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Which circuit to use to automatically turn off the backlight of a GLCD after 30s? Have you ever seen smart phones automatically turn their screens off after a short time to save the power? I want to do it on the back light of my GLCD (ks108sed). Descriptions: Back light will be turned on by a tactile switch. Back light will be turned off after 30s. The circuit shouldn't have a microcontroller. What circuit should I use ?? <Q> Oh, you could use something like the below, but the micro would be a more elegant solution for sure. <S> You can play with the value of R1 to get closer to the 30s if it matters. <S> If you use a CMOS 555 <S> then you can use a lower value capacitor and higher value resistor. <S> This will directly drive up to about a 100mA backlight. <S> If you need more current, add a p-channel logic level MOSFET and you can drive a blindingly bright light. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> NE555 ? <S> or even a transistor, resistor and a cap. <S> Alternatively, you can put AND gate (or just few transitors) between ! <S> CS/!SS and any data line <S> so when LCD is not getting any commands it can be used to control backlight (basically off for short moment you send data to LCD, and PWM when not sending any data to LCD) <A> So when I look at this I think of one circuit that you can trigger, push of a button, and it will output for a defined period of time and then turn off again until triggered again! <S> The chip needed for this is a 555 Timer chip and will need to be set up in a monostable configuration, I used this circuit many times in my college days and it is very basic with not a lot of calculations to it as there are plenty of online tools to help you out. <S> This is the circuit diagram for the circuit and you can work out the timings and values with a simple formula: T = <S> (1.1*R <S> *C). <S> Again this formula can be found along with online calculators to work out the values that you specifically need, for an output time of 30 seconds a value of R = <S> 123K and C = 220uF are required, (found using one of these mention online calculators), there are other arrangements of values that you could use but I feel these, or values close to these would give you your desired on time, after this time has elapsed the circuit output will go low and remain low and only requires a low to be triggered at the input (use the button in a pull up configuration). <S> This hopefully would be a solution to your issue! <A> Edit: So the author already accepted an answer while I was still working on this. <S> What this circuit offers over the 555 timer is higher timing precision and no dependance on component tolerances. <S> Simplest way I can think of that uses easily sourced and cheap components <S> is the following implementation of this diagram. <S> 32.768kHz crystal into a TI CD4060B <S> binary counter with oscillator circuit. <S> This can give you a 2Hz pulse if you use the Q14 output. <S> $$\frac{32768Hz}{2^{14} counts} = <S> 2Hz <S> $$ <S> Then use that for the CLK input to a TI SN74HC4040 binary counter. <S> You really only need 6 bits, but more bits give you more flexibility and the cost is nearly the same. <S> Then you can tailor the outputs to an AND gate and get any time pulse from every second to hundreds of seconds. <S> To get 30 seconds I did the following math. <S> 60 decimal is the same as 111100 binary. <S> If we use the four bits that are 1s as inputs to an AND gate then the AND gate will only be high once every 30 seconds. <S> You can then do as you like to this pulse to make it shut off your screen. <S> You didn't give details on what kind of control signal you need <S> so I'll leave that up to you. <S> Maybe an inversion is needed, maybe a latch of some kind too. <S> The nice thing about this circuit is if you decided you needed a shutoff after 15 seconds or 100 seconds implementing <S> it is simple, just encode the bits to an AND gate. <S> Total cost: <S> ~$3 <S> and ~7components. <S> Don't forget the necessary passives to make the crystal oscillate!
| Since the clock input is running at 2Hz then to get 30 seconds we will need to toggle something every 60 clock cycles. I would suggest adding some zero ohm resistors on different bits so you can tailor the shutoff time if need be.
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Can a proximity sensor read through fabric? I am trying to build an arduino project with the following requirements: Proximity sensors detect human presence and its distance The sensors are hidden behind fabric Range has to be 1-3 meters or higher It is for indoor use After a bit of trial and errors, here are my initial conclusions: Available light changes, and sensors being behind the fabric, photo sensors are out of the question Even though they're not supposed to be very sensitive to soft material, Ultrasonic sensors such as the HC-SR04 don't fit. Their echo bounce right off the fabric Pir Motion sensors don't read through fabric either I am starting to wonder if this is even feasible. Is there such thing as a proximity sensor that reads through soft material? <Q> You could try a Doppler radar sensor such as the ~10GHz motion sensor modules which are available on the surplus market and from China. <S> As unlicensed and likely unapproved intentional emitters, they are probably of dubious legality to use, if that bothers you. <S> They're probably safe to use (aside from possible interference with other electronics) but caution is called for- some people get sensitive if they think their DNA might get scrambled or their 'nads nuked. <A> I'd bet this'll do the trick: Parallax X-Band Motion Detector ($40) http://www.parallax.com/product/32213 <A> Keep in mind that not all materials are opaque to near-IR light and that there is some fabric that can easily pass Near IR while blocking visible. <S> So for us humans, we see the pattern and the device sees - nothing. <S> Look to synthetic materials. <S> Exposed and developed photographic film is transparent to Near Ir as an example. <S> A loose mesh inside of a darken box will block less of a view and prevent humans from seeing inside the box. <A> Inductive sensors spring immediately to mind, along with low-frequency (sub-ultrasonic) "click sonar" sensors. <S> Even for ultrasonic, try speaker cloth - it's good to about 20KHz. <A> Remember, if inductance will solve a problem for you, there is probably a way to solve that same problem using capacitance... <S> Also, sound penetrates most fabrics nicely. <S> It takes a sharp difference in acoustical impedance to get a reflection. <S> You know that an infrared detector can sense humans well clothed, and animals that are nicely fured. <S> (Or, is it furred? <S> Or, is that even a word...?) <S> If you asked a third grader to solve this problem, the third grader might say, "just throw a rock. <S> If you hear the word 'ouch', then someone is there. <S> " i.e., we must think out of the box...
| Even infrared penetrates most fabrics nicely. Thermal IR (which the PIR sensor operates at) will not work as the fabric will be at a temperature that emits, it is like trying to look through gauze that is lit up.
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Op-amp based audio amplifier not working I have a problem with audio amplifier I am trying to build. Given my ignorance, I cannot figure out how to fix the issue. There is absolutely no sound on the output. This is what I have assembled: EDIT1 Kaz, Venny, Perhaps I misunderstood something you said, but trying my best I modified the circuit into this form: And sadly it is not working yet. I would like to apologize for my utter nescience of the topic. <Q> You need to convey a reference voltage to this input. <S> (This has to be done through a reasonably large resistor; if you just tie the input to a stiff reference voltage, the input will have no impedance to work against.) <S> Additional problems stem from the choice of part for the circuit. <S> Firstly, although the LM358 can run on as little as 3V, you neglected to look at the common mode input range: according to the data sheet that range is from \$0\$ to \$V^+ - 2\$ or \$0\$ to \$V^+ - 1.5\$: depending of which of two conflicting tables you believe. <S> This means that on such a low supply voltage, you are better of building an inverting amplifier which keeps the common mode input voltages fixed. <S> If you stick to the \$0\$ to \$V^+ - 2\$ recommendation, this means you would build the inverting stage such that it is biased to 1V. <S> Next, look at the terrible output voltage swing figures! <S> On nice, big 30V supply, the op-amp will only swing up to 26V: a bad sign: the op-amp struggles to come within 4V of the upper power rail. <S> No VOH (high output voltage swing) figures are given for any lower voltage, but for 3V, it will probably be next to nonexistent. <S> So the fact that the op-amp can "work" on 3V does not mean much: it has poor output voltage swing, and a crummy common mode input range. <S> If you need an op-amp that works meaningfully on 3V, you need something more modern: something with rail-to-rail output ( <S> if not input, too). <S> Lastly, the LM358 is not very suitable for driving headphones (even when adequately juiced); that application requires an amplifier with an output stage that has decent current driving ability: a power amplifier. <S> (Among the operational amplifiers, one example of a part that is acceptable for driving headphones is the the NJM4556 ; however it is not low power, and will not run on 3V.) <A> You have no path for bias current. <S> Add 1Mohm resistor from +3V to in+. <S> Also the ring of audio input must be referenced to half of supply voltage, so add voltage divider consisting of two 10k resistors between +3V and 0V with the centre connected to ground (the grey line). <A> You have a serious input and output voltage range problem : the input signal must stay within input voltage range (V- to V+ -2V) and the expected output voltage must stay within output voltage range (V- to V+ - 3.5V). <S> Here is what I suggest : simulate this circuit – <S> Schematic created using CircuitLab <S> This way : <S> The input voltage swings around ground, and stays far from the power rails. <S> The output can swing from -9V to about +5V <S> As Kaz noted, the output drive of LM358 is probably not strong enough for headphones ; you may want to add an output booster stage between the LM358 out and the pot. <S> If you want to stay with single battery, you will need to create a ground at an intermediate value (I'd suggest around 3V since that op-amp output swing is limited from V- to about 3.5V below V+). <S> For that you can replace the two batteries with a single and create the ground from a resistive divider and an op-amp (the second half of your lm358) as buffer : <S> simulate this circuit
| The most obvious problem in the circuit is that the + input of the amplifier is floating; it is not biased to any DC voltage level.
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My thought on the reverse polarity protection I am now designing a power reverse polarity protection circuit for my system. I found that most methods used for the protection would cause voltage drop in the power path. Even the MOSFET, which has maybe 0.025ohm Rdon, would cause about 5mV drop(0.2A). Some may not cause voltage drop but would be a little complex or not resettable. http://www.wa0itp.com/revpro.html Then I have a new thought(maybe new or maybe not). My design is showed below. My circuit has three power sources(+-12V and 0V). I add two diodes between +12 and 0V and between 0V and -12V with big enough series resistors. Normally, the two diodes would turn off.No current would go through R1 and R2. So no voltage drop when normal operation. If the polarity is reverse,D1 and D2 would turn on. Current would go through the diodes. Is there any problem in this circuit? Can it work? simulate this circuit – Schematic created using CircuitLab <Q> This DPST NO <S> (2 form "A") relay solution will work: <A> The diodes will have no effect to speak of on reverse polarity, and will provide no protection. <S> If the device is a chip, it will act like a forward biased diode, and will take almost all of the current compared to your parallel diode with 1K in series. <S> Edit: You can play with simulations below. <S> With the left one I've simulated the 12V source (reversed) with 10m\$\Omega\$ source resistance (2mV drop at 200mA), and no load. <S> The right circuit is the same except D6 represents a chip that you're trying to protect. <S> Voltage on the left circuit is -12.0V, with or without D1+R1 <S> Voltage on the right circuit is -11.8V, with or without D1+R1, but now D6 (the chip that's supposed to be protected) is conducting about 19 amperes. <S> It is not long for this world. <S> In both cases D1 conducts little current (about 11mA). <S> simulate this circuit – <S> Schematic created using CircuitLab <S> There is no reason generally to worry about a few tens of mV drop ona power supply rail- <S> usually the tolerances are much more. <S> The MOSFET method is the best way for very low drop unless you know the source has limited current (in which case, consider a shunt Schottky), a series diode can work in some cases. <A> Thinking only the positive side of your circuit: assuming you use a normal diode with a voltage drop of 0.7V, GND and +12 reversed, the voltage across thq 1k resistor is U= <S> RI = <S> > all remaining voltage will be over the resistor. <S> Thus the protection doesnt work. <S> One way of doing this (better or worse) would be remove he 1k resistor and using a diode with as small voltage drop as possible, eg schottky diode. <S> Also you need a fuse wich prevents diode from burning up. <A> Could I interest you in building a bridge rectifier with Schottky diodes instead? <S> Then polarity would be immaterial.
| Voltagedrop across the "reverse protection line" has to be lower than voltage drop across reverse connected IC.
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Why do they usually make comparators open collector? I never see open collector opamps while I frequently see open collector comparators. What is reason for preferring open collector output in comparators? (Inside of LM193 .) <Q> The main reason for using open-collector is so that several comparators can have their output connected together in a wired OR gate. <S> All the open collectors can be tied together to a single resistor without any conflicts between comparators. <S> This would not have been easy to do when your comparator has also the capability to source current at its output. <S> This wired OR gate is useful for detecting ranges of voltage, for example, when you want your output to be low when the signal is below a value or above another value. <A> I always thought of comparators as the simplest A-D converter. <S> You may want a different supply voltage on the analog side ( <S> maybe +/-15V or something) and the output going to a different digital voltage. <S> (+5 V perhaps.) <A> An opamp is driving an analogue voltage. <S> To be useful, it needs to control the voltage swing completely. <S> A comparator is a binary switch. <S> Its output is effectively in only two states. <S> It can work correctly with an external resistor. <S> It doesn't need to drive the signal high when used with an external pull-up resistor. <S> An NPN transistor can pull down when it is on, so it could pull down on a voltage divider, making the device much more flexible than if it set the output swing internally. <A> The difference is that the comparator output is a logic output that can only be on or off, and the opamp output is analog. <S> An open collector output is simple and flexible but relatively slow compared to other logic outputs. <S> Since most comparators are slow, it is OK to use open collector outputs for them. <S> Fast comparators may have other types of outputs. <S> For example, the extremely fast ADCMP580 has ECL outputs. <S> Some people use it to generate fast edges .
| The open collector lets you easily adjust the output voltage reference.
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What is the gold-coloured area on this circuit board and what is it for? The below PCB belongs to a wireless access point. Could anyone explain what are the golden-like part of the circuit that is indicated by the red circles(ish)? What material is it? What is the the functionality/use of it? The left two circles have the same layout on the other side of PCB, and nothing is placed on top. The right circle is just under the wireless chipset. P.S. I could not come up with a good title as I don't know the proper name or technique that is used on the PCB, please feel free to edit it. <Q> The exposed areas could have a couple of uses: <S> As a grounding/earthing point, for example an RF screening gasket or can might make contact there. <S> Heat-sinking for a device. <S> We have very similar footprints on boards for Allegro A4988 stepper motor driver chips and switch-mode power supply chips, but they can be anything that requires heat sinking. <S> There may be other reasons, those are the two <S> I'm most familiar with. <S> Edit: Another reason is to do with RF performance or when RF devices such as filters or antennas are made purely using tracks on the PCB, in those cases the solder mask (and every other layer of the board) has an effect on the fabricated device, but I'm not an RF engineer so <S> can't really expand on that other than being something I've seen in equipment we've designed. <A> It looks like the exposed copper is gold plated. <S> e.g. <S> Wikipedia Electroless nickel immersion gold <S> Unlike other finishes, gold plating doesn't tarnish, or oxidise, so the PCB can be stored for months without it deteriorating. <A> That plating is part of a process called ENIG, or Electroless Nickel Gold <S> and it's purpose <S> is to increase shelf life as stated in an earlier response. <S> Also, it is flatter than other methods of finishing boards such as HASL (Hot Air Solder Leveling), which makes it more suitable for fine pitch SMD parts. <S> This process step occurs after solder mask and is only applied to the exposed copper. <S> Fun fact: The gold is only a few microns thick and will "dissolve" into the solder when the board is reflowed.
| The gold colour is just the surface finish of the PCB, called Gold Flashing. That is a very common for SMD PCBs because it gives a lovely flat surface to solder SMD to.
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PCB, pads with holes but with different nets on top and bottom sides I need pads with holes in by PCB, but the top and bottom side of the pads need to have different nets i.e. the top and bottom sides must be isolated. Here is the picture Is this possible to fabricate for the PCB manufacturers? If yes, how do I design this in Altium? <Q> I don't see any reason why you can't create a footprint with overlayed pads on top and bottom going to different pins. <S> If you avoid drilling one of the holes (make it an SMT pad <S> ) you may be able to avoid all DRC errors with little fuss (otherwise it may complain that the two drilled holes have insufficient clearance). <S> Altium definitely understands that pads on top of each other separated by an unplated hole are not connected. <S> Edit: It only seems to understand this when the unplated hole is part of a single pad. <S> Otherwise it thinks they are connected, even with an unplated hole. <S> I tried this with a simple pad on the top and a multilayer on the bottom, unplated hole and padstack edited to have copper only on the bottom and there is still a short circuit error in the DRC check, despite the unplated hole. <S> The 3D view shows that the copper is not there, so Altium's DRC is not picking up on it. <A> In addition to the other answers, regardless of what you can do in your software, you may have trouble when you go to have your board fabbed. <S> Some shops interpret copper at the edge of, or under a hole as being a plated hole. <S> OSHPark for example does things that way. <S> You'll have to be careful to ensure that the copper layers don't intersect the edges of the hole, and leave a teensy tiny margin. <S> If the fab has a different system of indicating which wholes are plated and which aren't, then you're home free. <S> Otherwise, you could put a hole much smaller than what you need in the spot, ignore the plating, and ream it out up to size yourself. <A> I'll add that in Eagle you can add mounting holes which have no electrical connection between the different layers in a board or in a package. <S> HOLE - Function Add drill hole to a board or package. <S> Board houses can do this but you should check with them before ordering. <A> That is basically an unplated hole drilled through two suface-mount pads. <S> Yes, it is possible to fabricate but may be for additional cost, because it is done in separate manufacturing step. <S> From what i have seen this will leave bare copper on sides of the hole (not a big deal). <S> In Altium, it will probably cause DRC errors which you can ignore.
| Just specify the hole as unplated.
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Is electronics an expensive hobby? I recently just finished my Eagle schematic for a simple robot that moves and plan to order the parts and the PCB soon. Unfortunately with the soldering iron, helping hands, lights, solder, parts, and PCB all add up to $125. Is this a lot of money? Or is my perspective not correct because I'm 14, entering high school, and have no sustainable income? I've ordered all my stuff from adafruit.com, is there a cheaper option? <Q> Tools can get pretty pricey in this area, ranging from a simple $5 soldering iron, all they way up to more than I will probably ever make in my existence. <S> They should serve you well for your purposes for sometime until you can get real income. <S> I would say since you are 14 and have no sustainable income, you might want to be very careful when purchasing parts to make sure you get the right ones on the first order. <S> Electronics has a different purchasing structure than, say, a grocery store, where the more you buy, they just add up linearly. <S> When you buy your components from a good distributor, like digi-key.com, you could get a single capacitor for say $0.60 <S> OR you could get 100 capacitors for $5.00 because the price per component would go down as you buy more. <S> Adafruit is a good website, although I wouldn't recommend them for basic components. <S> They have really nice breakout boards and are great for getting started on a project quickly. <S> For basic discrete components though, I would go somewhere like digi-key, mouser, farnell, etc.. <S> These are really reputable distributors and have a wide selection of many components. <A> Several years ago, during a move from one house to another, I gave away about $5,000US worth of electronic stuff <S> I'd collected over the course of the previous probably two or so decades. <S> You CAN start on what I think of as a shoestring and stay there, or you CAN go sky-wild and dump vast sums into it (in some cases shaking doubly vast sums of money back out of it), or anything in between those. <S> In your case... let me put it this way: <S> $125US wouldn't fill my pickup's fuel tank twice. <S> It wouldn't buy more than a couple dozen hamburgers at the local diner. <S> If I accidentally dropped that much money on the ground, I wouldn't be horribly upset about never finding it again. <S> When you establish your own personal resource supply, you'll look back at that initial investment you've made & think that it's pretty minor. <S> And in exchange for your investment... you get real hands-on experience, which is more valuable than money. <S> You gain better understanding of things, you gain the ability to design new things that've never existed before, you get exercise for previously unexercised parts of your imagining brain, you grow healthier. <S> Just... don't try to build an artificial girlfriend. <S> THAT's not healthy. <S> 8) <A> This is great! <S> Hey, you may spend a bit of money now, but think about turning your avocation into your vocation. <S> It could pay off big time. <S> (And let's you look forward to going to work.) <S> Parts sources: <S> I use to take apart all sorts of electronics junk to scavenge the pieces parts out of them. <S> That's a little harder today with all the surface mount, but still doable. <S> There are several parts distributors, Digikey, Mouser, Newark (if you are in the US). <S> There may be a minimum order. <S> If you live in/near a big city there maybe a maker space/ group that you could go visit. <S> You also may find a nice technology teacher at the high school that you can hook up with. <S> Good luck and have fun!
| Getting all of that for $125 isn't too bad, and will be a one time cost until you decide to upgrade to better tools.
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Data transmission through the human body As we all know, data can be transferred through cables using electrical pulses transferred into binary codes. My question is, is it possible to transfer data in a similar manner but through the human body? Basically, can we act as a cable? The answer I am looking for is if it is theoretically possible or not? <Q> Sony has patented that for "wireless" headphones, using conductive fabric as electrodes and a frequency between 500kHz and 3MHz: <S> Sony sends sound through your skin Full description <A> If we can send data to a space craft a billion (maybe less, maybe more) miles away (namely voyager) <S> undoubtedly we can pass data thru the human body. <S> The answer is YES. <A> Given that peope serve under some circumstances as ad-hoc antennae... <S> it's not a question of theory at all. <S> It happens. <S> It happens a lot . <A> For example, from the brain to muscles following intricate pathways including the brainstem, which connects to the spinal cord. <S> In the end, cells respond only at a certain threshold called the action potential: <S> Here's a model of a neuron: <S> In addition, chemicals travel from neurons to neurons transmit information. <S> These "neurotransmitters" are slower than the electric signals and their effects are often less specific. <S> In addition, some chemicals can travel from the brain to other organs. <S> These are generally known as hormones. <S> One common example of hormonal signaling is the HPA axis . <S> The electrical circuits in the brain can be hacked through genetic alteration of neurons, making them respond to light. <S> The potential of optogenetics for future therapeutic applications has been considered by scientists. <S> Today, patients with Parkinson's disease can have a kind of brain pacemaker that provides deep brain stimulation through electrodes, basically compensating for the low neurotransmitter count by plugging into the information circuit for posture and movement.
| Neurons transmit information all the time.
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Faster quadrature decoder loops with Python code I'm working with a BeagleBone Black and using Adafruit's IO Python library. Wrote a simple quadrature decoding function and it works perfectly fine when the motor runs at about 1800 RPM. But when the motor runs at higher speeds, the code starts missing some of the interrupts and the encoder counts start to accumulate errors.Do you guys have any suggestions as to how I can make the code more efficient or if there are functions which can cycle the interrupts at a higher frequency. Thanks,Kel Here's the code: # Define encoder count functiondef encodercount(term):global counts global Encoder_Aglobal Encoder_A_oldglobal Encoder_Bglobal Encoder_B_oldglobal errorEncoder_A = GPIO.input('P8_7') # stores the value of the encoders at time of interruptEncoder_B = GPIO.input('P8_8')if Encoder_A == Encoder_A_old and Encoder_B == Encoder_B_old:# this will be an error error += 1 print 'Error count is %s' %errorelif (Encoder_A == 1 and Encoder_B_old == 0) or (Encoder_A == 0 and Encoder_B_old == 1):# this will be clockwise rotation counts += 1 print 'Encoder count is %s' %counts print 'AB is %s %s' % (Encoder_A, Encoder_B)elif (Encoder_A == 1 and Encoder_B_old == 1) or (Encoder_A == 0 and Encoder_B_old == 0):# this will be counter-clockwise rotation counts -= 1 print 'Encoder count is %s' %counts print 'AB is %s %s' % (Encoder_A, Encoder_B)else:#this will be an error as well error += 1 print 'Error count is %s' %errorEncoder_A_old = Encoder_A # store the current encoder values as old values to be used as comparison in the next loopEncoder_B_old = Encoder_B # Initialize the interrupts - these trigger on the both the rising and falling GPIO.add_event_detect('P8_7', GPIO.BOTH, callback = encodercount) # Encoder AGPIO.add_event_detect('P8_8', GPIO.BOTH, callback = encodercount) # Encoder B# This is the part of the code which runs normally in the backgroundwhile True: time.sleep(1) <Q> Note that the BBB has three hardware quadrature decoders on-chip, and two appear to be accessible via the connectors. <S> Look for eQEP functions. <S> Some support software appears to be available now for these, search for Beaglebone black quadrature, etc. <A> A small Spartan FPGA could keep track of a large number of encoders running at high speed, and the processor can simply read out the counters as necessary. <S> GPIO is not designed for this application. <S> Unless you want to dedicate an entire core just sitting on the pins and polling continuously with a while loop, the best solution is to offload this to an external device. <A> You have quite a few logical tests joined together, which will slow things down: if (Encoder_A == 1 and Encoder_B_old == 0) or (Encoder_A == 0 <S> and Encoder_B_old == 1): <S> That one line requires SEVEN tests (==, and, ==, or, ==, and, = <S> =). <S> Also, this doesn't seem like a true interrupt routine as you are explicitly having to check if the pins really have changed state at the top. <S> If this was a real interrupt triggered by a pin change then that test is redundant. <S> Print statements in an interrupt routine are a major no-no in situations like this as they slow the whole routine down, printing on the console takes time - in fact, ANY call to an external function takes time. <S> Set a flag and move on, run the print routine somewhere else by checking the flag, and don't let it weigh your interrupt down. <S> I don't know how efficiently Python works and which patterns may be "most optimum" <S> but I would (in C) be inclined to something like this: if(a != <S> old_a) // <S> A has changed{ if(a == 1) { if(b == 0) { count++; // <S> Encoder moved <S> fwd <S> } <S> else // <S> We don't care what else, anything is an error { error; <S> // b cannot == <S> a } } else // <S> Must be 0, <S> no need to check { if(b == 1) { count--; // <S> Encoder moved back } else { error; <S> // b cannot == <S> a } }}else if(b != <S> old_b) // <S> B has changed{ if(b == 1) <S> { if(a == 0) { count++; // <S> Encoder moved <S> fwd } else { error; <S> // b cannot == <S> a } } else { if(a == 1) { count--; // <S> Encoder moved <S> bbck } <S> else { error; <S> // b cannot == <S> a } }}else // neither has changed{ error;}old_a = a;old_b = b; <S> You may notice that this is nested so that the processor does the minimum number of tests to arrive at the result, and we try to fall out to an error as a by-product of being unsuccessful rather than explicitly catch errors by careful (and time-consuming) tests. <S> Without knowing how efficiently / smartly Python compiles it's hard to guess on more efficient ways of doing this.
| I would recommend offloading it to a small FPGA or CPLD, then reading the counts out via a serial port of some sort - e.g. SPI.
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Do I need a Reset Switch? I'm trying to make a ATmega328 run on a breadboard. I am stuck on deciding if I need a reset switch. Do I need one? Can I just unplug and plug in the battery instead? If I do need one, how would I go about doing this? <Q> The ATmega can work without a reset button. <S> Powering down the controller will reset it, as the O.P. expected. <S> In-circuit programmer will also reset the controller every time new firmware is loaded. <S> If it becomes apparent that a reset button would be a nice thing to have, it can be wired like this: Source: <S> Atmel application note AVR042 (AVR Hardware Design Considerations) <S> p.6 <S> There is also a simplified approach , although be sure to read the comments made by @vaxquis below. <S> Since the circuit is built on a breadboard, one can always take a wire and touch between RESET# and ground. <S> That will momentarily bring the RESET# to ground, which will reset the controller. <A> Reset switches often fall in the category of things which aren't needed, but are nice to have, especially if a device will be communicating with a PC using something like an FTDI USB-to-serial converter. <S> If an FTDI chip is powered by the board, communications will not be possible for the first few seconds after the board is powered on. <S> If it's powered by the PC, the signal wires from the FTDI may power the board even when no other supply is connected. <S> That can a very useful ability to have, and a reset switch is one of the easiest ways to achieve it (another option may be to use the DTR wire from the FTDI to control the reset line, though most reset lines are active-low and the default behavior of most programs drive DTR low during communication). <A> Generally MCU itself does not require RESET because power cycle do it too with some exception as is register indication what is reason of reset (power, reset, watchdog, brown-out,...). <S> But if you have another statefull ICs then they are affected but power cycle too. <S> Typical example is USB ICs mentioned by supercat but also e.g. encoders, sensors, multi PCB project powered via bus etc.
| Adding a reset button will make it possible for the PC to have an open connection with the FTDI chip at the moment the board comes out of reset.
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How to deal with devices connected via a connectors in schematics? I have a few sensors (potoresistor, temp sensor...) that connect to the board indirectly. The PCB will have a connector such as the molex 4pin fan connector that will in turn connect to the sensor/device via a short cable. How do I show this in the schematic and the PCB layout? If I select sensor schematic I would have to manually add the pcb footprint to the component. If I select schematic for the connector then adding a resistor to a random pin of a connector doesn't seem right either. <Q> In general, place the connector symbol to schematic and connect it to appropriate nets - it does not have to be in the correct order, because later you can adjust it from layout editor with pinswap command. <S> For clarity, 'dummy' symbol with no footprint can be placed to schematic, to indicate what will be connected to the wires. <A> (For me.) <S> On the pcb there's the connector footprint and label. <S> If there are some important pins I might add text. <S> (+V, GND, sig) to help with testing/ trouble shooting. <S> The schematic tells the story and I'll add in what's on the other end of the connector. <S> (maybe it's a thermistor that is one leg of a Wheatsone bridge.) <S> I figure I want anyone else who looks at the schematic to be able to understand out what's going on. <A> In Altium, there are different types of components defined: <S> "Graphical" parts are included on the schematic but it isn't checked by DRC and isn't added to the BOM or netlist, so it doesnt' appear in the layout. <S> "Mechanical" parts are included on the BOM, but not the netlist. <S> You can make your part either Graphical or Mechanical depending whether you want it included on the PCBA BOM or not. <A> I normally choose one of these two options: <S> Draw the sensor as a schematic symbol and have a footprint for the desired connector. <S> The only problem is that the BOM is not complete then (you can either have the sensor or the connector). <S> Draw a symbol for the connector with two connections for each lead in the schematic symbol <S> (so you draw one "symbol" with both the jack and the plug). <S> Finally you only give the connector a footprint, and leave the footprint for the sensor out. <S> An example of a schematic symbol for a connector can be: <S> Good luck designing your circuit!
| You can then make a schematic symbol for the sensor and connect it to the other side of the connector.
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Is it possible to power an Arduino Uno with a Raspberry Pi? I have an Arduino Uno and a Raspberry Pi that I'm using to control some LEDs in a computer case. Power is coming from a Molex connector from the computer's PSU. I want to place the Arduino and the Pi at the top of the case, but the power supply is at the bottom. In order to try and reduce the length of wire runs, I'd like to power one of the devices from the other. When I supplied 12V to the Arduino through the Vin pin and tried to supply 5V to the Raspberry Pi via the 5V pin, I ended up frying the Arduino (too many amps?). I was wondering if I could do the inverse: supply 5V to the Arduino through the GPIO 5V pin, then provide 5V to the Arduino through the other GPIO 5V pin. According to the diagram here: http://www.megaleecher.net/Raspberry_Pi_GPIO_Pinout_Helper there are two 5V pins and five ground pins that I can use. I'd like to prevent against frying something else though. Will my idea work, or is this a foolhardy idea? <Q> You likely fried the Arduino by supplying a high input <S> voltage <S> (you say 12V). <S> The voltage regulator on Arduino UNO steps down voltage, e.g. 12V, by converting power to heat. <S> Usually it work okay as long as there isn't much current being drawn from pins. <S> I <S> R-Pi GPIO pins are rated at less than 20mA, and Gert Van Loos recommends using under 3mA/GPIO to remain within the design assumptions that were made for the R-Pi. <S> is less clear how much current the 5V supply on the R-Pi GPIO socket can supply. <S> In the absence of information, I'd assume it might be best to avoid it. <S> The R-Pi USB sockets are not supplied via the R-Pi's on-board power supply. <S> Instead they come from its external power supply, and so it very much depends on the R-Pi's external power source on whether the Arduino UNO would overload anything. <S> Edit: R-Pi's USB and GPIO 5V are all connected to the external power supply via a 1A resettable fuse (polyfuse). <S> That has the benefit that it protects the external power supply. <S> It also limits the entire current drawn from the external power supply. <S> In the case of powering an Arduino from the R-Pi, and nothing else, it should still allow the Arduino to consume up to 300mA, which should be plenty. <S> I'd suggest using a USB cable plugged into the RPi and the Arduino to power the Arduino, however the schematic shows the GPIO 5V has exactly the same power-supplying capability. <S> Be careful, some of the R-Pi external power supplies don't seem to have much 'headroom'. <S> Where I work, students have problems sometimes because of this. <A> YES just use the raspberry pi USB port and connect the nano via micro USBI'm using the nano instead of an ADC(analog to dc input <S> converter).works like a charm :-) <A> Yes. <S> The power consumption of a Arduino UNO is approx. <S> 40-50mA/5volt. <S> The 50mA is without shields and such. <S> The power consumption of a Raspberry Pi is approx. <S> 700mA/5volt. <S> 700mA is without USB devices, USB devices will add to the total power consumption. <S> So theoretically, a 750mA/5volt power supply will cover the consumption of a Arduino UNO and a Raspberry Pi. <A> You probably fried the Arduino from excessive heat on the regulator. <S> The regulator on the Arduino is a 7805 linear regulator. <S> The linear part of the name means the it acts as a variable resistor turning the excess voltage into heat. <S> You can calculate this by the voltage you are putting in, 12V - 5V (Output voltage) which equals 7V. <S> Then 7V <S> x <S> (The current you are drawing in amps). <S> If you are powering a raspberry <S> Pi which draws roughly 1.5 to 2.1 amps so by estimating 2 amps draw, 7 x 2 = 14W of heat. <S> For a chip smaller than a your fingernail, that's a lot of heat. <S> Most likely the 7805 overheated and fried leaving your Arduino and Rasberry Pi without power. <S> A better option is taking a 5V line from the PSU. <S> No regulators, no heat dissipation. <S> If you have no option <S> but 12V, a switch mode power supply module based on the LM2596 is a good option, it can provide up to 3A, enough for a raspberry pi and Arduino.
| So NO do not try to supply power to the Arduino via any R-Pi GPIO pins except 5V.It
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Connect USB to microprocessor lacking a output port? Sorry I'm sure this is a very simple answer but I know very little about how it works beyond serial ports and USB adapters for interacting with PCB circuits. I've taken apart an old junk flip camcorder, which runs various programs etc. and would like to program it to display whatever such as C code or reconfigure what the play button does. I've been able to locate the microprocessor and searched its datasheet but I still cannot find out how to connect it to my computer by any means. Beyond that and the large chips, I have no idea where to look. Here is the device itself: see below photo please And the USB connection: I can plug the USB into the PC but the issue is camera is buggy and doesn't register as a USB device (otherwise I'd just reformat it) so that's why I was trying to connect to the board itself and not mess with the USB. I know the device works because the load screen shows and a red light appears with only the USB attached. Any ideas? http://imgur.com/zykkmR9 <-sorry! this is the picture with all the active connections, the other side does not have much edit: ive got it to appear on the computer somehow, and can read all the .nib files with the variables etc. If say the usb was broken or something else the next way to go about it would be the JTAG as everyone is saying? <Q> There is no guarantee that the manufacturer put any way to program the MCU onto the PCB. <S> MCU's could be programmed before assembly. <S> Most MCU's have built-in mechanisms to allow a user (in this case camera manufacturer) to prevent a chip being reprogrammed, or the program read. <S> It is likely that the camera manufacturer used those mechanisms. <S> It is sometimes possible to defeat those mechanisms, but those forms of 'attack' may damage the chip, and make it unusable. <A> You will have to find a JTAG adapter suitable for this MCU, and, obviously, the relevant JTAG connection on the PCB. <S> Most consumer products cannot be programmed or debugged through the USB connection. <A> I could be wrong, but I strongly suspect that your better use of the device is as a parts supply - scavenge the display & learn how to hardware-hack it, scavenge the button switches, maybe grab a few other internal parts if they look handy.
| It may be infeasible to program the MCU even if you could get at the right pins.
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Why do 74 series IC's have two ENABLE pins I am using 74LS154 4 to 16 decoder Link to *.pdf here . It has two ACTIVE LOW 'ENABLE' pins at the input. What is the use of two ENABLE input pins is the question. Most of the 74 series IC's used in the lab has two ENABLE pins.. <Q> It's just simply to reduce the "glue logic" needed to implement the device. <S> If you look at the package you realize that without the extra /enable that package would have an NC pin. <S> I suspect the first designers decided that having a little extra functionality for "free" (i.e. in pin count) would be an enhancement. <A> The enable pins are ANDed so you can make a 32-output decoder with two inverters and four decoder chips. <S> Arguably it might be more useful in smaller systems if one of the inputs was inverted, but imagine they had a meeting on some Monday in 1965 or whatever and, over coffee and maybe cigarettes, decided to make an it a symmetric enable input. <A> That's interesting that they would have two active-low enable pins and no active-high enable pins. <S> I'm not entirely sure what you would need that for. <S> I suppose it acts like a free AND or OR gate on the enable input, depending on how you want to look at it, which could be uselful in some applications. <S> However, the 3 to 8 line decoders generally have 3 enable lines - one active-high and two active-low. <S> You can use the active high pin on one chip and the active low pin on another chip to cascade two 3-to-8 decoders into a 4-to-16 decoder as the 4th input line drives the complementary enable lines. <A> The two inputs required only two emitters of the same transistor and there was a free pin of the case.
| It's simple enough to just tie the one input to low and to use the other pin as a single input and this gives more flexibility in implementation.
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How does the Brita LED work? Specifically, I'm talking about this product here... Brita Faucet Filter It has an LED that blinks when the water is running. It blinks red, yellow, or green to indicate filter quality (maybe) - but it seems to have no power source. So, my specific question is what powers the LED? And, I would also like to know what it's measuring? It is frustrating to Google this because people don't seem to care how things work, and the phrase is equal to "What does it do" and all I can find is "It shows you the water quality" - well, that's not how it works. Brita claims the filter lasts for 100 gallons, but I'm not sure about that, it seems as though the LED turns to red based on the amount of time, but to do that it would need a constant power supply and I don't see one. <Q> If an average faucet flows at 2 gpm (according to the US EPA), and the filter does not significantly reduce that flow, then 100 gallons is about 50 minutes. <S> An inexpensive CR2032 is rated at 230mAh, which is more than enough. <S> Chances are it's something even cheaper like a stack of a few LR44 button cells. <S> That can also be used to retain information. <S> The other options (electrochemical action, paddle wheel generator, miniature nuclear reactor) would be too expensive or would cause contamination. <A> That latter seems very likely - the unit could be storing its count in semivolatile memory while it's still powered by water flow, so the count would still be available later. <S> Power consumption may be extremely low. <S> Nobody can know for sure without disassembly, though. <A> Spehro Pefhany has explained that a small battery is sufficient to power the filter. <S> This Amazon Brita Faucet Filter <S> review says: ... <S> We just purchased a new one with the brass threads because the old plastic one's battery has died and does not show the filter life status. ... <S> Which seems like a strong indicator. <S> The circuitry may be as simple as a 'timer'. <S> The Brita Reading your indicator <S> site says: <S> Green: <S> Your filter is working. <S> Half Red and Green <S> : Replace filter soon. <S> Your filter cartridge has two weeks or approximately 20 gallons of life left. <S> Red: Your filter cartridge has reached the end of its life and needs to be replaced now. <S> So it only needs to measure how long water is flowing. <S> That is my expectation. <S> However this patent for a Water conductivity monitor for a water purification system suggests that their is an electronic comparison of tap water with filtered water. <A> Look for the patents. <S> CA2573448 A1 (Moen Inc., 2007 Canada) describes an LED, printed circuit board, and connection to a battery. <S> This patent did not specify what the LED color is determined by, other than vague possibilities such as time elapsed or gallons passed. <S> US20050279676 <S> (Zuhair Izzy et al, 2004) - ah, this one is juicier! <S> The description goes into detail about an impeller, counting of its rotations, and changing the LED color based on a preset cumulative flow quantity. <S> This patent also refers to a battery. <S> WO2005032690 A1 <S> (Brian Boyd et al, Water Pik, 2005) explains the use of a magnet on a turbine blade in the unit to activate a sensor. <S> A microprocessor keeps count, and sets the LED color according to total flow. <S> No mention of a battery or other power source, though. <S> There are other patents that may be of interest. <S> Some focus on mechanical or system design, and only lightly brush over the topic of interest.
| My guess as to the power source would be one of two: either replacement filters also contain a battery for the electronics, or else the power comes from a tiny water-flow-powered generator (which would also serve as a "pour detector"). Brita's documentation says that it counts time and "pours", apparently detecting water flow.
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Connecting diodes to a motor in H-bridge I'm attempting to design a three wheeled toy that is controlled by a remote connected with it via a cable. I first designed it with relays, however I had to change to transistors due to contrast between low power required by the toy itself and a high power demand of relays.I know that if there are vulnerable components in the circuit with a coil such as a motor, you should connect a diode in parallel with the coil in the reverse bias direction. Here's a problem, in my H-bridge, current will flow both ways through the motors and I'm confused as to how to connect diodes to my motors to protect the transistors . simulate this circuit – Schematic created using CircuitLab Also, if you have any idea how the design could be improved further, I'm all open for criticism.Note that on the schematic there are no specifications of the devices as these are liable to change with the choice of components available. <Q> So put them across the transistors. <S> The current-both-ways problem isn't present at the transistors. <A> Firstly I must point out that your circuit will NOT work as you have wired the transistors but assuming this is just a picture rather than an actual schematic read on... <S> You connect them across all four transistors in the H bridge. <S> For instance Q1 emitter has a diode anode and Q1's collector has the diode's cathode. <S> For Q2 the emitter has the anode and the collector the cathode. <S> Here's a typical circuit <S> : - Note that this uses PNP transistors for the upper devices but the diodes still remain connected the same way as if you were using NPN devices in the upper positions. <S> Note also that this circuit is likely to work (for a low power motor). <S> For higher powers I'd use this FET version: - What you will also find is that if there is a lot of energy stored in the motor, when the transistors are deactivated, that energy is dumped back to the supply voltage <S> so you may want to consider a large capacitor to soak up that energy. <A> One option would be to not use any protective diodes, but to parallel the winding for each motor with an MOV and a smallish capacitor, while making sure that your transtors' VCE is higher than the MOVs' clamping voltage. <S> The capacitor will help level the inductive flyback, and the MOV will prevent that spike from getting very tall.
| The diodes aren't protecting the motor; they are protecting the transistors.
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Method to lower voltage of a laptop charger I bought an IMAX B6 LiPo Balance charger that needs as input 11-18V but I have an unused laptop charger (the laptop is gone) having 19V with 4.74A. What is the easiest method to lower the voltage to 18V or even a little lower? (if I connect it as it is to the IMAX B6 it says something about input voltage is too high). Thanks! Sorin <Q> Connect four rectifiers in series, each rated to carry the DC input to the IMAX, and connect the string in the forward direction between the laptop charger output and the IMAX input. <A> I've done this before with good success. <A> The easiest method is to buy ready-made buck regulator board from fleabay like this . <S> The proper method is to get another power supply with suitable voltage.
| If you're comfortable with circuits and soldering, you could open up the laptop charger, find the resistor divider or zener diode on the feedback circuit, and solder on an alternative value. Otherwise, I like the diode-drop solution mentioned by EM Fields (though I would just use diodes, not bridge rectifiers -- and I don't think you'd need more than one or two of them).
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Snubber causing more problems than benefits When designing and building a simple motor controller i used 2 snubber circuits, one to protect the main triac and another one the smooth the communication between the phototriac and the main triac. See schematic below. simulate this circuit – Schematic created using CircuitLab The Phototriac is a VOM3052T and the main triac is a BTA137-600 . When testing the circuit however, no matter if the Microcontroller's pin is high or low, the motor always turns on. I found that when removing C1 the problem did not occur and the circuit perform as it originally was supposed to do. now that C1 is removed, the phototriac does not have a snubber anymore, but seems to work fine The questions is: Is removing C1 from the circuit a viable option, or will this cause the circuit to fail in the long run? why does the circuit not work when C1 is placed? ==================================================================== Based upon community feedback (thanks!) & more online research I decided to update the circuit: simulate this circuit Where the resistors are 2W the the capacitors are film. <Q> I don't know where you got that circuit(r2,r3,c1), but IMO it is rubbish. <S> Think of what happens when the foto-triac is not conductiong. <S> In effect you can remove it and R2 from the circuit. <S> Now you have the gate of your triac triggered via R3/C1, which delivers a phase-shifted current through the gate, 'conveniently' non-zero when the triac voltage is zero, thus switching the triac on after each zero-crossing, eactly as you observe. <S> But at the same time the snubber circuit should not conduct so much current that it causes the load to be switched on. <S> The trouble with your 'snubber' for the foto triac is that it conducts so much current that it always triggers the triac. <A> Let's talk order magnitude for ease of understanding. <S> Say your high voltage side is 50Hz mains power. <S> In that case the reactance of the capacitor is: \$X_C = <S> \dfrac{1}{2\pi fC} = <S> \dfrac{1}{2\pi\cdot50\cdot 47\cdot10^{-9}} \approx <S> 68\text{k}\Omega\$ <S> As the capacitor's reactance is much higher than R3 and therefore we can safely ignore R3 for sake of this discussion. <S> Also the maximum gate voltage will be in the order of 1V, which can again be ignored with respect to eg. <S> 100V mains voltage. <S> If the mains power is 100V then the gate current of the TRIAC will be approximately: \$I = \dfrac{U}{X} = <S> \dfrac{100}{68\text{k}}\approx 1.5\text{mA}\$ <S> For an average TRIAC 1.5mA is sufficient gate current to trigger it and the photo-TRIAC has no part in it at all. <S> When using a 230V mains, the current will even peak at approximately 5mA. <S> As Wouter mentions in his answer aswell, the current calculated is shifted in phase with respect to the voltage across the TRIAC, causing the TRIAC to reliably trigger. <S> I am unsure why C1 was included in the circuit. <S> The TRIAC doesn't act as an inductive or capacitive load, so there is little need for a snubber arounc C1 anyway. <S> The other snubber C2/R4 can be much more benificial, although a snubber should ideally be dimensioned for a specific load. <A> The snubber circuit improves the triac behavior butit imposes to the device stresses which limit its use. <S> http://www.thierry-lequeu.fr/data/AN437.pdf
| The function of a snubber circuit is to suppress rapid voltage transients, which would otherwise cause the triac to start conduction.
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Determine break in long circuit I have a circuit of switches like this. The switches are a long way from the supply and a long way from each other. S1 S2 S3 S4 S5[24VDC+]----->>-o=o----o=o----o=o----o=o----o=o--+ |[input ]-----<<----------------------------------+[24VDC-]--| Assuming the power supply is not accessible at all — it must always be on, and you cannot get access to 24VDC reference, is there a simple way to test at each switch with a multimeter to determine between which switches there is a break in the line? If you test with a multimeter, voltage across a closed switch is 0. Voltage across an open switch is also 0 as the input is high impedance. Any ideas? <Q> You will have to test the output of each switches in reference with the ground, and starting by the first switch. <S> If your voltmeter reads 0, it means the switch is open. <S> Prepare some cables you can carry around, connect one side to ground (or zero volt) and the other to your voltmeter common. <S> Example: <S> You measure the output of sw1, it's ok. <S> Then you measure the output of sw2 and you figure out it's zero. <S> So you close sw2. <S> sw1 sw2 sw3 sw4 sw5VCC-o <S> =o----o/o----o=o----o <S> o----o= <S> o- <S> +24 <S> 0 <S> And going on like this you should find the good position for all switch in one pass. <S> sw1 sw2 sw3 sw4 sw5VCC-o <S> =o----o=o----o=o----o/o----o= <S> o- <S> +24 +24 +24 0 <A> Inject a high-frequency AC that rides on the DC carrier, and "sniff" for that AC signal. <S> At each switch along the line, an FM receiver tuned to the same frequency should be able to pick up that modulated signal provided that all the upstream switches are still closed. <S> Expect low efficiency and considerable distortion, but it should be good enough to detect. <A> Put a resistor between the INPUT and the 24V output to deliberately draw some current, just for testing. <S> For example, at 2.4 kΩ resistor will cause 10 mA to flow in the loop. <S> Note that it will also dissipate 240 mW, <S> so it should be at least a "1/4 W" resistor, but "1/2 W" would be better. <S> Now you have a much lower impedance circuit and any ordinary multimeter should be able to see the voltage accross a open switch. <S> If more than one switch is open simultaneously, then this method won't work directly either. <S> Each open switch will read 0 V since there is no current flowing. <S> In that case, use clip leads to temporarily bypass all the switches with a short. <S> Verify that the load resistor is seeing the full 24 V. <S> Then check each switch by removing the bypass and checking the voltage <S> accross it with a meter. <S> Don't forget to replace the bypass around each open switch you find until you are done checking the whole chain.
| All the open switches will read close to 24 V. One simple means of doing so would be to clip a small FM transmitter's antenna connection to one end of the wire (no other connections) and drive it with a simple oscillator.
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How to make a RF additive mixer for two inputs in same frequency band? One of the most common designs for a SSB modulator ( phase shift method ) involves an additive mixer as the very last stage. Here is a block diagram for reference: This additive mixer must provide isolation between the two inputs. I am only aware of one method for doing this, which is to "couple" the signals onto each other using well-chosen filters. Here is an example in the case of a multiplicative mixer: However, this method only works when the two input signals are vastly different in frequency. How would I isolate two input signals when they occupy the same frequency band? Given that I am dealing with RF frequencies, op-amp mixers are out of the question. Surely there is an active circuit design which provides sufficient isolation, with low distortion and able to run on a single ended supply. <Q> I don't know what sort of carrier frequency you are considering <S> but if you feel it may be too high for conventional addition via op-amps etc. <S> , you might consider using two small wideband transformers; <S> one for each modulator and place the secondaries in series - now you get A+B and isolation. <A> Let's use proper terminolgy here, what you want is an 'adder', or 'signal combiner', to add two signals together. <S> A 'mixer' multiplies signals, and is a component in the larger diagram you have shown. <S> A single stage Wilkinson works very well over a 10% bandwidth, and multi-stage Wilkinsons can be designed to work multi-octave. <S> image from the above linked wikipedia article <S> It's easy to understand conceptually what is happening. <S> A signal entering P2 will arrive at P3 by two routes. <S> One is through a half-wavelength of line, inverting the signal, the other is through the resistor, without inversion. <S> The two components combine out of phase to prevent transmission from P2 to P3, and vice versa. <S> It's not immediately obvious that the sum port must be well matched to preserve the isolation between the component ports. <A> How about using an isolator ? <S> Also see this link . <S> You only need one of them on either the modulator 1 or 2 branch, but using two identical units simplifies the 90° phasing requirements. <A> Just a note, it is near impossible to make a single audio 90 degree phase shifter, <S> but there are dual phase shifters which accomplish the task by phase shifting both signal paths, one keeps 90 degrees spaced from the other. <S> Complex, but very doable. <S> As the differential phase is the key, this works. <S> As to combining, another solution: active combiners could easily block leakage. <S> In microwave range this could be difficult but not impossible. <S> Even isolation amps followed by pads and a passive combiner would accomplish this reasonably well.
| The most convenient circuit is a Wilkinson Power Combiner (wikipedia) which is used to either split one signal into two, or combine two signals into one, with isolation between the pair of component signals.
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Custom package in EagleCAD not aligning to grid I'm trying to make a custom package in eagle, and somehow the pads of my part got misaligned w.r.t. the grid. Is there any convenient way to correct for this when it happens? I know about the cmd + left click for an individual component, is there a way to do something like that for a whole group? <Q> If the pads you are placing are off the grid when they are supposed to be on the grid, you can select the pad using the Info button and type the desired X and Y coordinates manually. <S> If you have lots of pads that are all mis-aligned by the same amount, there's a trick you can do using the grid. <S> Say, for example, all of the pads are off by 0.015 inches. <S> Change the grid spacing to 0.015 inches, select all of the pads using the Group tool, select the Move tool, right-click and select "Move Group", and slide all of the parts over by one grid spacing. <S> Then change your grid spacing back to whatever you had before. <A> <A> In case of SCH mismatching grid in the sch, Run the snap-on-grid-sch. <S> ULP in the schematic. <S> the components automatically fix to 100 mil /0.1 inch grid in sch. <S> Then you can run ERC and fix the unconnected errors. <S> You must check the ERC after that ULP.Its working 100%. <S> In package. <S> you need to align the pad as per the datasheet. <S> you can use 1.27 mm <S> is the default center grid for your tplace construction. <S> Otherwise you can change the grid as much low as you can.
| You can also change the ALT spacing to a very small value and after you group the pads and hit the Move Group command you can press ALT and move them however you want according to the ALT grid spacing set below the normal grid spacing (the same window).
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Improving RTD Sensor filtering and readout For an application a temperature sensor is required. an accuracy of around +-3 degrees is acceptable. For cost reasons I have chosen for this RTD . The sensor is sampled by a 12 bit ADC of an Microcontroller. The RTD is placed in an wheatstone bridge with a regulated 3.3v supply (looking at the supply with an scope I did not find much noise/ ripple). The Wheatstone Bridge is connected to an MCP6N11 Instrument amplifier. simulate this circuit – Schematic created using CircuitLab The RTD is mounted on a separate board with about 50cm of wire, which runs through a noisy environment. Therefore the filter stage is placed between the bridge and amplifier. The desired temperature range is between 50 and 200 Celsius. The gain of the amplifier is set to 15, and is supplied by a single 3.3v supply. Unfortunately, the measurements are constantly off by 30 to 50 percent. As well as there is still some noise sneaking through the filter. The amplifier is connected directly to the 12 bit ADC. What can i do to improve this circuit? <Q> Consider getting rid of the bridge and instrumentation amplifier. <S> Feed the RTD (via 10k) from the same reference voltage that your ADC uses - this is called ratiometric measurements and removes one of the big error sources i.e. because RTD and ADC are both ultimately fed from a common reference voltage, it doesn't matter if that reference voltage drifts. <S> So, from memory a 1k RTD will have a resistance of 1000 ohms at 0 degC and about 1690 ohms at 100 degC. <S> If your ADC reference voltage is 3.3 volts and the RTD is fed from 10kohm, <S> the voltage change between 0 degC and 100 degC will be 0.300 volts to 0.477 volts. <S> This is a change of 177 milli volts for a temperature change of 100 degC <S> Your ADC is 12 bit, therefore it has a resolution of 3.3 volts/ 4096 = <S> 0.80566 <S> mV. <S> So, for a 100 degC change, the input voltage changes 0.177 volts with a resolution of 0.80566 mV <S> - this means your ADC can resolve 219.7 steps in a range of 100 degC OR, its actual resolution is about 0.5 degrees C. <S> This should easily obtain the accuracy you want. <S> By the way, you need to calculate the full range up to 200 degC <S> but i don't see any problems in achieving that to the accuracy you want. <S> In short, reduce your errors by getting rid of stuff you don't need. <S> When it comes to filtering, sure use an external low pass filter but also filter in software by oversampling the signal and averaging. <A> However, you must remove C4- <S> it should not be present, or no more than a few hundred pF. <S> The way you've shown it, it is highly likely to cause instability. <S> If you can add maybe 1K between the output and C4 without causing ADC errors, it can stay. <A> You have a bug in your schematics. <S> You have to connect the upper wire between R1 and RTD to positive output of 3.3V power supply (there shoul be a joint instead of crossing). <S> And also disconnect the positive power supply from the R5. <S> And also for to have the input not floating, you should ground the negative output of 3.3V power supply to analog ground. <S> PS: <S> And as Spehro Pefhany mentioned before, the C4 cant be conected directly to tth op-amp output. <S> This will cause oscillations os op-amp. <S> Place some resistor in between output and C4 (which i did not repair in my picture).
| If you are using the same supply for bridge energization as the amplifier and the MCU, I don't see that much wrong with your circuit,
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summing op-amp: 2 voltages in parallel I didn't quite get how this circuit from falstad works: http://www.falstad.com/circuit/e-amp-sum.html Specifically, how can 2 voltages in parallel add up when negative input (ideally) has infinite resistance? <Q> There is no current flowing into the inverting (negative) input, because the input impedance of an ideal op-amp is infinite, therefore the current flowing through the feedback 1k\$\Omega\$ resistor <S> (let's call it \$i_3\$) is the sum of the two currents flowing through the two other resistors (\$i_1\$ and \$i_2\$). <S> The inverting input is a virtual ground, as the op-amp with negative feedback always tries to make the voltage between its inputs zero. <S> Let's name the two inputs (the 200Hz and the 20Hz ones) \$u_1\$ and \$u_2\$, and the output voltage <S> \$u_3\$. You can now use a nodal equation for the non-inverting input <S> (let's say every current is flowing to the right): \$i_1+i_2=i_3\$ <S> \${u_1-0 \over {1k\Omega}} + <S> {u_2-0 \over {1k\Omega}}={0-u_3 \over <S> {1k\Omega}}\$ <S> Now multiply the equation with 1k\$\Omega\$: <S> \$u_1+u_2=-u_3\$ <S> There you have the sum of the two input voltages on the output (multiplied by minus one). <A> Actually not the voltages add up but the currents. <S> Since currents are proportional to the voltage across resistors, the voltage of the output must be equal in magnitude (but opposite in sign) to the SUM of the voltages at the left terminals of the two resitors. <A> Specifically, how can 2 voltages in parallel add up when negative input (ideally) has infinite resistance? <S> You can make the reasonable assumption that an op-amp has really high gain. <S> With really high gain there can hardly be any difference in the voltage between the non-inverting and inverting input. <S> If there were a measurable difference then, due to the op-amp's massive gain, the output from the op-amp would probably be hard against one of the power rails. <S> But it isn't, therefore the voltage at the inverting input is "virtually" the same as the voltage at the non-inverting input With non-inverting input at 0V <S> , the inverting input is "virtually" at 0V because of the high op-amp gain. <S> This means that your input resistors can be said to connect between signal and "virtual" ground. <S> This means that the current that flows in either input resistor is subject only to its input voltage and its resistor value. <S> This in turn means that both signal-input currents (due to input 1 and input 2) flow also through the feedback resistor as I1+I2 <S> This in turn means the op-amp's output voltage is feedback-resistor*(I1+I2 <S> ) - don't forget the virtual ground when you read this last sentence.
| All currents flowing to the negative input (which is assumed to be high impedance, i.e. "open" and at ground level) must add up to 0.
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Crystal or Crystal Oscillator for time sensitive device A device that I'm working on is controlled by ATmega328P microcontroller.This will be some sort of a watch, therefore I want it to be as precise as possible. For instance if left to run for a whole year I want the time deviation to be not more than absolute value of +/- 10 seconds by the end of the year. (I'm not saying the battery power will be enough for the whole year..) Is crystal oscillator more precise than crystal with caps driven my micro? If so, are crystal oscillators more commonly used in time sensitive applications? Or just a regular crystal will be sufficient and more suitable? (Especially when they will not consume power like oscillators) I don't need atomic precision of +/- 0.000001 sec/year. <Q> 10 seconds / year is 0.3 parts per million error (or as the time-nuts would usually put it, an error of 3 * 10^-7). <S> A decent watch crystal is usually specified as +/- <S> 20 ppm (2 * 10^-5), which is far worse than what you're asking for. <S> The higher-frequency crystals used for clocking CPUs are usually worse than that. <S> Temperature-compensated oscillators (TCXOs) are better, in the 10^-6 range, but still worse than your spec. <S> Ovenized oscillators (OCXOs) can be better than 10 <S> ^-7 <S> but require watts of power, likely more than you can afford running on battery. <A> A GPS receiver is an excellent timing source. <S> There are small SMD modules which require between 30 and 100mA from a 3.3Vdc source, depends on the specific module. <S> Beside the power requirement, it also needs an antenna and the GPS signal (Can't work well underground or inside buildings). <A> <A> A high precision real time clock like the DS3231 is another option. <S> It can get you up to ±2ppm. <S> That is pretty nice, but would still result in around ±60s/year. <S> Are you sure you need ±10s/year?
| If the device is not portable, you could use the AC power line as a timing reference, as the average frequency of the power distribution system is held to extremely tight tolerance.
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Is it bad if I held solder with my mouth? When soldering, I have on a few occasions, used my teeth to hold a strand of solder while my hands were busy. I do not do this frequently, but sometimes I forget that solder is made of lead and I instinctively use my mouth. How bad is doing that for my health? Is it very dangerous? I use manly lead based rosin-core solder. <Q> It's a question of bio-availablity, which is facilitated by acids and lead that is more "absorbable" when it is in certain compounds. <S> Certainly Lead accumulates in tissues but even sub toxic doses would take years to accumulate. <S> The Romans used lead based wine mugs because the acidic wine would dissolve the lead which apparently made the wine more palatable (sweeter). <S> No one is willing to do the taste test of that one, that certainly would be unethical for a tester to ask someone to do that. <S> It took the Roman years to get to toxic levels. <S> As long as you weren't chewing down on the lead and having a good floss with it, it is very likely that very little transferred into your system. <S> I know this because I have myself had blood tests. <S> The Dr. told me it was unlikely that they could detect anything as the lead tends to be gathered into tissues and does not tend to be available in blood. <S> the next option was a biopsy which is more dangerous that any possible lead given predicted exposure. <S> TLDR; don't worry about having done this, but don't continue. <A> Heavy metals don't really flush from your system, they accumulate- <S> so its probably a really bad idea. <S> I try to use helping hands or blutac or piles or desk clutter to hold my bits and pieces in place. <S> I have also seen people stab components into old erasers and other bizarre things. <A> Good solder is only 37% lead, but I still wouldn't be doing that. <S> As well as outright lead poisoning, there could be subtle effects below the poisoning threshold. <S> If it drops your IQ by 10 points would that be acceptable? <S> Edit: As @Blrfl says, loss of mental acuity is a common symptom of lead poisoning . <S> It's not clear that there is a threshold below which lead is safe, especially for kids . <S> Really, "there is no safe level of blood lead in children," said Christopher Portier, of the Centers for Disease Control and Prevention . <S> He heads the agency's environmental health programs. <S> Also, if you don't care about your body, please think of the electronics - your viscous organics-laden spittle and drool won't do much for the integrity of the joints. <A> Even if the probability turns out to be very small, there is no need to settle for "small exposure" when you can easily keep it at zero! <S> Obviously, if you have (or get) <S> a "parts holder" it would be easier and much better to do the soldering.
| Although your exposure to lead poisoning may not be too great, if you are a young person and do it frequently, it could become significant. What I do, instead of holding the solder with my teeth, I hold it with any pair of objects I have on hand - pliers, hammer, blocks of wood, etc.
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Why would a switching power supply fail with a battery supply? I have this power supply circuit I have been using for a couple of years: I have put it on a lot of boards, and ranging input voltages up to 50 volts (as long as the capacitors are rated that high). I had never been able to get it to fail, and I even load tested it drawing the maximum off both rails, and it shutdown before it failed. However, I had always powered the input from an AC/DC converter. I recently began using a board with this supply on it in a battery powered application (24 volt - 2 12 volt lead acid in series). Everything worked fine for months. I was charging the batteries separately using a car battery charger. I then bought a 24 volt charger , which seems to work well. After charging the batteries (there are two supposedly identical systems) I powered up and one of the power supply circuits blew on the board. In the second system all was fine. To summarize my troubleshooting from that point forward: that one board that didn't fail on the original power up after charging is the only one that I can get to work on this charged battery system. I have replaced the LT3988 and tested it in the lab off a lab supply (up to 40 V) and AC/DC brick, and it works perfect. I take it back to the fielded unit and plug it up to the battery (only the supply connected) and the switcher immediately smokes. I populated an entire new board, the switcher works on a lab supply, and immediately smokes on these batteries. The one that works, works on both systems...but I cannot recreate it. So...what problems can you guys think of when using a battery supply that are not encountered on an AC/DC converter? I can think of a lot of more problems powering a switcher with a switcher, so I need your help. Edit: When powering this circuit from an AC/DC brick, I usually have the brick plugged into AC and just plug the barrel jack connector into the power input on my board. When powering from a lab supply, I obviously just press the button. When powering from the batteries there is a switch in the line between +VDC and my board. <Q> The LT3988 data sheet mentions the use of ceramic capacitors at the input: A final caution is in order regarding the use of ceramic capacitors at the input. <S> A ceramic input capacitor can combine with stray inductance to form a resonant tank circuit. <S> If power is applied quickly (for example by plugging the circuit into a live power source), this tank can ring, doubling the input voltage and damaging the LT3988. <S> The solution is to either clamp the input voltage or dampen the tank circuit by adding a lossy capacitor in parallel with the ceramic capacitor. <S> For details, see Application Note 88. <S> You might want to add some extra input impedance to your source. <A> The available current is much higher from lead-acid batteries, and it could have a very fast rise time. <S> I would expect problems if the C37 4.7uF is a tantalum. <S> They are very intolerance of high currents and will tend to produce fireworks, burning, smoke, small explosions etc. <S> If the chip is failing, that's more of a mystery. <S> Maybe some extra inductance somewhere. <S> A larger aluminum electrolytic cap in place of C37 and perhaps a series polyfuse (you should have a fuse somewhere as the batteries can produce enough current to melt wires and cause a hazard as a result). <S> If the incoming power return is fed into the wrong side of the circuit, for example. <S> I don't like that the schematic is 'flipped' left-right. <A> In addition to other suggestion to deal with the switcher's own current at startup, consider increasing the value of C38 and C39 to increase the soft startup time. <S> This will ramp the voltage, and thus current, on the output up over a longer period of time, reducing the startup current. <S> If you can, connect an oscilloscope in the field with a newly built circuit. <S> Measure the input voltage and current, and the output voltage and current while connected to the load, and from the AC power supply to the battery power supply. <S> I expect the difference between the two will become immediately obvious, and the solution won't be far behind.
| Edit: Another possibility is that there is a layout issue that is resulting in transient reverse bias of the chip. It's likely that the start up current is causing the issue.
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Current drawn by a permanent magnet DC motor I have been working with PMDC motors for a while now. In these motors, as the load increases current drawn by the motor also increases. I was just wondering what is the theoretical reason behind this change in current? <Q> As the motor spins, the magnetic field induces back EMF in the windings. <S> The induced voltage has opposite polarity than the supply voltage. <S> Theoretically, when no load is present, these voltages cancel out and no current can flow. <S> When the motor is loaded and slowed down, the BEMF voltage decreases and allows more current to flow. <S> When the motor is stalled, there is no BEMF and current is limited only by winding resistance. <A> Every electrical machine has two key constants associated with it K\$_E\$ = <S> BackEMF <S> constant \$\approx\$ <S> V/\$\omega\$ (machine topology dependant) <S> K\$_T\$ = <S> Torque constant \$\approx\$ <S> Nm/ <S> A (machine topology dependant) <S> Theoretically they are the same, practically K\$_T\$ is defined at maximum current and thus it contains aspects of stator saturation. <S> Now as K\$_T\$ is defined as Torque per Amp <S> ( Nm/A) so more amps equates to more torque. <S> But why does current equate to torque? <S> essentially the stator is one big electro-magnet. <S> If you pass current through it a magnetic field is created. <S> Increase the current and a stronger magnetic field is created. <S> A PMDC machine is essentially a synchronous machine (with its corePack machined & the rotor magnets aligned such to provide a trapezoidal-like BackEMF) <S> so... if an AC signal is applied to the stator windings, an AC airgap field will be generated and the rotor will attempt to "lock on" and rotate with the field. <S> As to the original question though I have been working with PMDC motors for a while now. <S> In these motors, as the load increases current drawn by the motor also increases. <S> I was just wondering what is the theoretical reason behind this change in current? <S> This actually isn't unique to PMDC machine, or PMAC machine, Induction, etc.. but motor-controllers. <S> An increase in load at the rotor will attempt to slow the rotor down. <S> With a BLDC machine connected to a speed-controller, it will attempt to counter the slowdown in rotor speed by increasing the current demand and thus increase the torque generated to oppose the load torque. <S> Now simple BLDC controller which only manage commutation rather than current control will equally experience this because the commutator will energize the coils in sequence and the rotor will just accelerate ( \$\omega = \int T(t)/J\$ ) until the BackEMF = <S> Applied voltage <S> As the rotor load is increased this maximum speed can no longer be maintained as there is a need to provide higher current & thus: Terminal voltage <S> > <S> BackEMF <S> + \$L_w + R_w\$ <S> So... the speed must decrease to facilitate an increase in current == increase in torque <A> That power has to come from somewhere. <S> Power in = <S> voltage x current. <S> Power out = torque x rpm. <S> Torque results from passing current through the stator windings at right angles to the magnetic field created by the permanent magnets in the rotor ( Lorentz Force Law ). <S> If the motor had superconducting windings (ie. <S> zero resistance) then it would not slow down at all under load. <S> The back-emf would remain equal to the supply voltage, and the motor would draw just enough extra current to create the torque required to keep it spinning at constant speed. <S> However normal wire <S> does have resistance, and this drops a voltage proportional to current ( Ohm's Law ). <S> That voltage subtracts from the back-emf, so the motor slows down.
| As the load increases it draws mechanical power out of the motor.
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How fast should a Triac be switched for a dimming application? I want to create a dimmer using a Triac - it will have a duty cycle where it is on for x% of the time, and then off for Y% of the time - my question is how often should this on/off cycle happen? I was thinking of using a BT137 - this is for a 120V / 60Hz / ~300W max circuit, however if there is an alternate/better option, I would be willing to switch. I would like to avoid any "humming" noise if possible (which I seem to get from commercial dimmers). <Q> So, you're stuck with the mains frequency (times two, since you can turn it 'on' on either the positive or negative half-cycle). <S> You would normally switch the triac on at some point after the zero crossing (a pulse of something like 100usec is sufficient if the load is resistive) and then wait for the next zero crossing. <S> You should keep the timing similar on positive and negative half cycles. <S> Note that it's significantly harder (2-3x the current) to trigger when MT2 is negative and <S> the gate is positive. <A> Normal dimmers 'cut out' the first part of the half sine wave; the TRIAC turns on when the voltage reaches a certain voltage, stays on until current drops below some value and it turns off again. <S> So the switching frequency is twice the mains frequency. <S> The usual circuit is like the one described here http://en.wikipedia.org/wiki/Dimmer <S> Once a TRIAC is on, you cannot turn it off again with some control signal like you can with a transistor. <S> So there not much choice in the control frequency; at most you could switch at a integer fraction of the mains frequency like 30 or 15Hz <S> but the power delivered to the lamp would be irregular and not much of dimming capability left over. <S> The buzzing sound is mostly generated by the inductor that's part of the dimmer, it's there to supress EMI and required to pass certification. <S> The filament of the lamp can buzz too. <S> You could use some MOSFETs to PWM at a higher frequency, but that won't be a simple circuit. <A> Do a quick Google search for "diac triac dimmer". <S> Dimmers don't get much simpler, nor much more reliable, nor even much more effective, than that.
| Triacs do not switch 'off' except when the current drops to (almost) zero - a few mA in the case of the one you've linked to.
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Does CKDIV8 save battery life when programmed? On ATmega328P there is a CKDIV8 fuse bit that when programmed divides clock by 8. I'm wondering if the power consumption is equal on these two setups: micro with 8MHz external crystal and CKDIV8 fuse bit programmed micro with 1MHz external crystal and CKDIV8 fuse bit not programmed In both cases I get 1MHz clock which is sufficient for my application.The reason is I can't find any cheap 1MHz crystals, for some reason they are incredibly expensive $15+ Digikey Store 1MHz crystals .Comparing to 8MHz crystals with better ppm (+/- 10ppm) for less than $0.50 Digikey Store 8MHz crystals <Q> The datasheet has a graph that shows the standby current at different clock frequencies and supply voltages. <S> The difference is around 160uA. <S> So yes, it consumes slightly more power, but the difference is much smaller than the difference of running the MCU at 1MHz or 8MHz. <A> It's possible that lowering your clock wouldn't give you a lower consumption. <S> It depends on how heavy your peripheral use is. <S> But if you're using peripherals that need to be constantly running (e.g. counters) or if you don't use sleeps modes at all (i.e. do idle loops instead) you'll probably benefit from a slower clock. <A> I'm wondering if the power consumption is equal on these two setups <S> The CKDIV8 Fuse determines the initial value of the CLKPS bits. <S> These bits just define the division factor between the selected clock source and the internal system clock. <S> The power consumption is (almost) proportional to the internal system clock, not the clock source frequency. <S> So we can ignore this factor. <S> However, the data sheet diagrams do not allow a good comparison. <S> Further (as jeroen74 pointed out), the oscillator circuit may require slightly more power with increasing frequency (about 150uA from 1MHz to 8MHz). <S> But the data sheet values refer to the standby characteristics, so its not clear if that translates to the active mode. <S> If that is the case, it is clearly relevant with respect to the 900uA active supply current (@5V).
| If it were to be only the CPU, you would be doing the same work in less time, so perhaps the consumption is lower at 8 MHz, provided you use your power saving modes wisely. It seems to make a small difference if the clock source fuses are set to either "Full Swing Crystal Oscillator" or "Low Frequency Crystal Oscillator".
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Solder Mask and Pad Area (SMD vs NSMD) I have been trying to figure out the "right" way to size pads for surface mount components, and I have found some conflicting wisdom floating around the internets. Lets say I have this hypothetical part that requires two square pads, the centers of which are 15 units apart, and the edge that are 5 units on edge. This SO post suggests that I should have my copper be the exact size of the pads, and the mask be slightly oversize to account for shrinkage and/or misalignment. This seems somewhat reasonable, as relative shift/shrink/misalignment can be tolerated. Under this scenario I might want the the masks to be 6x6 holes with the same center-to-center spacing as the underlying copper. Conversley, in a different scenario if you were to oversize the copper to 6x6, you would gain the same tolerance to misalignment,and additionally some extra adhesion holding the pads down. This approach is suggested in some BGA/QFN packages I have been looking at. ( Example 1 ) ( Example 2 ) ( Example 3 ) I realize that there isn't really a "Correct" answer, but what motivates one arrangement verses the other? <Q> It depends to some extent on where the part is metalized. <S> If the pad is restricted to the bottom such as on a QFN package the optimal pad will be different from that for a ceramic capacitor, say, that is metalized on the ends- where you'd want a fillet angle to the pad. <S> If you can get ahold of the IPC standards (maybe at a University library), I've found them to be very helpful. <S> Some software (eg. <S> Altium) has an IPC wizard that takes into account the IPC recommendations (with options to edit the program's selections) and (for some parts anyways) has three levels of density to select from (nominal, high density, low density). <S> I agree with @Vladimir that smaller pads can be a pain, and in fact can make the boards hard to inspect because you can't see the fillet easily through a microscope that is vertical. <S> For what it's worth, the solder mask used in a recent board with QFNs used a 0.106mm (4 mils) default (rule-based) expansion factor, which left slightly less than 4 mils solder mask sliver <S> between 0.35mm wide QFN pins (0.65mm pitch). <A> For regular non-BGA components, non-soldermask defined pads (where the mask opening is larger than the pad) are generally preferred. <S> The copper can be made to tighter tolerances than the solder mask. <S> Also, the thickness of soldermask that creeps onto the pad can prevent components (e.g. QFNs) from touching solder paste on the pad. <A> Different sources give conflicting advices because they refer to different needs. <S> You need three sizes to take care of: the copper area, the solder mask and the paste mask. <S> Their sizes are usually determined by: <S> The component itself: contact position and tolerances must be taken into account. <S> Fabrication tolerances. <S> Fabrication houses have different values for these, which you need to learn in advance, or at least estimate. <S> With these you can calculate what will your pads look like if everything goes off-spec. <S> The soldering method: <S> SMD components that you might attempt to hand solder (e.g. in a prototype) should have pads that go a little over the component so you can place your soldering iron tip. <S> If you are going fully automated, this is not an issue. <S> The PCB finish (hot air leveling, gold flash, etc.) affects the "wettability" of the pad, which affects you if you're going into very small stuff. <S> The available space in your board. <S> If you need to fit many tracks and components in a tight space you will want your pads to be at their absolute minimum. <S> The thickness of your stencil determines the amount of solder paste that goes in your pad. <S> You don't want your pad to overflow ( <S> e.g. big pads like center pads in SOIC chips), so your paste mask needs to be adjusted to avoid that if your your thickness is high. <S> Mechanical properties. <S> With large components like aluminium capacitors, the firmer the solder joint, the better. <S> This list is most probably incomplete, but it should give you an idea of what values you need to take care of in your particular project.
| The solder mask is important to avoid solder bridges, so if your solder mask tolerances are poor and your pads are too close to each other, you might need to make them smaller to make room for the solder mask.
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How to detect Resonance? I've made many RC circuits and RLC circuits ( series and parallel resonance).I want to make sure that these circuits work and make resonance, but I don't have CRO device. Is there any way to detect the resonance using Multimeter or something like that ? I tried to hear the resonance. Human ear can hear 20 to 20K Hz, so I used 100uF coil and 1 uF capacitor to make Resonance of 15KHz frequency. ( 12 volt / 50 Hz AC power source ).when I connect the 8-ohm speaker, I didn't hear sound. actually, it was a very low sound. It looks like the sound that I hear when I connect the speaker to the power source directly. <Q> When the coil of the oscillator is placed close to a resonant circuit, and the oscillator is tuned to the resonant frequency of that circuit, the resonant circuit will draw power from the oscillator and cause a lowering (dip) of the grid current. <S> Grid dip meters were constructed so that the inductor was external to the meter and could be easily coupled to a another circuit and also could be interchanged to allow a wide tuning range. <S> Newer versions used a transistor oscillator instead of a vacuum tube but the principle remains the same. <S> These meters typically covered a range from about 100 kHz up to 250 MHz and were calibrated with an accuracy of a few percent. <S> They were made by companies such as Eico, Heathkit, Miller and Kenwood. <S> You should be able to obtain one at a very reasonable price on Ebay. <A> What basically happens with resonance is that you insert a tiny bit of energy in a circuit at given times (resonance frequency) and because of the properties of the circuit this tiny amount of energy has nowhere to go other than swinging up an down from capacitor to inductor. <S> What happens if you keep supplying this tiny amount of energy is that the total amount of energy in inductor and capacitor builds up and as a result the voltage and current in the circuit increase until an equilibrium is reached. <S> At that point the load resistor (or your oscilloscope, or multimeter, or ... ) drains just as much energy from the circuit as is being fed. <S> This means: <S> For a parallel circuit you want a alternating voltage source with a high impedance (or even better a current source), so you feed a tiny amount of energy into the circuit and at the same time the source doesn't load the resonant circuit too much so it will build up energy. <S> The same goes for your meter, you want a device with a high impedance to check for resonance, otherwise the meter will drain too much energy from the circuit to measure the resonance. <S> The same goes for a series circuit but with low series impedances and a low impedance voltage source. <S> The 100 ohm resistor and the speaker will dampen the circuit too much to notice too much of the resonance. <S> The easiest way to prove resonance in this circuit is to use a voltage generator which you can vary the frequency of. <S> That way, despite the load resistors, you should be able to hear the volume change with changing frequency. <S> Do remember though that your speaker will have a certain volume/frequency response too (probably including resonance frequencies of its own). <A> You have a PC (because you are posting here). <S> Presumably it contains a sound card - use it for measuring the resonance using free software that can both analyse a waveform and produce a stimulus. <A> If your circuit generates enough power, you could connect it through a coupling capacitor to a full-wave bridge rectifier with a parallel filter capacitor. <S> If your circuit is oscillating, you should be able to read DC across the filter capacitor with your multimeter. <S> You may want to use Schottky diodes for that bridge rectifier, maybe not, if your circuit generates enough voltage and you're only trying to detect oscillation.
| A relative low cost device to check resonant circuits is what used to be called a grid dip meter because it measured the grid current in a vacuum tube oscillator circuit.
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Using an SPST knife switch to power on/off a PC The title is it in a nutshell I've got an antique knife switch that I'd like to use to power up (and down) an embedded PC. The PC power switch is essentially two wires, a ground and +5v (measured with a voltmeter). Normally, a momentary pushbutton is used. You press it for a short time and the PC powers up. Press and hold while the PC is on and it powers down. I'd like to use a knife switch, so the action would be 1) knife switch disengaged, pc is off2) engage knife switch, PC powers up, which required a +momentary+ contact, but the knife switch is a constant contact.3) PC is on while knife switch is engaged.4) disengage knife switch, PC powers down, which requires another momentary contact, but a little longer duration. I've found a simple circuit with a 5v reed relay and a 470uf capacitor that could work. But.... I was really hoping to only connect the circuit to the PC's power switch wires, and not require a separate power supply line. Is something like that even possible? EDITI don't believe I can add images to the comments so I'm adding the circuit I mention below here. I was mistaken about the 2 capacitors though. It's two relays to control 2 separate buttons, one for the engagement of the toggle, and one for the disengagement. see this post http://www.simprojects.nl/toggle_to_momentary.htm I originally thought I could use something like this, till I realized I'd have to run a completely separate power (and possibly ground) line for it. I +could+ run the additional lines. I'm really just wondering if the circuit I'm describing is even possible. It seems like it should be, but my EE background is pretty limited. EDIT: Switch is SPST. <Q> I was really hoping to only connect the circuit to the PC's power switch wires, and not require a separate power supply line. <S> Is something like that even possible? <S> Converting a toggle or latching switch, into a momentary output, typically requires some active parts, like logic ICs, a 555/556 timer, a microcontroller or dedicated IC. <S> A microcontroller with a simple interrupt code would be easy to setup for your requirements, and only need the microcontroller, a capacitor, and power. <S> Honestly though, you are already running two wires, one of which is either ground or VCC, a third wire would be trivial . <A> I love knife switches and I dislike solutions involving adding circuitry, <S> so... <S> If you are willing to modify the switch subtly you could probably make it give out a momentary pulse as it is slid home - something as simple as putting some kapton tape or something insulating over or inside the contacts so that only the leading edge makes contact as it first touches the contact, and is then insulated once it is fully seated. <S> You could also cut, file, stamp or otherwise create a notch, dimple, dent or cutout in the contacts or knife (harder to reverse) to give a similar effect. <S> Going a stage further, you could put a momentary switch under some part of the mechanism, for example near the hinge point so that it gets pushed momentarily by a bump or something on the knife switch as it's moved from open to closed. <S> Adding a fairly minimal amount of components (1) you could use a small beam-break type opto-sensor where the knife cuts the beam, if there was a small hole or slot drilled in the knife it would give a momentary pulse as the knife broke the beam & then the hole allowed it back. <A> This depends on a SPDT or DPDT knife switch, not a simple SPST knife switch. <S> Thinking about it, since you are willing to modify the knife switch (not visibly) or electronics, you could do it, if you add a battery. <S> Since the battery provides the power, no need for the extra cables. <S> A slight variation to the Cap + Relay circuit by replacing it with a transistor or opto coupler (A led + light sensitive transistor really). <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The first requires that one of the two wires of the power switch be ground. <S> The second does not. <S> I am not sure if the second will work on both sides of the toggle switch, might require a second opto-coupler in parallel (with the led side reversed) A coin battery, knife switch, and three smallish parts. <S> Can be made the size of a dime with a smd capacitor. <S> Otherwise, with a SPST Knife Switch, you can go with a microcontroller, again, powered by the coin cell . <S> simulate this circuit Internal pullup on the knife switch. <S> Any microcontroller can work, and since all it does is listen for the switch then pulse for a second or two, battery life will be years.
| Your capacitor + relay circuit is one of the simpler methods, but the pulse will be equal length on both connect/disconnect of the knife switch.
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Trying to understand negative voltages If I were to have an IC (Just some standard 5v microcontroller), and for some reason I only had a -5v supply, could I technically connect -5v of the supply to the GND pin of the micro and the ground of the supply to the 5v pin on the micro? I've always thought this would work, but everybody I've talked to has said it's a bad idea. I assume they say it's a bad idea because if you had a device on the same system that was spitting out voltage spikes down the central ground, it could overvolt your micro. So, more of a 2 part question Is powering a 5v micro with a negative voltage on the GND pin possible? If so, what are the downsides of doing it this way? <Q> It is totally possible to power the MCU off a -5V rail as you describe. <S> The person who told you about spikes in the "GND" which becomes the MCU VCC pin should be aware that spikes could just as well show up on the -5V rail. <S> The MCU can be over voltaged simply by applying too big of voltage difference between the MCU VCC and GND. <S> The main downside of using the -5V supply is the implication that it comes into a system that also has a +voltage rail or two. <S> Any circuitry in the system that uses such +V and the "GND" as its supply will not be able to interface directly with the MCU. <S> Special level shifting circuitry would have to be applied to translate the negative domain voltages of the MCU to the positive voltage domain of the other circuitry. <S> If the system power supply is just the single voltage rail that you describe as -5V <S> then just switch the leads around and consider it a +5V supply. <A> Voltage is a potential between two points; therefore, a -5 volt supply would work when connected as you mentioned (PSU -5v to IC ground and PSU gnd to IC +5v input). <S> If this were your only connection, there would be no issues (assuming that the PSU was supplying stable power). <S> The problem would be if you have other components in that circuit that are treating the power supply ground as reference (zero) voltage, since their ground potential is now 5v different from the IC's ground potential, leading to potential (sorry, I couldn't resist) undesired current flow or voltages between various points in the circuit. <S> In the worst case stuff would start burning up. <S> There is nothing special about ground; it is just a reference for other voltages throughout the circuit. <S> Ground in a given circuit could be at 10kv relative to earth ground and you could still be supplying only 5v to your IC if the other rail were at 9995v relative to earth ground. <A> VCC is still going to be +5vDC above ground. <S> If you are using a regulator, like a 7905 (-5v regulator) you can do plenty with your uC. <S> BUT if in the circuit, you use something that uses a 0v GND reference, (for an ADC for instance) <S> you are then in for a mess. <S> If you choose -5v as GND, and stick with it for the entire circuit, there shouldn't be an issue. <S> Just do not connect anything back into the circuit that uses 0v as GND that the uC could see. <S> if your transient polluted 0v was your ground reference or your VCC. <S> Make sure the VCC pin is <S> +5v more than the GND pin, <S> regulate and stay consistent for the entirety of the circuit. <A> If the IC and the peripheral components it connects to (exclusing the power supply) are connected to nothing, the connection you propose is fine. <S> That would be the case if the IC is a microcontroller and its peripheral components are leds, and isolated buttons. <S> However, as soon as you connect some external devices, your negative power supply is likely to become a problem. <S> That's especially true if your power supply is connected to the electrical grid (mains): perhaps the ground of your supply is connected to earth, and the ground of the other supply of that external device too. <S> Even if the ground output of one or both supplies is not earthed, you might run into problems due to capacitive leakage from mains to ground output, and power supply noise, which turns into a signal; e.g. in an audio or video context. <S> So in summary: for an overall system with at most one mains-connected power supply, what you propose is fine (and very justifiable if a negative regulator has better characteristics than a positive one). <S> Problem starts when your power supply is mains-connected and you connect the GND of your IC (the -5V output of your PSU) to the GND of another system that has a mains-connected power supply.
| Now when you connect the GND of your IC (the -5V output of your PSU), and the GND of that external device (also the ground of its power supply), you actually short-circuit your power supply, thru the earth connections of both power supplies. If you are getting voltage spikes on your "ground", it wouldn't matter how it was connected as your uC would see it
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How to select and use Strong Magnet together with hall sensor and amplifier? With reference hall effect IC sensor (hopefully on topic as normal EE work) used together with these silver color magnet, as this example of 20 by 10 mm from ebay, Do they have different formulation (chemical composition) with different property, strength of magnetic field and cost? What are the popular (most used, low cost, widely available) formula? What is the likley formula as in the photo? How is 'radiated' (field caused by magnet with 0 to 30mm) magnetic field strength measured? Unit and measurement procedure (distance, orientation). For antenna and loud speaker, they produce sound or wave over distance. We measure them by signal level and directional property diagram. What is the equivalent for these magnet? What typical mag field in G or T at distance 1 to 20mm for the 10 by 20mm magnetic as photo? Rough range needed for choosing sensor chip which rated at mv output per G or T. How is the magnet shape, rectangle, round, etc., affect the'radiated' field strength? How is the magnet shaped? Is is like ferrite coil core, that can practically make into any shape and size? Is that by 'mold', pressed or machined? For photo magnet, can the pole be made, North South at top andbottom of the 'flat biggest area side' or NS has to be at the twoends (20mm biggest length in this photo). <Q> The most popular types of magnets are Neodymium-Iron-Boron (sintered or bonded), Samarium-Cobalt, ferrite (often bonded), Aluminium-Nickle-Cobalt (often called Alnico), and ceramic magnets. <S> They all have different properties in terms of magnetic field strength, resistance to high and/or low temperatures, resistance to demagnetization, cost, etc. <S> The magnet in the photo is most likely a sintered Neodymium-Iron-Boron (NdFeB) magnet with a Nickel-Copper-Nickel coating to help protected it from corrosion. <S> NdFeB magnets are produced in a variety of different strengths and temperature grades. <S> This will be called out as a number and a letter, like 35UH or 50M. <S> The number tells you the energy product of the magnet (which gives you a rough idea of the strength of it) and the letters tell you what the Intrinsic Coercivity of the magnet is (which gives you a rough idea of the maximum temperature the magnet can be used at). <S> The equivalent for a magnet is called flux density and it is measured in Tesla, Gauss or sometimes "lines per square inch. <S> " <S> Flux density is a measure of flux per unit area. <S> "Typical" magnetic field strength at a certain distance is going to depend on the specific grade of NdFeB and the thickness of the magnets. <S> Once you have those, you can find calculators online that will give you the answer. <S> Once you have a material picked out, the main things you need to know are shape (square, rectangle, circle, etc), area and thickness of the magnet. <S> Sintered NdFeB magnets are made in the following way: Raw materials are melted down and then formed into a fine powder. <S> The powder is pressed into a shape (this can be done in a couple different ways -axial pressing or isostatic pressing). <S> Then they are sintered (heated). <S> Then they can be and often are machined, which could be slicing a bigger block into smaller parts or grinding in order to meet a certain tolerance. <S> Then they are coated to prevent corrosion. <S> The coating could be an epoxy, Nickel-Copper-Nickel, or some other method. <S> Once this is done, they are magnetized by being placed in a strong magnetic field. <A> Strong magnets come in two general varieties. <S> The most common is Neodymium/Iron/Boron and the less common <S> is Samarium/Cobalt. <S> Less strong ones are usually ferrite materials. <S> Neodymium/Iron/Boron Measure in Gauss, or <S> more commonly Tesla. <S> 1 <S> T = 10,000G Speaker magnets are usually either ferrite or neodymium Typically powder, compressed and heated <A> I'd suggest that if this is important for your design needs, that you avoid the ebay route and buy well-specced magnets from a reliable source. <S> For example, http://www.magnetsource.com/Solutions_Pages/NEOMAIN.html will give you great parameters for a variety of Neodymium grades. <S> From there, you can then transform all your units to what they need to be and use magnetic field calculators (or look up the equations). <S> Couple this with poring over the spec sheets for your Hall sensors, <S> and that should help you come up with a viable electro/mechanical/magnetic system that's reliable. <S> Another good source is https://www.kjmagnetics.com/fieldcalculator.asp
| There are many different grades of all the metals that can be used for magnets.
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How do I determine the distance from a magnet using Magnetometer? I have a triple-axis magnetometer and a magnet, and I am trying to calculate the relative position of the magnet from the magnetometer. The magnetometer outputs the strength of any fields detected in each of the (X,Y,Z) directions, and I understand that I'll have to compensate for the Earth's own field, as well as how to determine the relative location of the magnet, but what I do not understand is how to determine the distance that the magnet is from the magnetometer. So my question is: assuming that I have a magnetometer pointed exactly straight towards a magnet, and the magnetometer is outputting the field strength, is there an equation that I can use to determine the distance that they are apart? I have attempted doing some research into this, but all the results I am finding vary from one another incredibly. Any help/advice would be much appreciated. Edit 1: I feel like I should note that I'm using a cylindrical magnet. Would the calculations be different if I were to use a rectangular one? <Q> However, you can derive distance from two readings at different distances from the magnet. <S> Magnetic field strength falls off with the square of the distance. <S> By moving a small and known amount towards or away from the magnet, you can calculate the distance to the magnet by how much the field strength changed over the know distance between the two measurements. <S> For example, if you moved 1 m closer to a magnet and the field strength quadrupled, then the magnet must have been 2 m away from the original measuring point. <S> If the field strength only doubled, then you moved 1 <S> /sqrt(2) closer to the magnet. <S> The 1 m closer was therefore .29 of the distance to the magnet, which means the magnet was 3.4 m from the original measurement. <A> It all comes down to how strong the magnet is. <S> If you don't know the strength of the magnetic field close up then use the magnetometer to get a "short (or zero) distance reading". <S> From this you can compute distances against field strength measured. <S> The shape of the magnet can also influence how the field reduces as you back-away from it and the formulas can be difficult to decipher given the various shapes of magnet that you could have and the direction it is polarized. <S> This online calculator may be helpful <A> In reality, there is no equation that can calculate this. <S> You have to measure it in situ. <S> There will be offset from Earth's field and also offset from the sensor itself, which has to be corrected at every power-on. <S> For example, LIS3MDL has zero-gauss level of \$\pm 1\, \mathrm{G}\$, which is pretty attrocious. <S> And that is only typical value. <S> But once it is zeroed out, it drifts little over time (tested for several minutes). <S> Sensitivity may be also off by tens of percents. <S> That is why (among other reasons) compasses in mobile phones have to be calibrated by figure 8 before use. <A> Your magnet has a magnetic dipole moment mu (u). <S> For a current loop the dipole moment is the (area) <S> * (current) <S> * (number of turns). <S> (u=N*I*A) If you knew the dipole moment of your magnet, then "in theory" you can calculate the B field it produces at all points in space. <S> For points very close to the magnet.. say some distance that is about the size of the magnet, this dipole approximation will break down. <S> But other than that <S> , you are all set, the exact shape of the magnet won't really matter, only it's magnetic moment. <S> (A good freshman physics text will give you the field along the axis of the dipole.) <S> For points along the axis the field will decrease as the third power of the distance. <S> Now in practice you have to deal with other B fields, (like that of the Earth), and also any pieces of magnetic material that are in the area will distort the dipole field of your magnet. <S> Edit: <S> I just wanted to add that the B field from a magnetic dipole looks just like the E field from an electric dipole. <S> I don't think you should have any trouble doing the measurements. <S> (Well find a non-magnetic table and hold all the orientations constant.)
| No, you can't derive the distance to a magnet from a single field strength measurement without knowing the strength of the magnet.
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Can I series windings of separate transformers to increase the voltage rating of the whole? I need a 1:1 transformer, 25VA 600VAC. I'm not having much luck finding such a thing. It occurs to me that I could take five smaller 1:1 120VAC transformers (like this ), series all the primary windings and all the secondary windings, and get the same effect. It seems to me that this is little different than taking a single multi-winding transformer and connecting its windings in different arrangements. Is this valid? Recommended? Are there problems? <Q> I think it will work fine if they're connected properly (if they share voltage unevenly the one that gets more voltage will begin to saturate and the voltage across it will be limited). <S> However the transformers may not be sufficiently rated to be safe on 600VAC, so isolation and creepage distances may not be good enough to meet requirements. <S> At least they're split-bobbin type which is inherently pretty good for isolation. <S> Since 600VAC is a common industrial voltage in Canada, there are a lot of control transformers available that will step down from 600VAC to 120VAC or 240VAC. <S> I'd suggest using two of those transformers <S> (600:120- <S> > <S> 120:600 or 600:240 - <S> > 240:600) <S> to get what you want. <S> For example, a Hammond B356633 is a 50VA 600VAC:120VAC transformer. <A> If any one of those transformers develops a short across its primary winding, the fault would cascade until it ate all five transformers; lose one and suddenly the remaining four would each get 150V (and get much hotter than before), so a second would fail soon afterwards. <S> The second would fail, and now the remaining four would each get 200V. <S> A third would fail, and the remaining two would get 300V each. <S> Might get pretty spectacular. <A> In the diagram below, I show how the connection of the transformers must be performed, for two transformers. <S> simulate this circuit – <S> Schematic created using <S> CircuitLab <S> I recommend you previously Determines homopolar terminals (marked with a dot) by using an oscilloscope, entering a known signal phase and verifying it. <S> You must check all transformers before connecting. <S> Edit: <S> After writing this response, I consulted with the staff of transformer test of where I work. <S> The answer was that no matter whether the phase relationship if the transformer is single phase, since in that case there is no concatenation of magnetic flux. <S> I can not assure this assertion. <S> Personally I connected groups of up to three transformers as described above, and I can assure it works. <A> Your solution will work (is valid), except for the safety aspect of it. <S> To take care of this, you must add, in series with the input windings, a 15 ma 800V AC fuse. <S> This will prevent a possible "runaway" condition which could destroy the transformers and possibly cause a fire or an explosion! <A> With identical theoretical transformers there is no problem. <S> With supposedly identical real transformers there could be problems. <S> I can envisage that the primary inductance (this drives the transformer action) could be different by up to 30% compared to another and if 4 were 15% low and 1 were 15% high, this transformer would see significantly more primary voltage when wired in series. <S> Think of it like 4 resistors of value 85 and one resistor of value 115 - <S> together they add up to 455 and the fraction of voltage across the 115 resistor will be 25.3% or, as a fraction of 600 VAC, will be 157 volts. <S> Not a vast deal more on the face of it <S> but, if that transformer starts to saturate then it's going to get warm and may in fact burn out.
| It is perfectly feasible to implement the series connection of the transformers, but you should be very careful with the phase relationship.
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High precision soldering of through hole components I have a through hole component with many leads spaced closely together, see picture: I'm looking to unsolder it from the PCB and then solder it back again. Unfortunately, the leads are spaced so closely and the beads of solder are so tiny, that I'm afraid I wouldn't be able to replicate this detailed work with my soldering iron. What technique was used by the manufacturer of this board to achieve such high precision in soldering this component? What advice do you have for replicating this work at home while achieve the same quality result? I'm open to buying additional equipment if necessary. <Q> That is not tiny, and not even close to close (pun intended). <S> I'd say the spacing between any two leads is abou <S> 1mm, so you can use your soldering iron if you can buy a thinner tip. <S> With a steady hand and less than 10$ of additional equipment (tip+solder) you can make a job that will look (and be) much, much better than it is now. <S> Desoldering the connector will be much more trouble. <S> These are 13 pins, you will need to remove some of the solder using a copper desoldering wick braid cable or a small desoldering suction spring pump, then carefully pull out the component as evenly as possible, probably pulling out one side of some half mm, then the other side of a bit more, and so on. <S> As suggested that looks hand soldered because of the bent tips, through hole components are normally soldered with wave soldering anyway. <A> This is very simple to solder with a small tip. <S> I use one like this: "0.031 <S> " OD x 0.75" L (0.79mm x 19mm)" conical tip. <S> I think you'll have considerably more trouble de-soldering the part without ruining the board. <S> Sometimes it's easier to destroy the part (cut it apart) then individually remove the pins from the board. <S> Even with an expensive ($1K+) desoldering station (a carbon vane pump sucks the solder out leaving the plated through hole mostly empty) <S> it's sometimes not that easy to do without damaging the board. <S> If you have to make do with a hand solder sucker and wick it won't be easy. <A> This is not very high precision soldering. <S> It may be either hand or wave soldered. <S> No equipment is necessary. <S> With proper soldering iron and generous amount of flux, you can make it much nicer than it is now. <S> Cover a pad by blob of flux, then heat it with soldering iron touching both the pad and pin, then start pushing in solder wire (use very thin about \$0.5\, \mathrm{mm}\$) until a fillet forms. <S> If you end up with too much solder, drag it to another pad or suck it away with solder wick. <A> You should be able to do solder work like that with a smaller tip like this one: http://www.gotopac.com/product_p/t15-bll-hak.htm . <S> But I would make sure that you find one that is compatible with your soldering iron. <S> To remove the piece it will be tough, but very possible with a desolder braid like this one: http://www.gotopac.com/Easy_Braid_LF_A_100_p/lf-a-100.htm . <S> If you have a very cheap solder station you may want to look into a rework station of some kind. <S> That may help you out too. <S> Hope that helps.
| Thin solder can help a lot too, a bit of flux would be a great help but a waste for this kind of work in my opinion.
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Turn a load on with a pulse of power, keep on until (NO) button is pressed I have a device that sends a pulse +12V when access is confirmed. I would like to keep a relay live until a button is pressed to close the circuit. This is used to start up a cars (acc) and allow it to start. Then when I exit I will press the button and it's all off. I am lost as to what to search for and the ways to make this happen. <Q> Here's a circuit that will work for you: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> When a pulse drives D1 the relay turns on. <S> The first relay switch connects the coil to a +12V rail through SW1 <S> , that is a normally closed push button. <S> The second switch gets closed too and can be used for any purpose. <S> When you want to turn off your rig you simply push SW1 thus disconnecting the relay coil and opening both switches: when SW1 will be released the relay won't turn back on. <S> D1 is there to prevent current from the +12V to flow back to the pulse generation circuit, that might or might not like it. <S> If the pulse generator circuit is just a momentary switch connected to the same +12V you can safely remove D1. <S> Please note that I am assuming that the pulse will last long enough to turn on the relay and toggle the switches. <S> some 100ish of ms should be enough. <S> If your pulse is very short, i.e. 1ms or less, you will probably need a more sophisticated circuit. <S> I know that this topology has a name, in italian it sounds like "relay with auto hold". <S> If someone can point it out I'd gladly add it. <S> Thanks to Passerby <S> I've learnt that this is called (Self) Latching Relay Circuit , that is not a latching relay that normally includes two coils. <S> edit <S> I see that in the title you say your switch is a normally open one. <S> The easiest and safest way to use a NO <S> that comes to my mind <S> is to use a NC SPST or a SPDT relay to invert the behaviour of the disengage button as in this answer from Passerby. <S> Or substitute it with a NC button already. <A> You need a bistable relay such as this one . <S> It has two coils, one coil permanently switches to one position and the second coil to the other. <A> A simple latching relay circuit requires a NC button (or a NC/NO button or switch, one that has both states depending on how its pressed/switched). <S> As Vlad has shown. <S> But if you add a second relay, you get the pulsed on, and NO button setup you want. <S> simulate this circuit – <S> Schematic created using CircuitLab Trigger the circuit, Relay 2's Normally Closed contact allows Relay 1 to activate, latching it from the 12V power source through Relay 1's Normally Open contact. <S> Circuit will stay on until 12V power is lost, or until Relay 2 is activated by grounding the coil through the Normally Open Button. <S> D1 and D2 are protection diodes for the Relay, while D3 is there to prevent backfeed to the trigger.
| There is also a latching relay that depends on Acc/On switched power, where it turns off once the car is turned off, and only turns on when you press the button when the car is on.
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Miniature heater with more-or-less constant temperature. Need tip on element I'm a software engineer with very basic knowledge of electronics. I have a project in which I want to make a small (cylinder of length about 100-150mm and diameter 20-30mm), localized heater that stays at an absolute temperature of about 50-60ºC. I researched beforehand, but I'm not sure I understand everything I've found.Would a thermal cutoff work? I know I could also use a resistor, but doesn't the temperature produced vary depending on the ambient air temperature? Basically, my question is how can I control the max temperature of the heating element with the smallest possible solution? <Q> You can use PTC heating element like this . <S> They come in all sizes and shapes, including flat flexible ones. <S> The best thing about them is that they are self-stabilizing - temperature increases their resistance and therefore causes negative feedback. <A> The normal way to do this would be with a temperature sensor and a close-loop control. <S> It could be a PID control, proportional or something else. <S> The heater wattage would be chosen in order to allow the operating temperature to be held easily under worst case conditions (low ambient, high heat loss etc.). <S> Obviously it cannot cool the cylinder <S> so if the ambient got close to 50 <S> °C it might not be able to maintain the temperature constant Since your temperature is within the normal operating environment for electronics, the controller and heater could be contained within the cylinder and power could be supplied. <S> There are other ways such as bimetal (Klixon) thermostats and so on, <S> and +/-5 <S> ° <S> C is possible. <A> How accurate does the temperature need to be? <S> Only +/-5 <S> ° <S> C @ <S> 55°C?
| You could use... a nichrome wire as your heating element, and a tiny thermistor to detect the temperature, and a tiny comparator with a tiny 10-turn trimpot to adjust the temperature, with a power MOSFET to switch the nichrome wire on & off.
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Ok to use one of the two 12V batteries of a 24V system? Coach bus with a 24V system of 2x12V 120Ah batteries for aux load. I need 12V for IT stuff. Is it ok to take it from a single battery or is it a bad idea? Right now I have a crappy 12V->24V converter that I would need to replace if this is a bad idea. Update: Great answers. Sadly I could pick only one although all 3 are very useful. Thanks!! <Q> That's probably a bad idea. <S> Hooking your... <S> IT stuff to a single battery <S> will discharge it faster than the other, especially if you're planning on using your devices when the engine is off. <S> If your devices power consumption is low compared to the battery size then you are good to go, but a 24V->12V DCDC converter (that's what you meant, right?) is still the best solution. <S> If you have multiple devices and you can balance the load between the two batteries that's fair enough too but keep in mind <S> that your devices will not be sharing the same ground , so connecting them together would probably result in harmful (but spectacular) failures . <S> Like batteries exploding failures. <A> If you plan using the two 12V batteries again in series, I'd not recommend it. <S> Because you will generate charge asymmetry between the two. <S> The one you use alone will have less charge than the idle one. <S> So if you plug them back together, the one you used will be depleted earlier than the other one, and even be undercharged by the other battery, which might destroy it. <S> A way out may be to separately recharge them to 100% after your using one of them, so they are both at the same charge level when being used for the 24V system again. <A> As the system charges by simply putting ~28v across the batteries in series, it's not hard to imagine why there could be an issue if one battery develops a higher internal resistance than the other. <S> You're best off using a 24->12v converter or a separate 12v battery <S> + 12v alternator for heavy 12v loads.
| As others have said, it's generally a Bad Idea as you end up with unequally charged and unequally "worn out" batteries with different internal resistances, I am aware of at least one occasion of someone running this sort of system having a battery explode.
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What should I do with multiple 5V to 3.3V voltage regulators? I'm trying to put a PN532 breakout board , BLE "shield" daughterboard , and a PIC microcontroller all onto one board. The PN532 breakout board has an ADP121 (5V to 3.3V) regulator, the BLE breakout has a XC6204 (5V to 3.3V regulator), and I also have an AZ1117T (5V to 3.3V) regulator for a level shifter and a RS232 transceiver chip. I'm curious about what is usually done with multiple regulators with the same input and output voltage? Should I cascade them, or choose the best one and eliminate the rest? <Q> Since each board was designed separately odds are that each regulator is rated only for the chips each board includes. <S> The first board might need some 100mA to work and have a regulator that can go up to 150mA, and so on. <S> It's possible that none of the regulators you already have is capable of supplying enough current for all the chips you are planning to put on the same board. <S> First of all you need to compute the total power consumption of all the stuff you want to put on the board. <S> You find the values on the chip's datasheets of course, use the higest value if you are not sure. <S> The sum of all the currents needed by the various chips will of course be the max current the board will require. <S> Add another 10 to 20% as a safety coefficient, then look the regulator's datasheets: probably none of them can supply that much current. <S> Finally search for a "stiff enough" regulator and use it. <A> If you are simply packaging the three existing boards into a system, without modifying the boards, I would leave the existing regulators alone, and use the boards as purchased. <A> As rightly pointed out by Vladimir Cravero, having only one 3.3v linear regulator is the best way to go ahead. <S> 3 linear regulators will unnecessarily waste 3x the power of your input supply. <S> Plus, if you need a 3.3v rail on your PCB or circuit separately (for supplying more ICs or sensors), extending a 3.3v line from one of the 3 regulators will not be a good idea. <S> so, how to decide the specifications of the single 3.3v regulator? <S> Check the datasheet of all the sensors/devices you are using. <S> Note down the peak current, Ipeak and peak power dissipation,P(peak) of the devices. <S> Now select a linear regulator which can supply Ipeak current and P(peak) power in normal operating conditions (aka recommended operating conditions).
| On the other hand, if you intend to design a new PC board incorporating the functions of these separate boards, then I would use a single 3.3 volt regulator, making sure it can supply enough current (and is on a suitable heatsink) for the full system.
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What are some problems with running a heating element in PWM mode? I want to be able to control a heating element's (which runs on 230V mains) power level, and I'm thinking of running it using a solid state relay and PWM. However, I'm unsure of the implications of doing this. Current draw - The heating element is 2000W, so ~9 amps. Will the switching this amount of current at high frequencies cause adverse effects to the rest of the power circuit in my home (light flickering, etc.) Noise induced in the rest of the house circuit - I have read on EE.SE that SMPS cause high frequency noise in the mains circuit, and this can be avoided using a choke. Is this a valid solution to PWM'ing a heating element? If PWM isn't a valid solution to controlling the power of a heating element, please suggest alternatives. <Q> Heating elements are designed to handle the mechanical stresses from thermal cycling. <S> Turning them on and off many times doesn't usually cause problems. <S> Most likely this is much longer than a power line cycle. <S> This means the PWM can be quite slow but still be much faster than the system can respond. <S> Often you can arrange to have whole power line half-cycles either fully on or fully off. <S> Look thru solid state relay offerings, and you will see there are two basic types. <S> One switches immediately according to the input signal, and the other switches at the next power line zero crossing. <S> You want the latter. <S> Switching at a zero crossing greatly reduces radiated and conducted noise. <S> I did a project once where a PIC 18 had to control 24 heaters driven from the power line and controlled by solid state relays. <S> For each relay, you only need to calculate whether it needs to be on this power line half-cycle. <S> That takes very little computation, and multiple heaters can easily be managed by a small microcontroller like a PIC 18. <S> Instead of a traditional PWM with a fixed period and variable duty cycle, I used a Bresenham algorithm to decide the on/off state each half-cycle. <S> The rest of the system provided a 0-255 value for each heater to indicate how hard it should be driven, with 0 being full off and 255 being full on. <S> For each heater, keep a 8 bit accumulator. <S> Each cycle (of the algorithm, which is each 1/2 power line cycle), add in the 0-255 desired drive level. <S> If no carry, then keep the heater off for that cycle. <S> On carry, turn the heater on and subtract 255 from the byte, which is the same as adding 1. <S> That's it. <S> Yes it really is that easy. <S> The worst case frequency content is still 255 cycles, as it would be with PWM, but intermediate values have less low frequency content due to the inherent dithering nature of Bresenham's algorithm. <S> In any case, assuming 50 Hz power line frequency, the pattern will repeat every 2.6 seconds regardless of which method you use. <A> It may cause objectionable light flickering if the lights are on the same circuit, in the same way that laser printers and copiers sometimes cause flickering. <S> It should not cause any noticeable effect on things that are not on the same circuit. <S> Dimmers and high frequency PWM can cause noise/EMI issues, but that's not a problem with zero crossing switched low frequency PWM. <A> Normal PWM is not suitable for switching heating elements. <S> Simply because heating elements are very slow to respond to changes in current. <S> It takes time for them to heat up, and cool down. <S> Much much longer than things like motors or LEDs. <S> So you have to use a technique known as "Slow PWM", which is kind of like PWM in that you have an on/off duty cycle, and the ratio of one to the other defines the average current, but the period (or timebase) of the PWM is considerably longer. <S> Instead of switching at 500Hz, 1KHz, 20KHz, or whatever, you need to switch at fractions of Hz - for example 0.25Hz, or a timebase of 4 seconds. <S> Longer timebases can also be used, such as the 30 seconds mentioned in Sphero's answer. <S> Also to be taken into account <S> is the fact that even when a heating element is "off", it is still very hot and heating the area around it. <S> Consequently the temperature of the thing you are trying to heat up continues to rise after turning the heating element off. <S> I myself have a home-made reflow oven with a pair of heating elements. <S> These I am switching at no more than 1 second intervals, but I'm not using PWM for them. <S> Instead I am using a predictive algorithm which tries to estimate what the temperature will continue to rise to after the current has been removed and remove the current at a suitable point before the target temperature has been reached. <S> It utilizes the slope of the recorded temperatures over time, along with a manually determined heat extension constant. <A> I are using a cheap pwm controller on my water heating element (3kw) tuned down to 600 watts. <S> This is a way I can maximise my solar use from a grid tied system over a period of time. <S> There is a change over relay that will boost power if necessary earliy in morning and late at night if needed. <S> I have being running this for over 12 months without any problems. <A> I would like to point out that many temperature controllers REX 100, SESTOS D1S-VR etc use a standard SSR and standard PWM to control elements. <S> It works fine for me using both of these controllers.
| Turning it on and off using a zero-crossing SSR and a timebase of 10 or 30 seconds will not cause significant EMI (because it's switching at the zero crossings). One thing to consider is the time constant from applying power to a heating element to the temperature changing in whatever is being heated.
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Any good method for the power down sequence? In my circuit, +12V would be supplied by an external power source. After two regulators, two more voltage sources (+5V and +3.3V) would be produced and supply power to the AD7682 ADC. Because of the maximum rating limit of ad7682, 3.3V needs to be down before 5V when I shut down the external power source +12V. Is there a good method to achieve this? (I learned that some sequencer ICs can monitor the power sequence process. However, when I shut down the power source +12V, no power can be supplied to the sequencer ICs. So I think ICs are not suitable in my case.) simulate this circuit – Schematic created using CircuitLab <Q> Get 5 V and 3.3 V regulators that have enable, sometimes called shutdown , pins, then sequence them accordingly. <S> Properly sequencing the shutdown lines is not as trivial as it might appear at first glance. <S> Probably the simplest way is to use a tiny microcontroller, like the PIC 10F200. <S> This would have a single shutdown input, which then causes the individual shutdown signals to be driven accordingly in the right sequence with the right wait times between changes. <A> Depending on the current and and quantity of bypass capacitors, the secondary voltages can 'fall' at different paces. <S> Sometimes, for sensitive components, Schottky diodes are placed between power supplies for avoiding inversions, which could occur in your example if the 5V crumbles faster than the 3.3V. simulate this circuit – Schematic created using CircuitLab <A> You can use a Programmable Power Supply Monitoring, Sequencing and Margining Controlleras POWR1220.
| You can search for some device that supports 12V in your power supply or make a specific source from 12V to 5V to supply the POWR1220 and use it.
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What is the collector-emitter resistance of NPN Transistor? the question may look ridiculous since I'm not sure if the collector-emitter resistance exists or not.Here is a simple commom emitter circuit As I learn that when the Vb increase that will make Ib increase so Ic must increase too. When Ic increase as there is Load Resistor but Vcc is constant and Ic = (Vcc-Vc)/RL (Load Resistor) then Vc must decrease and vice versa. That how common emitter work Now, what I concern is the Voltage Drop between Vcc and Ground is constant as well as Load Resistor value. Suppose that there is not thing between Emitter and Ground that make Ve = 0 and Vb = 0.6-0.7 while Vc is much larger (that depend on load resistor). So, there must be something that wastes the energy to make Ve=0 that cause voltage drop between collector and emitter. Is there something act like varying resistor between collector and emitter to make that. In other words, to make voltage drop between collector and emitter there must be something act like resistor between them, right? If no, what make the difference in voltage? In other configuration does collector-emitter have resistance as well? <Q> The BJT collector current equation is $$i_C = <S> I_S\cdot <S> e^{\frac{v_{BE}}{V_T}}\left(1 <S> + \frac{v_{CB}}{V_A}\right)$$ <S> where \$V_A\$ is the Early voltage . <S> But, this formula is often written as $$ <S> i_C = I_S\cdot e^{\frac{v_{BE}}{V_T}}\left(1 <S> + \frac{v_{CE}}{V_A}\right)$$ <S> Thus <S> $$\frac{\partial i_C}{\partial v_{CE}} = <S> \frac{I_S\cdot e^{\frac{V_{BE}}{V_T}}}{V_A} = <S> \frac{i_C}{V_A + v_{CE}}$$ <S> This is clearly a non-linear function of the collector-emitter voltage and collector current so this cannot be interpreted as a conductance. <S> However, for small changes around some fixed value of collector current \$I_C\$ and collector-emitter voltage \$V_{CE}\$, we can write $$I_C + i_c \approx I_C\left(1 <S> + \frac{v_{ce}}{V_A <S> + V_{CE}} \right) <S> = I_C + \frac{v_{ce}}{r_o}$$ where <S> $$r_o = \frac{V_A + V_{CE}}{I_C}$$ <S> We call \$r_o\$ the collector-emitter dynamic, or differential or small-signal resistance . <S> It is not a true resistance since it is not constant but, instead, varies with the operating point of the transistor as can be seen by the formula. <A> You've got a couple of good answers. <S> I'll try to add some intuitive insight. <S> When the transistor is biased such that that it is not saturated, it behaves like a current sink (recall that a perfect current sink has infinite impedance), so the collector-load junction looks like a voltage source with a Thevenin equivalent source impedance equal to the load resistor. <S> The voltage is dependent on the base current and beta. <S> This is equivalent to what Alfred wrote, but with an infinite Early voltage. <S> The collector impedance due to the Early voltage is in parallel with the load resistor, so to get an answer that is realistic without the load resistor you must include it, as Alfred did. <S> When the transistor is saturated, it behaves more like a voltage source of << 1 volt with a fairly low small-signal source resistance. <A> To answer in simple terms: the collector behaves like a current sink, and the collector voltage settles to whatever value lets that amount of current flow (though it can't go any lower than approximately V e +0.2V). <S> It also heats up like a resistor: <S> I c <S> * V c = the amount of heat generated in Watts, heating up the transistor. <A> If the supply voltage and the load resistance remain constant, then as the base current varies, the collector voltage and current will vary. <S> Such being the case, then for any collector current there must be a resistance between the collector and emitter such that: <S> EDIT: $$R2 = \frac{E2R1}{E1 <S> - E2}$$ <S> Where R2 is the transistor's collector-to-emitter resistance, E1 is the supply voltage, E2 is the collector-to-emitter voltage, and R1 is the load resistance. <A> It is not really the proper question to ask. <S> While a semiconductor does have resistance to the flow of current, so does a capacitor. <S> The way to start is to ask, what is the voltage drop across the transistor. <S> This is a value that is typically published for each component. <S> This way, when you know the particular operating conditions, you can easily calculate the voltage and appropriate resistances to place in the other parts of the circuit.
| In your example circuit, the collector-emitter junction can be thought of as a variable resistance whose value depends on the electronical situation present at the amplifier's output.
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Why is there a potential difference between neutral and ground (earth)? I was surprised to find that bridging neutral and ground in a socket at home tripped the RCD. Measuring with a multimeter, there is indeed 0.1 V between them. If the RCD trips at 30mA that would mean that the wiring in the house must have resistance lower than 3 ohms. (Sounds likely if household wiring is around 0.01 ohm/meter.) So the questions are: 1) why is there a p.d. at all? 2) isn't it a problem for doing electrical work, since switching off the circuit breaker for the ring only disconnects the live wire, meaning the RCD can still trip the whole house if the neutral and ground wires touch? EDIT: since there are different earthing standard apparently, this question relates to the UK and to a normal urban setup (2-wire 1-phase mains). Also, the live wire is disconnected on the ring. I think the RCD trips because the (small) potential difference between the neutral and earth creates a small but big enough current through the neutral wire, which is not balanced by current in the live wire. <Q> 1) Why would you expect a potential difference not to appear? <S> N and earth are tied together only in distribution/transformer boxes <S> , that's a long way till your house, the N wire is usually carrying some current so its potential might slightly differ from ground. <S> 2) <S> If you call tripping the whole house an issue, well that's an issue then. <S> I'd say that safety is not compromised at all, it is instead enforced by this behaviour. <S> Disconnecting also the N through the circuit breaker would work. <A> An RCD checks for a balanced current in the live and neutral . <S> If you have paralleled live and earth <S> there will be an imbalance in the current and thus the RCD trips. <A> 1) Because the currents in the neutral and ground are different. <S> 2) <S> Switching off the live wire means that there will be no current in the neutrals served by that breaker, effectively floating them, so - other than a momentary spike caused by charge accumulated on the neutrals <S> - touching them to ground shouldn't affect the RCD's balance.
| With earth and neutral being tied at a distribution panel but the neutral being used to close the power circuit with live , there will be a potential difference but that will not be why the RCD tripped.
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How to actuate 24V solenoids from 3.6V batteries I want to build a battery-powered device that actuates a 24 V solenoid when a button is pressed. How could I power it from a 18650-sized lithium-ion cell? It should work down to about 3 V, when it's nearly flat. Or I could put two cells in series, if necessary. In my case, the the solenoid seems to be rated at 900 mA, since its coil resistance is 26.6 Ω measured with a multimeter. Only a short pulse is required; there's no need for a hold current. I'd like to keep the device small, cheap, and robust. If I got a little boost regulator module rated for 30 V / 1 A output, is there any chance it would work from 3 V (or 6 V)? An input current of at least 7.2 A (or 3.6 A) seems like a lot, for both the regulator and the battery. I could slowly charge capacitors to 24 V (or even 30 V). But I've already experimented with it using a 20 V supply (which IS enough for the solenoid). At that voltage, 6000 µF wasn't enough to actuate the solenoid, but 7500 µF worked. The capacitor bank was annoyingly large though. My favorite idea is this: build up a current of 900 mA (or more) in a coil and let that discharge into the solenoid. Could this work? It would have to be a big enough coil to power the solenoid long enough for it to actuate. Or would there be sparks or something because the current in the solenoid coil needs time to ramp up from 0? I have included a schematic below (ignore the coil values). A logic circuit would close both switches first, and very soon after that, open SW2. <Q> You want to power a 24V 900ma solenoid (21.6W) from a battery rated at 3.7V 3.4AH (12.58WH)? <S> You could build a charge pump to make 24V from 3..3.7V, but (a) your battery certainly won't last very long between charges, and (b) you should expect significant losses from the charge pump, so <S> your battery should be expected to last an even shorter time between charges. <S> Your thought of storing energy in a coil would consume a lot of extra energy, too, and as @JohnD pointed out would give you unpredictably high voltage spikes. <A> This page has a great example and schematic for charging up a capacitor to trigger a 24v solenoid: http://rayshobby.net/minty-water-valve-controller <S> It looks like he used a 3.6v battery and a 2200uF capacitor. <S> The schematic from his post: <S> And out of curiosity, what did you end up doing? <A> (Probably the switch, and would likely damage the contacts.) <S> The boost converter idea would work if you could get a battery that could support the large input current. <S> Otherwise the capacitor bank, or a redesigned solenoid would seem to be the best alternatives.
| You can't change the current in in inductor instantaneously, so your idea will result in the voltage on the solenoid flying to very high levels, likely high enough to cause insulation breakdown and/or arcing somewhere.
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Minimal pulse width to actuate a solenoid I'm powering a solenoid from a battery, so I want to spend the minimum amount of power to get the job done. In this case the solenoid doesn't even have a return spring, so as soon as the armature has moved a couple of millimeters, the job is done. Can I measure the current draw or another electrical property to see when the armature has moved? I haven't been able to find a plot of how the current draw behaves in response to a step voltage. As I understand it, the inductance of a solenoid's coil changes significantly as the armature moves. If it doesn't result in a clear change in the current draw as I feed it DC (or a falling voltage from a charged capacitor), would it be feasible to measure the momentary inductance by modulating the supply current with some AC signal? If all this fails, I can certainly fall back to driving it with just a timed pulse and experimenting to see what the required time is. <Q> The energy you need for the move of the solenoid core depend on several factors, such as: mass to move, its way, initial friction and time to move. <S> As you see there are too many variables and these could be voltage and temperature dependent. <S> Thus a feedback is needed to signal that the move is accomplished. <S> This could be a reed or Hall-switch or optical switch and could be quite an effort and maybe if you can define the maximum force needed it is simpler to have a driver circuit with fixed current profile. <S> Since the initial force needed is highest(due to friction) <S> an intelligent solenoid drive can be applied. <S> You find more details here: http://www.ichaus.de/wp8_whitepaper_en . <S> This intelligent drive can also involve a microcontroller if needed. <A> First I don't know much about the details of solenoids. <S> But I think what you describe is possible. <S> (In theory.) <S> When the solenoid is down, the magnetic loop is closed and the inductance should rise. <S> You could AC couple a little signal into the coil, measure the current and there should be a drop when it closes. <S> (Coil inductance vs coil resistance.) <S> You'd have to pick the right frequency.. <S> that means knowing the various L's and R's of the coil. <S> In practice I'm not sure how well it would work. <S> Because during this AC measurement you'll be banging on the coil with a big pulse to close it. <S> Do you have any idea of the time frames involved? <S> What the L/R time of the coil? <S> How long are the pulses? <S> How much synergy (battery current) will you be saving? <A> One of the common ways to operate solenoids is to store energy in a large capacitor, then dump that energy into the solenoid when needed. <S> There are many benefits to this. <S> 1) <S> the battery doesn't have to supply all the current to the solenoid. <S> This can keep your circuit working because the battery voltage doesn't sag. <S> 2) you can use a dc-dc boost converter to increase the available energy for operating the solenoid. <S> The downside is the extra cost and extra size. <A> One way of driving a solenoid actuator is indeed with a boost converter. <S> But the reason for that is a bit different than described. <S> It allowsproviding short pulse of high current (and voltage to overcome BEMF of moving actuator!) and then remain with low hold current that is just enough to keep the actuator in place. <S> It's an open loop operation! <S> The capacitor is big enough and the pulse is wide enough not because they are accurately calculated, but because they have descent margins. <S> It will require feedback (current sense) and a PI controller. <S> You will be able to make a current loop, which will go through three states: idle, drive, hold. <S> Idle- obviously do nothing. <S> Drive- apply high current to move the actuator (will make exactly the pulse that is required, not wider). <S> Hold- reduce the current to minimum. <S> The trick is to go to the Hold state, once voltage outputted by Drive state is minimal. <S> This is clearly most energy efficient way, although i seriously doubt that anyone uses it.
| If the energy is really so critical, you may want a closed loop operation, which is similar to driving a motor.
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Moisture Sensitivity Level - More than one chip in sealed bag I've purchased three WM8804 MSL 3 chips, which came in a sealed moisture-sensitive package with instructions to mount the chips on the board within 168 hours of the bag being opened. However, it is not possible for me to mount all three of them within that time (I'm doing prototyping right now, not production). Basically, I know that the first chip will be fine (I'll mount it right after I open the bag). However, the other chips will not necessarilly be mounted 168 hours after opening (there may be problems, I might need to order new boards, etc.). I don't have the equipment necessary to "bake" the chips, and I'm not doing reflow (just hand soldering). Are there any ways I can prolong the 168 hours (storage techniques, etc.) in order to prevent moisture damage to the unmounted chips? Or is it fine to simply reseal the bag after use? <Q> Open the bag, take one IC out, place a teaspoon or two of white rice into the bag, and make sure to reseal airtight. <S> The rice will take up most of the mosture that you've introduced with the atmosphere exchange when opening the bag. <S> If you want to do it professionally, use moisture absorbing silica gel, but rice works perfectly <S> (Don't cook rice or silica gel afterwards). <S> I've had a part which said it should be soldered within four weeks after unpacking, lying around for about a year in the open, and successfully put it to operation (hand soldering). <S> Nothing popped off or cracked. <S> So it can't be too critical, and maybe relevant only for reflow or wave soldering. <S> By the way: Why can't you bake the devices? <S> An oven in the kitchen will probably do just fine. <A> My understanding is that IC's are moisture controlled because excessive moisture can do bad things during reflow, causing the chip to explode or just move prior to soldering being finished. <S> If you are not planning on reflowing, you have nothing to worry about. <A> If you leave the remaining chips in the bag and reseal it, all you have to do prior to soldering them, is dry them with a hair drier. <S> Five seconds with the hot hair drier should be more than enough <S> (no need to "bake" them). <S> I recommend using a heat resistant glove, to hold the chip.
| The moisture in the air does not damage the chip, static does!
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Do cells phones put out a signal when charged through data out? Do cells phones put out a signal when fully charged through data out? I am wondering this because there is a competition to see who can build the best and most efficient portable cell phone battery backup <Q> Typically, no. <S> The USB charging standard has the Data Pins tied to ground or each other through resistors to indicate what type/capacity charger the phone is plugged into, to let the phone know how much current it can pull. <S> While connecting the phone to a regular USB Host also does not have that information by default (phones using non-standard drivers other than USB Storage might have it, varies by manufacturer). <S> While phones have charging circuits that are already built for efficiency by switching into trickle charge mode once done charging , you can either passively or actively sense this. <S> One way would be to have your charger measure the current draw, to figure out when the phone went from high speed charging to trickle charge mode. <S> Another is to use bluetooth (preferably Bluetooth 4.0 Low Energy) with a custom app/service that signals the charger when the phone is fully charged. <S> Both options would, IMHO, require that the phone be unplugged and replugged in (or a button pressed on the charger, or on the app for bluetooth) to start charging again, so that means it would discharge as if it were unplugged. <S> Hope that helps. <A> Mobile phones use Lithium based batteries. <S> Mostly Li-Ion. <S> They are voltage limiting battery. <S> For more on such batteries pl. <S> see http://www.cellphoneshop.net/usbdacaforsa59.html <S> Since we understand that is the voltage that needs to be sensed for perfect charging do see some voltage sensing circuits. <S> Mobiles, such as old Nokia 100, would give a "fully charged" indication on screen. <S> There is no "data out" as the circuit for data is not connected to charging. <S> If you see the basic USB connection links pins are 1:Vcc 2: <S> Data+ 3: <S> Data- 4: Ground. <S> Data is isolated from charging points Vcc and gnd to maintain data integrity. <S> However, for your purpose you may take up the software aspect which indicated that the display "battery full" and use data ports to display it outside the phone. <S> For this understanding of the cell phone model, the battery specifications and the OS running on it are necessary. <S> More on phone battery here <A> When the phone if switched off and is being charged there should be no communication with the network. <S> If you want to make doubly sure, switch it into airplane mode before switching off and charging. <A> It sounds like what you're building is a power bank, or external recharger battery pack for a (smart)phone. <S> These usually consist of a USB input jack, battery charger, lithium-ion or lithium-polymer battery, battery-to-5VDC regulator, and USB output jack. <S> The phone then gets powered from 5.0VDC on the USB output jack of the power bank, and since the phone acts the same whether this 5V supply comes from wall adapter, laptop, power bank, etc., there is no way to optimize how anything inside the phone performs. <S> The charge time or power supply efficiency inside the phone is completely unaffected by the power bank design. <S> This is assuming the pack isn't underpowered and can put out as much current as the phone would like to have as input. <S> Presumably what you're trying to optimize is the efficiency of the power regulators inside the power bank. <S> At the least, this would entail choosing switching regulators over linear or LDO regulators for the battery charger and battery-to-5V regulator. <S> From there you could work on optimizing the regulators for the (known) input & output voltage ranges to try to get the converter efficiency as high as you can. <S> You may also need to optimize any battery protection circuitry in the pack so that it wastes as little energy as possible. <S> As to other features making it "better," you don't specify what this means, but we could guess some possibilities. <S> First would be what size battery you wish to use in the power bank <S> - you might want to design the electronics to work over a broad range of battery sizes, or be highly optimized for one specific size. <S> For the battery itself you can trade off between capacity, power capability, cycle life, size, weight, cost - what chemistry or flavor of lithium-ion <S> do you choose and which cell manufacturer. <S> Other features might be adding protection against ESD or a short circuit on the output pins. <S> Multiple output USB jacks with individual protection on each. <S> Multiple power input sources such as adding a wall adapter input. <S> How beautiful is the case. <S> In a real product these would be traded off against the price you believe the consumer would be willing to pay.
| If you mean do they put out a radio signal ie connecting to the mobile network when charging via USB, then the answer depends on whether it is switched on or not.
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Why did my homemade PCB turn out like this? I did a first experiment with the UV-lamp + Transparency + Presensitized-Board method. Using a randomly drawn pattern, this was my (promising but unsatisfactory) result: What could typically be the culprit behind such speckles, i.e., the patchy-nature of the copper rectangles/lines? For the record, this is my method: Print pattern on transparency Print pattern again on same transparency (double toner = double mask darkness) Place transparency on presentisized board (positive photoresist), and hold down flat using plexiglass UV exposure (with 13-watt CFL bulb) for 8-9 minutes Develop in NaOH + Water for about a minute (at end, there was clear contrast between the pattern and the rest of the board) Etch in FeCl with periodic agitation by hand <Q> I get excellent results every time I use presensitized boards. <S> Probably it depends on the PCB you use. <S> I buy Kingston boards from eBay and they seem to be good quality. <S> At least I've never had any issues so far. <S> For etching I use ammonium persulfate. <S> It's a white salt which dissolves in water to clear solution. <S> Therefore it's very easy to control the etching process visually. <S> Also it doesn't leave nasty stains everywhere. <S> You need to warm it a little bit though (40C is probably enough). <S> You might want to look at this video. <S> It's explanation of whole process I do every time. <S> http://www.youtube.com/watch?v=f_MCRcQdBQk <A> Completely unrelated to your question, but you have removed a LOT of copper. <S> Adjust your mask to fill the currently empty areas so <S> the only copper you need to etch out is the 1mm space around the desired traces. <S> If possible, connect to circuit ground. <A> My best guess is that there was something (grease / mask residue) blocking the etching of some parts of the board. <S> To compensate, you etched too long, which caused over-etching on the other parts of the board. <S> In optimal case it would appear totally black at the marks, but with usual printers it is not the case. <S> If it is more like gray, you will need to be careful with the exposure time. <S> If you get smudges in the transparent areas also, try another brand of transparency or another printer. <S> After developing, dip the board for 10 seconds in the FeCl, pick it back up, rinse with water and inspect closely. <S> The FeCl will change the color of exposed copper, and you can verify what is being etched (sometimes the mask may look like it is removed but has left some residue that stops etching). <S> If some point is not etching even though it should, a little bit more time in the NaOH can help. <S> If something is etching and should not, you can fix it with a permanent marker. <S> When etching, pick up the board a few times during the process and check the status to avoid over-etching. <S> Ideally your etching time should be less than 30 minutes; at 1 hour or more the mask may start peeling off at details. <S> Warming the etchant usually helps, though hot FeCl smells really bad. <A> Transparencies warp when heated up during printing, more so if you print twice or run it around a corner (as most laser printers do). <S> If you look at the errors in the top half of the second and third stripe, these are indicative of a fold that hasn't laid flat on the PCB surface. <S> The same issues exist in offset printing, basically the plates I've made using a laser printer always turned out rather light, but mitigated by the ink spreading out. <S> In your case however, the etching solution spreads out, so the opposite happens. <S> Things <S> I'd try: (non-DIY) commercially prepared masks. <S> Ask a local offset printing shop <S> if they can do this for you, I've typically paid around 10 EUR for an A3 mask that was good for 4 minutes exposure time with a 4kW lamp. <S> Shorter exposition time. <S> Print the transparency only once. <S> Last but not least, a friend of mine had great results taking apart his printer and replacing the image drum with a mechanism that transports the PCB through the machine, so the laser would directly shine on the PCB, so you can skip a whole step.
| The "bubbly" kind of error happens because laser toner is not really that opaque. It seems that you have both over-etching and under-etching on your board. It is difficult to know the reason based on just the results, but here are steps that you can use to troubleshoot: Take your transparency mask and look through it towards a light or the sky. See if your printer has a paper flow that exits before the last bend (we kept an old HP LaserJet for that reason).
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Designing a circuit to record timing of power cut I live in a place which is plagued by frequent power cuts (load-shedding). I have to design a circuit by which I can record the timings while power is gone or vice-versa. I also need an equivalent computer application which can be used to print daily/weekly/monthly reports of these power-cuts. How should I go about doing this? <Q> Use a microcontroller (For beginners, the arduino platform is very nice) and power it off a battery. <S> Maybe an ATtiny is a good fit here, because it is small, simple and has low power consumption. <S> The device continuously monitors (directly via its ADC) <S> the voltage of an AC-DC wall-wart (e.g. mains to 5V), and if it notices a drop to below some threshold, it saves a timestamp to its EEPROM (non-volatile memory). <S> That's basically it. <S> Reading out the device can be done in any way, e.g. Serial/USB. <S> If you want it fancy <S> you could make the wall-wart recharge the microcontroller's battery. <S> But such simple circuits can usually work for a long time off a battery. <S> You could also get a µC that can go into a very low power consumption sleep mode, and is wakened by a falling interrupt (not sure if that applies to the ATttiny). <S> You might need to make a clear falling edge by throwing in a schmitt-trigger between wall-wart and interrupt pin. <A> It's almost similar question from this thread <S> How can I detect a power outage with a microcontroller? . <S> Just put 100k resistor on transistor collector and 1uF cap across GND and collector and it should give you smooth signal. <S> Just remember that signal will be inverted. <S> Edit: I thought my answer was clear but just to make it more obvious I've added schematic. <S> Edit: You absolutely must make sure that optocoupler has dual diodes inside. <S> If it has only one, just add one external diode in reverse across it's terminals. <S> Otherwise every other cycle full mains voltage will be applied across single diode in reverse. <S> If it is not rated to that voltage it can emit magic smoke. <S> In any case working with mains voltage is dangerous and you need to understand what you are doing. <A> Take a laptop with a good battery and run this tool: http://www.nirsoft.net/utils/battery_information_view.html <S> It logs when your laptop is disconnected from the power supply and re-connected again. <S> And it can put this log into a file automatically. <S> Drawbacks: <S> The log is not event based, so with small time intervals it will get very large <S> If the power cut lasts longer than the battery time, some information may get lost <A> Going old-school, a clockwork chart recorder sort of arrangement using a cheap battery-powered (or battery-backed) <S> clock movement would remove the need for microcontrollers etc. <S> All you need is a drum (if you don't mind the recording looping) or roll of paper (EG a shop till receipt reel), some sort of indicator or marking device, and mains-powered solenoid or similar to move it so that when it's powered, it marks in one place, and when the mains drops it moves to another place. <S> It could be easily done using a dish of sand, where the hour hand of the clock sweeps a line in the sand (like a robot zen garden project I saw somewhere), that would give you 12 hours endless recording. <S> Adding features like timestamps etc. <S> is left as an exercise for the user ;)
| If you want precise timekeeping and don't want to sync the internal clock via the PC every few days, you'll also want an external realtime clock IC.
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Selecting digital componentry in bulk There are a ton of electronics retailers and suppliers on the internet, all selling components at (essentially) retail prices. But when big companies need to use digital components (batteries, cameras, motors, etc.) in their products, they certainly aren't going to, say, Adafruit and paying $40 per TTL serial camera. I have a hobby project (involving swarm robotics) that will require me to build hundreds of little tiny independent electronic devices. Each device will be its own "SoC" (system on a chip) and will be complete with its own power source and TTL serial camera. I'm wondering what big electronics manufacturers do when they are designing some new product to sell in stores: where do they go for bulk rate componentry? As a simple example: when Tyco makes a new RC helicopter toy to sell, I'm sure they don't buy the individual parts from online retailers like JAMECO and pay retail prices. So what do they do? Where do they go? What kinds of "bulk deals" are available to them? <Q> From my experience, distributors are prefered for small to medium production quantities. <S> The main reasons for this would be: <S> No lead time. <S> Distributors stock components and can deliver them in a day. <S> When buying directly form the manufacturer, lead times of 14-weeks are not unusual and can get much longer for more obscure parts. <S> Safety. <S> If you want to buy cheap components from China, you have to worry about counterfeiting. <S> This involves having extensive testing procedure, involving x-ray scans and thorough functional tests. <S> Minimum order quantity (MOQ). <S> You would go directly to the manufacturer if: You need a part that's under an NDA and is not sold via normal distribution channels. <S> You need a special service as a part of your purchase (e.g. factory programming of chips, custom marking etc). <S> The lead time is short and the price is better. <S> Also worth mentioning is the fact that Big Three (Digikey, Mouser, Farnell/E14) will not alsways be the cheapest option. <S> EDIT: <S> Adding some info on counterfeiting. <S> Counterfeit components can be placed in the following categories: <S> Completely fake and non-functional parts. <S> You basically get an IC package and nothing inside. <S> You can often use x-ray to check this. <S> Night shift. <S> Basically the factory runs an extra shift and makes components for the grey market. <S> They may skip testing to save time and costs. <S> Semi-"Functional" parts that pretend to be original. <S> See this FTDI example . <S> Recovered parts. <S> These have been desoldered from scrap boards, cleaned up and repackaged. <A> You might try a slightly different tack. <S> Make do with distributers, but do 'comparison shopping'. <S> Creating one Bill of Material (BoM) is quite a lot of work. <S> Finding the right thirty or forty components can take a day or two. <S> I tend to do most of this at a favourite distributer who I have already confirmed carries the key components. <S> This first pass creates an order or 'basket' with all of the manufacturers part numbers, and distributers part numbers. <S> Even armed with this initial BoM, generating an order for a second distributor can still be quite time consuming. <S> For example, some of the parts might be unique to a distributer ("own brand" parts, like a supermarket), or from different manufacturers, or different families from the same manufacturer. <S> I do this second BoM when I am concerned about cost or part availability. <S> There are services which do some of the work, and can help to create a basket for each of multiple distributers. <S> Those baskets give the opportunity to compare part costs more easily between those distributers. <S> For example Sandsquid will take one BoM, and help to create a BoM specific to each of several different distributers (seven I think). <S> They claim to optimise the order, by choosing the lowest cost alternative. <S> They have some demo examples. <S> You don't need to sign-up to see how they work. <S> It is quite a lot of work to create a BoM for more than a couple of distributers, so this seems like a useful service. <S> So, for modest quantities of parts, they will take some of the hassle out of searching through each distributer, and let you compare prices for similar parts for each of those distributers. <S> This might save some cost, and also identify the availability of alternative parts. <A> Hundreds of pieces is a very small quantity- <S> you'll be dealing with distributors such as Element 14/Newark, Avnet, Digikey, Mouser etc. <S> in North America. <S> Maybe Farnell and others in Europe. <S> Even up to reels of components at a time (a couple thousand to 10,000 parts in a reel), they're not that bad. <S> Prices will probably be several times what a volume manufacturer would pay, but you're paying them to hold stock, to get shipping from a single source and also to guarantee that the supply chain has some integrity (that the parts you buy are not counterfeit <S> , didn't fall off a truck, are not 'seconds' etc.) <S> At higher quantities you may find you (or your contract assembler) dealing with manufacturers' agents or the manufacturers themselves, but still some distributors. <S> You'll probably be waiting some weeks (8 weeks is common) for the parts to ship. <S> The prices are more likely to be negotiated (perhaps several parts from a single manufactuer will be designed in for a package deal) and might cover a 1-year contract for tens or hundreds of thousands of dollars or more with a single manufacturer. <S> Sales agents will buy you (or your overworked purchasing gal/guy) <S> a decent lunch, trade industry scuttlebutt and the sun will shine a little brighter on you.
| There are a lot of smaller distributors that offer the same services (and more) at better prices. Manufacturers normaly sell components in quantities of 2500-4000.
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Implementation of coulomb counting How do I implement coulomb counting algorithm ?Once I read the current using a ADC , what processing do I do ?say for 10ms load draws 1mA, next 5ms it draws 7mA. So am I supposed to just add the current for 15ms i.e 8mC is my charge removed from battery ?For this do I require a real time clock,other than initial SOC of the battery what other specifications are required, charging current, discharging current ?Once by battery is plugged in and being used by some applications do I need to integrate throughtout this period ?could someone just give an example of the equations that need to be implemented <Q> Current (I) = <S> rate of change of charge (\$\dfrac{dq}{dt}\$) so integrate your current with respect to time. <S> 1 amp flowing for one second is the delivery of 1 coulomb of charge. <S> If you are sampling current at (say) <S> every 10 milliseconds I'd consider using an analogue integrator and resetting it every 10 milliseconds just after sampling the integrator's output. <A> I think if you built a summing integrator and fed it timed pulses from a reference voltage you could make a simple Delta-Sigma converter. <S> Just count the pulses you have to feed it to drive it back near zero. <S> Of course it could also be done with a monostable multivibrator rather than a timer in the MCU and trigger an interrupt with the comparator to count the pulses. <A> I will explain by example. <S> Rated capacity of battery given by manufacturer = 65Ah. <S> so convert this in Asec = 65 <S> *60 <S> *60.Capacity of battery is(SOH <S> ) = 2,34,000 Asec. <S> So now you know the capacity, start calculating SOC. <S> first time <S> SOC = SOH + current. <S> use AS8510 IC for calculating the current and voltage which is drawn by battery. <S> if current increasing then its a +ve otherwise -ve(if drawn). <S> Check the current every sec, and do the addition with SOC which was previous one. <S> so now use formula SOC = <S> SOC <S> + <S> current(every sec).like wise you will get the results in Asec, convert that again in Ah and then in percentage.
| You wouldn't even need an ADC, just a comparator and a hardware timer on the MCU.
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Trying to understand Peltier module specification I baught this module The specifcation says: Product Specifications: * Operational Voltage : 12 VDC * Current Max : 6 Amp * Voltage Max : 15.4 VDC * Power Max : 92.4 Watts * Dimension : 40 x 40 x 3.6mm * Power Cable Length : 280mm I am trying to understand each of these values, and have written what I understood: Operational Voltage : 12 VDC This is the input power supply that should normally be used Current Max : 6 Amp This is the maximum current this device will draw from the power supply source. So it means my Power supply ratings should be 12 v Dc , 6 A (or greater than 6 A for safetly) But under Recommendations it says: RECOMMENDATION: * Use standard Power Supply 12V, 5A to power the TEC1-12706. Is this 5A rating safe? Voltage Max : 15.4 VDC This is the maximum voltage that this module can handle. Power Max : 92.4 Watts This is maximum power or energy dissipated per second by this module. This energy will be used in cooling oneside of the module. Kindly let me know if my understanding is correct. <Q> This is a somewhat poorly written spec, but I think it's trying to say that you should aim to operate the Peltier at 12 V, at which point it will draw 5 A. <S> The maximum specs are what you should never exceed or the device could be damaged. <S> It's not clear whether 12 V or 15.4 V is the maximum cooling point. <S> However, it is clear you should design the power supply for 12 V nominal operation <S> , it's OK if it occasionally goes a bit higher, but never over 15.4 V. <A> In general, it is the current that needs to be limited by altering the voltage. <S> The overall power is determined by voltage multiplied by current. <S> From the spec it looks like you can run this from 12VDC without a problem. <S> However, when first connecting it up check the current with a multimeter to make sure current is within limits. <A> Peltier effect modules really ought to be powered by a constant current source. <S> If you try to power them from a constant voltage, the current will increase when the temperature increases. <S> The increased current results in the diode heating up... <S> so the current increases even more. <S> Depending on the characteristics of the supply and the diode, this vicious cycle can continue until the diode burns out. <S> The same is true for LEDs. <S> That's why you will nearly always see low power LEDs in series with a resistor. <S> The resistor (approximately) converts the voltage source into a current source. <S> The resistor swamps the diode's impedance change. <S> A series resistor works fine for a diode that draws only a few mA, and it's really inefficient for higher power devices, like Peltier effect devices or power LEDs. <S> People seem to think power sources are always constant voltage (perhaps because they got used to working with batteries). <S> You can buy pre-made constant current modules (try Amazon or eBay). <S> It's not too hard to build your own constant current source. <S> For example, look in "Current Sources and Voltage References: <S> A Design Reference for Electronics Engineers" by Lindon T. Harrison. <S> Maybe your library has it. <S> Better yet, get yourself a copy of "The Art of Electronics" by Horowitz and Hill.
| It seems these specs are meant for running from a 12 V lead acid battery, with some operating margin above reasonable charging voltage.
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USB Circuit Failure After Potting I'm experiencing a strange failure of electronics after potting which I'm hoping to get some more insight and explanation to. The device is a custom, rugged USB drive in an Aluminium/Titanium body (see www.viud.net for pictures and more details). I'm potting the USB board (a variety of off-the-shelf boards) in Aremco 526N epoxy and curing in the oven. The USB drives are verified to work fine before potting but after potting I'm getting a high failure rate (17% Aluminum body, 78% Titanium body). Possible explanations for the failures I've considered are: Heat - Since I'm curing the devices in the oven I thought that maybe they were getting too hot. The first batch was cured at 90C for 2 hours but after testing I found the oven does get considerably hotter than that. The next batch I did at 55F for 4 hours and confirmed that nothing gets hotter than 80C but the failure rate didn't change. Epoxy - The Aremco 526N is not technically advertised as a "potting" epoxy although its electrical properties are better or similar to other potting epoxies I've used with no issues. USB Drive - The failures occurred with 4 different USB boards so it is unlikely related to that. None of these possibilities explain why the Titanium models have a much higher failure rate than the Aluminium. If it was heat related I would expect the Aluminium models to have a higher failure rate anyways (thermal conductivity of the Al is x36 greater than the Ti). The only possible cause of the failure I have left is related to internal pressure of the epoxy while it is curing in the oven. The epoxy is essentially sealed completely into the body with only a very small escape through/around the USB connector. Since the epoxy around that area likely sets first the rest of the epoxy in the body is likely under some pressure due to thermal expansion. This also explains why the Titanium models fail more than the Aluminium: the Al conducts the heat faster through the body which means the epoxy sets more evenly reducing the internal pressure. Unfortunately, I have no way of verifying this theory as to open the drives I have to destroy the USB. I also cannot believe there is enough internal pressure from the epoxy to actually damage the circuit board. I can try curing the epoxy at only room temperature but it then takes weeks to set and the epoxy is not as strong. Any ideas, theories or comments on my issue are more than welcome. <Q> On the initial batch the temperature was definitely a little too high which may have contributed or exacerbated the issue. <S> I've used several different models of USB flash drives and one in particular seems extremely sensitive to temperature and shock. <S> For example, one unit failed after just falling 1 meter to the floor and was confirmed to have a large air pocket at the top of the flash drive. <S> Another unit test with no epoxy at all similarly failed after a low speed drop test. <S> I've changed my potting procedure to try and eliminate air pockets and ensure the entire flash drive is covered. <S> This is a little tricky as the design is completely enclosed and there is no (easy) way of confirming how much or little air is left inside. <S> But recent tests seem to confirm that careful assembly to minimize air pockets leads to a more reliable product with no premature failures. <A> At an internship quite awhile ago, we would pot connectors with some kind of black two part potting compound. <S> We would outgas the epoxy in a vacuum chamber, the vacuum doesn't have to very good. <S> Maybe there is some trapped gas causing your problems. <A> At a 78% failure rate, I don't understand your hesitation to open a drive and see what's going on. <S> For two-hour soaks, I wouldn't think that the thermal conductivity of the case has anything to do with things -- though maybe thermal expansion does, creating an interesting shear on your PCB. <S> No way of knowing without getting in there. <S> Can you xray through the case?? <A> According to the datasheet for that range of epoxies, the 526N you're using has among the highest cure shrinkage in the range. <S> 0.01 in/ <S> in may not sound like much, but its more than enough to, at the very least, cause hairline fractures in joints on the PCB. <S> I have experienced this exact phenomenon when potting accelerometer sensor PCBs into an aluminum enclosure. <S> It you're set on using that epoxy, you could give the PCBs a conformal coating of some sort before potting - but make sure its flexible. <S> I used a thin layer of low viscosity RTV silicone (Dow Corning 3140) and my failure rate dropped from over 80% to well under 1%.
| If any significant portion of the circuit board is not covered by epoxy it seems to fail very easily. After a good deal of testing I believe the problem is actually a combination of three things: Temperature Sensitive USB flash drives Air pockets However, it seems the root cause of the issue is large air pockets.
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How important is "mm pitch" for a capacitor? I want to repair my boiler, so I found this howto. He writes I need a MKP X2, 26.5 x 10 x 19 mm, 0.47 µF 275 V/AC ±10%, 22.5 mm pitch . The closet one I can find in my country is this one that doesn't mention 22.5 mm pitch . Question What is mm pitch and is it important for my usage? <Q> The pitch is the spacing between component pins or, like with your capacitor, the spacing between the leads. <A> It's the distance between the leads in millimeters. <S> It's important if you want it to fit in the same space. <A> pitch <S> [in mm] is the distance between the leads (pins) of the capacitor. <S> However, this is not that important in the O.P.'s case, because the capacitor is mounted in a bent fashion. <S> The replacement capacitor, which the O.P. had found is smaller and has smaller pitch. <S> But the leads can be bent out to accommodate wider pitch. <S> Having said that, repair questions are off-topic on EE.SE <A> "Pitch" refers to how far apart the leads are. <S> For through-hole discretes it's not vitally important since the leads will have some give on most through-hole packages, but the more you have to bend them off <S> pitch <S> the worse the fit will be.
| This pitch dimension normally mates with the corresponding distance between the holes on the circuit board. It doesn't really affect the electrical characteristics of the part and are solely for mechanical fit.
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Calculating energy in battery Let's imagine a regular deep cycle gel lead-acid battery from the building market. Capacity: 100Ah Nominal voltage: 12V Is it really correct to say that the energy in the battery is 100Ah*12V = 1200Wh? Or should I rather consider that the voltage when fully charged is more like 13.5V and when fully discharged much less? <Q> As a first very rough approximation, yes you can multiply the capacity and nominal voltage to get energy. <S> However, batteries are quite complicated with lots of variables. <S> The output voltage will certainly vary over the discharge life. <S> What does the 100 Ah spec actually mean? <S> What is the output voltage at the end of those 100 <S> Ah? <S> To truly get delivered energy, you have to integrate the instantaneous product of the voltage and current over one full discharge. <S> But wait, it's not that simple either. <S> That was one cycle at one temperature. <S> Batteries can generally deliver less energy when cold, sometimes significantly so. <S> Overall capacity generally decreases as a battery ages or as it is cycled. <S> How deeply the battery has been cycled in the past can matter. <S> The output voltage and capacity are a function of the current. <S> 100 <S> Ah doesn't necessarily mean that you can draw 100 A for one hour. <S> It might be specified to deliver that charge at 10 A for 10 h, and if you draw 100 <S> A you only get 80 <S> Ah, for example. <S> How does your load actually use this energy? <S> If you have a switching power supply, then maybe you can actually use the integral of voltage times current. <S> If you have a linear regulator, then you really only care about the delivered current as long as the voltage stays above some mininum. <S> Any extra energy the battery delivers in the form of higher voltage will just be dissipated as heat by the regulator. <S> In that case, the Ah capacity figure is more relevant that a energy figure. <S> Good design will derate from there to a figure you can rely on over the range of temperature, age, and cycle lifetime <S> you want the battery to work over. <A> Is it really correct to say that the energy in the battery is 100Ah*12V <S> = 1200Wh? <S> It depends on how accurate you want to be. <S> That is a mathematically sound assumption, but it assumes that the battery is a constant voltage for the entire cycle. <S> For many needs, that is close enough. <S> However, battery voltage decreases as it is discharged. <S> What you really need to do is measure the current and voltage at every instant during discharge to calculate power, then integrate power to get energy. <S> Of course, both the voltage and current will depend on many variables, among them temperature, age of the battery, how fully charged it was initially, how far discharged you are willing to run it, the rate at which you discharge it, and so on. <A> The nominal capacity and nominal voltage aren't really enough to calculate total energy capacity, they are just a quick way of comparing two batteries with the same chemistry. <S> Full battery specifications usually include a discharge curve which you can use to calculate usable energy for your application. <S> For a quick example, see http://www.power-sonic.com/images/powersonic/technical/1277751263_20100627-TechManual-Lo.pdf <S> If you can only use the battery down to 11.5V, the useful energy will be lower than if you can use it when it has discharged further, and the rate that you take power from the battery will affect its usable capacity. <S> Some specification sheets have a more useful ' <S> Wh vs voltage' graph, so you can quickly see that the battery capacity is 1200Wh when the terminal voltage is 11.2V, but that is not widespread. <S> It also doesn't take account of the battery capacity being reduced at higher discharge rates
| Due to all these variables, just multiply the capacity by the nominal voltage to get a rough idea of the energy, and realize a rough idea is all you're going to get.
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How to identify mutual inductance? I have two coils covered in a box and I don't know the physical arrangement of the two coils and mode of each winding in a magnetically couple DC circuit. How can I make an experiment to determine the dot convention of the two coils? <Q> You can use an inductance meter and connect the two windings in series. <S> The configuration with the higher inductance has the windings connected dot-to-no-dot. <S> In other words if the inductance of one winding is \$L_1\$ and the inductance of the second winding is \$L_2\$ and the total inductance measured with the two in series is \$L_X\$ then <S> the mutual inductance is M = <S> \$\dfrac{L_X - L_1 - L_2}{2}\$. <S> By shorting one of the coils and measuring the inductance of the other (preferably with an instrument that gives you an L+R measurement) you can get a measurement of the leakage inductance. <S> Of course the placement of the dot is (usually) arbitrary- <S> if you reverse the dots on all the windings it is exactly the same thing for a typical inductor or transformer (there are some types of inductors that are magnetically pre-biased <S> so the are not symmetrical). <A> Applying a sinewave to one coil and measuring the output amplitude on the other might be seen as a good start. <S> If it were a perfect transformer (i.e. 100% coupled coils) then the ratio of the output coil voltage to the input coil voltage tells you the relative turn ratio <S> BUT I guess there won't be 100% coupling between coils <S> so it's a little harder. <S> To understand the dot convention is easy - if the voltage applied on one coil is in-phase with the voltage seen on the other coil then, the dots can be drawn on the signal-in wire and the o-scope signal-out wire. <S> If you wish to go further read-on... <S> I'd definitely start by understanding each coil in turn - use a signal generator, o-scope and tuning capacitor to measure the inductance of each coil whilst the other coil is open circuited and playing no role. <S> This is OK providing the coils each have a high self-resonant frequency where one can assume that the self resonant frequency of the "open-circuit" coil is not influencing readings on the coil being measured. <S> So, to avoid this try making the coil resonate at the lowest practical frequency. <S> Once you have a reasonable value of inductance for each coil you then have to delve into the mutual coupling properties. <S> For perfectly coupled coils with no flux leakage: - \$M_{perfect} = \sqrt{L_1 <S> L_2}\$ henries <S> But, for coils that aren't 100% coupled the formula becomes: - \$M_{real} = <S> k\sqrt{L_1 L_2}\$ henries where k is the coupling coefficient and has a value between 0 and 1. <S> Next is understanding what M is in a bit more detail. <S> This site shows a good picture: - <S> Note the formula <S> - it says that the induced voltage in a 2nd coil is: - Induced voltage = <S> \$-M\dfrac{\Delta <S> I_1}{\Delta t}\$ <S> This allows you to determine M by injecting a sinewave into the 1st coil and integrating the voltage sinewave produced on the 2nd coil. <S> Once you have actual M you can calculate the coupling coefficient, k using the \$M_{real}\$ formula given higher up in my answer. <A> You must connect a generator with a known signal (square, sine) at the input and use a two-channel oscilloscope. <S> With a channel you see the input signal and the other channel to see the output. <S> The phase relationship will tell you what the homopolar terminals.
| The actual mutual inductance is the half the difference between the sum of the two inductances (measured separately with the other winding open) and the total inductance with the coils in series and phased as above.
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How can I connect two different DC power sources together? How can I connecting two different DC power sources together? for example 15Vdc(5Amp) with 5Vdc(2Amp).I want to make one DC source (20Vdc,7Amp) by more powerful supplies. <Q> You can connect them in series to add the voltages, but the resulting current capability will be the lower of the two current ratings. <S> Think of it like tying a 5 foot and 15 foot rope together. <S> You get a 20 foot rope, but its strength is whatever the strength of the weaker rope is. <S> No, you can't put two supplies together in any way to add both their current rating and their voltage. <S> This should be clear just from the power output alone. <S> The 15 V 5 A supply can deliver 75 W. <S> The 5 V 2 A supply can deliver 10 W. <S> Both together therefore can't deliver more than 85 W. <S> However, 20 V at 7 A would be 140 W. <S> That's <S> a free lunch physics won't let you have. <A> If you want 20VDC from a 15VDC supply and a 5VDC supply, you would connect them in series. <S> However, you must make sure that they are floating with respect to ground. <S> If both supplies are grounded, then you cannot connect them in series. <S> Also, note that the maximum current that can be drawn from the series connected supplies is equal to the lower of the current ratings of the 2 supplies. <S> In your case, connecting a 2 amp supply and a 5 amp supply in series results in a supply that has a maximum rating of 2 amp. <A> Power = volts x amps. <S> So you want to get (20*7) 140W out of (15*5) <S> = <S> 75W <S> + (2*10) = <S> 10W. <S> 75+10 is less than 140, so it cannot be done. <A> In present case, you can't obtain the desired output. <S> There are three types of battery (DC Source) connections you should remember. <S> Series <S> Parallel <S> Series - Parallel Series connection Such connection is used for adding voltages. <S> In your present case 15 V + 5 V = 20. <S> But remember current always remains same in series. <S> If you'll connect two indifferent sources (in terms of current) in series. <S> It'll damage. <S> Figure below displays series connection: <S> Parallel connection Such connection is used for adding amps. <S> In your case 2 Amp + 7 Amp = 9 Amps, but remember voltage sources with different values should never be added in parallel. <S> In case of same values voltage, you can do so. <S> See figure below: Series-Parallel Connection Such connection is used for adding amps, as well as volts but remember the rules. <S> Both of above are applicable here. <S> Hopefully, the info above is quite enough and it helps you. <S> For more details check this link on 3 different types of battery connections with mathematical calculations. <A> You can connect the +ve of one supply to the -ve of another supply to get 20V, but you will be limited to the smallest supplies current (2 amps in this case).
| You make more volts by connecting in series, and you make more amps by connecting in parallel, you can't have more of both! How to connect supplies together is complicated and depends on the type of supply, so don't just connect supplies together without finding more.
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How do Wi-Fi controlled light switches feed themselves? There are smart light switches in the market that are Wi-Fi controlled like Plumlife (formerly ube) and Belkin WeMo . I wonder how they control the electricity that they operate on. When I unplug my old fashioned light switch, I see that there are two wires. I believe that they work in this way. But in smart switches, there must be some inner circuit that contains a relay inside of the smart light switch. If current goes through that circuit, that means smart light switch is closed too; therefore bulb is on all the time. How do these products run on and control the same electric wire. If there is an external power source that feeds smart light switch like a fellow stackexchanger mentions in this question , I understand. But there is not. So what I ask is how is it possible to feed a circuit that controls it's own electricity. <Q> Most likely it uses a transformer and a relay. <S> The transformer brings it down to a DC current with usable voltage for the wifi and anything else that is needed. <S> The relay then turns on and sends power to the bulb. <S> I would think since the wifi uses such a low power and there are only 2 wires that there is always a small amount of power going through. <S> It must use a current limiter to not let enough power through to turn on the light bulb as it is using the power it needs to let the wifi work properly. <A> The WeMo web page specifically says that a neutral wire is required! <S> Another method, which I think 'smart' thermostats use, is to have a battery to maintain power (the battery charges when the switch is off). <S> The batteries will eventually wear out, and Belkin lists the lack of a battery as a feature. <A> The simple answer is that the switches' electronics are in parallel with the load, which it controls via a SSR or triac, solid state devices, instead of a mechanical relay. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> It is in parallel because these switches normally require the neutral wire be connected, and typical house wiring has both wires going to the light switch box, then run to the light fixture.
| The power is always going to the switch and the transformer to keep the wifi working.
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Are USB's Data+ and Data− needed when just charging? Are the data + and − wires in a USB cable needed if you just want to charge something? Also what are the benefits and drawbacks of removing the data + and − wires? <Q> Many USB powered devices use these lines to recognize charger. <S> Most Android devices detect a charger looking for direct electrical connection between D+ and D- pins (DC resistance 200 ohms maximum). <S> If they found it, the device draws the maximum current allowed and battery charging is quite fast. <S> If not, they assume they are connected to a computer USB port and charging current is usually quite low, taking an eternity to charge the battery or even no charging at all. <S> Source / more information: link <A> Yes, something has to be done with the D+ and D- lines to guarantee you get more than 100 mA. <S> In practise, USB ports on wall-powered machines with large power supplies often don't bother actively controlling current. <S> On such machines, you can usually draw 500 mA. For just supplying power when data communication is not wanted and the host complies with the new USB high power standard , you have to pull the data lines certain ways with particular resistances. <S> See the USB power spec for details of how to ask for high current, and how to know you actually got it. <A> it depends of your device. <S> many cheap devices (powerbanks, small bluetooth headphones and speakers, other USB powered devices...) come with a cable that has only 2 wires inside. <S> so do whatever you want with D+ and D-, there will be no effect. <S> Those devices are usually drawing up to 500mA For more power hungry devices or smartphones, there are different config possible : pins shorted together or not, with a resistor or not, with a potential or not... USB spec gives a standard, but the market is only following it halfway.
| So, for some USB powered devices you can remove D+ and D- connection, but some may require it to enter "charging mode".
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Can I prevent copper traces from delaminating due to soldering heat? I'm making a PCB that has a 100pin QFP pacakge microcontroller. After gaining experience with the UV photoresist process and reliably making the small, 0.5mm traces, I have a new problem: the copper traces are delaminating when soldering them. Specifically, they get stuck to the solder tip and peel off the board. It seems this mostly happen when a large part of the trace is heated simultaneously and kept heated for > 1s. I cleaned my solder tip repeatedly to prevent a big blob of solder from accumulating, but that didn't completely make the problem go away. The boards I'm using have a copper thickness of 1 oz. I'm also using a Weller WPS18MP, which has a tip temperature of ~900F. Is that too high? Is there anything I can do besides using sheer soldering discipline? <Q> I use a small 'toaster oven' (cost about 20GBP) solder paste in a syringe, and a lighted magnifying glass. <S> Look for all the solder paste to 'go shiny', and its done. <S> The toaster oven needs to be <S> at least 1.3kW. Folks <S> I know have failed with attempts to use a less powerful toaster oven. <S> Other people use an 'electric skillet', or pizza oven with very good results. <S> Apply the solder paste in a very thin line across the PCB pads for the QFP, place the chip, and bake it in the oven. <S> Providing there is very little excess solder, capillary action 'sucks' the solder under the pads, and there is very little bridging to clean up. <S> Use fine (e.g. 1mm or narrower) to remove solder bridges between pins. <S> We have done LQFP64, 0.5mm pitch, this way. <S> We mostly use manufactured PCBs, but people have used this successfully on DIY FR4 PCBs. <S> There are a lot of videos on the details of the process. <S> Maybe start at Sparkfun, which had some videos. <S> Edit: A couple of techniques which improve the success rate are: Use a fine 'needle' on the syringe. <S> When the hole is fine, it is muchless likely you'll apply too much solder. <S> Several vendors sell fine plasticnozzles which can be cut to make a hole to suit the job. <S> Don't panic about making solder bridges. <S> Fine solder braid removes itreliably. <S> I'll add, this technique was taught to me by a local (UK, West Midlands) <S> Secondary School teacher. <S> He started teaching Surface Mount Technology soldering to 13yo's. <S> They had much higher success rate than through hole parts, which tend to have more difficult to diagnose failure modes. <S> Also, the students found it quicker to build boards; with three toaster ovens a class of 24 could all build a board in less than two hours. <S> They are not using such fine technology as QFP though, typically small PIC's. <A> Yes - 900 degree <S> Fahrenheit is way too hot. <S> Your tip-temperature should be between 650 and 750 degree Fahrenheit for lead-free solder. <A> This is a sign of too much heat in the copper strip. <S> That you're seeing this with thin traces suggests your soldering iron temperature is too high, you're using too large a tip, or your holding it on there for too long. <S> Without more information, I can only suggest my parameters for your situation: 700F temperature controlled iron Small conical tip (1mm or smaller at the tip) <S> Lots of flux Small diameter solder, or paste <S> You should be able to solder a SMT joint in a second or two without lifting. <S> You shouldn't keep reheating the same trace. <S> If you must use solder wick for rework or blob removal, try to keep the iron on the wick pressing into the joint for under 15 seconds. <S> If after all this you still experience the problem, you might consider asking your pcb supplier about it. <S> Poor quality copper adhesive or fabrication practices could be the culprit as well.
| One way which works well, is to use a different soldering technique. I also have a multi-meter with a temperature probe to monitor the process, but in reality a small torch, shone into the oven gives very good feedback.
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Why is it so expensive to produce cameras for non visible light? Typical consumer cameras can capture wavelength of 390-700 nm 400-1050nm . But why is it so difficult and expensive to produce cameras for infrared, ultraviolet, hard x-rays, etc.? The only thing which differs them are the wavelength and the energy eV. <Q> It comes down to market size. <S> Where is the demand for such cameras and do the number of sales justify the production set up costs? <S> You can get an infra red conversion to standard type DSLR cameras (eg <S> Do It Yourself Digital Infrared Camera Modification Tutorials ) <S> and you can convert the camera to a 'full spectrum' type which takes in some ultra violet. <S> (see Full-spectrum photography ). <S> These, by their specialist nature and low volume production, tend to be very expensive. <A> First of all: standard CCD sensors are sensitive to wavelength far beyond 700nm. <S> As much as I know Si-sensors are even more sensitive for near-IR light than for visible light. <S> Of course it changes for much larger wavelengths: One condition for light being detectable is that photons have enough energy to create a hole-electron-pair. <S> This energy threshold is the band-gap of the particular semiconductor material (e.g. for Si: ~1.1 eV). <S> Since photon energy is inversely proportional to wavelength <S> ( E = h * c / lambda) <S> there is a maximum wavelength that can be detected with a given semiconductor material (e.g. for Si: ~1100 nm). <S> For cameras the lens is also relevant: Most types of glass are less transparent to UV light. <S> Lenses optimized for UV transparency <S> are very expensive (although a cheap alternative could be plastic lenses). <A> Both your existing answers are valid, but may be taken in combination: Simple Si sensors are good for visible and NIR and are common and therefore cheap. <S> Modifications to the imaging system are required in many cases as the IR is normally blocked because it's undesirable. <S> See for example <S> Canon's EOS 20Da . <S> Silicon sensors are fairly easily adapted to UV use by means of a phosphor coating (I wanted to try a homebrew version of this on a webcam I'd modded with a B+W CCD but never got the chance). <S> Even X-rays use is possible with a scintillator (which is normally fibre-optic-coupled). <S> To go beyond ~1µm further into the IR requires other semiconductors - which are expensive. <S> InGaAs is a popular choice, but is ridiculously expensive as you say - <S> but that's not surprising as you need dedicated production facilities. <S> InGaAs and other NIR cameras are also regarded as military technology for the purposes of US export regs (which are also imposed on many NATO countries in effect); this adds cost to the camera manufacturer in terms of compliance. <S> Cameras which have any sensitivity at all to thermal radiation, or which are made from narrow bandgap semiconductors, will need significant cooling to remove thermal noise that could be greater than the image you're trying to measure. <S> That often means a Dewar of liquid nitrogen (material cost + operating cost). <S> There are newer technologies (even uncooled) coming on the market - in particular for thermal imaging, but the resolution is much less than for Si CCD or CMOS sensors. <A> For both visible and bolometer type, the reason they are cheap is because they can leverage the economies of scale in the silicon business. <S> As soon as you get out into wavelengths (i.e. energies) that need other technologies (InGaAs as mentioned, InSb) <S> you're talking 2" and 3" wafers at best, nothing like the pizza sized silicon wafers used to make chips today. <S> Also, the transistors still have to be made of silicon, so you need a connection from each photodetector on the photo-sensitive chip to each detection circuit for that pixel on a silicon chip. <S> If you have a megapixel imaging array, you have a million connections to make. <S> But wait, it gets worse. <S> If you are depending on the photoelectric effect, say for mid-wave IR at 3-5 µm, you have to cool the camera so that you are seeing something more than the heat being generated by the camera itself! <S> Imagine a visible camera with a brightly glowing lens and housing <S> -- that's the world a thermal camera lives in. <S> Cooling adds a lot of expense, and usually noise as well, since the most power-efficient coolers are refrigerator type. <S> Peltiers can't take you down to liquid nitrogen. <S> Oh, and BTW, glass is not transparent to wavelengths past about 2 µm, so you need a different lens material than what the last five centuries of optics has been working on. <S> At the other end of the spectrum, X-ray is a pain because it's hard to deflect X-rays. <S> They like to go right through. <S> Big imaging arrays for medical X-rays work because there is no lens, but take a look at the mirrors on something like Chandra space telescope - the "lens" is a series of glancing angle mirrors arranged in cones.
| For smaller wavelengths you'll need different sensors.
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Multiple leads on capacitor? Beginner question, but at least my Google skills didn't yield any results. Why do some capacitors have 5 terminals? Such as this one What are they used for? <Q> Check out this data sheet of the Vishay capacitors that look alike the one you showed. <S> It seems the extra pins are for mounting, since it's a big capacitor. <A> The legend states "1 <S> = 220 <S> uF". <S> Are there also markings like "2 = xyz uF", and are the connections by any chance marked -, 1, 2, etc? <A> For completeness, motor "capacitors" are often more than one actual capacitor bundled into a single package, and thus have more than two leads. <S> A very common example is the start/run capacitor usually found in air-conditioning units. <S> I don't think your image is one of these though.
| This could very well be multiple capacitors in one package, probably with a shared negative (-).
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Problem about superposition theorem For the circuit given below the voltage across the resistor is definitely 5V. If we add more than two sources the voltage will remain same as before. If we try to analyze the circuit with superposition then first we have to make one of the source inactive. If we make one voltage source inactive then we can find the voltage 0V. After that we have to make inactive the another source. At that time the voltage remains 0V. So if we add the two voltages according to superposition theorem the voltage across the load is 0V. How??? <Q> You are misapplying the superposition theorem. <S> When you calculate the response to a source, other sources must be passivated. <S> In the case of a voltage source, the same is replaced by a short circuit. <S> If in this particular case, try to apply the theorem, both sources will be connected in short, implying that the voltage on the resistance would be zero, while each voltage source should deliver an infinity current. <S> Clearly, there is a singularity for which circuit theory of lumped parameters does not apply. <S> If the voltage sources they were replaced by real models, considering its internal resistance simulate this circuit – Schematic created using CircuitLab <S> you can apply the Superposition Theorem smoothly. <A> Two condition you must consider to apply superposition theorem- <S> The number of sources of the network must be two or more. <S> Sources are not in series (current sources) or parallel (voltage sources). <S> The second condition says that you cannot apply super position theorem in a circuit where two voltage sources are connected in parallel. <S> Therefore, the circuit you are showing doesn’t give you a valid result when you use superposition theorem. <S> We can change the circuit to something else so that superposition theorem is applicable but that wouldn’t be the given problem. <A> There is a singularity in your circuit. <S> You connected two voltage sources in parallel. <S> When you connect voltage sources with different values, \$\frac{\text{value}} {\text{zero}}\$ singularity occurs when calculating the current. <S> And \$\frac{\text{zero}}{\text{zero}}\$ singularity with same ones. <S> And you can't use the superposition theorem in a circuit with a singularity like this. <S> But don't worry. <S> Just assume that there is only one 5V voltage source in cases like this.
| In practice, you can connect two identical voltage sources in parallel, and it will work in theory if you assume that there is a single source instead of them.
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Power supply for 144 LEDs in parallel? I appreciate any help as I try to figure this lightbox project. I need to wire up to 144 3V (built-in resistor) LEDs from flickering tea lights (normally battery powered). Not sure how many to wire in parallel for a 120V outlet, and what power adapter to get. Suggestions welcome. <Q> 3V LEDs can't be wired in parallel and plugged into 120V AC socket. <S> They will all be destroyed. <S> They can only be wired in parallel and powered from (approximately) 3V DC <S> This article "Hacking a Candleflicker LED" suggest that it might not be practical to wire a few in series either as they appear to modulate their power supply. <S> Edit: You could drive them in parallel with a 3-5V power supply. <S> Assuming each LED is no more than 30mA (0.03A), 33 LEDs would be <S> 1A. A 'wall-wart' would be cheap. <S> (The LEDs might be even less power.) <S> Edit2 <S> : While we haven't got a datasheet for your tea light LEDs, you might want to compare them to this range of Candle Flicker LEDs to get their current rating. <S> For example, these Yellow 5 mm Clear Candle Flicker LED are rated at forward current of 30mA. <S> They need a current limiting resistor on each LED. <S> Some have quite significant forward current (abut 3x nominal), so decoupling capacitors would help. <S> I would build the chain in sections and test them using a multi-meter to measure current, to ensure it stays within the capabilities of a power source. <S> Be careful when measuring current because the LEDs are not on all the time, so the reading might be wrong. <S> Give yourself some headroom. <A> Also, you should account for voltage drop if this is going to be a long run of any sort. <S> 40 ft of 22 guage wire would cost you 2 Volts for instance <S> so you should consider running a 5V power supply to compensate if you are running any lenght. <S> Here is a voltage drop calculator. <S> http://www.calculator.net/voltage-drop-calculator.html?material=copper&wiresize=52.96&voltage=5&phase=dc&noofconductor=1&distance=40&distanceunit=feet&eres=1&x=64&y=21 <A> You have shorted all LEDs. <S> This is not a parallel connection of the LEDs yet. <S> Again 3V LED can't tolerate 120V. <S> One more thing is that you have to keep in mind the current rating of the supply. <S> If you want to connect too many LEDs in parallel then it requires more current.
| If you want to connect those LEDs in parallel then you have to use voltages near 3V and some resistances in series to fix the current. Also the power supply should be increased significantly to ensure it has some headroom.
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Is my PSU design safe, what should I change? Normally I would just go and breadboard the design out, but since this is using mains power, I would rather ask experienced people first. I want to build a desktop 170V DC power supply to fiddle around with nixie tubes.I came up with this circuit: The Transformers will be(looking at the cheaper end parts, both will be around 7-10 VA): T1 : 230V -> 9V T2 : 230V -> 12V but in reverse. The capacitors will be: C1 : 1mF 25V C2 : 47uF 450V So will this circuit work and how should I improve it? EDIT: OK, here is an updated circuit taking into account all the replies. I missed a lead on that regulator, also a resistor on the HV warning led, ignore, please Some highlights of modifications: 7.5k Bleeder resistor, which should dissipate the power in a second or so after turning the HV off L7805 regulator for keeping the +5V clean HV on warning LED Fuses on all positive terminals It will be in either a plastic or wooden enclosure, to eliminate risk of touching anything inside All work on the circuit will be done after the HV is turned off, the warning LED unlit and voltmeter reads zero on panel T2 will be 18V output terminals will be this kind or similiar: all connections inside will be maniacally heat-shrink tube wrapped to counter any chance of a short will probably etch a PCB for that <Q> I don't agree with the accepted answer. <S> so it's less likely to be in a position to deliver a lethal shock. <S> You will get about 11-14V at the point marked "9V". <S> Too low to be reliably regulateddown to 9V and a bit high for (linear) regulation down to 5V (but okay for 5V if you don't draw much current). <S> The point marked 170V will actually be at about 240VDC with a light load. <S> That's because the AC voltage is RMS <S> so you get about 1.4 times higher. <S> You're using a 450V cap so it won't hurt the capacitor. <S> You might be better to use an 18V transformer. <S> 1000uF/25V means you if you draw 100mA you'd get about 1V p-p ripple on the 11V line, which is fine for regulation down to 5V. <S> Nixie tubes typically take a few mA. <S> If you have (say) 12mA, the ripple on your 240V line with 45uF will only be a few volts, which is quite acceptable. <S> I agree that (say) <S> 200K 1/2W bleeder resistor across the 45uF would be a good idea, and a suitable fuse on the input (there is no need to make the fuse more than a few hundred mA). <S> Take appropriate precautions when dealing with high voltage. <S> Typical precautions include, but are not limited to, not probing the circuit when power is on, using appropriately rated test equipment, keeping everything neat, using a GFI/RCD, discharging the capacitor manually before touching things, even if you think there is a bleeder, working with (at least) one hand in your pocket, etc. <A> You may not find a transformer which outputs 170V from either 9V or mains. <S> In this case, use two transformers. <S> Connect primary sides parallel, secondary sides in series. <S> Resulting voltage will be sum of secondary voltages of two transformers. <S> But you have to be careful on the polarity of the secondary sides. <S> You are using only a capacitor for filtering. <S> Notice that there will be to much ripple in your DC voltage. <S> No problem though, if it is OK with your project. <S> You may add a fuse to the DC sides as well. <A> The 170VDC output is no joke. <S> It is lethal and not to be toyed with. <S> In fact, you need to take measures to ensure that it cannot be touched. <S> Consider a mechanical enclosure around your nixie tubes, something like a hinged plexiglass box with a mechanical 'dead man' switch. <S> When the cover is down, power is applied and the nixies can do their thing. <S> When the cover is up, the power is disconnected and some bleeder circuit is connected to discharge that big 450V capacitor in the secondary. <S> We do this sort of thing in production enviromnents - our initial power-up fixtures are exactly as I describe (a clear box with a contact switch that cuts the power when opened).
| As to your questions: No it is not completely safe, since it involves high voltage, but it won't burn up unlessyou short it, and it does provide galvanic isolation
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Why do circuit analysis methods work? Both the KCL and KVL equations are required to completely model a resistor network. However, the two main methods to analyze these circuits Nodal analysis and Mesh analysis only use KCL and KVL respectfully. In other words,these methods don't use the "complete model of the circuit" (i.e. Nodal analysis ignores the KVL part of the circuit model). Yet somehow they're able to completely analyze the entire circuit - how? How is something that doesn't take the entire model into account able to completely analyze the circuit? <Q> In other words,these methods don't use the "complete model of the circuit" (i.e. Nodal analysis ignores the KVL part of the circuit model <S> That is not correct. <S> Nodal analysis appears to be using KCL only, where it actually uses KVL implicitly , absorbed in Ohm's law in the currents equations. <S> By the way, KVL and KCL are just a consequence of the lumped-element abstraction of linear circuits at low frequencies <S> (when the wavelength of the signal is much more larger than the physical lengths of the wires and components). <S> You can derive it easily from Maxwell equations (which are more accurate models of the circuit behavior.) <A> Both the KCL and KVL equations are required to completely model a resistor network. <S> This is true, but often we don't need to know everything about a network to solve our problem. <S> The complete solution of a network would give the voltage at every node and the current through every branch. <S> If we use the nodal analysis, we get the node voltages but not the branch currents. <S> If we use the mesh analysis we get only the branch currents (or mesh currents, from which we can easily find the individual branch currents), and not the node voltages. <S> If we have one set of variables (currents or voltages) it's trivial to obtain the remainder. <S> For example, if we know the node voltages we can get the current through any particular branch by applying the characteristic rule for the component on that branch. <S> But we often don't even need to do that. <S> If we are only interested in one set of variables (voltages say, and maybe one or two of the currents) it wouldn't make sense to solve a larger matrix to get the currents we're not interested in. <A> I am using a simple circuit here to apply KVL and then prove to you that KCL can be derived from KVL with the help of ohm’s law. <S> First we apply Ohm's law: <S> \$V_{R_1} = <S> I_1R_1\$ <S> \$ or, I_1=\frac{V_{R_1}}{R_1}\$ <S> \$V_{R_2} = <S> I_2R_2\$ <S> \$ or, <S> I_2=\frac{V_{R_2}}{R_2}\$ <S> \$ <S> E =I(R_1||R_2 <S> ) = \frac{IR_1R_2}{R_1+R_2} \$ <S> \$or, I=\frac{E (R_1+R_2)}{R_1R_2}----(1)\$ <S> Now we apply KVL: For loop at the left: <S> \$E =V_{R_1}\$ For loop at the right: \$ <S> V_{R_1}=V_{R_2} \$ <S> Therefore, we get- <S> \$E=V_{R_1}=V_{R_2}\$ <S> Now we use the equation (1) <S> \$ <S> I=\frac{E <S> (R_1+R_2)}{R_1R_2}\$ <S> \$or, I=\frac{ER_1}{R_1R_2}+\frac{ER_2}{R_1R_2}\$ \$or, I=\frac{E}{R_2}+\frac{E}{R_1}\$ \$or, I= <S> I_2+I_1\$ <S> If you apply KCL to node “a” you will get this same equation. <S> Therefore applying KVL only doesn’t mean we are neglecting KCL.
| Mesh analysis appears to be using KVL only, where it actually uses KCL implicitly , absorbed in Ohm's law in the voltages equations.
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Motor controller - transistor taking up 3/4th of the voltage I am trying to make a motor controller for a DC brushed motor. The design is fairly simple, see below - motor represented by inductor The problem I am having is that if we imagine A to be HIGH and B to be LOW, the motor is not dropping 12V (or close to). If I use a multimeter to measure voltage between drain-source on Q5 I get about 8.4V, on Q8 I get close to 0V. The voltage drop across the motor I measured to be the remaining 3.6V. Would anyone have an idea why my circuit is not operating as I had imagined, and if so, what improvements could I make? <Q> You're not turning Q5 on properly. <S> It needs a certain Vgs to turn on fully. <S> Measure Vgs on Q8 which works and arrange for the same Vgs on Q5. <S> (Start by working out the source voltage on Q5 when it is fully on). <S> This usually means arranging for Q5 to be driven from a different signal (generated by a level shifter). <S> Which means a supply voltage greater than 12V to power the gate drive. <S> Or using a different FET for Q5 (usually a PMOS with source connected to 12V, drain connected to load, and pulling the drive voltage down from 12V. <A> Lots of problems here. <S> First, to turn the high side transistors on, the gate voltage would have to be higher than 12V. How high depends on your choice of transistor. <S> Second, what FETs are you using? <S> Depending on their gate threshold spec, they may not be turning on all the way anyway. <S> A FET H bridge is best done with an H bridge driver IC. <S> They have the bootstrapping for the high side transistors and gate drivers built in, and add the appropriate dead times to avoid shorting the power bus. <A> You have two basic problems. <S> First, the top FETs should be P channel, not N channel. <S> That means a gate voltage below the source will turn them on. <S> This is necessary with your drive arrangement since the gate voltages will always be between ground and the supply. <S> This prevents the N channel FETs at top from turning on. <S> Second, you have to consider break before make. <S> Even if the top FETs were P channel, it is quite possible that there is a gate voltage where both FETs will be significantly on at the same time. <S> That will cause large current spikes, which is not good for the FETs and probably not for the power supply either. <S> If you drive each of the 4 gates separately, you can arrange for some dead time between turning off one of the transistors on a side before turning the other on.
| Fortunately many microcontrollers are available with PWM outputs meant for exactly this application, including a programmable dead time.
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Does it matter the order of amplification and rms squaring? If in a circuit a signal is first amplified and then its rms value is found, would we get the same result if first the signal is rms found and then amplified? <Q> I read your question as meaning that if instead of feeding a real signal into the amplifier you fed into the input a DC level representative of its RMS value, will the output DC level be the RMS of the amplified signal. <S> If this interpretation is correct then theoretically, on a perfect amplifier with no DC-offsets to worry about and providing the amplifier is DC coupled in its stages then the output DC voltage would be OK but there are a lot of things that could make this wrong. <S> For instance if the real signal were somewhat attenuated by frequency shaping circuits (even just a regular 3dB roll-off as seen in op-amp circuits) then the real signal at the output may be significantly less than what you'd expect. <S> This of course means the DC version would give a falsely high representation of the RMS. <S> This is just one example. <A> No, it doesn't make a difference. <S> RMS is a linear transformation itself, so it doesn't matter, <S> provided you multiply the RMS and Gain as well. <A> If you meant measuring the rms value of the signal before and after amplification, then in general the value will not be the same. <S> Measuring before amplification, and after amplification will only give an equal answer with unity gain. <S> In all other cases they are different.
| The operation "its rms value is found" is typically a measurement; it should not change the signal in any way .
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The Art of Electronics - differences between book versions? http://www.barnesandnoble.com/w/art-of-electronics-paul-horowitz/1100491782?ean=9780521370950 There appears to be a couple of versions of the 2nd edition of this book. I noticed US and International editions and a Hardcover or Paperback versions. Are there any contents differences between the versions? [editor's note: The third edition of this title is now available] <Q> They are printed on lower-quality paper, with smaller print, and black-and-write ink. <S> (But AoE has no color sections anyway.) <S> Hardcover and paperback versions refers to the binding. <S> I don't think there are any differences in content. <S> Before buying the book, take note that a 3rd edition will be published Real Soon Now: <S> http://www.eevblog.com/forum/chat/the-art-of-electronics-3rd-edition-finished-writing-and-copy-editing/ <S> [Editor's note: The third edition is now available] <A> Since this is one of the first results coming up on Google while searching for the book, I believe it might be useful for the casual Internet zapper to know that the third edition is out now (April 2015) in Europe and will be soon out in the US. <S> The book is some 250 pages bigger than the previous edition and a sample chapter (9: Voltage regulation and power conversion) along with the detailed table of contents is available online <S> (you can have it emailed from Cambridge University Press , or find a copy in someone's Dropbox account - hey, it's the sample chapter, not the book!). <S> The above chapter has massively changed from the previous edition. <S> In the third edition, the chapters are now 1. <S> Foundations2. <S> Bipolar transistors3. <S> Field effect transistors4. <S> Operational amplifiers5. <S> Precision circuits6. <S> Filters7. <S> Oscillators and timers8. <S> Low noise techniques and transimpedance9. <S> Power regulation10. <S> Digital electronics11. <S> Programmable logic devices12. <S> Logical interfacing13. <S> Digital meets analog14. <S> Computers, controllers, and data links15. <S> Microcontrollers <S> Looking at the detailed ToC <S> it appears some chapters have been split (Filters and Oscillators, for example) while others have been rehashed. <S> Some others appears to have been dropped. <S> In an email by W. Hill shared online , it appears that some five further chapters are still in preparation as additional 'modules' (perhaps only in electronic form? <S> It's not clear to me) <S> All in all, it appears the third edition is a brand new book, with the 'basic chapters' still the same, but most 'application chapters' profoundly revised in content and depth. <S> So, if you - Internet zapper - stumbled upon this page by chance, my advice to you is "go for the third edition". <S> I did. <S> And I have the second edition sitting on my shelf. <A> I have several paperbacks of the 2nd edition floating around my lab. <S> To me paperback is better because of slightly smaller size. <S> It's easier to carry it around. <S> My advice is to buy the 2nd edition now. <S> It's still 95% good, as the main point of the book is to teach you the right design principles, and these don't get obsolete. <S> I guess this is the reason for the lack of rush with an updated edition. <A> Some subsection like the subsection 15.08 "Biological and chemical voltage probles" was removed. <S> I check several times and didn't find it in new edition.
| International editions are substantially cheaper, in both meanings of the term. There is no difference in the content that I know of.
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Playing with this SoC I have done a few projects on Arduinos. Now I want to step up my embedded skills. I have found this SoC with embedded Linux and video encoding/decoding capabilities. I am interested on running OpenCV on it. My question is how do I play around with something like this? Do I have to make a PCB using Eagle? Do I have to buy an external programmer? Maybe use a converter PCB so I have DIP interface? I know I will probably get downvoted for this question, but I want to buy it and start playing with it as soon as possible. That's why I ask. <Q> If you're simply interested in using a SoC processor that can run OpenCV, I suggest the Raspberry Pi . <S> There was just a new version released, B+. <S> It uses a Broadcom BCM2835 SoC which includes an ARM1176JZF-S 700 MHz processor, a VideoCore IV GPU and 512 MB RAM, and also four USB ports (so generally you don't need an external hub for a keyboard and mouse), HDMI in several resolutions, audio (so-so quality), Ethernet, and a microSD card. <S> Debian and Arch Linux ARM distributions are available for download from the main site. <S> All this for $40 (plus cables and power supply). <S> Here's the official page for Raspberry Pi and OpenCV. <S> And here's a OpenCV project on the Raspberry Pi that does facial recognition . <S> So forget about going to the trouble of making your own board. <S> You could be running one of these tomorrow. <A> The link from the TI site was broken, but Googling the name of the company brought it right up. <S> Before you plunk down your money, make sure you understand exactly what you're getting in terms of software licenses for the chip. <S> The chip is really intended for high-volume OEMs, and the development tools and other support is priced accordingly. <S> But if you're only just getting your feet wet, you might want to consider seeing what you can do with one of the hobbyist-oriented SoC boards, such as RaspberryPi or BeagleBone, first. <S> You'll get a lot more community support. <A> The easiest way is to buy an 'evaluation board', 'evaluation kit' or 'development board'. <S> The Arduino is a popular development board for the Atmega328. <S> A quick search for 'DaVinci Digital Media Processor evaluation board' brings some up. <S> What you need in terms of external programmers will vary with what board you get. <A> Normally I would explicitly recommend learning how to use ARM processors first but since this evaluation board comes with linux, you don't absolutely need to <S> but it is still advisable. <S> This thing runs on an ARM Cortex-A8 processor. <S> Arduino runs off an 8-bit AVR core. <S> Needless to say Arduino is much less complicated. <S> Check out TI's range of lower powered ARM microprocessors. <S> I'd advise starting with a Cortex-M processor to get the jist of 32-bit programming <S> so you understand what is going on underneath the linux environment you'd be working in.
| For "playing around" with this specific chip, you go to Z3 Technology and purchase either the bare OEM module or the complete Rapid Product Design System . To expand upon Petes answer, you will most likely need to get an evaluation board since laying out and paying for a PCB with this will be very expensive.
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Reverse engineering old serial devices and cloning with Arduino I have a piece of music equipment that uses a remote control. This remote control communicates bi-directionally with the main unit. It communicates over RS-422 protocol (mainly for the cable distance, I believe). I have purchased a couple of RS422 to USB adapters with the intention of snooping the data in both directions using some serial port monitoring software. I haven't quite gotten that to work properly on its own yet; I think it's most likely the connection settings I am attempting as the data I see looks messed up. From what I can tell from the service manual, this should be 8 bit characters with no parity and 2 stop bits. Anyway, as this device is now archaic in terms of technology (1982, I believe), I think it would be a great candidate for cloning with an Arduino. The remote itself uses an old Intel 8749 microcontroller. This was an MCS-48 based device (I think) which also had a small EEPROM built into it for program memory. This was 8-bit with 3 I/O ports. The remote unit basically has 12 old Litronics DL-1414 alphanumeric 4 digit displays (which receive an ASCII character and hold it in their memory until refreshed), a bank of buttons and faders (which are connected via an ADC) and a dB level meter. I have the service manual of the unit which goes into much more depth about the bit addressing scheme of the devices, which I can upload. I might be thinking about this the wrong way but my idea was to sniff and capture the data being sent and received to figure out what is sent, when for instance, a fader is moved. This I would imagine would then be easy to strap to a slide pot in Arduino and then have that movement sent to the serial bus possibly with some control bits added and then the value. I'd really appreciate some thoughts and maybe advice, on how such reverse engineering can be tackled? <Q> What you seek is a tool named logic analyzer, such as this one <S> http://www.dx.com/p/logic-analyzer-w-dupont-lines-and-usb-cable-for-scm-black-148945 <S> A software which goes with it should be able to decode UART as well as many other protocols to faciliate the reverse engineering. <A> You can do it either Hardware or Software. <S> Hardware <S> : Buy something as mentioned by Spark. <S> Or build your own; In general use some thing like and Arduino to echo between SoftSerial (on an UNO or Serial on a Leo) to the Serial. <S> Where you wire the SoftSerial to your 422 bus through an appropriate interface drivers. <S> Software: <S> On windows something like http://freeserialanalyzer.com/ looks good enough. <A> You can monitoring the data RS232/422/485 COM ports and also displays, logs and analyzes all serial port activity in a system by utility <S> https://github.com/eltima-software/RS232-Data-Logger <S> From my firsthand experience <S> it’s free and does all it claims to do with no restrictions.
| Or simpler use software to Sniffer the messages to the serial port to capture the data.
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Quantifying the piezoelectric effect of ceramic capacitors In comparing ceramics vs tantalums one of the oft cited disadvantages is the tendancy for ceramics to exhibit a piezoelectric (i.e. microphonic ) effect. Old-school technicians talk about ceramic disk capacitors "singing along to the music" in audio circuits. I am having great difficulties trying to quantify this effect. The closest I have is a Kemet paper . So, is this effect still significant in modern X75/X5R capacitors (for example)? Does it vary by package size and manufacturer? What frequency range will cause problems? What sort of noise (microvolts? millivolts?) can be expected to be generated and at what frequencies? Most papers discuss audio circuits, but primarily I am interest in applications with vibrations under 200Hz (sometimes much less). <Q> The answer finally comes as two blog posts from the TI Precision Hub blog. <S> It isn't a laboratory-grade test, but the post looks at the effect of dropping a known mass from a known height on a PCB with differing capacitors. <S> Accordingly, it's dealing with a "unit impulse" test rather than low frequency vibration, but the results are nonetheless interesting. <S> A standard X7R capacitor caused a spike of 120mV on the output. <S> Owch! <S> Special "soft-end" ceramics exhibited about 80mV, whereas tantalum and film capacitors caused no measurable effect. <S> Missing from the test were C0Gs, which I would have really like to seen. <S> This adds to documents from AVX <S> ( and more ) as well as more qualitative documents from ADI , AT <S> Ceramics <S> as well the aforementioned Kemet paper. <S> This isn't quite over the frequency spectrum <S> I was interested in, <S> but I think it's as close as I am going to get without attempting an experiment myself. <S> If someone has this information, I'll un-accept this answer and happily give them my vote. <A> This generally becomes noticeable when they are used in power supply circuits and the load transients are in the high audio range. <S> Mounting the cap to a PCB amplifies the sound as the dimensional changes in the cap couple to the PCB as a small warp causing it to act as a speaker. <S> The physical size of the capacitor is definitely a factor, as is the dielectric. <S> To mitigate this, if possible ceramic caps should be placed in pairs on opposite sides of the PCB in the same exact X-Y location. <S> (I.e. if you need 200uF use 2 100uF caps in parallel, one on the top side and one on the bottom.) <S> If the caps are just for decoupling on DC lines or there is no large ripple in the audio band there is usually no issue. <S> Some manufacturers claim to have ceramic cap products that mitigate this, either by mounting the cap inside a case with poor acoustic coupling or by choice of dielectric. <S> I've tested some of those and was not impressed with the results, but they may have improved over the last several years. <A> (This started as a comment but grew too long.) <S> Nice Kemet paper. <S> Thanks. <S> (Figure 1 is interesting a "Moore's law" in ceramic cap size.) <S> I don't know about the piezo behavior of ceramics. <S> I would expect that size change of the ceramic would be independent of frequency for these low frequencies. <S> (You squeeze it and get a voltage.) <S> The comparison of 100Hz and 1k Hz in the paper seems to show this. <S> As for quantifying, that sounds hard since a lot will depend on the layout and the acoustic coupling between the cap and the pcb. <S> And then the coupling of pcb to outside world. <S> For the size effects one might guess that the large dielectric materials would show a larger piezo effect. <S> (I have no idea if that is true.) <S> But then they will be smaller <S> is size and so the deformation caused by motion of the pcb will be less. <S> It's hard to know which would be the dominant effect.
| In a practical sense modern ceramic caps can certainly be a source of audible noise or mechanical vibration.
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Why is the linearity of a sensor a desirable feature? I mean, a sensor can have a characteristics like Y=X^2 (X = the input, Y=the output).That is, I can easily find X if I know Y.Why is the linearity so important? <Q> Yes, you could have a sensor that responds like that (Y=X^2) <S> Your sensor must still reliably do that. <S> In a typical sensor, Y=cX+d where c and d are ideally constants. <S> Due to various factors, they usually aren't simple constants. <S> How close c and d are to really being constants is what is referred to as linearity. <S> In your example, you would probably have something like Y=(cX+d)^2 , which would cause all kinds of fun. <S> Linearity as in "linear response" isn't the problem. <S> Linearity as in "the Y values closely fit the expected linear response" is what you want. <S> In your example, you wouldn't be concerned with whether the a plot of Y and X is a straight line. <S> You would want your plot to closely match the expected parabolic curve. <S> It most likely won't perfectly match. <S> For a sensor with a linear response, you would say that the linearity is poor. <S> For one like your proposed sensor (Y=X^2) <S> I don't know what you'd call it - maybe just "poor response." <S> if you know Y. With real sensor with good linearity, you can (with in reasonable limits) determine X from Y. With a real sensor with poor linearity, you have to go to greater lengths to determine X when you know Y. <S> Maybe its response varies greatly with temperature, so you have to track temperatur as well and use it when determining X from Y. <S> Maybe it just isn't very linear, and you have to use some funky curve to relate X and Y. <S> In either case, your job as an engineer is much easier if the sensor closely approximates the expected straight line. <S> Often times this works out to making a system simpler to calibrate and more reliable - but also more expensive since a better sensor usually means more expensive. <S> Sometimes you'll want to go with a cheaper and less linear sensor, and compensate for its failings in software. <A> A lot of stuff still uses relatively low-powered or cheap microcontrollers, 8-bit micros haven't gone away. <S> Having to do any of this means having to include maths libraries (more code, more space, more RAM, more storage) <S> If you look at the hoops a CPU has to jump through (in machine instructions executed to achieve one mathematical operation) to do signed floating-point maths, and the resulting output precision, you might decide that just using a linear sensor is a more attractive prospect. <A> Many monitoring circuits for sensors are purely analogue and therefore the complication of mapping a value measured via a square root law is really not trivial. <S> If the measurement is digitized then using math inside the CPU is fairly trivial in comparison but, like I said, many circuits are purely analogue. <A> Linearity minimises the influence of imperfections later in the circuit. <S> For instance, say there is a small constant level of noise introduced by a preamplifier. <S> If you need to correct for the response of that Y 2 sensor of yours, the effective noise level ( SNR ) is proportional to the inverse function's derivative, i.e. ∂ X /∂ <S> Y = <S> d <S> / d <S> X <S> √ <S> Y = <S> 1 / 2 √ <S> Y <S> ∝ <S> 1 <S> / X . <S> At low levels of X , this effective noise tends to infinity (equivalently, the SNR approaches zero). <S> In reality, it's not quite as dramatic since the pertubations aren't infinitesimal, but the problem is real: where the output voltage is only weakly correlated to the measured value, the maximum attainable accuracy suffers. <S> You might now think to filter the noise somehow, but there comes another problem: frequency filtering is basically linear, and to work well it <S> assumes <S> the signal itself has come linear – to use the physics terminology, you'd like filtering to commute with measuring. <S> That's not given with a nonlinear sensor, for instance <S> if you have a high-frequency signal in your X quantity around some X 0 , and filter the resulting Y signal, you'll get a constant offset above the corresponding Y 0 value, because the square-nonlinearity "bends the pertubation upwards". <S> Now if you say then let's first digitally correct the nonlinearity . <S> Next problem: you can only take discrete samples. <S> PCM-sampling is very well under control mathematically , but guess what: it assumes everything is linear! <S> Nonlinearities cause aliasing artifacts. <S> To wrap up: yes, you can somehow correct it if sensors aren't linear. <S> But each such correction brings new problems with it; if everything is linear in then first place you can be most confident to actually get the signal you want.
| Poor linearity in a typical sensor means that you have to do more work to get X if you know Y. With an ideal sensor with a linear response, you can calculate X eaisly The upshot is that: Maths is expensive (in CPU cycles) Floating-point and/or signed maths is even more expensive Doing maths introduces errors / adds to error bounds Type conversion can introduce coding errors Having to handle more bits than your CPU is natively capable of gets very expensive in maths terms.
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Can inductor voltage and capacitor current change abruptly? I understand that inductor current and capacitor voltage cannot change abruptly, but can inductor voltage and capacitor current change abruptly? I have a feeling the answer is no but I cannot explain why. <Q> but can inductor voltage and capacitor current change abruptly? <S> Yes, in the context of ideal circuit theory. <S> Indeed, this is often the case when there is a switch in the circuit. <S> Mathematically, if the slope of inductor current (capacitor voltage) changes abruptly, the inductor voltage (capacitor current) is discontinuous. <S> So, for example, consider the case that a charged capacitor, an open switch, and a resistor are in series ( as in problem 2 here ) <S> At the instant the switch is closed, the voltage across the resistor instantaneously changes from zero to the initial capacitor voltage \$V_0\$. <S> Thus, the capacitor current discontinuously changes from zero to non-zero and is given by $$i_C(t) = <S> \frac{V_0}{R}e^{-\frac{t}{RC}} \cdot <S> u(t)$$ <S> where \$u(t)\$ is the unit step function <S> The dual of this is an inductor, with non-zero current \$I_0\$, in parallel with a closed switch and a resistor. <S> At the instant the switch is opened, the current through the resistor changes instantly from zero to the initial inductor current. <S> Thus, the inductor voltage discontinuously changes from zero to non-zero and is given by $$v_L(t) <S> = I_0R\;e^{-\frac{tR}{L}} \cdot u(t) <S> $$ <S> In physical circuits, voltages and currents cannot instantaneously change but depending on the characteristic time scale, they can effectively change instantaneously. <A> Note that the reverse is not true: <S> the voltage over a capacitor, and the current through an inductor, can not change abrubtly (unless you allow for non-finite currents or voltages, like a Dirac-shaped pulse). <S> Note that this ideal world is an mathematical abstraction, you can't buy such components. <A> I = <S> C*dv/dt. <S> So a step of voltage on a capacitor (infinite dv/dt) leads to infinite current. <S> The response to a step of current is 1/C times the integral of the current. <S> The integral of a step is a ramp, so the capacitor voltage will ramp linearly in response to a step of current. <S> (From whatever initial condition voltage is on it.) <S> An inductor's response is analogous with current and voltage switched.
| In an ideal world, where a capacitor has no series inductance and an inductor has no parallel capacitance, and voltage and current sources can provide voltages and currents with a step-shaped profile, the current into a capacitor and the voltage over an inductor can change abruptly.
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How to detect a change in resistance? I would like to have a circuit that can be controlled by two variable resistors, and it should use the value of the "most-recently-changed" resistor. How can I find out when one of the variable resistors is changing in value? I would like to figure that out and then flip a latching relay such that it uses that resistor for the input. Is this feasible using a circuit without a micro-controller monitoring the two resistors? what would such a circuit look like? scenario: imagine a lightbulb that is controlled via a three-way switch. Now imagine the switches replaced with dimmers - the most recently adjusted dimmer should be the one used to control the lightbulb, the other one exists simply to allow adjustment from a different physical location. <Q> I can't think of any easy way to do it without a microcontroller. <S> Use firmware to decide which is "most recently changed". <S> Drive a digital potentiometer from the microcontroller to operate the actual circuit. <S> This is a tiny, two-chip solution. <S> Anything else would require many more parts. <A> I would also recommend to go for a micro-controller based solution using two ADC channels. <S> Most probably a very small µC (with regard to foot print, memory size, electrical power usage and price) would be sufficient. <S> Depending on the circuit to be controlled it might even not be necessary to drive a digital potentiometer - as the µC might control the load e.g. by PWM. <S> The description of the task leads to the guess, that any simple circuit "switching" those resistors might suffer a certain usability flaw. <S> Consider both potis to be set to 100%, now somebody plays around with number 1, setting it to 20%. <S> Later on somebody turns number 2. <S> Your circuit now switches immediately from poti 1 to poti 2 - <S> that is from 20% setting to 100% resulting in an uncomfortable flash of your light bulb (if that's the load). <S> The µC based solution could easily be programmed to overcome that issue. <A> OK, so you have two variable resistors and you want to use the signal from the variable resistor that last moved? <S> OK so far? <S> You don't want to use a micro so the only real option is to have a circuit that detects rate of change of each resistor value and the one that last moved i.e. created a \$ABS(\frac{dR}{dt}\$) signal <S> , would "trip" a SR flip-flop. <S> The flip flop output drives <S> DPDT analogue switch thus selecting either resistor A or resistor B. OK so far? <S> You'll need a fairly sensitive differentiator circuit (probably formed around an op-amp) for each resistor and, each resistor would be biased with a current so that it produces a voltage signal to feed its differentiator. <S> That gives you the ability to generate positive and negative pulses when the resistor value changes - the size of the pulses are dictated by the speed that the resistor changes value. <S> So then, and accepting that small slow changes will not be detected, you can use a window comparator that looks for "going above" and "going below" thresholds - either one will set a SR flip-flop. <S> The other resistor drives a similar circuit that resets the flip-flop. <S> That should work I think.
| Find a small microcontroller that has (at least) two ADC channels, and use these to measure the two real resistors (potentiometers, rotary encoders, whatever).
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Is there any way to interface 4x4 matrix keypad using 2 IO lines I have two IO lines and wants to interface 4x4 matrix keypad. Using MUX ICs I need 4 IO lines. Is there any method to achieve it using 2 IOs? <Q> I have done it by using an I 2 C GPIO expander chip. <S> I used the Microchip MCP23017 , but there are others out there as well. <A> A serial interface could do the trick. <S> An I2C-like interface could operate on only two <S> I/O lines (Data and Clock). <A> Use one GPIO as an Output and toggle it while timing the rise and fall times of the other GPIO set to input. <S> Choose resistor values to separate the timings for each key as much as possible while accounting for component variations and uController logic input threshold variation with temperature.
| Utilizing shift registers would be possible using three I/O lines (Data-in, Data-out and Clock) in an SPI like mode (assuming to skip any chip select lines).
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What are the bare minimum parts needed to operate a magnetron? I recently took apart a microwave, and I managed to get a magnetron out of it, as well as a transformer that appears to be the power source for the magnetron, but looks a little small for the advertised amount of power this magnetron can handle (~1 kW). Also, the magnetron does not have any leads for supplying the filament voltage, just for the HV. I would like to get that magnetron running, I plan to put on a waveguide and do a bit of experimentation with microwaves. 1) Does the magnetron have something built in to step down a fraction of the power recieved to the 3.3V required to heat the filament? 2) Is it possible to run the magnetron simply by powering the microwave transformer with mains and hooking up the hv output to the input of the magnetron, or do I need a microwave capacitor? I was not able to locate a capacitor in the microwave, I think that might be because it is an inverter-style microwave. <Q> I believe the way microwave oven magnetrons are wired is that there are two coils on the transformer more or less in series, one to supply the HV and one to supply filament. <S> The filament doubles as the cathode and it has some series inductors to keep the RF from leaking out the power terminals. <S> The HV return from the anode could just be the metal case as it is less than an amp. <S> An inverter-driven microwave likely has the same setup, however the transformer will be smaller as the switching frequency will be much higher (60 Hz vs several kHz). <S> The inverter should have essentially the same connection to the magnetron as the older style single transformer solution. <S> Also, you won't be able to run the inverter transformer by itself, you'll also need the rest of the inverter drive and control electronics. <S> I'm not sure if it will work without the front panel/user interface; it may be required to turn the inverter on and possibly select the power level in some way. <S> However, turning on a magnetron outside of a case is very dangerous. <S> Not only could you kill yourself with the high voltage, the high power microwaves could give you serious burns as well as possibly causing interference and damage to electronic deivces. <S> The magnetron could also overheat due to reflected power if it is not properly coupled to a waveguide. <S> Edit <S> : Looks like the inverter microwaves operate on exactly the same principle as the older ones, just at a much higher frequency. <S> You'll either need to keep the front panel or figure out what signal it sends to the inverter to turn it on. <S> Image from http://www.electronicspoint.com/threads/microwave-inverter.234684/ <A> Acknowledging that this is an old question... <S> Does the magnetron have something built in to step down a fractionof the power received to the 3.3V required to heat the filament? <S> Sort of... <S> The power circuit generates the low power DC for the heater and the high power AC for the tube and the magnetron separates them. <S> On the bottom of the magnetron, just below the input terminals, there is usually a square crimped-on cover. <S> With that cover removed you should see two heavy gauge leads that run to a pair of fat RF chokes. <S> These filter out the high voltage AC and leave the low voltage DC required for the filament heater. <S> Is it possible to run the magnetron simply by powering themicrowave transformer with mains and hooking up the hv output to theinput of the magnetron, or do I need a microwave capacitor? <S> I was notable to locate a capacitor in the microwave, I think that might bebecause it is an inverter-style microwave. <S> No. <S> If your goal is to eliminate the microwave control circuitry you'll need to keep several components. <S> Most microwaves have a schematic in them. <S> Pay attention to the relay circuit that turns the magnetron drive circuit on and off. <S> Connecting AC mains power to the magnetron will very likely destroy it in very short order. <S> This link can answer many of your questions. <S> Do be careful, an operating magnetron is like a 1,000 Watt Lightbulb. <S> Close up or as a directed beam it can burn you and make you blind faster than you can blink. <S> Additionally, it can destroy electronics at obscene distances. <S> I have yet to see one equipped with an undo button. <A> It is an inverter type. <S> The transformer will not work on the mains. <S> It needs a much higher frequency and the current needs to be controlled, using pulse width modulation, which is done by the control circuit. <S> The magnetron acts like a 4000V zener and needs to be run off a current source or high impedance source. <S> According to the diagram, there is a small extra secondary on the transformer to generate the 3.3V for the filament, which is also the cathode (known as a directly heated cathode). <S> The 2 inductors block any RF coming out through the cathode. <S> In one magnetron, they appeared to be tuned, as the turns were separated, which leads me to believe that they may have a second purpose: to smooth the cathode current & improve efficiency by preventing current surges as the spoked electron cloud inside switched from 1 set of vanes (which are now negative, to the next ones, which are positive). <S> They have nothing to do with generating the 3.3V, which is between the lugs. <S> Click on this address to see how the magneton works. <S> You can use a mains frequency transformer from another microwave. <S> The mains frequency circuit has a high leakage transformer which has magnetic shunts with air gaps between the primary & secondary, so it acts like it has an inductor in series. <S> This gives it a high impedance output so it can accommodate the square wave across the magnetron. <S> The capacitor will have an odd looking value like 1.19uF. As well as forming a voltage doubler with the diode, it is designed to resonate with the secondary to bring the voltage back up, while retaining the high impedance. <S> Magnetron Working Principle
| For a non-inverter microwave that would be at least a fuse, transformer capacitor, and diode.
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How many volts can a 1 Watt 8 ohm speaker take? I am working on an Arduino project and the sound is too low because the output pins on my Arduino UNO is only 40 mA. I can hook up an NPN transistor to amplify the sound, but I don't want to blow the speaker. How much voltage can a 1 watt 8 ohm speaker handle? <Q> It can take 1W of power. <S> Voltage is not the problem. <S> Any more than 1W and the coil will overheat and melt. <S> It is <S> 8Ω. Look at it from the point of view of DC. <S> That means we can use simple Ohm's Law to examine it. <S> You have 1W and 8Ω. <S> There are two formulae that incorporate those two values: <S> \$P= <S> I²R\$ and \$P=\frac{V²}{R}\$ <S> We're interested in voltage, so rearrange the second to give: <S> \$V=\sqrt{P×R}\$ <S> So 1W through an 8Ω load must be 2.83V. Rearrange the current one, so it is: <S> \$I=\sqrt{\frac{P}{R}}\$ <S> and we get a current draw of .354A, or 353.55mA. <S> The fact that your IO ports are limited to 40mA <S> (That's the absolute maximum by the way <S> - Atmel don't recommend more than 20mA), means: <S> \$P= <S> VI = <S> 0.2W\$, which is why your speaker doesn't melt and isn't very loud. <S> So what do you want? <S> Well, you want 2.83V across your speaker with unlimited current available, or unlimited voltage available with 353.55mA current. <S> The former is more achievable, so we'll do that. <S> A simple voltage divider can limit the voltage to 2.83V. <S> The formula \$V_{OUT}=\frac{R_2}{R_1+R_2}V_{IN}\$ can be re-arranged <S> to give: \$R_1= <S> R_2(\frac{V_{IN}}{V_{OUT}}-1)\$ <S> We know R2, that's 8Ω <S> , Vin is <S> 5V and Vout is 2.83V. <S> So substitute the values and we have: \$R_1=8(\frac{5}{2.83}-1)\$ which gives us 6.134Ω. <S> The closest E24 would be 6.8Ω, which would be ideal. <S> Of course, you need a nice chunky resistor, at least 1W, preferably a little more. <S> Your schematic could look like: simulate this circuit – <S> Schematic created using CircuitLab <S> Or, for the more traditional class A amplifier arrangement: simulate this circuit Of course, your 6.8Ω resistor would then have to cope with the full 5V across it, so would need to be a minimum of 3.6W. <A> Power = <S> voltage x <S> current current = voltage / resistance power = voltage <S> x (voltage / resistance) <S> voltage^2 = power x resistance voltage = <S> sqrt(power <S> x resistance) = <S> sqrt(1 * 8) = sqrt(8) = <S> 2.83 V <A> It's not an easy question, because speaker ratings are sometimes specified as peak power and sometimes as RMS (average) power: http://www.bcae1.com/speakrat.htm <S> Either way, for calculating the maximum current or voltage, you can assume the speaker acts like a resistor, <S> so P=U^2*R. <S> For U, you will have to plug in either the amplitude or RMS value, depending on the speaker rating. <S> Also, amplifying with a single transistor can lead to lots of distortion, except if you're using a square wave signal. <S> Read up on some basic amplifier circuits such as the "common emitter amplifier" or operational amplifier circuits. <A> and then it will sit around 2.2 volts in quiescent conditions (in order to maximize the undistorted AC signal applied to the speaker). <S> This 2.2 volts forces a dc current thru the speaker of about 370mA <S> - this is based on the likely dc resistance of the 8 ohm speaker being about 6 ohms. <S> This is generating a power (heat) of 0.806 watts therefore the "spare wattage" remaining for audio is slightly less than 200mW. <S> This is equivalent to a sinewave amplitude of 1.265 volts RMS or about 3.6 volts p-p. <S> If you used a push-pull circuit and a capacitor decoupler, a 1 watt speaker at 8 ohms impedance could be expected to handle about 2.828V RMS or 8 volts peak to peak. <S> The better circuit would be about 7dB louder and have less distortion.
| If you are dc connecting the speaker to an NPN transistor's emitter and feeding the base from an arduino (presumably 5v logic), it's likely you could see a peak DC voltage of around 4.3 volts across the speaker so it needs to be biased up correctly
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Is there any pattern or standard to through-hole resistor sizes? I've long been under the impression that a quarter-watt resistor has a standard package, and a half-watt resistor has a standard package, etc. But I've recently gotten through-hole resistors which are not in the expected package size. I've done an analysis of a few thousand of the top stocked through-hole resistors at Digikey, and I see no pattern. 3.5 +- .3 mm length seems to be one grouping, which I would normally have thought of as a 1/8W resistor. But Digikey has resistors listed in this size range of anywhere from 1/8W to 1/2W. 6.3 +- .3 mm length is another grouping, which I'd normally call a quarter-watt resistor package. But the wattage is anywhere from 1/8W to 1W. Similar groupings can be observed around 9mm, 12mm, 15mm, 18mm, 22.2mm, 26mm, and 45.2mm. This makes creating our internal part numbering scheme problematic; we can't just assume a quarter-watt resistor is in a "quarter-watt package", if such a thing there be. So is there a set of size standards for through-hole resistors? And is there any relationship between those size standards and the wattage of the resistor? <Q> Almost all manufacturers still make them in the standard sizes, but not all of them refer to that DIN size any more in datasheets or product numbering, as they used to. <S> This is a four character number. <S> The most common of these are 0102, 0204, 0207, 0309, 0411, 0414, 0617 and 0817. <S> (although the numbers do go higher).Sometimes these are prefixed with DIN-****.As stated size no longer really equals power, as there are different ways of building up the resistive film that will damage at any temperature between 75 and 200 degrees Celsius, where each build-up can be equipped with advantages and disadvantages in other domains. <S> If you want to implement DIN compatible size numbering <S> I'm sure someone somewhere on Google has scanned an old book about the regimen of sizing using DIN numbers. <S> They have become so normal/standard to me, that I hardly remember the meaning of the numbers. <S> Probably something to do with those silly inches. <S> ;-). <A> "This makes creating our internal part numbering scheme problematic" - IMHO part numbering is a very complex issue. <S> Smart, sane, Engineers in manufacturing companies have "pie fights" over it. <S> Approaches which can work: <S> Use non-significant part numbers (i.e. almost random numbers) and make it easy to see key attributes 'at a glance'. <S> The unique numberensures you can manage stock, part changes, etc. <S> Providing keyattributes helps humans understand what the part is. <S> This carried theproblem that people make mistakes, or think the attributes are thepart identity, and really, they are not. <S> Use significant part numbers, i.e. something made up from type ofpart, and <S> values of key attributes, and make it easy to add newattributes fields without breaking any existing systems. <S> This isoften hard to do because computer systems were not designed to beflexible enough. <S> I have worked with companies which did neither, and it was often painful. <A> My experiance is that there are conventional body sizes for each power rating of carbon film resistor. <S> However, metal film resistors can tolerate higher temperatures and some but not all manufacturers use this to advertise a higher power rating for a given body size.
| Standard Through Hole Resistors are very commonly, though unfortunately not absolutely always, referred to in the size domain by their DIN-size.
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Why is there a resistor inline with switch in blender motor? Our hand-held blender stopped working and I took it apart to see if I could fix it (no, as it turned out). The circuit was simple; a switch and a 0.47 Ohm (R47) resistor in series with a full bridge rectifier whose DC output was connected directly to the motor. The wattage is 200 W and the voltage 120 V, so the voltage drop across the resistor is small. Question: Why bother with the resistor? Note : I am not looking to fix the device, I know one of the windings is burned out, but I am curious about the resistor! (I am an EE, and 'should' know these things, but...) Added : The labeling is R47 10%, and UTM 206-8. A quick look at the Jameco catalog gave me 'Power Wirewound Ceramic Resistors, axial'. <Q> It may be a fusible resistor designed to protect the wiring in case of a short in the bridge rectifier or the motor. <S> If it is open, then it either has opened early or it has done its job. <S> The 200W rating would seem to indicate a higher amperage of fuse, but perhaps that's a peak rating or something like that. <S> That item would be the first thing I'd check, and the second would be the bridge rectifier. <S> Then the motor brushes. <A> It could be a ntc - for softstart. <S> It could also have a capacitor parallel to it as an energy saving or a speed control <S> mechanism: the capacitor shorts out the resistor in start up but after the resistor is there to lower the current to control speed for example. <S> The 2nd approach is quite common for relays. <S> Btw. <A> R47 is 0.47 ohms <S> so it acts like a fuse, depending on size, a 5w fuse. <S> Common design decision - than glass and metal, not designed to be user replaced.
| It may also be a simple fuse .
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Supercap charging circuit for RTC backup on STM32 I'm asking for a reliable/simple/cheap circuit for the V BAT section of a STM32 design, powered at V DD =3.3V, that will safeguard the built-in RTC (based on 32768 Hz Xtall) for like 10 days. Based on C = t⋅I/ΔU, its looks like a 1F supercap will do (I'm taking t = 864000s, I = 1.2uA, ΔU = 1.1V, see below). I need a charging circuit that does not fry the supercap; does not fry the STM32 or prevent it from starting-up reliably (in particular: I do not see a specification for what happens when V DD =3.3V, V BAT <2V) does not add significantly to the STM32 power drain on V BAT ; delivers proper V BAT quickly (0.2 s) including after full discharge, so that the RTC is always available soon after power-on reset; brings the supercap to near full charge soon enough after V DD is applied (perhaps 80% after 50 s), so that testing leaves a decent reserve. The simplest I can think of is Any criticism? Suggestion on the diode (perhaps a Schottky would be better) or other component? Alternative? A most relevant section of the STM32(F100) data sheet is: <Q> I find the circuit proposed in the picture perfectly fine. <S> Your worry about VBAT being lower than VCC is mostly mitigated by your diode right across those two rails. <S> The power-on delay introduced by this diode would be so minuscule, that it is not worth considering. <S> Also, frying the supercap mostly depends on a particular type. <S> Some of these have rather low allowed currents. <S> This is mainly due to their high ESR. <S> During current flow, this causes heating and damages the part. <S> Some types don't have their maximum currents explicitly stated. <S> In such case you need to rely on the ESR value and a wattage the package can reasonably deal with (sometimes also given in datasheets). <S> In the end, you adjust your \$R_1\$ to be a compromise between charging time and charging current. <S> \$22\Omega\$ sounds good for starters. <S> Keep in mind that you will almost never discharge to 0 volts. <S> Also, charging current in this case drops exponentially, so you will only draw \$100\tilde- 200 mA\$ for a few seconds. <A> I've put together a circuit very similar to this in the past <S> and I think, based on that, you'll be just fine with something like this, especially if it's not a critical application. <S> A couple of considerations:Check out your specific supercap to see how best to charge it. <S> You'll want to avoid charging to too quickly if it has a high ESR. <S> Also, certain types of supercaps hold more charge if you charge them slowly (I'm thinking of Panasonic's "Gold" electric double layer supercaps). <S> Although you'll be losing more as you discharge your capacitor, these two considerations could mean it's worth increasing the value of R1. <S> For maximum lifetime you'll have to balance these factors. <S> Also, I think a Schottky is a great idea if you're looking to get a little more charge out of your cap. <A> does not fry the STM32 or prevent it from starting-up reliably (in <S> particular <S> : I do not see a specification for what happens when VDD=3.3V, VBAT<2V) Looking at the Device power supply scheme of STM32F3XX you can see there is a power switch. <S> I guess it's all the same for all the ST32 families, but you better checkIt's controlled by Vdd, if it's present Vbat <S> is only "monitorable" with ADC but does not take any action for what concerns RTC etc. <S> So just for this question, I think with your circuit you can play safe
| The one thing I'd look out for there is that Schottky's can have horrendous reverse leakage current at high temperatures, so if this would be in an environment that gets toasty you could lose a lot of battery life.
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Switch on computer with microcontroller I'm making a simple microcontroller based circuit to turn on a computer when certain events are triggered. I am interfacing it using the PCs soft power switch pins on the front panel header: There are two relevant pins on this header, I'll call them SW+ and SW- . SW- shares the PCs system ground, though I'm not sure this is always the case. SW+ is, in my test system, 3.3v above ground, although I suspect this could be more or less. In a typical computer, a NO momentary switch on the front panel shorts these two pins when the power button is pressed. I want this to instead be controlled based on a logic level signal from a microcontroller. My first attempt at a test circuit is as follows: I tested the polarity of my header and made sure SW+ really was the positive pin. My microcontroller was powered from USB, therefore sharing system ground, so connecting SW- and controller ground was not a problem. The circuit worked as intended, but it's obviously not very great. I want to make this circuit as universal (not system dependent) as possible. This poses the following problems: The polarity is not indicated - the user can not be relied on to insure a specific polarity. The voltage is not exactly known - it's better to assume it will be somewhere between 2-12v. The controller should be isolated - it might share the computer's ground, but this is not guaranteed. The simplest solution that comes to mind is using a relay. I may end up doing this, but I really would like to avoid it due to reliability, size and price. Is there any transistor based solution (essentially like a relay) meeting the criteria above? <Q> My first reaction is to use a opto-coupler. <S> That isolates your microcontroller from the PC ground, which is a good idea. <S> This solution still requires that the device be installed with the correct polarity, which personally I don't think is a big deal. <S> I also don't think that worrying about the voltage being high makes any sense. <S> The front panel button is going to be a logic level signal. <S> However, the output of optos will be a bare transistor. <S> Most can withstand 20 V at least, which will surely be enough. <S> Check the maximum allowed reverse bias voltage of whatever optos you use. <S> These are good to 6 V, which will be fine for a PC power switch which isn't going to operate on more than 5 V logic, most likely less. <S> This example puts about 10 mA thru the LEDs, and draws around a mA from the logic output. <S> Nothing is being pushed to the limit here, and the output is polarity independent. <A> If you put a diode in series with transistor collector it is likely that this would work. <S> If someone wired this up in reverse, placing a high value leakage resistor across the transistor (maybe 1Mohm) would protect the transistor. <S> Duplicate this circuit so that one of the transistors would always work no-matter what the polarity and you have a bipolar solution. <S> Then get rid of the transistors and replace with opto-couplers. <S> Now you have a biploar and isolating solution. <S> I did think of using a single opto "encased" within a bridge rectifier <S> but I believe the overall volt-drop (when activated) may be too great <S> but it's also worth a try. <A> So, I tried the suggested avenue with a relay. <S> Additional detail <S> [the relay is a Wi-Fi controlled device by eWeLink from Amazon. <S> The eWelink is then paired with an Amazon Alexa which then allows me to control the PC from anywhere.] <S> I found that the PC is ok with a fast momentary pulse to start (eg. <S> 0.5s), <S> but it requires a long pulse to shutdown (eg. 5s). <S> The problem with just using a relay is that the shutdown works fine, but the slow pulse when used for startup causes the PC to start and then immediately shutdown. <S> This seems to be a debouncing problem. <S> This is not enough when the used with the 5s timing to start the machine. <S> So using the 5s relay pulse causes multiple pulses which start it and then starts the long timer which ends after the 5s causing the PC to shut back down. <S> I have not tried this, but a debounced relay contact at 5s with the dual opto circuit proposed above would be the Cadillac solution for universal use - particularly with the web based wi-fi relay setup.
| I surmise that the PC may have a very short debounce circuit for the power switch. If you really insist that the output should be polarity-independent, then use two optos with outputs wired opposite but in parallel: As mentioned by @gbulmer a solid state relay would work but availability and cost may be an issue.
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Options for semi-permanent memory in a low-energy setting (Hopefully this will be broad enough to be on-topic.) I'm looking for options to store several megabytes of data over the period of a few weeks in a coin-cell powered application. The data doesn't need to survive a power outage (dead battery, etc.), but I do expect to perform several hundred reads and writes over the life of a single battery. What technology type would be the most power efficient? The flash options I'm looking at (such as the m24px80 ) have fairly substantial current draw on write (15mA @ 3V). What other options are there? <Q> Some of TI's MSP430 microcontrollers provide persistent storage in FRAM , which has some very interesting characteristics: <S> It's non-volatile - no power is required for data retention. <S> Fast read and write times (TI claims 50 ns) <S> High endurance (TI, again, claims 10^15 cycles) <S> (The numbers I'm quoting are from TI's FRAM FAQ .) <S> Of course, this is only useful to you if the MSP430 is an option — it's not available as a standalone part at this time. <A> An SRAM might do it. <S> Microchips 23A1024 for instance, it draws 3mA at READ at 5.5V 20MHz, they can be driven by voltages down to 1.7V depending on model. <S> Unfortunately they are quite small, only 1Mbit. <A> You need to look at devices that are specified for low-power or "ultra low-power". <S> For long battery life, low standby power can be especially important. <S> Here is <S> an example from Cypress: CY62167DV30 which lists a typical standby current of 2.5µA A supervisory circuit for controlling SRAM power and chip-enable is often also used. <S> One example is the TPS3613-01 from TI <S> Regarding the Micron M25PX80 NOR <S> Serial Flash device you referenced: For only a few hundred reads and writes over the life of the battery, the 15mA write current may not be such a big concern if you can take advantage of the "Deep power-down" mode when the device is idle. <S> Specifications for that mode indicate a maximum current of 10µA (Device Grade 6). <A> Another option : MRAM : <S> Nonvolatile magnetoresistive memory. <S> http://www.everspin.com/ <S> High density chips are available (>1MB) <S> As the supply current is quite high, you would have to power-off the chip between accesses.
| It sounds like you are looking for a low-power Static RAM (SRAM).
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Noise cancellation for electromagnetic signals I am working in the oil field with tools which I need to connect by cables (which "should be shielded") to computers. Most of the time when I communicate with my tools, I face grounding issues. I tried to ground the computer but it didn't always work, even when am trying to use the laptop to minimize that problem. Does anybody know how I can reduce or cancel that noise to get better communication with my tools? <Q> It is difficult to address your problem without the exact setup and attempts to solve the problem, but I would like to add to the above comments with some good shielding practices, since that is the solution you are using. <S> When shielding, grounding on one end will only help against interference from electric fields. <S> Grounding both ends will help against both electric and magnetic fields, but will introduce ground loops, so extra measures must be taken to prevent/compensate <S> these(you should try to ground everything to the same reference point in <S> a 'star' grounding configuration andyou should try to keep the enclosed surface between shield and ground as small as possible). <S> Prevent grounding your shield with a long wire (pigtail grounding), because this wire itself can radiate EMI. <S> Shielding done wrong can actually backfire, so I would personally use twisted pairs as a starting point against EMI. <A> If you have an electrical device ("tool" in your description) that is grounded at a distant point but is then also connected to a computer that is grounded separately (or not earth grounded at all), that can be the source of a lot of ground noise. <S> It's called a "ground loop". <S> Google it. <S> To fix a ground-loop, you need to run a separate wire from the same grounding point as the computer out to each device. <S> Meaning all devices and the computer all converge on the same grounding point. <S> More specifically, use a separate power-strip just for the computer and devices connected to it. <S> So you will have a power-strip plugged into one clean high-power supply and then all of the power cords fan-out from that one point. <A> You should have a look to common and differential mode noise in EMC. <S> If the cable is shielded, the signal still may be polluted as you experience it. <S> You might try to twist your wires as a first step. <S> Maybe use coax cables instead of two shielded signal cables. <S> Finaly, you should find many ways to reduce the noise in EMC lessons on the internet, like : http://www.murata.com/~/media/webrenewal/products/emc/emifil/knowhow/26to30.ashx
| Twisted pairs(usually with differential signaling), common-mode chokes, ferrite beads, decoupling capacitors, shielding are common solutions.
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What can be the consequences of charging a lead-acid battery with an excessive voltage? In my field-operating device I use a simple PWM step-down to charge a 6V 3.9Ah lead-acid battery from a 5W solar cell with a voltage of 7.2V. Unfortunately the DC regulator got damaged today and the battery is charged at 8.6V, that is 1.1V above the limits specified by the battery manufacturer. At the sunny days the charging time is approx. 8h.I have no access to the device, I cannot fix it right now. My question is: what can happen to the battery, can it explode, set on fire? I use three 1.3Ah batteries connected in parallel. <Q> This information can be found at various lead acid battery manufacturers, such as Power Sonic. <S> Here is an excerpt from their FAQ : <S> Q: Does overcharging damage batteries? <S> A: <S> This will cause decomposition of the water in the electrolyte and premature aging. <S> At high rates of overcharge a battery will progressively heat up. <S> As it gets hotter it will accept more current, heating up even further. <S> This is called thermal runaway and it can destroy a battery in as little as a few hours. <S> For further information about charging please refer to our Technical Manual pages 12-19. <S> To access our Technical Manual please go to the Literature - SLA Batteries section of our website <A> A doubt 5 watts of heat is enough to even get hot or explode the battery unless is was poorly vented such as in a sealed box. <S> What happens is the sulphuric acid electrolyte (H2SO4) liberates Hydrogen easiest from excess energy wasted and if there is a spark with H2 in a container it can be dangerous as <S> 4% H2 plus any amount of oxygen is an explosive condition with a tiny spark. <S> But with vented H2 it dissipates in the atmosphere quickly and rises (lightest gas) out of reach, so is not a problem. <S> But in large car batteries under a hood with enclosed gas and an spark from a loose jumper cable, it could detonate. <S> (big boom) <S> Most likely the battery has now dried its electrolyte and if after a long time the battery is worn out. <S> FYI Sealed gel cell batteries MUST have a vent . <S> In external cases it must have an added Teflon plug or "H2 vent" to allow it to H2 leak in case of over charging as Teflon allows tiny hydrogen gas to escape but prevents moisture from getting into a sealed box. <S> In an open room, it is no sweat for such a tiny battery. <S> If you had a small room full of big batteries , I would be cautious about venting. <S> You should be able to hear bubbling of H2 in a lead acid battery being created but watch out for ESD static discharges in an enclosed space. <S> H2 gets easier to trigger above the lower explosive limit of 4%. <S> Hydrogen is odorless, colorless and tasteless, so unless you can hear it, you cant see or smell it , but you may smell the sulfur-oxide gas which is often added to H2 tanks or from the sulphuric acid. <A> I know this is an old question but since I stumbled onto it while investigating the effects of overcharging a Lead Acid battery I'll contribute my findings for whom it may be of interest. <S> As described in this Battery charging basics Article from PowerStream . <S> Overcharging a battery causes hydrogen gas to be released. <S> Sealed lead acid batteries can recycle the generated gasses as long as they are being overcharged at less than C/3. <S> However, leaving the battery to be overcharged even at C/10 will corrode the plates if left on for weeks at a time . <S> The amount of hydrogen gas generated is an important factor to take into account , especially in your case where your battery is in an almost closed box only with a vent hole for ventilation. <S> As hydrogen is lighter than air it will move upwards, so if your ventilation hole is facing upwards or downwards will have an effect on the gas concentration accumulated within your box. <S> The Lower Explosive Limit for Hydrogen gas is 4% which means that Explosion is a possibility above this concentration. <S> Battery Outgassing Math is a great post describing how to calculate the gas generation and the theory behind it. <S> Calculate Industrial Battery Hydrogen Gas emission is another article allowing you to determine if your charging environment requires forced ventilation or not. <S> Batteries - Explosive gasses and Ventilation : <S> Another article on calculating ventilation requirements. <A> You say 3 batteries connected in parallel? <S> First off, parallel charging is a good idea considering a broken charge controller! <S> Each parallel connected battery has they're own resistance, I've noticed! <S> Like right now, on charging two 115 ah batteries, with 19, 8.8 amp volt solar panel! <S> (160 watts) <S> I haven't seen voltage go over 15 volts yet! <S> If I charge 1, volts would go about 16, which is lethal for a battery! <S> 14.5 volts is ideal most for 12v battery! <S> Depending on outside temps of course! <S> Hotter lower voltages! <S> I'm reading your post, noticing error! <S> Considering 7.2 volt charge for 6 volts battery that's ideal! <S> How I've got 8.6 volts?
| As a result of too high a charge voltage excessive current will flow into the battery, after the battery has reached full charge.
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Noise from switching fans nearby disrupting the I2C bus I have a RPi controlling a solar system which uses temperature sensors on the I2C bus, the sensors are away from the RPi (20m). The problem is that I have cooling towers and cooling compressors which have high switching currents. In the beginning I tried the I2C and it did not work at default frequency. So I reduced the frequency which got it working for a while. However whenever some of the fans start working I lose some readings for a couple of seconds and sensors start acting weird. I think it is a noise problem, especially that my 230VAC power supply is coming from the same 3~ph power supply of the fans. What would be the best way to protect the signals on the I2C bus? I use now CAT5 cable to wire the sensors. I wanted to use CAT7 but then I read that this will actually make it worse since CAT7 has higher capacitance which is worse for the bus at longer distances. More Info: I'm using 10kHz currently since the default 100kHz did not work.On a similar setup however (but without the nearby fans) i got away with 100kHz and more than 20m without extenders or anything. System has been running smoothly for more than 2 years now. That is why i'm thinking it is a noise issue. <Q> According to this source which is an FAQ on I2C bus protocol. <S> This suggests that typical lengths are 3-4 meters, and with the serial clock running as slow as 500Hz in applications of 100 meters are known to work. <S> I suggest you lower the clock speed significantly from the standard 100Khz to something like 8-10Khz or something and do some tests. <S> Also, what I2C bus pull-up resistance do you have? <S> With such a long distance, the bus capacitance must be very high so your resistors should be quite low, maybe 1.8-2K Ohm (as shown further down in the FAQ actually!). <S> EDIT: <S> May I suggest going away from wires sensors? <S> You can put together a very simple little package that can act as wireless sensor nodes. <S> Then you can place them 80m away if you really wanted to. <S> And you can have lots of them without worrying too much about running cables everywhere. <S> Depending on your level of skill/embedded systems knowledge, you may want something simple like a Synapse like those sold at Sparkfun which supports all sorts of cool mesh networking too if you wanted. <S> Otherwise you can investigate other more in-depth designs... <A> IIC is going to be problematic. <S> You are sending a clock signal out to the remote devices and that clock signal (when it arrives) clocks back data to the master and the master reads the data by clocking it back in but the clock the master is now using bears little resemblance to the delayed clock at the sensor and the data coming back is delayed even more. <S> This has to mean running as slow as you can get away with. <S> but it might stop interference giving you a hard time. <S> I had to send SPI bus to a remote device at 10m - data <S> was being sent from the master only so both clock and data were properly aligned (cable length same distance) <S> - I was using 2MHz clock <S> but I decided that the only way it would work reliably was with differential drivers and receivers on both clock and data. <S> It worked of course and I might have got away with coax <S> but I'm sure regular unshielded cable and single-ended data and clock wouldn't work reliably. <A> Go diffrential on the I2Ccheck out LVDS. <S> In practice it is quite easy, all you need is an IC on each side and you are set. <A> I would recommend switching to a different bus and then placing small microcontrollers alongside each sensor to bridge to the I2C bus. <S> I2C is really not designed to work over that distance, especially since it is an open drain bus. <S> RS232 (standard serial port, make sure to use a level shifter to get the actual +/- <S> 12 volt levels) and RS485/RS422 (differential, half or full duplex signalling, again make sure to get the right driver chip) will be better able to cope with the interference over a long run. <S> Also, it would be possible to buffer readings at the sensor so you can just ask for a retransmit if something gets lost. <S> Wireless is another option, but that brings with it another can of worms - RF interference, multipath, fading, reflections, path losses, etc. <S> It can sometimes be dicey to get RF working properly when there is a lot of machinery and/or large metal objects where you are trying to communicate. <S> And you will still need a microcontroller with the sensor to act as a bridge.
| Given that you have interference issues I'd suggest running clock and data down coaxial cables - at least try it to see if it helps - it won't help the data be more aligned to the clock at the master
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Can voltage be controlled in a fixed DC circuit? If resistance is fixed in a DC circuit, can voltage still be controlled and changed from one value to another? Imagine this circuit, that has a fixed resistance of 1 Ohm, and the desired current is 20A, If the power supply is 100W, what would the voltage be? 5V?If so, what about ohms law? V = IR? And P = IV? I'm confused because I thought that voltage and current can be controlled to whatever value required, using a power supply. The resistance of the circuit is the most important thing isn't it? Another example: A 20,000W power-source, and the resistance of the circuit is fixed at 0.001 Ohms, can the power supply control the voltage to any value required? 2V, 10,000A? Or any other combination that does not exceed 20,000W? 20,000V at 1A? The same resistance of 0.001 Ohms? <Q> Usually power supplies are designed to try to output a constant voltage (to approximate an ideal voltage source). <S> If the resistance is also constant then you can predict the current from Ohm's law. <S> Usually if a wattage is specified, that's a maximum input or output wattage. <S> It's also possible to design a power supply that attempts to output a constant current or even a constant power, but that's not the norm. <S> If the load resistance is too low for the constant voltage supply, or too high for a constant current it won't be able to deliver the set voltage or current. <S> Constant power is a bit more tricky because if the load is too low resistance or too high resistance it won't be able to deliver the desired power (it will run into a current limitation at the low resistance end and a voltage limitation at the high resistance end). <A> You cannot put 20A through the resistor, since this would require \$20\text{A}\times1\Omega = <S> 20\text{V}\$ <S> and hence a \$20\text{A}\times20\text{V}=400\text{W}\$ power supply. <S> The most current you can put through the resistor is \$\sqrt{100\text{W} \over 1\Omega} = <S> 10\text{A}\$. <S> Any deviation above or below \${100\text{W} \over 10\text{A}} = 10\text{V}\$ will cause a lower current to flow due to the limit of the supply and the limit provided by Ohm's Law. <S> Note that only part of the voltage will be dropped by the resistor if it varies above 10V, and the power supply will have to drop the rest internally. <A> If resistance is fixed in a DC circuit, can voltage still be controlled and changed from one value to another? <S> Yes <S> but, since the resistance is fixed, the current is not independent of the voltage and, indeed, the current will be proportional to the voltage. <S> If the power supply can provide 100W of power, the maximum voltage it can produce across the resistor is $$V_{max} = <S> \sqrt{100W \cdot <S> 1\Omega} = 10V <S> $$ <S> At that voltage, the current will be $$I_{max} = \frac{10V}{1\Omega} = 10A $$ <S> You can put in different values for the power rating and circuit resistance to find the maximum voltage and current. <A> In your example, if the resistance is 1 ohm and the desired current is 20A, then ohms law can be applied in the form: <S> V = IR. <S> Thus, V = 1X20 = 20 volts. <S> If your power supply applies 20 volts to the 1 ohm resistor, the current through the resistor will be 20A. <S> The power delivered by the power supply and dissipated in the resistor will indeed be VI = <S> 20X20 <S> = 400 watts. <S> If your power supply is only able to supply 100 watts, then it won't be able to apply 20 volts across your 1 ohm resistor. <S> The best it might do (from 100 = <S> V^2/R) is 10 volts. <S> Then the current will be 10/1 = 10A and the delivered power will be 10X10 = 100 watts. <S> In general if you have a fixed resistance load, you cannot independently specify both the voltage and current.
| Ohms law says that V = IR, so if R is fixed, you can specify V or I and then calculate the other from the equation.
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Continually reading an RFID Tag I am attempting to build a system which stays active when an RFID tag is present and then deactivates when it's removed. RFID readers seem to have an output that fires once when the tag is in range and then doesn't run again until the tag is moved away then then rescanned. My initial idea was to use a standard reader and then reset it forcing a new tag to come through each time. The ID-20 reader has a reset pin which I believe supports this. I would prefer to use a cheaper reader for this project as I have quite a few to build so I am currently working with this device. The cheaper reader doesn't have a reset pin so I was going to toggle the power to the reader when I wanted a new read. Sadly this didn't work as planned, this cheaper reader only seems to read when a tag is moving through the field. Has anyone done anything like this before and can offer advice or would anyone know why the cheaper reader wouldn't work when the tag is static? <Q> This is a reader firmware problem. <S> Most access control readers default to emulating old-fashioned magstripe readers, so they only report once per "insertion" and mask the rest of their reads, as long as the tag is still visible. <S> Internally, the tag is constantly repeating its ID -- otherwise, the system would be much less reliable, since there would only be one chance to catch the message. <S> The "right" solution is to find a reader that's configurable for repeating reports, such as the Olimex MOD-RFID125 , but I don't know if any meet your price point. <S> It's strange that the cheaper reader doesn't report a tag on powerup. <S> It's possible that it has a sloppy RF powerup ramp, so the tag gets into a weird state because it didn't cleanly initialize. <S> If you have an oscilloscope and a loop of wire, you can try to put it near the reader and tag to try observing the signal. <S> You're working with a 125 kHz carrier, which should be slow enough for most scopes to capture. <A> The RFID tag you are using is most likely powered through electromagnetic induction. <S> Therefore, the tag will only be powered when it is exposed to a varying magnetic field. <S> In your case, the magnetic field is probably constant in time but varies in space. <S> The RFID tag will be powered when it is moved through the spatially varying magnetic field. <S> When you turn on your reader, my guess is that the magnetic field produced by the reader doesn't vary fast enough to generate power in the RFID tag. <S> That is, although you are transitioning from 0 magnetic forces (when the device is off) to some constant magnetic field (when the device is on), this transition does not occur fast enough. <S> If you want the RFID tag to produce a signal while sitting in 1 place, you will need to generate a time-varying magnetic field. <A> You have to design a new module or use programmable module. <S> The easier way is to have an IR sensor or similar behind the reader, so it can detect card's presence. <A> I have actually been able to make this kind of system work with an RFID-RC522. <S> I don't know how or why it works, but it pings the card every time the loop goes around. <S> Will attach code, hope this helps others. <S> CODE: <S> #include <SPI.h>#include <MFRC522.h>#include <Wire.h> //include <S> Wire.h library#include "RTClib.h" //include <S> Adafruit RTC library#define SS_PIN 10#define RST_PIN <S> 9#define RED 3#define <S> GREEN 2RTC_DS3231 rtc; //Make a RTC DS3231 objectMFRC522 rfid(SS_PIN, RST_PIN); // Instance of the classMFRC522::MIFARE_Key key;String cardnum = "";// <S> Init array that will store UIDbyte uid[4];void setup() { Serial.begin(9600); pinMode(RED, OUTPUT); pinMode(GREEN, OUTPUT); digitalWrite(RED, HIGH); digitalWrite(GREEN, LOW); SPI.begin(); // <S> Init SPI bus rfid. <S> PCD_Init <S> (); // Init MFRC522 for <S> (byte i = 0; i < 6; i++) { key.keyByte[i] = <S> 0xFF; }}void <S> printHex(byte *buffer <S> , byte bufferSize) { cardnum = " <S> "; for (byte i = 0; i < bufferSize; i++) {// <S> Serial.print(buffer[i] <S> < 0x10 ? <S> " 0" : " ");// Serial.print(buffer[i], HEX); cardnum += buffer[i <S> ] < 0x10 ? <S> "0" : " <S> "; cardnum += buffer[i]; }// Serial.print(cardnum);}void loop() { if (rfid. <S> PICC_IsNewCardPresent() && rfid. <S> PICC_ReadCardSerial()) <S> { for (byte i = 0; i < 4; i++) { uid[i] = rfid.uid.uidByte[i]; } printHex(rfid.uid.uidByte, rfid.uid.size); <S> if (cardnum == "144123239121"){ Serial.println(); Serial.print(""); digitalWrite(RED, LOW); digitalWrite(GREEN, HIGH); } else if (cardnum == <S> "64126193128") <S> { Serial.print(""); digitalWrite(RED, LOW); digitalWrite(GREEN, HIGH); } delay(20); } <S> else { digitalWrite(RED, HIGH); digitalWrite(GREEN, LOW) <S> ; } }
| The RFID reader you are using has fixed function(read tag when tag approaches).
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What type of motor is this? So I recently bought a motorised pedicure set thinking it would be some amazing thing. However, I used it once and the batteries (4 AA) died after about half an hour and not before getting ridiculously hot. I decided I would take it apart to see what mad electronics have gone into it, and it turns out it is literally just a button and then wired straight into the motor. So I have decided I would like to replace the motor with one that has a higher torque (was really not very powerful so didn't have much effect) as well as possibly a higher RPM. After taking it apart and finding the motor, I tried to find out what type of motor it currently is but it has no markings or part numbers on it at all so I couldn't find out. I wanted to know this so I could know the characteristics of it and things like that so I know how to get a better motor that would provide a better product! Here are some pictures of the motor in various stages of assembly and I was wondering if anyone could help out with regards to what type of motor it is, and if anyone has any comments/solutions to some of the issues I mentioned they would also be greatly welcomed! As it might not be clear from the pictures, inside the 'drum' of the motor are two magnets that the coiled wire sits in. I have taken the voltage and current readings of it and it is as follows: Voltage: Anywhere between 4V - 12V it seems to work, towards 12V it starts going a little bit mental. Current: Around 500 - 600mA. <Q> <A> It is a DC motor , like those used in radio-controlled toys or models. <S> This in particular is a permanent magnet motor. <S> You should try to determine the supply voltage. <S> Append : For a 6V DC motor, maybe this is a good reference. <A> You don't want a motor with higher torque. <S> The torque is low on purpose. <S> If it had higher torque, you could grind right through a finger nail or the tip of your finger - ouch. <S> Even a buffing wheel could rub hard enough to get painfully hot. <S> If the motor is really horrible (real cheap trash) then you might need a better motor - <S> but then I'd suggest a current limiter to keep the torque down.
| It's just an ordinary brushed DC permanent-magnet motor.
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What is the significance of 0V output from a regulated power supply? I am mainly interested to know the application of 0V output from a lab power supply. Which linear regulator IC one can use to get 0V output? Edit :The LT3080 datasheet says that it is adjustable to 0V. How is it useful? As far as I know, the well-known linear regulator LM317 can output as low as 1.25V but nothing less. However, this page 0 – 24 Volt, 2 Amp Bench Top Power Supply shows a linear power supply circuit that uses LM317 which can output 0V. Is it true? <Q> You can get 0V from an LM317 trivially if you have a negative supply. <S> Just make a stable -1.25V supply and return the 5K pot part of the divider to that rather than to ground. <A> You can use a power supply as a general purpose analog source, not just for supplying a VDD pin. <S> For example, say you want to test the DC response of an ADC. <S> I've also sometimes used a power supply as just a general purpose digital output (0 V or 3.3 V to an enable pin, perhaps). <A> Yeah the posted circuit will not go down to 0 volts. <S> It has a 1.25 Volt minimum as you observed. <S> (Don't believe everything you read on the web : <S> ^) <S> Also be careful using that circuit at low voltage and high current, the LM317 is may overheat and shut down.
| You could connect the power supply output to the Vin terminal of the ADC and sweep it starting from 0 V.
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Does the value of a resistor matter when doing voltage division to cut it into a half? I have a 7.4V (8.4V when fully charged) battery that is connected to an Arduino and I need to monitor the battery. To do so, I need to plug the battery into an analog pin. However, that pin only accepts voltages from 0-5V, and returns a value from 0-1023. Therefore I want to cut that 8.4V when fully charged into 4.2V. I understand I need to use voltage division where the resistors are both equal, does it matter which resistors? \$10\text{k}\Omega-10\text{k}\Omega\$ vs \$1\text{k}\Omega-1\text{k}\Omega\$? <Q> The Atmel data sheet says "The ADC is optimized for analog signals with an output impedance of approximately 10KΩ or less. <S> If such a source is used, the sampling time will be negligible". <S> To have an impedance of 10K\$\Omega\$ or less, the resistors in the divider should be 20K or less. <S> As others have pointed out, lowering the resistors consumes more power, so using 20K resistors makes sense to me. <S> simulate this circuit – <S> Schematic created using CircuitLab Edit: To explain the source impedance looking into the "middle" of the divider and the top: <S> If the top of the divider goes to a 'stiff' voltage (a battery in this case), the impedance looking into the center point is 20K||20K. <S> You can think of it as 20K||(20K+Rs) where Rs is the source resistance of the battery (or whatever the top of the divider is connected to). <S> Since Rs << 20K, it's very close to 20K||20K = <S> 10K. If you were to disconnect the battery, (Rs \$\rightarrow\infty\$) it would be 20K. <S> The impedance from the point of view of the battery (looking down into the divider) is about 20+20 = 40K, so the drain is only a couple hundred uA. <S> That is because the input impedance of the ADC is very high and is in parallel with 20K, so it's about equal to 20K, and it's in series with another 20K. <A> It doesn't fundamentally matter - you'll get half the input voltage regardless of resistor value. <S> However, it should be obvious that if you use extremely large values, the amount of current the voltage divider will be able to source <S> /sink will not suffice for the analog in pin, as it does have some, if very little, capacitance and leakage current. <S> So the goal is to find the maximum resistor value that will reliably interface to the arduino pin. <S> From my own experience <S> I guess 10k resistors will do fine without wasting too much power. <A> The rule of thumb to size the resistors is to make sure the bias current of the unloaded divider is about \$10\times\$ the load current to make sure the divider isn't loaded down too much <S> (but the resistors are still as large as possible). <S> This gives you two equations and two unknowns: <S> $$\frac{R_2}{R_1 + R_2}V_{\text{IN}} = <S> V_{\text{OUT}}$$ $$I(R_1 + R_2) = <S> V_{\text{IN}}$$ <S> where \$R_{2}\$ is the lower divider resistor and \$I\$ is the bias current of the unloaded divider (which you set to \$10\times\$ the load current using the rule of thumb). <S> An improvement to the divider would be to add an op amp buffer to the output of the voltage divider: simulate this circuit – <S> Schematic created using CircuitLab <S> The op amp non-inverting input at the output of the voltage divider has a very low bias current so you can use very large resistors in the divider. <S> If you choose an op amp with a very low supply current you can actually use even less power than you would need with the divider by itself. <S> The trade-off is, of course, the added complexity of the op amp. <A> Mathematically, it doesn't matter. <S> Two 1K resistors or two 10k Resistors will both divide the voltage in half. <S> Practically, you should use the higher values, maybe going to 33K or 47K. <S> The two 1K resistors will draw around 4 milliamperes. <S> Two 47K resistors bring that down to less than .1 milliamperes. <S> If run time is important, use the higher values, else use what you like or have handy. <S> I would put a 100nF capacitor (maybe 10nF for the larger resisitors) from the middle point to ground to filter out noise. <A> To determine the optimal value, you must know the input impedance of the A / D converter. <S> Suppose that it has a value of 10k. <S> Why? <S> Because the input impedance of the A / D is comparable with the resistance of the divider. <S> Then, following the example, if your A / D converter has 10k input impedance, the voltage divider in question, should be implemented with resistors 1k or even lower, so that when you connect in parallel the converter impedance of 10k, this value not appreciably affect the value of the divisor resistance. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> As schematics show, without the A/D connected $$V_O = \dfrac{V_{in}}{2}$$ <S> but if \$R_{AD}\$ is comparable with \$R\$ <S> $$V_O = <S> \dfrac{V_{in}\cdot (R \vert\vert R_{AD})}{R + R\vert\vert R_{AD}}$$ <S> A rule of thumb is that the divider resistor would be 10 times lower than the impedance of the converter . <A> Also, in addition to using resistors that won't load the battery too much, also consider the tolerance of the resistors in the voltage divider which influence the accuracy of the measured voltage. <S> Tighter tolerance resistors will permit a more accurate measurement.
| If you make the voltage divider with two resistors 10k, it will work fine ... until you connect the A / D converter. In short, the value of the divider resistors, should be as high as possible, but which is not affected by the value of the input impedance of the converter .
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How greatly can heat increase resistance? A conductor has a resistance of 0.001 Ohms, at room temperature, by increasing the heat x100 degrees Celsius how greatly would that effect the conductor's resistance? I'm trying to find the correlation of heat and wire resistance. <Q> It depends dramatically on the material of the wire. <S> A temperature coefficient for resistance can be calculated. <S> See here. <S> Also, this related thread . <A> This means that when the temperature increases, the resistivity increases as well. <S> Here you can find a table with various coefficients, and an online calculator. <S> The relationship can be described by $$\Delta\rho = <S> \alpha\cdot \Delta T + \rho_0$$ where \$\Delta\rho\$ is the resistivity variation, \$\alpha\$ is the thermal coefficient for the material, <S> \$\Delta T\$ is the temperature increase and \$\rho_0\$ is the original resistivity. <S> The resistance of a conductor is $$R = \rho\cdot\dfrac{L}{A}$$ where \$R\$ is the resistance, \$\rho\$ the resistivity, \$L\$ the length of the conductor and \$A\$ the cross-section area. <A> The temperature linearly increases the resistivity of the materials assuming it doesn't get to cold or becomes so hot the material begins to liquify (as shown in figure below) <S> Lets say the resistivity of a material is given by$$\rho_{Total} = \rho_{T} + \rho_1\\where ~ <S> \rho_T <S> ~ is~equal~to~resistivity~due~to~temperature~ and~\rho_1 <S> ~is <S> ~the <S> ~normal ~resistivity$$ since resistivity is inverse proportional to the mobility of a material$$\rho_{T}= \frac{1}{\sigma_T}=\frac{1}{en\mu_d}$$where <S> \$\sigma \$ is conductivity, n is free electrons per unit volume, and \$\mu \$ is the mobility of material <S> The key point is that $$ \mu_d <S> ~inversely~ <S> proportional ~to <S> ~ <S> temperature <S> ~$$which <S> then implies that the resisitivity is directly proptional to temperture. <S> $$\rho_T = <S> AT$$where <S> A is a constant. <S> Different materials of course will have different constants for A, but the underlying theory that temperature increases resistivity still holds Here is a picture showing what I mean <A> Elemental metals (e.g. Copper) generally increase in resistance at about +0.4% per Kelvin near room temperature. <S> Alloys are often less. <S> For hundreds of degrees C the temperature coefficient will not be a constant and you'll need to find tables or graphs to get an accurate answer for the particular material. <A> Resistivity tables showing how the resistance of various materials vary with temperature are readily available online.
| Typically, for a conductor, the coefficient that relates the resistivity of the material with temperature is positive. It depends on the material of the wire.
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LCR meter and iron core inductors I am trying to wind some coils for a RF transmitter, so to get the values right, I though of using a LCR meter, but there is a problem with the measurements. If I use air core inductors the measured values don't differ that much when switching from different frequencies of operation. But if I want to measure an inductor with a iron core (from a nail) I get the following results: 100Hz: 16uH 120Hz: 16uH 1kHz: 11.1uH 10kHz: 4.59uH 100kHz: 1.708uH I've read somewhere that inductors below 2mH should be measured with 1kHz, so the 100Hz and 120Hz are off, that I can understand. But 10kHz and 100kHz measurement frequencies shouldn't be that much off from 1kHz, shouldn't they? What am I doing wrong? <Q> A simple LCR meter will not give accurate measurements if the core is very lossy. <S> It <S> it's truly an iron core (not iron powder) <S> then the eddy current losses will be significant- like a resistor in parallel with the coil. <S> That translates to a lower impedance which the LCR meter will interpret as a lower inductance since it is not measuring the phase angle of the result. <S> Iron powder is also lossy. <S> An LCR bridge would sort this out for you (and show you the equivalent parallel resistance as well as the inductance), but the bottom line is that your core is going to be way too lossy to use at RF frequencies and you should use a ferrite core or air core that is suitable for RF. <A> When exposing a conductor (such as the iron nail) to a changing magnetic field, eddy currents develop. <S> The iron core acts like a short-circuit transformer winding. <S> This means that some energy is dissipated in the core. <S> This worsens as frequency increases, hence massive iron cores are rarely chosen for any transformer/inductor related applications. <S> Grid transformers working at 50/60 Hz are frequently built with laminated (isolated) sheets of iron, so as to prevent eddy currents from developing. <S> High-frequency inductor and transformer cores employ ferrite ceramics to minimize core losses. <S> You should consider using a ferrite core. <A> Schematic created using CircuitLab Feed noise into the circuit and look at the notch frequency with some audio interface software that can to an FFT. <S> Adjust the capacitor so that the notch is in a spot that can be accurately measured like 1kHz - 10kHz. <S> Then take the values of the resistor and capacitor and guess as to what the inductor value might be and plug them into a spice simulation. <S> Adjust the inductor value in the sim until the frequency of the notch in the sim exactly matches the frequency measured in your hard circuit. <S> Then you know the inductor value. <A> For iron core inductance measurements, drive the coil with a 60Hz voltage; read the voltage and current simulaneously and make the following calculations. <S> $$Z = \frac{e}{i}$$ <S> $$X_L = \sqrt{Z^2-R^2}$$ <S> $$L = <S> \frac{X_L}{2 <S> \pi <S> 60}$$ <S> R is the d.c. resistance of your coil.
| Another way to measure inductance that I think should be quite accurate would be to breadboard a trivial inductor capacitor notch filter circuit like: simulate this circuit –
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Will standing on a non-conductive surface with wet feet prevent grounding and therefore electrocution? It is a well known fact that one should not handle electrical equipment in general and electrical home appliances in particular when barefoot, or when one's feet are wet (even when wearing flip-flops/crocs). I was wondering, could one stand on a non-conductive surface to avoid the grounding / electrocution even when on'e feet are a little wet (say after the shower)? Perhaps a couple of newspapers? A rug? <Q> Power could flow through your body from one part of an electrical item to another. <S> ( I DO NOT RECOMMEND THIS ) <S> For example, if you stood on a plastic chair (a very bad conductor), in perfectly dry conditions, and put your fingers into a live light socket (i.e. with the power on) you will get electrocuted with the power flowing through your fingers. <S> If you touched the live power with fingers on either hand, you might die. <S> ( I DO NOT RECOMMEND THIS ) <S> It is extremely dangerous for power to flow through your heart. <S> That could happen by touching two different live parts of electrical equipment with two different parts f your body. <S> You might, for example, touch some part of your bathroom which is itself earthed, such as water pipes, a water-heated towel rail, a metal sink drain, etc, while touching a live part of a piece of electrical equipment. <S> Your homes electrical distribution board should detect power running to Earth, and switch off, but I DO NOT RECOMMEND <S> YOU TRY THIS. <S> Summary: You can be hurt or killed without having a connection through your feet to ground. <S> However, consumer electrical equipment is designed to be safe to handle. <S> They are either encased in plastic, or the case is connected to an electrical ground connection. <S> When the case is connected to ground, and the equipment is damaged , the flow of power via the case should trip the fuse or the circuit breaker. <S> This protection mechanism can trip even when you are not touching the equipment. <S> Hence I do not agree " <S> It is a well known fact that one should not handle electrical equipment in general and electrical home appliances in particular when barefoot". <S> I do not believe it is a "well known fact" because it is not adequate and is misleading . <S> It is an illusion of safety . <S> The only time this might be true is when equipment is disassembled or broken (Edit: which is an extreme meaning of "handle"). <S> The danger through feet is not significantly worse than through your hands/arms. <S> Insulated boots <S> might save your life, which would be great, but they might not. <S> Assuming safety because of them, or because feet are not bare, is IMHO a dangerous assumption. <A> As soon as they get at all wet, the water in them provides a path for electricity. <S> Thus you are not safe from electrocution. <S> You say "only a little wet". <S> Then you are in a gray area where you might wind up "only a little bit dead. <S> " <S> Is it worth risking your life? <S> No. <S> Don't do it. <S> You would need to stand on an insulating surface that was impermeable to water and rated to handle the voltage you're exposed to. <S> Even then you'd need to be concerned about puddles of water on the insulating floor surface making a path to ground <S> (Say there's a puddle that is touching a drain pipe, metal radiator, or an electrical cord with an exposed earth ground connection.) <A> Given that people use speciality items, like gloves, to handle live wires directly ... <S> I've personally seen 330 V (not sure of the exact voltage) being disconnected and reconnected live by hand. <S> It's possible , and is done every day by professionals. <S> The equipment being used was fully tested by a special testing machine for pinholes that the human eye can't see, and the equipment had it's own carrying case (so there would be no damage in transit). <S> And also, special boots which had under gone the same techniques. <S> I'm not sure I'd classify newspapers and rugs as good insulators even when dry. <S> And do you know there are no conduction paths? <S> even in your brand new Crocs? <S> Have you tested them? <S> All it takes is one micron sized hole in the proper material to make it a bad day.
| You can get electrocuted while fully insulated from the ground. The two examples you gave of non-conducting surfaces (newspapers and rugs) are both water absorbent. (Edit: Sitting on a metal framed chair might be as lethal).
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How hot can MOSFETs get before blowing up? This is different from a lot of other stuff here, but I have a motherboard that has no MOSFET cooling, and when I touch them they burn my hands; they are probably 70-80C. I just want to know how long can the MOSFETs last under these conditions? Mobo: Gigabyte 990FXA-D3 <Q> Try to get the part number of the MOSFETs, and search the web, to get their datasheet. <S> As Ignacio Vazquez-Abrams says cooler lasts longer. <S> However, I would be surprised if a motherboard from a reputable manufacturer failed due to power MOSFETs first. <A> It depends on the heat rating of the MOSFETs. <S> If they are getting that hot, then they can probably withstand that temperature also. <A> Probably they'll last longer than the viability of a PC motherboard! <S> If the internal (junction) temperature isn't much over 100'C <S> they're in relatively conservative territory. <S> The electrolytic capacitors are usually the first parts to die, and that's usually after the motherboard is obsolete (5 years at least).
| Power MOSFETs are often rated for operation upto 125C or 150C, i.e. they could boil water and still be within normal operating conditions.
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Finding current (I) in a basic circuit This has been bothering me: If I am told \$v_1 = -36\,\mathrm{V}\$ and \$v_2 = 18\,\mathrm{V}\$, how am I meant to find \$i\$? I figured since \$v_1\$ is giving off a negative current it would be flowing in the opposite direction of \$i\$, and wouldn't the equation to find \$i\$ then become \$18-(-36)=i\$? I thought about loop or nodal analysis but then I realized there aren't really any nodes, just loops. <Q> I figured since v1 is giving off a negative current <S> That doesn't make any sense. <S> \$v_1\$ is a node voltage and can't be described as 'giving off current' one way or the other. <S> I realized there aren't really any nodes, just loops. <S> There are three nodes. <S> Honestly, I think you're in over your head for even this simple circuit. <S> Hint: <S> the voltage across the current source is \$v_2 - v_1 = <S> 54V\$. <S> If you can find the equivalent resistance of the resistors, the current \$i\$ is found by Ohm's law. <S> Since another answer has been accepted, the solution is, by inspection: $$i = <S> \frac{v_2 - v_1}{R_{EQ}} <S> = \frac{54}{6||(8+4)}A$$ <S> Note: <S> you can exchange the position of the current source and 6 ohm resistor to more clearly see how to combine the resistors into the equivalent resistance. <S> I must comment that node voltage analysis isn't applicable here since the node voltages are given . <S> If you don't see the equivalent resistance method, then simply use Ohm's law and KCL: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Using the voltages and resistances given, the resistor currents are easily calculated as shown. <S> Apply KCL at node 2 to find the current \$i\$: $$ <S> i = <S> 9 + 4.5 <S> = 13.5A$$ <S> which is identical to $$i = <S> \frac{54}{6||(8+4)} = \frac{54}{4} = <S> 13.5A$$ <A> There're nodes. <S> Just use ohm's law/nodal analysis. <S> If you find \$ \frac{V_1}{8} = <S> I_8 \$ <S> you have one current flowing out of the node where V1 is. <S> Then \$\frac{V_1-V_2}{6} = <S> I_6 \$ is another current flowing away from node V1 (and into node V2). <S> Then you do \$\frac{V_2}{4} = <S> I_4 \$is <S> a current flowing out of node V2. <S> Now you have all the currents flowing into and out of both nodes except for i itself. <S> But \$i\$ flows into one node <S> and out <S> the other node <S> so you could write it like this: <S> At node V1: <S> $$ I_8+I_6+i = 0 <S> $$ <S> At node V2: $$ <S> I_6 <S> + i = I_4 $$ <S> Then sub in the voltage/resistance equations into the node equations. <S> Also note that you can simplify this if you realize that \$I_8 = <S> -I_4 <S> \$. <S> That's one less variable to have to track down. <S> Use algebraic substitution or any other method you prefer to simplify your equations to solve for i. <A> The crude way: 1)Let's call the 6 ohm resistor R1, the 8 ohm resistor R2, and the 4 ohm resistor R3. <S> 2)If v1 is -36V and v2 is 18V <S> then the total voltage across them is 54V 3) <S> With 54 volts across R1, the current through it must be I = <S> E/R = <S> 54V/6R <S> = <S> 8A 4)Since R2 and R3 are in series, their resistances add up to 12 ohms 5) <S> With 54 volts across them, the current through them must be 4.5A <S> 6)Since <S> the 54 volt source is providing 8A into R1 and 4.5A into R2 and R3, all at the same time, the total current, i, will be 12.5 amperes.
| The solution is just a straight forward application of Ohm's law and combining series and parallel resistances.
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Are control systems and motors typically driven from the same source in robotics projects? My control system is a simple microcontroller between 3.3V and 5V, but the stepper motors I require are 12V. I have ordered a 14.8V battery that I intend on regulating down to 12V to drive the motors, but should I also regulate down to 5V for the control system or use a separate power source? In short, are the control systems and motor drives typically powered by the same source or are they independent? I know that high series resistance batteries can show significant output voltage drop when high current is drawn, but my question is really regarding common practice in power design. Assuming the voltage drop from high current is low (not sufficient to affect the 5V regulated bus), would it be reasonable to use the same power source for both systems? Using something like this, for example: http://www.hobbyking.com/hobbyking/store/__24579__Turnigy_5V_6V_5A_Heli_UBEC_for_Lipoly_6_16V_.html <Q> You'll usually have one main power source to the project. <S> Having multiple sets of batteries (A batteries for the filaments, B batteries for the plates) went out of style decades ago. <S> You want LOTS of decoupling between the microcontroller supply lines and the stepper supply lines. <S> Ditto the microcontroller and the stepper driver board supply lines. <S> Stepping motor coils are good inductors (DUH!), and they throw BIG spikes back at the controller when they are switched. <S> You do NOT want those big spikes to get back into your microcontroller supply. <S> Bypass capacitors and filter capacitors, conservatively rated, are your FRIENDS. <S> Stepping motor driver design is a Black Art. <S> It is not unusual to see a 12V stepping motor being driven from a 50V supply, as this allows creative design to minimize settling time. <S> The old Airpax stepping motor catalog had some REALLY good design information on this. <S> In this day and age, decent quality cheap switching power converters are easy to come by. <S> For a battery-powered system, these days, I'd look for one of these, in preference to the old 7800-series linear regulators, for the microcontroller supply. <A> Both the batteries and controller uses same power source except they'll have separate voltage regulators. <A> I would say absolutely. <S> You may need to ensure your circuit can handle the fluctuations in current(smoothing cap) and protect the circuit from back EMF. <S> How you step down the voltage is where you will see different practices. <S> Especially considering that microcontrollers can "sometimes" run off a variety of input voltages and uses very low current. <S> For running "just" the microcontroller a linear regulator should work fine. <S> I think the poor efficiency will be nothing compared to that of the motors.
| In your case you can use 7805 IC which takes 12V input and delivers 5V output which can be connected to your controller.
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