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The reason for using air coils in LC tank circuits I noticed that in LC tank circuit designs very often people use air coils as the L element as opposed to an axial inductor. I understand that the problem with the axial ones is that they tend to self-resonate. But I also thought that using an air core coil allows for much smaller inductance (on a scale of nH), as opposed to the usual uH scale of axial inductors. This in turn allows for a larger capacitance of the capacitor in the tank so that the other capacitances in the circuit don't influence the resonant frequency as much. Would it be reasonable to conclude that using larger capacitors and smaller inductances (air coils) is a more robust approach to designing LC tank circuits, or should the effect be negligible and the main concern be reducing self-resonance of the inductor? <Q> The main reason why air core inductors are used in RF is that non-air core is nonlinear, it can saturate and it does not behave uniformly over wide bandwidth. <S> I wouldn't make an erratic conclusion like : "I see there are almost anywhere air cores inside, which means low inductance, therefore we should put big caps to have the same resonant frequency." <A> All inductors are self resonant <S> so choose an inductor that has a self resonant frequency sufficiently higher than the operating frequency of your circuit. <S> Or, make use of the self resonance if appropriate to do so. <S> However, for a tank circuit in a quadrature detector (used in FM demodulation), higher Q occurs with a higher capacitance/inductance ratio and this is usually desirable for FM demodulation. <S> In other words, the target application dictates whether you want a higher L/C ratio or a higher C/L ratio. <A> There are really two things that you are trying to specify when designing an LC network, centre frequency and Q factor, which makes sense when you think about it because you have two unknowns, the inductance and the capacitance. <S> There are also typically non negligible secondary effects like self resonance and saturation that have to be considered, but in the first instance the required centre frequency and Q will set the required inductance and capacitance via trivial maths, and the technology will then be driven by secondary considerations (Size, mechanical limits, availability, self resonance and required unloaded Q). <S> You usually see air coils at real power, or VHF where the core losses from using a ferrite start to become painful, or where very high unloaded Q is desired.
| For a tank circuit in the collector or drain of a transistor, higher Q occurs when the inductor/capacitor ratio is bigger and this may be desirable in some applications.
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Using AC rated relay in DC applications Something that confuses me is why AC and DC relays are not swappable in an application. For example, I am looking for a 12V (DC coil voltage) relay that its contacts can withstand cold-switching of a 40V/2A DC signal . I found some RT 314 SPDT 12V relays in the lab bin, and I want to use them for this DC application. But nowhere in its datasheet, it has mentioned anything about contact specification in a DC application. Does this mean this relay can not be used in any DC application? I mean if they can switch 16A,250VAC....why can't they switch 40V/2A DC? Isn't the relay contact just a conductor? Thanks to @SteveG This is the DC-LOAD graph in the datasheet that I did not spot myself: If not, Do you know a pin compatible relay that can be used in a DC application 40V/2A contact rated? <Q> A relay is not just a conductor, it's a switch. <S> When opening and closing a switch under load, there will be an electric arc. <S> When switching off a AC load, the arc will go away in the next zero-crossing. <S> With DC loads, the arc stays there longer. <S> Therefore, most relays have lower DC ratings if they are rated for both, AC and DC. <S> This relay will probably work with DC loads, but due to the missing rating, you don't know how long it will work under which conditions. <A> As Manu310us said... <S> it's all about arcing. <S> Using the wrong relay with a high DC current can quickly set your relay on fire . <S> Nothing demonstrates this better than this video ... <S> So don't use an AC relay for switching DC loads. <S> NOTE: <S> the same rules also apply for mechanical switches. <A> When relay contacts make or break a circuit there is an arc. <S> When AC current makes or breaks the arc <S> - the arc sometimes happens near the zero crossing of the sine wave, which equals minimum arcing. <S> When the make or break happens above or below the zero crossing the arc trades a small amount of metal between the contacts. <S> Where the arc occurs randomly during the positive and negative part of the sine wave the trade of metal between the contacts goes back and forth. <S> When <S> D.C. Current makes or breaks in a relay contact there is most always an arc ( no zero crossing). <S> The arc and trade of metal on the contacts is always in the same direction - one relay contact loses metal and that metal builds up on the other contact. <S> Well designed <S> systems that use D.C. Relays - will try to close the relay before conducting high currents. <S> I see this practice in high horse power D.C. Drives.
| Using DC relays for switching AC loads is much less problematic, but you can usually find a cheaper AC relay which does not include the mechanical arc interrupter mechanism.
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Unknown inductor in power delivery I repair laser tag vests and on the phaser board is a strange component on what appears to be the phaserboard. See the black box on the center of the bottom, close to the left edge of the ribbon cable. It has no reliably measurable resistance and as far as I can tell is involved in power delivery, as when it falls off (which it does often) the phaserboard no longer gets power. It has a marking on the top saying "100k" but nothing else. I would like to know what it is so I don't depend on trying to find it when it falls off and can just order new ones. Edit: It would appear from answers that it is an inductor of some kind. During my next shift I'll try to get some closer, higher quality photos to see if we can narrow down, since I've combed through DigiKey and haven't found any matches I'm comfortable with. Edit: Here are more pictures (I can post more if I get more reputation, sorry!): <Q> Inductor (hence the "L" designator), 100kuH, i.e. 100mH. <S> Part of a power regulator <S> no doubt, with U3 being the controller and switch. <A> I would surmise that it is a small inductor of some kind - 4 attachment points. <S> So if you think it might fall off the board, then I suggest you go proactive and solder it now, for reinforcement, while it is still attached. <A> Is that a partially hidden marking of "L1" on the PCB in your photos, at the end of this component near the ribbon cable? <S> If so, the designation of "L" is confirmation of this component being an inductor. <S> To be sure of its inductance (although that isn't the only part of its specification which might be important, especially if it is part of a switching power converter) <S> then you need to remove and measure it using a suitable meter. <S> Again, based on its size, I respectfully disagree with other answers in the likely interpretation of the "100K" marking. <S> I can find several sources which back-up my usual interpretation, which is that the units for reasonable size power inductors like this, are in µH with the "K" suffix being a tolerance value of +/- <S> 10%. <S> (Values on much smaller inductors can be in nH - but this inductor is too large for that to apply IMHO.) <S> Therefore I expect its likely value is 10 µH <S> The marking breaks down as follows: <S> 100 = 10 (value) with a multiplier of 0 (i.e. no additional multiplier) K = tolerance of <S> +/- <S> 10% <S> Here are some sources which confirm that interpretation: Randomly chosen manufacturer's inductor datasheet showing 100K = <S> 10 µH +/- <S> 10% Talking Electronics page about inductors, explaining that 100 = 10 µH <S> How To Wiki showing that 101K = 100 µH +/- <S> 10% (so 100 would be 10 µH <S> +/- <S> 10%) <S> Another random manufacturer's datasheet, Würth this time, showing that a marking of 220 would be 22 µH <S> (so 100 would be 10 µH) <A> Awfully small for 0.1 H. Plus, that's a pretty large value for a switching power supply component. <S> My vote is 10 nH, which, similar to capacitor codes, is noted as 100: the number of nH (10) followed by the number of zeros (0). <S> K indicates 10% tolerance. <S> A datasheet is here: http://www.delevan.com/seriesPDFs/1210.pdf
| It can't be a resister, and can't actually be marked 100k Based on size, shape and marking, I agree with others that it is an inductor. Also, maybe mark it for orientation in case it does fall off.
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Needed PCB surface to cool a SOIC-8 EP package I have got a part in a SOIC-8 EP package. The "EP" indicates it is a package with a exposed pad which can transfer heat to the PCB. I would like to have a better understanding how much pcb surface I need to cool the part at different power consumption levels. Lets say 1 Watt 1/2 Watt and 0.1 Watt. I did read some white papers. They basically tell: PD = (TJ − T A)/θJA WhereθJA = theta ja (junction to ambient) in C/WTJ = junction temperature rating in CT A = ambient temperature in CPD = power dissipated in watts θJA can be split into three parts which add up: θJA = θJC + θCS + θSA Where:θJC = theta JC (junction to case) °C/WθCS = theta CS (case to heatsink) °C/WθSA = theta SA (heatsink to ambient) °C/W The datasheet of the part tels me:θJC = 10 °C/WTJ = 150 °C I can think of a ambient temperature lets say 22°C But then I am still missing the following: θCS and θSA. I could imagine that θCS is negligible, is this true? θSA I find difficult I plan to use via's to get the heat to the other side of the PCB but I can not find any data giving me a idea what number I can use for θSA. I also find it hard to figure out if I need 35um (1oz) or 70um (2oz) PCB. <Q> This is actually a pretty deep question. <S> Fortunately, there is an extensive layer of literature on this subject. <S> You are basically on right track, but, for some reason, didn't get to the right articles. <S> Yes, if the pad is soldered down, you can assume the θCS as zero. <S> Regarding vias, typical thermal resistance of via over a 1.6mm FR4 PCB is 130 to 250 ° <S> C/W, depending on stack-up and via size. <S> So you would need few of them to have any effect. <S> Or you can make one 2-mm diameter via, and fill it with solder. <S> There are plenty of calculators on this subject, Google for "Thermal resistance of via calculator". <S> All details are perfectly explained with formulas and practical examples in this Application Note AN-2020 . <S> Final results will depend on details of ambient conditions, whether the board is vertically oriented or horizontally, are there obstacles to natural convective airflow, or there might be some forced ventilation around. <S> A thermal image of PCB will help tremendously to evaluate the board thermal condition, and, if necessary, design corrections should be made. <S> But for a 1W dissipation and 3x3mm thermal pad soldered to 1.5 oz PCB <S> , I wouldn't worry much, given TJ = <S> 150 <S> °C. <A> The standard copper foil --- 1 ounce per foot square, 35 micron thick, 1.4 mil thick --- has thermal resistance of 70 degree Centigrade from one edge of the square to the opposite edge of the square. <S> Thus 0.1 " long trace, of width 0.01", aspect ratio of 10:1 and 10 squares of foil, has thermal resistance along the trace of 10 * 70 = 700 degree Centigrade per watt. <S> A heat spreader like this has 70/8 = 9 degree Centigrade per watt simulate this circuit – <S> Schematic created using CircuitLab <A> This is a good resource: <S> HeatSinkCalculator <S> They do have limited functionality free use. <S> Also you can do thermal analysis using TI's WebBench. <S> Cree: Optimizing PCB Thermal Performance <S> Regarding thermal vias <S> I did a lot of research and the consensus was to use 15 mil holes spaced on 35 mils centers and a maximum of 15 holes. <S> And use 2 oz copper. <S> I found the thermal via method to be inadequate. <S> What I did was extend the thermal pad on the PCB top layer out one end of the chip with a larger pad with a 4/40 or 3mm screw hole to mount a heatsink. <S> My thinking on this method was the thermal resistance would be lowest on the top layer. <S> Another thing I did was to use bare copper PCB surface area when a heatsink was not necessary. <S> A oxidized copper pad has much higher thermal emissivity. <S> LED
| The LED manufactures have some good documentation on PCB design.
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How low an insulation resistance is too low for a residential electrical system? I have had occasion recently to test the wiring of a residential building for damage from a lightning strike. I opened all the breakers, removed all the loads, opened the panel, and meggered each hot circuit to neutral, and 220V circuits from hot to hot. Many of the results maxed out my meter at 275 megaohms, but some did not. I got quite a few results ranging from 20 megaohms to 200 megaohms. I got a few as low as 1 megaohm. Since those were on kitchen circuits, I suspect contamination of the receptacles. But this leads me to ask: how low is too low? Much below one megaohm I'd start to worry about heating. But I can't find any standard saying "thou shalt". The best I've seen is ANSI/NETA MTS, which I can get an older 2011 copy of. http://www.iemworldwide.com/pdf/ansi-neta-mts-2011.pdf Table 100.1 recommends a resistance of at least 25 megaohms on circuits of 250 volts. But it also says that there's no consensus standard, or at least there wasn't in 2011. Is there a better answer than that? <Q> UK perspective: <S> I just checked the 17th Edition and it still is. <S> That's with the proviso that if you get a result that low, or basically not very high, you should check what's causing it. <S> This is also with the proviso that you're testing it at twice the rated voltage, so on 240V (as it was then) that meant using a 500VDC insulation tester. <S> A low voltage test (e.g. with a multimeter) is not considered valid, whatever reading you get. <A> I don't have access to the whole DIN norms <S> but I found a document citing DIN VDE 0701 (09.2000) and DIN VDE 0702 <S> (06.2004). <S> Which contains a table: Limit value: <S> Applicable for: 0,3 MΩ class 1 with turned on heating elements 1,0 MΩ <S> class 1 2,0 MΩ class 2 and non-earthed metal parts on class 1 0,25 MΩ class 3 <S> So this is for the devices and not the house installation. <S> I found a reference to IEC 60364-6 (limits in chapter 61.3.3, table 6A) and IEC 61557-2 in this document which has: Limit value: <S> Test voltage: <S> Voltage of the circuit: 0,5MΩ <S> 250 VDC <S> SELV / <S> PELV <S> 1,0MΩ <S> 500 <S> VDC <S> up to 500 V including FELV 1,0MΩ <S> 1000 VDC <S> over 500 V <S> I'm not 100% sure if this also counts for the house installation, but the document made it sound like it does. <S> For solar panels I also found something in this document which references DIN EN 61646 and DIN IEC 61215: <S> The insulation resistance has to be larger than 40 MΩm². <S> So a panel with 1 m² has to have at least 40 MΩ, one with 2 m² needs at least 20 MΩ. <S> This is the stuff I found, everything in German <S> so my translation might not be correct <S> and I certainly take no claim that this information is valid or even applicable outside of Germany. <A> In industrial manufacturing environments the rule of thumb is 1 Meg ohm is sufficient insulation per 1000 volts. <S> I currently have a motor with less than 1/2 Meg ohm to ground, that has run 24 hours a day for over 5 years at 300 volts D.C. <S> With no signs of failure. <S> For home purposes I would megger at the 500 volt scale and would consider anything over 1/2 megohm to be ok.
| when I trained as an electrician (early 80s) the pass/fail for an insulation test on an installation was 1 megohm.
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Any thing better to start learning with other than an arduino? I've been wanting to get into electronics for a while, I've seen loads of people say that arduino is the best way to go for starting out in electronics. After saving some money I bought a knock-off arduino board with an LCD and started programming. I found it horribly boring. It was just too easy in my opinion, I was hoping to move bits around and interact with hardware, but all I did was call a function to print to a line of text to the LCD. I'm hoping for something a little harder, Ive got 2-3 years of C/C++ experience and was wondering if there is any 'true' electronics I could get my hands dirty with. As in actually interacting at a low level instead of using a function call. <Q> As a lifetime electronics engineer, I understand the sentiments in the other answers. <S> However, I don't know what you mean by electronics , and I'll wager that you don't either (yet). <S> If you want to make stuff that 'does things', flash LEDs, read knobs and buttons, then for all the purists cringe, an Arduino is a great way to start. <S> As soon as you come to drive a LED off board, or a relay, or want to read a thermistor, you'll be forced into learning at least the basics, so Ohm's Law, and with any luck, some 'proper' electronics. <S> Statement of interest. <S> My first Arduino project was to change the flash rate of the on-board LED from 1Hz to 2Hz, that's the 'Hello World'. <S> I mainly use them now as the cores of automatic bench instruments that I talk to from Python on my PC. <A> Two things: <S> First, if you're trying to learn electronics in general then don't start with an Arduino. <S> Arduino is a type of embedded system and doesn't really teach you anything about electronics . <S> Just embedded software. <S> Start with basic circuits like the 555 timer, try configuring it in astable mode, then bistable, then monostable. <S> Identify the differences. <S> Also try logic gates (AND, NAND, OR, NOR, XOR, NOT) and see if you can configure the gates to perform a predetermined task. <S> That's probably the best way to get started in electronics . <S> Second, if you're really set on embedded systems, then try using AVR Studio to program the Arduino, not the miserable excuse for an IDE it comes with. <S> AVR Studio will give you much more freedom and will allow you to work more closely to the hardware level, rather than just using basic library functions. <S> You'll be able to work with the registers directly, manipulate bits, etc. <S> That's my recommendation if you really want to work with embedded systems first, but like I said, you'd be better off starting with discrete ICs rather than with a microcontroller. <A> Forget Arduino for now. <S> Instead, build lots of simple circuits. <S> LOTS. <S> There are key circuits that you will encounter in nearly every larger circuit. <S> Start with voltage dividers. <S> Build all three modes of a 555 circuit. <S> Use a simple transistor as an amplifier. <S> Use it as a switch. <S> Construct some op-amp circuits and comparators. <S> Play around with component values to understand what happens. <S> So get yourself a breadboard, some 22 gauge wire for jumpers, and a bunch of components and have at it. <A> The best Thing to learn electronics would be a kit of reusable devices and components together with written instruction of how to make functional experiments on it. <S> Examples: "RadioShack Electronics Learning Lab", "Maxitronix Electronic Lab 500 in 1", etc. <A> per1234's comment is totally correct. <S> You've focused on a programming environment as an example of electronics design. <S> Why - because you're a beginner programmer? <S> Countless people around the world have used the Arduino platform to learn new skills and have fun. <S> It's a wonderful and revolutionary platform that's taken the hobby world by storm. <S> You need to find something that you want to build, be it a drone, micro satellite or 400V valve amplifier. <S> There seem to be commitment /focus issues if you can't identify a piece of electronics to develop. <S> I'm concerned with your quoting of the word true . <S> Are you using it as a pejorative? <S> I read your profile. <S> You like game development and are interested in computer science. <S> And you want to get into electronics. <S> Then do both and get an Arduino and develop a game for it. <S> Don't mess with a LCD if you've mastered that. <S> You're 14. <S> Imagine what your CV /resume would look like with this running on one of these using this and a whole pile of really true electronics (micro switch inputs, digital to analogue converters and oscillators.) <S> You can get a duff scope for £50. <S> Get your parents to buy it for school . <S> Everything you need to help you build this is available here and other places on the internet. <S> You might even be able to get space invaders running. <S> But you might not be good enough for that. <S> Or do you think you are? <A> "Arduino" (as distinct from the Atmel ATmega328 itself) is aimed at serving artists rather than engineers, so there's more of a focus on providing lots of essentially finished example programs. <S> In some ways that does kind of take the fun out of it, though it is less frustrating than a normal microcontroller development kit that provides little or no example code. <S> But you don't have to use the Arduino IDE, you can dig deeper and use it as an Atmel ATmega328p development board. <S> This is a good place to start, since you've already walked through some of the examples. <S> I recommend looking into the avr-objdump and avr-readelf utilities, these should already be installed along with the Arduino IDE package -- check around the Arduino's hardware\tools\avr\bin directory. <S> These utilities let you look at the raw ATmega328 instructions generated by the compiler, and see the resulting memory usage. <S> Take a look at the Atmel <S> ATmega328p <S> datasheet -- at 442 pages, it's typical for a small microcontroller. <S> One nice thing about this one is that everything is in this one datasheet; many others I've worked on have one manual for the CPU core instruction set, another manual for each of the peripheral modules, and yet another datasheet just for the IC itself. <S> When delving into a deep technical document like this, it's important to page through the entire document and skim over the major headings first. <S> Trying to read it as the world's most boring 442-page novel is the wrong approach. <S> Treat it like a roadmap: find the major landmarks and then navigate down to the relevant local detail.
| The right way to learn electronics is to build simple hardware circuits.
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Tiny shock from 230V AC? So first at all this question isn't really about electrical engineering, but about electricity in general. However, I think this is the best place to ask, correct me if I’m wrong. Also I’d like to mention that my knowledge about electricity is very limited, especially my knowledge about alternating current. While working in the garden today, I used an electrical mowing machine which is powered by standard 230 V AC. What I didn’t see, is that the extension cable I used, had a crack in the insulation wrap and some wires were exposed. However when I touched one of the exposed wires by accident, I felt a little “shock”, but it was way less hurtful than I would have expected from 230 volts. My guess is, that I only touched one of the power wires, so the current tried to flow through my body to earth, but because I was wearing shoes with rubber soles I was insulated from earth very well and only a tiny amount of current was able to flow through me. Does this make any sense? Also, if I had touched both power wires, or one power wire and the earth wire, I would have shortened the circuit with my hand and thus I would have gotten the “full load”, right? <Q> Your skin has resistance. <S> Dry skin has a relatively high resistance resulting in a lower current. <S> If you would have had wet hands, or one of the strands punctured the top skin layer, you would have had a painful arm. <S> You were probably also wearing nonconductive shoes. <S> Further limiting the current path. <A> You may have been fortunate and only touched the neutral wire. <S> The neutral wire carries the return current and is generally close to zero volts with respect to ground because at some point on the distribution system it is connected to the earthing system. <S> Note that this is done at an approved point such as the utility distribution transformer or at the entrance to your property. <S> You should never connect neutral and ground on any of your circuits. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Cable resistance results in a voltage drop on the live wire and a voltage rise on the neutral wire. <S> If significant current is flowing in your extension lead or elsewhere on your property the neutral voltage will rise above zero. <S> This may be what you (fortunately) felt. <A> A "full load" depends on a lot of factors, not least of which is whether you are cold, or hot and sweaty. <S> Not to mention whether the current path is through a single hand, arm to arm or arm to ground. <S> Touching a live piece of metal can result in anything from a mild sting to death. <A> Yes, probably you were wearing well isolating shoes, so you got only small schock. <S> Yes, touching both wires would give you more current trough. <S> It depends on how you touch these wires. <S> Let's say that current over 40 milliamps through heart would start to cause irreversible damage. <S> Touching both wires with two fingers is not so deadly, even if the fingers would be burnt to charcoal. <S> Even touching with two arms is better than having passed the current from arm to leg.
| As the other answers explain the shock current will depend on many factors. The most dangerous is to touch with arm and leg, or just live wire and no shoes, since the current would pass through the heart.
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What is causing this Op-Amp Distortion I am new to analog electronics and I am still learning about op amps. I am driving an 8ohm 15 watt speaker with just an op amp and my phone. What I do not understand is that I have a dual supply, 10V and -10V, more than enough voltage to handle audio being doubled from a cell phone. If the speaker is 15 watts , and there is sufficient supply voltages on both ends. How is there distortion when I turn up the volume or heavy bass occurs? When I run a sine wave from my function generator through the circuit I have no problems at all, I can turn up the amplitude until the output is clipped by the supply voltage. Does anyone know what is happening with my phone input/output though? <Q> Since you're new to electronics <S> I simplify here a little bit: <S> Speakers usually have a impedance in the low ohms. <S> 8 Ohm is typical. <S> A ordinary OpAmp won't drive much more than 50mA to 100mA. <S> At 10V Signal that is roughly equal to an impedance of 1000 to 2000 Ohm. <S> Much higher than your speakers. <S> If you connect speakers to your OpAmp output, the OpAmp will likely try to drive the speakers nonetheless, but after reaching the current limit it won't drive it any further. <S> That will clip the waveform, and this is what you hear as distortion. <S> You need a power amplifier stage to drive your speakers. <S> These can be built using an OpAmp and some extra transistors to handle the higher load. <S> On the other hand it's more economic if you just use an audio amplifier chip that handles the 15 watt directly. <S> The TDA2030 chip for example will happily drive 10 to 14 watt and is really cheap. <S> In short: Think of OpAmps as pre-amplifiers, to drive anything bigger you need a power amplifier. <A> There are several things wrong here. <S> You say you want to drive a 8 Ω speaker with 15 W. <S> That means you need sqrt((15 W)(8 Ω)) <S> = <S> 11 V. Note that this is RMS. <S> For a sine wave, that means that the peak voltage is sqrt(2) higher, or ±15.5 V. <S> Clearly ±10 V supplies aren't good enough. <S> Then there is the issue of providing enough current. <S> The RMS current needed is sqrt((15 W)/(8 Ω)) <S> = 1.4 A. <S> And again, that's RMS, so <S> the peak current needs to be at least ±1.9 A. A typical "opamp" <S> can't provide that kind of current, not even close. <S> What you need is a real power stage that runs from around 20 V and can source and sink 2 A. <A> Op-amps don't have the grunt for this <S> An ordinary op-amp is designed to drive small-signal loads such as small headphones, the inputs of other amplifiers, or other circuits. <S> A 8Ω speaker simply requires too much current (amps vs millamps) for an op-amp to drive it alone. <S> But, boosting the output current is simpler than it sounds <S> Fortunately, it's possible to build a power amp around an op-amp, as follows: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> In this circuit, Q1 and Q2 act as a class-B emitter-follower current booster for OA1's output. <S> Normally, a class-B or AB stage requires a bunch of biasing to keep crossover distortion down, but by placing the stage within OA1's feedback loop, the op-amp takes care of most of that funny business for us by way of feedback action, although it does require a fairly fast-slewing op-amp to do this. <S> As long as the power supplies have enough grunt, and Q1 and Q2 are beefy enough and adequately heatsinked, this circuit is adequate to drive whatever speaker you throw at it, and cheap as chips to build. <S> Of course, the sonic performance is mostly dependent on what op amp you're using, and if you need to, you can configure this as something other than a unity voltage gain amp by changing the feedback network, just like you would with a small-signal op amp stage. <A> You should post more info about the opamp, but if it is one of the usual suspects like NE5532, you most likely missed the "maximum output current" line in the datasheets. <S> Some opamps will choke on 2k ohms loads (like TL072) because their output stage can only handle a minuscule amount of current, others will work fine in 600 ohm loads and go up to 60mA or even 100mA... <S> But for a 8 ohm load you really need something designed for the job. <S> If you're a beginner, try a LM1875 for example, it's a cheap and easy to use amplifier chip (just an opamp really, but designed for the kind of current your speaker requires). <S> It also needs a heat sink. <S> If the thing runs on batteries, consider purchasing Class-D amp module from aliexpress, like IRS2092 or the TI chips.
| OpAmps don't have the current drive capability to drive speakers directly.
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How to use off board components in Altium and have them show up in net list & BOM? For example, let's say I have a power jack which is mounted on a panel and connected to the PCB using wires. I might also have a few 1/4" audio jacks also mounted on the panel and connected to the PCB using wires. On the PCB they have place for the wires and not a footprint. How can I have these parts in the schematic and show up in the BOM and if possible the net list as well? <Q> You can choose "Standard", "Mechanical", "Graphical", "Net Tie (in BOM)", "Net Tie" or "Standard (No BOM)". <S> Show them on the schematic- using text etc. <S> so they stand out if you need to verify or change them. <S> They will then show up in the BOM automatically. <S> You can do a similar thing with mounting holes, except they would be "Standard (no BOM)". <S> This is useful for parts that you want to appear in the 3D model (and BOM) but have no electrical connections on the PCB. <A> If you are just using wires soldered into holes on the board, rather than connectors, to connect these items to the board, I would use any suitable footprint for the holes on the board - the footprint needn't bear any relation to the actual part, other than having the right number of pads. <S> Associate this footprint with the schematic symbol for the off-board part. <S> You should probably have a note on the schematic that these parts are mounted off-board. <A> I use the same method Spehro described, except I create a schematic "component" with the type "Mechanical" selected, but I also add a graphic so that the person viewing the schematic can see what the external "mechanical" component is. <S> I use this for fuses, socketed modules, hardware, etc. <S> Example: Schematic: <S> Component properties: <S> Note <S> the "Component Type" is set to "Mechanical" <S> The headers have footprints but the module does not. <S> Both the sockets and the module still show up on the BOM though. <S> In your case you can either create a footprint specifically for your power jack that is simply made up of the pads you solder the wires to, or you can set the component type to mechanical and place pads on the PCB manually.
| Create symbols for the off-board parts and define the type as "mechanical" parts. You can also create parts and footprints that have no pins, which means that if you put them on the schematic they will automatically be transferred to the 3D PCB layout. There is no footprint associated with this part, instead it is socketed in two headers.
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How can I measure low power energy usage of bluetooth device? I need to measure the energy usage of a small bluetooth device which is normally operated by a coin cell 3v battery. Background:Using an oscilloscope I was able to get fairly good pictures of the energy used during standard connection and advertising events (measuring current through and external 10ohm resistor and integrating), but that is in a steady state condition when not much is going on and the event is easily captured with oscilloscope. More complex interactions happen between the device and the other bluetooth device it is connected to (typically a smart phone) during re-connection events and when characteristics are being read or written. Questions:Is it possible to use a simple capacitor to measure energy usage by charging the capacitor initially, then running the device off the capacitor power for a few minutes during the test and then measuring the remaining charge? How accurate would that be? What other techniques would you recommend? (edit: note I'm actually interested in mAH usage not energy since ultimately the goal is to estimate battery life and battery is rated for 220mAH) <Q> You can power your device with a micropower LDO, and put the current sense resistor before the LDO. <S> Thus you can use a much larger current sense resistor. <S> You can also bypass the resistor with a diode to have a better look at very low currents, but still allow the device to draw more current when needed. <S> You can also bypass the resistor with a cap for averaging. <S> Using a LDO means you can let the voltage on the cap drop a lot more (like 5.5V down to 2V), than if you were powering the device directly (like 3V to 2V). <S> But you need a rather accurate electrolytic cap. <S> Well, you can always calibrate it with a known resistor as a load... <S> You can also use an IV converter: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The opamp replaces the LDO and will make sure your device is powered by a constant 3V, while giving a current reading on the 1k resistor. <S> The cap keeps the opamp stable. <S> The diode should be enough diodes in series to allow the opamp to supply the device directly, bypassing the resistor, if needed. <S> It is optional. <S> If your device has onboard caps, you'll need to add a resistor in series because the opamp won't like the capacitive load. <A> Instead of a capacitor, consider for instance using a Battery + <S> Coulomb counter IC (or just any voltage regulator + <S> Coulomb counter <S> IC).Many battery monitors have a Coulomb counter. <S> The capacitor has the drawback that its voltage is not constant. <A> What you want to do is integrating the current consumption over time to get the average value. <S> Your device might most likely be in advertising (or connected) mode only. <S> The biggest issue is the dynamic behavior of the current profile: While most of the time the bluetooth chip will be in low-power mode (no Tx/Rx) <S> its current consumption will be in the range of (very few) micro-amps, but (every now and then) it will wakeup for just a few milli-seconds for processing the ADV/CON packets that have peak consumption up to tens of milli-amps. <S> The amount of charge (or average current) from the low-power mode might be in the same range (or even higher) than that of the pulsed packets. <S> So, a single measurement with a single measurement range will give totally wrong results: if the measurement range covers the peak pulse current, there'll be so much noise and offset error in the low-power phases; but if the measurement range is good for the low-power phases the pulse currents will be clipped (ignored). <S> SMUs might sound great for such a job, as they can record current profiles; but they won't switch the measurement range fast enough. <S> If you have something to recorded two analog channels with sufficient speed and memory (>=100, better 1000 kSps, for the timeframe required <S> - e.g. oscilloscope with deep memory), I'd propose to use two current shunts (with instrumentation amplifier, e.g. AD8422) in series, one for the small currents (e.g. 20µA range - with a bypass diode in parallel to limit voltage drop) and one for the packet pulses (e.g. 20mA range), capture the two current signals, and calculate together, based on a simple rule (use the low range when not out-of-range , otherwise use the high-range). <S> The idea with the capacitor discharge is also not bad, but you must consider its internal leakage - and (esp. <S> for ceramic caps) there's voltage-bias effect reducing capacitance, so do not judge from delta-voltage, but from the required recharging back to the initial capacitor voltage level.
| You can also use a capacitor as power source for the LDO.
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LM2596 and other DC-DC converters why oscillate if the output capacitor ESR value is too low? I would like to use LM2596 converter. Datasheet says:'If the selected capacitor's ESR is extremely low, there is a possibility of anunstable feedback loop, resulting in an oscillation at the output.' Why? Why not good a too low ESR value? <Q> The ESR of the capacitor forms a zero at frequency <S> fz = <S> \$\frac{1}{2\pi\cdot\text{ESR <S> } \cdot C}\$, which tends to reduce the stability excessively if the ESR is too low. <S> The ESR also should not be too high, of course, so usually there is a recommended range. <S> Older parts that were designed before high-capacitance ceramic parts were ubiquitous may omit specifying the lower limit. <S> This is equally true of LDO regulators. <S> You can either add some series resistance to the capacitor (maybe a few hundred m\$\Omega\$) to keep it within the recommended range, or use another type of regulator that is designed to be stable with ceramic capacitors. <S> To confirm that you've got it right, you can measure the phase margin, which ideally might be 60 or 70 degrees, for example by injecting a signal and measuring the response. <S> If you roll the dice and just throw the cap in there, hoping for the best, you may find that it oscillates at temperature extremes or it will have such a low phase margin that it exhibits excessive overshoot/undershoot. <A> Without that, charge might just be stored in the capacitor without significant slew, and then the control might (probably) overshoot quite a bit, leading to a suddenly increasing voltage, leading to swift shutoff, leading to aforementioned oscillations. <S> I wouldn't worry about it until it actually happens. <S> If it happens, add additional trace length, thin your trace, or use a cheaper capacitor. <S> Think about it as a minimum resistive loading. <S> If you're generally worried about that: Pick a different SimpleSwitcher. <S> TI's product page lists viable alternatives right at the top. <S> Rule of thumb: <S> Higher switching frequencies mean lower measurable oscillations after an output filter, so that might be advantageous. <A> This simple regulator demands that both input and output cap are low ESR for low ripple out. <S> This means ESR*C < 20us . <S> This breakpoint effectively attenuates the ripple but also reduces the AC feedback signal which is amplified with high gain needed to regulate PWM on each cycle. <S> When this happens, the comparator used on the feedback signal ends up squaring an AC ripple feedback signal with more phase noise used to regulate the PWM output switch. <S> Effectively feedback become a low pass filter with high Q peaking from the high Q or ultralow ESR output cap. <S> When the ESR*C < 5 us reduces to "ultra"-low ESR , now shorter than the typical switching period, the capacitor becomes an ideal integrator (-90 deg). <S> When you add the phase shift in the internal gain block (GB) and comparator, you dont have much margin left at unity gain where the 180 deg Barkhausen criteria is satisfied for oscillation. <S> (an unsatisfactory result ;) <S> To solve this , a small shunt cap is used on the feedback R to give a phase lead and high frequency boost to feedback ripple. <S> Then you improve step response ringing from better loop phase margin. <S> Hope this hand-waving description is understood. <A> No proof given, but there's also some inductance in the capacitors. <S> When the ESR is low enough, the resonant circuit property caused by the parasitic inductance can jump to remarkable effect. <S> High ESR make the resonant circuit too lossy to cause the unwanted effects. <S> This story could be inspected by simulating.
| As it seems, the dampening effect of a series resistance (remember what ESR stands for!) is required to make the control loop, that a DC/DC switching converter controller actually is, stable.
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Can I use a dual primary transformer to generate isolated 110v output? Given a transformer with a dual primary winding (generally used to step down a series 220v or paralel 110v), can I instead hook up one winding to a 110v input and use the second as an isolated 110v output? What kind of power restrictions or undesired behavior might this cause? EDIT: This transformer, for example: http://mouser.com/ds/2/410/media-1067445.pdf EDIT2: Added 24v to the image secondary coil for clarity <Q> You can if you only need functional isolation. <S> Since you are only powering half the primary you can only put in half of the primary current so you only get half the VA rating of the transformer, so compared to a real isolating transformer you end up with twice as much iron and an unused secondary. <S> Safety: <S> These units are designed with 4000VAC isolation between the primary and secondary, and also,between each winding and the core. <S> If there had been a comma after "winding" it would be a different story, as it is there's probably only about 250V isolation between the two secondaries, they may well be wound bifiliar (intermixed) <A> If you are planning on using the isolated primary to do something like supplying Nixie tubes, and the secondary supplies the logic, and safety depends on mains isolation of both, I would not suggest using this kind of transformer. <S> It will work, but it will not have adequate isolation for safety. <S> There are dual split-bobbin construction transformers and they would be much better, but I suggest checking with the manufacturer to see if they Hipot test between primaries and if they would recommend this service. <S> The below illustration is from a Hammond 229A12 6VA part: <S> As you can see, each coil is independently isolated by the bobbin structure. <A> Why not connect the two "110" windings in series? <S> (Do observe polarity - connect the "dotted" lead of one to the "not dotted" of the other.) <S> Then apply 110 across the two of them, and you get 110 out, isolated, across the "220" secondary. <S> Or turn it around: Put 110 into the "220" winding <S> and you'll get 110 out across the pair of series windings. <S> 110V <S> in, 110V out, and no loss in rated power. <S> There is no rule that says that a "220" winding absolutely must be operated with 220 across it.
| If safety isolation is required you need a transformer designed for that.
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Is this a delay circuit? / Which resistor is responsible for load impedance? I just stumbled upon this circuit of a microphone preamp. However, I`m not entirely sure about two of its components/ properties. My first question regards R3 and its closely located capacitor. Someone online claimed this was in order to reduce any clicking when phantom power is switched on, and after some googling I found that this might be a simple delay circuit. Am I correct in thinking that? My second question regards R4/ R5. I´m aware that these together with C1/ C2 form a high-pass filter, but someone told me these would also present a load impedance of 2.2k to the microphone. Since impedance is still quite the conundrum to me, I thought I´d ask here wether that is correct.I was under the impression that R1/ R2 not only limit the current to be drawn from the 48V supply, but also play a role in the input impedance of this circuit. This is because that´s what I gathered from the diagram here http://educypedia.karadimov.info/library/mic1dat.pdf where a 2.2k resistor in series with the 2V supply indicate a 2.2k load impedance in the table above said diagram. I apologize for the convoluted question and many thanks in advance for your help =) <Q> R3 simply connects R1,R2 to ground when phantom power is switched off, and the C simply provides some decoupling on phantom power. <S> No delay, and I would expect quite a thump when switching on phantom power. <S> R4,R5 provide a 4.4K load (leg to leg) on the microphone, or 2K2 on each leg - and you are absolutely right that this impedance is further reduced by R1,R2 which are effectively in parallel (for AC) with R4,R5. <S> And phantom power is always supplied from 48V in series with 6.8k on each leg <S> - it's a de-facto standard that studio mics are designed to. <S> It's not primarily current limiting (though it does that too), but to keep the impedance on the audio signal reasonably high. <A> Chances are the INA needs a DC reference for the input pins. <S> R4/R5 provide that. <S> Datasheet says the input bias current is 2uA nominal, 12uA worst case. <A> Well no, the mid point is grounded so it doesn't really form a circuit <S> the capacitor looks like supply decoupling <S> C1,C2, R4,R5 do form a high pass filter, have you calculated the corner frequency? <S> it'll be sow low as to be inaudible) the purpose of those two capacitors is to block the 48V DC so it doesn't get into the amplifier.
| the resistor is possibly to stop the switch from arcing, 48V DC is quite a lot to switch and having ground opposite power on a change-over switch is asking for trouble, 47K limits the current to about 1mA, which is not enough to sustain an electric arc.
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5 volt regulator not getting desired output I have been trying to assemble a simple 5V regulator circuit using the L7805CV and have had no luck at all getting the output to 5V. I have tried many circuits and multiple L7805CV regulators, thinking that maybe it was the regulator itself that wasn't doing its job. So far the closest I've come to 5V is 6V, which won't work for what I'm trying to power. First, I tried the simple circuit with a 10uF cap from input to ground pin and 1uF cap from ground pin to output, and was getting 3.8V output with an input of 9V. I was getting 11V out with a 12V input. I tried another one that was similar but with a 1K resistor connected to an LED on the output, which gave me 6V. I thought maybe the capacitors were the cause so I tried going with 100uF from input to ground and a 10uF from ground to output and another 100nF on the output for filtering and still no luck. I basically played with cap values all the way up to 2200uF and didn't get anything close to what I needed. I either got really low (about 3V) or really high (close to 12V) on the output and also tried different input values of 6V, 9V, 12V, and even 18V and nothing worked. I used a heat-sink on the l7805CV and am stumped how this is happening. I eventually went back to one of the first designs that had the output of 6V and tried using a diode on the output for its voltage drop and was getting almost the exact reading on the output (actually a few mV higher) and am stumped at this point. After a while I got bored and tried out a few different transistors I had around to see how they would affect the circuit and was surprised to find almost exact results. What am I doing wrong? How can I successfully achieve 5V on my output? <Q> Here is a L78 05CV datasheet. <S> Note that on page 7, for the 5V version, Vout is specified with loads of 5 mA to 1A and there is a footnote that says "a Minimum load current for regulation is 5 mA". <S> To meet this minimum current requirement a load of <S> no more than R = <S> V/ <S> I = 5V/5mA = 1 <S> K is required. <S> That said, "usually" a 7805 will "more or less just work" if connected correctly. <S> Even Vout is liable to be "about right" without a load. <S> (These older style regulators power themselves between Vin and ground <S> so load current is less important.)(The LM317 family power themselves from Vin to Vout <S> so a minimum Iout is needed for correct operation). <S> SO Try it with a 5mA or higher load. <S> (1 K or lower). <S> Look at the data sheet(s) and ENSURE that you have the pinout correct. <S> || <A> An open on the GND pin will cause the output to rise to a volt of two of the input. <S> 100 ohms resistance will cause the output to rise by about half a volt. <S> Solderless breadboards are prime offenders, and this can expose the circuitry connected to the regulator to damaging voltage, so it's best to do that part with solder. <A> From looking at the internal schema tic and example test circuits, I'm pretty sure it works with input at 6V min with no load and 7V min at 1A . <S> 7.5 V is ideal. , but 9V drops too much power across regulator. <S> It must use a good heatsink to keep cool. <S> It will shutdown when too hot. <S> If it is too hot to hold, its too hot. <S> Expect the heatsink to be at least 1 sqin per watt. <S> For a rectified input, the input cap must be large enough to ensure the Vin min does not sag below minimum <S> so you must measure AC pp and DC voltage. <S> My guess 7.5Vdc avg in with 1Vpp ripple @ <S> 1A needs C=80ms/5 ohms = <S> 1.6mF not 100uF. <S> If you raise Vin more than Vmin, it must dissipate the drop voltage x load current, so 9V is not good. <S> Recheck your connections and shorting the input will damage the part, so reverse power diodes are suggested between in and out. <S> There is no need to put series diodes on output, but if you only have 9V, use diodes on input to run cooler to reduce the drop power across series regulator to just above Vmin at 1A.
| The usual reason for an unexpectedly high output voltage on a properly wired (as in pinout correct) 7805 regulator is a flakey ground connection. Caps as shown help but usually are not essential to make it work.
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How to choose capacitor for an IC I'm a beginner in electronics from the software industry. With some self-taught things, I'm trying to implement some basic Arduino circuits. My confusion is about capacitors mainly. My understanding about capacitors is that they act as power storage for few seconds or milliseconds. I found that most IC's must have capacitors connected to their pins. My confusion is how to find out which pins need a capacitor, and how to find the correct capacitor for a circuit or a capacitor for an IC. Finally, why are capacitors necessary in a circuit in such situations? <Q> What you are referring to is called a decoupling capacitor and is used to decouple the IC supply pins from the bus. <S> In other words, it prevents a sensitive IC from being "starved" if another device on the bus turns on quickly and draws significant current, which would drop the bus voltage for a period of time. <S> The capacitor supplies the extra current required to start up the device, as well as to prevent its chip from suffering the effects of a sudden loaded bus. <S> This is generally required for high-speed devices that switch very quickly, as this tends to draw significant current. <S> The capacitor is not necessarily chosen by its capacitance, but by its ESR (equivalent series resistance) and its ESL (equivalent series inductance). <S> The most common decoupling capacitor value is probably 0.1uF <S> but for faster circuits you may require 0.01uF or 0.001uF <S> (again, depending on their ESR and ESL at those speeds). <S> If multiple devices with different speeds exist on the same bus, you may need more than one decoupling capacitor, one for each speed. <S> 99 times out of 100 the datasheets will tell you exactly what value decoupling capacitors to use on which pins, so read the datasheet. <S> This tutorial from Analog Devices is also a great resource. <A> It's all due to inductance: <S> Say your microcontroller draws supply current which ramps up from 1mA to 11mA in 5ns then back to 1mA every time it processes an instruction. <S> di/dt = 10mA/5ns = 2 000 000 <S> A/s <S> Now, the voltage across an inductor is <S> v = L di/dt and the trace from the power supply to the microcontroller has, let's say 50nH inductance... <S> v = <S> L di/dt = 100mV drop on the supply. <S> OK, it doesn't crash yet, because it's a slow micro, doesn't use lots of current... but a faster micro, or other chip drawing faster/higher current spikes needs to have its power come from a low inductance source to avoid voltage sag when it draws current pulses, and a capacitor placed close is a good way to achieve that. <S> Just as important is the fact the capacitor keeps the noisy current drawn by your micro in a small local loop. <S> Loop antenna efficiency is proportional to area, thus amount of radiated noise will be much less when the capacitor is close. <S> Also if you have other components, say an opamp on the same supply, then the capacitor at the micro will prevent the micro's noise from screwing up the opamps' supply, which tends to cause some garbage at the output... <S> So here you have it, the caps do: power integrity: caps serve high di/dt supply current locally EMI: reduce loop antenna area EMC: <S> keep the noise out of the other sensitive devices <S> Now, how to choose the value: <S> A roll of 100x <S> 25V 0805 <S> X7R costs €1.40 for 100nF and €5.40 for 1µF. <S> So, buy a roll of 100 of 1µF. <S> Every time you got to put a decoupling capacitor on your circuit, remember if you spend 10 minutes to read the datasheet and you discover 100nF will work, well you just lost 10 minutes and saved 4 cents if you only build one unit... <S> I just put in 1µF, guaranteed to work every time. <S> Also it has less ringing, works better with lowish-ESR electrolytics, etc... <S> Also I use 25V caps <S> so I only have to stock one value for 3.3V to 15V... <A> So my confusion is how to find to which pins we need a capcitor to connect <S> For each chip you use, there will be a data sheet that tells you and if it doesn't tell you it's because the chip comes from a particular logic family (for instance) and there will be a manufacturer's generic data sheet for the family that will tell you. <S> Also how to find the correct capacitor for a circuit or a capacitor for an IC. <S> See the above - it's in the data sheet. <S> And finally why capacitor is necessary in a circuit in such situations? <S> A lot of chips will "consume" pulses of current and the capacitor will provide those pulses of energy so that the whole of the power supply wiring (or tracks on a PCB) don't have to handle those instances. <S> This means better reliability and less radiated and conducted emissions to other chips and systems. <S> Some ICs such as op-amps will rely on capacitors for maintaining performance and avoiding instabilities on the output especially when driving some loads. <A> Level 1 (often good enough. <S> not always.) <S> : Just slap on >10uF and 100nF paralleled, the latter with as short leads as possible. <S> Level 2: Just read the datasheet, as suggested. <S> Level 3: <S> Read Linear Technology Appnote 47. <S> Also, consider using ferrite beads in your decoupling circuits. <A> For what I know, the capacitance is not so important, it's just for some 'too much' energy between VSS and GND. <S> That's why normally very low capacitors are used. <S> I use mostly ceramic ones with 104 marking (meaning 10e4) which is 10e4 pF <S> which is 0.1 uF. <A> Give each power pin a 0.1 µF ceramic cap,preferably size 0805 or smaller, in parallel with a 10 µF tantalum or ceramic. <S> You can probably omit the 10 µF cap, or replace it with something smaller,if you are concerned only about high-frequency noise. <S> The location of larger capacitors intended for low-frequency bypassing is not quite as critical,but these also should be close to the IC—within a half-inch.
| Ideally you would determine the speed at which the device would turn on, and pick the capacitor with the lowest ESR/ESL for that speed.
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Equal and opposite Voltages applied across 2 resistors of equal Value Take a look at this picture here: Please note that the two resistors are equal in size. Here it says that the voltage at Point TP3 is zero volts. I do not understand how that is.The creator of this image says in a video that this is because the Voltage at Point TP2 (-6V) and TP8 (+6V) cancel each other out. Is he right? Do they cancel each other out? Here is the link to the youtube video: watch?v=tsrAP8EgcbQ ****EDIT**** So here is my scribble: My understanding is that the Voltage "AB" (the voltage above R2) will be feed to the amplifier. In the first picture i posted that would be the voltage "TP3TP8". So the voltage between the point A and the point B will be 0 volts? Edit Thanks for the help. Something must be wrong with my understanding of voltages. I have now something to work with. Again thank you! <Q> Rather than "cancel each other out", I would say that the two resistors form a voltage divider between TP2 (-6V) and TP3 (+6V). <S> If the two resistors are of equal value, the mid-point must be half way between the two end voltages. <S> This assumes that the amplifier input does not source or sink any current. <S> If it does source or sink current, the current in the two resistors, and therefore the voltage across them, will be affected by that current, so the mid-point will not be zero volts. <A> No the voltages do not "cancel out", it's a a bad way to think about it. <S> But we can show that the voltage at TP3 is in fact 0, assuming the two resistances are equal. <S> Look at my poor scribble below. <S> Let's do a KCL analysis at the TP3 junction. <S> I3, the current going into the OpAmp is 0, because that is an assumption of ideal op-amps. <S> We are only left with two currents I1 an I2 which must be equal. <S> Now substituting the equations with voltages using I = <S> V/R we can see that TP3 voltage must be 0 volts. <A> Think of it as both resistors being between 12V and 0V - simple voltage divider, you get 6V at the node TP3, which is thus 6V apart from both rails (TP2 and TP8). <S> Equal voltage across both equal resistors is a given when they both have the same current through them (which is an absolute given in the circuit shown). <S> Whatever voltages you put on TP2 and TP8 - TP3 is always going to be an equal amount apart from both. <S> At 6V and -6V, 0V is the value where it will be 6V apart from them. <S> BTW, if you are building this circuit in practice: This will fail if you use high value (say 100s of Kiloohms) resistors and an Opamp that needs a high input bias current.
| Since the end voltages are equal, but opposite polarity, the mid-point must be zero volts.
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Connecting a microcontrollers ground pin to a reference above ground When I was at work the other day I was pulling apart an LED fault detection unit. I noticed they were using an Atmel ATtiny13A as the brains. The interesting thing is this unit was multi-voltage and so they were connecting 9-33V to the microcontroller's Vin and then using a zenor diode to reduce the Vin by 5V then connecting the microcontroller's ground pin to Vin - 5V. Lastly they dissipated the remaining voltage with a resistor connected to ground. The reason for them doing this was so it could interface Vin on its digital pins without any additional circuitry. My question is, is this a normal thing to do (The microcontroller still had 5V between Vin and Ground pins)? Is there some limiting factor stopping you from using the same logic and connecting the micro to 1000V at Vin and 995V at ground? At what point does the voltage become too high to do this? Is the only thing stopping this at higher voltages arcing out to other components etc Thanks! :) simulate this circuit – Schematic created using CircuitLab <Q> This is equivalent to deriving the 5V from a much higher voltage, and the same problems apply. <A> But from the perspective of the microcontroller, it's within specification. <S> This is typically done to cheap out on parts, but you have to pay attention when designing your circuit. <S> Is there some limiting factor stopping you from using the same logic and connecting the micro to 1000V at Vin and 995V at ground? <S> The 1000V (above mains earth as 0v) will be common mode for the microcontroller. <S> Microcontrollers don't care about this because nothing has to connect to 0V. <S> If you do connect something that has 0V, like your programmer with mains earth, magic smoke will be released if you do not use isolators (eg: optocouplers). <S> Optocouplers do care about this, since they connect to both 1000-995V and 0-5V. Voltage is always relative. <S> You can flip the labels and use -5V and 0V to supply your microcontroller. <S> It will be an unusual circuit diagram, but the electrons don't care. <A> There are quite a few high-voltage highside switches. <S> Those are put in high voltage connections and derive power by a small voltage drop between connectors. <S> The control signals are electrically decoupled, so you can use voltages on completely different potential to switch the unit. <S> A simple example of such a switch would be an optocoupled transistor.
| For the power, any fault in the voltage-reduction circuitry can destroy the uC. For the logic, it must be within the 0-5V range as seen by the uC. This is unusual, since it complicates what other circuitry you connect to the microcontroller.
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Current limiting for constant current source I have 6 LED strips, each of them requires 1080mA & 36V DC. I've found a Constant Current driver which can supply up to 8.9A & 36VDC. As there is no resistor on the strip so I think it should be current driven for high efficiency. The driver is: Meanwell HLG-320H-36B LED Specs:- Forward voltage: 33.6V - 36V- Forward current: 0.9A - 2.7A I planned to wire all 6 LED strips in parallel and use a current limiter circuit to protect each strip. So if one the strips is down, the others won't sink more current. I want to ask is this a good approach ? And how to build that current limiter ? Any suggestion or instruction ? As the driver is being shipped to me, I have to use it anyway. So is there any other solution that can help me light up my LED with the driver ? Thanks. p/s: I don't want to use separated driver for each strip because of the price & shipping fee. <Q> Going for a separate driver for each strip will increase reliability and prevents all sorts of issues. <S> If the strips are not 100% equal then the current will not divide equally between all strips. <S> The strips might be equal now but they will not be after some weeks of usage. <S> This will result in: <S> The strip which gets the most current will wear out first Strips not being equally bright once a strip develops a fault either it will take all the current and destroy itself or fail open and <S> all other strips will get more current (the total current remains 8.9 A) putting stress on them making them fail sooner. <S> A complex current monitoring system could be made which would shut off everything in case of a fault <S> but I guarantee you that that will cost much more than an individual driver for each strip. <S> So all in all, using one 8.9 A driver is asking for trouble . <A> The best solution would have been to add 6 Mean <S> Well LDD-1000 ($3.50 ea) to drive each string and they all can be powered by the HLG-320-36. <S> Another solution is to use TI's LM3466 Multi-String LED Current Balancer for Use with Constant-Current Power Supplies <A> The driver linked above has an output voltage range of 18-36V, in which it will behave as a current source whose current is set by a pot between 4.45A ~ <S> 8.9A. <S> The 36V limit is a problem. <S> If the LEDs are "36V nominal" then it is likely their voltage will vary in an interval like 32-38V so without seeing the LEDs datasheets, it is unclear whether they will all light or not. <S> Now let's suppose you're lucky: all your LEDs have a Vf of 36V. <S> You wire them in parallel... <S> but there is no voltage headroom to install a "current limiter" (whatever that would be) on each LED. <S> With the supply set to over 6 amps, then it will simply turn into a constant-voltage supply (instead of constant current) and deliver 36V (with unspecified precision)... <S> So... this whole arrangement seems a bit half-baked... please give the LED info. <S> EDIT <S> OK, the LED Vf is 33.6-36V which means that: If they are in parallel, the one with lower Vf will hog all the current and burn, so an individual current limiter should be installed on each... <S> But you will not have enough headroom since the max Vf is 36V and <S> the max voltage delivered by the supply is also 36V. <S> So you need a low dropout current source of 1A, something like that with a RdsON 0.02ohm MOSFET and a resistor like 0.1R <S> you should have a dropout of 0.12V at 1A which should be just fine, MOSFET dissipation would be quite low (max 2.4W per FET <S> so still need a heat sink TO220) and no efficiency problems... just need to select proper component values and FET.
| Now, if your LEDs have built-in resistors or drivers, and are able to regulate their own current, then it should work. Current protection: that's not a solution as disconnecting the current to one strip will force the other strips to take more current and making them fail sooner.
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MOC3021 appropriate input resistance value for microcontroller application I have this design. Mounted on "Project Board" and "Universal PCB" it works great. But soldered in a specially designed PCB only works fine for the first 10 tries, every 2 seconds, high work load. All 3 versions use same kind of componentes. So... I tried to cut off the led attached at input (Led1), and the performance increased 20%, but again fails in 14th try. It looks like a saturation in moc3021. I assume: R5 ( 220Ohms) is to high? R6 (1.5KOhms) is not needed? <Q> According to the datasheet MOC3021 requires 15mA at T=25 degree C to be sure of triggering. <S> You should add some to that to account for aging and lower temperatures, so <S> maybe 20mA. <S> If your microcontroller can safely source 20mA you can look at the minimum Vout(high) while sourcing 20mA to see what the resistor should be. <S> Otherwise, either use a transistor to drive the LED or use a more sensitive optocoupler. <A> You have approximately 16 mA through the MOC3021 LED, and another 8.5 mA through your indicator LED. <S> This would seem to be much higher (25.5 mA) <S> than I'd expect you should source from an MCU output pin. <S> As you draw current from an output pin the voltage will drop, and some heat will be dissipated in the device. <S> While I'm not sure that this is your problem, you certainly can improve the situation. <S> Most MCU's will sink slightly more current than they will source, I'll use the Atmega328p as an example here: <S> Here is the output voltage versus source current and at 20 mA you lose about 0.55 V. <S> Here is the output voltage when you sink current and at 20 mA you lose about 0.45 V. <S> This voltage loss will modify your currents slightly, but more importantly the current you are drawing is just too much. <S> I'd recommend that you could run both LEDs in series. <S> If the indicator LED is a RED LED then Vf is likely about 2.2 V, and the MOC3021 LED is at most 1.5 V. Setting the current to say 18 mA (the maximum for the MOC3021 at 0 degC) would mean a series resistor of about 72 Ohms. <S> You can see from the MOC3021 datasheet that the current requirements do go up for lower temperatures or narrow pulse widths: <S> If your current levels go up beyond 20 mA I would use a buffer to drive the LEDs and provide isolation from the MCU <S> I/O pin. <A> Remove R6, has no functionality more than to increase the current drawn from the MCU's D2 pin. <S> Calculate the value for R5. <S> OK1 LED = 15mA and 1.25 volt. <S> R5 = <S> (D2 voltage - OK1 LED <S> Voltage)/OK1 LED current = <S> (5v - 1.25v)/15mA R5 = 250 ohm (so 220 ohm in the picture is spot on). <A> The problem was the printed circuit. <S> and I solved using a cutter. <S> Thats caused changes in resistance that affects the output voltaje.
| It has a bad connection in a solder point
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Why is this circuit isolated? From Linear Optical Isolation for Safe Sensor Operation : Are the three grounds indicated all connected? Are the four Vcc indicated all connected? If the above is all true, how could this be an isolated circuit? The grounds aren't isolated. If the above is false, wouldn't U1 fry at the first hint of significant common-mode potential to ground? Are U1 and R1 considered pre- or post-isolation? <Q> That is not a very good illustrative diagram, and it's copied directly off the IL300 datasheet : <S> Here is how it should be connected: <S> The LED has an output that varies with temperature and decreases with time, whereas the photodiodes are quite stable with time and will track each other well. <S> So one photodiode is used in a feedback loop to control the LED brightness, and the other responds similarly on the other side of the isolation barrier. <S> As the LED decreases in efficiency, the feedback loop will increase the LED current to maintain similar emitted light, which is seen by both photodiodes. <S> The photodiodes are in reverse bias. <A> If the optocoupler is to be useful, the ground at the bottom of R2 must be isolated from the ground on R1, and the Vcc on U1 and U2 must be isolated. <S> Pin 3 of the opto should be connected to U1's Vcc, and pin 6 of the opto should be connected to U2's Vcc. <A> The grounds between isolated sections may be different. <S> You have also to make it different on PCB with a groove or enough distance between traces. <S> What's the point? <S> With such, you break the ground loop. <S> For example: the sensor can be grounded at distant point, while your DAQ is grounded in the cabinet. <S> Also all the EMI or overvoltage spikes on the sensor side won't go further to the DAQ side. <S> VCC and GND are different on each side.
| The labelling of that diagram is very misleading if it is intended to show optical isolation. The grounds on Vin and at the bottom of R1 must be connected, or U1 won't work.
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Why is it that "a single supply voltage is seldom used for audio mixers"? I'm reading " A Single Supply Op-Amp Circuit Collection " and in the paragraph about summing (page 9) I read " a single supply voltage is seldom used for audio mixers. Designers will often push an op amp up to, and sometimes beyond its recommended voltage rails to increase dynamic range ". Does this still apply? Do contemporary single-supply op-amps have limited dynamic range for mixing applications? If this is the case for how many channels does summing signal dynamic range need split-supply op-amps? <Q> Why is a single supply voltage is seldom used for audio mixers” Because in audio circuits the circuits become much simpler when used with a symmetric supply voltage (like +/- <S> 15V). <S> If one were to use a single 30 V supply (for example) then you have to take measures to DC bias <S> the inputs of the opamps to a voltage somewhere "in the middle", with a 30 V supply that would be +15 V. <S> You could also create a +15 V "AC ground" rail to DC bias all the inputs from <S> but then it is often just a small step to make the power supply deliver +/- <S> 15 <S> V. <S> Then all the inputs can be DC biased to 0 V which is "neat" and "easy". <S> Then the signals will also be centered around 0 V (ground) instead of having +15 V DC added to them. <S> Designers will often push an op amp up to, and sometimes beyond its recommended voltage rails to increase dynamic range <S> Does this still apply ? <S> Yes, in principle it does. <S> Using the maximum supply voltage does indeed give the maximum output voltage swing which is what is needed to reach the largest dynamic range. <S> Feed an opamp with +/- <S> 5 V and even the best opamp is limited to 10 Vpp output voltage swing. <S> Use +/-15 <S> V <S> and you get 30 Vpp output voltage swing. <S> But usually there should be no issue unless you'd exceed the maximum. <S> All circuits have a limited dynamic range, depending on the opamp you get more range or you get less. <S> You cannot say that the number of channels is limited by the dynamic range. <S> As long as you take care that the summed signal (current) is below what the circuit can handle (the output of opamp will not clip) <S> then there is basically no limit on the number of channels. <A> Because single supply designs suck. <S> On a dual-supply you can bias your signals around ground. <S> Coupling capacitors may still be used but they only have to deal with the small DC offsets coming from imperfect op-amps, not large deliberate DC offsets. <S> In a single supply system you must bias your signals around somewhere else. <S> This creates a couple of issues. <S> It is likely that any fluctuations in the bias voltage will become noise in the signal. <S> There will almost certainly be a "thud" at power on as coupling capacitors charge to their steady state levels. <S> Unless great care is taken there may also be similar thuds when gain controls are adjusted. <A> Audio signals are AC only. <S> Everything below 20 Hz is noise and can be discarded. <S> It is therefore often convenient to design the circuit so that the quiescent level of all the signals is 0. <S> That means using a ± supply. <S> Pushing any part beyond it <S> specified <S> voltage range is irresponsible design. <S> Since there have been and surely still are irresponsible engineers out there, you can probably find products where this is still done. <S> Opamp bandwidth has little to do with supply voltage. <S> The bipolar supplies of some audio circuits is for convenience of dealing with signals symmetrically about ground. <S> It has nothing to do with bandwidth. <S> No, we're not going to Google something to get information that should have been in the question. <S> That's squarely your job.
| However, not using the recommended supply voltage (but instead, using the maximum) might result in issues which might not be listed in the datasheet.
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Why is the feedback of op-amp oscillators connected to inverting input instead of positive input I'm new to the subject of oscillators and learned that sustained sinusoidal oscillations are made possible by positive or regenerative feedback . Is that right? If that's true then I can't quite grasp why, especially in op-amp LC or RC oscillators, is the feedback network terminated to the inverting input while the non-inverting input is usually grounded. Doesn't it make it negative feedback ? My circuit analysis ability is not yet fully developed so please bear with me. Also, I would highly appreciate if someone patiently traces out the operation of this particular colpitts oscillator for me in relation to this question. One more thing: does the LC network act as the feedback or it's just a tank circuit while Rf entirely serves as the feedback? I know. silly questions. Thanks everyone. <Q> Making this as simple as possible, when oscillating one end of the inductor will be rising in voltage while the other end is falling. <S> The current in the inductor charges one cap positively while it charges the other negatively. <S> As those charges accumulate the current falls in the inductor till it reverses and the whole thing repeats in the other direction. <S> The resultant see-saw in currents is what makes the whole thing ring at it's designed frequency. <S> Since, as I mentioned above, the voltage polarity is reversed across the inductor the op-amp needs to be connected as an invertor, i.e. using negative feedback as shown. <S> Note: <S> Given an initial stimulus the circuit will ring at the oscillation frequency on it's own, with no op-amp in the circuit, till it decays due to losses. <S> The op-amp overcomes those losses and keeps the thing ringing as long as power is applied. <A> The difference here lies in the resonant LC circuit that is present. <S> In such a parallel arrangement, the capacitors and inductor create a 180° phase shift in the signal. <S> This means that if you start by applying a 0° input on the negative side, the inverter op amp arrangement will produce a 180° shifted (inverted) output. <S> This output gets shifted by another 180° by the LC circuit, and you end up with 360° or effectively, 0° again. <S> This signal adds constructively with the input signal, and you have positive feedback. <S> Trevor's answer explains a little bit more about how the circuit specifically works. <S> Hope this helps, cheers. <A> Any Colpitts oscillator uses two capacitors and an inductor AND one other extra passive component - a resistor. <S> This may seem strange that I mention this but, with a perfect op-amp driving C1 (as per your circuit diagram) you don't quite get 180 degrees phase shift from the two capacitors and inductor. <S> Here's the general idea: - <S> So now the output of the op-amp is additionally phase shifted by R3 and C1 (in my circuit) <S> a few degrees and, this ensures that the frequency selective feedback network can produce a phase shift of 180 degrees with ease. <S> Your circuit doesn't show this so you may be wondering <S> how yours can work - op-amps are not perfect and at some arbitrary high frequency there will be a delay through the op-amp and that will be equivalent to a few degrees phase shift. <S> This ensures that the phase shift from L and C2 (maximum of 180 degrees at infinite frequency) can be used with the additonal phase shift from R3 and C1. <S> If you built the circuit you might want to put the extra resistor in circuit and initially choose a value of 10 ohms - if you get an overly distorted sine wave try increasing it <S> but, be prepared, the sine quality is not that great from a basic colpitts oscillator because there is nothing that takes care of ensuring gain/amplitude stability. <S> Nonetheless it's a fairly useful circuit and despite the plethora of information out there on how they work, hardly any of it mentions the extra resistor or the delay incurred by the op-amp in order to make it oscillate. <A> This circuit is getting positive feedback at the inverting input, because the feedback network adds 180 degrees of phase shift at a certain frequency. <S> I simulated the feedback network with some arbitrary values. <S> You can see that the phase of the feedback network is 0 across all frequencies, except for the oscillation frequency. <S> This means at that frequency (in my sim ~191 MHz) the feedback is inverted before coming back to the amplifier. <S> Typically, if the op amp input was a 191 MHz sine, the output would be inverted (180 deg phase shift) and added back to the input - <S> this is negative feedback. <S> Since the feedback network adds ANOTHER 180 deg of phase shift (360 deg total) <S> the sine being added back is in phase, resulting in positive feedback. <A> " sustained sinusoidal oscillations are made possible by positive or regenerative feedback. <S> Is that right? " Not always. <S> Sustained oscillation occurs when the Barkhausen stability criterion are met, and one of the criteria is that the total phase shift through the the amplifier and its feedback loop is 0 degrees, or 360, or 720, etc. <S> A common way to achieve this is with 180 degrees in the feedback loop and 180 degrees in the amplifier. <S> The amplifier part is met by making using an inverting gain circuit. <S> https://en.wikipedia.org/wiki/Barkhausen_stability_criterion
| In a typical inverting amplifier circuit, you are correct that the output feeding back to the negative input would be a form of negative feedback.
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1GHz digital oscilloscope with 50MS/s sample rate Can someone please explain to me why can't I measure a non-periodic pulse signal \$T_r=1ns\$, with a digital oscilloscope \$1GHz\$ with a sample rate of \$50MS/s\$? I may have to solve similar problems in an entrance exam for Faculty of Electrical Engineering, so I would like to really understand the problem and the calculations. Any help or hints are welcomed and appreciated. Thank you. <Q> Nyquist Theory! <S> Here's a great app note on Nyquist Theory and Oscilloscope sample rates <S> I put together a quick 3m video a while back explaining the difference between oscilloscope bandwidth and sample rate . <S> In the case of your question, the oscilloscope bandwidth is fine, but the sample rate is so low the resolution wouldn't be accurate enough to give you a good idea of the signal's characteristics. <S> Like other comments mentioned 50 MSa/s = <S> 1/50,000,000 = 20 ns/sample. <S> To accurately sample a 1 ns rise-time signal, you'd theoretically need at least 500 ps resolution. <S> In the real world, though, you'd still want faster than that. <A> As indicated in the comments: <S> 50 MS/s = 1/50,000,000 <S> s between samples = 20 ns <S> If you are measuring one point ever 20 ns <S> how could you tell the difference between a 1 ns rise time and a 19 ns rise time? <S> And to be honest if you can't answer that sort of question on your own then you need to do a lot more work before your exam. <S> This isn't the sort of thing that you should need to memorize, it should be obvious to anyone who understands the underlying concepts. <S> Learn the ideas rather than the answers to questions, that is the only way you can hope to answer questions you've not seen before. <A> Put a 49MHz square wave into a sampling scope running at 50MHz Fsample. <S> Examine the scope display, and think about what does on.
| Basically, to get an accurate depiction (and measurement) of your signal you have to have a high enough oscilloscope bandwidth and a high enough oscilloscope sample rate.
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Tapping SPI communication with 328U I'm trying to extract data from a 4x14 segment alphanumeric display controlled via 2*595 in series and 4 transistors. Serial data is fed from STM32f103, its firmware should not be changed.Problem is that Atmega running 16MHz can't cope with 2.2MHz SPI clock from ARM. And it is a 16bit continuous word, Atmega has only 8bit SPDR buffer. Is there some one-chip solution that would allow me to store data from dataline, and read it on slower clock as master between transitions?I don't have enough pins to do it in parallel.Will or-wired (diodes) clock on 595's work? <Q> I worry that an 328p is no match for a 72MHz Cortex-M3 with DMA. <S> You can easily get one byte, SPI peripheral will give you an interrupt. <S> You then have to process/store that byte and wait for the next one. <S> If the STM32 is using DMA then there is literally no delay between bytes, so no matter how you optimize ISR code you'll loose data. <S> I would attack the problem with a SN74LS674 - 16-bit paraller in, serial out register. <S> Connect it to all the outputs of the 595s and read over SPI. <A> If you have access to the 595s <S> then you may be able to clock the serial data out of them after they have been loaded by the STM32. <S> The ATmega would monitor the latch signal (perhaps using pin change interrupt), then clock the data out via the second 595's QH' pin once it has been stored in the output registers. <S> You would need to switch the 595 clock input between the STM32's SPI clock and a clock output from the ATmega. <S> If the SMT32 clock idles low then a simple OR gate would suffice, or you could make a multiplexer from 4 NAND gates (one 74HC00 IC). <S> simulate this circuit – <S> Schematic created using CircuitLab <A> One solution is a small 8-bit micro with an SPI subsystem that can handle this speed. <S> 2.2 MHz is not very high for SPI, so most "modern" micros should do it, and have enough RAM to buffer everything. <S> Reading it out can be done in any way convenient for the Arduino. <S> You will probably not find a custom IC for this application, and if you do, it will be more expensive than a small 8-pin PIC or AVR. <S> Microcontrollers are today sometimes cheaper than the 595's. <A> I'd suggest you don't need to work with the high speed output shift register at all. <S> What you describe is a simple multiplexed display where the individual digits (14 segments) are turned on for relatively long periods by the 4 enable signals. <S> The data outputs are stable for the complete display period for each digit. <S> Take the four digit enable signals to 4 328p input pins using the (I assume here) falling edge to trigger an interrupt. <S> You now have an ISR for each digit as it is enabled. <S> Since it's likely that the scan rate for the digits is less than 2 kHz or so, so you could expect that each digit is enabled for about 500 us. <S> Feed the 14 segment data bits into 2 * 8:1 multiplexers (74HCT151), with the output bits each going to one input pin on your 328p. <S> Use three output bits on the 328p to set the bit being received from the low/high byte of the segments. <S> For the code, when you receive an Interrupt from the digit enable: <S> Set select to zero (0x000) for the muxes Read LS bit on each byte and store value in 2 memory locations for digit. <S> Inc select to loop through and read each bit in the low/high byte. <S> Ret <S> Since the interrupts never conflict, you can use a single routine to read/loop and simply set up a pointer for storing the low/high bits. <S> This would end up being a 2 chip + MCU solution, with 6 input pins and 3 output pins used on the 328p. <S> You might also need pullup resistors on the scan transistors. <S> (you'd need to provide some schematic information to be able to ascertain the need)
| SPI clock speed is not the problem (328p can go up to F_CPU/2), but the inter-byte time.
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How to reverse engineer controller outputs of a sprinkler control box? I want to replace the controller of a sprinkler control box with my own design, while keeping the backplane and module(s). The controller has a 14 pin ribbon cable connection to the backplane and 4 of those pins are feeding power to the controller from the backplane. The remaining 10 pins aren't powered without the controller connected and show 24VAC when the controller is connected. Using an oscilloscope, I've monitored the pins when changing the sprinkler zones, but haven't been able to figure out the control signals (the pins are always 24vac when I measure them regardless of what zone is on). Since it is AC, I really don't know where to go from here (if it was DC I might have a better shot). I'm assuming the signal changes too fast for me to see on the oscilloscope. Is there something I can use to 'watch' the connections? (Maybe cut the cable and add a circuit/monitor??) Is there a better method? I've also searched for schematics of the controller, but haven't come up with anything useful. Controller: The back of the board has traces and a reset switch, but no other components. On the ribbon cable: Pin 1 is one phase of the 24VAC and pins 2,3,4 are the other phase. There's a 9V backup battery that isn't present and isn't required for operation. The 6 pin connector goes to a rotary switch that makes controller selections. Backplane: A 24VAC wall transformer feeds the backplane with the two orange and single green wires. The yellow wire is a jumper for an uninstalled rain sensor (rain sensors are a normally closed switch). I've (poorly) soldered the module to the backplane so it stays connected while it is out of the cabinet. Module: The module has 7 outputs. There is a single solenoid valve wired to the module (the black wire with white stripes going to the screw terminals). One of the outputs of the module is always on and the other 5 are switched. The seventh connection is for the unswitched phase of the AC. Three more modules can be added to the backplane. The controller 'knows' when a module is not installed, but doesn't sense whether or not a solenoid is installed. The smd (transistor maybe??) is labeled 5D with a sideways 'p'. The triad is labeled PJ 600E BT134 A1449 D6. simulate this circuit – Schematic created using CircuitLab Update: I tried to make my first schematic with the module soldered to the backplane and the solenoid still attached. I've updated the schematic with my second attempt (I probably have the triad oriented incorrectly). I'm using continuity on my multimeter to figure out connections. Now that I might have the module schematic close to correct, what makes the outputs turn on (trying to avoid the randomly applying power and burn it up method)? Just to clarify what I mean by common output , is it is the second leg of the AC. All other 6 outputs are the opposite leg of the AC. I'm not sure if I am using common output correctly. <Q> Reverse Engineering 101 <S> Look at return wire for ground switched solenoids from relays <S> measure DCR of solenoids for surge current rating <S> (V/DCR) use a dummy load (Triac outputs will not conduct without a return load resistance or sufficient holding current flow. <S> map out continuity with ohmmeter to make a wiring table. <S> label everything instead of a schematic e.g. P1-1 to J4-7 p.s. <S> The only questions in this bounty are : Is there something I can use to 'watch' the connections? <S> (Maybe cut the cable and add a circuit/monitor??) <S> Is there a better method? <S> The board with 5 transistor switches and 7 screw terminals implies you have 5 outputs and 2 inputs for 24Vac power and return . <S> My assumption is you know how to use Ohm's Law and trace circuits and do design at this level. <S> Otherwise, much to learn. <S> With a load, it is trivial to measure when power is switched on and locating the common screw voltage to which the power is connected to. <S> There is no need to cut cable and monitor current unless you wish to diagnose connection faults in the solenoid, then use an ommeter with power off. <A> I think it will be a lost cause trying to retain use of the modules. <S> Get some DIN rail terminal blocks and don't look back. <S> Edit <S> : I know you really want to, but it just seems like a slow expensive process. <S> You're sure to let some magic smoke out as you design/test, and what? <S> You call Rain Bird and ask for part 635575 and pay full-boat parts price with shipping? <S> That could be $65. <S> The fun and craft, if any, would be in the reverse engineering. <S> You say you see 24V on all the lines. <S> Can you measure the current on an active being-operated solenoid, and then figure the resistor value to make a dummy solenoid? <S> (or just another solenoid lol). <S> Seems like you'd need 2 or more solenoids to even troubleshoot this <S> , just so you can be sure you are commanding them to be on or off. <S> If the controller isn't doing what you think, you'll never figure it out. <S> I'd also watch for phantom voltage, it may show 24V on all the lines, but will a 22K resistor across the test points (1.1ma@24V) make that voltage go away? <S> I'm deleting the part about how to do it because you've already done solenoid control projects. <A> This is just a helper and not a complete answer. <S> I found the image below on an eBay ad for "MMBD4148 1N4148 Diode 200mA 100V". <S> Figure 1. <S> SMD diode. <S> Three pin package used for automated assembly. <S> See BAS16 / MMBD4148 / MMBD914 datasheet for a similar diode package and pinout. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> Less unlikely schematic? <S> I've rearranged your schematic into a less unlikely arrangement. <S> The varistors would be across the solenoid coils to absorb any inductive kick and protect the triacs. <S> In this arrangement a negative pulse would be required on P5 to trigger the triac. <S> (This seems unusual. <S> A positive pulse would be more likely.) <S> Things to check: <S> Use the diode test function on your multimeter to see if the three-terminal device is just a diode. <S> See if my wiring diagram is better than yours! <S> Figure 3. <S> A micro PLC with keypad and LCD. <S> These are available for the price of one of your output modules but include the controller! <S> They typically feature eight inputs and four outputs on the base module with expansion modules available to plug in or network connection. <S> The one in the photo is from Crouzet but most of the big manufacturers - Allen-Bradley, Siemens, Mitsubishi, etc., have a version. <S> You would require the relay or triac output version for your application.
| The 3-pin SMD package might be just a diode, explaining why you can't find where the third terminal goes. Alternative solutions: As is evidenced in the comments, you should consider replacing the control system with an off-the-shelf solution such as a micro-PLC (programmable logic controller).
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Use diode in parallel to increase current I just realized that some diodes I have are rated for 1A only. I'm planning to wire a l298 chip to some stepper motors that are rated for 1.7A. The datasheet seems to recommend to connect each phase to VCC and GND in reverse order. Something like this: simulate this circuit – Schematic created using CircuitLab My guess is that the current passing in those diode should never be really that high and must happen when phases are reversed as it seems to feedback current into the input. Though the question is: Will the 1A diodes be enough (IN4004). If not, can I wire 2 diodes in parallel to spread the current accross 2 diodes instead? In other words, instead of using 8 diodes for 4 outputs I'd use 16. <Q> Those diodes are peak catchers and hence can have a momentary rating far in excess of average current rating. <S> Read the appropriate data sheet. <S> Also, the L298 is fairly crappy and I reckon you'd burn that first before damaging a diode. <S> Same advice read the data sheet and note the power dissipation OR go to my profile page and find a question and answer I raised on low voltage H bridges that explains the problems. <A> if your motor DCR caused surge currents to exceed 2A , you will exceed the spec. <S> , which is most likely with a full RPM rating of 1.7A typically 8 to 10% of surge rating. <S> The external bridge of diodes D1 to D4 is made by four fast recovery elements (trr ≤ 200 nsec) that must be chosen of a VF as low as possible at the worst case of the load current. <S> Choose better drivers using MOSFETs. <S> Tons out there online for PWM operation for motors.... <S> cheap... <S> why build when these are cheap? <S> [10][1]$ for 15A or less $4 for 5A. Update Stepper Motor? <S> Make vs Buy is every designer's choiceI built this kit, last week using these drivers in CNC shield over an UNO with 1.2m square gantry for laser or magnetometer head. <S> Cost $350 Sale price $4k <S> hi velocity mode <S> 0.1 mm <S> full steps <S> https://m.youtube.com/watch?v=I7DCYEM3dW8 <S> Video doing my macro for a 4 corner sweep the log spiral test that follows intentionally made fast jogs with accelerate and brake in 192 G code steps rather than arcs which would have been perfectly smooth to see what the limits were for speed on a light rail gantry. <S> A crude servo test. <S> 1st test before tightening up everything... <S> [1]: https://www.ebay.com/p/6v-90v-15a-Pulse-Width-Modulator-PWM-DC-Motor-Speed-Control-Switch-Controller/1960103853?iid=301390515744 <A> Although the 1N4004 will handle considerable current for short periods, if you are driving that stepper motor at any appreciable speed those short periods will add up very quickly. <S> Further, 1N4004 are quite slow. <S> I'd also not put diodes in parallel unless they are very closely matched.
| You really ought to be using fast diodes for this application.
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Why did my solder joint fracture? I've been in the habit of making vias on homemade pcb's by putting the leg of 1N4148 diode into a 0.5 mm hole, cutting it off to length and soldering both sides. However, now my led controller broke and I traced the problem into a broken ground via. On closer inspection, the wire has separated from the solder: There was about 2 ohms of resistance between the wire and the ground plane where it connects. During operation, there has been about 1 ampere of current going through it for a few hours each day. The device internal temperature stays below 50°C at all times, and has always been stored and used indoors. Are there any obvious things that I'm doing wrong? I doubt it is a cold joint, because a bit of solder remains also on the wire. <Q> Looks like a cold joint to me- the slightly crystalline appearance, for example <S> (though it's hard to tell with all that flux on it). <S> It was probably fractured from the beginning and corroded a bit so it stopped working. <S> You can avoid this by preventing motion as the joint cools. <S> You can catch such defects by inspecting the board with a 10:1 magnifier, and to some degree, by burn-in at elevated temperature. <A> I know that annular ring failures are common with any strain, mass and vibration and worse with crystalline solder joints. <S> So even the major players have these quality issues. <S> Corrective Actions by design and process improvement. <S> Strain relief with rigid polyurethane , to prevent micro-vibration motion of attached parts. <S> improved solder process with adequate heat source with a quick response time rather than slow for crystalline growth which is structurally weaker. <S> ie cold solder joint verify board flex does not aggrevate vibration from resonance. <A> Looks like it has got a thump one second before the joint had cooled enough to stand that thump. <S> Another possiblity is vibration. <S> Check, are there any other sources.
| Sometimes a coil vibrates enough to break its own joints, but this is not a component that generates vibrations. Relay THT solder joints are common in stoves, flyback transformers on large Apple monitor mobos etc etc.
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Circuit breaker for single outlet I have several 3d printers. One of them keeps tripping the breaker for the room. I'm trying to debug (obviously will not use it regularly until I can determine what is wrong) but in the mean time I need to keep my other printers running. A power strip's breaker isn't enough for it to trip, so it's drawing less than 15 amps when it spikes. It also draws 50 watts when just running normally. The breaker for the room is 20 amps, but I'm running an AC and various other appliances as well. The combination is what causes it to trip the 20 amp. It's absolutely not the AC causing it to trip. I have eliminated variables and it's one of the 3d printers for absolutely certain. I'm guessing I need between a 3 and 5 amp resettable breaker, one that works for a single outlet only. 120V. Where can I buy such a breaker? Would be even nicer if I could switch between a few different preset values, ie, 3 amp, 4 amp, 5 amp, etc. My workshop only has one circuit, just how the building is set up. (rather each room has one circuit, I only have access to the room I pay for) <Q> This is complicated since the downstream breaker must be faster then the upstream breakers. <S> It needs to have different time characteristics. <S> This is called breaker selectivity. <S> Without replacing upstream breakers, you'd probably need a breaker that is so sensitive that you can't even enable the breaker. <S> The inrush would trip it immediately. <S> Also, a 20A breaker won't trip immediately at 21A. <S> This will take minutes . <S> See the image in this post . <S> I think you have a larger overload. <A> You can buy low current panel mount circuit breakers from Digikey and other electronic suppliers. <A> I don't know about the US, but here in the UK I would go to the local hardware store where I could find a consumer unit (I believe the standard term for such an item in the US is a distribution board?) <S> along with a 6A breaker that's designed for lighting a garage, such as this one . <S> Wiring it up is pretty simple, and you can then enjoy protected outlets that won't cause your main circuit breaker to trip. <S> You also get an individual RCD (ground fault breaker) for your circuit too.
| You could mount one of these breakers and an outlet in a box to produce individual low-current outlets for your printers.
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Can I use a VFD to send power of a specific frequency to a frequency meter? I have a beautiful old frequency meter from a power station, a "System Frahm" type panel based on vibrating reeds. It has two rows with 21 reeds in each, each going from appx. 45 Hz to 55 Hz. It says that it accepts 110 volts. I want to be able to provoke the instrument to indicate frequencies of my choice, between 45 and 55 Hz. If possible, different frequencies on the two rows, but it's not mandatory. I have access to a VFD, the Delta VFD004S21A 220V 1-Phase model. I know it can generate 3 phase power at the frequency and voltage that I need, but will it be able to run without driving a motor, or any other load? Would it work if I added a small 3 phase motor? Would I be better off with some kind of audio power amp + step-up transformer solution? <Q> You can get stereo sounds synthesized in Audacity (free) and sweep each channel then playback with a PA and stepup XFMR using a stepdown wallwort. <S> I 've never seen these ABB historic instruments <S> but I see they have medical interest in cochlea demonstration, perhaps in your area of interest. <S> You probably only need 10 watts at rated voltage. <S> (guessing) <S> They actually are tuned at 2f each using relay armature mechanical pulse being broad spectrum. <S> (I think) <S> https://peerj.com/articles/1333/#fig-1 <S> Alternatively you could easily measure the armature R to estimate current and feed it from a transistor and clamp diode from a sig. <S> gen with 10Hz <S> +/-10% with 10% duty cycle <S> using lower as this may be enough energy with pulsed DC voltage and 5th harmonic. <S> or many other methods... the Q of 50~100 of resonant reed wil reject any PWM noise or VFD carrier noise. <A> You may need a step up transformer based on your voltages. <A> The VFD should have no problem operating without a load. <S> If that is easily available, I see no reason to use something else for a one-time demonstration. <S> If you want a permanent setup, you might want to construct something. <S> You might want to run a test to see how much power is required. <A> Very cool meter by the way. <S> Given that it says "Kristiania" on the faceplate, that dates the meter to pre-1925 when Kristiana's name was officially changed to what we know today as Oslo, Norway.
| Yes, it is not only possible but very doable.
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What is the purpose of a resistor between the and gate output and ground? I found this image of a transistor-based and gate: and I can't understand what the purpose of the 4.7K resistor is. If the component controlled by the gate is placed between "Out" and ground, then that resistor is just a waste of energy since the circuit should work with or without it. What am I not understanding? <Q> First, let me tell you that this AND gate is one of the worst AND gate implementations. <S> It has the several problems. <S> Two of them: 1) <S> It lacks regeneration (ok, this is a common issue on other discrete-component implementations too). <S> 2) Under certain circumstances, it could not even work. <S> If B is high (6V in your circuit), then the output will be: ( Vb - Vbe <S> ) * R2 / (R1 + R2), where R2 is 4.7k and R1 = 10k. <S> In other words, 1.69V. In some cases that value could be too high to be considered a "0". <S> Similar approach (just with a higher input current) could be achieved with a diode-resistor AND. <S> Still, this would not have the issue of point 2. <S> That said, the 4.7 kOhm serves as pull down resistor, i.e. to create a "0" output signal, when A or B is "0". <S> Furthermore, a "low" resistor value (few kOhms) is mandatory for the case of B = " <S> 1" and A = "0" to work correctly (see point 2). <S> In fact, if it were, let's say, 470 kOhm, with B="1" and A <S> = "0 <S> " <S> (Vb = 6V, <S> Va = 0V) <S> you would get Vout= 5.2 V, as the base-emitter junction would act like a diode... <A> <A> The "primitive" "AND" gate needs an emitter pull down with inputs A or B at 0 V <S> then Vout = <S> High if both inputs are high (5V) <S> and then R reduces the turn off time for a given xx pF load when goes low. <A> If the output is connected to a modern 'normal' gate (TTL or CMOS) it will not work without the resistor. <S> It will work with RTL if you have access to such museum pieces from the 1960s. <S> So it really depends on what you imagine might be the characteristics of whatever you will connect it to. <A> There are several pros and cons for using such a resistor. <S> I would recommend to use google for further reading
| This Resistor causes "emitter-degeneration". If both transistors are on, and there is no resistor, you are shorting out the power supply
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What is the role of shunt in a Power Meter? I just got a PZEM-051 Power meter and I can see that in order to measure the power, one needs a shunt. At least that's what I can see from different YouTube videos and I was wondering why would I need one? By looking at the scheme on the back of the meter, I can see that it has a shunt integrated (please see picture). Do I understand the scheme wrong? What would be the role of the shunt and do I need one to read the power and the voltage from a power supply and a small pump? <Q> The role of the shunt is to enable measurement of the current drawn by the load. <S> The shunt is a low value resistor that provides a small voltage proportional to the current (Ohm's Law). <S> Power is calculated by multiplying the voltage and the current. <A> Power is the product of the voltage applied to a load and the current consumed by it. <S> The shunt is low value resistor that may only be a small fraction of an ohm (depending upon the load current). <S> This example is designed for currents up to 100A and so the value of the shunt may be just 500 micro-ohms. <S> At 100 Amps that would cause a voltage drop of 50 milli-volts. <S> It is desirable to avoid dropping too much voltage across the shunt because any voltage dropped <S> doesn't get to the load. <S> Also the power dissipated by the shunt maybe excessive. <S> In this example with 50mV drop and 100A flowing the shunt resistor will dissipate 5 Watts. <S> It looks like this power meter expects the user to supply an external shunt, wired according to the diagram. <S> It should say somewhere on the meter or in the manual what the maximum voltage is expected from the shunt. <S> Something like 50mV would be common. <S> The display is scaled to 100A when the maximum shunt voltage is presented. <S> It is up to the user to calculate the required value of the shunt. <S> Note the '4-wire' connection to the shunt (also called 'Kelvin' connection) that is used to avoid the resistance of the connecting wires causing an error. <S> The shunt resistor should have 4 terminals. <S> I haven't found a manual online <S> but there is a YouTube vide explaining the connections Connecting the PZEM-051 <A> Sensing power consumed by a load will require measuring the current that passes through it. <S> If the current-measurement device were located within the power meter, then all current that flows through the load would need to flow through the power meter. <S> A power meter capable of handling a large load would thus need to be connected using very large connectors and wires. <S> If one instead uses a power meter that is designed to be wired as shown using an external shunt, the wires that connect the supply to the shunt and the shunt to the load will need to be large enough to handle the load current, but the wires that connect to the meter will only need to be large enough to supply current for the mater. <S> Even if a load would consume thousands of amps, the meter itself might still be connected with 24ga wire (though there should probably be fuses to protect the wires going to the meter in that case). <S> Note that there is no limit to the amount of power that may be sensed by a power meter using an external shunt. <S> If a meter is designed to measure up to 200mA using a 0.1ohm shunt, the same meter would work just as well to measure up to 200A using a 0.0001ohm shunt. <S> If the meter has a numeric readout along with annunciators to select power, current, etc. <S> one might label the "current" LED as "mA" when using a 0.1ohm shunt, and "A" when using a 0.0001ohm shunt, but the meter wouldn't otherwise need to know or care what kind of shunt was being used.
| The shunt is used to measure the current to the load.
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Controlling a 4 wire servo I'm trying to control a 4 wire servo that was used in an RC car for the steering. It has two power connections, but I can't make sense of the other ones. If there is power on the red and black wires, the motor turns into one position and stays there, no matter what I put on the other wires. I also tried to vary the voltage of the power, no difference. How can I control its position? The motor: https://drive.google.com/file/d/0B8Ox1li7YRxoa3ZBY2RNakxCT0k/view?usp=sharing The board it was connected to: https://drive.google.com/file/d/0B8Ox1li7YRxoREt1Z3RNcXp4OU0/view?usp=sharing <Q> Usually RC servos work on a pulse principle. <S> The input idles low. <S> A pulse (TTL-level) of width 1.5ms puts the servo at mid-position (using feedback from an internal pot), 1ms brings it to one extreme and 2ms to the other. <S> The pulses need to repeat at some frequency, 50Hz for example, the exact frequency is not too important. <S> The same receiver can supply, in turn, pulses to several servos since the maximum pulse time of 2ms is much less than the 20ms period. <A> I'd suggest from the photo that Grey/White are the DC motor power, and in an RC car will most likely be 7.2 V (a 2S Li-Ion). <S> The Black/Red wire is the signal wire. <S> You should find that one wire of each pair is ground ......measure from Grey to Red and Black ... <S> then White to Red and Black. <S> Whichever pair are shorted those represent the ground. <S> That will then leave you with two wires .... <S> one of the Grey/White will be the 7.2 V motor power and one of the Red/Black (most likely Red) will be the input signal. <S> As Sphero pointed out the signal is a repeating frame at 50 Hz with 1.5 ms period. <A> But it has to be an external power source, connecting them to the same 6V source in parallel didn't do anything.
| So I figured it out, all I had to do is connect grey and white to a power source, and depending on the polarity the motor turns either left or right, and centers itself otherwise.
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Why this code wasn't written in a much simpler way? I have came across a question while working on assembly language. Here is the question: Assume that bit P2.2 is used to control an outdoor light and bit P2.5 a light inside a building. Show how to turn on the outside light and turn off the inside one. Solution given: SETB C ; CY = 1ORL C, P2.2 ; CY = P2.2 ORed w/ CYMOV P2.2, C ; turn it on if not onCLR C ; CY = 0ANL C, P2.5 ; CY = P2.5 ANDed w/P2.5MOV P2.5,C ; turn it off if not off I just felt like it would do just the same job to code: SETB P2.2CLR P2.5 What's wrong with that? <Q> You are right in that it appears the code you show is silly. <S> Perhaps whatever machine this runs on can't do immediate operations to set bits on <S> I/O ports, and that that's why something like SETB P2.2 isn't possible. <S> Still setting the CY bit to 1, then ORing anything into it is just plain silly. <S> The same goes for setting the CY bit to 0, <S> then ANDing something into it. <S> Clearly the CY bit can be directly copied into a I/O pin bit, since the code does that. <S> At most this should be 4 instructions, certainly not 6. <A> The code is almost certainly for a processor using the 8051 instruction set. <S> On that processor, the code variation you give would have the same effect as the original except that it would run faster. <S> Executing "ORL C,P2.2" when carry is set will have no observable effect except to waste some number of cycles (two CPU cycles totaling 24 clock cycles on an 8051 if I recall correctly; likely a different number on some other variants). <S> Likewise with executing "ANL C,P2.5" when carry is clear. <S> Although there may be some kinds of processor where a request to read some I/ <S> O locations would have some observable effect, I don't think any 8051-style processor has ever had such behavior for any bit-addressable I/ <S> O locations, much less for bits of P2. <A> The assembly code given is likely compiler-generated. <S> It is the unoptimized version of the following C statements, where P2_2 and P2_5 are the bit-addressable objects: P2_2 |= <S> 1;P2_5 &= 0; <S> This may seem equivalent to <S> P2_2 = 1; and <S> P2_5 = 0 <S> ; , but it isn't if the bit-addressable registers are volatile objects. <S> A read-modify-write operation on a volatile object must perform the read and the write, in that order. <S> This ensures that any side-effects from reading or writing the register actually occur. <S> Although I know of no 8051 bit-addressable register with side-effects, a compiler cannot assume there isn't or won't ever be one. <A> The real difference between these may be subtle. <S> In your simplified answer the logic is read the port, set or clear the bit value and then write it back out to the port. <S> Note the whole port could be re-written here. <S> The solution on the other hand uses the MOV bit instruction which may operate in a rather different manner. <S> Without going into the details of the particular part used here it is difficult to determine if there is a difference, or if it matters. <S> Or it could just be the instructor decided to make you think.... which is after all.. his real job. <A> The only answer is that the processor does not support 1 bit instructions directly. <S> However, when the carry bit is used, it knows it is only one bit being manipulated.
| Perhaps the purpose of the code was to demonstrate the ORL C,bit and ANL C,bit instructions, but this seems like a weird example to demonstrate them.
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If I connect a schottky-diode to the DC converter output, why doesn't sense the voltage? Why doesn't work the second version? If I try this version no voltage on the output. Why? I would like to make a reverse polarity protection, but the problem with the first version: in idle mode there is a voltage e.g: 12.00V, but when I connect a variable load the voltage decreases when the current increases, because more voltage drop on the schottky-diode. I would like to use a reverse polarity protection but I don't want that voltage drop vs current increase.I can't use a mosfet because I use 1V output voltage too. <Q> The MP2307 is a synchronous buck converter. <S> At light or no load, the part will draw current from the output cap to maintain regulation. <S> (The low side FET turns on, reversing the current direction in the inductor, and pumping up the input caps.) <S> By putting the diode in the control loop before the output caps, the part has lost its ability to pull negative current through the inductor. <S> This likely causes the output caps to rise each switching period until the part reaches OVP protection and shuts down. <S> Or, it could cause a control-loop instability which shuts the part down for other reasons. <S> Look at the startup with a 'scope to see what happens if you want to know, but this topology (especially without a light-load PFM or diode-emulation mode) can't work with a diode where you want to put it. <A> Try moving the top of C2 left of the diode in the right circuit, keeping the resistor chain as is. <S> ADDITION: <S> Generally the idea of preventing back drive like that can be problematic. <S> On light loads with any substantial capacitance after the diode, the voltage may creep up higher than you want past the diode. <S> Some additional over-voltage protection may be prudent. <A> SW output (pin 3) is a half bridge FET driver going between Input (p2) then to ground (p4) at some PWM clock rate and duty cycle. <S> Therefore inserting a diode as you have shown blocks current in one direction. <S> With a Pch FET on high side output or Nch FET on return side, you can get better reverse protection with low If*RdsOn= Vdrop. <S> N.B. <S> This also protects the FB input from reverse voltage. <S> See here for details. <S> http://www.ti.com/lit/an/slva139/slva139.pdf
| As John D mentions you can't have a diode before the regulator cap since the device will try to pull charge from the cap when the voltage is detected as being too high.
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Figuring out a substitute for an old amp capacitor #2 What type are the yellow caps in the picture and what do I substitute them with? <Q> These are "film capacitors", usually rated for voltages suitable for vacuum tube amplifiers, 60V-200V-600V. <S> They should be substituted with similar film capacitors, with any kind of dielectric, check with DigiKey . <S> As I see, the caps are 0.047 uF 400VDC. <S> So this one would be a perfect match . <A> This PP (polyprop) is smaller length 10mm , better performance than PE and cheaper. <S> 0.58 or $5.88 for 10pc's https://www.digikey.com/product-detail/en/kemet/R75LF24704000K/399-13014-ND/5765881 <S> Although radial, I believe this is a better Form, fit and function to drop in place (if needed) <A> If they are not leaking DC, I would not concern myself with replacing them.
| If they are leaking DC, a typical replacement component is the Cornell Dubilier Type 150 Series Film Capacitors (formerly Mallory 150M).
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Are RJ45 jack sockets universal regarding CAT-X cables I've noticed some trade/wholesale websites list CAT-X jack sockets separately as the individual numerical version, i.e., (CAT-3 jack socket). I'm aware CAT is a standard, and RJ45 is the socket itself. So for example, if I were to buy one labelled as a CAT-3 jack socket, and stick a CAT-6 cable in it, does the contact material of the jack really make that much of a difference, that the cable would then only have the throughput of CAT-3? Or is it just that the materials used for a CAT-6 jack are GUARANTEED to give the full throughput, whereas YMMV with the CAT-3 jack socket. Thanks! <Q> Cat6 cable generally has thicker wires. <S> Take this image as an example (from similar serverfault question ): <A> CAT is a cable categorization. <S> So, strictly speaking, there's no "CAT6 jack sockets". <S> Sockets aren't cables. <S> Anyway, since the important thing about cable categories is the maximum bandwidth that the cable can transport without losses that are too high, it's fair to assume that the vendor wants to say "this is a socket that matches the capabilities of a CAT-X cable". <S> Connectors typically are pretty critical for the bandwidth of systems. <S> I've not seen a CAT-3 RJ45-connected cable ever . <S> In my experience, it's either not <S> CAT-X anything, CAT-2 (typical ISDN line), or CAT5 or more. <A> Actually RJ45 is how it is electrically connected, the connector is a "Modular 8P8C", but its a normal misunderstanding. <S> And Yes it is the same connector except there is also variants that are shielded. <S> Ref. <S> : <S> https://en.wikipedia.org/wiki/Modular_connector
| Connectors labbled as "Cat6" likely had provisions for dealing with these thicker wires. There is also a difference in what thickness of cables the connector can handle.
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Is there a 16 to 4 one-hot simple encoder IC chip? I'm looking for a chip that has 16 inputs where if only one of the inputs go high then it encodes the data to a 4 bit output. If one doesn't exist what would be the simplest way to build such a circuit? <Q> The function you are looking for is normally referred to as a Priority Encoder. <S> A priority encoder can cope with more than one signal being active and still give the correct result for the highest number active device. <S> If your system can guarantee that one and one input only will be active you can use a simpler solution as suggested by Dave Tweed in the comments, however such solutions can give a completely incorrect result if two or more inputs are active. <S> An 8 bit version has been around for many years. <S> One of the current versions is SN74HC148 . <S> @Marku points out that the linked data sheet has a 16-bit example using two such devices together with 4 2 input and gates for three packages in all. <S> These days it is more common to perform function like that either in software in a micro controller or as Verilog/VHDL in programmable hardware. <S> Can you describe your system in more detail to see which solution would fit better. <A> I don't know whether such a device exists, but: Approaches in lieu of a 16x "one hot" -> 4 bit output IC Serialize, count clocks until the 1 appears <S> Use a parallel-to-serial shift register(s) to convert your one-hot to a sequence of 15 zeros and 1 one in the right place. <S> Clock that shift register and feed its output into a 4 bit or more (you'll probably find 8 bit) counter with parallel output that you clock with the same clock, but latch on the data input <S> Programmable logic <S> Example IC Simply store right 4-bit value at right 16bit address <S> ROM in <S> diode logic <S> You can basically do a lookup table matrix in PCB traces. <S> Will require lots of discrete diodes (4 output bits being ORs of 8 input bits, each, so 32 if my math is right). <S> But it's "logically" easy. <S> Assume you have an 8-input "OR" gate (that would be 8 diodes with a common cathode). <S> Then, the highest output would be "Is it any of the second half of inputs", right? <S> So, you just connect that OR to these. <S> Similarly, your lowest output would be "any of every other input", so you connect these to another 8xOR gate. <S> Fun fact: there's 8-diode common-cathode components, and they're pretty cheap! <S> Just use TVS clamping diode arrays ( example ) and an elegant PCB layout. <S> If your input voltages aren't higher than the TVS' clamping voltage, you'll have to go for 8 quad common-cathode diode arrays . <S> ROM using discrete gates Same as above, but use ready (probably 74-series) logic gates (ORs, or actually just buffers with output resistors). <S> Resistor ladder DAC + parallel output ADC <S> Assuming constant voltage inputs: build a resistor ladder: <S> simulate this circuit – <S> Schematic created using CircuitLab Hint: there's resistor networks (many resistors in the same package) that make this less of a pain to solder. <S> Vout is proportional to the input that is currently hot. <S> Use an ADC with a parallel output to read that out. <S> Strangly, this might be a relatively simple and cheap approach, considering 8-bit parallel-out ADCs are readily available and not that expensive. <A> There are 16-/20-key encoders available in an IC package: 74922 and 74923. <S> These are "add-on" <S> 74-series ICs, available since 1995. <S> Intended use case is a numerical touchpad e.g. your phone, with the layout intended to provide a 4x4 matrix or 4x5 matrix. <S> Datasheet is here
| As far as I know there has never been an off the shelf device to implement a 16 input priority encoder. Your problem might very well cheapest be solved with a CPLD or even a GAL Memory IC Use Memory IC with parallel address bus and parallel data output bus.
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Help with using 4N25 optoisolator I'm using an open collector comparator LM339, and I want to isolate V2 from the circuit by a 4N25 optoisolator. I want V(out) to change its state from at least around 1V or less up to 10V. But I get the following result in LTspice IV where Vout swings between 8.5V to 10V: I cannot change R8, and I want LED to blink for every pulse. I might be using the coupler wrong way. I also noticed R7 also have big effect on the coupler; if I set it to 10k it doesn't work at all. How can I modify this circuit so that Vout can swing more(from around 1V to 10V) and LED blinks? Edit: Suggestion from a comment: EDIT 2: <Q> You have two things that need changing: 1: A 4N25 only has what is called a Current Transfer Ratio of as low as 20% ( 4N25 datasheet ). <S> This means that the current out of the photo-transistor may be as little as 1/5 of the current the is injected into the LED. <S> You have less than 2 mA being put into the LED, but require about 10 mA from the 4N25 phototransistor. <S> Also from the data sheet the absolute maximum current you are permitted to inject into the LED <S> is <S> 60 <S> mA. <S> You may be able to inject enough to get the output current you require, but it would probably be better to put amplification after the optocoupler. <S> The LM339 is only guaranteed to be able to sink 6 mA (typically 16 mA), so it will have difficult time directly driving the 4N25 with a high current. <S> 2 <S> : You have the output LED on the ground side of the photo-coupler. <S> This will prevent the output (Out) going below about 2 V (LED voltage plus the voltage across the photo-coupler). <S> Again if you put a transistor after the photo-coupler to amplify the current it should be no problem to meet your requirements. <S> You could also do something like using two couplers with the LEDs in series driven by about 40 mA, but put their outputs in parallel. <S> Drive the output LED in parallel with R8, it could be operated with only 1 mA or so as modern LEDs can be bright with only a small current. <A> As others suggested and you've changed your schematic to get rid of the LED in the output .... <S> here's a simple modification that should provide your required output voltage swing. <S> Choose an appropriate PNP to sink your required current: <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Your choice of location for the output LED is actually rather silly -- it's floating the emitter of the opto output up to 2V for no good reason. <S> Put it in series with the optocoupler's LED instead if it can be moved to the other side of the optocoupler and reduce R7 accordingly; if you can't, put it in series with R8. <S> As to reducing R7 anyway, it's probably a good idea as the 4N25 is an older, not-so-hot part, and really needs all the LED current it can get to produce a decent output.
| You could also use a FET to achieve the same result. That LED is in a silly place
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Audio Transformer Orientation on PCB I am working on a project using two audio transformers for two separate circuits mounted very close to each other on a pcb as shown in the image. With the transformers as mounted, there is a bit of inductive crosstalk, as would be expected. My question is: ignoring shielding for the moment, is there a particular orientation of the transformers (both cores vertically parallel, each perpendicular to each other) that would improve this? I am able to separate them to some degree, but not more than a few centimeters max. To get an idea of the field strengths, with one signal active I placed 2 ungrounded scope probes' tips at different distances from the active transformer as a test. I get a higher induced voltage on the coil side (above and below in pic) than on the core side (left and right in pic). The iron core seems to keep the flux from straying as far outside in that orientation. However, I was told that it is was common in old radios to mount power and audio transformers with cores perpendicular to reduce 50/60hz cross-talk, so curious to get some other opinions. EDIT: A schematic with a better description may help: With music on IN1, and relay on T2 closed, I can hear quiet music at out of T2. When shorting IN2 primary, it reduces the music volume, but still faintly there. Correct me if I'm wrong, but I'm generally getting the impression that audio transformer circuits should not be left open when un-used, unfortunately I haven't been able to find much info about this. <Q> As per Brian Drummond's suggestion, I mounted one transformer on wire and played around with the orientation. <S> I found that at about 1" separation between the transformers, I can crank up the volume all the way with no audible induced voltage from the other. <S> The more separation, the quieter the induced sound. <S> Rotating them parallel/perpendicular to each other didn't seem to make a noticeable difference. <A> The suggestion that these are 'the worst audio transformers' means thatit is possible that there is no conductive outer wrap around thetransformers. <S> For audio, to keep signal leakage low, it is commonto see a copper foil tape wrapped around the outside of the core(so <S> it encloses both the windings and the steel core outside thewindings), with the copper soldered to form a conductive belt. <S> Instead of just foil between the cores, a conducting wrap aroundone or both might be an improvement. <A> Only the flux that cuts a conductor at an angle (any angle) will induce an EMF in it; with 90 degrees being the angle which gives the max magnitude of EMF. <S> Therefore if you want to have zero have the flux parallel to the conductor. <S> As far as capacitive coupling distance is the answer. <S> This is why you see on audio amps that have power trans and audio trans on the same mounting plane; with their cores at a 90 to each other. <S> I know this is simple compared to all the other "fancy" answers; but sometimes going back to the basics is overlooked. <S> Your original question is about orientation; so here is the answer. <S> Yes all the other things said are true, but get the obvious out of the way before you approach the esoteric. <S> Not knocking the "fancy" stuff just trying to help. <S> If you really want to get exact, inject a signal of the median frequency of your device into one transformer <S> and then your oscope probes connected to the other and play with the angles to obtain minimum inductive coupling; the angle might be slightly off 90 but close to it.
| Placing the two transformers with one core at a 90 degree angle to the other will reduce inductive coupling.
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Does modern market has IC chips which can be fully reverse-engineered with optical microscope? I'm planning a new course on digital electronics and semiconductors devices for my university. I think, it would be great to go to the lowest available level of abstraction, or, at least, slightly touch. So, I naively think it would probably be a good lab assignment for students to grab microscopes and some acid, and try to completely reverse-engineer some simple chip. By “reverse-engineering” I mean not only looking at the picture of naked gates and conducting lines, but restoring the original model of the chip operation, without probing it electrically. Back in history, it was quite common to have micrometer-scale integrated circuits which had only a dozen of isolated logic gates in a single package. Is it possible to buy something similar today? <Q> I think the 4000 series chips remain at the micron level of fabrication (\$3\:\mu\$ metal gate and not self-aligned) technology. <S> As a percentage of operating FABs, there are less of them still operating at this fabrication level. <S> But they exist. <S> (I like to think of it as a "hand-me-down" kind of process, where older FABs still perfectly operational and more cheaply supported now, get passed on to those who can make good use of them.) <S> Some of these FABs wind up making things like detector diodes (which of course need to be 'large'.) <S> Some for other reasons. <S> Many ICs destined for use in satellites in space must also have large feature sizes in order to tolerate radiation events. <S> Even some (semi-) higher-end microcontrollers, for example, are still made with large feature sizes for those purposes. <S> (I know because I worked on them, some time ago, as part of writing operating systems that are tolerant to such soft error events.) <S> I'd recommend contacting a few of the major companies and simply asking them for some good ideas, here. <S> I think you'd find the response welcoming and positive. <S> That's been my experience in the past. <S> If you explain your goals, which seem quite reasonable to me, I'm sure you'd quickly find someone who'd help you think about the best way to develop the curriculum for it and to suggest some good alternatives to consider. <S> (Besides, if you involved them you would have zero worries about someone later giving you a phone call and asking about why you are exposing their "proprietary" devices in a classroom setting.) <S> So, to start, you should grab up some 4000 series devices and just "do it." <S> See what you find. <S> If that satisfies your need, you're golden. <S> (I think they will, as there is quite a variety of devices there.) <S> These are so-called "radiation-hardened" <S> but part of that hardening is about the feature size they use! <S> So look for that, if it is of interest. <S> Third, contact a few companies and just ask for some ideas and help. <S> I think you'll get it. <A> Is it possible to buy something similar today? <S> Anything more advanced and that might already be the simplest and cheapest of microcontroller, then no, these mostly use processes with such small feature size that you probably would need an electron microscope. <S> Also there's not much to learn if you do not know what you're looking at. <S> Also, getting a usable die photograph takes a lot of effort, time your students can spend in better ways. <S> Focus on digital: well the digital chips are also the ones with the most boring dies. <S> The ones with like 100 k gates or more on them I mean. <S> Have you looked at ZeptoBars yet and also at Electron Update ? <S> If not do so and judge for yourself. <S> The most interesting parts are the Analog parts. <S> I know that as I design such structures myself <S> ;-) <S> In old, simple, "vintage" chips the digital was designed by hand, those are interesting. <S> In modern chips it is all syntesized and thus boring. <A> You could consider the venerable LM555 timer chip. <S> You can find detailed schematics of the internals, high resolution die photos , and lucid explanations from the original designer. <S> Of course if you wish the students to discover it for themselves, having the die photos etc. <S> freely available on the internet may be a disadvantage. <S> There's a bit of analog and a bit of digital in there- an SR flip-flop and two comparators with a 3-resistor divider across the power supply. <S> All together only a couple dozen transistors. <A> Yes, you can still buy "old" designs, and there is little chance that would change soon. <S> For the very popular 74xx series of ICs, check out https://project5474.org/ <S> which is a project to decap all of these "classic" chips, which are still being built into billions of devices to this day, and document the die shots. <S> I think this would allow you to pick an interesting IC before doing your first own decap. <S> You might want to just visually compare different kind of ICs qualitatively - simple, purely analog ICs like the NE555 and the (personally, probably easier to explain if you're into frequency mixers) <S> SA612 simply look very different. <S> An idea to avoid dozens of students messing around with dangerous chemicals: <S> Look at LEDs that integrate a digital controller. <S> I'm thinking of the neopixel things. <S> They come in clear package, because the light has to exit somewhere! <S> You might be able to polish the surface of the polymer that covers the die to be flat. <S> I also would like to encourage you to follow jonk's advice and ask manufacturers. <S> They might even be giving you cut dies without the package around them - after all, that is a shape that they offer commercially for a lot of their "low end, high volume" devices for companies which directly bond the dies to the PCB and just encase them with a blob of epoxy. <S> If that doesn't work out: I'm sure someone on aliexpress will happily sell you a couple hundred "music playing postcard" dies. <S> Also, a lot of ICs are available in lfcsp, lead-free chip-scale packages, where you'd only have to remove a thin layer of lacquer instead of a whole millimeter-thick package
| Second, if you want to expose a microcontroller for classroom demonstration purposes (and not as a difficult assignment, which it probably would be), search out microcontrollers that are still being used for satellite use in space. Yes, if you restrict yourself to simple chips like opamps, timers, standard logic.
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SMD LED for microprocessor output without resistor I have a very simple problem to solve and I'm surprised there is no common solution for it. I need to connect several indicator LEDs to microprocessor outputs and I know I need a serial current-limiting resistor. To save space I'd like to have LED that takes 3.3v without resistor. There is a couple out there rated 3.3v but I understand it is not safe to connect it without resistor even though it is 3.3v rated. Also they are 20mA, I'd prefer 10mA since they are driven directly by MPU with 200mA total output limit. Since this is such a common problem, I'm sure it must be solved and I might be just looking into wrong direction. Please enlighten me! <Q> LEDs have a very steep voltage verses current curve as see in the following plot taken from this page : <S> Notice how fast the current rises once you forward bias the diode (as seen in the green colored quadrant of the plot). <S> This is why designs include a current limiting resistor in series with the diode. <S> It is true, if you monitor the voltage very carefully you can likely get by with out a current limiting resistor. <S> But using a resistor is the easy & safe way to design the circuit. <S> Also, different color LEDs operate at slightly different voltages. <S> It would be much easier to design a circuit using different current limiting resistors instead of a circuit using different power supplies, one for each differently colored LED. <A> The internal Equivalent Series Resistance of a microcontroller output, as well as the higher forward voltage for a target forward current of some leds (3.3~3.6V Blue, Green, White, etc) means you can directly connect some leds to the output without damage. <S> The output voltage of the pin will fall as the current increases, due to the ESR, and the two will balance out. <S> A low forward voltage LED, such as red, will not play as nicely. <S> Frankly, you should just play around with it to see how the voltage and current through the microcontroller pin and led work out. <S> You are really just risking the microcontroller and the led, a few bucks at worst. <A> Understand that the colour, physics and chemistry of the semiconductor determines the "zener-like" forward voltage, Vf <S> and it's tolerance <S> depends on the ESR variations from process controls at the test current used. <S> The general solution can be simple. <S> If you choose Red or Yellow with a 3.3V drive signal on a 2V LED, you need a Resistor. <S> If you choose Blue, white or green, you may not need a resistor. <S> You must examine the choice of both uC driver and LED to ensure this meets your power budget tolerance. <S> Let's examine the ATtiny20 which has <S> typical CMOS design RdsOn rated for <=5V <S> Then look at a tiny SMD Green LED. <S> What current would you expect? <S> for a high side LED turn on current at 3.3V with no resistor? <S> Overlapping the a 60 ohm load line from 3.3V , where it intersects w2ith the LED curve determines the resulting current. <S> Answer? <S> About 7mA <A> The forward voltage of "bare" LEDs depends on the LED colour. <S> LEDs are not at all fussy about their operating current, as long as you don't exceed their Absolute Maximum rating. <S> LEDs will be dimmer at lower currents, but may still be bright enough for your use at 5 mA or less (I have some green LEDs that I had to run at somewhat under 1 mA to get them dim enough!) <S> These may be advertised as "5 Volt" or "12 Volt". <S> A "5 Volt" red or yellow LED may be OK at 3.3 volts, as a bare red or yellow LED will want a little less than 2 volts.
| There are some LEDs that include a current-limiting device, so don't need an external resistor.
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What is the purpose of these P+ and P- connections? I took apart my LED light fixture to clean its diffuser, when I noticed something odd: As you can see, it has a DC power source connected to 4 daisy chained circuits. I expected that these 4 circuit boards to be connected in parallel, with the daisy chaining forming a shared + and - bus. However, there are these P- contacts connected to P+ , and I cant figure out what they're for. Any ideas? Edit: I've laid out the traces of the first two boards. It appears that the first two boards form a pair, with 2 current paths. The green current path goes between the + and - connections of the first board, passing 11 LEDs along the way. The magenta current path passes from - through 6 (only 5 are numbered on the picture, I missed one) LEDs, to P- , then from P- through all 16 LEDs of the second board. The cyan indicates the etched portions of the board. The red and black are the power rails, chained straight through the boards. <Q> You missed the common point on the right board. <S> Figure 1. <S> Second board has two strings of LEDs. <S> I've marked up the 11-LED string in orange. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> Layout of first two boards. <S> It's a weird layout alright. <S> Each board has eleven LEDs in series between <S> + and - but spread between the two boards are another ten LEDs also wired in series between <S> + <S> and -. <S> Whoever designed the board mustn't be a regular reader here! <A> P- and P+ (if they represent power supplies terminals for each module) are probably routed in and out on each module and hence all modules are in parallel. <S> Try connecting a DVM and measuring resistance between P+ nodes and do the same for P- nodes. <A> Here's the final schematic I ended up with, based on @Transistor's answer. <S> simulate this circuit – <S> Schematic created using CircuitLab
| My guess is that they all connect due to internal copper on each module.
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Using two switches to each send current in opposite directions I have two momentary switches, one to activate a solenoid and one to reset it. It is only possible to press zero or one of the two switches at the same time. Using a single battery, how can I wire the circuit so one switch triggers the solenoid (sending current one way) and the other resets it (sending current the other direction)? <Q> You could use a reversible motor driver circuit using a few transistors. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> (a) Using changeover switches. <S> (b) Ugly but passive. <S> Figure 1b uses two capacitors to hold both ends of L2 at ground potential. <S> Pressing either switch raises that end suddenly to V+. <S> The other capacitor will rise up to follow it but with the correct value the time delay will be sufficient to kick the solenoid over. <S> R1 (and L2) allows both capacitors to discharge. <S> This also sets the limit on how frequently the solenoid can be toggled. <S> It might be nicer to put a resistor of twice the value on both capacitors instead. <S> (Symmetry is usually good!) <S> A little series resistance might be prudent. <A> You didn't limit the selection to SPST switches, so the obvious solution is to use DPST switches: simulate this circuit – <S> Schematic created using CircuitLab <S> Obviously, Bad Things™ will happen if you ever press both switches at once! <S> Put a Polyswitch (resettable fuse) in series with your power supply to be safe.
| In this example (from here ), just tie the momentary switches to A and to B, respectively and replace the DC motor in the circuit with your solenoid. The switches might not last long due to arcing caused by high capacitor current.
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ATtiny13a base resistor for 2N2222A I have a small magnetic buzzer which needs to be turned on and off at 10Hz speed via an attiny13a mcu io pin. Since the buzzer is a magnetic one I am using a 2N2222A BJT NPN transistor for the switching part according to what the datasheet suggested along with a 1N4148 diode as flyback just like the diagram suggests. The diagram base resistor 180 ohm is given for a 3.3V output pin but my mcu will give 5V 40mA on it's pin so what resistor should I choose to substitute. Will a 330 ohm do the job? <Q> Interesting. <S> I'd assumed it was a piezo for a moment. <S> Then the datasheet said otherwise. <S> Actually, it's quite similar to some cheap earphones they used to make -- circular, very thin disk of metal over the top of a circular magnet with a coil present. <S> It'll work over a modest audio range. <S> (I used to dismantle them to see how they worked when I was a kid.) <S> The datasheet says they require \$\approx 35\:\textrm{mA}\$. <S> But I think that's a mean value, since they are operated typically with 50% duty cycle. <S> (And because the datasheet actually says "mean current", too.) <S> The resistance is given as \$42\pm 6.3\:\Omega\$. <S> So worst case collector current should be \$\frac{5\:\textrm{V}}{35.7\:\Omega}\approx 140\:\textrm{mA}\$ with your \$5\:\textrm{V}\$ rail. <S> Let's look at the PN2222A curves for a moment: <S> Entering into this (typical) set of curves, I can see that the \$150\:\textrm{mA}\$ curve flattens out starting at a base current of about \$I_B=5\:\textrm{mA}\$ and is pretty solid by \$I_B=10\:\textrm{mA}\$. <S> So I'd decide to drive it with \$I_B=10\:\textrm{mA}\$ as a reasonably safe choice. <S> This curve tells us something about \$V_{BE_{SAT}}\$: <S> I see that \$V_{BE_{SAT}}\approx 850\:\textrm{mV}\$, typically. <S> Let's call it <S> \$V_{BE_{SAT}}= 900\:\textrm{mV}\$ and be safe. <S> The missing bit is the drop from your I/ <S> O pin, when high. <S> But my own experience says that you shouldn't expect a drop of more than \$500\:\textrm{mV}\$ when applying \$10\:\textrm{mA}\$ out of a \$40\:\textrm{mA}\$ output. <S> So your resistor value is \$R_B=\frac{5\:\textrm{V}-900\:\textrm{mV}-500\:\textrm{mV}}{10\:\textrm{mA}}= 360\:\Omega\$. <S> I'd go with a \$390\:\Omega\$ resistor and then take a voltage measurement or two, just to be sure. <S> Keep in mind that this is a pulsed application. <S> So it's not solely ON and the datasheet's current is listed as "mean current" which does NOT MEAN that this is the actual peak current when turned on. <S> The BJT power will be \$900\:\textrm{mV}\cdot 10\:\textrm{mA}+100\:\textrm{mV}\cdot 140\:\textrm{mA <S> } < 25 \:\textrm{mW}\$ at 100% duty cycle. <S> So at 50% duty cycle, there's just no worry about using a TO-92 packaged device. <S> You are fine with the PN2222A here. <S> Similarly, the resistor power is \$390\:\Omega\cdot \left(10\:\textrm{mA}\right)^2 < 40 \:\textrm{mW}\$ at 100% duty cycle. <A> Let's calculate all thing backwards, from the idea that the buzzer should be dreiven full swing. <S> Your speaker has the coil resistance of 42 Ohms. <S> Therefore at +5VDC it would need about 120 mA with fully opened transistor. <S> The 2222 transistor has the hFE at least 50, so to get it into full saturation <S> the base current should be about 120/50 <S> = 2.4 <S> mA. <S> Let's take it at 3 mA <S> Since your driver will drive to about 5V (minus pocket change), and Vbe is about 0.7V,you need a resistor that makes 3 mA over 4.3V, which gives 4300/3 = 1433 Ohms. <S> Therefore you base resistor should be about 1.3k, and the round number of 1k will do just fine. <A> Use NMOS instead of NPN transistor. <S> For your application BS170 is cheap and very simple to operate. <S> If you are using SMD components, you can use 2n7002. <S> Source of image: http://design.stanford.edu/spdl/ME218a/pastprojects/0708/pennybowling/links/schematics.html With MOSFET transitors <S> you don't have to calculate value of base resistor. <S> They actualy don't have base <S> but instead they have gate. <S> If you are more interested in knowledge of NMOS check wikipedia article.
| Again, most any resistor package will be fine.
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Why do we use step response? We tend to identify systems more often by using the step response. Why? Especially when the impulse response is directly related to the transfer function? <Q> Step response and transfer function are interchangeable via Laplace transform (step response -> differentiate to get impulse response - <S> > laplace -> transfer function), <S> but you can't run a Laplace transform on a graph pulled out of a datasheet... <S> For example, the user of a LDO will usually be interested in things like voltage sag/overshoot when load current steps up and down, so in this context a graph of the step response is useful. <S> Same <S> if you design the control system for something like a mechanical actuator: the transient response is often more important than transfer function, you'd want to avoid overshoot, avoid excessive slew rate, have good damping, <S> things like that. <S> Step response shows all this in a way that is easy to understand, whereas transfer function will not (instead, it provides more insight into stability, etc). <S> Also testing the step response is much easier than measuring the transfer function. <S> And the transfer function is only valid when the system is linear, not when it is slewing. <S> If slew limit is involved, or other large signal conditions, then step response is no longer the Laplace transform of the transfer function, because the system is no longer linear time invariant. <S> Opamps' datasheets usually give both, since we're both interested in the frequency response, and settling time/ringing/overshoot, clipping recovery, etc. <S> EDIT: I just realized you asked why step response was used instead of impulse response: <S> It is easy to generate a fast step, very difficult to generate an "infinitely short" pulse (or an approximation of). <S> The step contains much more energy and will generate a much larger response. <S> Whereas a pulse would only generate a small blip on the output. <S> Signal-to-noise ratio is much better with a step. <S> An impulse would not allow to test for slew rate. <S> It would provide little information. <S> Thus the step is easier to use in practice. <S> If you want the impulse response, do a step, then differentiate. <A> An impulse, i.e. hit it with a hammer, is not very friendly, particularly in systems that have mechanical bits and pieces. <S> A perfect impulse cannot be generated, so you have to tailor the pulse duration to the system. <S> A pulse that gives meaningful response data needs to be relatively strong, and that means high amplitude (strength = area = <S> height x duration, and duration=small, therefore height=large) <S> To get the step response from the impulse response you have to integrate, which means you need to know initial conditions. <S> An impulse response does not give DC gain information readily; a step response does. <S> A step isn't as violent as an impulse. <S> A step is a step regardless of the system dynamics. <S> You can get the impulse response from the step response by differentiating - and doesn't require initial conditions. <A> It's possible to describe an "almost perfect" step function using positive parameters δ, ε, and ε', such that: For all values of t<-δ, the output will be between -ε <S> and +ε. <S> For all values of t>+δ, the output will be between 1-ε and 1+ε. <S> For all other values of t, the output will be between -ε' and 1+ε'. <S> Graphically, such functions may be described by saying that their output is always within an "envelope" that looks like: : █ : █▀▀▀▀▀▀▀▀▀▀▀ <S> ▀▀▀ y=1 : <S> █ : <S> █ :▀▀▀▀▀▀▀▀▀▀▀▀▀▀ <S> ▀▀█▔ <S> ▔ <S> ▔ <S> ▔ <S> ▔ <S> ▔▔ <S> ▔ y=0 : <S> ▀ <S> Any application that would work with all functions that would meet the requirements for some particular values of δ, ε, and ε' would be satisfied by any function that met the requirements for smaller values of those parameters. <S> The exact position of the function within the envelope may be poorly defined, but most applications wouldn't be interested in that level of detail beyond having upper limits for δ, ε, and ε'. <S> This thus makes it easy to describe a set of functions that will be "good enough" for a particular purpose, and ensure that the behavior of a real-world system is a member of that set. <S> An "almost perfect" impulse function, by contrast, cannot be described in such a fashion. <S> Behavior may be well-defined at the points where nothing is happening, but the part of the domain where behavior is least well-defined is also the part of the domain where everything interesting is happening.
| The step response is often the inherent integral of the impulse response (eg motor velocity to motor displacement) and integration has noise-rejection characteristics.
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charger in series with light bulbs I mistakenly connected a power charger in series with my ceiling led lights... oh... whell... the reality is more complex, but essentially:1) when the switch is off, the phone power charger is in series with the lights, the lights do switch off in this case (I presume because of too low voltage), and the charger works correctly 2) when the switch is on, the phone charger is shorted and goes off, but the led lights go on correctly. The scenario 2 is of little interest to me. Of course when in 1) they are in series they will not get 220V* each as normal/parallel connection, but a smaller portion. (*I am in europe) It is my understanding that an electronic device may try to draw more current if working under lower voltage than what it should get, in order to generate the same amount of Watts and operate correctly. In short: I am concerned about the power consumption in scenario 1).This is 99% of the time of course, and the main concern is that electricity will have to go through the led bulbs evenduring the 99% of the time while they are off. Can anyone help? Leds are non dimmable if that makes a difference, the mobile charger is of course 220/120v , this is hardwired in behind the wall, so no risk of someone coming in and connecting their hair dryer in place of it :D Thanks for any help!! <Q> When the switch is off, power is applied in series to non-dimmable buck regulated LED and buck regulated phone charger. <S> The impedance of the Bulb must be lower so it sees an undervoltage conduction but may still be passing current and getting conduction losses. <S> You may check to see if the bulb gets warm even after a while indicating efficiency loss and beware that you have no power off condition or switch to enable both. <S> Fix it properly. <A> When the two devices are connected in series, the current through each of them will be the same, <S> no matter how the voltages split. <S> When the phone charger is not actually charging , it will do its best not to pass any current (except for the small quiescent current the regulator circuit needs to work), which means that there is no current for the LED lights to work with. <S> Thus the charger will actually see close to the full line voltage. <S> This goes no matter <S> whether "not charging" is because there's no phone connected, or because the battery in the connected phone is already fully charged. <S> When you actually start charging, the charger will start drawing a current, which gives the lights a chance to consume power. <S> If the lights were a good old-fashined ohmic(ish* load such as incandescents, we could compute how much power they would get if the charger does its best to maintain, say, a steady 5 W consumption. <S> However, LED lights are very much not ohmic, and will usually contain their own switching regulator to make sure each individual LED gets the current they want (since LEDs are much more picky about their current than filaments are). <S> What do you get when you put two switch-mode power supplies in series and let them compete for the same supply voltage? <S> Chaos and instability, that's what -- potentially with more than one possible equilibrium. <S> What happens when the dust settles could depend heavily on the internals of each of the regulators. <S> It sounds like in your case it ends with the lights being starved and the charger still seeing close to full line voltage. <S> (It looks to me like this <S> is the likely outcome when the native power demand of the lights is several times that of the charger). <S> Because of the low voltage, only a small amount of power will be delivered to the lights. <A> however the way the apartment is cabled, the cables run through the ceiling to the lamp, and then only the live cable comes down to the switch... <S> I don't know exactly what is the reasoning behind this, I do understand it may have some good reason such as saving on cables and/or allowing to have separate general switches for lights and power , <S> etc... it still seems dumb to me, but it's beyond the scope of this conversation I suppose! <S> Anyway... <S> I now measured the voltage between the 2 poles of the charger and it goes like this: 1) <S> Lights off - charger connected and charging phone <S> ~192V <S> (hence ~30v are on the lights)2) <S> Lights on - charger connected to phone (but not chargin as described aboveg) ~0V <S> (as expected, as it's shorted with lights on)3) <S> Lights off - charger not charging <S> ~200V (hence ~20v are on the lights)4) <S> Lights on - charger not charging <S> ~ -5V <S> (negative... weird, but more or less as expected) <S> Now... the charger operating @ 190V/200V cannot be using all that much extra power compared with when operating @ full 220V... <S> so it seems my concerns can be settled from that point of view :) <S> Thank you everyone for all the help, it was greatly appreciated and extremely useful to getting this understood!
| In this situation the lights will still see too low a voltage to work correctly, but will pass enough current to allow the charger to work. I totally agree that I "should" fix it...
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Why does the underside of this flip-chip BGA have small notches in it? I am reverse engineering an embedded system which has an ARM SoC on it. I have no datasheets at all, so I am going fairly deep with the investigation. It is packaged in a lidless flip-chip BGA. The carrier substrate that the die is mounted on provides clues to the function of pins, so I was investigating the SoC under the microscope. I have noticed that there are a number of notches cut through the solder mask and outer layer of copper. They cut traces between the balls. Overview of the underside of the BGA: View of some of the notches: Oblique view showing depth: Traces being cut by notches: My initial thought was that these were used to configure the device after they had been binned. There seem to be too many though - well over 50 on a 452 pin BGA package. What are they used for? I'm also intrigued as to how they are made. They have very square sides and no undercut given that they are only 0.25mm long, pretty much ruling out etching and laser. I can't see how a mechanical method would get such a uniform bottom. <Q> I think the links were to tie the pads together for plating. <S> Every single pad once had a connection to the outside if I am correct. <S> You can see traces going off the pattern on the top. <S> After plating they CNC route out the connections. <S> Many of the ball pads are tied together in groups for ground and supply rails so only single traces would be required for each group. <S> Something similar is often done for edge connector pad plating- <S> the connections are milled off later. <A> I think you are on the right track about device configuration. <S> They appear to be laser fusible links, some of the "slots" show bare substrate, others have copper intact, while some show oval slots in the copper. <S> There could be a number of reasons why they are used: <S> Increase yield of the SoC. Chip has n+1 of a component e.g. RAMbanks, and n good components are selected during test. <S> Use a common BGA footprint for different devices in a family. <S> Program <S> a unique identity e.g. MAC address into the SoC. Change the BGA footprint of a standard device for different markets. <S> Since the slots appear to open-circuit between BGA balls rather than on traces going into the substrate I would suspect the last reason. <S> Manufacture a single device <S> then cripple features such as external memory bus, interface ports, etc, on a low retail price device. <A> As we can see there is some tracks routed on a visible layer (first layer under the solder mask). <S> The transmission mode of those lines are microstrip or coplanar grounded types. <S> Designers problably needed to make those lines pass here because of no more area available into other layers of the PCB stackup ( How many layers ? ). <S> Why impedance matching ? <S> In order to minimize errors and signal losses in digital ( and analogical signals ) and open the "Eye Diagram" , impedances should be matched. <S> Therefore we can see thoses "notches" <S> are only between connected Pads. <S> They are forming a "Matching Network" designed for a specific frequency band (probalby a small bandwidth above GHZ). <S> Looking at the design from a ball Pad, it goes with a very thin line which actually behave like an inductance in high frequency, and then followed by a larger line (the rectangular notches) which behave like a capacitance in high frequency, and again a thin line until you reach the other pad. <S> To me, it looks like a typical Inductance - Capacitance- Inductance matching network. <S> Not taking into account those matching notches would result into a less performant digital transmission, with higher binary error rate.
| In my Opinion, might be to optimize impedance matching between the metallized tracks. I have also seen this on boards that are punched with holes to break the temporary connections. Use a common BGA footprint with die from different manufacturers.
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If the voltage at ground is 0, why is there current flow? From what I understand, voltage is required to motivate electrons to flow through a circuit. With KVL we have that the voltage will essentially be depleted at the return point (ground). So, if this is the case then why can we have current flow (at the return point)? simulate this circuit – Schematic created using CircuitLab <Q> Voltage cannot be depleted, it is simply a difference in potential between two points. <A> If you study physics voltage actually derives itself from the strength of the electric field of a charge distribution. <S> The electric field from a setup charge distribution causes current to flow with energy proportional to the the strength of the electric field. <S> Voltage actually represents energy per unit charge. <S> So basically voltage determines the amount of energy 1 coulomb of charge delivers. <S> You see the electric field generated imparts energy to the flowing electrons in a wire that we call current. <S> So when it is zero voltage <S> it just means the electrons have zero energy relative to when it first started because it was used up after flowing through devices that used the energy. <A> If the voltage at Ground is 0V, why is there a current flow? <S> The ground potential is one node in a circuit where we define the potential to be 0V. <S> This gives us the basis to express voltage potentials for every other node in our circuit since they are always expressed relative to the ground node. <S> When we can choose any node in our circuit for that purpose at own will, we surely do not influence any current with that decision. <S> And indeed: choosing the ground potential does not say anything about any current in the circuit. <S> With KVL we have that the voltage will essentially be depleted at the return point (ground). <S> KVL tells you that in a closed loop the sum of all voltages are 0. <S> Depending on how you draw the loop, you will have a positive value for your voltage source(s) and negative values for your consumers -- or the other way around. <S> The result would be the same: the sum of all sources and all consumers will equal 0V.The word "depletion" should not be used here. <S> But feel free to comment what you mean with it. <S> So, if this is the case then why can we have current flow (at the return point)? <S> As said above: the voltage potential of a node has nothing to do with the current flow at that point. <S> In theory you could strip the cord of the power cable to your running computer and touch the wire which is connected to neutral (0V). <S> You would not get electrocuted since there is ideally no voltage difference between you and the neutral wire. <S> Still you know that there is a current running through that wire to power your computer. <S> (PLEASE DON'T TRY THIS AT HOME. <S> ONLY A PROPER MEASUREMENT REVEALS WHICH CONNECTION IS CARRYING GROUND POTENTIAL. <S> DON'T <S> EXPERIMENT WITH LINE VOLTAGE!)
| Conventional current flows from higher voltage potential (the positive of the battery) to lower voltage potential (ground, or 0 volts).
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Why execute code from RAM? I've just come across some macros for my microcontroller compiler to force (or suggest) a function be executed from RAM. https://siliconlabs.github.io/Gecko_SDK_Doc/efr32mg1/html/group__RAMFUNC.html#gac6abbc7f869eec9fb47e57427587c556 http://processors.wiki.ti.com/index.php/Placing_functions_in_RAM https://www.iar.com/support/tech-notes/linker/controlling-placement-of-the-section-where-__ramfunc-functions-reside-ewarm-5.x--6.x/ https://community.nxp.com/thread/389099 In what cases is this valuable? Why wouldn't I just always execute from RAM if the benefit is only increased speed? Does this generally cause higher current draw? <Q> In addition to the speed & other features which others have already mentioned, executing code from RAM can be useful in bootloaders where you need to reprogram your micro's flash - you can't execute code from flash which you're in the middle of erasing & reprogramming. <A> I didn't look at the datasheet for that micro. <S> However, it is often the case in this situation that fetching from RAM is faster than fetching from the flash the program memory is implemented from. <S> The advantage of flash is that large amounts can be relatively cheap. <S> Microcontroller manufacturers therefore sometimes put a lot of flash on a chip, then provide a more limited RAM space that code can execute from. <S> This allows copying time-critical routines into RAM, then executing them from there. <S> The compiler switch you refer to probably works with the linker and flags <S> that section of flash to be copied to RAM by the compiler run-time code that runs from reset. <S> Different implementations will vary on the details. <A> When you want to execute in RAM because it is faster, it's usually because that RAM is on-chip SRAM. <S> This is a scarce resource, which you will probably want for data that requires read/write access. <S> Using it for code when you already have the code in ROM/flash means that you need X amount of flash and an additional X amount of RAM. <S> It also requires an extra copying stage at boot or when you want to run it, although it is mostly insignificant. <S> Traditionally this is solved with an instruction cache, but in a microcontroller it may make more sense to keep the internal SRAM generic, because you don't use a microcontroller because you want the fastest execution speed. <S> There is also a reliability issue - code executing in actual ROM is hard to modify by buggy code. <A> In addition to all good answers: Why wouldn't I just always execute from RAM if the benefit is only increased speed? <S> Because in an embedded system, usually you don't have the required amount of RAM. <S> For example a STM32 having 32kB or RAM and 512kB of EEPROM. <S> To be able to execute entire program in RAM you would need a RAM size bigger than EEPROM. <A> Other answers don't seem to have discussed power consumption much, which you specifically asked about. <S> A typical use would be to run a low power "sleep" function from RAM, with the flash memory powered down. <S> Not only is power consumption reduced, but if the microcontroller needs to wake up quickly (e.g. in response to an external interrupt) there is no delay while the flash memory is powered up again. <S> Some parts, such as some of the Atmel SAM range, have special extra low power RAM that can be used for this purpose. <S> This allows a small amount of code to be loaded to the special RAM, while the bulk of the available RAM and all of the other memories is powered down and the microcontroller enters a deep sleep mode. <A> Other than the potential speed benefits mentioned by others, RAM code is also dynamic and can be modified on the fly by some tailoring code in the FLASH as required. <S> This could be as simple as changing a few parameters or could be entire handler routines uploaded remotely. <A> Executing code from RAM is significantly faster than executing it from flash memory. <S> Most CPUs are heavily optimized for the fastest possible RAM access, and even the fastest flash memory only reaches a fraction of the speed of RAM. <S> However keep in mind that moving the code from flash to RAM also takes time. <S> If code is executed only once, you only need to read it once, and therefore you would actually loose time for copying it into RAM first instead of executing it directly. <S> If code is executed occasionally (so copying it into RAM would increase execution the second time it is called), but the system is generally idle, then you would execute that code faster by copying it into RAM, but no one cares, because the system has enough time to spent. <S> So such optimizations are only worth the effort, if the code is executed frequently, and you have measured it to be a choking point of the system. <S> On the other hand RAM needs to actively keep the data stored, while flash memory does not, so the total power consumption increases, if RAM needs to be kept active. <S> This is however only relevant, if the RAM is otherwise not used at all, but most modern systems will - in one way or another - use the available RAM already and therefore already keep it active. <A> There are two very common reasons to execute code from RAM: <S> Some microprocessors cannot execute from flash during flash programming - though many can do this as long as the code is in a different block from the flash being written. <S> Flash writes might be reprogramming the application (boot-loader case), or when flash is used to store non-volatile program info (configuration, calibration, etc.) <S> On many microprocessors, RAM is much faster than flash. <S> For these devices, small speed-critical routines may be executed from RAM, though usually RAM is in much shorter supply than flash. <A> Another usecase for RAM only execution security against random bitflips. <S> We use this model on our small cubesat because the main computer board has an ECC ram that tolerates bitflips due to radiation. <S> The entire OS is loaded into ram as a ramdisk on startup runs completely in an ECC enviornment. <S> The flash is not ECC protected(standard off the shelf micro SD cards) however we have other methods to check for corruption(multiple images, checksums etc.)
| The answer is that it depends somewhat on the microcontroller, but often execution from RAM can reduce power consumption because it requires less energy to read instructions from RAM than from flash memory.
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Altium Designer - Move an object only horizontally Is there an easy method to move an object only in the horizontal direction? I know what there is the "Move Selection by X, Y, ..." command but I am looking for something that uses the cursor, since I want to be able to visualize the change in real time. For example, sometimes I need to move designators around but I still want it to be in line with other designators . <Q> Note: This also works in the up/down and 45 degree angle directions. <A> Usually when I need to do this I just click and hold whatever I want to move <S> and I use the arrow keys on the keyboard to move it only in a certain direction. <S> If it's going too slowly using the arrow keys, hold down shift while you press the keys and it will move faster (by a factor of about ten, I believe). <A> There are two approaches I use for this. <S> The first is alignment tools .Note <S> the object you select first is often what the other objects align for. <S> Alternatively, I double click the object and modify its x or y directly. <S> This is a little cumbersome but its very accurate. <S> Knowing exact x and y coordinates helps keep everything inline. <A> You could also just change the grid in the non-travel dimension only to something large, so that a mouse-move won't be able to drag it in that direction. <S> Doing this (different grid size for X and Y-directions) and then changing the reference point is a technique I often use when designing component footprints.
| Press and hold the "Alt" key and then move the object with the cursor in the left or right direction.
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How exactly falling/rising edge works Maybe my question will be very basic but I am newbie and I have stucked... Supoose we have uC and one pin (voltage is in range 0-3v3) connected to external interrupt. Which of these sentences is true? The falling edge interrupt would fire when voltage goes down from 3v3 to lower value (for example from 3v3 to 3v1). The falling edge interrupt would fire when voltage goes down from 3v3 to 0 volts Which of them is true? Thanks! <Q> Neither. <S> See the microcontroller datasheet for what the appropriate \$V_{IH}\$ and \$V_{IL}\$ voltages. <A> It will fire a falling edge interrupt when it goes from logic 1 to logic 0, this will depend on the uC (hardware), for ex some uC may consider from 0V to 0.7V a logic zero and from 2.3 V to 3.3V a logic high, while the rest is floating <A> can be considered true (if it's a test), it will definitely fire when the input voltage drops to 0V. <S> At exactly which voltage will a falling edge be detected, can vary. <S> There are usually two guarantees. <S> It will never happen before the input voltage gets below V IH - V hys . <S> It will definitely happen when the input voltage gets below V IL . <S> Where V IH , V IL , <S> V hys <S> are the logic high, logic low, and the hysterese voltage in the datasheet, respectively. <S> They may be given as absolute values, or derived from the supply voltage. <S> When there is no hysterese mentioned, assume 0V. <S> Between the two given voltage levels there is a grey area <S> , the exact threshold may vary between parts, or with temperature, wear, moonphase, etc.
| The falling edge interrupt would fire on the transition from logic high to logic low.
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Creating DIY ON/CHARGING switch for an Arduino Currently I'm using the newly released WEMO LOLIN32 (pictured below), and it's a great board, save for the lack of one thing, a power button. Initially I was thinking of a simple ON/OFF switch that disconnects the battery, but I realized I need it to be able to charge a Li-Po battery while the device is powered down. Now since I'm charging a Li-Po I need to output 4.2V to the charger (at say 50-100+ mA or something comparable to the current from the battery terminal). Sadly as this is a new board I don't have much in the way of a spec sheet to guide me on the capabilities of each port. Two questions Will I fry the board if I try to run power through analog out ports? Is there charging behavior that is built into Li-Po charging that I need to emulate on the analog out that I use? <Q> Q2 is a mosfet used to disable power draw from VBAT when 5V or VUSB is present. <S> Disconnect the trace from pin 3, and place a switch between pin 3 and the trace from VBAT. <S> Based on what I see on the schematic. <S> This means it can still charge from 5V. <S> But will only power the 3.3v regulator when the switch is on. <S> Edit: <S> based on https://macsbug.wordpress.com/2017/05/08/battery-and-battery-interface-of-lolin32/ there are two revisions. <S> The original uses a diode or bridge, while the newer uses a mosfet there. <S> The same applies to either revision. <S> Disconnect the VBAT from q2 or the diode bridge and insert a switch. <S> Edit 2: <S> here <S> https://macsbug.wordpress.com/2017/05/31/power-switch-to-wemos-lolin32/ <S> they mod a power switch. <S> Similar to what I suggested, this would also keep the board mostly off even of you use a usb or 5v power source. <S> It will allow you to charge and have the board off, but not board on no charging. <A> I'm not certain <S> you actually need to cut the power. <S> As you stated in the comments, the board already supports charging, you just need to shut your application down when the switch is off. <S> I'm sure this chip has a "sleep" mode you can put it into. <S> When the interrupt is triggered, you can stop your application and put it into low power mode. <A> Yes, you will fry something if you try to run power through an I/O port on any microcontroller (and I don't see any analog outputs on the schematic, anyway.) <S> The schematic and spec page indicates that the board has built-in battery charging capability. <S> (see U5 on the schematic).
| Wire your switch into a digital input pin and use an interrupt to monitor its state.
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Altium Designer - Drag a component and all traces attached to it In the PCB editor, I am trying to move a component and also the traces that it is connected to. I tried using the Drag command but the component still becomes disconnected from the traces. <Q> Edit>Move> <S> Drag Track End works best for me (Altium 18.1.7), and after a bit of use <S> I actually like it pretty well. <S> I found that solution in another thread with many more options to try to get the traces to move with the component. <S> Try a search for the phrase "rubber band traces altium". <S> I know this is an older thread--hopefully you found an answer to your question before now-- <S> but I figure if I'm reading this now then someone else might benefit too. <S> Happy routing! <A> Altium is not good at dragging a part while also keeping the tracks from crossing each other. <S> I came from Orcad which did as you desire. <S> I was very surprised that Altium did not do it, since I thought of it as a basic concept. <A> It is possible in Altium Designer 19. <S> Watch the below video. <S> https://www.youtube.com/watch?v=wfML_NWr2sI
| The best you can do in Altium is to slice the tracks that are attached to the part, un-route the component, drag it to its new location, then re-route the tracks manually or auto-route the component.
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DC plug pins. Which is positive? I am trying to convert my old 9V battery operated Wah-wah pedal to a DC converted one, so I can use AC power. I bought a DC plug for it and was told that the center pin is for ground. However there are TWO additional pins. I only expected one since there's just a positive and negative wire. Which pin on the DC plug is for the positive wire? And is it true that the center pin is for the negative wire? <Q> The pin with the tab on the bottom left is the center pin. <S> The tab at the very top is the sleeve. <S> The middle pin is a break contact on the sleeve connection. <S> Most old guitar pedals use center (-) - but not all. <S> Check for continuity between the center pin of the connector and the battery connector. <S> You can either ignore the break connection or, if you still want to have the option of a battery inside the pedal, connect whatever lead corresponds to the sleeve to the break connection. <S> This will disconnect the battery when external power is plugged into the pedal. <A> For most barrel jack/plugs, the center pin is usually positive terminal, but there are exceptions, which cause severe confusion and a lot of burned gadgets who use this reversed polarity. <S> This signal usually used in devices to switch between external and internal power sources. <S> SO you need to check carefully the arrangement in your device before changing the jack. <A> The sleeve or outer contact often includes a switch which will open when the mating plug is inserted. <S> This switch may be used to disconnect an internal battery when external power is used. <S> I think (but don't quote me!) <S> that the center pin is usually positive, but both ploarities can be found. <S> If in doubt, measure with a DVM or other voltmeter. <S> An ohmmeter or continuity checker can be used to determine which switch contact is which.
| The plug polarity is usually marked on a body of AC-DC adapter, by approximately this sign, The power barrel jacks usually have three pins, one for center, and two other form a normally-closed switch, see this illustration: When a plug goes in, it breaks the contact, so the wire connected to #3 gets disconnected from ground.
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Are waves inside conductor wires considered longitudinal waves, just like sound waves? Are signals transmitted through electrical wires considered longitudinal waves (signals) or electromagnetic waves? I think it would make sense for them to be longitudinal because the electrons are pushed then dragged back by the force of voltage or potential difference. <Q> Interesting question. <S> This is one of those questions that gives most EEs a bit of headache because it is rather difficult to get one's brain wrapped around what is theoretically going on with electromagnetic waves. <S> The truth of the matter is it is not quite as simple as we first imagine. <S> Here is how I rationalise it. <S> First and foremost you need to separate the notion that electrons have anything directly to do with electromagnetic waves. <S> They don't. <S> EM waves propagate without the need for any material. <S> In a vacuum they propagate at the speed of light, when the wave encounters a material they are slowed by the materials physical properties. <S> Having grasped that bit of wisdom, it is easy enough to understand that when you apply a voltage to one end of a bit of wire, it takes time for you to be able to detect that voltage at the other end of the wire. <S> That voltage propagates down the wire at the speed of light for that conductor material. <S> In effect you created, or more accurately, changed the E-Field at one end and the change wave takes time to get to the other end. <S> Now consider instead modulating the applied voltage, that is, applying a signal to the wire. <S> If we break up that signal in the time domain into infinitely small periods you can see from what we just discussed that there will be that propagation delay for the instantaneous change at one end to reach the other. <S> The E-Field must "carry" those changes to the other end. <S> Again, it is important to remember, this has nothing to do with electrons. <S> So, to summarize so far, the electromagnetic waves are the carrier for your signal not the signal itself. <S> EM waves happen to be transverse, and that does not change. <S> The signal, or whatever voltage wave you are transmitting, is effectively a modulation of that carrier, and is usually longitudinal. <S> You are setting up, or transmitting, a difference in the electrical field. <S> It's those local differences that excites the electrons in your conductor and make them move in reaction. <A> Conductors have very high absorption, so EM-waves propagating inside a conductor do not reach very far. <S> An ideal conductor has infinite absorption and EM-waves cannot propagate inside an ideal conductor. <S> EM-waves propagate in waveguides, which may consist of conductors (e.g. coax cable or twisted pair), however not all waveguides need conductors (e.g. fiber optic cable). <S> Also the energy the wave carries is contained in the field which is mainly in between the conductors. <S> The conductors are only needed for confinement of the wave, i.e. that the wave is going where it is supposed to go and not arbitrarily dispersing into free space. <S> In most cases waves are a mixture and have both longitudinal and transversal components. <S> Note that in free-space no longitudinal EM-waves can exist (but in waveguides this is possible). <A> Electromagnetic waves are Transverse in nature. <S> The E and H field oscillate perpendicular to the direction of propagation of wave. <S> If you consider a transmission line such as a coaxial cable, there are two conductors that carry currents in opposite directions. <S> One current flowing from source to load and other from load to source. <S> It is true that these two conductors carry currents. <S> They are separated by a dielectric layer of foam or polyethylene. <S> Your Electromagnetic wave propagates along the gap between the two conductors. <S> The electromagnetic wave is produced due to the jiggling of electrons. <S> As a result of which a self propagating wave is produced for long distance transmission. <S> In a coaxial cable like RG59, RG6 etc the outer conducor is kept to confine the EM wave inside the cable. <S> This is done to tackle skin effect at high frequencies. <S> Moreover, Ron Schmitt's "Electromagnetics Explained" tells that the electrons flowing thro the inner and outer conductors just serve to guide the EM wave inside the dielectric region. <S> For, dielectrics allow EM waves to pass thro them. <A> The presence of a longitudinal wave requires an extension of Maxwell equations to incorporate this extra piece. <S> In addition, the energy conservation equation should include a term that gives momentum and energy to the longitudinal wave besides the transverse wave. <S> In that case one can say that the longitudinal wave exists at a fundamental level.
| Whether the waves are longitudinal or not (or have some longitudinal component) depends on the waveguide geometry and the type of mode the EM-wave corresponds to.
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What does an oscilloscope's input resistor and capacitor do? I watched a YouTube video on how to use an oscilloscope, How to Use an Oscilloscope . It says there are 16 pF of capacitance and a 1 Mohm resistor connected in parallel on each input port of the oscilloscope. However, I still don't understand why there's a capacitor and resistor inside, and what the purpose of those things are. Why are those things in there on the input port? What do they do? <Q> It would be really nice if a scope input had infinite resistance and zero capacitance but that, unfortunately is impossible. <S> Sensitive input amplifiers will always have a small amount of input capacitance and, there will always be a little leakage current from an amplifier's input. <S> Don't forget the scope lead too <S> - it might be a metre long and easily introduce 10 pF. <S> A 1 Mohm resistor might be enough to convert the leakage current to an offset of a few millivolts i.e. sufficiently small so as not to give a false measurement of any significance. <S> So, with 1 Mohm and 1 nA leakage you get a millivolt offset change in the scope when you connect the probe tip and earth together. <S> There is also the issue of noise - you would be unlikely to be impressed if the probe was unconnected and you saw 100 mVp-p of ripple on the display. <S> The 1 Mohm resistor and (say) <S> 15 pF capacitor form a low pass circuit when the probe is unconnected and, subsequently have a noise bandwidth of about 15 kHz. <S> Given that your scope analogue channel might have a noise of (say) 10 uV/\$\sqrt{Hz}\$, the ripple will be about 1 mV RMS or about 6 mVp-p (six sigma calculation). <S> It's much more complex than this to analyse <S> but hopefully, my simple calculation hints that there are other things to consider that might give the impression <S> the scope's performance isn't that good when the probe is unconnected to a circuit. <S> Added to this is the need for all scopes to standardize between manufacturers <S> means 1 Mohm is commonly accepted. <A> The input impedance of oscilloscopes is limited for a special reason, to accommodate wide range of input signals. <S> In general, input sensitivity (voltage range) is limited to 5-10 V. <S> In today's electronics it is plenty, but in the past people were working on vacuum tube amplifiers with 100 - 200 - 600 V signals. <S> So there must be probes attenuating the signal by 10X - 100X. <S> This was done in so-called "passive probes", which are voltage dividers. <S> Therefore, to get a divider, you need to have a limited input impedance, so 1 Mohm was a reasonable value, and for a 10X attenuation the probe resistor must be big 9 Mohms. <S> For user convenience, there is a 1-meter long cable as well. <S> All these necessary components have parasitic capacitances, as well described in this nice article , and the picture within: <S> So, 9 Mohm:1 Mohm resistors provide a 10:1 voltage divider, for DC signals. <S> However, for AC signals the parasitic capacitance of probe head leads to effectively lower impedance than 9 Mohm, which must be compensated to maintain the same attenuation for high-frequency signals and keep the real shape of AC signals. <S> And it should be done for a wide range of frequencies. <S> This is done by ADDING some input capacitance, so the divider is "frequency agnostic". <S> As a matter of fact, this capacitance is not universal, and is individual for each manufacturer and even scope model. <S> As result, passive 10X probes are not completely interchangeable, and their AC compensation may fail. <S> I've seen 8 pF, 10 pF, and 13 pF inputs on various scopes. <S> In summary, the input impedance values of oscilloscopes are designed to accommodate frequency-compensated 1:10/1:100 probes. <A> In order to have a balanced 10:1 simple divider, the cable capacitance is tuned in the probe to match the cable capacitance which is lower than the standard 75 Ω coax and probably use 100 Ω (custom) coax, perhaps 10 pF/ft (33 pF/ft). <S> Each design of the scope preamplifier and coax feed has a different rating for capacitance, but a resistance of 1 MΩ is standard. <S> Thus oscilloscope probes and oscilloscopes must be calibrated with a square wave test port on the front panel to give a square response. <S> In better probes, there is also an inductive and two-stage RC balance. <S> However the ground lead inductance is not compensated, so for measurements with f > 10 MHz or rise times <S> < 30 ns, the ground strap length must be reduced significantly or eliminated using tip and barrel between the two pins. <A> One of our ADC designers just did a whole talk on oscilloscope front end design, you should check it out here: <S> https://www.youtube.com/playlist?list=PLzHyxysSubUmxGOMVpiKLxouweh2AAlG1 <S> It went into a lot of the details about how the oscilloscope's front end and ADC work together. <A> The resistance and capacitance in the probe form one section of a voltage divder, the resistance in the scope and combined capacitances in the cable and scope form the other section. <S> With a square wave source, the variable capacitor is adjusted to show a square wave on the scope. <S> With too much capacitance in the probe, you will see overshoot (pointy spiked corners) in the square wave display; with too little capacitance, you will see undershoot (rounded corners). <S> This happens when the RC time constant of the probe is the same as the RC time constant of the cable+scope. <S> Of course, if you are probing a very high impedance source at high frequencies, you can expect trouble. <S> In that case, some kind of isolation amplifier would be required for you to see a true representation of your waveform.
| The goal of the system is to make the signal at the scope be representative of the signal you are probing.
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Connecting an LED to a "incompatible" voltage source I'm band new to electricity and electronics and am confused by something I saw with a recent video that explains how to connect a battery, LED and resistor together. In that article (see link for further details if you're interested), its a 9V battery, a 3V LED rated at 20mA and a resistor that are being used. Is it OK/safe to connect the LED to a voltage source providing more voltage than it is rated for? What potential side effects are there? I think the answer here is yes , because it seems perfectly fine to hook a 3V LED up to a 9V battery. Is it OK/safe to connect the LED to a voltage source providing less voltage than it is rated for? What potential side effects are there? Pretend I only had, say, a 1V battery. <Q> As far as over voltage, read here: <S> How can voltage burn out an LED? <S> Under voltage, it will not light or be very dim. <A> Applying more voltage than the LED is rated for will cause it to overheat and burn out / catch fire if you really overdo it. <S> Of course that is assuming the battery can source that much current. <S> You may just find the battery voltage drops off... <S> Applying less voltage will make it dimmer. <S> ADDITION: <S> To address your voltage/current confusion look at the simple circuit below.. simulate this circuit – <S> Schematic created using CircuitLab R1 is your drop or ballast resistor, R_Led is the effective resistance of the LED when it is on. <S> We do not normally think of LEDS as presenting a resistance, but in fact they do have a complex effective resistance. <S> For the purpose of this description it makes things a little clearer. <S> Can you see that the current through the LED will simply be <S> \$I = V_{BAT}/(R1 + R_{LED})\$ <S> And that the voltage across the LED is <S> \$V <S> = V_{BAT} - (I * <S> R1)\$ <S> OK <S> so maybe the math confuses you.. <S> Can you visualise what happens when you INCREASE R1 ... <S> When you do, you increase the total resistance of the circuit so the current must be less. <S> At the same time, since R1 got larger compared to R_LED, the voltage <S> V <S> MUST drop. <S> As such, the ballast resister sets both the voltage AND the current through an LED for a particular supply voltage within the tolerances of the LED's forward voltage. <A> One does not regulate an LED by regulating the voltage applied to it: <S> One regulates the current . <S> The reason is, an LED's IV Curve is very steep near its normal operating point. <S> Small fluctuations in voltage cause large changes in current. <S> The cheapest way to regulate the current* is what they showed in the video: put the LED in series with a ballast resistor (a.k.a., "current limiting resistor"). <S> That makes the current/voltage relationship much more manageable. <S> The most efficient way is to use a constant current power supply . <S> (often called an "LED driver" when designed especially for LED lighting applications.) <S> * <S> Actually, there's a cheaper way: <S> In keychain flashlights that use tiny button cells or coin cells for power, they often just connect the LED directly to the battery. <S> In those designs, the battery itself acts as the ballast resistor. <S> All batteries have internal resistance , and smaller the battery, the more resistance it tends to have. <A> Lets examine the behavior of 2 circuit components connected in series: a resistor & an LED, that series circuit connected across a voltage source. <S> We keep the same resistor value, and increase the voltage from 1 volt to 9 volts, in 1 volt steps. <S> simulate this circuit – <S> Schematic created using CircuitLab
| If you try to regulate the voltage, you risk exceeding the maximum current rating, and burning out the LED.
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Can a controller have a negative gain? I'm working on a problem that involves a Ćuk converter . The question asks "find k_c (a gain for a controller) such that the system is marginally stable". So I got the properly space-state model with matrices A and B. Then, I set the real part of the eigenvalues of A equal to zero and try to solve for the unknown k_c. After that I got that kc must be equal to -1.8 approximately. That is, a negative gain. Is it physically feasible to implement in real life? Does it make a difference to use a positive gain or a negative gain? <Q> 1) Sure, this inverting amplifier circuit has a negative gain. <S> Gain = output/input, so a negative gain simply means that the polarity of the signal is inverted between input and output. <S> 2) <S> OH <S> YES in a system with negative feedback, like this <S> : The sign of the feedback signal (running through (H) ) cannot be changed sign-wise or the feedback would not longer be negative feedback but positive. <S> That results in a completely different system behavior. <S> Imagine cruise-control in your car, when going up a hill the cruise-control compensates to keep the speed constant. <S> That's negative feedback, the speed becomes less (we're going up a hill) power is increased. <S> In a positive feedback system, power would be decreased so you'd slow down ! <S> That's not what you want. <A> But if not then the negative feedback would cause positive loop gain and if loop gain = <S> >1 @ 360 deg <S> then it is unstable. <A> Plot the voltage conversion ratio as a function of the control variable (in your case this is the duty cycle). <S> You will see that the slope (DC gain of your small signal model) is always negative. <S> You must therefore have another negative gain in the forward path (controller as you suggested). <S> The same occurs with an LLC converter operating in the inductive region.
| If it an inverting converter, it must have negative gain.
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Alkaline batteries used in Ni-Cd battery rechargable power outage LED emergency light I have a Vector Rechargeable Power Outage LED Emergency Light that plugs into the wall. When the electricity goes off, it comes on. The instructions state to use three AAA Ni-Cd batteries and I only have alkaline batteries. Hurricane Irma is headed my way and I was unable to find Ni-Cd batteries anywhere.Is is safe for me to use alkaline batteries instead? <Q> Cadmium is obsolete due to toxicity for disposal. <S> NiCd is 1.2V/cell while Alkaline is 1.5Vcell thus 4.5V vs 3.6. <S> Meanwhile White LED,s are around 3.1V <S> so they have internal resistors selected to limit the current ( and brightness) and extend the duration. <S> Since the luminaire is obsolete due to NiCd availability, you have nothing to lose by trying it in the dark with no power. <S> It is safe, but the internal resistors will see a bigger power loss and get hot inside. <S> But this is still very low power, but may fail if poorly designed. <S> Portable flashlights are cheap now and some come with Lipo cell charger and some using Alkaline. <S> These much brighter than what you have. <S> My preference is LiPo hand torch 5W. <A> Ni-Cd are rechargable batteries, and your emergency light has charger built in. <S> You cannot put non-recharchable batteries in. <S> It is not safe. <S> Charger (when plugged into the wall) will eventually try to charge those batteries. <A> As long as you don't use the charger, the light will have no clue that the batteries are of a different type. <S> In fact, you will most likely get a longer runtime with alkalines, as they have higher mAh capacity than NiCd. <S> A fully charged NiCd is 1.4V, a fully charged Alkaline is 1.5V... <S> not much of a difference. <S> Don't use the charger with alkaline batteries though, they could leak and ruin the lamp. <S> They are not designed for charging. <S> If you still have NiCd batteries though, you should bring them to a recycling.battery disposal place because Cadmium is extremely toxic. <S> Make sure they don't end up in a landfill. <S> Well, you have other worries at the moment, but keep it in mind. <S> Stay safe ;)
| It is a fire hazard to charge non-recharchable alkaline batteries.
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How to compute electric bill for a server netbook I've taken an old netbook and started using it as a tiny game server. I have removed the screen, wifi card and reproductors, but the rest is there (battery, fan, sata HDD). It runs 24/7 and charger is plugged in at all times. I wonder whether it is possible to compute a fairly accurate estimate of an electricity bill for this machine in both idle and maximum usage scenario. The computer is Asus EEE 1201HA and adapter has 19V~2.1A output. Also is the energy consumption affected by having battery plugged in? <Q> Get a $20 Kill-a-watt and measure the actual wattage during load and idle. <S> Then determine the average amount of time you'll be in each mode and from that you can get a good estimate as to how much power it'll use. <S> At the upper end, you know that it can't use more than 19V*2.1A ~= 40Watts <S> (ignoring powerbrick inefficiencies). <S> 40 <S> Watts * <S> 24 hours * 30 days = 28.8 kW*hrs per month. <S> Lets say electricity costs 0.15 cents per kWHr: the monthly maximum charge from using the netbook is $4.32. <S> If it's idle most of the time, it'll likely be a lot less than this. <A> Considering ASUS choose their chargers to both operate the unit and charge a fully drained battery at a slower rate, you may assume it uses <50% of rated output of 19 <S> *2.1 = 40 W. <S> Then since the LCD backlight typically drains 1/2 of the total power at max brightness, you are left with 10W, which is puny compared to running an air conditioner or electric stove. <S> The battery drains nothing substantial when charged and should be cool if in good shape. <A> Asus 1201 HA has a 6 cells Li-ion battery, being the expected duration around 7 h (see here ). <S> This page is unclear, but we can estimate battery power around 50 <S> Wh (see, by example, this replacement of 62 Wh). <S> That means that the unit consumes in one hour around 50 Wh / 7 h = 7 W. <S> 7 W * 24 h/day = 170 <S> Wh/day. <S> At a price of 0.15 e/kWh it means 0.025 e/day. <S> These numbers can be improved if replaced by the ones of an specific running mode. <S> These two numbers (capacity and discharge time) must replaced the ones given by the manufacturer. <S> Moreover there are software that shows power consumption for most of modern operating system. <S> In particular, see here for several options in Debian/Ubuntu/Mint: powertop, powerstat, ... . <A> Energy consumption is probably slightly affected by having the battery plugged in, but its a fairly minimal amount.
| If the battery is bad you might want to leave it out though The only reliable way to calculate the usage of the device is by connecting one of those home power meters to it, you can get these for around $10-25 at your local electronics store. The method is: first, ask the battery the current power capacity (it decreases with the time), there are software for that; second, fully charge the battery, execute the application/s until discharged and measure the time.
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Linear regulator using emitter follower and operational amplifier I have understood that you can create a linear regulator using a high-current BJT power transistor in the emitter follower configuration and one operational amplifier (and a voltage reference, which can be a voltage source or some kind of Zener diode or similar). How does such a circuit work? What is the minimum supported voltage drop of such a circuit? Is it worse or better than dedicated linear regulator ICs? Are there any reasons to use such a circuit in the modern days where adjustable-voltage linear regulator ICs are widely available at a very cheap price? <Q> If the op-amp can be supplied with a higher supply voltage than input voltage you can reduce that to tens or hundreds of mV, which might be competitive. <S> I have done this in a situation with +/- <S> 8V rails and a 5.6V rail regulated down to 5V using an emitter follower, with the op-amp supplied from the 8V rail. <S> Most commercial regulators do not have a separate supply pin for the internal circuitry <S> so this is not an option. <S> If you use a PNP or p-channel transistor rather than a follower the drop an be less, but ensuring stability is more difficult, as with commercial LDO regulator chips. <S> If you design your own you are free to use a better reference, a more accurate op-amp etc so you could get better performance. <S> You might have a free op-amp out of a multiple package. <S> Generally though, you will be better off using commercial parts in almost all situations. <A> Yes, you can... <S> https://tangentsoft.net/elec/opamp-linreg.html <S> How does such a circuit work? <S> The opamp acts as a follower (or an amplifier) which takes the reference voltage as input. <S> It drives and controls the pass transistor (the opamp handles the feedback). <S> What is the minimum supported voltage drop of such a circuit? <S> If you make it a LDO (with PMOS/PNP pass transistor) you can get low dropout like a good LDO. <S> If it's a NPN pass transistor, then you'll get a few volts. <S> Since the load is a capacitor, it is essential to understand control theory... <S> Is it worse or better than dedicated linear regulator ICs? <S> The main reason to use it is low noise (by using a low noise reference and low noise opamp) and high PSRR. <S> Also it has very low output impedance. <S> But it's much more expensive, complex, and large than an off the shelf regulator. <S> So before doing such a circuit, you first must be sure that you need the ultra-low noise and other benefits it offers. <S> Otherwise, it's a waste of time and money. <A> If you look inside any LDO, this is how it is done usually either emitter followers for BJT type or Pch FETs for + <S> ve ultra-low dropout FET type LDO's. <S> Since they use precision band-gap reference diodes and a circuit with excellent PSRR at low cost, there is no advantage for a discrete circuit these days. <S> e.g. LM7805 and LM317 both use Darlington NPN emitter followers with the lowest possible Zout. <S> These cost only 3 cents at low current and the same topology is used with bigger FETs for 1A and 10A , just lower RdsOn. <S> In some due to Ultralow cap ESR , FET types are less stable to regulate low ripple feedback, so they advise the range of ESR for the load cap. <S> But they perform better with lower I-O voltage drop. <S> The is one exception with a PNP LDO rated for 10A from Sharp. <S> There will always be exceptions to the rule that affects cost increases, but these parts define some of the topologies in low cost LDO's , each with tradeoffs for cost and performance. <S> Ultralow Vce(sat) <S> PNP's are very expensive with special processing. <A> To avoid oscillation, do this simulate this circuit – <S> Schematic created using CircuitLab
| An emitter follower has a minimum voltage drop below the Vout of the op-amp of about 700mV.
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Faraday cage with a RF WIFI router inside - does the electricity "evaporate"? I have a Faraday cage and inside there is a WIFI router whose RF I am trying to contain. The cage is made of aluminium mesh with hole sizes of 3 mm or so. What happens to the RF electricity coming from the RF WIFI router and hitting the Faraday cage (from inside) if I don't ground the cage? Would the electricity "build up" inside/on the cage and in what form since it will cease to be RF? I have had the RF WIFI router emitting for few hours however I cannot detect any voltage increase on the cage's surface. I measured for DC and AC using a multimeter and the values remain constantly the same - about 4 volts AC. <Q> Electromagnetic waves are produced by electric charges moving inside a conductor (the antenna), but the charges never actually leave the conductor and go somewhere else. <S> These items will get slightly warmer as a result. <S> The router itself consumes a significant amount of electrical power while operating. <S> A small fraction of this power gets emitted as radio waves, but the vast majority of it gets turned directly into heat within the components of the router. <A> RF doesn't "build-up" on conductive objects. <S> The RF will bounce between walls with finite reflection coefficient, with wave scattered and randomized, and each act of absorption will create so-called eddy currents in the bulk of surface and around your holes, and these currents will dissipate into heat. <S> Eventually all RF will dissipate. <S> If you have a 1W radio, your cage should behave as a small heater, and ambient air will cool it off, so you might notice no difference. <S> If you don't connect the cage to ground, it will "float" and might get charged by accidental touching by charged hands or by convective air flow, to any potential, theoretically, just as one part of a capacitor. <A> You need to step back and understand some basic theory. <S> This site is about electrical engineering, where we expect people to know the basics, or that be the topic of their question. <S> "Electricity", whatever that really means, doesn't come out of the WiFi router somehow. <S> The WiFi router emits radio waves. <S> When those waves hit the wall of your Faraday cage, most will be reflected, some absorbed, and some will cause currents in the metal of the cage since it acts in part like a antenna. <S> Some of the radio signal being conducted by the metal of the cage will be re-radiated by that metal. <S> The fraction that gets re-radiated to outside will make it appear as if the cage leaks. <S> The amount that leaks out this way will be significantly less than the amount that would be radiated past the cage if the cage wasn't there. <S> The net result is that the cage will appear to significantly attenuate the radio signals both from and to the WiFi router. <S> This means the effective range of the router will be significantly dimished. <A> Insert an insulated straight wire through one of the 3mm holes, the wire of length <S> wavelength/2. <S> Push the wire in 50%. <S> Thus wv/4 is inside and wv/4 is outside. <S> You'll have a fine internal pickeup and fine external radiator. <S> Use insulated wire. <S> A prior question, quoting Feynman, showed the attenuation being 2.718^6.28 or about 500:1 (54dB) for every 3mm you back off the holes in the mesh. <S> Why are many IR receivers in metal cages? <A> If you are asking where does the "Power" that the router consumes actually go, absolutely all the electricity consumed by the router is turned into heat, radio waves, and light with the vast majority being converted directly to heat (Conservation of power, there is no other magic place for the power to go). <S> Ultimately all of the power consumed by the router ends up as heat (Unless it escapes the earth as light or other EM radiation). <S> Of the tiny amount of power that actually goes into the radio waves--without the cage it would go out to heat antennas and metal conductors ever so slightly (and a tiny fraction would even wander out into the universe), but with the cage in place, most of it would heat the cage itself or be reflected back to heat the air inside the cage and the metal components in the router.
| The RF power emitted by the router reflects from the inner surface of the cage, and bounces around until it encounters something that can absorb it, such as the plastic case of the router and other nonmetallic items inside the cage. No, radio waves are not "electricity" in that sense.
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220->12 AC voltage transformer into 120 -> 12-20 AC I bought a budget headphone amplifier Objective 2 (Mayflower manufactured), yet just realized that even power supply has a North American plug, it expects 220 V 50Hz (european voltage). I cannot return due Massdrop sales policy. Apparently compatible/recommended power supply ( 120 to 13-20 AC with 2.1 barrel plug) are not so expensive per se (<15$), yet rare and expensive to deliver (about 30 CAN). The amplifier also supposed to work from two 9V rechargeable batteries (I guess about $20), though I do not see where to insert them, probably need disassembly the device. I also have a 220 -> 110 V converter but not wise versa. So is there reasonable way to make it work besides ordering new power supply from Mayflower? For some reason unit is not expected to work with normal AC/DC converter, and requires AC-AC power supply. The amp designer stresses out "The transformer output can be anywhere from 14 VAC to 20 VAC and at least 200 mA. Don’t use more than 20 VAC or less than 14 VAC (except for the 12 VAC transformers listed in the parts list). Do not try to use a DC adapter . " Schematics is on: https://drive.google.com/file/d/0B52Awjeyc5zKMjRlYjlhNGItNGJlNC00ODlmLWIwM2MtNDI4ZWU4YWRjY2Y4/view 2.5 x 5.5 output plugs seems to be more easily available yet I guess it would need some plug adapter or find soldering iron. <Q> The reason you are directed not to use a DC source is because that would concentrate all the current in only half of the rectifier, potentially overloading it. <S> You could try supplying DC to the battery terminals, but that might introduce noise from the power supply/line that the battery side lacks as much filtering for, resulting in more audible noise. <S> Your best choice is to get an AC adapter with the right voltages. <S> This doesn't mean that you have to buy from the manufacturer of the amplifier, though. <S> This type of adapter is literally just a transformer in a nice case, and given that you have the specs (14 VAC, 200 mA) <S> that is all you need to match. <S> If you have a local electronic parts store (new or used/surplus) it is worth checking with them, and if they don't have it they might know where else to look. <S> It is also possible that you could find a local seller of similar equipment that could supply the power adapter. <S> For example, I'm in the US <S> and I could get a matching adapter (except for the plug which is 2.5 mm I.D.) from Schiit Audio , for (currency converted for comparison purposes) <S> 22 CAD. <S> Other usual sources for generic cheap stuff including the direct-from-China ones. <S> It looks like 16 VAC transformers meant to be hardwired <S> are a fairly common item you might be able to find at hardware stores. <S> I would recommend this budget option only if you are confident in doing line voltage wiring and could build the transformer into a suitable case to protect the splices that would be required — and it probably adds up to more in total than getting the proper transformer shipped given the transformer, case, AC plug, and DC plug, unless you have access to an extensive junk-box. <A> Solved my problem. <S> According to O2 Power Supply Wiki , as "last resort" one may use a "travel" voltage transformer in conjunction with an O2 power supply, even though wiki addresses US power supply use in Europe. <S> I checked local Chinese dollar shop and bought the cheapest $6Can 50w "travel" transformer 120->220V. <S> It looks extremely ugly, yet apparently does the job. <S> The headphones sound almost as good as directly from the sources <S> : Spotify on my cell phone, Sansa Clip or Walkman mp3 players. <S> I'd say sound from amp is <S> more resolving/detail rich yet less pleasant, it feels like <S> minor distortions creep in - could be cheep cable fault as well) <A> The current can be upscaled higher it wont disturb the device and if all fails you can simply recoil the transformer . <S> In recoiling the transformer you remove the iron core carefully and the simply calculate the no of turns that will perfectly step down the input voltage while retainig the output parameters . <S> All you need do is dont tamper the output secondary windings which supplies 12v <S> but you should un boil the primary which is set to take 240v european voltage . <S> After un boiling it you should fold it exactly into two thereby half ing the lenght but doubling the thickness <S> and then you recoil again and put back the iron core carefully back and connect up your new transformer <S> will be able to use 120v american supply and give the exact specification of output the only con this has is that the method is tedious but it works perfectly well when done correctly
| Try and get an adaptor with the right voltage and a slightly higher current if possible . Based on the information you have supplied, we can conclude that the amplifier has an internal bridge rectifier.
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Driving a 44Ω Speaker off a board that expects an 8Ω speaker I'm turning a Western Electric 500 into a mobile phone using a FONA Feather . I see the Feather want/expects an 8Ω speaker, but the speaker in the handset measures 44Ω. I presume if I try to drive the handset speaker, the FONA won't have enough power and I'll barely be able to hear. Is that assumption correct? What would be the best way to work around this? An amp in between the FONA and the headset? Or might it work OK without? Or another method? <Q> Given the SIM800H data sheet, if you do go with a replacement you might look at a 32 ohm element, say from unpowered personal audio speakers that plug into a headphone jack. <S> (If you want to figure out why Adafruit listed an 8-ohm speaker on their product page you could try asking on their own forums - it's entirely possible that was what they had on hand or in stock, they tried it, and it seemed to work) <A> “Enough power ” is not the issue. <S> If we ignore the efficiency of the transducer itself, the electrical power in is the acoustic power coming out. <S> What a too-high load impedance does is it doesn't draw enough current to produce the output power that the source is capable of. <S> On the other hand, a too-low impedance draws too much current, resulting in audio distortion or damage to the source. <S> Whether or not the specifications you have say so, in practice there must be a range of acceptable impedances. <S> 4 Ω and 8 Ω are common full-size speaker impedances and they can be considered fairly low. <S> Headphones can have a very wide range of impedances, commonly 25 to 600 Ω, but in general higher than typical speaker impedances. <S> Additionally, simply for acoustic reasons, headphones/headsets/earpieces always need much less power than a speaker, in the sense of something that you don't hold directly to your ear, requires. <S> Thus, a particularly high impedance is not necessarily a bad thing. <S> All this together means that it is highly likely that an output that says it can drive an 8 Ω speaker will almost certainly be just fine — possibly even too loud! <S> — <S> when used with a 44 Ω transducer. <S> If you hook it up and it turns out it <S> doesn't work well, and you confirm that it is for impedance reasons <S> , then the solution is to use a transformer with a suitable ratio and frequency response. <A> The earpieces used in old telephones were driven directly by the line signal, so they had to be efficient. <S> However the impedance at voice frequencies is probably much higher than 44Ω. <S> It would just try it direct, and if it is too soft then replace it with a modern element (which should have significantly better fidelity).
| Actually the Simcom data sheet for the SIM800H found on that board assumes a 32 ohm element in its output figures, pragmatically speaking your 44 ohms (if that is indeed an accurate figure at audio frequency - doubt has been raised) will probably work, though you could also replace it with a different element, since the earpiece cup readily unscrews.
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Declare signed numbers in Verilog There are so many resources online talking about how to represent and extend signed numbers in Verilog, but I still can not get it. Let's say I have a number 244, which is 'b1111_0100, or 'hF4. If I want to represent this number in signed decimal, should I need to declare the size with one extra bit for the sign? 8'sd244 or 9'sd244 // signed 244?8'sb1111_0100 or 9'sb1111_0100 // which one is correct 244?8'shF4 or 9'shf4 // Do I need 0FA, or it is assumed? Even more confusion comes with negative numbers: do I need to represent them in 2's complement format? How about the size? It would be nice if someone could give an explanation with example in the <size>'<signed><format><value> format. <Q> To declare a negative number in 2's complement form, you place the negative sign in front of the width specifier, for example <S> -8'H10 <S> Would be the value of -16, and would have the same bit pattern as the unsigned value <S> 8'HF0 <A> You need to separate the bit-pattern value expressed by a numeric literal from signed or unsigned type. <S> Signedness only comes into play when a value gets used in another expression, and it also determines the interpreted minimum/maximum values. <S> The <size> , <radix> , and <value> parts of a literal give you an unsigned bit pattern. <S> If you are working with an 8-bit expression, the largest signed value you can represent is 127. <S> So if you have 8'sd244, that will be interpreted as a signed negative <S> number(-11, I think). <S> If you are trying to represent -244, you need at least a 9-bit wide value. <S> Verilog has tricky rules when mixing signed and unsigned data types. <S> But in general, the MSB of a signed expression gets sign-extended when used in a larger width signed expression. <A> For representing signed numbers, you definitely need 1 extra bit, like if +250 in unsigned could be represented in 8 bits but <S> both +250 and <S> -250 when declared unsigned would need 9 bits. <S> You can try out this small program and see yourself `timescale 1ns/1psmodule tb; reg signed [8:0] a; reg [7:0] b; initial begin <S> a = 9'b011111010; <S> b = 8'b11111010; <S> $display("a is <S> %d and b is %d", a, b); a = 9'b100000110; b <S> = ~b + 1; $display("a is %d and b is %d", a, b); endendmodule <S> So, +250 when represented in 8 bits (unsigned), if done 2's complement does not produce -250. <S> To get -250, you need 9 bits and that when 2's complemented produces -250 correctly. <S> So <S> 9'sb011111010 = <S> +250 <S> 9'sb100000110 <S> = <S> -250 <S> 8'ub11111010 <S> = <S> +250
| You do indeed need to include the sign bit in your width considerations, and it is of course essentially up to you to keep track of which vectors are to to be interpreted as signed and which as unsigned.
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Is this negative regulator really meant to be an inverted positive regulator? Project: make a power supply that can replace the 240V AC supply shown, so that I don't die when messing with the board that this connects to . I'm only interested in the part inside the yellow box; the +40V output goes to the power amp section which I'm not working with. So I went out and bought a 7812 and 7815, but then I got confused because of the general consensus that you shouldn't use a 7815 with inverted output pins as a negative voltage regulator, but it seems like that's what they're doing here. I've ordered a 7915, but my question: (a) Is this circuit doing something smart by connecting the 7815's positive output to ground and using its output as -15V? (b) Is this circuit doing something weird, and I should use the 7915 to source -15V as the gods intended (c) it's impossible to tell, needs more info Note that I'm sourcing ±30V DC from a preamp circuit in this project, which is higher than what would be at the input in this schematic but well within tolerance for the 7812/7815/7915. My schematic, using a 7915: simulate this circuit – Schematic created using CircuitLab Version with 'reversed' 7815: simulate this circuit <Q> This means the 15 V output is fully floating, and any terminal can be grounded. <S> If you want to use a centre-tapped transformer (with some slight savings in hardware and diode drops) which produces non-isolated rails, or the non-isolated rails from another piece of equipment, then you need to use 79xx series regulators for the negative rails. <S> The reversed 7815 you've drawn will not provide the -15 V you want. <A> Look at the 1st picture, the photo. <S> Note how the 7815 based supply is completely separate from the 7812 one except for the ground connection. <S> In your 7815 schematic V2 (-30 V) is wrongly connected, it should not connect to ground. <S> It must be floating with respect to ground. <S> Change that and your schematic will be the same as in the photo. <S> Although the solution as in the photo looks strange, it is OK and will work just like using two lab-supplies to make +12 V and -15 V. <A> To add to Neil's answer, it is best to think linear regulator as variable resistor (or transistor if you prefer so) between IN and OUT terminals regulated by some "smart" circuit inside. <S> Ground terminal is (mostly) a reference only, not a power terminal. <S> If you take a look on your last schematic, there is V2 generating fixed 30 V voltage between In and OUT terminals of U2 regardless of current through the regulator. <S> Redrawing the regulator as a transistor + error amplifier maybe helps, see below. <S> Your source is connected between IN and OUT while load is between OUT and GND, so you would need load current flowing between IN and GND terminals in order to make things work. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> On contrary, situation in the yellow box is different. <S> Voltage source (secondary winding + rectifier) is connected between IN and GND terminals and load is connected to OUT and GND. <S> That is, current from source to load flows through In and OUT terminals of regulator <S> and it can do its job by controlling this current. <A> You are over complicating this quite a bit! <S> Remember that voltage is relative! <S> Turn that -15 in to 0 and the gnd in to 15V and the +12 in to 27V and see if it makes sense! <S> The 7812 is using the 7815's output as a voltage reference for it's ground which raises the output by 15 voltages. <S> By treating the middle "tap" as the ground, you can get what appears to be a + and - swing and a negative rail... but note that it really is all floating so there is no ground <S> , it's just that you've chosen a point of reference to be ground and <S> so anything below it will be negative and anything about it will be positive. <S> If you chose the -15V as ground, then you'd have 0V, 15V <S> and 27V <S> like I said earlier <S> and you've simply design your circuit that way. <S> Of course, your circuitry will have to be reversed to handle the reversed current you through at it <S> but it will all be the same in the end. <S> The two regulators are exactly the same circuit. <S> But without that connecting, they both are floating! <S> (and simply have 12 and 15 Voltages across the outputs. <S> By connecting the "ground rail"(which isn't 0V, it's just point of reference) of the top one to the "power rail" of the lower one <S> you then just couple them, but the whole combination is still floating. <S> I think if you just stare at it a bit and realize it's not complicated and that voltage is relative, you'll see what I mean. <S> Remove that coupling leg and then think about it, if you still have trouble. <A> Using a voltage regulator upside down is not totally uncommon. <S> This article deals with it in the context of a single voltage source: http://www.edn.com/design/power-management/4458675/Turn-negative-regulator--upside-down--to-create-bipolar-supply-from-single-source <S> Your circuit does not actually need such a solution because there are two secondary windings which could be used to create a +12V and a -15V supply in the usual way, but maybe the manufacturer of this circuit had 7812 and 7915 regulators already in stock and did not want to add another component type.
| What you really have is two stacked voltage regulators Your "reversed" 7815 (last schematic) is incorrect, it cannot work. What you have shown in the yellow box is OK, because the 18 V secondary that's sourcing the 7815 is fully isolated from the other rails. Regulated current flows always (for both positive and negative regulator) between IN and OUT and regulator can control resistance between these two terminals only.
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How to measure the inverse saturation current of a diode? I'm trying to measure the inverse saturation current of a diode, for using it in the schockley equation. I would like to know what is the best circuit for doing this. I already tried the following circuit, but can't detect anything. simulate this circuit – Schematic created using CircuitLab I'm using a 0-30V direct current power supply, a resistance of 230 ohms and a IN4001 diode. Do you have any suggestions? <Q> The last time I made some diode leakage measurements, I used a cheap (distributor's own brand, <£10) DMM, with 1999 full count, 10M input impedance, and a 200mV range, as the current meter. <S> On that range, the full scale is 200mV/10M = 20nA, with the nominal current resolution 100uV/10M = <S> 10pA. <S> You can use an external shunt to get a higher current range for big diodes, use an external 1.1M resistor for 200nA full scale, and 100k for 2uA. <S> It's worth using a variable power supply, and inching it up from zero, so as not to embarrass your meter's 200mV range with a leaky or failed diode. <S> I found (and the Schockly equation will tell you) that once above a couple of volts, the reverse leakage current is more or less constant. <S> FWIW, I measured about 35nA for a BAT42 (schottky) <S> , 4nA for a 1N4148, and failed to measure (so <10pA) <S> the current for a BAS116 which is advertised as a 'low leakage' diode. <S> If you want to measure lower currents, then you need to build a pico-ammeter round a low bias op-amp, next-hack's answer shows you how. <S> Two cautions when working like this. <S> (1) You might want to verify that your meter is actually 10M input impedance on the 200mV range, put an external 10M in series with it, and check that halves the reading when supplied with a voltage. <S> (2) <S> When using non-native ranges like this, the decimal point will usually be in the wrong place. <S> I am skilled at moving it the wrong way in my head, so I simply enter the voltage I read into a spreadsheet, and the relevant resistance in another column, and let it do the sums. <A> The output voltage is proportional to the current. <S> To get a 1V/nA you should still use a 1GOhm resistor. <S> With a 10MOhm resistor, you have only 10mV <S> /nA. <S> However any 3.5 digit DMM has the 200mV scale, i.e. with 100uV resolution. <S> I.e. 10pA resolution @10MOhm, 100fA @1GOhm. <S> You must apply externally "Vtest", while the other terminal will be forced to ground, due to the negative feedback through R1. <S> Notes: <S> The circuit is dual supply (not shown). <S> That particular OP amp has aVcc-Vee max of <S> only 15V. Still I expect that the reverse current doesnot vary a lot with the reverse voltage. <S> You might need a suitable compensation capacitor in parallel to R1. <S> Also the decoupling capacitors are not shown. <S> Be aware that, since the LMC6482 has 20fA of leakage current, the overall leakage will be likely determined by the rest of the system. <S> If possible use the dead-bug style mounting. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> To be able to find the diode saturation current (scale current) <S> you need to take two measurements in the forward biased region. <S> simulate this circuit – <S> Schematic created using CircuitLab Notice that \$V_{F2} <S> > V_{F1}\$ and \$I_{D2} <S> > <S> I_{D1}\$ And the diode equation: $$\Large I_D = I_S \left(e^{\frac{V_F}{n*V_T}} <S> -1\right) <S> $$ And from the Shockley diode equation we have: $$\Large nV_T <S> = <S> \frac{V_{F2} - V_{F1}}{\ln\frac{I_{D2}}{I_{D1}}}$$ <S> And finally the \$I_S\$ current: <S> $$ <S> \Large Is = <S> I_{D2}*e^{\left <S> (\frac{-V_{F2}}{nV_T}\right)}$$ <S> And we done.
| A "simple" way to measure small currents, is by using a low input bias op-amp.
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Advice on connecting batteries in series I have a 4 NiMh AA HR6 batteries each with a capacity of about 2000mah. And am working on a LED project which requires 500ma but I need to power it up with a voltage of around 5v. So I was planning to hook these 4 batteries together in series to achieve that level of voltage. But here are my concerns these 4 batteries are not of same make. Two batteries are GP Recyko and two are Kodak Digital camera batteries. Does it poses a problem when i connect batteries of different make in series. Moreover I have recharged the battery two days back and when i measured the voltage none of them seems equal even the batteries of same make. The batteries measured 1.24, 1.31, 1.30, 1.27 respectively. If i do connect these batteries together what will be the resultant output voltage? If this is good to proceed how long can i safely draw 500ma from these batteries? <Q> If you connect those batteries in series, you'll get the sum of the rated voltages: 1.24 + 1.31 + 1.3 + 1.27 <S> = 5.1VDC <S> but this is true on unloaded state. <S> Since you claim that each battery has a capacity of 2000mAh, total capacity will be 2000mAh when they are connected in series. <S> Practically, it can be said that the lifetime will be approximately 2000mAh / 500mA = 4 hours, but please consider the discharge curves. <S> Like the one below: According to the graph, one battery voltage will be about 1.25VDC for a 500mA load. <S> Then the total voltage will be 1.25 x 4 = 5VDC. <S> If the load current remains constant then the lifetime will be about 4 hours. <A> Battery voltage is a 'low class' measurement, you can't tell a lot from it, so those four measurements are more or less the same. <S> You do the sums for the present series voltage, it's a bit more than 5v. <S> Of course during discharge the voltage will drop, many people choose an endpoint voltage of 1v per cell, so the final voltage will be around 4v. <S> More important than the make is the history of the batteries. <S> If two are new and two have had years of use, it would be unwise to run them in a series pack. <S> If they are all the same sort of age, and nominal capacity, then go for it. <S> NiMH is a fairly forgiving chemistry. <S> They tolerate low current overcharge, and self-balance during that. <S> They tolerate deep discharge, and while it's not recommended, they are not fatally damaged by being taken well below their endpoint voltage, though voltage reversal will damage them. <S> On the first discharge of the series battery, monitor all the cells individually to make sure there isn't one weak cell that will run out early. <S> How long will a 2Ah battery last at 500mA? <S> The naive calculation is 4 hours. <S> However, the rated 2Ah will be at the 10 hour rate, discharging them faster <S> will give you a lower capacity. <S> The capacity will also vary with endpoint voltage. <S> Check the specs for those cells to see how they're tested. <S> In summary, no more than 4 hours, but probably the majority of that. <A> The voltage of the pack will be the batteries combined voltage, 1.24 + 1.31 + 1.30 + 1.27 = <S> V pack . <S> The current load is 500mA and batteries is specified in mA/H. <S> So the time it can supply current to the load is 2000mA/500mA = <S> X hours <S> Regarding different brands. <S> Don't mix 'em. <S> Even if they are all NiMH cells the differing characteristics would mean they wouldn't charge properly, and the weaker cells could be damaged during discharge in use.
| When the batteries are connected in series, the capacity will be practically the minimum one of the rated capacities, not the sum.
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How to turn Flashing LED into constantly glowing one So I have one detection circuit with LED that flashes continuously when something is detected (1-2 sec interval). That's a bit nerving while it happens when you drive, especially at night. I don't have any option to view or modify the code, so it would be nice if I can "hack" this LED to glow continuously....can I do it somehow with some active components? Tnx in advance I don't have any schematics of this thing. It is a Sensor device that detects when a car is in your "blind spot". So it blinks slowly when something comes in the zone, and rapidly if you put your turn light on....so that's basically a problem for a Cap while it will soften than all the impulses? right? The LED is packed in a case where just two wires came out, so i need to look later onto the PCB to make it clear if it is with GND or Vcc regulated And it is definitely a LED controlled with a micro controller <Q> In principle, this is a job for a retriggerable monostable (CD4047 or 74HC123). <S> To avoid voltage level problems with sensing the LED voltage, you could use a photocell to sense the flashes; this then drives the monostable with its period somewhat longer than the flash period, and that drives another LED which stays on as long as the flashes continue. <A> ALTERNATIVE SOLUTION <S> That's a bit nerving while it happens when you drive, especially at night. <S> Surely that is the whole point of the LED, to be visibly irritating enough that you notice it BEFORE you change lanes. <S> Making it stay on you may never even notice it till it is too late. <S> It seems to me your real issue here is not the flashing but rather the fact that it is too bright at night. <S> As such, I'd recommend you don't mess with the functionality, but rather switch in a resistor or add a pot that lets you adjust the brightness at night. <A> Assuming you have access to a constant voltage somehere, something like this could work: <S> simulate this circuit – <S> Schematic created using CircuitLab Slow to turn off and probabaly overly complicated, but should work.
| If you have the ability, there may also be the possibility of wiring up some circuit that adjusts the brightness based on the ambient light, or ties into the existing dashboard lighting levels.
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How to limit current for voltage regulator/decrease power dissipation? I need to power a 5V device that needs more or less 1.2A in order to work correctly. I'm using a 13.8V power source with two diodes in series (1N4007) to limit the voltage to roughly 12V. I tried using the voltage regulator AZ1084T, since it has 5.0A of maximum output current, but even with a large heat sink attached it was heating up a lot, like 117°C (and the maximum admissible temperature on the datasheet is 125°C). So, just to test, I replaced the AZ1084T by the old and good LM7805CV, which has 1.5A of maximum output current. It worked just the same, heating up just the same (and that's a good thing, because this one is much cheaper). But how can I avoid the heat? I'm pretty sure it's the power dissipation from 12V to 5V that's causing it, but this voltage regulator was supposed to work with that input voltage. I tried this scheme, using TIP127 to "split" the current, but my device won't even turn on (the measured output current was showing peaks of 2A~3A, which kept turning off the power source which limit current is 1.7A). Anyway, any suggestions on how to split the current while maintaining the output voltage or on how to decrease the power dissipation are welcome. <Q> over it's entire input range - especially not without a good amount of heatsinking (potentially with fan). <S> The reason both of the devices heat up the same is because they have to get rid of the same amount of power, given by $$P = <S> (V_{in}-V_{out}) <S> \cdot <S> I$$which in your case is about \$(12\ <S> V-5\ V)\cdot1.5\ A = <S> 10.5\ <S> W \$, which is a lot of heat! <S> (there is a reason computers have big heat sinks with fans on top). <S> You have a few options. <S> You can look at going to switch-mode powersupplies, which are far more efficient (because they don't lower the voltage by turning the excess into heat). <S> Alternatively, you can look at using multiple regulators in series - Use a regulator to go to 9V, then to 7V, then to 5V. <S> This way, the load is spread out over multiple regulators and each has less power to dissipate <S> (of course the total dissipated power is the same!).Finally, you can look at using external components like you already were doing. <S> You mentioned having issues - I would suggest trying that topology with lower loads and a good amount of protection on the powersupply end to see what is going on - could be that you are getting some kind of instability. <A> Your power dissipation is caused by the amount of voltage you drop to get to 5V times your load current plus the whiff of current <S> the regulator uses to do it's thing. <S> The latter is usually negligible. <S> You pretty much have three choices as shown below in order of preference. <S> Use a lower voltage source. <S> If that 13.8V is coming from a transformer, buy a different transformer that is closer to the lowest voltage the regulator needs. <S> Build a switch mode regulator circuit. <S> There are many devices and circuits out there these days that will convert voltages for you at high efficiencies. <S> There is some cost in complexity and noise though. <S> Split the power <S> : Add in additional power diodes or power resistors before the regulator to soak up some of that heat. <S> You may still need to heat-sink all that <S> but you may find the heat-sink requirement is quite a bit less to keep the active component within tolerance. <S> Of course this method does NOTHING to improve efficiency <S> and so comes last in the list. <S> Note: <S> The current splitting technique you show is really intended to allow you to use the regulator at higher currents than the regulator itself is designed to provide. <S> That approach does not really help you in this situation. <A> I assume you need noisefree 5V signal, since you use a linear regulator. <S> I would drop voltage with a decent quality buck converter (>85% efficient) to 6.3V, and feed that to a LM350 (3A) or 7805 <S> (1.5A) <S> - both drop a nominal 1.25V regardless. <S> As already mentioned above, it is pointless to dissipate and lose so much power. <S> W_dissipate = <S> ( Vin - Vout <S> ) * I_load <S> is pretty clear. <S> I always use step down converters -> linear regulators if there's more than 2-3 volts difference, and my load is moderate.
| Also, you can get better regulators called low dropout regulators that need even less input voltage. Just because a regulator is able to output 5 A of current, does not mean it can do so As long as you are using such a large voltage source you will have to deal with that power if you are using a linear regulator.
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Capacitance to Voltage converter - Water level sensor I have to design by myself a capacitive level sensor. To create my "capacitor", I use two copper strips that I take from a copper tape and put it on the tank where I have to take the measurement. However, the measurement must be a voltage or a current as my PID controller can only "read" this kind of value. Therefore, it means that I need something to change my measured capacitance into a V/I and I do not find an easy way to achieve the conversion.By easy way, I mean that I am in a city where there are not a lot of electronic devices that are available so easy means "without too many components hard to find". The specifications of my system are the following ones: Measured Capacitance : [5pF(low level) --> 35pF(High level)] This is the full range but I could use a small range inside this one Input of Controller: 0-10V or 4-20mA I hope that someone can resolve my problem, or at least give me an idea.Thank you in advance. Sincerely. <Q> By DC-blocking the electrodes with C2, corrosion should be minimal. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The NOT inverters are 74HC14 Schmitt inverter sections, intended to be powered either from 5V, or the emitter of Q1. <S> If the ammeter isn't of the moving-needle type, some filtering will be required. <S> R2 <S> * Csenseshould be larger than R3 * C3. <S> I've used this circuit with a shielded cable from the Q1 emitter,so the oscillator can be adjacent to the sense capacitor, but thepower supply and meter at a more convenient location. <A> First, you probes should use stainless steel. <S> Copper will corrode. <S> There is a “simple” capacitance to voltage converter below. <S> It’s been around a long time. <S> You can make a 1kHz osc. <S> with an op amp or a 555. <S> The waveform is not critical. <S> Active rectifier circuits abound (instead of an AC meter). <S> R is for bias current and should be high for best linearity. <S> I've used an LF356 because I had some laying around <S> but I'm sure there are better modern choices. <S> That means you want a very low bias current op amp. <S> Cgain sets the gain, if Ctest=Cgain the AC gain will be 1. <A> Since you live in a city where components are hard to find, how about starting with a simple LC oscillator with a single transistor than can run from two AA cells? <S> If you choose the right frequency range, you should be able to detect the radiated carrier on any nearby AM radio. <S> It does not deliver the 4-20mA you ultimately require, but perhaps it could stimulate some ideas! <S> Here is an on-line simulation of the circuit. <S> The hardest parts to find are probably the 1mH inductors, but you can absolutely make your own - a 1mH inductor is around 200 turns of thin magnet wire on a 1.5 inch diameter coil form (e.g. cardboard) that is also 1.5 inches long. <S> You can use two AA cells or even a 3V coin cell for the 3V power. <S> The current draw is less than 2mA. <S> Your calibration could be water level versus where on the AM dial the carrier is detected. <S> (You will not hear a tone, just a distinct "quieting" of the AM noise"). <S> Note that most RF emissions are subject to laws in your country, but based on your project description I'm guessing you are not going to manufacture this for sale. <S> If you have an oscilloscope (or Arduino) to measure the frequency, even better! <S> Finally, don't build the circuit on "proto board" because there is a lot of capacitance between adjacent rows of contacts - lay down soome of the copper strips you are already using an insulator and solder everything together while keeping all interconnects as short as practically possible.
| An oscillator can be fashioned from the capacitor, and it is simpleto either count the frequency, or convert to a controlled current. The circuit below does just that. Almost any general purpose NPN transistor would work.
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How do I use this Strain Gauge sensor? The load cells here: http://www.cnloadcell.com/mlc700-postal-scale-load-cell and here: http://www.cnloadcell.com/mlc714n-personal-scale-weight-sensor have 4 wires (red, white, black, green). The diagram is quite confusing, I think I can see it is somehow part of a wheatstone bridge . But from reading up on Wheatstone bridges, each strain gauge should only have 2 wires and these gauges have 4? Is the diagram showing how to use just one gauge or many? Two wires coming out the strain gauge: <Q> Red and Black are for your excitation voltage, white and green is your differential output voltage that's scaled by that excitation voltage you provide. <A> This is from an Analog Devices reference design for a weigh scale, showing how it is used in practice <A> Found a bigger explanation: https://learn.sparkfun.com/tutorials/load-cell-amplifier-hx711-breakout-hookup-guide <S> These pins are labeled with colors; RED, BLK, WHT, GRN, and YLW. <S> These colors correspond to the conventional color coding of load cells, where red, black, green and white wires come from the strain gauge on the load cell and yellow is an optional ground wire that is not hooked up to the strain gauge but is there to ground any small outside EMI (electromagnetic interference). <S> Sometimes instead of a yellow wire there is a larger black wire, foil, or loose wires to shield the signal wires to lessen EMI. <S> In General, each load cell has four strain gauges that are hooked up in a wheatstone bridge formation as shown above. <S> The four wires coming out from the wheatstone bridge on the load cell are usually: Excitation+ <S> (E+) or VCC is red <S> Excitation- (E-) or ground is black. <S> Output+ <S> (O+), Signal+ <S> (S+)+ or Amplifier+ <S> (A+) is <S> white O-, S-, or A- is green or blue <S> Some load cells might have slight variations in color coding such as blue instead of green or yellow instead of black or white if there are only four wires (meaning no wire used as an EMI buffer). <S> You might have to infer a little from the colors that you have, but in general you will usually see these colors. <S> So basically the load sensor internally already has the wheatstone bridge and these 4 wires are the output from that <S> (not that the four wires are expected to be used inside a wheatstone bridge.
| Wheatstone Bridge style strain gauges have 4 wires, as shown in the circuit diagram you posted.
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How to reduce the speed of a single-phase induction motor I have a 1 h.p. single phase induction motor driving a dust extraction blower. The air flow is about 600 cubic feet per minute. The device that the dust collection duct is attached to requires only 300 CFM. I don't want to waste energy; is there an inexpensive way to reduce the fan speed to 300 CFM? Update; I measured the flow in the duct yesterday with a hot-wire anemometer - it's actually only 245 CFM. 600 CFM is what's printed on the blower, but I guess it's either an inflated number, or the conductance of the duct system is limiting the flow. Thank you all for the informative replies. <Q> First be aware that the flow rate of fans relies on the square of the turn speed. <S> So you had to reduce the speed to 70%, not 50%. <S> Unfortunately, the speed of an induction motor is mostly a function of the frequency and it's expensive to change that. <S> The second option would be a resistor in series. <S> That would flatten the M/n ramp at the working point. <S> But for a 30% speed reduction, this is a huge waste of energy. <S> The third option is reducing the voltage but you cannot apply it here because you usually only get at max 10% speed reduction unless the motor is specifically built for that. <S> Unlikely for a fan drive. <S> So, think of the fourth option and mount a smaller fan on the same motor. <S> That would waste no energy at all. <A> As already pointed out single phase motors are hard to speed control. <S> It won't save any energy, but you could simply put a bleed opening in your inlet to control the CFM for your extractor. <A> If you just reduce the flow rate (using an obstruction) that will reduce the load on the motor and thus the energy consumption. <S> Blowers a weird like that: try it with a vacuum cleaner some time, when you block the inlet (or outlet) <S> the motor speeds up indicating that the mechanical load has been reduced. <S> another option is to use a smaller fan. <A> There are multiple types of single phase true AC motors, only two of them can have their speed controlled without risking damage to the motor; Shaded Pole motors and Permanent Split Capacitor (PSC) types. <S> Shaded Pole motors can be controlled by varying the voltage only, which lowers the torque and increases the slip, but since they are designed to deal with it, the motor itself does not overheat, it just does less work, but Shaded Pole motors typically stop at 1/4HP. <S> PSC motors can be controlled with voltage control as well, but ONLY if the load is a centrifugal machine in which the load on the motor drops at the cube of the speed change. <S> If you simply reduce the voltage of a PSC motor on something other than a centrifugal machine, it overloads and burns up. <S> Not knowing what type of motor you have means the best advice is to not try this until you know for sure. <S> Also, "blowers" come in both centrifugal and positive displacement varieties, so from your simple description we can't determine if any of this is possible even if you do have a PSC motor. <S> A third type is a "Universal motor" that is actually a brushed DC motor designed to run from an AC input. <S> Universal motors are the type found in portable power tools and small kitchen appliances with variable speeds. <S> Again, not typical for that to come in a 1HP variety. <S> You can also control a PSC motor using a Variable Frequency Drive, but for 1HP 120V, the few that are out there that will work on a single phase motor will cost a LOT more than a new motor. <S> If your motor has a starting switch inside, i.e. when you turn it on, you hear a "click" right after it starts spinning, then it's one of the types that cannot be speed controlled.
| another is to reduce the supply voltage using a transformer or series capacitor, this will work with a blower because the mechanical load is not constant, but insted dependant on the motor speed.
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Piezoelectric effect in cables? I am using a very sensitive transimpedance amplifier on a 3m long cable to a photodiode. Signal levels are about 70 nA - 700 nA. Bumping into the cable causes a large voltage spike on the output of the amplifier. The insulating material is FEP. Could this be the piezoelectric effect in the cable insulaton? Edit: Here's a typical voltage spike. This happens when I bend, shake or tap the cable. <Q> Triboelectric effect probably. <S> You can find cable made for musical instruments that is screened with a low resistance carbon impregnated layer (They call it semi conducting!) between the screen and the insulation of the centre conductor, this is helpful when dealing with high Z sources as it minimises the noise due to movement. <A> Adding to Jack's answer: Triboelectric effects are just like every day static electricity: rubbing dielectrics together rips some electrons out and creates charge. <S> This creates a current (i=dq <S> /dt)/ <S> How this will affect your measurement is proportional to the impedance of the circuit driving the cable. <S> The error voltage created is Z*i <S> according to ohm's law. <S> If the cable is driven from the low impedance output of an amplifier, then the effect will usually not matter at all. <S> But you are measuring tiny photodiode currents, so here it matters. <S> There is another effect: your coax is a capacitor. <S> If it is charged to a DC voltage (not 0V) then bending it will modify its capacitance. <S> Since the charge inside the cable's capacitance is constant, changing the capacitance changes the voltage. <S> Again, this only matters in a high-impedance setting... like yours. <S> Solutions: <S> Everything <S> Jack said ...and the obvious one, although not always possible <S> : Put the amplifier next to the photodiode and have its low-impedance output drive the cable. <S> If you want non-triboelectric coax, the keywords you have to google are "low noise coaxial cable". <S> This is a type of coax with a graphite/carbon based resistive layer between the insulation layers, which dissipates any charge created by rubbing. <S> However, they are hard to find and expensive. <S> They are used in electrocardiograms, or for vibration measurement using piezo sensors, for example. <S> I also second the use of stage microphone cable: musicians like cables which don't produce a THUMP in the speakers when someone steps on the cable. <S> If you walk in any music/guitar shop and ask for such cable, the sales person will know what you're talking about. <S> They may not know about the physics involved, but they will know about clients returning cables because thy're microphonic... <S> These cables usually have a cotton filler around the wires, which does not generate charge when bent. <S> I use balanced stage microphone cable for audio measurements: they're cheap, easily available, and work very well. <S> No need for fancy audiophile stuff. <S> If you want coax, try guitar cable. <S> They're designed for the same setting: tiny signals, lots of spastic movement, high-Z setting... <S> These won't be specified for HF impedance though. <A> This sounds like microphonic effects in the cable. <S> It arises from several sources, some of which have been mentioned in other answers, including Surface charge building up on the insulators, then being discharged when the cable bends Triboelectric effects (i.e. friction) <S> Piezoelectric effects (in the insulator, not the copper) <S> It could also be capacitive or inductive pick-up. <S> Though this is less likely if you already have good quality shielded cable with a grounded shield. <S> Stiffer cables which are less likely to bend and pick up noise as a result. <S> Larger diameter cables with lots of filler, which reduce the bending on individual cores. <S> Running cables where they are not likely to be bumped/disturbed. <S> Covering or protecting cables with foam to avoid acoustic pick-up from the environment. <S> Nailing or otherwise fixing them down so they can't be bumped/moved, though be careful not to crush or damage them. <S> Cables designed for on-stage audio use might be good here, as they are often non-microphonic. <S> There is a lot of overpriced garbage in the audiophile market, but there are also some nice products which - though vastly overengineered for audio use - are great for precision lab equipment. <A> What is probably happening is that small charges are getting moved around on the outside of the insulation of the cable. <S> The insulation is basically the dielectric of a capacitor, with the signal wire one side of the capacitor. <S> As the cable gets moved around and touches different things that may have various static charges on it, small currents are caused thru the capacitor. <S> Normally these currents are too small to matter, but in your case you are specifically amplifying them. <S> The solution is to use a shielded cable. <S> The insulation capacitor is then between the shield and whatever is outside. <S> The center conductor sees a relatively constant capacitance to the shield, which is held at a constant voltage by your circuit. <A> If the cable has a high DC voltage, then any capacitance change in the cable will result in a charge flowing and, therefore, a signal. <S> To test this, disconnect your transducer and see if the effect is still present. <S> If so, perhaps your photodiode is ALWAYS disconnected, so check if it is sensitive to an optical signal. <S> To see this kind of signal due to capacitance change in cable requires an abnormally high impedance, so perhaps something is not connected. <S> If everything is connected and the photodiode responds to optical signals, you can try shunt the photodiode with a load resistor <S> collocate the amplifier with the photodiode <A> What you are seeing, is not piezoelectric effect. <S> It is a capacitance effect due to the " distortion " of the dielectric material. <S> To fix the problem, one could try to reduce the charge generated by the bending (expensive), or one <S> can prevent the "distortion" by "encasing" the twisted pairs inside metal conduit (like EMT), which gives them rigidity, protection from bends and bumps, and additional shielding. <S> The conduit would, of course, need to be anchored so that it does not move.
| Common solutions include: Cables with slightly conductive packing material which bleeds out any accumulated charge Cables with different insulator materials which are less susceptible to charge build up or triboelectric/piezoelectric effects. Copper doesn't exhibit any appreciable piezoelectric effect.
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When would I not want to replace a PN diode with a Schottky diode When repairing or maintaining equipment, or even during design revisions, why would I, for electrical reasons (not interested in the potential financial aspects) not want to replace a PN junction with a Schottky diode? Three reasons I can come up with: Leakage: Schottky diodes have, in general, larger leakage Biasing: If the PN junction is used to bias other devices (such as is sometimes done in BJT output Class AB amplifiers) Switching applications: We might desire to use the slow reverse-recovery in RF switches. Are there any others? In my specific case, I am replacing diodes in a older HP 6253A powersupply. The powersupply has several 3A rated PN junctions in the signal path which look very corroded, and I have a number of 6A Schottky diodes I am looking to replace them with. Is there any reason why I wouldn't want to do so? <Q> In my experience a Schottky diode will fail at a lower reverse voltage (or during a transient) that a standard silicon diode would handle without a problem. <S> It all comes down to your application, what the diode is doing, and what sort of electrical events you expect it to see. <S> There is no right or wrong answer here. <A> If the design has to function over an extended temperature range, leakage becomes a very big issue. <A> You may as well ask why would you not want to replace all 1n4001s with 1n4007s? <S> There is no boilerplate answer to this question, and you can never dismiss cost. <S> The answer, in reality, is just the same as every other part you might chose.. <S> "If the shoe fits and you can afford them, then use them if you want to."
| While Schottky diodes generally have a lower forward voltage drop and faster recovery time, they are also more susceptible to reverse surges.
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Using Bench Power supply to charge NiCd pack I have an 18vdc battery powered drill hat works fine but the charger has died and a replacement is $54. Almost double what I paid for the drill plus battery plus charger two years back. Battery was still holding a good charge. The label says, 18vdc 1.2Ah Battery pack I did some searching and am using a bench power supply charging it in Constant Current mode at 0.750A but the voltage to get that current rate is 24vdc and rising slowly. I have it right alongside me and cannot feel any appreciable heat build up after 30-minutes. The rising voltage bothers me, is it OK to let it charge like this or should I clamp it at a set voltage. I can't do both this power supply as one setting limits the other. <Q> Don't be fooled by early results when checking temperature. <S> They heat rise past 70% charge accelerates quickly from internal gas pressures and charge transfer inefficiency dissipates more heat. <S> from batt <S> U. Figure 1 shows the relationship of cell voltage, pressure and temperature of a charging NiCd. <S> Everything goes well up to about 70 percent charge, when charge efficiency drops. <S> The cells begin to generate gases, the pressure rises and the temperature increases rapidly. <S> To reduce battery stress, some chargers lower the charge rate past the 70 percent mark. <S> Therefore do not exceed 1.46V per cell <S> * 13 cells = <S> 19V <S> The fact that you need to drive 24V just to get 0.7A indicates your batteries have aged significantly and are on their last legs. <S> I know from past experience manufacturing and testing burp chargers this can rejuvenate NiCd batteries. <S> Can you make a pulse switch with low <10% duty cycle at 24V ? <S> Then track your Amp-seconds of charge and useage and see if that improves it. <S> http://batteryuniversity.com/learn/article/how_to_restore_nickel_based_batteries <S> Which reminds me I have to fix my Hitachi LiPo cordless drill after left out in the rain. <A> Your NiCad batteries will likely have a terminal voltage (fully charged) of 20.8 V. <S> This is 13 cells in series. <S> Set your power supply to 20.5 V <S> (there are other complexities such as temperature and voltage drop on charge, but I'll ignore them for the moment). <S> Set your current sense OC/CC to around 120 mA (as WhatRoughBeast suggested). <S> When you connect a discharged pack the OC/CC will trip and reduce the voltage. <S> When the terminal voltage of the pack approaches 20.5 V the current will reduce to a trickle. <S> Under these conditions the pack will NEVER fully charge, and to explain that here is the typical charge graph for a single NiCad cell: <S> Notice that to get to 100% charge you have to manage a terminal voltage DROP as the battery approaches fully charged. <S> Using your simple power supply charging, we need to stay on the left of the peak. <S> I've said above to use 20.5 V, this represents about 1.6 V per cell in your pack. <S> At this terminal voltage you can expect to be about 75% charged. <S> You could select any voltage between about 20 V and 20.5 V to get reasonably consistent results with little chance of damaging your pack. <S> You could also of course build yourself a simple charge controller to connect to your power supply and get closer to 100% charge ... <S> however that's possibly not in your DIY capability. <S> For example, this might be all you need: <S> simulate this circuit – <S> Schematic created using CircuitLab <A> I would not use a bench power supply to rapid charge any chemistry of batteries. <S> NiCd however usually respond well to a trickle charge at C/20. <S> They absorb about one half of the supplied charge, and can dissipate any over charge and hold a stable temperature and remain undamaged for several weeks. <S> For a 1.2A.h (1200mA.h) NiCd battery, a constant current charge of 1200/20 = 60mA for 40 hours should fully charge the battery.
| The charger from Dewalt is only rated at 18V with a current rating of 2A. Dewalt DW9116 7.2V - 18V NiCd Battery Charger New for DW9057 DW9094 DC9071 $34.90 Ebay. You should NOT set your power supply to 24 V as this will severely overcharge the batteries.
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VL6180x sensor breakout stays on despite removed VIN I'm experimenting with a couple of VL6180x distance sensors on breakouts ( http://www.ebay.com/itm/172823051918 , the blue ones). Because the sensor resets its i2c adress when the power is removed, and because I want to use more than one, I need to be able to set the desired adress after boot. My idea to solve that was to simply turn on the VIN to the sensors one after the other, and set the adress before the next one turns on. Unfortunately it seems like the sensors get power somehow even without any VIN connected. I'm guessing it's through the pullup resistors on the i2c bus. How can I solve this? Do I have to add something to break the i2c bus connection together with VIN, or are there other better ways? (Since the project they're intended for have limited free space, I'd prefer a solution with a low component count) <Q> So that's just a breakout board for this? <S> http://www.st.com/content/ccc/resource/technical/document/datasheet/c4/11/28/86/e6/26/44/b3/DM00112632.pdf/files/DM00112632.pdf/jcr:content/translations/en.DM00112632.pdf <S> If so, you can hold the GPIO0 input (the 0 on the breakout board) low to put the chip into a reset state. <S> This means you will need to dedicate an I/O line to each board, but that way you can bring them up one at a time without doing crazy things to the I2C bus itself. <A> This is a common problem, often referred to as "back-feeding". <S> Where an active signal can connected to a powered down board can upset its operation. <S> It can even cause the powered down circuitry to continue operating <S> The usual way to avoid this issue is to place some form of isolator between the main bus and the circuitry on the board, often they will accomplish level shifting at the same time. <S> There are commercial integrated solutions as well as ones using discrete devices. <S> This application note describes the operation as an isolator in section 2.3.4 on page 11 (It is not required to perform level shifting, the two power supplies can be the same voltage) Bidirectional Level Shifter/Isolator <A> Some years ago I had a similar problem with 6-axis movement sensing modules based on Invensense MPU6050 chips - they also "talks" through I2C bus. <S> My solution was rather "brute force" but works excellent: for every module I've made a simple circuit based on HCF4066 quad bilateral switches. <S> Two switches are used to switch the I2C lines, and another two connected in parallel are used to control supply voltage for MPU6050 modules. <S> simulate this circuit – <S> Schematic created using CircuitLab
| The usual culprit within the IC are the ESD structures used to avoid electrostatic damage - often they consist of diodes between the input signal and the supply rail.
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Is there an adapter for a TO-220 part to go into a TO-3 We have an old board that we do not want to redesign because it is near end of life. However, there is a TO-3 5V regulator that we can not source and when I do find it, it is for $13. I am wondering if someone makes an adapter to solder a TO-220 5V regulator to a TO-3 adapter as that would be the easiest solution. The next easiest is to just make a small pcb that does this which we may end up doing. <Q> I'm pretty sure I've seen this before: <S> Just cut the middle leg off the TO-220, bend the remaining leads 90° at the right distance, and mount the TO-220 to one of the existing TO-3 mounting holes. <S> It isn't hard to get the two remaining TO-220 legs to to line up with the TO-3 pins reasonably well. <A> Looks like someone makes a switch-mode replacement for a TO-3 5V linear regulator. <S> Will this work for you? <S> http://www.ezsbc.com/index.php/psu5.html#.WbmuWBNSyuU <S> If not, you can make your own using an off-the-shelf design from Linear, Maxim, etc. <A> One thing to strongly consider is the voltage regulator current capability. <S> Many TO-220 package regulators are limited to 1A maybe 1.5A at most. <S> A TO-3 regulator may have been used in the original design due to greater current capability. <S> In the past I've used TO-3 parts that were rated for 3A. <S> By the way I really like the idea that was proposed elsewhere in this question <S> /answer posting to make a circuit board adapter that uses a small SMT regulator to create a high efficiency buck switching regulator to replace the antiquated and inefficient linear regulator. <S> A high frequency PMIC type part with a small size inductor, a few caps and few resistors should easily be able to fit the outline and maybe even rid you of an existing heat sink. <A> This biggest issue is <S> max Pd depends on the heatsink on the board. <S> If none, then you may have to add one. <S> Consider that the TO220 has a thermal resistance from junction to case of 60'C/W and with 3W it will fry. <S> So you must define; Pd Max ( Vdrop*Imax) <S> Tj max ( usually 85'c ) for high MTBF <S> T amb max ( maximum internal ambient ) Air flow or convection restrictions ( helps or hurts Rth ) <S> space available around one mounting hole that is grounded for heatsink footprint and height fastener size ( may need smaller size ) <S> Then compute the max Rjc + Rca for a TO220 + heatsink to meet your above requirements. <S> Then consider one of these options in order to complete this simple exercise to prepare an ECN with assembly and BOM changes. <S> Use thermal resistance @ <S> natural and add 5'C/W for Rjc of the TO220 <S> (j=junction, c= case) <S> and you do the math. <S> Both use case ground, so bending the leads to reach the other 2 holes is easy at right angles with 2 needle nose pliers. <S> Then like Ohm's Law for voltage, compute the junction temp rise and heatsink Rca @ natural needed to not exceed 85'C ( or higher if approved) at max interior ambient temp. <S> Pd*(5'C/W + Rth_ca(heatsink)) <S> = <S> < (85'C-Tmax) (internal ambient) <S> solve for Rth_ca
| Installing a TO220 into a TO3 design is simple with a different size fastener for a smaller hole size and 5 in-lb max torque until you get into thermal design ( essential for any EE to learn)
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Why does the capacitor exist on this breakout board but not others that have same function? I am looking at using either of these parts to connect to a server power supply. I realize a capacitor helps to regulate voltage. Why would each of these parts that have similar function not both have a capacitor. That makes me think that a capacitor is not needed.Could it possibly be due to inductance. But it still does not answer why one has and one does not. Any input is appreciated: With capacitor: http://www.ebay.com/itm/DPS1200FB-A-Power-Supply-Breakout-Adapter-Board-10-PCIE-Cable-fr-Ethereum-Mining-/322705408187?hash=item4b22bdc0bb:g:sS4AAOSwcQlZqoHk Without Capacitor: http://www.ebay.com/itm/DPS-1200FB-QB-A-12-ports-Power-Supply-Breakout-Board-Adapter-For-Ethereum-Mining-/253139686137?hash=item3af04d12f9:g:hlAAAOSw23lZsQsL <Q> You mean this one? <S> The power supply and the devices it powers should both already have filtering capacitors of their own, so one more small capacitor placed this close to the power supply won't do anything useful. <A> There are dozens of server power supplies, and dozens of modifications, with extra voltages and without, with variety of standby voltages and power-on circuitry. <S> The two boards apparently have different design and layout. <S> What is so strange that one does have a cap, and another doesn't? <S> It could be some power-on delay, or delay of enable of some secondary voltage. <S> Without details and schematics it could be anything. <A> I haven't noticed any capacitors on either of these boards. <S> one shows molex type connectors only, while the other shows: The same molex type conectors, A dip switch (red plastic box with white levers) 4x 4-pin IC's (which look like optocoupler packages) Miscenaleous resistors One KK type header, probably 2.54mm pitch. <S> If you're wondering what the white rectangular marks on the silkscreen are, My guess is that they would be jumpers. <S> No capacitors on sight my friend.
| That capacitor looks pretty pointless to me.
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Getting Unwanted Transitional Values from Push Button Switch When I toggle a push button switch on and off, instead of getting a pure on/off continuity levels I sometimes get "transitional" values. I need it to be purely on/off with nothing in between. What components do I need to add to the switch to make that happen? There is +5v being applied at what is labeled "TIP". Here is am drawing of the original circuit I am trying to replace: Hopefully that helps explain what I am trying to do. <Q> Mechanical switch contacts "bounce" both when making and when breaking. <S> Bounce is when the contacts make partial, higher resistance contact or bang into each other repeatedly and very quickly. <S> It is common for contacts to make and break 5 or 10 or even 100 times in a few milliseconds. <S> Depending on the circuit the switch is driving, this can be taken as an intermediate voltage or as 100 really fast clock pulses. <S> Two common methods are 1) <S> any one of a group of circuits that essentially act as a lowpass filter; or 2) have the switch input clock a flipflop. <S> Once a "1" is clocked into the flipflop, multiple successive clocks have no affect. <S> With this method, you have to reset the flipflop before it will respond to another button press. <A> All you need is cap. <S> such that the RC= <S> T exceeds the bounce time using the large internal pullup R. <S> The contact closure gives a low RC attack time and the release a slow decay time to act as a Sample and hold for as many milliseconds to cover your transitional time. <S> Logic and software can also do the same thing with edge detection with a timeout. <S> After new info <S> But if you introduce a series string of binary weighted R's, the the MSB R weighted value will swamp the value of all the others. <S> You then have to deglitch by averaging or look for successive values in a 5% window of full scale for a duration > max transitional time. <A> The issue you are facing (or more accurately the solution) is called 'debouncing'. <S> Google it <S> and you will find answers. <S> When a physical switch closes, and applies a voltage that goes from (let's say) <S> ground to 10V (or whatever) <S> , at first there is intermittent conduction. <S> Think of it as microscopic sparking, until a solid circuit is formed. <S> Then it is solid at 10V. usually, a capacitor (with or without a resistor) across it will do the trick.
| The standard way to fix this is to "de-bounce" the switch with an electronic circuit. The only solution then is to use software averaging and choose analog thresholds to discriminate which combination of switches are active.
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Why motherboards have two pins for LEDs, led+ and led- Very often you can find that there are two signal pins (+) and (-) for the LED in the motherboard.1) Why not use one signal (+), and minus replace the ground?2) What signal can be considered a common (+) or (-)? <Q> If the LEDs are driven using the most common method, as shown below.... <S> you have no commons, no Vcc and no Ground. <S> If you have more than one LED you can not join any of those wires together either. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Note: <S> you could swap the LED and resistor to generate a common plus side. <S> However it is usually better to have the resistor on the high side to prevent inadvertent shorting of the Vcc to ground, or elsewhere, in a bad or faulty cable connection. <S> Also, if the cables are any significant length, attaching a long antenna directly to the power line adds EMI, EMC, and static discharge issues. <S> However, either way you would label the pins LED+ and LED- so whomever is creating the cable or attaching the LED knows which way round it goes. <A> In addition to the other answers, note that the format for connecting LEDs to computer motherboards became standardised quite a long time ago, long before the currently popular integrated case front panels. <S> When these connections became the de facto standard at some stage in the early 90s, the LEDs would literally be separate LEDs, mounted directly in the case with no connection between them. <S> This would probably be more expensive than just providing an extra couple of wires to support the 3 LEDs that most cases had back then. <A> Pin 1 is probably 5V and -ve is low with a series R to direct connect a LED. <S> just a SWAG, that you can easily test. <S> simulate this circuit – <S> Schematic created using CircuitLab
| In this case, using a common connection for multiple LEDs to the motherboard would be counterproductive: additional circuitry would be required in the case to connect the LEDs together in order to be able to use the common connection for all of them.
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will my linear regulator overheat? I am making a new design which I use a 4S 18650 batteries with total voltage of 14.8V in my design I need to have 9V and 5V using 7809 and 7805 regulators are these regulators going to overheat ? here is a schematic of my design <Q> 1 amp from a 5 volt linear regulator fed with a supply of 14.8 volts quite simply means the power dissipated by the device will be 1 x 9.8 volts = 9.8 watts. <S> Power dissipation in the device is defined by the volt drop across it (9.8 volts) and the 1 amp flowing through it. <S> I'd definitely consider a switching regulator - power efficiency is going to be about 90% so, with 5 watts out, there will be about 5.5 watts in and, from a 14.8 volt supply <S> this means it will take a current of about 375 mA i.e. a lot smaller drain on your battery. <A> 1A will be pulled from the 9V regulator only .. <S> the 5V regulator is used for sensors and low current applications.. <S> That is 5.8W of heat to dissipate in U1. <S> The L7809ABD2T that you are using is a surface mount part with a practical power limit of around 1W. <S> Therefore it will overheat, badly. <S> If you are to use a linear regulator you need to use the TO-220 part (L7809ABV) with a small heatsink. <S> The better solution is to use a small DC-DC switching regulator which is far more efficient, will dissipate far less heat and use less battery power. <S> If you like ST parts try a L5973A. <A> Assuming this is a hobby one-off, suggest you use an adjustable LM2596-based module for the 9V 1A supply and hang an LM7805 off the 9V input for the low current 5V supply. <S> You can get 150mA @5V without a heatsink conservatively.
| You may be OK using the L7805ACD2T for the 5V supply so long as you draw less than 100mA. You could also consider replacing that series diode with a MOSFET to reduce the voltage drop for reverse-voltage protection.
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Analog circuit to maintain a minimum current? I am looking to get a USB battery pack for use with both my phone and my bluetooth headset, but my bluetooth headset has a drain of only around 50mA when charging, and it seems that most battery packs have a minimum current draw of 50mA, below which they will shut off (presumably to prevent vampire currents from draining the battery pack). This is essentially the same problem posed in this question , but the answer there relies on the fact that the person asking the question is trying to power an Arduino (and can use the GPIO-boards to modify the current draw). Is there a more generic, preferably analog circuit that will provide a minimum current draw on a battery? I have a preliminary design using one op-amp and sense resistor as a current sensor, then using the voltage difference between the current sensor and a fixed voltage to drive an NPN transistor, but I think I've messed up the feedback somewhere, because once the load current drops below the threshold, the fallback current goes to the rails. See the circuit diagram: You can download the Qucs .sch file from this gist . My questions are: is there a better approach to an analog circuit that provides a minimum current? If so, what is it? If not, how do I fix the feedback on this so it provides exactly the desired current? Note: For simplicity's sake I used traditional op-amps with negative and positive rails, but once I nail down the general approach, I'm going to need to adjust this so that it can be powered with the battery itself, so only 0V and 5V rails. I don't think it should make a big difference, but it's important to know I suppose. Edit: For practical purposes, I think that this answer on a similar question (HT Ali Chen) actually makes a lot of sense - since the battery packs tend to stay on whenever the momentary current draw is above the specified value, it seems to me that some circuit with a low duty cycle would likely be the simplest and most power-efficient, but that can always be implemented in addition to the answers below, so that doesn't change this question. <Q> Add a Resistor in parallel. <S> For 50 mA at 5V you need a resistor of 5/0.05 = 100Ohm and it will dissipate 0.25W. <S> In other words a quarter Watt 100 Ohm resistor. <S> Though using 2 200Ohm resistors in parallel will reduce the power dissipation per resistor. <A> How about a 3 transistor solution? <A> Here's a poor man's 2-BJT version. <S> POT+Q1 set the voltage at the base of Q2. <S> (V(R2)-Vbe(Q2))/R1 sets the current. <S> This circuit, however, is too sensitive to V1 variations. <S> Furthermore (depending on the resistor values), there could be an excess current, for heavier loads (smaller YOUR_LOAD values in Ohm). <S> simulate this circuit – <S> Schematic created using CircuitLab <S> A better solution is the "not so poor man's" 6 components solution <S> (capacitors excluded).Some remarks: <S> You must use an OP amp capable of going down to 0V (both input or output). <S> You can change it so that it regulates the positive rails (you must use another op-amp and modify the circuit accordingly) <S> You need an OP amp capable of driving 50mA. (LM358 can't so this circuit will work only if you load has a minimum consumption). <S> You can add an output npn BJT or MOSFET (toward Vdd), instead of changing OP amp. <S> Watch out for OP Amp (or BJT/MOSFET) power dissipation! <S> A decoupling capacitor is recommended. <S> simulate this circuit <A> Use a high side current sense chip like the LTC6101: - And use a comparator, transistor and resistor as mentioned above in red.
| A filter capacitor is recommended between the non inverting input and ground.
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When do I connect to negative terminal or ground? Easy question (or at least I think). I've installed a siren in my vehicle which calls to pull power direct from the 12 volt battery. I currently have the circuit completed by connecting to the negative terminal on the battery. I've been told that sometimes it is best to connect to a grounding point on the vehicle like the frame. Is this necessary for my situation or ever? Thanks in advanced! <Q> You should avoid connecting cables directly to your battery. <S> There's no point at all. <S> There's no reason to have multiple cables on your battery's terminal. <S> Cable harness will be more organised. <A> In a car the amount of iron (even though not as conductive as copper) is so substantial that a short wire to the nearest chassis point is the best (lowest resistance) ground for an accessory in almost all cases. <S> Running a wire to the battery minus is also fine for low current loads as well. <S> High current and long earth wire can cause earth potential issues but <S> these seldom matter unless there is a further connection to some other device at the end of the long power cables. <A> The negative of the battery is connected to the chassis so run your postive to the battery (make sure it is fused) and connect the negative of your siren to the chassis. <S> Look for any bolt nearby, use a ring terminal and fix it down. <S> However do make sure that the chassis surface is bare metal (sand it down), any significant amount of paint will result in a bad connection and cause problems.
| The negative terminal of the battery is connected to your car body , so you can connect your siren's negative terminal directly to your closest ground point (car metal body).
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How do i measure watts on single phase power? With a clamp on amp meter, I have 8.2 amps at 123.6 volts on one leg and 7.2 amps at 123.9 volts on the 2nd leg. When I touch the volt meter to both legs I get 247.8 volts. What is the total wattage? I get 3816 watts total if I add the amps together and divide by 247.8 or 1905.5 total if I calculate the watts (v x a) for each leg and add together. What is the correct way to do this. The measurement was taken from the two main wires that feed the house panel box from the transformer, so everything that was on in the house was showing in the measurment, there can be an amp difference since single pole breakers only draw from one of the wires. <Q> What is the correct way to do this? <S> The correct way to measure AC power is using a wattmeter. <S> This is because you cannot know (or reasonably estimate) <S> the phase angle between voltage and current and measuring amps and volts and simply multiplying them together gives a falsely high reading in all but simple cases of a purely resistive load. <S> So, if you had two wattmeters you can measure the wattage entering one circuit to neutral and also measure the wattage entering the 2nd circuit with respect to neutral. <S> If you loads are resistive the powers will be: - 8.2 <S> x 123.6 = 1013 watts <S> 7.3 x 123.9 = 892 watts <S> Total power (if resistive load) = <S> 1905 watts. <S> However, if the load is resistive AND non-linear there will be harmonics in the current waveform that don't contribute to power loss but will register on an RMS measuring current meter <S> so here's another error source. <A> Residential Power Meters <S> only the real component power a product of only the RMS V*I that are in phase. <S> This method cannot measure phase or power factor and thus only measures the apparent power not the "real power." <S> The phase of the voltages in your case is assumed to be split phase or 180 deg. <S> The neutral current may account for the difference in L1 and L2 or not if each current has a different phase shift. <S> There are other sources of errors with out of phase harmonics. <A> utilizing a one phase transformer with a central common--neutral-- terminal in order to get 110/220 VAC <S> the aproximate total power is: P =( <S> V max) <S> I (prom). <S> I(prom)=(I(line <S> A)+ I(line B))/2. <S> If you are using standard meters -calibrated for sinusoidal signals- the difference between line A and line B is normal because of the electrical connection of neutral and ground but is really harmful if major to 10%.(this disrupts and distorts the sinusoidal low voltage signal) <S> We are talking about VA -reactive power-. <S> Watt or active powerdepends on power factor: <S> W= VA (power factor), and oscillates between 0.7 and 0.87 for AC equipment-HVAC- and one phase motors. <S> You can add a capacitive bank to raise power factor. <S> Check your electrical bill and note if a charge exists by this--low power factor-.
| Best way - use a proper wattmeter.
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Why exactly is it discouraged to gap the ground plane? From time to time I hear (and read) that it is not good to make separate Gnd planes for digital and analog circuit parts. It's all summarized in this rule of thumb: "Don't split the Gnd plane, don't make gaps in it." Usually this comes without clear explanation. The closest I got to an explanation is this link: http://www.hottconsultants.com/techtips/tips-slots.html .The author points out that return currents will bend around the gap, such that the surface areas of the currents get large (borders of that surface area are defined by 'departing' and 'returning' current): The return currents of the different signals are squeezed together at the corners of the gap, leading to cross-talk. The larger surface area of the current loops will emit and pick up EMC. So far, so good. I do understand that no signals should be routed over such gap. Presuming you keep that rule in mind, would it still be bad to make gaps in the Gnd plane (eg. making a split between analog and digital circuit parts)? <Q> High frequency return currents want to follow the outward currents due to inductance. <S> If you force the return currents to take a different path then a couple of bad things happen. <S> You create a loop that can receive and transmit magnetic interference. <S> You introduce extra inductance in the signal path which can reduce signal integrity. <S> Note that digital signals with fast edges can produce strong high frequency spikes even if the switching rate is low. <S> Note also that the outward path may not always involve just tracks, it may be inside a component. <S> Even if a component has separate analogue and digital power and ground pins there are likely to be some signals crossing over the boundary inside the chip. <S> OTOH at low frequencies currents take paths determined primerally by resistance. <S> So splitting planes can be a useful technique to influence the path return currents take and avoid shared impedance. <S> If you have exactly one place where signals cross the mixed-signal boundry then splitting the plane makes a lot of sense, it forces <S> analog return currents to stay on the analog side and digital return currents to stay on the digitial side. <S> If you have multiple places where signals need to cross the mixed signal boundry (i.e. multiple ADCs, multiple analog switch chips etc) then the benefits of splitting get much more questionable. <S> Each mixed signal chip needs a connection between the two planes but once you put multiple connections between the planes you lose a lot of the benefits of splitting them in the first place. <A> The reasoning is very similar to the trend away from split grounds for digital & analogue. <S> Its all about return current <S> There has actually been a trend away from split ground planes and instead concentrating on placement separation AND consideration for the return current path. <S> Use large area groundplanes for low impedance current return paths <S> Keep over 75% boardarea for the ground plane Separate analog and digital power planes <S> Use solid ground planes next to power planes <S> Locate <S> all analoguecomponents and lines over the analogue power plane and all digitalcomponents and lines over the digital power plane Do not routetraces over the split in the power planes, unless if traces that mustgo over the power plane split must be on layers adjacent to the solidground plane <S> Think about where and how the ground return currentsare actually flowing Partition your PCB with separate analog anddigital sections Place components properly <S> Mixed-signal design checklist <S> Partition <S> your PCB with separate analog and digital sections. <S> Place components properly. <S> Straddle <S> the partition with the A/D converters. <S> Do not split the ground plane. <S> Use one solid plane under both analog and digital sections of the board. <S> Route digital signals only in the digital section of the board. <S> This applies to all layers. <S> Route analog signals only in the analog section of the board. <S> This applies to all layers. <S> Separate analog and digital power planes. <S> Do not route traces over the split in the power planes. <S> Traces that must go over the power plane split must be on layers adjacent to the solid ground plane. <S> Think about where and how the ground return currents are actually flowing. <S> Use routing discipline. <S> Remember the key to a successful PCB layout is partitioning and the use of routing discipline, not the isolation of ground planes. <S> It is almost always better to have only a single reference plane (ground) for your system. <S> (pasted from the below links for archiving) <S> www.e2v.com/content/uploads/2014/09/Board-Layout.pdf <S> http://www.hottconsultants.com/pdf_files/june2001pcd_mixedsignal.pdf <A> The #1 priority is placing stuff in the right place on your board. <S> For example, if you have the power entry connector on the left, the motor controller and its output connectors on the right, and the sensitive analog bits in the middle, you're off to a bad start. <S> Better place the power connector right next to the high current outputs, which makes high currents flow naturally in a manner that makes your job easier. <S> Also the best IMO is to use split planes (AGND, DGND), then place all components on the corresponding plane then, in the end... <S> remove the split and turn it into a solid ground plane. <S> This forces you to make a good placement. <S> For the rest, this question is more or less the same, I advise reading the answers. <A> This is a difficult topic often with contradictory information. <S> One common example where this comes up is when laying out the copper for analog to digital converters. <S> Often the datasheets specify keeping the analog ground return separate from the digital portion and only tying them together at one point. <S> The datasheets often specify that specified accuracy can only be achieved when the chip is grounded in this way. <S> If the entire board was one AtoD chip then this would be easy but when you start mixing DtoA's, Op amps, comparators, and digital circuits, this quickly becomes impractical. <S> I won't rehash what others have said about good layout practices. <S> Similar to resistors in parallel, current will flow in the path of least resistance. <S> At high frequency, the inductance of boards can contribute significant reactance. <S> The path of least reactance for the return current would be right underneath the signal trace in the ground plane. <S> When there are gaps in the ground plane, the return current has to take a longer path back to the source which results in a larger loop and higher inductance. <S> For more detailed information on this subject, I would recommend Electromagnetic Compatibility Engineering by Henry W. Ott. <S> It is the bible on EMC.
| Do not split the ground plane, use one solid plane under bothanalog and digital sections of the board
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What is the practical advantage of using mirrored signal outputs eventhough what matters is balancing the wires? Some devices output differential outputs since both wires carry varying signals which are mirrored. Most devices on the other hand output one signal carrying and one GND wire. There is also differential-ended wiring and single-ended wiring. My question will be about "external" interference type of noise(EMI or RFI or maybe capacitive coupling) impinging upon both wires. After some reading, if I'm not wrong I came to understand the fact that what we really want is "the same noise should appear upon each line" regardless of it is a differential or single ended wiring. And the noise must be common(common mode) so that the differential amplifier can cancel it out. As far as I understand, there are two main ways to make the noise common. One of them is twisted pairing. The idea here is the wires are very close to each other because they are twisted and the same magnetic filed or electric field hits upon both wires. The second method to make the external noise common is to make the total output impedance(from source to receiving end) of both wires equal, so if an EMI happens both wires would induce the same voltage at the same time. Upto here was just a summary of my understanding to show/check to whom read this. Here is a simple illustration of a single-ended and a differential-ended wiring: If they are both using twisted wiring and they are all balanced, they would both reject the common mode noise. So what is the point to send the signals in mirrored fashion if one can balance and twist a single-ended wiring as well? Is that because something is practically easy to establish in differential/mirrored outputs? I really couldn't find a good answer to this. edit illustration for a comment: <Q> As has been mentioned by peufeu, if you use balanced differential drivers you create twice the p-p voltage and hence the SNR is doubled because noise can be assumed to remain as it was previously. <S> But, there is another subtlety that was alluded to in the question and this relates to impedance balance. <S> Quite simply, if the balanced driver had an output impedance of 20 ohm (ignoring the output feed resistor R5) <S> you would need to add 20 ohm to the balancing drive resistor (R6) that connects to 0 volts to keep balance. <S> But it's a bit harder than this because the output impedance of a driver chip will likely vary across its bandwidth <S> so, it's much easier to use a dual inverting driver and equal values for R1 and R2 (right hand diagram). <A> Good answer are already posted but I would also add another point. <S> Single ended systems usually rely on ground as the differential signal. <S> That is fine if the connection is short and you can guarantee the grounds at both ends are the same <S> and you do not have too many of them. <S> Differential transmission removes this reliance and provides some degree of ground isolation and prevents ground loops. <S> Also, from your question you seem to imply that balancing the impedances and using twisted pairs are alternative solutions. <S> They are not, for best performance you need to do both. <S> In fact, if you don't balance the impedances, using twisted pair is almost pointless. <S> Also you might find my answer here interesting. <A> So what is the point to send the signals in mirrored fashion if one can balance and twist a single-ended wiring as well? <S> 2x <S> more voltage swing, thus dynamic range and SNR doubles. <S> For high-speed signals, driving the cable differentially also reduces the amount of radiated noise. <S> On a PCB, it keeps return currents out of the ground plane. <S> For logic signals (LVDS) it also reduces the amount of noise and ground bounce at the driver side, by making the power supply current of the driver relatively constant and independent of signal. <S> Consider USB3: <S> if the drive wasn't differential, you'd need 2x more voltage swing for the same noise margin, also you'd need more elaborate common mode filter (ferrites) to reduce radiated noise.
| On the receiver side, equal and opposite signals cancel as they couple through parasitic caps, so this also reduces noise.
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Isolated way to detect 240V contact closing I want to create a small project with an ESP8266 that makes a server call when a thermostat is open or closed. My main problem is detecting the thermostat status. The electro-mechanical thermostat has two wires going to it, one 220V line and a return line. When closed, the switch passes 200mA 20mA through it, to keep a relay closed. There is no neutral line available. I don't have access to the relay. For safety and legal reasons I want to have the ESP8266 battery-powered. I could be smart and "leech off" current but charging a battery once a month is not that bad. I also want the detection to be isolated and non-contact, if possible. One thing I could do is: add a kind of light indicator and use a light sensor to detect that. Not sure if I could make it light in the on-state, but should be straightforward to do for the off-state, with a resistor, LED and diode. I don't have enough place inside the thermostat housing for a complicated circuit and I want to keep the ESP8266 module outside the thermostat housing, as a clip-on solution. TL;DR: an easy non-contact way of detecting whether there is mains voltage without a neutral wire, to read by ADC or digital logic. Edit: Inside one of the thermostats there is a neutral wire available. I guess with a little digging the other one has it as well. The available space inside the thermostat is around 20x20x20mm, so a module like TSP-03 would not fit. I might switch the scope of the project from contactless to low-voltage, since it's a bit more reliable and easier to use. The idea is to provide two pins, that either have a voltage difference or close a contact (optocoupler). Whatever choice, it has to fit inside that enclosure and provide a safe output. Edit2: FYI: The relay is in a remote location (fuse box) and triggers a 3-Phase heating element. No use going that route, even if I could.The uploaded picture shows the insides of the thermostat. The available space is on the left side. The outside dimensions are 65x65x30mm.Another solution would be to completely replace the thermostat with another model that provides WiFi or access to the signal. A very comprehensive discussion about a similar topic can be found here: https://www.raspberrypi.org/forums/viewtopic.php?f=45&t=9617&start=25 as well as throughout StackOverflow. I've read most of those. <Q> If the current passing through the thermostat is AC i.e. 50 or 60 Hz then you could use a small clamp-on current transformer. <S> The current flow will produce a voltage output across a burden resistor and this can be simply amplified to generate a "detect_current" signal. <S> but it's costlier and a little more tricky sometimes. <A> You could perhaps use a hall sensor to detect the magnetic field of the relay. <A> You could use something like this (connected in series with the thermostat contact). <S> BR1 can be a 1A bridge. <S> C1 something like 1000uF/6.3v. <S> R1 is chosen so that the 200mA current provides a 5V drop from R1 in parallel with the relay coil resistance. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You could use an optoisolator instead of the mechanical relay. <S> There is plenty of current and enough voltage. <S> Reduce R1 to 30 ohms and put a 130 ohm resistor in series with the LED. <S> You can use something cheap like a 4N35. <S> R1 will need to be rated at 2W. <S> If the relay current does not turn out to be exactly 200mA you may need to adjust the value of R1. <S> Edit: You have revised the problem description by reducing the relay current from 200mA to 20mA. That is still enough to operate a low current relay, such as the V23026D1021B201 or an optoisolator. <S> Modify the value of R1 appropriately, and C1 can be reduced to 220uF or so.
| If the current is DC then use a hall-effect sensor (which you could still use if the current is AC)
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Convert an analog voltage into digital and relay it in serial over a single wire I have Engineering background but not Electrical. I am working on a project where I have to integrate a sensor which has an analog voltage output to a LoPy module for IoT applications. Another person is taking care of the LoPy module stuff but I do need to provide him the output in 8-bit digital serial form over a single connection so that it can act as an input to the module (I have been told that that's how the input is provided to the module). Is there a way to achieve this without using microcontrollers? EDIT 1: I meant 2 wires. I understand that a ground wire would also be needed. <Q> It's a bit unclear, but it seems you have a analog voltage from a sensor, and need to send that information in 8 bit digital form to a module using whatever protocol that module requires. <S> The first thing to do is to get the specs for what exactly that protocol is. <S> That includes the low level electrical interface (IIC, SPI, UART, etc), and exactly how the bits are transmitted. <S> Only after you know that does it make sense to decide what you need between the analog voltage <S> you are presented and the digital information you need to send. <S> Trying to avoid using a microcontroller for some strange reason is just plain silly. <S> That's quite obviously what this is going to take, at least without a ridiculous and cumbersome pile of hardware from the Pleistocene. <S> The micro reads the A/D and sends on the data digitally. <S> The flexibility of a micro lets you do additional things that might be useful. <S> For example, you might want to apply a little low pass filtering on the digital values from the A/D. <S> It might also be useful to not just map the whole analog range to the 0-255 digital range. <S> You can trivially add a offset and apply a gain to map a part of the analog range to the digital range. <S> Non-linear mappings can also be done. <S> Again though, before you do anything else, get proper specs for the digital interface you must send to. <S> Note that "single connection" might not be possible. <S> If "connection" means ground and a signal wire, then that is doable, depending on what the digital interface is. <S> If you are thinking a single wire, then you need to re-think. <A> According to the datasheet of LoPy it has "Analog channels: 8x12 bit ADCs" <S> so it will read analog voltages without trouble. <S> Depending on your sensor, you may need some amplification, conditioning, etc, or maybe not. <S> More info about the sensor? <S> And, you can't communicate over one wire. <S> You need at least one wire and ground, so two wires (or a cable)... <A> There are several approaches of serial data bus ( half-duplex) depending on number of slaves and bandwidth; GPIO (1pin), SPI(3pin) , I2C, (2pin) <S> GPIO <S> Pros: <S> Simple, Requires a single pin, can operate from 1.8 <S> ~ 5V <S> Cons: <S> Not good for sending complex data , slow ~ <S> 1kbd, short haul <S> aka: bit banger Needs signal conditioning, shielding , twisted pairs with power and ground if remote power needed. <A> You can buy ADCs that implement 2-wire serial interfaces, among many others. <S> For example, TI's site allows you to filter ADC parts by interface , including UART and 12C. <S> You need to find out from your colleague what this '8-bit digital serial form' is. <S> It could refer to any number of protocols. <S> It could even refer to the AMC1304-Q1's odd but neat time-averaged 1-bit stream encoding. <S> Its impossible to identify a part with any but the most generic capabilities (i.e. a microcontroller) without knowing. <A> Could this just be a simple V/F converter? <S> Something like the KA331 or the LM2907 converts your voltage into a frequency and at <S> 0.3% accuracy is just better than 8 bits. <S> You'll only need one signal line. <S> Of course you'll need another line to power the far end and a ground. <S> Use the pulse count function of the ESP32 to determine the frequency. <S> You don't mention the frequency response that you require for analog voltage. <S> The low operating frequency of the F/V means that your sensor voltage can't vary quickly. <S> There is a bit of code on Git (esp32-freqcount) that facilitates accurately measuring the frequency of square pulses on a GPIO using Pulse Counter, RMT and Interrupt.
| Many many micros are available that have integrated A/D converters, and integrated hardware to help send on the data digitally, like IIC, SPI, or UART peripherals.
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Resistors before every led? I'm working on lighting up an entire LEGO city from my nephew. But the problem is I need a lot of leds with different amperes and voltages ( I'm gonna put them in parallel). So my 2 questions are : Do I need to put a resistor before every led? or can I just put a resistor in front of a bunch of leds ( I have around 15-25 leds per resistor)? I need for most of my leds about the same value of the resistor (around 150 Ohms). Can I put the resistor of exact 150 Ohms (1% tolerance) in front of the leds? or do I need to set the value a little bit higher with an extra resistor? EDIT: also in the comments: 41 yellow and red leds have 2 volts and 20mAAbout 30 white leds have 3.5 volts and 20mA and about 10-20 leds have 2.6 volts and 70mA <Q> You should use a resitor for every series string of LEDs. <S> Put as many LEDs in series as can be driven by the available supply voltage, minus a little. <S> Then size the resistor so that you get the right current thru the string. <S> Only string together LEDs that you want to run at the same current. <S> For example, let's say you have a 12 V power supply. <S> You have a bunch of LEDs that you want to run at 20 mA, and that drop about 2 V in the process. <S> 6 in series would come out to 12 V, leaving nothing for the resistor. <S> Therefore, you'd put 5 in series. <S> Those should drop 10 V. <S> The resistor drops the remaining 2 V. <S> You find the resistor value by Ohms law. <S> (2 V)/(20 mA) = <S> 100 Ω. <S> Since 20 mA is the maximum spec, and you therefore don't want to exceed it, the next larger common size of of 110 Ω would be appropriate. <A> LEDs voltage at rated current has a rather large dispersion, so if you wire them in parallel, some will be a lot brighter than others. <S> It is better to start with a high enough yet save voltage (like 12-24V) and wire the LEDs in series with one resistor per group. <A> You can put leds with the same current requirement in series and use a single resistor or current regulator to drop the last bit of voltage. <S> Though the total voltage drop over the leds may then become too large to be safe. <S> (keep it under 30V).
| You can also put sets of leds with the same forward total voltage drop in parallel and sum up the current of each set and size the resistor to that.
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How to identifiy this "2222 063 90025" vintage capacitor? I am restoring an old stereo valve amp and would like to test (and replace if needed) some components. I don't know the history of the amp, but it's made in New Zealand and looks to be from the 1960's or 70's, but I could be wrong. There are a couple of capacitors that need identifying. The first and larger one is clearly marked 50 \$\mu\$F, however, it has three legs. What I would like to know about this capacitor is: Do the three legs signify a 2-in-1 capacitor? What does the product code of 2222 063 90025 signify (tried here , but no luck)? What do the triangle, square and circle symbols indicate? How do replace this capacitor? Capactior product code: Capacitor from below, with 3 pins: <Q> The three capacitors share a common negative terminal (the can) which you will observe is grounded by virtue of being bolted to the chassis. <S> I see soldered lugs on the can to reinforce that connection. <S> Commonly used as the power supply reservoir capacitor in a vacuum tube amplifier. <S> Note that one section has a lower rated voltage <S> : you'll see it's connected to the others via a high-ish resistance, to provide a well filtered (quiet) supply to the sensitive preamp stages. <S> Circuitry will be easy to trace out and will look roughly as below... simulate this circuit – <S> Schematic created using CircuitLab <S> If you can't find a replacement, you can use three separete 50uF 450V capacitors. <S> Bonus points for assembling them into a lookalike can... <S> (A bit more reading suggests this was a Philips part number) <A> This appears to be what is known as a "Triple Section Electrolytic Capacitor" which seems to be a fairly apt name. <S> The three legs have symbol stamped into them which correspond to each of the three markings on the capacitor. <S> The shell is presumably then connected as the common terminal. <S> Essentially it appears to be three capacitors wired up in a Wye configuration, each 50uF in capacitance. <S> Two are rated for 400V, and the third rated for 350V. <S> Although the third one having a different rating seems to be weird. <A> It is a twist lock multi section capacitor. <S> Seldom designed into new equipment but pretty common in still functioning and restored valve (audio) gear. <S> You could try and see if your circuit could cope with one of the replacement types from TEDSS or from Amplified Parts , failing which you could insert 3 x 50uf 400V contemporary radial electrolytic capacitors under the chassis or into the gutted old can. <S> The can will be the common negative terminal.
| It's a three section electrolytic capacitor, with the three symbols indicating which pin is associated with each capacitor positive lead.
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Solar Panel not heating heat 30 AWG SS304 Wire i try to make simple wire heating using solar panel.Problem is the wire not heating at all.I assembly solar cells 52mm*26mm in parallel 11.38 volt, amperage should be at least 2 or 3 amps. Some experiments that i have tried using Solar Panel assembly as power source, 11.38v 775Motor: 0.09V, 0.11A, motor not running. ToyMotor: 0.34V, 0.1A, motot not running. SMD LED: 2.83V, 0.1A Cree LED: 2.56V, 0.14A Single 30AWG SS304: 0.74V, 0.1A, 7.1Ohm, not heating. Multiple 30AWG SS304: 0.05V, 0.11A, 1,4Ohm, not heating. <Q> Since motor resistance (DCR) starts at 10% of full power impedance, its a poor match without a battery. <S> An impedance matching MPPT optimizes the charge current according to input solar power by tracking or some other algorithm based on Voc and/or Isc. <S> Consider solar power as a current source, limited by some Voc voltage just the opposite of a zener which acts like a voltage sink. <S> Both have a knee in their curve and ideal for PV is 72% to 82%Voc from morning to mid-day full direct sun, with some variations for PV chemistry etc. <S> Looks like a smoggy day or overcast at least. <S> Try a battery like with a Schottky diode to protect the PV so you can run 3 White Power LEDs in series with a series R or LM317 current limiter at night. <S> WHite LEDs are like 3V zeners. <S> The higher the power rating, the lower the Zzt where Zzt or ESR ~1/ <S> Pd rated. <A> You show two different wire configurations, one at .74 V and .11 A, and another at .05 V and .10 A. <S> The fact that the current is essentially the same in both cases is an indication that your measurements are correct, since photodiodes (of which a solar cell is an example) act as current sources when shorted. <S> Now calculate the power dissipated in the wires. <S> .74 <S> V <S> x .11 <S> A equals about 81 mW, while .05 <S> V <S> x .1 <S> A equals 5 mW. <S> The ratio, 16 should be equal to the ratio of the resistances, which is 5 in this case. <S> I suspect you failed to zero out the probe resistance. <S> Now. <S> Look at the actual powers. <S> 80 mW should be detectable, but 5 mW - not so much. <S> So, how are you "detecting" your temperatures? <S> I doubt you're using a thermal camera, and any physical thermometer may well be too big a heat sink to allow easy measurement. <S> Assuming your power source is well-isolated, you might try letting the wire run for a few seconds, then touch it to your (dry) lips, which are very sensitive. <S> Note that, if you do this, the temperature will almost immediately seem to cool to neutral temperature - again, heat sink effects. <S> Of course, this requires absolute certainty that your power supply is isolated, and is not remotely a good idea if you don't know what this means. <A> You have built a panel with 23 solar cells in series, and 6 in parallel. <S> For each cell of 52x26mm in size, you have total of 0.187 m^2 of area. <S> The cells you seem to be using are coming from eBay China, they claim 17.8% efficiency of their products (which seems to be reasonable in 2017 technology). <S> Now, assuming that the panel is under direct sunlight in absolutely clear sky day and oriented perpendicular to the direction of light, the incoming power will be about 187 W, and, after the assumed efficiency of 17.8% this panel should produce about 33 W at about 0.5*23 <S> = 11.5 V. <S> Since there are serious indications in your pictures that your panel is not under direct sun, assumptions about how much sun does your panel get can vary wildly. <S> It can be minimum 1/10th of direct sun, so you should expect no more than 3 W out of this conditions. <S> Your measurements with LED indicate that you have about 0.35W, which is pretty normal for indoor light conditions (1/100th of direct sunlight). <S> Try to get your panel into afternoon direct sunlight, maybe you will get a better results, assuming that your soldering is good, and wires are think enough (which doesn't look true from the pictures).
| THe maximum power will be about 80% of the short circuit current and 80% of the open circuit voltage, ( more or less) at whatever ratio of resistance that works out to be. The error is probably due to resistance measurement error using your meter.
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Capacitance on the output of an op amp I have read a lot about capacitive loads on the output of an op-amp and the possibility of unwanted oscillations and instability. my knowledge on phase margin and frequency domain analysis is a little weak and I can't figure out if my circuit bellow is gonna be unstable or not: The Op-amp supply is connected to 5vdc and GND. and here is it's datasheet: http://www.st.com/content/ccc/resource/technical/document/datasheet/bd/fc/46/43/26/8f/40/7f/CD00001046.pdf/files/CD00001046.pdf/jcr:content/translations/en.CD00001046.pdf I read in some application note that placing R19 would help the stability issue by increasing phase margin or something like that and fortunately I already had that in my circuit. but as you can see C10 capacitor also has a very high capacitance value. is there an analysis or rule of thumb which would determine if this circuit is safe? should I change my op-amp IC? or should I use other methods as well to insure my circuit's safe operation? I can provide additional information if you need it. Thanks. <Q> You do not have this issue as long as R19 is present and larger than 500 ohms or something in that order. <S> When an opamp has a large capacitor to ground or supply directly at the output and the output is also used in a feedback configuration <S> then instability can occur or the circuit might even work as an oscillator. <S> Both opamps do have feedback loops but the large capacitor C10 is not directly on any opamp output <S> so I do not see any potential stability issues with this circuit. <S> Here R19 saves your day as it separates the opamp circuit from the large capacitor C10. <A> Most OAs have limited capacitive load capabilities. <S> The origin is quite intuitive. <S> Consider the simple circuit below. <S> On the left there is a real OA. <S> On the right, the OA has the same characteristics (poles, etc.) <S> but the output resistance is externally shown. <S> This will have the effect of creating a pole, with time constant \$R_{out}\cdot C\$. <S> Such pole will reduce your phase margin of your system. <S> Limiting case: <S> C is so large that it can be considered a short. <S> Then you're actually removing the feedback, and you get a comparator (assuming non zero feedback Z)! <S> Instead, if there is a "large enough" resistor in series to C (R19 in your circuit), the series R-C will add a pole (with time constant \$(R_{out}+R_{19})\cdot C\$ ) and a zero (with time constant \$R_{19}\cdot <S> C\$ ) <S> .If, as said, R19 is large enough, the pole and zero time constants are close enough and the overall effect is negligible (limiting case: R19 is infinity). <S> Intuitively, after the zero frequency, the series R 19 -C will act as a load, and it will not open the feedback (it will change the feedback attenuation factor, but if \$R_{19}>>R_{out}\$ then this variation will be negligible as well). <S> EDIT: <S> In your case, the circuit will be stable. <S> EDIT2: <S> the following consideration is no more valid with the new edits to the schematics made by the OP: <S> I assume that "Input Signal" in your schematics is a current signal (i.e. not a connected voltage source), otherwise the circuit won't work. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Tim Green wrote a quite nice and hands-on tutorial on stability of opamps. <S> There are very little formulas and he tries to give an intuition by using lots of graphs and drawings. <A> Lets cut through the hand waving, and provide a specific answer to Rseries, so as to dampen the ringing. <S> OpAmps have excellent Rout at low frequencies, with that Rout increasing continually above the open loop <S> gain rolloff frequency, often 10Hz or 100Hz. <S> The increase in Rout also shows a 90 degree phase shift. <S> Result is the OpAmp has inductive Zout, above that 10Hz or 100Hz corner frequency. <S> That inductance, given a capacitance on Vout, will ring unless adequately dampened. <S> Can we predict Lout (output inductance)? <S> Some opamps provide a chart of Rout versus freq; if so, grab the Rout at UGBW (you may have to extrapolate), and compute the Rout = 2 <S> * pi * <S> Freq <S> * L, solving for L. <S> An opamp with 100 ohm Rout at 1MHz (many ordinary opamps), has inductance of L = 100 / ( 2 * pi * 1e+6) <S> = <S> 16 microHenry. <S> How to dampen? <S> Use the formula Rdampen = <S> sqrt(L / C), derived from consideringthe various equations for Q=1 for R and L, and R and C, and combining to eliminate Frequency as a variable. <S> Given 1uF Cload, and 16uH output inductance, the proper Rdamp is sqrt(16/1 <S> ) = 4 ohms.
| Some opamps can have an issue when driving a capacitor to ground or supply directly meaning, without any additional series resistance.
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Why is the positive terminal of the DC source connected to the primary of the transformer? Why is the positive terminal of the DC source connected to the primary of the transformer? Link to random inverter circuit I found online <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> (a) SW1 conducting. <S> (b) SW2 conducting. <S> Figure 1 shows a simplified version of your inverter circuit with switches replacing the transistors. <S> When SW1 is closed current flows through the centre-tap through SW1 inducing a voltage on the secondary as shown by the arrow. <S> When SW1 is opened and SW2 is closed a similar current flows in the other half of the primary winding but note that the current direction in the coil has now been reversed. <S> The induced voltage on the secondary is now also reversed. <S> If this cycle is repeated the phase reversal on the secondary will generate alternating current, AC, which is the purpose of your circuit. <S> Why positive terminal of dc source is connected to primary of transformer? <A> The centertap of the transformer is connected to the positive supply because this is a push-pull configuration. <S> It is a circuit which can make an AC voltage from a DC supply. <S> Only one of the two NMOS transistors is on at any time. <S> When the upper NMOS is conducting (on) the lower one is off etc. <S> When the upper NMOS is conducting the current through the primary side of the transformer will introduce a magnetic flux in a certain direction. <S> Then the upper NMOS is switched off while the lower is switched on. <S> This causes the current in the primary side of the transformer to reverse direction and also the magnetic flux will reverse direction . <S> This introduces an AC voltage at the secondary side of the transformer. <A> Because the Drain of the N-channel FET transistors are switching on and off by turn and pass a negative source to transformer. <S> That`s why a positive source is required as primary connected to transformer.
| Because the transistor switches are connected to negative and a positive connection is the only way to get current to flow.
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Switching a CAN bus I am designing the controller for a needle test station intended to do functional tests of a 20-PCB panel in a production line. The controller has to sequentially interface over CAN to 20 devices mounting a MAX3051EKA+T transceiver. Do you have any idea of an elegant way to switch the CAN bus connecting the controller to the 20 devices? Would it be possible to use some kind of simple MOSFET-array-based solution? I'd rather not using relays. <Q> You can use repeaters then disable the transceivers to the nodes you don't want to talk with. <S> There are tranceivers with standby mode. <S> A repeater is as simple as two tranceivers. <S> ( source ) <S> If you use DIP for the test board side transceiver, you could quickly fix damaged tranceivers due to fatal failures of test boards. <S> Note that repeaters limit the maximum length due to latency. <S> - If one node is shorted, the entire panel fails without you knowing on what board the problem is. <S> The solution ready in the shortest time is a bunch or small dual pole relays. <S> The repeaters would require a custom board, or a large investment in off-the-shelf repeaters. <A> The repeater above is a fine electronic solution if you choose transceivers which can be disabled, but as you mentioned in the comment, can become costly and <S> I'm not sure they give you any real advantage over what I'm about to suggest below. <S> What is wrong with a pair of analog switches or two 8:1 analog <S> MUXes? <S> It would seem this would achieve logical disconnection and minimize parts, although if you absolutely need physical separation (e.g. you're actually trying to match network capacitive loading or otherwise doing more serious testing) <S> then relays are probably the best way to go. <S> If not, though, keep in mind that you'll have to worry about voltage levels, so go for the older CMOS technology devices. <S> Speed really isn't an issue here, and the devices are plentiful, offered by many vendors, and best of all, cheap . <S> A pair of MC14051s, for example, would give you 8 DUTs to a single CAN bus on the tester. <S> CD4067s would do 16:1. <S> These may even work if you need stronger separation (see above) <S> but it sounds like you're after functional testing. <S> Two or three CAN buses (depending on the tester) could handle 16, 24, 32 or even 48 DUTs at a time. <A> Some options: Use relays. <S> This seems like the conceptually simplest solution. <S> Most likely the few 10s of ms time for switching between different boards is much less than the test time per board, so shouldn't slow things down. <S> Multiplex on the other side of the CAN bus transceivers. <S> This means dedicating a MCP2551 to each unit, which should be no big deal for a test system like this. <S> Now you use logic gates to multiplex the CANTX and CANRX signals of the microcontroller to one of the MCP2551. <S> Only power on the unit you want to talk to. <S> Transceivers like the MCP2551 are specified to not get in the way when powered down. <S> You should keep all units connected to common ground. <S> Only the one unit that is powered up will respond on the CAN bus, and the other units just let the bus float as it wants to. <S> I don't like multiplexing the CAN busses directly with transistors due to the impedance requirements of the bus, and the fact that both lines are bi-directional.
| Relays are the obvious choice but limit testing speed. You could also put them on one bus, but that has drawbacks: - If all boards are powered together with the same firmware, you have problems with the arbitration.
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RJ45 connector on breadboard For my project which for the next (many) months I want to make it on a breadboard. I ordered RJ45 connectors, but these do not fit on a breadboard. What would be a good way (or is there a way) to connect these on a breadboard? Of course I can solder 8 wires but than still the connector is hanging loose. Maybe there is a (not necessarily electric solution) how to keep these connectors in place somehow, but I don't have experience in it. <Q> I have seen ready-made "breakout boards" being sold for RJ45 connectors <S> (RJ45 connector + header pins on 0.1" pitch already fitted to PCB). <S> Search the usual places. <S> Alternatively, use hot glue to mount the RJ45 socket on a small piece of normal 0.1" "perf board" <S> prototype PCB yourself (for mechanical stability). <S> Then solder wires between the RJ45 pins and header pins which you have mounted in the "perf board". <A> Solder wires on, terminate the other ends of the wires with “dupont” 0.1" header pins, glue/tape them together. <S> Put your breadboard on a larger board <S> (that's what the double-sided tape it comes with is for) and fasten the RJ45 connector to it however you like. <A> Do a quick search for "RJ45 breadboard" and lots of options come up; something like this:
| Several ways: Make a small PCB which holds the connector and also 0.1"-aligned pins, or buy one ( example, not the same footprint as your connector ).
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Reversing the Direction of An Induction Motor? I have this motor I salvaged from a dryer, and I'm interested in reversing the direction of spin because it spins in the direction opposite of typical machine threads. The Capacitor is shown at the bottom of this picture(relative to the picture, just above the white letters): The cyan and white-red wires are AC lines. I can't figure out what the purple wire is, and the dryer wiring diagram says nothing about the motor aside from how it is wired in the dryer. Additionally, the connector that connects to the terminal block does not have a wire for the purple pin, leading me to believe that it is used in starting the motor. Using an ohmmeter, the purple wire is connected to all of the ac lines. <Q> If you don't have access to both ends of the windings, the motor can not be reversed. <S> That is likely the case for a dryer. <S> Some motors that are designed for reverse operation have two identical windings. <S> In that case, the capacitor can be connected to either winding for either direction. <S> That is certainly not the case for a dryer. <S> If the motor runs in the "wrong" direction for its original use, the capacitor may now be connected to the "wrong" winding. <A> It looks like Charles' point that you can't get to the windings is the case. <S> There looks to be three wires coming from the windings and no starter switch. <S> so you have this case: <S> So it looks like the white/orange is tied to the two windings, main and aux, (the lower line). <S> So unless you can reach into the windings and reverse one of those, you can't reverse it. <S> If you can, you want to bring another wire out so you can separate the connection from main and aux. <S> from the white/orange. <A> A tumbler dryer motor needs to run in both directions to tumble the laundry as intended. <S> Usually the schematic is as follows: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> So one coil is connected to AC directly while the other is connected through the cap. <S> By selecting which one the tumbler control logic can change the direction from time to time. <A> Single-phase motor rotation direction is based on the initial direction given by the starter winding, nothing significant about the main winding will favor initial movement.(180 degree pole reversal has no rotational aspect and the magnetic field is zero twice per cycle) <S> Direction can also be reversed by starting it with a mechanical spin in the desired direction. <S> Did this dryer have an auto start or manual starter button? <S> Most old dryers that I have used require the user to hold a starter button until the motor has enough momentum for the main winding to operate. <S> Just in case this is a commercial dryer motor(only induction was mentioned). <S> 3-phase induction motors do not have starter windings because the rotation direction is inherent and can be reversed simply by reversing any two of the phase wires.(120 degree pole changes are clearly rotating either clockwise or CCW, and the magnetic field is never zero at any point in the rotation.)
| In order to reverse a single-phase motor, it is usually necessary to reverse one winding with respect to the other.
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Why are there not more cores in modern CPUs? Many often confuse the meaning of Moore's Law... it refers to the number of transistors on a chip, not performance. A while back, it became apparent that the gains from increasing clock frequency on chips was not worth the expense and chip makers started adding extra cores to CPUs. However, the increase in the number of cores on consumer chips has not matched the increase in transistors on each chip. I surmise that a lot of these transistors have gone into features such as prediction logic ect, because it is difficult for some workloads to be parallelized, or many programmers find parallelizing their programs too time intensive, or CPUs are optimized for existing programs. However, from my perspective, I would like to see transistors go into increasing core count and on-chip-cache as this would benefit my programs more than marginal increases in single threaded performance given that I have no trouble writing multi-threaded code for most of my particular goals. If I use the extra transistors for a really large cache, I will not have to make as many trips to memory, which can also be a big performance booster. Am I incorrect as to the reason core counts do not seem to be increasing at the same rate as the number of transistors? Or is there also some diminished return for increasing core count even for easily parallelized work loads such as memory bandwidth? Why have core counts not increased at anywhere near the rate as the number of transistors on a chip? Edit: Just because a workload can be run in parallel does not mean it is an appropriate task for a GPU ect which tend to deal with doing a lot of floating point calculations. CPUs have diverse general purpose capabilities which more specialized chips lack. An example of this could be, let's say I have a set of 50 heuristic functions I need to run against a large set of data that is already in memory. This is easy to multi thread, give each function its own thread, and you can multi thread it further by diving up subsets of the data for each function (if the data is not highly interdependent). You could easily satiturate all the cores of even a top end Xeon processor, but you won't be able to make much use of a GPU or SIMD. Or, just a common web application serving many different requests that do not need to be coordinated. Or, just several different applications running on the same server for political or administrative reasons. <Q> There's a number of technical and business reasons in no particular order: Memory bandwidth becomes an issue with scaling of cores. <S> Memorycontention can actually decrease your performance. <S> Xeon Phi isthe platform where core hungry (and cash loaded) <S> customers can go. <S> Most software has been designed to run well single threaded. <S> Thisforms a chicken and egg problem. <S> Why try and sell more cores whenmost customers can't use them? <S> Most customers won't use them becausehardware isn't built in a core-scaled fashion. <S> Many customersare more interested in IIO bandwidth. <S> In that case, you just needenough cores to service IIO. <S> Intel Xeon's do have many morecores as well, but you'll pay a pretty penny for them in general. <S> Inthat regard, it's simply supply and demand. <S> Because transistor count continues to scale (although not really by Moore's law anymore either), single threaded applications still dominate, and core processing power isn't the bottleneck usually, it's more effective to put those transistors to use making the cache larger and more efficient. <S> Basically, instead of parallelizing the workload by creating more cores, the cores are now getting fed better. <S> Lack of competition in the highly parallel compute segment prevents consumer level pricing. <S> Most mainstream programming languages are ill equipped to handle parallel code well. <S> Even those that appear to haven't seemed to find a way to easily debug parallel code. <S> Potentially a new programming paradigm is necessary to overcome this. <S> Certain common OS's can actually suffer exponential performance loss the more forks you make so even if you have the cores, the OS handling of it ruins the usage of them. <S> This is an extension of points 3 and 8. <A> If you want many cores, buy a GPU. <S> Or one of these 1024-core processors. <S> The main limiting factor then becomes memory bandwidth . <S> given that I have no trouble writing multi-threaded code. <S> With which tools? <S> What sort of algorithms? <A> Additionally to what has been already said, there is also an additional cost in parallelism. <S> And I am not talking about Amdahl's law. <S> The more cores you have in parallel the more complex the hardware becomes that has to negotiate between the cores. <S> Some of these scale with $O(n)$ others with $O(n^2)$. <S> This imposes an upper limit before it becomes more economic to insert another layer and use more parallel CPU's instead if more cores. <S> You can see these trade-offs quite nicely with Intel: there are multi-core CPU's with higher number of cores, like the Xeons that go up to 28 cores. <S> But the Xeons are limited to about 3GHz clock speed. <S> Most of them just do around 2.5GHz. <S> The consumer processors go higher up in clock speed, but are limited on the number of cores. <S> If you have an embarrassingly parallel problem you can get one of the 8 socket motherboards and eight of those 28 core Xeons to get a whooping 224 cores in a single computer. <S> At this point you are deep in NUMA territory and your application has to be tailored to the computer architecture you have, in order to get the full performance.
| Basically as discussed: most people buy Intel-compatible processors to run their existing programs, which have minimal parallelism or may be entirely dependent on single-threaded performance.
|
Why do some EMI/RF shields have holes on the top and some don't? I am not talking about cutouts for tall components. I don't think they are for ventilation as they are often covered with manufacturer labels. <Q> Advantages of holes in shield: Allows some air flow for better heat dissipation. <S> This is the primary reason. <S> Less weight. <S> Small holes don't really compromise the shield, as long as the holes are significantly smaller than the wavelength of what you want the shield to attenuate. <S> As a aside, you won't ever see long slots in RF shields. <S> If a larger overall opening is desired, it will be accomplished with a array of holes. <S> The shield is then still a mesh in that area, which is mostly as good as solid, as long as the individual holes are small compared to the wavelength. <S> A single long and thin slot is actually a antenna. <S> Imagine a conducting sheet with RF current flowing in one dimension. <S> A slot perpendicular to the current flow has the same characteristics as a dipole antenna. <S> In fact, such things are called slot antennas . <S> Obviously it would be bad to add slot antennas to something intended to be a shield. <A> Good answers here already <S> but I would also add, holes also significantly change the thermal/mechanical properties of the shield. <S> As you know, when metal gets hot it expands, similarly, it shrinks as it cools. <S> If a "can" type EMI shield is soldered down to the PCB, and said shield is solid, that will introduce a significant difference in expansion rates between the PCB and the shield. <S> This can cause effects like: Failure of the solder joints holding down the shield, Tearing the solder down pads off the board, Warping of the board, with possible resultant intermittent/failing connections elsewhere, Audible popping of the shield as internal stresses are redistributed. <S> (This can also introduce a concussion shock to the joints and PCB.) <S> The shield to dislodge. <S> This can be a significant issue if the EMI shield is soldered down during normal manufacturing where the boards are pre-heated before the solder flow phase. <S> When the board cools again, a residual stress will be introduced. <S> Boards can actually come out with quite the curve or warp in them. <S> Shields with nicely laid out holes also look much "cooler". <A> Providing holes will provide shielding while saving on material costs. <S> Presence of holes doesn't mean that RF signals will pass un-attenuated. <S> There is a cut off frequency for the given perforation dimension. <S> In terms of wavelength, it becomes: Cut off wavelength = <S> 3.142 <S> * radius of hole (for circular perforations) <S> For a 2.4 GHz wave, wavelength = <S> 12.5 cm <S> Thus a hole smaller than 12.5/3.142 cm = <S> 3.98 cm in dia will attenuate the RF signals. <S> In a lot of cases, shielding is required against 50/60 Hz line noise or some hundreds of kHz noise coming from a switching regulator. <S> In this case, even a much bigger hole can provide shielding while effectively saving on material costs and making the system light-weight. <A> They may be for cleaning. <S> I have designed a few small RF shields like this. <S> We always use small round holes similar to those shown in some of the pictures above. <S> The shields are soldered in place during the normal reflow process at the same time as all of the other components on the board. <S> After reflow the boards are cleaned using high pressure water jets (or sometimes solvents) to remove flux residue and other contaminants. <S> Without holes in the lid, areas underneath the shield would not get properly washed. <A> An un-holey shield will obviously provide even better shielding and avoid problems with something shielded being nearer to the shield than the hole diameter (which is said to impair the shielding effect) - but will make any forced air or convection cooling ineffective (excepting whatever heat is transferred to the shielding material by convection within the shielding enclosure).
| Also, larger holes allow positioning adjustment facilities (trimmer caps and pots) beneath a hole so they are accessible without removing part of the shielding - which is important since some circuitry will be inherently out of tune with the shield gone and/or hard to adjust because it will catch massive interference.
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Current distribution for parallel zeners Suppose you've got two 5V1 zeners in parallel. Due to slight differences between the two parts (tolerance on zener voltage), one of them will conduct more current. simulate this circuit – Schematic created using CircuitLab The one consuming most current will heat up more. This will probably affect its properties. What kind of equilibrium will be reached? Will the current distribute evenly between them? I couldn't estimate myself what equilibrium would be reached because... I got confused by the temperature coefficients. Some zeners tend towards negative, other towards positive temperature coefficients. Negative temperature coefficients would be the worst-case scenario: the zener voltage would drop when the part heats up, drawing even more current. Nevertheless, while drawing more current the voltage over the diode will increase a little because the zener I-V-Curve is not ideal. This would push some current to the other zener(s) mounted in parallel. But would that be enough to overcome the detrimental effects of a negative temperature coefficient? <Q> Low voltage Zener diodes (about 5V) feature true Zener breakdown, i.e. band-to-band tunneling. <S> The larger the temperature, the more likely the tunneling occurs because the larger the carrier energy (and the smaller also the energy gap). <S> Therefore, there coud be a thermal run-away. <S> In fact, unlike normal diodes (or LED diodes) in forward conduction, small mismatches can lead to large current variations as the curve is very steep in breakdown voltage. <S> The zener carrying more current will likely heat more (Which will decrease the zener voltage etc.). <S> High-voltage Zener diodes ( <S> e.g. 10V or more), instead, feature avalanche breakdown. <S> The carriers are accelerated by electric field, and if they reach enough energy, they will produce more electron-hole pairs, due to impact ionization (which are in turn accelerated by the electric field, etc.). <S> However, the larger the temperature, the larger the lattice vibration and thus the smaller the probability that an electron can gain enough energy between impacts, therefore the less likely they can ionize (i.e. produce another electron-hole pair). <S> In this way, there will be a negative feedback, that will make the diode initilly carrying more current to be less conductive (therefore the current will spread more equally). <S> The datasheets shows in fact different temperature coefficient for different Zener values. <A> You could look at a datasheet and notice it mentions a temperature coefficient for zener voltage, which answers your question. <S> In the case of a 5V1 zener tempco can be positive or negative, so anything can happen... <S> For a 3V zener, tempco is negative, so you'll have current hogging. <S> For a 12V zener, tempco is positive, so they would tend to share the current. <A> You didn't state whether you're using zeners for overvoltage (OV) protection or a voltage source. <S> I'm going to assume you want OV protection because, if you were making a power supply, you would want to minimize the power wasted by running a semiconductor in its active region. <S> For the purpose of OV protection, a better circuit would be to put several resistor-zener circuits in parallel. <S> Then you have control over both the sign of the coefficient and its magnitude. <S> Think about it from a feedback perspective. <S> As the current increases through a resistor, its voltage increases proportionally. <S> Using a series resistor, you can always force the system into negative feedback* regardless of the sign of the diode's tempco. <S> Each diode should be in series with its own resistor and in parallel with nothing at all. <S> Put as many pairs as necessary in parallel in order to achieve an acceptable risk profile. <S> *Do not confuse sign of feedback with sign of tempco.
| In a resistor-zener series circuit, now that the resistor is eating up more of the total voltage, there is less voltage for the zener to eat, therefore its current will be lower.
|
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