge diode protection. What else can cause this kind of reverse bias voltage? From Various datasheets: With the LM7805, the output capacitor should not exceed 1mf, as largervalues could damage the 7805 due to backfeeding of current when poweris switched off. When a surge voltage exceeding maximum rating is applied to the inputterminal or when a voltage in excess of the input terminal voltage isapplied to the output terminal, the circuit may be destroyed. 4. Flyback Does the diode on the above picture also protect against flyback current of an inductor? Or is it better to add a separate diode across the inductance load? 5. What else? What are the other possible causes of LM78xx failure? <Q> Failure to short- crowbar + fused input (a thyristor and trigger circuit) or at least a TVS. <S> The TVS might be able to limit the voltage to something like 8V which your delicate equipment might have a fighting chance of surviving (newer chips with 5.5V abs max supply may not live). <S> An open GND connection will cause the output voltage to rise so make sure the soldering is solid. <S> Don't depend on the thermal protection as a matter of course. <S> D1 is unnecessary for a 7805 or 7806, only for higher voltage regulators (and only where the input voltage can be actively discharged). <S> It provides another possible path for failure to short, so I suggest leaving it out unless there is some way the output can (say) get connected to a 12V battery <S> and you don't have a TVS or crowbar on the output. <S> The power supply does not care about load inductance- <S> the inductance will only tend to make the current continue to flow in the same direction. <S> Any flyback voltage appears across the switch so you need a flyback diode to be placed across the switch or the load. <S> You might want to put a reverse-biased diode across the output in case two supplies are connected in series and one is driven negative. <S> If you use a unipolar TVS that can serve both functions with a single part. <S> TL;DR: <S> Put a polyswitch fuse in series with the input (before the capacitor) and add a TVS to the output, and lose D1 (if 7805/6). <S> For extra points mount the polyswitch close to the regulator <S> so it sees the heat. <A> Crowbars are virtually mandatory on switch mode power supplies to prevent a greater output voltage than what would be allowed under the European Low Voltage Directive (LVD). <S> Failure to open is lost on me - maybe you can explain that. <S> Reverse biasing you have covered fairly well except for the scenario of a battery charger - usually extra precautions need to be made to prevent damage when primary charging power is removed. <S> The flyback situation totally depends on the target circuit you have in mind. <A> If you are open to using other regulators, you might want to consider more robust automotive linear voltage regulators, such as Micrel MIC2940A or MIC2941A.
| If you have an application where the failure of the linear regulator causes a higher output voltage then either use a fuse and zener diode (5V6 maybe) or design a small crowbar protection circuit. Failure to open- bad solder connections, destroyed chip.
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Adding 90 ohm resistors to DIY USB cable? I just started with electrics a few months ago, but now I am obsessed but still have MUCH to learn. What I am doing is essentially trial and error. I needed a longer USB cable to a printer so I took a USB cable I had, cut it close to each end of it and then took the thickest 4 cables I could find and connected them to the USB connectors/ends. My idea was simply the thicker the better, the lower resistance the better. But it didn't work! The computer did not recognize the printer. I read some more on USB and then I saw something about a need for 90 ohm impedance (imped... what?). Read some more and learned that impedance is sort of like AC resistance. After that I connected a 10 ohm resistor to D+ and a 20 ohm resistor to D- (not 90 ohm I know but that's all I had). Then I tested the cable again and it miraculously worked! I was quite confused that lower resistance is not always better in a cable. My question is if my thinking is correct when I think that when measuring a DIY USB cable with a multimeter the resistance (sort of impedance) should be as close to 90 ohms as possible? Some additional info added after asking Cable length is about 2.5 meters (which is less than maximal 5meters). I measured with a multimeter before and after the adding of the resistors and + connected to +, - to -, D+ to D+, D- to D-, so I think the "wiring" is/was correct. It is not entirely unlikely that the connection was loose though and something fell out, but not that I noticed. My extended cable works almost all the time except that sometimes when starting the printer then a USB network card connected to the same computer (as the printer) stops working/is disconnected, so I have to remove then reinsert the USB network card and it works again (so no permanent damage ;) ). And sometimes (after a long time of inactivity) it my computer looses connection to the printer (at least it doesn't print on command) and that is fixed by switching off the printer then switching it on again. <Q> USB cables are not simply "4 wires". <S> They have certain properties that need to be met so that the data signals can travel through the cable. <S> Your thick wires will not have the correct properties causing the data signals not to travel properly, instead they are suppressed and reflected resulting in the USB connection to fail. <A> You may find this previous question helpful - Custom USB Impedance question . <S> Essentially, your problem is that impedance is not resistance (in the most general sense). <S> Rather, resistance is impedance at DC and slightly higher frequencies. <S> Since USB2 sends data at 480 Mbit/sec (peak) you need to consider questions of inductance and capacitance at GHz frequencies. <S> Not only that, these issues have to apply over the entire cable length, so they become questions of geometry and spacing, rather than just adding a couple of components. <A> For your USB cable, you need a differential impedance of 90 ohms between the lines and a 45 ohms on single-end wire impedance. <S> I think it is a hard project for a beginner, a more simple project could be to implement USB track on a PCB, but still that is pretty hard and not trivial.
| For PCB, you can use the Saturn PCB toolkit that let you calculate differential impedance.
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What is the part name / description for these test points? What is the name/description for the test point pins on this board? I've tried Digikey for all combination of binding post, test pins et, but no luck. These are through hole parts, height ~6mm, pin dia 1mm. <Q> I have found this one very similar to the test point. <S> it will be impossible and not worth the time to search by only image as reference. <S> This link gives a lot of similar looking parts you are looking for. <S> Please feel free to choose .. <A> In Digikey's hierarchy look for: <S> Product Index > Connectors, Interconnects > <S> Terminals - PC Pin, Single Post Connectors <S> For example, this one (P/N 3137-1-00-21-00-00-08-0 ): <S> There are many other types! <S> Depends on the exact dimensions you need. <A> Below testPoints may suits your requirement 1 <S> MULTICOMP <S> TEST-27 <S> TEST PIN, PCB, <S> 1.5MM <S> 2 WEARNES CAMBION 460-2970-02-03-00.. <S> TEST POINT, <S> PCB 3 MULTICOMP <S> TEST-4 <S> TEST PIN, PCB, 0.95MM <S> 4 <S> MULTICOMP <S> TEST-22 <S> TEST PIN, PCB, 1.0MM <A> There are many kinds, not necessarily you will find the specific part on digikey.
| Definitely those are test points.
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Is there a "faster" photoelectric beam sensor? We've been experimenting with some Velleman PEM10D sensors. (So, something like this .) The main reason chose, was they have a 10m length (we need a good 6m or so). I have found that, the response is a bit slow for our needs. So for example: if you drop a basketball through the beam from a height of two feet, it is travelling slow enough it will register. However if you drop a basketball through from 6 feet height, it is too fast, it will not trigger the device. I surmise, we need the device to be (let's say) a good ten times faster. Now, the spec of the device suggests .. "Response time: 5 - 100ms" I don't really know what that means - is that the minimum time of cutting the beam which will register? If so, what does the 10x spread mean? Perhaps someone here will know. Secondly, in general is this a well-known problem, are there "much faster" photoelectric beam sensors? Or no, or ...? Thirdly, indeed could the whole thing be a fubar on our end ... say, poor programming on the Arduino, wrong ... power supply or something, or some other mess-up by us? So, to be perfectly clear, the thing cutting the beam in our setup is a fast-moving ball-like object, which can be moving quite fast .. some 10 or even 20 m/s. Finally I apologise if this is not in fact the optimum forum. Thx <Q> The manual for your sensor - briefly - mentions "response time adjustment" via a screwdriver, which would explain the spread. <S> Some applications, such as sensing someone coming through a doorway, require longer response times, so as to not trigger separately for, say, each leg. <S> In general, the response time of this sensor is going to be limited by the fact that it uses an electromechanical relay, which takes time to physically switch on and off. <S> It's likely the internal electronics were designed with low speed operation in mind as well. <S> Broadly speaking, if you need high speed measurements, you probably want to look for a sensor that has either a transistor output (open collector, open drain, or logic level), or speaks a communication protocol like I2C or SPI. <S> Sensors like these are likely to be designed around higher data rates and response times. <S> Many of these more specialised sensors may be orientated around measuring distance, rather than just detecting beam breaks, but it's certainly possible to construct a beam break sensor with a fast response time; I'm just not sure if any exist commercially. <S> In passing, Sparkfun have a Lidar sensor that ought to meet your requirements pretty well. <A> Nick's answer is good. <S> That'd give you about 1ms response times with a CMOS-compatible output that can go directly to your AVR. <S> You might want to add some simple optics if you're getting too much beam scatter and detections of indirect beams; even just a baffle (black tubes around transmitter and receiver) would be a good start. <A> The manual is a bit vague as to the actual detector used ("retroreflection" is not a sensing method, it's simply indicating that the beam is being reflected back at the unit, so the emitter and detector are in one housing.) <S> In general (of the sensors I know) <S> photo-resistors are slowest (10's of milliseconds) <S> photo-transistors are faster (10's of microseconds) and photo-diodes are fastest (100's of nanoseconds, possibly less)
| If you want a very cheap and probably-fast-enough solution, you could consider using IR remote-control technology like a TSOP38238 receiver and a matching IR LED (950nm) modulated at 38kHz with a 555 or similar.
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Insufficient current to drive automotive relay? I'm adding a leisure battery to my van so I can run some stuff with out fear of draining the starter battery. My leisure battery is wired to my starter battery so it can charge from the alternator when driving but is separated by a relay to avoid draw while stationary. I have split a wire from my dashboard which becomes live with the ignition and reads 14.5 volts with the engine running to use to activate the relay, and grounded the other side. My Problem When I connect this wire to the relays trigger circuit (pin 86 I believe), it's voltage drops to 0.6 which doesn't seem to be enough to close the circuit. Any one know what I'm doing wrong? edit I'm using this relay I have wired it up like this; the switch pin 30 positive on starter battery pin 87 positive out of relay control/coil pin 86 live from dash pin 85 ground. Van is mk6 ford transit, wire is white and green one from display behind the steering wheel which I thought might be alternator warning light... hope that helps. <Q> You should be looking at installing a battery splitter instead of this relay. <S> The splitter is a pair of high current diodes mounted on a heatsink and equipped with nice stud terminals for connecting up the wires and a case cover. <S> The alternator output goes to the common anode terminal of the splitter. <S> The two cathode terminals connect off to each battery. <S> No special control wiring is required!! <S> Also no mechanical part to wear out. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The diodes block the leisure load from discharging the running battery. <S> Such splitters can be found at common RV supply outlets. <A> What you are doing SOUNDS correct. <S> Many examples here <S> Try connecting 86 via a wire directly to battery. <S> If that works then you need some other dashboard feed. <A> So after some investigation it seemed the voltage drop was caused by a lack of current(amps) which was needed to power the coil. <S> 85 went to ground but was controlled by a transistor. <S> The original low powered cable from my dash was enough the operate the transistor which opened the ground for the battery wire. <S> It works though its a bit convoluted. <S> Probably should have just brought a voltage sensitive relay to begin with...
| Soooo I added a new wire straight from the battery to pin 86, put a diode between 86 and 85. This should operate the relay without any drop in voltage. It sounds as if the pin 86 feed from the dash is not full battery voltage but has something else in series with it.
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How come 18650 Li-ion cells aren't meant to be sold loose? I've heard that 18650 Li-ion cells aren't meant to be sold loose. They're meant to be built into packs and the "good" manufacturers won't actually sell them for any other purpose. FWIW, there's the Panasonic NCR18650 - which I believe are sold loose. So, my question is: why aren't 18650 cells (perhaps LiPo cells in general, for that matter) meant to be sold loose? <Q> "Not meant" is not completely meaningful as there is not a consistent international regulation to that effect. <S> The norm is to either have one protection module per cell or to use a multi-cell module which connects to all cells in a pack directly. <S> While it is notionally possible to connect multiple cells in series and to "protect" the combination the notion is not a good one. <S> LiIon cells tend to best live long and prosper when the individual cells can be individually managed. <S> Vcell too high <S> , cell dies, maybe with flame or just degrades. <S> Vcell somewhat low, life reduces. <S> Vcell very low - will not recover and attempts at charging may cause flame. <S> I discharge too too high and flame will happen. <S> I charge too high and damage will happen BUT most protection cctc do not prevent this. <S> Selling protected cells to unqualified buyers may be a bad idea - destruction of cells and equipment is still "easy enough". <S> An 18650 cell is rated at perhaps 10C or about 25A and probably capable of 50-100A in short bursts (possibly VERY short & just before flame) depending on model and maker. <S> A 4S 18650 stack is an awesomely capable beast - maybe a kW+ burst output capable. <A> I think your premise is incorrect. <S> From what I can see, Sparkfun (online) and Jaycar (brick and mortar in Australia) sell these cells. <S> They're not very common <S> so I guess the local supermarket or gas station won't be selling them <S> but they're obviously readily available. <A> A high quality 18650 Li-Ion cell will have internal protection devices which limits discharge current in the event of an external short circuit. <S> I once did an investigation and attempted to replicate an explosion/fire by short-circuiting high quality fully charged Li-Ion 18650 cells. <S> Although the temperature of the cell increased I could not create a catastrophic failure. <S> (This was done in a controlled laboratory environment.) <S> There will be low-quality cells as well and the outcome may be different. <S> Sad state when a "manufacturer" is willing to place unrealistic (and unverified) <S> mAh figures on a cell just to get some sales.
| However, an 18650 cell is potentially rather dangerous and use without an accompanying protection circuit module is extremely unwise. Selling unprotected cells to unqualifed buyers is a bad idea.
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What makes an electrical device draws a current May be this is very silly question but i want to ask that if the system has 220v and we connect our mobile charger with this draws very little current and when we connect iron with this very same socket draws much higher amount of current. My question is that what make current flow through device and what mentioned the quantity? <Q> Current will flow when the circuit is closed. <S> If you have an open circuit it acts as infinite impedance, thus no current will flow. <S> But as soon as you close the circuit (connect terminals of a voltage source) current will flow. <S> Let's say you complete the circuit with copper wire (aka short circuit), this is essentially no impedance and the current will shoot to infinity. <S> Now of course, in reality voltage sources have current limiters and will not allow the current to shoot to infinity. <S> So open circuit equals infinity impedance (no current flow) and short circuit equals zero impedance (infinite current flow). <S> As you decrease the impedance, the current drawn from the source increases and vice versa. <S> Good luck! <A> When you connect a device , such as a charger or an electric iron, to a voltage source you complete an electrical circuit that will allow electrical charge to flow (i.e. current). <S> The higher the resistance or impedance the smaller the current. <S> The term impedance relates to a load that has <S> component(s) (capacitor,inductor,resistor) that alters the phase angle between the voltage and current in the circuit <S> The relationship between voltage, current and resistance (impedance) can be expressed using Ohm's law V = IxR or <S> U = <S> IxZ <S> The lower the resistance or impedance the more current will be drawn. <A> Voltage is a difference in potential, like pressure, or height. <S> When electric charges like electrons and protons are subject to a voltage, they experience an electromotive force. <S> Like the air in a balloon wants to get out, or a boulder wants to roll downhill due to the force of gravity. <S> Just because the electric charges are subject to a force <S> does not mean they move. <S> How much they are able to move depends on the impedance preventing them from doing so. <S> A rough analogy: your water pressure is approximately constant, as is the voltage of an electrical outlet or a battery. <S> Open the hose a little, and a little water comes out. <S> Open the hose a lot, and a lot of water comes out. <S> This is where the analogy breaks down, because electric circuits don't spew charge everywhere like an open hose does: it all stays in the circuit and flows in a loop.
| The amount of current in the circuit will be determined by the resistance or impedance of the device.
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Drop Voltage using Resistor: is it practical? I should probably say that this a basic question, I don't know much about this stuff, please don't be too hard on me. Ok so I kind of thought it would be fun if I could light up 12V halogen lamps using 220V power supply, so I did a few calculations:The bulb is rated at 60W, so according to P=VI, it should draw 5A current. Now since the input voltage is 220V and I need 12V, the voltage drop across the resistor will have to be 220-12=208V. The bulb draws 5A current so according to R=V/I, so I would need a 41.6 Ohm resistor, say approx. 40 Ohm. I wanted to ask if my calculations are correct and if it is indeed safe and practical to drop voltage using resistors. I would also like some details on what kind of resistor I should use. Thanks. <Q> Your calculation is correct, however you also have to consider the power rating of the resistor. <S> Power is \$I^2 <S> \cdot R\$ which is 1040W. <S> That's a physically huge and expensive resistor and your circuit would be wasting 95% of the energy that you put in before it even gets to the bulb. <S> Here's a typical resistor style capable of that level of power dissipation: <S> It's 300mm long, 60mm in diameter and costs more than $40 US. <S> Would it ever make sense? <S> Possibly- <S> if you needed a 1kW heater for some reason as well as the lamp then it could conceivably. <S> Another consideration is that the socket of the halogen lamp may not be designed to keep fingers away from the supply. <S> No big deal with 12V- <S> it's pretty hard to electrocute a human with 12V, but 220V mains could be lethal. <S> Most approaches to supplying low voltage lamps provide galvanic isolation which protects the user. <S> A better choice would be a 220:12 transformer or an 'electronic transformer' which uses switching power supply techniques to reduce the voltage. <S> Electronic transformer: <S> For a hobbyist, an old PC power supply could provide regulated 12V at 5A without breaking a sweat and it should be completely safe. <S> See articles on the internet on how to get it to turn on (you may have add a dummy load). <S> It provides DC rather than AC, however halogen bulbs of that voltage and power don't care much- <S> the life may be slightly reduced on DC. <A> The calculations are right mathematically but just see this: The voltage drop across the resistor is 208 V. <S> So power dissipation is <S> P= <S> V 2 /R <S> P= <S> 1.04 kW !!!!!!!! <S> Now that is some power you are wasting, 17.33 times what you actually need. <S> So this is not a practically feasible method. <S> Instead you can go for voltage regulators and transformers which are much more feasible and efficient . <A> For the lamp current: <S> $$I = \frac{P}{E} = \frac{60W}{12V} <S> = 5 <S> \text{ amperes} $$ For the ballast resistor value: $$ R = <S> \frac { <S> Vt-Vl}{I} =\frac { <S> 220V - 12V}{5A} <S> \approx <S> 42 \text{ ohms} <S> $$ <S> For the ballast resistor dissipation: $$ P = <S> (Vt-Vl) <S> \times I = \style{color:red} {1040\text{ watts}} $$ <A> If one is driving a 12V light bulb with a 12V source, the lower resistance will cause the light bulb to draw more current and thus consume more power until it gets hot. <S> This behavior causes light bulbs to turn on quickly--in some cases more quickly than would be optimal for the lifetime of the filament. <S> Adding the monster-sized resistor in series with the light bulb will mean that the bulb will effectively be fed with a five-amp constant current source. <S> When driving a cold bulb from a 5A current source, the low resistance of the bulb will cause it to drop far less than 12V and thus consume a lot less than 60 watts. <S> If it consumes enough power and generates enough heat to warm the bulb to the point where its resistance increases significantly, the bulb might be able to warm up to its normal operating voltage. <S> On the other hand, the hotter the bulb gets the more heat it will produce. <S> The rate at which power consumption increases will probably be slower than the rate at which radiative dissipation increases, thus preventing thermal runaway, but behavior will likely be far less stable than when driving a bulb with a stable voltage source. <A> Essentially all you calculated for was the required resistance on a 220V source to dissipate 60W. <S> I would try to avoid using 220V to light up 12V halogen bulbs without some form of an adapter. <S> The other thing to consider, is the 220V AC or DC? <S> I am assuming the 12V required by the bulb is DC, so if 220V is supplied in AC, you would need to do an AC to DC conversion. <S> This would simplify a lot if you used an existed adapter and just plugged it in. <S> Just make sure that the current supplied isn't over <S> it's current spec. <S> Hope that helps. <S> Also, could you post the model # or data sheet of your bulb? <S> And can you clarify as to whether the 220V is AC or DC?
| Another difficulty with using a resistor is that light bulbs have a resistance which increases with temperature.
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Clear material for UV Box I have constructed an UV light box using the innards of a commercially available finger nail drying unit consisting of 4* 9w tubes. (should be more than enough). I have had very poor results in exposing both pre coated and DIY board. My question is - Is 10mm thick polycarbonate too thick to transmit sufficient UV or should I be using Acrylic sheet or glass? <Q> Polycarbonate is almost perfectly opaque to UV, so it's no wonder your setup isn't working. <S> You can get special UV Transparent (UVT) plexiglass, which will gain you energy in the 300 - 400 nm (long-wave). <S> http://www.plasticgenius.com/2011/05/infrared-and-ultraviolet-transmission.html <S> You really need to find a data sheet for your resist, and see what wavelengths are required. <S> I would recommend that you don't use a plate for support of your board. <S> And I agree that I'm not being helpful, but I can't resist (get it?) <S> pointing out that sapphire will work excellently for your application. <S> http://www.valleydesign.com/sapppic.htm <A> Ordinary soda lime window glass works okay for longer UV wavelengths. <S> Use as thin a piece as practical. <S> The ideal material is fused silica or quartz glass <S> but it's probably not easily available or affordable. <S> It will transmit even UV-C germicidal light. <A> We used to make our own PC Boards using Dupont Riston dry-laminate film. <S> This process worked extremely well and we used it for years until a commercial supplier of prototype circuit boards became available (AP Circuits in Calgary, Canada). <S> Shameless plug: AP Circuits ships world-wide. <S> Get your board files to them before 11:00 AM MST and finished boards are shipped via FedEx at noon (MST) the following day. <S> Our expose lamp started off being a short-arc Xenon lamp <S> but we later changed to mercury-vapour lamp. <S> This would be very similar to the lamp that you are using but ours was much larger - 175 Watts instead of your 9W tube. <S> The negative film was held in contact with the sensitized circuit board with a vacuum frame made from ordinary window glass. <S> I have no idea <S> what the UV transmission properties of this glass is - we simply adjusted the expose time until we achieved proper results. <S> To do this, we obtained a commercial expose-time calculator film. <S> This is a simple piece of film that has several identical targets on it, with each target having a different piece of neutral-density filter over top of targets. <S> You would expose your test board for double the normal expose time using this film. <S> You would then choose the best target after developing and etching the test board and apply that multiplier factor to your doubled expose time to end up with the new best time. <S> It is now apparently possible to make your own exposure time test film. <S> One such is available at SCREEN PRINTING EXPOSURE CALCULATOR . <S> This is somewhat different from what we used but it may work well for you. <S> It appears that there are several such DIY exposure-time calculators available. <S> For what it's worth, our tests showed that we were able to achieve approximately 100 micron resolution with our circuit board process. <S> This amazing result was due to both scrupulous board prep (physical cleaning, chemical cleaning, pre-etch) and the absolutely-fabulous Dupont Riston dry-laminate film. <A> Looks like polycarbonate isn't a good choice <S> no matter what. <S> Also, check this out.
| Plastics degrade with UV exposure so they tend to have UV absorbing additives that will scupper your plans. Rather, make an adjustable framework which will just contact the board corners, and support the board that way.
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Sign of current through a voltage source simulate this circuit – Schematic created using CircuitLab I am a total newbie to circuits, so please forgive my naivety. My question is the following: When I simulate the above circuit in LTspice, the current across the voltage source is shown to be negative (-.05). That means in this case they mark the current coming out of the the positive terminal of the battery as negative. Is that just a convention or is there some logic behind it? If it's a convention, is that the general standard? <Q> the current across the voltage source is shown to be negative (-.05) <S> Before getting into the meat of your question, we normally say that current flows "through" a device, not a "across' it. <S> That means in this case they mark the current coming out of the the positive terminal of the battery as negative. <S> Is that just a convention or is there some logic behind it? <S> This is the passive current convention . <S> It is widely used (Wikipedia says "universally") in electrical engineering. <S> If you use this convention consistently, then when you calculate the power \$P\$ associated with the component, you will always know what the sign of \$P\$ means. <S> If you calculate $$P=IV$$ with the signs defined by the passive current convention, then a positive \$P\$ means that the component is receiving power from the rest of the circuit, and <S> a negative \$P\$ means that the component is delivering power to the rest of the circuit. <S> Therefore when we use voltage sources to deliver power to the circuit, we expect the current through the source to have a negative sign. <S> If it's a convention, is that the general standard? <S> This is the most common standard, in my experience, when somebody actually thinks about using a standard convention to define the current direction. <S> It's also common <S> , at least in one-off calculations, to simply do things ad hoc , and assign the sign of currents arbitrarily or according to what the engineer thinks will end up giving positive values in the solution. <S> But that doesn't really apply to the designers of LTSpice, who had to choose a convention for any circuit a user might come up with. <A> A 'positive' current is defined as flow from a higher potential to lower potential. <S> You should see that for the resistor in your simulator. <S> Within the voltage source however, the current flows from the negative terminal (lower potential) towards positive terminal (higher potential). <S> Hence the negative sign. <A> So if you have negative value, check if you use the right pin of your source.
| In simulation normally positive current is one that flows into a pin, while negative flows out of it.
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Is control theory too abstract? What do engineers think about the application of control theory? Is it exaggerated? Is the math too overwhelming for real life application? <Q> For most real-life applications (read 90-95%), a PID (proportional,integral,derivative) controller in a feedback loop is sufficient. <S> It can be the case that the calculations for some control algorithms is too complicated for the computer involved, but oftentimes the main reason that engineers stick with PID is that the improvements from more advanced solutions aren't needed- <S> they might only need to hand tune the controller gains once and then the system response is good enough that they can forget about it. <S> Source: I'm in graduate school for mechanical engineering/robotics. <A> No, it's not. <S> Theory in Engineering is devolped as applications become more complex and allows the design of things that otherwise would be impossible. <A> No - it is a basic tool also in the field of electronics. <S> Each working amplifier needs negative feedback (effective for DC only or for DC and AC). <S> Think of operational amplifiers which cannot work without negative feedback. <S> However, in this context it is important to know (a) that (and how) negative feedback degrades dynamic stability (phase margin) and (b) how the tendency to instability can be reduced - for example, with the help of a BODE diagram. <S> Here we make use of the classical rules and tools of control theory (loop gain, phase compensation, pole-zero cancellation, stability margin,...)
| Although there is a lot of research done on things like self-tuning controllers and advanced theory for specific applications like flying supersonic aircraft, a high proportion of controls problems can be solved without many of the advanced techniques.
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Help with dirt-simple led array I am making a real simple 8x8 led array (each in parallel). According to this, can I just skip the resistor?Also, I have a usb charging port capable of 2.1A 5v. Can I just use the generic usb cable to bus the power? Is there any problem that you can see? (last thing I want is burning out some thing) <Q> You require a resistor for each LED. <S> The LEDs are not guaranteed to have identical characteristics <S> so, if you connect them all in parallel, with a single current-limiting resistor, some LEDs will draw more current, and be brighter (and burn out sooner) than others. <S> Are you sure that the 30 mA/LED you show in your calculations is the recommended operating current, and not the Absolute Maximum current, beyond which the LEDs may be damaged? <S> You should never operate an LED (or anything, for that matter) at the Absolute Maximum rating. <A> Should work OK, but leave the resistor in place! <S> I would go about 20mA (assuming a generic LED), and thus ~1.5 ohms. <S> Three 1 ohm resistors, two in parallel with one in series should do the trick +---[===]---+Vin---[===]----+ <S> +---Vout <S> +---[===]---+where - <S> and + are wires, and [===] is a 1 ohm resistor. <S> The resistors will drop 1.8V at 1.3A though, so you'll probably want 5W resistors, 2W minimum. <S> Make sure that 30mA is actually the operating current, NOT the absolute maximum. <S> If so, then your calculations are correct, but I'd bet that it's 20mA continuous. <S> If it's 30, then you can use a 1 ohm resistor. <A>
| Paralleling LEDS usually works if they are all of the same type.(especially for small LEDs) but you need that resistor, either use resistive cable or a discrete part.
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PIC32 does not have Memory Managment Unit, why would a high end microcontroller not have such a peripheral MMU is needed to run Linux and perhaps for other tasks related to OS. However, the PIC32 does not have one of these. Why would someone make a high end microcontroller and not include a MMU? It may make sense for lower level PICs like PIC18F, PIC24 to not have MMU, but why did Microchip not put in an MMU into the PIC32 which is their high end processor? <Q> First, the PIC32 doesn't run Linux. <S> A cursory Google search shows a lack of working full-fledged Linux distributions or kernel builds for the PIC32 (for reasons such as insufficient RAM, lack of interest) <S> Second, while it may be a high-end offering, it is still a microcontroller, whose purpose is somewhat different from a general-purpose processor. <S> It only carries up to 32 KB of physical RAM, which is definitely not enough . <S> However, a reading of the PIC32MX family datasheet shows that the PIC32 family contains a "fixed mapping translation" unit for virtual memory. <S> This MMU-like feature does implement memory-mapping resources such as flash, as well as kernel/user segmentation. <S> Edit: <S> I can't answer why Microchip doesn't get involved with Linux/Android, as I don't work for Microchip. <S> I can only say that, based on the specifications of the specific microcontroller that you mention, it may be high-end for a microcontroller/microprocessor, but is not sufficient for a Linux-based system, let alone one running a Java VM, with a JIT (won't happen due to memory, and because JIT on a Harvard architecture is a questionable premise). <S> A microcontroller is designed to act in a simple embedded application in a low-latency and highly reliable manner, and running Linux or Android is not necessarily conducive to doing things cheaply, and in a reliable and realtime manner. <A> PIC32 does mot have Memory Management Unit ... <S> But it does. <S> The PIC32 uses the MIPS architecture, which normally is paired with an MMU. <S> The largest PIC32MX, in terms of memory, is the high-end PIC32MX795F512L which has 512KB of program flash and 128KB of RAM. <S> However the newer PIC32MZ family, introduced earlier this year, does have an MMU (or so the datasheet claims on the first page; I haven't used it yet). <S> The PIC32MZ2048ECM144 , for example, has 2MB of program flash and 512KB of RAM. <S> I'm pretty sure this is still not enough to run a real version of of Linux. <S> In any case I'm guessing Microchip will continue to come out with new PIC32MZ versions with increasingly larger memory spaces. <S> Ideally, they might come out with a PIC32 that has an external memory interface (appearing in the memory map as regular memory, not a peripheral). <S> PIC32s already can run code out of either flash or RAM. <S> A while back I saw a job listing from Microchip for a Linux drivers engineer. <S> Hmm. <A> An MMU is primarily useful in cases where a system will need to host multiple processes with independent arbitrarily-overlapped lifetimes. <S> Support for such actions is essential in Linux, but completely unnecessary in many embedded applications where all processes are established on startup, and continue to run until the system is powered off or reset. <S> In general, there are significant tradeoffs between a system's ability to adapt to dynamically-changing usage requirements and its ability to guarantee consistent timely responses to stimuli. <S> Something like an automatic guided vehicle's motion control system may need to perform some calculations 10,000 times/second consistently without hiccups, but may have no need to keep some processes active while others are changed. <S> It may be completely sufficient for it to have a command to "confirm that vehicle is motionless and brakes are set, then reboot. <S> " <A> I think you can use µClinux ; μClinux is a fork of the Linux kernel for microcontrollers (in embedded systems) without a memory management unit (MMU).
| When Microchip designed the PIC32MX family, they left it off. If there is no need to have a system's memory layout change while processes are active, there's little benefit to having an MMU.
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Exact definition of overshoot I am a computational biologist and am working on some aspects of control theory in biological systems. Since control theory concepts are not well known among biologists, I need a good standard reference for certain terms. I was thinking of citing a control theory textbook. Now I have a doubt regarding the definition of overshoot. The book — Modern Control Engineering by Ogata defines maximum overshoot as: The maximum overshoot is the maximum peak value of the response curve measured from unity. If the final steady-state value of the response differs from unity, then it is common to use the maximum percent overshoot. It does not say what maximum overshoot is, when the final steady state is not unity. In my analysis, at present, I am defining overshoot as the max value above the tolerance zone normalized by the steady state value. I am not sure if this fits the above definition (steady state being scaled to unity, however I do not report the steady state anywhere). When I google overshoot I see images that have labelled overshoot differently which is either of these two: The dynamic maximum The difference between the maximum and the steady state. Can someone please let me know what is the correct definition of overshoot (not percentage overshoot)? I need a good citable reference (a book or a review). <Q> When you mention an example where steady state is unity, that is normalized. <S> Typically we are only interested in the percent overshoot, so unity makes it easy to scale. <S> I.e. <S> If a system has 10% overshoot on a system that's steady state is unity, the max peak is 1.10. <S> Your definition is identical to the control systems approach. <S> Overshoot is simply the difference between the max peak and the steady state value. <S> If a system has a quick response, it will typically have a larger overshoot value. <S> If a system has a slow response, typically the overshoot is very small. <S> This image has a good breakdown of a response. <S> Hope <S> this helps. <S> If you have any other questions, please shoot them my way. <S> Good luck! <S> - Josh EDIT <S> This is a page out of the book I use: <A> WYSIWIG - as you need to cite a reference: <S> G.F. Franklin, J.D. Powell and A. Enami-Naeini: Feedback Control of Dynamic Systems , Prentice Hall, <S> 4th edition: <S> (Quote) <S> The requirements for a step response are expressed in terms of the standard quantities illustrated in Fig. <S> 3.27: ... 3.) <S> The overshoot <S> Mp is the maximum amount the system overshoots its final value divided by its final value (and often expressed as percentage) . <S> EDIT/ <S> UPDATE: <S> Karl Johan Astroem, Richard M. Murray (Princeton University Press) <S> "Feedback Systems": <S> Quote: <S> The overshoot Mp is the percentage of thefinal value by which the signal initially rises above the final value. <S> This usuallyassumes that future values of the signal do not overshoot the final value by morethan this initial transient, otherwise the term can be ambiguous. <A> You should review the references cited in some of the answers given here, but regarding what you should use <S> - I suggest you define your own. <S> There are IEEE, ASME Standards, but as far as I'm aware no one defines a 'standard' way to define overshoot. <S> You should define it as you see fit - for what works best in your situation. <A> When the step response is expressed as a non-dimensionalized equation, the definition of maximum percentage overshoot becomes easy. <S> For some second-order systems, the original equation itself is a non-dimensionalized equation (when it has wn^2 in the numerator) and the steady-state value will be unity for a unit step response. <S> That is not generally true for all second-order systems depending on the numerator. <S> Therefore, we must ensure that the given step response equation is converted into a non-dimensionalized equation by dividing by an appropriate value of constant that would give unity steady-state value.
| You can see that the overshoot is simply the difference between the max and steady state value.
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Why use Optcoupler + Mosfet? I saw this project: http://diyhacking.com/arduino-lamp-dimmer/ which I'm interessted to(except that I will have to use some 24V, still enough to hurt the Raspberry Pi). There is several things that I'm not sure to understand: The optocoupler is used to isolate the 5V from the 230V(24V in my case), ok The MOSFET is used to normaly used to command an higher voltage with a lower voltage. So what I don't understand is why we need the MOSFET since the optocoupler already receive some 230V(24V)? Thank you <Q> It adds another layer of protection even with 24V. If you screw up, screwdriver or test probe slips and shorts two connections together (or the wrong ground connection opens up) <S> you're much less less likely to ruin your micro board. <S> Your choice, there's not a human safety reason if the 24V has proper galvanic isolation as well as a common ground. <S> Since the 4N35 is so very painfully slow, using it degrades the switching of the MOSFET. <S> This could be dealt with by using a somewhat better part and/or squaring up the signal on the other side of the barrier, but I think this is a major downside. <S> In development, it's not uncommon to temporarily add isolation from mains or high power supplies. <S> Or maybe just add some 10K resistors and a buffer. <S> It tends to ruin your day if excessive voltage finds its way into the wrong place and destroys the test object and your computer motherboard at the same time! <A> someone mentioned not needing to use opto-coupler with a 24v circuit.it can be done, but it is best practice to isolate all LV control lines from higher voltages where possible to avoid damaging your control logic. <A> For RPi I use a ULN2003, strictly because I happened to have some laying around. <S> Coincidental, I'm going to 24V as well, and sharing gnd from 24v supple with RPi. <S> Real simple <S> Most pages say to use the opto-couplers, and probably best practice.
| Real product designs would probably not use the opto unless there was some reason to keep a noisy 24V supply completely separate. The collector current for the 4N35 Opto-coupler is only rated to 100mA,if the circuit you are controlling is going to consume more current than 100mAYou will need a to use a MOSFET to sink the current.
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Maximum voltage allowed in trigger pin of 555 bistable I want to measure the frequency of a bicycle hub generator, consisting of a variable AC voltage output. Nominally, the RMS voltage is 6V, but with no load attached the voltage peaks can be tens of volts. My plan is to use a 555 in bi-stable mode, attaching its output to an Arduino interrupt to count its "rising" event. What I need to know is if I need to, and how to protect the TRIGGER pin from over-voltage. Most probably the 555 will be powered from the Arduino (Vcc = 5V). So the question is: What is the maximum (positive and negative) allowed voltage on the TRIGGER pin of a NE555? How should I typically protect this pin if external circuit is expected to go over the voltage limit? I took a look at the datasheet, but I'm afraid my current knowledge is not enough to interpret it. <Q> Per the 555 datasheet, absolute maximum voltage on the trigger pin is equal to VCC. <S> The easiest way to protect the pin is to use a series resistor and a schottky diode to VCC: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The resistor should be sized sufficient to limit the current shunted through the diode; anything in the range of 10k to 1M is likely to be fine in your application. <S> In a pinch you can omit the diode and rely on the protection diodes in the 555, but it's generally considered a bad idea to do so. <S> Since your output is AC, too, you will need an additional diode to block negative voltages, or a resistor divider to shift the voltage appropriately. <A> The first answer to this question has the correct idea, except the diode suggested is NOT a Schottky diode. <S> This means that it is the static protection diode on the input of the 555 timer that will carry the current. <S> As the input voltage rises past Vcc, the current will flow through the diode with the lowest forward voltage. <S> Almost all chips have Schottky diodes on their input pins to both ground and Vcc - these are not intended to carry current, but are a last line of defence to protect chips from static discharge. <S> IMPORTANT: <S> Because the input is an AC signal (i.e. goes negative), you will also need a Schottky diode to ground! <S> The cathode on both diodes face Vcc. <S> For further protection, you can add a small (e.g. 100 Ohm) resistor between the external diode and the chip to further encourage the excess current to flow in the external diode(s), and not the internal one. <S> They also tend to have a large forward voltage, so they don't do a great job of protecting against negative input voltages. <S> Some brave (foolish?) <S> designers will rely on the input protection diode to limit the input voltage and they will simply install the series resistor. <S> But here's a question <S> - why not feed the signal from the generator directly into the Arduino? <S> If it has a Schmidt input (so it can deal with slowly-rising input signals), and you provide appropriate input protection, then you do not need the 555. <A> You are adding in a 8 pin component to do complex input conditioning that could probably be done in software. <S> You are just comparing the frequency of the pulse train. <S> The second circuit in this answer is all you need. <S> You just need to size the input resistor to protect the diodes and prevent currents higher than the circuit consumption so it will not raise the 5V rail, if you fear this happening then having a crowbar Zenner diode on the supply rail or your input clamp is a good idea. <S> You can select a pin on the Arduino that can interrupt or count if needed. <S> The frequency counter project <S> described here may give you more ideas. <S> Also note that you may want to be able to set the frequency setpoint <S> , this may pose difficulty or benefits depending on the software skills and requirements yo have.
| Another option is to use a reverse-biased zener to ground (which has the added advantage of also protecting against negative transients on the input line), but the problem with zener diodes is that the "knee" of the conduction curve is soft so, you are almost guaranteed to force some current through the internal input protection diode.
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Optoisolator with EL wire won't turn off, TRIAC works fine. Why? I'm pretty new to electronics, and I'm making a circuit to control EL wire with a DC logic control signal. Specifically, I would like to blink the EL wire with the output of a 555 timer. My first pass at this was with an optoisolator. MOC3031 data sheet The circuit looked like this: (C1 is a stand-in for the EL wire, which I'm told is a capacitive element, and the 555 outputs 5V) The good news is that if I turn the EL wire inverter on when the 555 outputs 0V, the EL wire will be off until the 555 jumps back to 5V. The bad news is that the EL wire never shuts back off, even when the 555 outputs 0V again (I've confirmed that the 555 circuit works w/ a multimeter). I ripped out the optoisolator and replaced it with a triac, so now it looks like: And everything works perfectly. I'm happy in the sense that I have a working circuit, but I'd like to know what changed. Can anyone explain why? EDIT: I want to switch some of the EL elements on and off while the inverter remains switched on and drives other EL elements continuously. The frequency is likely higher than 50-60 Hz and the voltage is probably not a clean sine wave. <Q> Something is being missed here. <S> The optotriac is quite capable of switching that load current, however the triac in the optoisolator is handling the inverter's high frequency AC differently from your 'triac'. <S> It's designed to control mains frequency AC - 50 or 60Hz, and is too slow to turn off at the very brief zero crossings when it is fed with ~1kHz AC current Edit: there is a good app note from Vishay here <S> which makes it a bit more clear. <S> You cannot just use the static dv/dt spec - <S> that is what prevents the opto from turning on when it is off <S> , rather you need the (lower) commutating dv/dt, which may not be given. <S> In fact, perhaps it's not actually a 'triac' but an alternistor that you're using, which would tend to perform better because the semiconductor structure is different. <S> The answer in that case is to use a more appropriate switching element, such as a OptoMOS SSR. <A> Based on your response to my question the answer is relatively simple. <S> A DC powered device draws DC current. <S> A triac or triac-opto coupler stays on as long as the current through it is above its minimum holding-current. <S> Holding current is just "stay-on current". <S> Once it's triggered as long as the current is larger than the minimum, it will stay on. <S> This is why they are only used for AC voltages or at least some other voltage that goes to 0 regularly, to make them switch off. <S> But,... why does the Triac actually switch off? <S> Simple: <S> One of those MOCs can only handle about 100mA depending on type, so their holding current is probably in the single mA's (I'm too lazy to Google a datasheet now). <S> Your triac might be made for a current of 10A or such (you don't tell us), at which point its holding current can easily be up to 50mA. <S> If you EL inverter then uses 25mA DC, the MOC will not turn off, because that's much more than its holding current. <S> The Triac will turn off, because you aren't at the holding current yet. <S> In general you want to not use Triacs and such for low voltage DC, because they also waste a lot of energy: They always have a comparatively large voltage drop, even when on. <S> You should probably just switch to a transistor or MOSFET: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Or if you want opto-isolation: simulate this circuit <A> The datasheet for the MOC3031-M specifies that it's meant to be a trigger and not directly drive the load. <S> As a result, I would imagine that the internal phototriac is much more sensitive than the eternal triac depicted in your second circuit and is probably why the phototriac was never turning off and your EL wire was always on. <S> Instead, you should drive an external triac using the optocoupler to switch the load. <S> See, for example, this circuit from the datasheet: If you put the optoisolator in-between your 555 timer and the triac from your second figure, you should get the desired result.
| The Triac is meant for a higher current, as such its holding current is higher.
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Unstable Feedback in Opamp+MOSFET circuit for Voltage Controlled Current Source I'm attempting to build a 1 Amp LASER driver that can be pulsed (at ~10kHz) and gives control over current with a voltage (from a DAC, for example). I'd like to scale it up to 10 Amp pulses eventually. I tried this common circuit below on a breadboard, but the feedback is unstable. Arrows indicate what should be (apprixmately) happening, which I confirmed on simulations. I'm using a 500 mOhm 10 W resistor to simulate LAESR diode on resistance, and a 1 Ohm sense resistor. Here is what happens at the opamp output: And here is what happens at the inverting node (voltage across sense resistor): I tried changing around the values but things only get worse. I thought the opamp was too wideband so I increased compensation cap C3. Feedback becomes stable when I make it 4.7uF (yes, 4.7uF), but pulsing Vctrl makes rise and fall times ~1s. Value of R3 doesn't seem to matter as long as it's there. This is the opamp I'm using (MCP6022, not 602), and this is the FET I'm using (IRLI3705N). Is the opamp too wideband for this application? Its correcting too quickly and overshoots? Am I stuck at the "knee" on the FET's I-V curve so a little bit of gate voltage changes current exponentially? Is my Vcc not large enough? Is this even the right VCCS solution for a 10 A pulsed load? <Q> Implementing this in a plug-in breadboard will probably not work, the parasitics are too high. <S> 2) Remove C4 - it is possibly causing phase shift in your feedback loop. <S> 3) You have a 22pF cap (C3) from the output of the amplifier to the input, but you have it directly connected to the 1 ohm sense resistor. <S> This will prevent it doing anything useful. <S> I would add a 1k restore between the sense resistor and the inverting input of the opamp to isolate the stabilizing feedback path (C3) from the lower frequency signal from R4. <S> You may need to try different values for C3. <S> Have you simulated the circuit? <S> kevin <A> A couple of points... <S> Monitor the 5V supply, <S> just in case you're inducing instability in it - or <S> the 300ish kHz signal you're seeing is actually its switching frequency. <S> Experiment with increasing R3 to at least 1 kilohm, thereby isolating the opamp output from a substantial load capacitance (Cgs of the MOSFET). <S> Now, even if increasing R3 cures the oscillation, it may not be the solution, because it decreases the bandwidth of the MOSFET drive (you can measure your pulse rise/fall times with and without the change, to determine if its effect is harmful). <S> But there is a substantial body of literature on stabilising opamps driving capacitive loads, which should help you find a better solution. <A> Add a resistor between R4/M1 and the rest. <S> 1K for starters, up to maybe 5 or 10K. Remove C4. <S> You can increase C3 rather than the resistor value. <A> I used a same circuit for driving a laser diode with up to 500 mA current. <S> I have some recommendations as follows: <S> Use <S> LM358 <S> op amp Remove C2 and C4 <S> Use fast bipolar transistor for better stability such as D828
| A couple of things can cause instability 1) With such low impedances you need to make sure you have a low inductance ground system.
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Frequency response of a series RC circuit The magnitude and phase plot of the circuit are as in the image below. I have trouble understanding the meaning of phase plot. The transfer function being a complex quantity has both magnitude and phase. The magnitude plot of the transfer function is simply provides the ratio of output to input. But what does the phase plot mean? Is it the phase of output voltage being plotted? <Q> Firstly,The plots and the RC circuit doesn't match. <S> The plots are for LPF and the RC circuit in the picture is HPF. <S> Secondly, coming to the answer. <S> Based on the operating frequency, the transfer function of the system alters the magnitude & phase of the incoming signal. <S> If \$V_{\text{in}}, \theta_{\text{in}}\$ is the input signal <S> and\$V(s),\theta(s)\$ is the system response to the input frequency. <S> then$$V_{\text{out}} = <S> V_{\text{in}} \times V(s)$$$$\theta_{\text{out}} = \theta_{\text{in}}+\theta(s)$$ <A> These graph are use to represent dynamic system. <S> When you apply a SIN to a RC circuit, it is a dynamic problem. <S> If you go into time analysis, you end up with equation that look like something like that: \begin{gather}C\frac{\mathrm{d} V(t)}{\mathrm{d} t}{} + \frac{V(t)}{R}=0\end{gather} <S> If you did a frequency sweep of a dynamic system, you will see that amplitude and phase are linked to frequency, <S> if you amply a high frequency signal to a low frequency system, you will see that the low frequency cannot follow the speed of the excitation therefore, the amplitude of the output is smaller than the input and the output is delayed in reference to the input (phase). <S> The opposite can happen, some system respond well to high frequency input and can't follow low frequency input. <S> So because of that, we can conclude that the differential equation is a bad tool to analyze circuit, because we would need to re-calculate everytime the response. <S> In this case, you call your friend the mathematicians, and they came up with something call Laplace Transform ( it is just Laplace that did it :P). <S> The laplace Transform is the following equation:\begin{gather}F(s)= \int_{0}^{ \infty }e^{-st}f(t)dt\end{gather} <S> In a nutshell, by performing a Laplace transform on a time function like the V-I relation of a cap or an inductance. <S> You end with a transfer function that connect the frequency to system.\begin{gather} Z_{L} = <S> sL <S> \\ Z_{C} = <S> \frac{1}{sC} <S> \\\end{gather}More precisely, when you only consider the permanent regime, you make a Fourrier Transform and you end with:\begin{gather} Z_{L} = <S> jwL \\ Z_{C} <S> = \frac{1}{jwC} \\\end{gather}So with the impedance you can find the transfer function of a circuit. <S> Random Example:\begin{gather}\frac{V_{out}}{V_{in}} = \frac{\frac{1}{RC}}{2\frac{1}{RC}+jw } \\\end{gather}and <S> you can make vary the w to see how the frequency affect the input vs output of the system. <S> The phase plot represent how much delay there is between the output and the input. <A> Again, the plots and the RC circuit do not match. <S> By using the principle of voltage division and then creating the ratio of the output to the input, it can be seen that the transfer function is given by: $$H(s)=\frac{sRC}{sRC+1}$$ where \$s=j\omega\$. <S> This would indicate that at low frequencies the transfer function approaches zero and at high frequencies the transfer function approaches 1.
| To answer the question, you can think of the phase plot as indicating how the system changes the phase of the input signal at different frequencies, resulting in the output signal.
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Is it safe to connect an Arduino to the human body? I want to create some project which uses an Arduino to collect and display ECG data. It uses two electrodes which are attached to the body (on the chest). The signal is then transmitted through op-amps to an Arduino ADC pin. The Arduino is connected just to a PC. Is it safe for both the human and the Arduino board to be connected this way? Should I make some safety pracautions and if so, which? <Q> It's not advisable unless the entire project is low voltage and battery operated. <S> Anything mains operated could be dangerous even if transformer isolated. <S> There are specific safety requirements for mains operated patient attached equipment to protect against excessive leakage currents and potential safety hazards due to equipment or component failures. <S> Those regulations apply to professional medical equipment, but they are there for a reason. <S> You shouldn't assume that it's safe to ignore them even if it's just a hobby project. <A> When doing an ECG test, the electrodes are much more conductive than dry skin, so the possibility of triggering fibrillation or other arrhythmia from small voltages is present and needs to be taken very seriously. <S> Most pacemaker pulses for implanted pacemakers are from 2mV to 250mV. <S> That's not a whole lot of voltage, and if your device accidentally drives that to the heart, you could be in a great deal of trouble. <S> Short answer: <S> Read IEC 60601-2-25. <S> Medium answer: To do it these days, you need to have all your digital signals opto-isolated across a physical isolation boundary, and your power source has to be isolated (i.e. transformers). <S> There are very detailed and strict requirements on what kind of protection <S> those need to be capable of, which involves being able to withstand being zapped with many kiloVolts and not crossing the boundary. <S> All your amplification and data processing needs to take place upstream of your isolation boundary, with basically nothing but <S> a UART crossing it. <S> Use something like the TI ADS1298 as an analog front end, which you can communicate to through SPI. <A> You are not going to be able to read ECG signals with the on-board A/D, it does not amplify enough or have enough common mode gain rejection. <S> You are going to have to use an instrumentation amplifier. <S> With proper design (opto-coupler) it should be fairly safe to connect a battery powered instrumentation amplifier to an arduino, at least that is how I would do it. <S> (I would wall power the instrumentation amplifier, but I'm a risk taking idiot) <A> is the laptop's battery voltage - <S> that should be safe enough). <S> I made a similar project some time ago using different parts - I chose to power the whole device from batteries and use Bluetooth to send data to the computer. <A> The easiest, and safe , design would be to use a battery powered amplifier to amplify the electrodes signal, connect the amplifier to a battery powered Arduino to capture, process and store the information. <S> After disconnecting the electrodes from the patient , connect Arduino to PC any way you want (can), and transfer the data to the PC. <S> Obviously, you don't use this apparatus on anyone that has (wears) a "pace maker" !
| It is safe if you connect Arduino to the laptop and run it only using batteries (then the highest voltage that could shock you In short, you basically have to make a custom PCB to do it properly.
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Connecting 2 male headers on a pcb together I have 2 male headers like this on 2 different boards: I need to connect 2 such male headers together. One way to do this would be to use something like individual female to female jumper wires. However, I am operating at a frequency of about 32-48 Mhz and don't want such long traces. I wanted something like a female connector on both sides, but I looked through all the types shown here and I still can't find any. Could someone give me any ideas as to what kind of connector I need to use? I dont want to make a PCB just for this. <Q> I would use two female headers like this... <S> 2X20 40Pin <S> Double Row Female Straight Header soldered back to back to make a female-female joiner. <S> If you overlap the contacts they only add 15mm to the overall length (compared to a normal male/female header combination) which should not be significant at 32-48MHz. <S> If any of your signals still have excessive ringing or ground bounce then you could try inserting small value resistors (10~50Ω) in series between the connectors. <A> These do require a fair amount of clamping force to snap the connector ends into the ribbon so if you have to make many of these <S> I'd advise you to have some custom-made by a cable assembly manufacturer. <S> This is so common that it should be quite cheap even for small runs. <S> Cable length, width, keying directions, pin pitch, and number of rows are all customisable. <A> I'd use a male header on one board and a matching receptacle on the other, something like this:
| Ribbon-type IDC connectors are available as a self-assembly kit from any vendor, just choose the number of connections for both the ribbon width and connector width...
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While using arduino to close a circuit (the output to transistor's base), is it necessary to connect negative of circuit to ground in arduino? Please refer to the following image: I hope the quality is not too bad. Anyways, as can be seen, the circuit has its own external power supply from a wall wart (AC to DC transformer). The arduino is being used only to close the circuit by means of providing the signal current to the base. My question is: is the only thing that needs to be done is to connect the Arduino output to transistor's base? Is there a need to somehow make use of the GND pin in the Arduino (for example, by connecting the -Ve of wallwart along with the emitter to the GND)? While I am at it, I have another question: how to find the +Ve and -Ve on the female DC jack? As can be seen, there are 3 pins sticking out of the DC jack, (it is centre positive I think. I bought the jack because it makes it easy to connect the wall wart to the motor, i.e., without having to cut open the adapter's male jack). <Q> You will need to ground the emitter/DC jack's ground with the Arduino. <S> Without doing this the base has no reference point to be driven from. <S> As for the DC socket. <S> Check first which contact of the plug is positive and which is ground. <S> Most often the internal contact is positive, but not always. <S> Once you've done this, either check the datasheet of the socket or use a multimeter in continuity mode to check which pin the middle is connected to. <S> MY guess would be the one on the back is connected to the center pin, and the other two will be the outer. <A> Yes you will need to connect the arduino ground and the GND of the supply to the dc motor. <S> I also second Jarrods opinion of using a diode (flyback) as this is an inductive load. <S> You can also use a PMOS (with a low Rds and relevant Id) to do the job. <S> Most of the present day MOSFETs have inbuilt flyback diode. <S> Only thing is you need to modify the software of the arduino as the PMOS will on when the arduino output is low and vice verca. <S> P.S - am assuming you are using a 5V compatible UNO or similar. <A> Since others have covered most of your questions, I'll address what I think is missing from their answers: <S> The third pin on the barrel jack you have is used to detect when the connector is inserted. <S> I rarely use the feature. <S> If you don't use it, leave it unconnected. <S> Here's a links to an article about it: SparkFun tutorial on Power Connectors and a one picture "datasheet": Make sure you have a series limiting current resistor in series with the Arduino output and transistor base. <S> The transistor doesn't limit current by itself, so without the resistor, you'll fry the transistor when you turn it on. <S> You should select the resistor value considering the transistor gain and the amount of current you want to put through the motor. <S> Here's a handy transistor base resistor calculator . <S> Motors demand much more current to start up than when running free or with a moderate load. <S> This may trip your wall-wart current limiting protection circuit if it's not capable enough making it shut down when you turn the motor on. <S> The symptom of such problem is that your Arduino may reset intermitently and your motor may run jerkly. <S> If that happens, it means you need a more potent wall wart. <S> Or you can solve that using a set of AA batteries to power your setup.
| Just so my answer is complete , yes, you must connect the Arduino ground to the DC jack ground.
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Why are two diodes used here? I'm following the wiring diagram for a digitech fs3x guitar pedal footswitch. For the third SPST switch, it's wired from the other two switches connected by 2 diodes, both in the same direction. Can anyone explain why a single diode wouldn't work here (with both wires connected to it) <Q> <A> The two diodes are preventing a high potential on EITHER the Ring or Tip from generating a current flow into the Tip or Ring respectively, when none of the momentary switches are depressed. <S> A single diode, say for example the diode between the Tip and the Up button, would not prevent a potential on Tip from generating a non-negligible current flow into the Ring when Ring was at a sufficiently lower potential than Tip. <S> This potential would be approximated by the forward voltage drop of the N14002. <A> Looks to be so you can allow three switch configurations that don't affect one another. <S> Mode pulls down tip, down pulls down ring and up pulls them both down. <S> Without the diodes you couldn't use the same spst switch for up. <S> Notice how they tie tip and ring together through the diodes, <S> so they never short to each other, but pressing up pulls them both down at the same time.
| The two diodes are to that the UP switch actuates both lines, while still allowing those lines to act independently and prevent the other switches on those lines to have the same affect as the UP switch.
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Phone travel charger for powering circuits dangers I am thinking of using a phone travel charger to power a circuit.Ratings: input:110-240VAC(my outlet has 220volts) 50/60Hz 0,3A output:DC 5,2volts at 400mA The problem is that 220 V are of course dangerous and I 'm not sure of how safe is my charger. My question is:do phone travel chargers have fuses inside them in case there might be a component failure and an electric shock would become possible?I also opened the case a few times ,but I didn't modify anything inside and before I put it back last time,the circuitry was intact. EDIT:Here are the requested images- <Q> An unmodified and genuine charger actually made by an internationally renowned cell phone company such a Nokia or Samsung will be very safe under normal operating conditions. <S> Apple ones are great too, but there are many, many counterfeits about. <S> Even if it is top-drawer, I still would not tempt fate by using it when water is about, or any other situation that could involve low impedance connections to a human. <S> There are counterfeits and all sorts of junk flying around. <S> Fuses won't necessarily save you from electrocution- <S> if there is an internal breach of the galvanic isolation barrier (say due to bad design, component failure, modification or debris inside bridging the gap) <S> then the fuse (if any) could happily pass enough current to kill you. <S> If the adapter power is flipped it may not even be in the circuit. <S> The purpose of a fuse is to allow the adapter itself to fail gracefully without excessive drama. <S> I have personally taken apart several chargers with counterfeit markings (both the company name/logo and the safety agency markings were counterfeit) and those particular ones were extremely dangerous. <S> They would never pass safety agency approval. <S> I would never give one to a kid or to any other human I cared about. <S> I might plug them in for test purposes, but would keep a careful eye on them. <S> Also junky electrically- noisy output voltage and a lot of conducted EMI. <S> Aside from reliability and the quality of the power (cheaply made ones may not have OVP, so <S> a failure of a solder joint could render your $1000 phone into trash), the two main dangers are electrical shock and fire. <S> Those dangers are mitigated by careful design meeting safety agency guidelines, listed or approved materials for every safety-critical application and thorough testing. <S> For example, the approval documents will list the approved suppliers of the polyester tape and wire used in the transformer. <S> They must use material only from those suppliers (who, in turn, follow QA procedures internally). <S> Junk adapters will use whatever is cheapest- <S> maybe even rejected material from top-tier suppliers. <A> If your charger is Cheap-Special-Offer-Crap from eBay, then it's anybody's guess how safe it is. <S> Even if it has a UL/TüV/ <S> Similar marking, because cheap crap means there was very likely no actual testing, it's probably just a sticker/stamp-mould <S> the manufacturer used without official permission to do so. <A> The problem is that 220 V are of course dangerous and I 'm not sure of how safe is my charger. <S> My question is:do phone travel chargers have fuses inside them in case there might be a component failure and an electric shock would become possible? <S> It looks more competent than many, but is a low cost hand built product. <S> That is not necessarily bad in itself but raises doubt about the things than cannot be checked visually. <S> They have isolated high and low voltage moderately well with the gap in the PCB copper area and have added a slot where the low to high voltage feedback optocoupler is to increase "creepage distance". <S> These two aspects show they understand what is needed and have made efforts in the right direction. <S> What cannot be seen is the quality of the switching supply transformer construction and this is crucial. <S> It is possible to "cut corners" badly on such things to save a small amount of money and you cannot tell directly if they have done so. <S> BUT there is another area where they HAVE cut corners to save money (or just do not care) <S> and this suggests that you may be unwise to trust them with your life. <S> There is NO input <S> RFI (EMC) filtering at all on the mains input and the supply would not pass modern certification requirements for this reason alone, let alone others which may be unseen. <S> The diodes at top left here <S> rectify the mains. <S> In the photos below, I have mirrored thge top PCB left for right so that the layout matches the underside view. <S> The only filtering for switching noise is the resistor at top let and the capacitor at bottom left. <S> The resistor also serves to limit the inrush current when 1st plugging ioj the power supply. <S> The amount of noise filtering would not be enough to meet regulatory requirements. <S> The high to low voltage PCB copper gap can be seen and the slot where the opto isolator is. <S> I've seen MUCH worse than this. <S> If you have used this with your phone for some while and the phone does not die <S> then they MAY have made a good job of the transformer. <S> Any such supply is a gamble but this is potentially better than many. <S> Whether potentially and " actually " match is unknowable. <S> DanielPSUV3.jpg <A> The big risk is that the transformer has not been constructed correctly. <S> There may not have been sufficient insulation in the first place and/or bad winding processes may have damaged the insulation that is there. <S> Unfortunately there is no good way to test or inspect said insulation, you can do high-voltage insulation tests but they are not conclusive. <S> The best you can do is buy power supplies of known-brand from large vendors in countries with strong product liability laws. <S> Your power supply's label has absolutely nothing on it to identify the manufacturer which is concerning to say the least.
| If you overload it, it is possible it may break, but with many 5VDC chargers that's unlikely, but it must certainly not create shorts or hazards on the 230VAC side. If your charger is branded (Siemens, Samsung, Nokia, Apple, HTC, etc etc) and/or has a UL/TüV listing that can be taken seriously you can use the charger's 5VDC output for whatever you want.
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Why do I first need to set a value and then direction of the GPIO in an embedded processor? I have a full time job as a firmware engineer. I've recently been given a task to review GPIO configurations and change the settings as needed. I found a few pins that were incorrectly configured so naturally I reconfigured them, however I was told I did it in the wrong order. Here is what I'm talking about: Before: GPIO1.direction = INPUT; After: GPIO1.direction = OUTPUT; GPIO1.value = 0; However during the code review I've been told that I need to change the order of initialization to the following: GPIO1.value = 0; GPIO1.direction = OUTPUT; In other words set the value first and then set direction of the pin. I've also been told that this is how it needs to be on the modern processors because they use two registers, one for input and one for output, however old processors use only one register, so the order of operations wouldn't matter. (Note: Modern = ARM Cortex M3 and above, Old = Intel 8051) I asked for a better explanation at work, but I couldn't get a good answer. That's why I decided to ask here. So here are my questions: Why does the order of initialization matter on the new processors? Why does the order of initialization not matter on the old processors? What two registers are they talking about in the modern processors? What single register are they talking about on the old processors? If someone could provide some sort of a diagram, that would be even better. <Q> The original 8051 used so-called pseudo-bidirectional output ports (open-drain with pullups), so there was really no port direction setting. <S> Of course for modern true bidirectional output ports it's better to have a known value set before enabling the port pin for output, because otherwise you could have a transient on the output that could do something undesirable. <S> See my answer here , for example. <S> Edit <S> : Here is the I/O pin structure for a (relatively) modern CMOS microcontroller : TRIS (TRIState) is called DDR (Data Direction Register) in many other micros. <S> In this case, if the TRIS latch output is high then both transistors are 'off', but the port can still be read. <S> Here is a slightly more complex <S> I/ <S> O pin structure for a newer Microchip micro . <S> Again, the TRIS latch disables the output. <S> This one includes a LAT latch that helps avoid read-modify-write issues . <S> On the PIC series you should write to the LAT register only (and read from the PORT register). <S> Here is the original 8051 and CMOS 8051 classic <S> I/ <S> O port pin internal circuitry (from this source ): <S> There's a bit of extra complexity in that there is a speed-up transistor in parallel with the pull-up that is briefly turned on to overcome external capacitance. <S> As you can see, there is no TRIS/DDR control at all. <S> The pull-up MOSFETs used in normal operation are 'weak'- <S> they are small enough (low Idss) that an external output connected to the pin can pull the pseudo-bidirectional port line low. <A> If you set the direction first, the pin will briefly be configured to output whatever its current output value is. <S> If you set the value first, this won't happen. <S> So, doing it the way you've been recommended avoids glitches on the output, which could range from harmless to catastrophic, depending on what the pin is connected to. <A> It is in fact necessary when your application requires that on the startup the value of the port won't be, say 1 . <S> Then you will set the value to 0 and then change the direction. <S> In this case you avoid the possible momentary "glitch" between the setting the direction and the value, which might result in a spike on that pin. <S> And it is true for all of the processors having such a logic, not only the new ones.
| Assuming the default direction is an input (i.e. High-Z, which is making sense as we don't want the MCU to force any value on the lines connected), this order of setting up the port is preferable but not necessary.
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How can I rename a Sheet in Altium Designer? I'm trying to rename a schematic sheet file inside Altium Designer 14 (in Project Panel). The only way I can do it is renaming the sheet file in Windows Explorer and relink the file to the project. This is very time consuming and unproductive. I found no way to do it directly in Altium. I tried the context menu, F2 shortcut and the File menu with no success. Does someone knows if it can be done directly in the program? And how to do it? <Q> As mentioned, you can Save As, which saves a copy of your document. <S> However, the correct way to do it is to open the Storage Manager panel (System-->Storage Manager): <S> I would show a picture of the storage manager but mine contains confidential files from my work, but I'm sure you'll be able to figure it out. <S> Hope this helps! <A> Right-click on the file in the Project panel and Save <S> As . <S> But then you will have to delete the file with the old name, anyway. <A> it´s really simple, you just have to rename on your folder and then open the project again and it will be with the new name. <S> if you want to change the name of a sheet into the project is the same process, when you open the project again you just have to move the sheet to the project again ;) <A> Files such as schematic sheets can be renamed from within Altium using the Storage Manager window. <S> The method for accessing the Storage Manager was changed in Altium 18. <S> To access the Storage Manager click on the Panels button on the bottom right of Altium. <S> Then click on Storage Manager <S> Inside of the Storage Manager <S> you can right click your file under the Project Files heading and select rename (or hit F2 ).
| From within the Storage Manager panel you can right-click on a file and choose "Rename".
|
Why does the intensity of light in LEDs not increase with current after a particular value? I read in books that light intensity from an LED does not increase beyond a certain value of current. The amount of light emitted depends on the combination of holes and electrons. If so, then as the electron flow increases in the circuit, the effective combination must also increase resulting in higher intensity. But generally why doesn't this happen in a LED beyond a particular value? <Q> Not all recombinations result in the emission of a visible-light photon. <S> Only ones that occur within the P-N junction of the LED itself have the energy for that, and this volume can become "saturated" at high current levels. <S> When this happens, some of the electrons and holes pass all the way through the junction before recombining in the bulk material on either side, where they do so with reduced energy, resulting in the release of longer-wavelength (heat) photons. <A> For what it's worth, Maxim claims a somewhat different mechanism (thermal) than that cited by Dave Tweed: As LED drive currents increase for multiplexing, internal temperatures within the LED also increase. <S> There is a point at which the temperature increase causes a drop in photon conversion efficiency, which, in turn, negates the effect of the increased current density through the junction. <S> At this point, increasing drive currents can result in a small increase, no change, or even decrease in light outputs from the LED chip. <S> The difference may be important if very brief pulses of current are being fed to the LED. <A> Like the current answers by Spehro and Dave state, the limiting factor is by the heat that is generated by the current. <S> The hotter the junction, the less efficient the LED becomes. <S> Thus you reach a point where increasing the current actually decreases light output simply because the LED becomes less efficient at turning electricity into light. <S> It is common practice to increase efficiency of an LED by cooling it via heatsinks. <S> (Also referred to by few as "heat plates" as some popular LEDs come pre-mounted on copper laden PCBs.) <S> To get the best light output / current ratio out of an LED set-up the general practice is to use more than one LED for the purpose and under-driving it. <S> By actually using less current per LED you are rewarded more efficiency, however this is at the cost of using more LEDs in any given design. <S> LEDs can also have more current pulsed through them compared to having constant current. <S> This is used to great effect in some stage lighting equipment as well as other products that use high-intensity strobing effects such as this Rescue Beacon . <S> Overall an LED is limited in intensity by the amount of heat it generates.
| As the current increases, light output increases, but as current becomes high the junction of the LED becomes hot.
|
Does wire gauge significantly impact the charging rate of smartphones? This article claims that wire gauge in USB cables significantly impacts the charging rate of smartphones: USB cables have a data wire and a charging wire within the cable itself. Most USB cables, probably over 99+% are 28/28. Buying after market micro USB cables will pretty much always result in getting cheap quality 28/28 cables and your device will barely break 500mah [ sic ] when it charges, maybe even less. The misuse of units makes me skeptical of the claims made by the article. Furthermore, the length of cables should also impact the total resistance, and the article does not mention cable lengths at all. Is the article correct in saying that increasing the wire gauge from 28 to 24 (which has ~40% the resistance) would allow the charging current to increase from 500mA to 2A? <Q> The contention is absolutely true. <S> AWG 28, often used in USB cables, at 2m is about 0.5 Ω either way, and charging at 1A thru such a cable would require the phone to draw 1A at 4v, which won't work. <S> At 2A it would be even worse. <S> The current situation is unacceptable, and USB cables offered should never have a resistance greater than 0.1Ω either way, regardless of length. <S> To check in reality, buy a USB current/voltage meter insert device. <S> Usually about USD4-5 on EBay. <A> It sounds very suspicious for multiple reasons. <S> 1) Resistance of the wire is not the limiting factor for power transfer. <S> Let's do a few calculations. <S> Based on the resistance for copper wire provided by this chart , 1000 ft of 28 gauge wire is approximately 64 ohm. <S> Say we have a 3 ft cable, the resistance of the cable is approximately 0.2 ohm. <S> 0.2 ohms combined with the 5 V power from USB would be able to provide many amps of current. <S> Ultimately the current is controlled by the charger itself. <S> See the next point. <S> 2) USB current is specified pretty well in the USB spec. <S> See USB Power . <S> By default, USB will provide between 150 mA to 500 mA based on the USB version. <S> Past that, the USB controller and device must negotiate for a higher current. <S> 2 <S> A most certainly is only possible with USB negotiation and the USB controller <S> will be the limiting factor, not the cable. <S> 3) <S> Based on the same wire gauge chart, 28 gauge can handle 1.4 A, 24 can handle 3.5 A. Copper wires don't fatigue over use. <S> I'm more inclined to think that there is physical wear from bending/ <S> inserting/removing the cable multiple times. <A> It is at least plausible that there is significant voltage drop in a 28 AWG cable used to carry <S> 2A. See this calculator, for instance: http://www.calculator.net/voltage-drop-calculator.html?material=copper&wiresize=212.9&voltage=5&phase=dc&noofconductor=1&distance=3&distanceunit=feet&eres=2&x=73&y=6 <S> Using a short cable with large conductor area will help. <S> Wether a 0.7 volt drop affects charging rate would depend on the voltage drop of the charging circuit in the phone. <S> The phone battery is probably lithium ion and requires about 4.2V to reach maximum charge. <S> To determine this one would have to make experiments :-). <S> Unfortunately I don't know of any such experiments.
| So if there is any voltage drop in the phone charging circiit, then thin cables could plausibly alter the charging rate.
|
Small form factor over voltage protection circuit for 5V power supply I would like to add small form factor OVP (Over voltage protection) circuit after my power JACK. Could you please any one help me on this, because we have similar kind of 2 adapters one is 5V and another one is 12V. So by mistaken we will connect and power-up 5Volt circuit through 12 Volt adapter. So i need to add OVP circuit. please guide me: my idea about series resistor with 5.1V zener diode, Voltage break down is 5.1V and power peak pulse max 75Watts it will accept max 5A current shall i implement this? <Q> You could add a 5V LDO regulator to the 5V circuit. <S> Something like the LD2981 , however it will probably appear to work and go into overtemperature protection if a 12V adapter is plugged in and there is substantial current flowing. <S> Something like a voltage divider and ADC or comparator. <S> That will also indicate to the user that they made a mistake and should change adapters. <S> The LDO will prevent even a momentary application of overvoltage, and your circuit has a bit of time to react before the LDO heats up too much. <A> Your easiest option is probably a crowbar circuit . <S> This is designed such that when the voltage crosses a threshold level, it shorts the input to ground, which will cause the fuse to blow. <A> You could use a P-channel MOSFET as a series switch, with its gate pulled to ground by a comparator. <S> Bypass the comparator's positive supply pin with a 0.1uF capacitor to ground. <S> Provide a reference for the comparator's non-inverting input with a 5.6V Zener diode and current limiting resistor. <S> Sample the DC supply at the comaparator's inverting input. <S> The comparator will switch the MOSFET off when the input rises above ~5.6V. <S> This schematic is an example using half of an LM393N comparator (1), a 1N5232B Zener diode and an IRFU5305 P-channel MOSFET. <S> The power and comparator elements of the LM393N are split into three sections in this schematic; just pay attention to the pin numbers as they relate to the 8-DIP package. <S> The MOSFET won't dissipate more than 260mW at 2A, so you shouldn't need a heatsink if the ambient temperature is reasonable. <S> There are lots of options in both through-hole and surface mount components. <S> MOSFET power dissipation in watts (W) equals current (I) squared <S> times <S> MOSFET on-resistance (Rdson): <S> W = <S> I <S> x I <S> x Rdson. <S> Use the maximum Rdson value in the datasheet. <S> MOSFET junction temperature rise (Tr) without a heatsink equals power dissipation (W) times junction-to-ambient thermal resistance (RΦja): <S> Tr = <S> W x RΦja. <S> MOSFET junction temperature (Tj) equals ambient temperature (Ta) plus junction temperature rise (Tr): <S> Tj = <S> Ta + Tr. <S> Keep junction temperature well below the datasheet maximum.
| Perhaps your circuit could detect the application of overvoltage and put itself into a low current state for the duration of the overvoltage. If you use different components, ensure that your MOSFET's gate-source voltage (Vgs) is well above 12V, and that your comparator can operate with a supply from 5V to above 12V.
|
Capacitor vs battery which cost less per capacity? I know that batteries have better density ie capacity per volume. But I want to get the most energy capacity for the money. Which one would would you get? I'm trying to store energy from solar for 1 day. <Q> I expect that the traditional Lead-Acid battery will be the best choice for this application at this moment. <S> Have a look here: <S> allaboutbatteries.com Scroll down a bit to the green table and notice that indeed Lead-Acidbatteries have the lowest Cost per stored WattHour. <S> In the coming years this might shift to Li-ion based batteries as these improve in cost and performance (there is less development in the Lead-Acid department I believe). <S> Li-Ion batteries are more difficult to use than Lead-Acid batteries though. <A> Batteries, by a long way. <S> Exactly which sort depends on how big a system and what you have available, but both lead-acid and li-ion/li-poly are widely used for this purpose. <S> Some larger systems use NiFe which has a longer lifetime. <A> Batteries have much better $/Joule initial investment costs. <S> However, the overall cost should also be considered. <S> Ultracapacitors have much longer lifetimes, and may survive better in outdoor environments. <S> Once you account for maintenance and replacement costs, ultracaps <S> can be more cost-effective. <S> In short, it depends on the details of your application.
| Lead-Acid batteries are more robust in general.
|
Why not use only a MOV as a snubber for relay control of AC motor? I'm building a circuit to control an AC motor using a relay. Since the motor is an inductive load, there will be a voltage spike when the relay opens. It seems that best practice for dealing with this voltage spike is to connect a capacitor across the load (or the relay), with a series resistor to limit inrush current to the capacitor. why not just put a cheap, beefy MOV in parallell with the relay contacts? One example of such a MOV is the "Panasonic ERZE14A391": https://www1.elfa.se/data1/wwwroot/assets/datasheets/ERZE_series_eng_tds.pdf What are the pros/cons of this solution? Is the main problem that the MOV allows the voltage to rise too high, thus putting more wear on the relay than a capacitor-based solution would? <Q> Metal Oxide Varistors (MOVs) are cheap but will wear out and fail shorted. <S> Properly rated capacitors as part ofa snubber will last indefinitely. <S> Both will allow a significant voltage spike. <S> See, for example, Electromagnetic Compatibility in Medical Equipment: A Guide for Designers ... <S> By William D. Kimmel, Daryl D. Gerke MOVs are more appropriate to deal with occasional spikes rather than continuous clamping applications, where they are appropriate at all. <A> MOVs do allow a lot of voltage for relay contacts which could experience arcing on opening or contact bounce. <S> An AC TVS would have a tighter breakdown without a wear-out mechanism as long as you don't exceed its ratings. <S> It might be a smaller solution that a snubber. <A> Beware of MOVs the way that they fail short circuit can be very bad <S> they can go on fire <S> in fact I set up a discharge using a halve sine current pulse and got a short duration flame to shoot out My associate digicamerad this showing the flame to be 20 cm long and posted it on the web some years ago SO using the MOV isn't simple when you have to take fire precautions
| MOVs also have a wear-out mechanism meaning that over time they are less and less able to clamp to the specified voltage with the specified current flowing in them.
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How to reduce the voltage dropped in a mosfet used as a switch? I'm trying to use a N-channel mosfet as a switch for controlling a LED, but the most of the voltage is dropped in the mosfet itself (drain-source) and the led are not turning ON. The voltage in the gate is 3.3[V] and in the drain is 4.2[V]. What is my problem? I've seen this circuit before for controlling LED's and it works, Should I use a P-channel instead? The circuit is already done, so I measure the voltages on my board. Attached is a screenshot of the circuit. Thanks! <Q> Let's analyze this. <S> Assuming the MOSFET is conducting and the LED is getting, say, 10mA with voltage drop of around 1.5V, the voltage on the Source pin will be 1.5V+220 <S> *10mA <S> = 3.7V. <S> While the gate voltage is 3.3V. <S> So your Vgs is -0.4V. <S> Which, of course, will never make the MOSFET conducting. <S> Contradiction. <S> To overcome this you need to make sure Vgs is sufficient by making it known and constant by connecting source to the ground and moving the load (resistor and LED) in series with the drain. <A> For a hobby circuit you could probably get away with using BSS314PE P-channel MOSFETs. <S> It will pass only a few uA when the gate is at 3.3V and will turn on fairly well when the gates are low. <S> If the few uA are visible, a 10K resistor across the LED will keep it off. <S> It's not a good way to do it- better to level-shift and use a p-channel MOSFET with gate driven from 4.2V to 0, but that's considerably more complex. <S> Or use n-channel logic-level MOSFETs and shunt the current away from the LEDs (and waste the power). <S> The problem with the circuit you have is that you can only get the input drive voltage minus Vgs(on) at the source, which will probably end up being a couple of volts. <S> It would be better if you put the transistors on the other side of the resistors, but still probably not good enough. <A> Although this is possible, the more common configurations (i.e. easier to drive) are P-channel high-side switches and N-channel low-side switches. <S> This question discusses the four possible topologies <S> so I won't repeat that here. <S> I have redrawn your circuit using a P-channel MOSFET as a high-side switch. <S> When the gate to Q1 is grounded, the MOSFET will be on, and so will the LED. <S> When the gate is high, then the MOSFET and LED will be off. <S> So the logic is inverted as far as the STAT_LED_R line goes; a 0 turns on the LED, and a 1 turns it off. <S> You indicate the gate voltage you are using is 3.3V. Because R1 provides the necessary high voltage (4.2V) for the gate drive, then if the output pin is 5V tolerant, you can use an open-drain configuration for the pin and connect the gate of Q1 directly to the STAT_LED_R line. <S> If the output line is not 5V tolerant, then you will need to add a second N-channel MOSFET (Q4) to switch ground to the Q3 gate as shown on the right side. <S> In this case, the logic is not inverted; a 1 on the output will turn on the LED.
| You are currently using a N-channel MOSFET as a high-side switch.
|
Why AC socket polarity is important? In North America sockets and plugs design makes sure you'll be able to plug only in a certain way: ( source ) Why is this important for AC? Moreover, many plugs, such as cellphone chargers, are in fact symmetric, i.e. not "sensitive" to the way they are plugged. <Q> The wide slot is supposed to be the neutral, the narrow slot the hot. <S> Neutral is nominally supposed to be near ground potential. <S> However, there's no guarantee the receptacle was wired up correctly. <S> If it is wired correctly, and if a correctly wired polarized plug is used, then the threads on something like an Edison-base light bulb will be near ground potential and there is less chance of an electrical shock than if the screw is at 120VAC with respect to ground. <S> So, it's 'safer'. <S> It's also backward-compatible with older non-polarized plugs that have two narrow blades, however newer plugs that are polarized are not compatible with older receptacles (barring the improper use of tin snips). <S> Edit: As kabZX points out, when one side of the line is switched or (most importantly) when one is fused , it should be the hot side only. <A> The "hot" supply wire is more dangerous to a person than the "neutral" return wire: Neutral is closer to ground potential. <A> Think that you have two wired communication devices whose ground is not isolated from the mains ground (i.e. supposed to be connected to the wall ground). <S> And the communication line is requiring the ground wire to run between the two parties. <S> Now assume you have reversed the polarity on one of the devices. <S> Boom! <S> Ground is shorted to the phase line. <S> Well, anyway it is not a good practice to make such a devices, but it may happen.
| For safety: Equipment that has potential for end user to come into contact with one of the wires needs to ensure it is the neutral return wire, not the hot supply wire.
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