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Supercapacitor with a rated voltage value greater than 2.7V feasible? I've been researching into the development of electrical power units for Automotives (motorsports in particular, as I participate in my Uni's Formula Student team) and have started by looking into the applications of supercapacitors for energy storage purposes. I'm seeing a pattern, were all supercapacitors currently on the market are rated at max, 2.7V (with some pushing close to 3.0V) without any real reason why. Compared to standard electrolytic capacitors (which I've viewed types that can withstand 1000V comfortably) I understand that the construction is different however I'm not exactly sure why this has such a drastic effect on the max voltage. Would there be anything on the quantum level that stops a supercapacitor having a rated voltage greater? My suggestion is that maybe the Helmholtz regions that hold the capacitance has an effect on the maximum voltage that can be but I'm not entirely sure (just an educated guess) <Q> Super Capacitors are a sort of cross between electrolytic capacitors and batteries. <S> see wikipedia article <S> The lifetime of the system is also somewhere between batteries and electrolytics. <A> If you google it, they do make 10v 2200F supercaps... beasts! <S> Afrotechmods on YouTube got ahold of a few of these <S> and they sure do let the sparks fly. <S> I only put this as an answer because I do not yet have 50 rep <S> so I cannot comment. <A> Similar to batteries, you can have whatever voltage you want by having someone put "cells" in series with a balancing system, most commonly you can get "18 volt" ones to sort of simulate a 12 volt battery. <S> Each cell is limited for in voltage for reasons similar to battery cells, as involves chemical reactions helping to simulate a conventional capacitor. <A> Super capacitors are mainly used for Real time application monitoring solutions. <S> means wherever you need RTD, you can use super capacitors to get long lasting energy. <S> Mostly RTD application uses microcontroller and microcontrollers are fairly rated for 3V applications. <S> I feel that's why most of the available mass super capacitors are rated for 3V applications.
| Unlike in electrolytics, where you can build the insulating film up to an arbitrary thickness by applying a high voltage, in supercaps the insulating film has more to do with the polarisation layer of a battery, so has a hard upper voltage limit.
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LED dimmer for high current applications Hi Guys .. I am designing a few panels using LEDs for backlighting. Some of these panels have over 20 leds, using a 5v power source. So only one of these panels can draw around 500mA alone. I am planning on using the same PSU to power them all, but need a dimmer circuit to dim them all at once, but most of the dimmer circuits I have seen can only really do lower current draw, i.e. 2N2222 can only handle 800mA. Any idea if this is possible with one high current dimmer, or do I have to build multiple dimmers ? <Q> Pulse width modulation (PWM) is the most efficient way to do this. <S> The basic principle is that you cycle the LEDs on and off at a speed fast enough that the eye can't detect a flicker - say 50 times a second. <S> By varying the % on-time from 0 to 100% the LEDs can go from 0 to 100% brightness. <S> The advantages are: <S> The power circuit is very simple. <S> No power is used while their off. <S> We need to know what is going to control the dimmer if we are to provide further information. <A> You can use a circuit like this one (from here ): <S> (operated from +5V only) <S> Use a CMOS version of the 555 (eg. <S> TLC555) rather than the NE555, and reduce R3 to 100 ohms. <S> Q1 must be a logic-level MOSFET such as an AOD476 . <S> The circuit as described will easily handle several amperes. <A> I am just including this to expand on the concept, especially if there is any harness involve. <S> The use of an inductor ensure a constant current to the LED's as oppose to a pulsed current. <S> By including a sense resistor you could include a simple control loop to maintain current as the load characteristics change (increased V due to temp) <S> Two sets of "freewheel" diodes are shown. <S> The 1st is the diode that is used to commutate the current from the FET (to maintain current flow through the inductor). <S> The second as part of a Zener+diode stack is for safety if the load goes open-circuit. <S> NOTE: <S> the zener/TVS needs to be sized to dissipate the energy within the inductor. <S> It would only conduct during a failure situation and thus it does not need to be sized for continuous operation. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> As far as using the same psi for everything, i would multiplex the banks of LEDs. <S> This way you only power one at a time, but they would all "appear" on but you would only power one at a time
| As others have mentioned, PWM is the most efficient method to drive the LED's.
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How many SMD 5630 LEDs can I power with 12V 1A power supply? I am new to this and have what is probably a newbie question. I am using SMD 5630 LEDs from a strip that I am cutting off and wiring each LED individually. They will be wired in parallel and I am trying to understand how many I can power with a 12V 1A power supply. (Edit: The power supply converts to DC current.) From the specifications I found each LED requires 3.6V at 50 mA and using an Ohm's calculator I found I need 22 ohm resistors. I just don't understand how to figure out the maximum number that I can support. Could someone explain, speaking slowly and using small words? :) <Q> IF the picture on the Amazon page is accurate to what you have, each segment has 2 27Ω resistors in series with 3 leds. <S> Using Ohm's law <S> I = <S> V/R to find out the current, Assuming 3V to 3.6V Forward Voltage drop on each led, that means the current at 12V is between (12V <S> Source Voltage - (3 Leds * 3.0V)) <S> / 54 <S> Ohms = <S> 0.055 <S> Amps (55 mA) <S> and (12V Source Voltage - (3 Leds * 3.6V)) / 54 <S> Ohms = <S> 0.022 <S> Amps (22 mA) <S> Since current is shared in series, this means each chip on that strip is only 22 to 55 mA. <S> So they are no where near your believed 150mA or 1/2 Watt. <S> To get the correct current, as far as the leds you have go, cut a single segment off, and using a ammeter or multimeter in current mode, measure the amperage of that single segment. <S> That will tell you what it is using on the strip. <S> Then I would adjust at least 10% down for a longer life. <S> If you can't measure current, then measure the voltage across one of the resistors. <S> Using Ohm's Law as above, take that voltage and divide by that resistance for the current flowing through it. <S> That's what the LED should be using. <S> Update: It was suggested that the resistors are in parallel, and that the normal 5630 leds take 150 mA @ <S> 3.4V. <S> That may be true. <S> (12V <S> Source Voltage - (3 Leds * 3.4V)) <S> / 13.5 <S> Ohms = <S> 0.150 <S> Amps (150 mA) Only way to be sure would be to trace the circuit and measure the current or voltages with a multimeter. <A> Adding to Passerby's answer: Replicate what's on the strip. <S> It wastes the least current. <S> The strip is made up of repeating circuits as shown below. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You should do the same. <S> Polarity of LEDs matters. <S> The resistors can go anywhere in the series circuit - they're probably evenly spaced on the strip to spread the heat. <A> if you divide that by 18 it comes out to 13.666666 ma for each led. <A> i run them each on 6.7 volts (6.2v nominal) 18mA. if you push them to 22mA you'll need some googles. <S> the whole led strips are less than 120mA each. <S> 80mA may be too dim but stable for a single regulation for each line. <S> with proper line regulation you can reach 180mA fine at 26 or 27 volts... <S> Thus. <S> if it ain't a special 12V package it may be two each 12v line but not precisely shiny because of lower voltage exitation...
| I have a strip of 18 5630 smd leds and by test meter it draws 246 ma when powered by 12 v. dc power supply.
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Is the laptop's power adapter output negative terminal grounded? Today I find my lenovo laptop power adapter's output negative terminal is not grounded (the input ground terminal is not connected with output negative terminal, measured with DMM). But another old one is just opposite, the output grounded. Why? There is any standard for power adapter? If the laptop is not grounded, is it safe? <Q> If the laptop is not grounded, is it safe? <S> Some laptop adapters don't have an earth pin and therefore they can never have a grounded output. <S> It's not a problem providing that the insulation and clearance/creepage distances inside the adapter are adequate. <A> My experience is that the power supplies for the older laptops in which the backlight was CCFL were usually grounded on the DC side as well. <S> I suspect it was because of the high-voltage inverter needed for the CCFL (approx. <S> 600V DC output), <S> but I'm not entirely certain <S> that was the determining factor. <S> There's a concrete example of an older HP I have that does have minus grounded; details on that in this superuser post . <S> The newer laptops with LED backlight tend not to have the minus pin grounded. <S> There can be static buildup and leakage currents on the non-grounded laptops [the latter particularly on metal cases] which can be irritating to some people. <S> See this discussion on MacBook and this video of an actual measurement; he only measured AC voltage from case to ground as 78V AC with no grounding, which is probably enough to feel a slight tingle on a leakage current. <S> A better test would have been to measure the leakage current as well... <S> but I guess you need to try harder finding EE pros among Mac fans. <A> The first thing to understand here is that the AC ground coming from the wall, does not serve the same purpose as the DC ground going to the laptop. <S> They both typically sit at 0V and are often physically connected, but they serve different purposes. <S> The DC ground is used to complete the circuit so power can flow, the AC ground is used for saftey, the power flows between live and natural and, under most circumstances, won't touch ground at all. <S> The AC ground is used when there is a fault. <S> For example, say the live wire came loose and touched the casing of a metal microwave. <S> If anyone touched the microwave, the AC power would flow through them and most likely kill them. <S> To prevent this the casing is connected to the ground wire, so as soon as the live wire touches the casing, a massive current flows through the ground, which blows the fuse / trips the breaker, isolating the device. <S> This ground connection is not necessarily needed if the device is suitably insulated. <S> Phone chargers often don't have a ground pin (or for me in the UK, they have a plastic one) because if a wire did come loose, there is very minimal chance of shock as the whole device is insulated. <S> In your example of the laptop charger, the AC side of the charger often has a ground pin. <S> I'm not sure if this is required or if some manufacturers are just being extra careful. <S> However the DC side of the charger does not require one, as it is insulated very well from the AC connection, if the live wire came loose inside the charger, there is no way it could make your whole laptop live, so there is no need to ground your laptop. <S> Some times the manufacturers simply ground the whole ground plane within the charger, which the negative terminal of the charger is connected too. <S> This is why sometimes you see a connection, and sometimes you don't, it's not needed at all, it just happens sometimes due to every 0V being connected together.
| If a laptop uses a grounded adapter and the adapter is unplugged from the laptop then, the laptop isn't grounded - it's clearly safe in this scenario and, just as safe as when plugged into a properly designed and manufactured adapter that doesn't need an earth pin in the AC plug.
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Powering multiple white LEDs in parallel Recently the idea struck my mind, that I wanted to build a simple but useful emergency light which can run off of any USB port (the specific use case would be to power it from an USB power bank). For that I wanted to design a simple circuit that drives 25 white LEDs (3.2 V, typ.) in parallel. The question I am facing now is how to achieve a simple design while not using 25 series resistors, as this is just wasteful and not space efficient, for me anyway, I just want to make it more clever somehow. Is it enough to just step the 5 volts (from the USB) down to 3.2 volts using an off-the-shelf linear regulator like the LM317 and then connect all the LEDs to it in parallel or will the non-ideal characteristics of aforementioned spoil my plan? I might be overthinking this, which is why I am asking if there might be a better, simpler or more efficient possibility to achieve what I want? :) <Q> I would never put the LEDs in parallel because there are chips that can do this job properly such as this one from Maxim: - <A> 1) USB can only supply about 500 mA. <S> You can get a single white LED that is rated for that amount of current. <S> 2) <S> You basically have to 'ballast' each LED to swamp the differences -- a R does this. <S> 3) <S> A regulator would perform the same function as a resistor in your plan -- both drop the V from 5 to 3.2 V. 4) In an emergency, how is the USB powered ? <A> You can use a backlight driver IC which steps the 5V up to 20V+ and power several LED strings (perhaps with I2C brightness control from a micro, such as with the Skyworks AAT1236), however I think your best bet is to simply use series resistors and keep it simple. <S> Especially in keeping with the function of the unit.. <S> parallel strings and resistors are more likely to be functional when you need them. <S> The advantage of the switching regulator would be higher efficiency- <S> with resistors you've got 64% efficiency- <S> a switching regulator might be more like 85%, so the battery might last 1/3 longer. <S> A linear regulator would simply gather the heat from the resistors into one place, you should use some resistors anyway, and it would degrade the current regulation, so I think you should forget about that. <S> but I would not stoop to suggest it openly. <A> "you are hereby also saying, that there is absolutely no way to power 25 LEDs more efficiently than through resistors" <S> Well, no, not necessarily. <S> It would be possible to control each LED separately with a switch-mode current regulator. <S> Of course, this means that you would need 25 regulators. <S> While this would be more power-efficient (for some regulators), it would be much more expensive than 25 resistors. <S> Likewise, it would be possible to make a boost-type current regulator which would provide a much higher voltage to a series string of LEDs. <S> In any case, I'd recommend a nominal LED current of about 10 mA per LED. <S> LED power dissipation is then $$P = <S> 25\times <S> 10\text{ <S> mA}\times 3.2\text{volts} = 0.8\text <S> { watts}$$which is well under the nominal 2.5 watts available from a USB port. <S> And under no circumstances should you even think about connecting 25 LEDs in parallel. <S> LEDs have a negative temperature coefficient for the voltage drop. <S> This means that if you keep all the LEDs at one voltage, whichever one dissipates the most current will get hotter than the others, which will cause it to get hotter, which produces a vicious circle. <S> This will cause the next-weakest to do the same thing, and then the next, etc. <S> This is called "firecracker mode failure", for obvious reasons.
| If you were making a toy light in China, putting the LEDs directly in parallel and using the most horrible smallest-chip white LEDs you could find with a single (or a few with a few parallel banks) series resistor(s) might be an option If you put multiple LEDs directly in parallel, small variations in the LEDs themselves will cause them to run different currents in each LED. Since the voltage source must, by definition, be able to provide 25 LEDs worth of current, the most likely result is that the weakest LED will enter thermal runaway and self-destruct, failing open.
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Inductive reactance: frequency vs current. Why current goes to zero as frequency increases? Suppose I have the circuit below with the frequency of the voltage source as a free parameter. Assuming the inductance is constant, the current flowing through the circuit shoud equal $$I(f) = \frac{V}{j2\pi fL + R}$$ (I'm using phasors). Then by increasing the frequency up to infinity the current goes to zero. Why is this happening? I mean it is probably clear to me from a mathematical point of view and by using the frequency response I guess the absolute value of the trasfer function is going down as f tends to infinity, but what is the physical explanation behind this effect of a high frequency? simulate this circuit – Schematic created using CircuitLab <Q> One way to look at this is that the inductor is a frequency-variable resistor (not quite since there's also a phase shift in there, but that's not important for this point). <S> The higher the frequency, the higher the resistance. <S> At the same voltage, the inductor draws less current at higher frequencies. <S> A totally different way to think of this is to visualize what the inductor is doing, going back to how inductors work. <S> If you apply a fixed voltage across a inductor, the current will rise linearly. <S> Put another way, it takes time for the current to build up. <S> Now imagine that you keep switching polarities of the voltage. <S> The current builds up linearly in one direction. <S> When the voltage flips, the current then linearly decreases to 0, then builds up to ever higher magnitude in the other direction as long as the voltage persists. <S> Put yet another way, the maximum current is a function of how long you keep the voltage before flipping. <S> Now consider AC is like the flipping voltage. <S> Higher frequencies means faster flipping, which means the current has less time to change between flips. <A> To also explain it on the actual physical level: any current in a coil causes a magnetic field. <S> In case of DC current, the magnetic field doesn't have any further effects, but an AC current will also cause the field to change all the time, and that causes an induced voltage in those same coil windingings, proportional to the time-derivative of the magnetic field, i.e. proportional to \$I\cdot <S> f\$. <S> But now, in your application the voltage amplitude is actually limited by the supply, so <S> the only way the law can be fulfilled at very high frequencies is if the current stays very small. <S> Any substantial \$I\$ would cause \$I\cdot f\$ and thus the voltage to blow up towards infinity, in the high-frequency limit. <A> Ideally, an inductor is a coil wire whose resistance is zero. <S> Therefore, at zero frequency (DC current) <S> the inductor no opposition to current. <S> The voltage on the inductor, assuming that the inductance thereof is constant, is given by \$v_L = <S> L\,\dfrac{di}{dt}\$ <S> that is, the voltage across the inductor depends on the rate of change of current through it . <S> The higher the frequency, the greater the rate of change and the greater the voltage across the inductor. <S> In the limiting case of infinite frequency, the rate of change is infinite and therefore, the inductor behaves like an open circuit. <A> If you understand that capacitors are pretty much the inverse of inductors , and can understand that for a capacitor the current goes to zero as the frequency goes to zero, a lightbulb may come on. <S> Capacitors seem to be more intuitive for most people.
| As the frequency gets higher, the inductor has less time to build up current between flips, and the average magnitude of the current will be less.
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Does (and when how) the lifetime of a LED depend on the PWM frequency Let's assume two options for driving a standard LED at its rated current. PWM set to 50% duty cycle at 10 kHz PWM set to 50% duty cycle at 50 kHz Technically both LEDs would produce the same amount of light and the 'blinking' won't be visible to the human eye or a camera (except maybe for a high speed camera...) <Q> Let me open my trusty MIL-HDBK-217F and see what it says about LEDs and their longevity: - The main factor affecting the failure rate per million hours is temperature. <S> Of interest, if I read the next section about laser diodes they do take into account duty cycle pulsing but their conclusion (on page 6-21) is that at 50:50 duty cycle the failure rate for laser diodes is about 25% of that when continually driven. <S> They also conclude (on page 6-22) that if you operate a laser diode at a light output power of 50% of its rating it will last ten times longer than operating it at 95% of it's rated output power. <A> LEDs are just diodes which don't "wear out" with frequency. <S> Also, your frequencies are low. <S> 50 kHz and 50% duty cycle means 10 µs on and 10 µs off. <S> That's a "long" time for a LED. <A> Personal experience: <S> I've driven a standard UV LED rated for 3.4V, 20mA with about 1A for 5ns at a rate of 87kHz (duty cycle: 1:2300) but didn't observe any "wear" in terms of brightness or pulse shape within 10^11 pulses. <A> No discernible impact. <S> The LED itself would only be sensitive to the total lifetime, but reliability is measured in 10's of years. <S> Thermal failures due to packaging or wire bond failures are more likely, but the probability of failure is still very low. <S> Most likely failure for a self-made system is the solder joints, or wires between the Led and PCB, or PCB and power source. <S> Thermal failures are caused by different rates of thermal expansion, and the resulting overstress this causes on the structure. <S> Small stresses or stress cycles have negligible effect. <S> Consider that the LED's plastic was probably molded and cured at +175 C -- it is always under stress. <S> The LED's thermal time constant is probably in the 10-100's of ms range. <S> Cycling faster than that leads to very small temperature excursions which don't cause problems, and cycling slower than that limits the total number of cycles to a very small number.
| The maximum current and average current do affect how the LED wears out, but frequency doesn't have any affect that I've ever heard about.
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Easiest project ever, but I still need straight answers. AA 1.5v x2 to RGB LED (trying to make white) Ok so I pulled out some old electronics stuff I have laying around. I have two AA batteries in a holder, in series. I have common cathode RGB LEDs where you can find all their specs here . So I plugged in the cathode in and started touching the other pins with a wire. Red works, blue works, and green works, rubbing blue and green is making teal, cool so how about white? all three connected just gives me red. After trying white green stopped working.. and eventually tapping blue does nothing or is very dim. I'm thinking maybe I did something wrong maybe too rough with the pins or something I pull out another and all are working again and eventually red's all thats lighting up. I start to think maybe I have the positive and negative thing backwards and switching them does the same thing. Is it normal that the red will work both ways? rubbing or tapping green or blue on and off will turn the color on for a millisecond. Seems to be a brighter flash of color when the electricity is jumping from the wire to the leg, but holding it on it wont do it. So holding it just far enough that the shock is constantly jumping seems to make blue/green flicker. Maybe too much information but I just think it's weird that it's doing that and doesnt seem to be burnt out or anything and that it works fine until I try to get white out of it. I was hoping that if I burnt something out it would be more obvious or instant instead of really having no sign of damage. So I'm thinking I'll need a resistor, I have a few in this box here. Batteries put 3.0v since in series, right? They're new from the store, Duracell MN1500 batteries. I'm now looking at the chart and it looks like I might need 3.2-3.5 volts to power green/blue? but it was working for a moment. If I'm trying to power white, do I need to add up the volts for each one or can I get all three up with the same voltage? So my simple question is what circuit do I need to create here to get this thing to work for one night? I'm starting to think I need another battery in series with a resistor. If I jump up to 9v battery and resistor, will I achieve longer battery life? Noob <Q> You need 3 batteries minimum, and 3 resistors. <S> With 3 batteries, try simulate this circuit – <S> Schematic created using CircuitLab <S> You have 2 problems. <S> The really fundamental one is tieing the LEDs in parallel. <S> They operate at different voltages. <S> The red has the lowest voltage, and it's hogging the current from the others. <S> You need a separate resistor for each LED. <S> Red has the lowest voltage, green next, and blue the highest. <S> The current (and brightness) is determined by the resistor, so for the same supply voltage you really ought to have different resistor values. <S> This is not entirely true, since different LEDs can have different efficiencies, but it's a good starting point. <S> The other problem you have is using a 3 volt battery. <S> This is just about the blue LED voltage. <S> When you disconnect the battery, it recovers and will briefly work again. <A> Having a look through your datasheet, you've got 3.5v drop, a 1.95v drop and a 3.2v drop, with a desired current of 20mA. <S> You'll need at least 3 AA's to properly work the RGB LED, so assuming that, you've got a supply voltage of 4.5v. <S> 3*1.5 for AA. <S> To get the rated current for the device you need to add a resistor. <S> Forgive me, I don't know how to insert proper equation symbols yet, but the math goes like this. <S> (V_supply - V_led)/R = <S> I <S> I = <S> 20mA. <S> So you can rearrange and get that you need a 128 ohm resistor for the 1.95v LED, a 65ohm for the 3.2v and a 50ohm for the 3.5 You can get all three diodes up off the same voltage level, however with all three on the battery will be supplying 60mA. simulate this circuit – <S> Schematic created using CircuitLab <A> Human vision needs 3 different "stimulus" values to see white: Blue 420–440 nm Green: 530–540 nm <S> Red: <S> 560–580 <S> You'll have to read your datasheet, install the correct resistors to produce the correct Vf (forward voltage) & current on each diode, in order to produce the right combination of light from all 3 diodes to see "white light". <A> These answers so far are unfortunately off the mark. <S> The reason you only see red is because the forward drop of the red LED is less than it is for blue or green, so ALL of the current will go through that one if it has the chance. <S> You have to keep it from getting that chance by using separate resistors on the three R G and B anodes. <S> The typical Vf values for each from the datasheet are <S> \$V_{f(red)}=1.95V\$ <S> \$V_{f(green)}=3.2V\$ <S> \$V_{f(blue)}=3.5V\$ <S> To get the basic resistor value for each leg, use the formula$$R = <S> (V_{bat} - V_f) <S> \div 20mA$$ <S> Now experiment with increasing various resistor values to get white output. <S> You will probably start out with a pale green or blue. <S> Note that \$V_{bat}\$ will change as the charge goes down, which will change the quality of the white balance somewhat. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> And as @whatroughbeast points out, the Vf exceeds the battery that you've specified, so that's another problem.
| You need to provide three separate resistors on the anodes to control the current to each LED to get your white balance. When you first connect the blue LED, the battery has just enough voltage to light the LED, but this quickly gets exhausted.
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Why the stable supply voltage present a ripple after connecting it to the testing board? I used a power supply to supply 1.8V as the Vdd of the tested-circuit,after I connected 1.8V to the Vdd pad of the testing-board,the originally "clean" 1.8V began to have a ripple (the total current is about 70mA). The phenomenon can be improved by adding 1uF cap betweenVdd and gnd Pad,but the ripple still exsited. But why the originally clean Vdd present a ripple after connecting it to the testing-board? By the way,the board only contain an OP amp that I want to test,and some I/O port(ex:header,SMA,switch......),and all inputs are shorted to ground. U1 and U2 is regulator,the others is I/O ports and the op amp being tested. <Q> It can depend from a lot of factors, to give an example, there might be something in the testing board that ruins the ground i.e. clocks, DC/DC converters etc. <S> How do you measure the 1.8 V? <S> How do you "see" that it is shaking? <S> How much is the "shaking" compared to the clean signal? <S> If you want more information, you have to provide a schematic or at least some details regarding the testing board. <A> Sounds like your regulator is going unstable -- depending on the pass element type and internal topology (I.E. <S> The addition of an internal zero), you may need some minimum ESR on the output of the regulator to ensure it remains stable. <S> Your schematic is a bit confusing, but I believe I don't see any output filter on that regulator. <S> There must be some amount of C on the rail. <S> Try a 4.7uF tantalum on the output of it, or if you have none, use a ceramic with a series resistor. <S> This will provide some capacitance, and the tantalum capacitor should have an ESR in the target range for you, maybe .3-1 ohm or so. <A> looks to me like you are missing MANY capacitors, on the output of the regulators. <S> Use at least one 0.1uf for each IC on the board (placed as close to the IC it is intended to decouple as you can reasonably get), and bulk capacitance (a few 10uf and/or 100uf in parallel) usually at the input header for the 1.8V test voltage supply you refer to. <S> The op-amp in question may also be oscillating <S> if the input and output are not within spec, oscillations from the op-amp without a decoupling capacitor nearby could perhaps affect the power supply if that supply is also not stable nor filtered correctly.
| Capacitors smooth out voltage ripple ("shaking"), and are required for the output of most standard regulator ICs (always check the datasheets and follow the requirements) in order to help them stabilize their regulated output, otherwise the internal control circuitry can oscillate ("shake") which affects not only the output, but also the input because it becomes an unstable load.
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Is it possible to pick up noise in an audio circuit through the ground / earth wire? In an audio circuit, such as a guitar's electronics (the only thing I ever dabble with!), I have always thought (probably simplistically) of every point along a continuous earth path being essentially all at the same voltage (a 'zero volt' reference voltage). However, I was wondering - if you extend an earth wire from some point along that path, is it possible for that in itself to introduce noise to the circuit by changing the '0V' of the earth reference? (Possibly I am wondering "What is the difference between a long earth wire and an antenna" - but I don't know enough about antennas to know if that's what I'm asking :) <Q> For single-ended signals, this ground offset voltage usually appears as part of the signal. <S> There are two common strategies for avoiding this situation: <S> Use differential signals. <S> The actual signal is encoded as the difference between two signals that are driven oppositely from each other. <S> Ground offset then looks like a common mode signal, which can be largely ignored by the receiver. <S> Pay attention to how things are grounded. <S> Make sure all the grounds are tied back to one place without other connections to elsewhere, like the local power outlet ground. <S> This ground net is then connected to real ground in one place only. <S> Also avoid having deliberate current flow thru any ground connection. <S> There should be separate returns for power current. <S> Never use ground as a power return. <A> 0V is only 0V at the point of reference - it gradually degrades as you move away from that physical position. <S> It may be nano volts difference up close but can rapidly become milli volts and, <S> if sensitive input circuits have 0V connections that are not at the same point, the several milli volt difference can be an AC noise-signal and is quite often very annoying. <S> This is why PCBs use ground planes <S> but, these are by no-means exempt from this problem. <S> Other systems use star-point wiring to keep any 0V connections at the same physical point but these can suffer from magnetic pick up of AC. <S> It's a difficult nut to crack sometimes. <S> Here's a nice picture of how a digital signal can create both ground bounce and power bounce: - The fast rising edge on an output will inevitably charge up a PCB track's parasitic capacitance - <S> this is seen as a small "bounce" on the power rail and associated distortion (side effect) on the outputted signal. <S> When the edge falls there is a ground bounce. <S> Now the signal is slightly corrupted by importantly, for this question, the ground plane and power plane have pulses of current injected into them and these can affect other circuits closeby. <S> Hears another example of where gaps in a ground plane can cause "bounce" because the return current of a signal has to "spread" out around the gap <S> : - Here's another idea of how mis-positioning functional parts can wreak havoc on sensitive analogue circuits: - The clever thing to do here is avoid common 0V connections for things that can cause ground bounce to each other <S> - this is a form of star-pointing i.e. separate power and ground connections from each individual circuit function only connect at a single "clean" pair of nodes (usually at the output of a voltage regulator or battery). <S> This appears to be quite a useful document that explains the phenomena <A> The answer is YES if your application depends upon the the difference of voltage between the two ends of the wire to be close to zero. <S> In other words if a functioning system is sensitive to signal levels at both ends of the wire at once then there can be reason to be aware that noise pickup can happen. <S> On the other hand if you connect two systems together using a cable such as a twisted pair (or a coax cable for high frequency) and make sure that the current traveling in one wire is equal and opposite to the current in the other wire and the target system is only sensitive to the signalling between the target ends of the two wires then any common mode pickup in the pair of wires will go unnoticed by the target equipment.
| Yes, long ground connections can pick up noise, which means the ground in one place is a different voltage from the ground at another place.
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Interfacing multiple rotary encoders to a microcontroller What, in your opinion, would be the most efficient way to interface muptiple (let's say, >20) incremental rotary encoders to GPIO-limited microcontroller (for example, ATxmega128A4U , 44 pin)? Requirements: It should be possible to operate multiple encoders at the same time. Encoders have integrated push switches. What I have tried: To search for multi-channel quadrature decoder IC. No luck. To search quadrature decoder to serial IC (Encoder-to-SPI, I2C, etc). No luck. To build a multiplexer-alike logic that would allow to connect as much encoders as possible while saving GPIO's. No luck. <Q> I am assuming each rotary encoder will have two signals, and may have one more for a switch. <S> Designate one microcontroller as the 'coordinator'. <S> Have the other MCUs ('handlers') communicate with it. <S> If the 'coordinator' actually runs as a communication slave for SPI, I2C or some other convenient interface, the 'handlers' can drive the communication interface interrupt. <S> That way the 'coordinator' will only get triggered when something happens. <S> (Otherwise, the 'coordinator' will have to repeatedly read the other MCU, called polling, which may be less convenient for your system) <S> You'll need to devise a simple protocol to communicate which encoder has changed, and by how much. <S> That will very likely be more than one byte, which may add a bit of complexity. <S> Depending on your experience, you might look at microcontrollers which containe quadrature decoder hardware. <S> As an example, ST Micros STM32F parts have timers which can also decode and track quadrature signals. <A> Ignoring the pushbutton switches associated with the encoders, a 50-step/second step rate suggests that you can use external multiplexers to read the encoder outputs. <S> You can find 16-to-1 multiplexers (74HC150, 74HC4067), and connect these to the two phases. <S> Drive the address lines of the multiplexers with 4 IO lines, and use 2 lines to read the phases. <S> You can add a 3rd multiplexer to look at the switch contacts. <S> You can expand the multiplexers as well, using 1 or 2 more lines as select lines. <S> This simulate this circuit – <S> Schematic created using CircuitLab shows a 32-line multiplexer, using 5 select address lines and an external inverter for the most significant address. <S> You can use two IO lines for the MSA, but you must be sure that only one is low at any time. <S> In practice, you would (at intervals less than 20 msec) read all of the encoder phases, then analyze each pair of bits to determine if the encoder has moved, and take whatever actions are required. <S> For an Atmel, 20 msec is a long time, but it would be important that the software not get involved in some other task and miss a read. <S> If the encoder has moved by 2 steps, there is no way to determine direction. <S> A complete setup would require 6 multiplexers - 2 per phase and 2 to read the switches, plus 3 inverters (1/2 of a hex inverter chip such as a 7HC04). <S> Total MCU lines would be 8. 5 address lines and 3 inputs. <S> Total IO operations is 64. <S> 32 address select writes, and 32 data reads. <S> Of course, if you're only using, let's say, 20 encoders, you would only need 40 operations. <A> US digital make the LFLS7184-S <S> which interfaces the quadrature A/B outputs to give up/down countes suitable for feeding a logic chip. <S> It's only a single channel but if you are into logic circuits rather than MCUs it might appeal to you. <S> You can even get a serial output version and a dual version. <S> It might be worth taking a look at their website. <S> Here's another dual device from Avago and ELM404 <S> is another device worth considering and, from memory I think TI do one. <S> Here's the link I used for googling. <A> You can simply poll the encoders at a reasonable rate (500Hz-1000Hz perhaps) and apply simple logic to the sequential readings. <S> This can be done in an ISR. <S> That means you have to read a total of 60 or so inputs and react to them every 1-2msec.. <S> so the routines shouldn't take too much time- preferably in the low 100's of usec. <S> The encoders can be addressed by using several multiplexers per input. <S> I suspect an AVR running at 20MHz and coded in assembly is up to this, and a single 4MHz PIC is probably not. <S> Or use a 32-bit micro (eg. <S> Atmel Cortex M7 running at 300MHz) and relax with everything coded in C. With a 144 PQFP (114 I <S> /O) you probably don't need any additional chips. <A> For anyone living in 2018+, try these I2C encoders: https://www.tindie.com/products/Saimon/i2c-encoder-v2/
| Use multiple microcontrollers (MCUs), with each one handling a subset of the rotary encoders. I would connect each A input to a 1-of-20 MUX and thence to a single port pin (say bit 0).
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What does NC mean in this pinout diagram for an ESP8266? The ESP8266 Datasheet has a Pin 5 labelled as VDD_RTC and it describes its function as "NC (1.1V)". I usually understand NC to mean no connection on most ICs, or on relays as normally closed . Neither of those seem to apply here. What does it mean in this case? <Q> Extract from datasheet: <S> It appears that if you keep RTC powered on, it will keep basic settings intact while in sleep. <S> Even though complete info is not present, I think pin 5 can be fed 1.1V externally to keep the RTC running even while your 3.3V supply is off. <S> So, usually, you aren't supposed to connect anything to it <S> but you can connect a 1.1V (or <S> maybe 1.1V is the minimum voltage required) power source to keep RTC running so that wifi configuration remains intact. <A> If you view the Hardware user guide on page 12: <S> It can be seen that it is left Not Connected , the standard interpretation for NC. <S> There is a chance it could be used to connect a supercap/battery (at 1.1V) to supply the RealTimeClock <S> but there is insufficient information to even consider trying this. <A> Nobody has a definitive answer. <S> I suggest you email feedback@espressif.com and ask them. <S> It is a very simple question to answer for them. <S> It looks like it would be safe to leave it unconnected (based on @JonRB's answer). <S> After you hear back from them you can answer your own question here (for the benefit of people searching for this answer). <S> --McKenzie <A> Usually this means 'not connected' or 'no connection' <S> This may mean that the pin will read 1.1v in normal operation, but it's very confusing to the engineer or technician.
| Since it's labelled as a type P, it should be a power input pin for RTC.
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How to program multiple STM32Fs on same PCB I work in a management roll and only have a very limited electronics knowledge (enough to make simple arduino projects etc). We are working on a product at the moment which has 3 STM32F303s and 2 STM32F103s and the main processor being a AM5718. I was just wondering how you would go about uploading and upgrading the firmware and bootloader one each processor including the AM5718. The 5 STM32s are connected to the AM5718 via UART and all are on the same PCB. I would like to get a good idea about how this would be done after the PCB is produced and the product is in the field. I would obviously ask the electronic engineer working on the project but we have finished for xmas. <Q> If hardware design is done correctly and assuming UART is used for programming, the steps would be roughly as follows: <S> The main CPU resets a selected MCU. <S> The CPU then brings the required BOOT pins to an appropriate state and releases the reset. <S> The MCU bootloader starts and waits for data to come in on the appropriate UART line. <S> The CPU send the binary data over the UART and resets the MCU when done. <S> The MCU runs with a new firmware. <S> Here's an image showing bootloader selection for STM32F10xxx parts: <S> The same steps would be repeated for each MCU. <S> The main processor can be updated in numerous ways, but it usually goes like so: <S> Firmware is downloaded into a local storage (FLASH, RAM, SD Card etc.) <S> The CPU reboots and the bootloader discovers the upgrade code. <S> The CPU bootloader reads the data, checks that the binary is valid and programs the appropriate FLASH sectors. <S> The CPU reboots itself again and loads with the new firmware. <S> Update. <S> Based on your current hardware, you'd have to execute the bootloader by jumping to the appropriate memory address (see page 20 of AN2606 ). <S> I don't know a great deal about this, but your hardware designer should be able to figure it out. <S> Not having the control of the reset line is a bit of an oversight though. <A> I suspect you need to load the AM5718 firmware via JTAG, however I'm not far too familiar with this CPU family. <S> The STM32 (like many other ARM CPU's these days) have a ROM bootloader built in by ST. <S> On some STM32 series this is done by holding BOOT0 pin high during reset. <S> To boot your normal program you need to hold the pin low. <S> This however is conveniently not always in the datasheet, but often in the user manual. <S> The ST bootloader often supports several protocols, and UART is a very common one. <S> Not all UARTs or pin locations on the STM32 chip are supported though - <S> so you need to choose the serial port pins. <S> This document is very helpful if you can find the right family. <S> The procedure Armandas describes is correct. <S> This does cost some GPIO pins on the CPU - in theory 1 additional reset line for every CPU added. <S> You may also need to consider how your board will boot-up in this setup while the AM5718 is not completely running. <S> A little caveat: on some STM32 parts the ROM bootloader is turned off once you enable read protection. <S> You can still access the chip via JTAG (after erase), but not via the bootloader. <S> Additionally if you cannot make the automated activation seem fit in hardware, you may need to do it manually via jumpers & a paper procedure. <S> This however is only practical to do in the factory - it is not a very field serviceable solution. <S> Both reasons may steer you towards an in-house bootloader which can be activated via a serial command. <S> It also adds the advantage of 'protecting' of your firmware images via encryption, given that you handle decryption inside the bootloader itself. <A> If all parts are JTAG-compliant, and are all in the same power domain (or otherwise in a state that guarantees chain integrity), you can construct a scan chain and use that to hit each part in turn. <S> However, it sounds like the project is done <S> and you're in management :) <S> , so I think the UART option is the best route. <S> To be clear, each STM has its own UART to the large processor, or do they share (which is a bit odd...)? <S> If the latter, as the other posters said, you'll need one GPIO per to select who is active.
| If you have some spare pins on your AM5718 CPU you could automate the activation of ST's bootloader via software.
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Energy wastage in wireless charging rotary toothbrushes? I'm not going to change my habits based on the answer, but I did wonder from a theoretical standpoint. I get idea of wireless charging. I still have flashbacks to my power generation & electric motor courses in undergrad (3 phase systems, lossy magnetic field circuits, etc.). So basic intuition, I have a vague sense of. What I don't know is whether today's wireless chargers for rotary toothbrushes are made with enough monitoring and control logic to be aware of when a load (i.e., a toothbrush) is docked and drawing power, so that it dials down the EM field when there is no toothbrush. As a related 2nd question, does the toothbrush switch loads when full charged so that the charger can tell, and dial down the juice? <Q> I can't give the detailed working information about how a toothbrush holder delivers power because I don't know that <S> but I can tell you that theoretically no power has to be consumed if there is nothing sapping power from the magnetic field. <S> Think about a coil of wire attached to a sinusoidal voltage source. <S> If the coil is made from low resistance wire and is reasonably "reactive" to the frequency of the sinewave, the current drawn will be near enough 90 degrees to the voltage waveform <S> i.e. it will have a power factor of virtually zero and very little power will be used up. <S> As the toothbrush (or whatever) comes up close its coil can convert some of that magnetic field into electrical energy using the principle of induction. <S> Therefore real power can be withdrawn from the mag field and the phase angle that was 90 degrees becomes somewhat less. <S> The power factor is no longer zero and <S> this signifies power is taken from the voltage source and delivered to the toothbrush. <S> It does look like the toothbrush uses AC power frequency like this: - Operating at a much higher frequency <S> is quite possibly a sensible thing to do and this is exactly what passive RFID tags utilize. <S> A lot operate at 13 MHz and you can make quite small coils and, to get extra distance <S> they are resonant tuned with capacitors. <S> All done without incurring hardly any off-load power: - I've built bigger ones to supply energy to telemetry electronics mounted on a spinning rotor in a turbine. <A> A robust modern product would probably have circuitry to detect the toothbrush load back in the base and turn itself off. <S> Unfortunately the competitive market and cost pressures to keep product prices down means that the products are often designed as simple as possible. <S> So likely no extra special circuitry. <S> Other base changers are just going to be a many turns primary winding that exhibits a good amount of inductance. <S> This winding is across the power line when the base is plugged in and will act as a transformer primary when the toothbrush handle is placed in the holder. <S> A coil in the toothbrush handle acts as the transformer secondary and induced current flow occurs and charges the battery. <S> The secondary load characteristics are reflected back to the transformer primary and cause more current to flow from the mains AC connection. <S> With a carefully designed transformer setup like this the primary current could be very low when the secondary in the handle is not in position. <S> Is it low enough to justify the simplest design where it is 100% connected to the power line? <S> In today's world probably not!! <S> In the past? <S> Highly likely. <A> EEVblog did a teardown of one of these a while back, mildly interesting and might help to answer some of your questions: http://www.eevblog.com/2012/05/30/eevblog-284-braun-toothbrush-teardown/
| Some tooth brushes may have a magnet that activates a reed switch in the base to activate the charge circuit.
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Why am I getting very different measurement results on a breadboard? I have an AC to DC multiple output adapter. It has the following output: I set the voltage to 5V. When I measure the supplied current, I get 3.26A. Then I connect an output to it: When I measure the supplied current now, I get 3.23A. Then I try to measure the current on a breadboard: And I get 0.02A! What is going on? Why is this happening? More interestingly, I have discovered that the LED on my adapter gets dim proportional to the current it supplies. However, in all three measurements, although the results differ (especially the last result), the LED gets exactly the same amount dimmer. So this is a curious point as well. Could you explain why the measurement on breadboard differs so much? <Q> You are short-circuiting the adapter, first with the ammeter, and then with the breadboard. <S> Both are not good for the adapter. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The dashed boxes represent the equivalent circuit of your supply and your ammeter; numbers are very approximate. <S> On the right hand side you see the ammeter shorting out your supply and causing it to overheat and probably fail in short order. <S> On the left hand side you see the same thing happening except most of the current is flowing through the solderless breadboard and only a small amount ( <S> the voltage drop in the conductors) is getting to the ammeter. <S> If the conductors were perfect the meter would read zero since you've shorted the meter (as well as the adapter). <S> Don't ever do it. <A> The 3.2A is now flowing through the short you have on your breadboard. <S> Try putting black wires on the blue line of breadboard pins. <S> Another bit of advice. <S> Unless your power supply is current limited you might do it damage shorting it out like this. <A> In your first setup the ammeter was in series with the source, that's why it correctly measured its short-circuit current. <S> In general, 3A through a breadboard is not a good idea. <S> And neither is shorting a supply that's not designed for it. <A> What you are measuring is the short circuit current of the power supply. <S> This is rarely a useful figure. <S> Normally, you would just trust the maker's specifications to give the maximum useful current that the supply can deliver. <S> If you don't trust the specs, you can (maybe) determine the maximum current by measuring the supply's output voltage while drawing increasing current from the supply - when the voltage drops too much, you've exceeded the maximum permissible current. <S> This method may work OK with a regulated power supply, but with a cheap unregulated supply it may be difficult to determine how much voltage drop is too much. <S> If you have an oscilliscope, you can watch for ripple on the output voltage - when the ripple voltage is "too high", you are drawing too much current. <A> It looks like you have the + and - shorted out in the same track on the breadboard. <S> The breadboard track will probably be getting hot. <S> It is not a good idea to put an ammeter directly across the power supply like you have done in any of your examples anyway. <A> You have created a massive short circuit.
| In your breadboard setup you have put the ammeter/DMM in parallel with the source, which is shorted through the breadboard. This is bad practice and you will eventually damage something.. putting an ammeter across a voltage source can cause serious injury or death if the source is capable of a lot of current, and certainly can damage the meter or the electronics.
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Do buck converters increase the amp hours of a battery? I've read that they are 90% efficient compared to linear regulators so given the relationship between volts and amps and I want to power the project off of a 12v battery that has 18Ah and is charged using a small solar cell. The project itself only needs 5v, so if I buck the voltage down to less than half the required voltage, wouldn't the amp hours close to double giving me 36+ Ah? Or am I missing something about the relationships between volts and amps and the mechanism of the buck converter? <Q> You are kind of on the right track. <S> The amp-hour rating of the battery specifies how many hours you could pull a certain rated current from the battery. <S> The buck converter to 5V won't change the amp-hour rating of the battery at all. <S> But if you were to only pull half of the current from the battery then the battery will last twice as long. <S> So let's make an example here. <S> As you stated you have a 12V 18AH battery. <S> Let us consider, for sake of example, that your load at the stated 5V required 500mA (i.e. 0.5A) to run. <S> If you were to use a linear regulator to drop the battery 12V down to 5V the linear regulator would pull 0.5A from the battery and radiate (12V-5V)*0.5A <S> = 3.5 Watts of energy as heat whilst delivering the 5V to the load. <S> The 12V battery delivering the 0.5A would supply the regulator for 18AH/0.5A = 36 hours. <S> The efficiency of such a system is EnergyOut / EnergyIn = <S> (5V * 0.5A)/(12V * 0.5A) = 41.6% efficiency. <S> So if instead you used the buck switching regulator at 90% efficiency we can evaluate the battery life. <S> The buck regulator operates as: Energy Out = 0.9 <S> * Energy In. <S> With a load of 5V @ <S> 0.5A <S> the energy being delivered out of the buck regulator is 5V*0.5A <S> = 2.5W. <S> Applying the 90% efficiency formula the regulator input energy would be <S> 2.5W/0.9 = <S> 2.78W. <S> This wattage at 12V corresponds to 2.78W/12V = 0.231A draw from the battery. <S> At this lower current the battery would last 18AH/0.231A = 77.9 hours. <S> As you can see the battery usage is hugely more efficient. <A> Amp hours are not a great way of measuring battery capacity. <S> Watt hours is a much better figure. <S> If you stick a buck converter after the battery, the equivalent watt hours of the battery will decrease by the loss in the converter. <S> However, since the converter will draw fewer amps from the battery, the equivalent amp hours will increase. <S> This isn't necessarily the greatest comparison, though, as in this case the equivalent amp hours are at a different voltage after the converter. <A> To expand on what alex.forencich said, since it was apparently not well understood, what you are missing is that amp-hours is a good way of comparing batteries at the same output voltage. <S> It's not a good way to compare batteries at different output voltages ... or once you buck or boost the voltage. <S> E.g. 3Wh battery (1.5V*2Ah) can become 0.2Ah if we boost the voltage 10 times [to 15V] or can become 20Ah if we buck the voltage 10 times [to 0.15V]. <S> And to address what Nick Alexeev said; this is as the battery is seen through the other (=output) side of the converter. <S> Of course the Ah of the battery itself doesn't change. <S> But no practical DC-DC converter has 100% efficiency, so the converted power needs to be adjusted (multiplied) by the converter's efficiency. <S> E.g. if the converter has 90% efficiency, the 3Wh rating becomes 2.7Wh after conversion. <S> That's regardless of what the voltage you convert to is. <S> And regarding "90% efficient compared to linear regulators", you probably mean they can have 90% absolute efficiency. <S> Linear regulators have maximum theoretical efficiency given by Vout/Vin; so for 12V to 5V conversion, a linear regulator cannot exceed 41.66% efficiency, and in practice will be less than that. <S> And indeed a linear regulator cannot boost amp-hours because of this. <S> Let's say you have X Wh at the Vin (inpt) voltage. <S> This becomes X Vout/Vin <S> Wh at the Vout voltage; Vout is always less than Vin for a linear regulator. <S> Now divide this X Vout/Vin Wh by the output voltage to obtain <S> X/Vin <S> Ah at the output <S> ; that's exactly the same Ah rating as at the original/input voltage [for an ideal linear regulator], but note we have a [usually large] loss of energy/power because Vout is [significantly] lower than Vin. <S> So to summarize; no converter [switching or linear] can increase the Wh rating. <S> A linear one cannot increase the Ah rating either.
| When you pass the output of a battery through a theoretical DC-DC converter that has 100% efficiency, the Watt-hour rating stays the same; the amp-hour rating being Watt-hour/volts can go up or down depending whether the converter is buck or boost.
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What is the application of the SMA connectors on the DE2 115 FPGA board The SMA connectors are expected to be used for clock IO. However, the board already has oscillator and the PLLs can be used to generate a variety of clock frequencies. Besides, for any daughter board connected to the extender socket or the HSMC interface, the clock signal can be sent via the connector itself. Why then would someone need to use these SMA connectors? <Q> It's a development board, so it is designed to be flexible. <S> Maybe you need a clock that can't be generated from an onboard oscillator with a PLL. <S> Or you need a very low phase noise clock. <S> Or perhaps a variable clock from an RF signal generator or ARB. <S> Or perhaps you need multiple boards to run from the same clock source. <A> It would help if you stated what the DE2 115 board was, and who made it. <S> However, I'll guess that it is an FPGA demo/evaluation board. <S> If so, the manufacturer probably put the SMA connectors on the board assuming that some potential users may require them. <A> There are a variety of applications for SMA clock I/ <S> O. <S> Additionally, the clock output may be important for chaining device clocks or simply connecting to test equipment to check frequency/jitter/etc. <S> In all these examples and more, you might not want to use the other connectors; SMA is nice when you need a high frequency signal run over a long length of coax. <A> The SMA connector is an arbitrary choice for a clock connector and is quite often used on fpga dev boards. <S> It can be misleading/confusing if your a RF person like me; I've not seen one designed to drive/match 50 ohms, so I would verify any comments or answers that imply that you can do so. <S> In the likely case that it is not matched, tee with a 50 ohm termination or keep cable short: for example, a 10 MHz clock in coax has roughly 20 meter wavelength <S> so you longest run should be less than 2 meters. <S> A 100 MHz clock should be less than 20 cm. <S> Despite the designed frequency limit of a SMA connector, you will be limited by the performance of the fpga.
| You might want to generate a specific clock externally (syncing to an external reference clock, for example; or a variable clock that can't be easily generated via PLL).
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Computer PSU 5v Output direct with Raspberry Pi USB I'm using an old power supply unit from an old computer (2011 HIPRO) to power my Raspberry Pi 2 Model B V1.1 (2014). I have measured the voltage across the terminals of the 5v output (Red and Black) and have proceeded to hook this up to a USB adapter which goes straight to the Pi. My problem is that using this PSU the red "on" indicator light does not turn on, versus when I use a USB charger it does. Was there a problem with this red light not turning on? Is this bad for the Pi? BTW everything works fine using both power supplies, the Pi turns on and the screen is shown, able to login etc. Please see my photo of my setup below: Closer View of the PSU pin outs: <Q> Your five volt power isn't really five volts. <S> The newer Pis will run but turn off the LED if the power in drops below 4.63 volts. <S> Check the 5Volts where it connects to the Pi. <S> Given the way you have things wired <S> (multiple plugs and jacks and a breadboard in the middle,) it isn't just MAY have losses, it is rather WILL have losses. <S> (Thanks to @pipe for pointing it out.) <S> Info from here. <S> There are links in that post that lead to more info. <A> Red light off on a modern Pi (B+ or newer) indicates undervoltage, this won't damage the Pi but may well cause reliability problems. <S> You need to go through the circuit with your multimeter measuring the voltage at each stage (including at the test points or GPIO header on the Pi) and find the cause of the low voltage. <S> My guess is that the wires in the USB cable you are using are too thin <A> As @JRE mentions you have losses on the wires. <S> Instead of connecting to the 20/24 (can't see in the photo) <S> pin header, use several of the black and red lines in the ATA (or SATA) power lines, probably soldering them together. <S> Using a similar methods I used and old PC PSU to power a LED tape with 6A@12V with no problems.
| You may have losses in the wires from the power supply, and the power supply itself may just put out a slightly low voltage on the 5Volt rail.
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Why does my resistor consume so much power? I've created a low battery indicator circuit which lights an LED when the voltage drops below 3.75V. in this circuit, the 62ohm resistor is consuming about 240mW of power. why is this? is there any way to reduce it? Here is the circuit when the battery voltage is below 3.75V: <Q> This circuit will place less than 0.2mA of load on the battery until the battery voltage falls below about 3.75V. <S> At that point the LED turns on and the load on the battery becomes primarily the LED current which can be adjusted by changing the value of R5. <S> The 10meg feedback resistor around the comparator adds hysteresis so that the detection point does not cause flickering of the LED. <A> In your circuit there should be no need for there to be a resistor as small as 62 ohms as the collector load of the left most NPN. <S> Reduce the power and current at the same time by making this resistor much larger. <S> A starting point might be 3.3K ohms. <A> Just get rid of the 62 ohms and move the 10k like this: - <S> Now, when the 1st transistor is on it will take a few hundred uA through the 10k <S> and when that transistor is off that current will be allowed into the base of the final transistor and turn it on.
| As indicated in comments I have prepared a low battery detector circuit that uses a comparator to sense the low battery state.
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Is voltage measurement of a battery done in series? This might be a stupid question but I can't find an explanation. Voltage measurement is done in parallel. But when we measure the voltage of a stand alone voltage source, say a battery, it looks like we are doing it in series and yet we get the correct reading. simulate this circuit – Schematic created using CircuitLab So we connect the voltmeter like above to measure the voltage of an isolated voltage source. Although it is connected in series, we get the correct reading. Am I missing something here? <Q> This situation is a something of a paradox, yes, but you need to understand that 'series' does not mean 'not parallel', and vice versa. <S> Any loops which share the same potential difference across them are in parallel . <S> In your example, current flows from the one battery terminal to the other through the voltmeter, which acts like a very large resistance (10M-50M, typically). <S> Current is flowing in a loop, therefore the two components are in series with each other . <S> The voltage across the battery terminals is equal to the voltage across the voltmeter, therefore the two components are also in parallel with each other . <A> The voltmeter has a high ohm resistor (e.g. 10 MΩ) which is placed in series with the battery. <S> The voltmeter is actually measuring the voltage across this resistor using an ADC (analog to digital converter) and using that voltage which is then displayed. <S> I replaced the voltmeter in your circuit by the one in the diagram. <S> Right-click on the image and choose View Image to see a larger version. <S> In this circuit, the input resistance is a little over 11 MΩ. <S> There is a voltage divider made up of R1 through R4 which is connected to a selector switch to select one of several voltage ranges (20 mV, 2V, 20V and 200V) to be input to the ADC (which is contained inside IC1). <S> In this case, you would set the range switch to 20V since you are measuring a 5V battery (and the next lower range is 2V). <S> This creates a voltage divider with 10MΩ on the top leg, and 1MΩ+110K+1.1K on the bottom. <S> So with a 5V input, the ADC will see $$5V\times\frac{1MΩ+110K+1.1K}{10MΩ+1MΩ+110K+1.1K} = <S> 5V\times\frac{1.1111MΩ}{11.1111MΩ} = 0.1V$$ <S> When you measure the voltage across the red and black terminals, you are really measuring the voltage across a resistor divider, which is also the voltage of the battery. <S> This is then scaled for the ADC input. <A> Look again at your picture. <S> In this simple case the meter is in parallel with the voltage source from the perspective of voltage measurement. <S> The meter is also in series with the voltage source from the perspective of current that wants to flow around the loop. <S> Hopefully the internal impedance of the volt meter is very high so very little current flows. <A> It may help you to think of it a little differently as in your example you have a power source (the battery) and a load rather than, say, two resistors in parallel. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Here we can clearly see that the two voltages are in parallel . <S> What may not be so clear is that the currents are in anti -parallel - i.e., they are forming a loop and <S> so, as your hunch led you to believe, the current is running in series through the loop. <S> Another way of looking at it: on the battery the current is exiting the '+' terminal. <S> On the load (voltmeter) <S> the current is entering on the '+' terminal. <A> The 'voltmeter' we normally use is really a very sensitive ammeter with a large series resistance. <S> What the 'voltmeter' is doing is measuring the small current that flows through it (a serial measurement) and converting this current value back into a 'voltage' reading (i.e. the voltage across the terminals = <S> voltage across the 'voltmeter'). <S> By Ohm's law <S> (V=IR) <S> we know that V is directly proportional to I , so its just a matter of how big R (the constant of proportionality) is to convert one to the other. <S> It's much easier to measure a current directly (even very small ones) than it is to measure voltage directly. <S> Examples of 'voltmeters' that are really ammeters: <S> A moving coil multimeter : <S> A low cost moving coil ' voltmeter ' is typically a 50uA movement ( 1000 ohm coil). <S> For a ONE VOLT reading (full scale) we need to have a total resistance of 20K. (you'll sometimes see this marked on the meter as 20K/volt) i.e 19k (external) + 1k (meter). <S> The range selector switch adds suitable parallel/series resistance into the basic meter circuit A digital multimeter - requires/measures a much lower current something like 1M0 per volt and is closer to an 'ideal' voltmeter which would take no current from the circuit (infinite resistance?). <S> The resistance of the 'voltmeter' is only really important if the current taken from the circuit significantly interferes with the reading. <S> In the case of a battery (very low internal resistance) this is highly unlikely.
| Any components which form a loop in which current is allowed to flow are in series .
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Mains LED light bulb flickers when switched off We have LED bulbs in our apartment. In the bathroom, the bulb flickers on and off when it's switched off. Same when it was a CFL bulb before I switched to LED cause I hate CFLs. Our mains voltage is 220v. The bulb we have looks like this If I remember correctly, I should add a capacitor in series to the bulb to fix this. But what kind of capacitor? and what rating? Update: The switch is a combined switch and plug in one part. Looks like a normal one line switch. This apartment and the whole building doesn't have grounding. We don't have this in requirement in our building codes. I'm in Lebanon. Only this one bulb in the apartment does this. Other bulbs are also LED and two more CFL not changed yet. Any help appreciated, thanks. Resolution: I fixed it with an incandescent bulb! I'm too lazy to go electrician detective for now. <Q> You might find that a different brand of bulb is much less prone to this kind of behavior. <S> The internal circuit varies quite a bit from bulb to bulb. <S> If you have a store with a generous return policy that applies to such products you could simply buy a few different types and try them. <S> In particular, I would expect bulbs based on Microchip's CL8800 to not suffer from this problem. <S> See this excellect article if you are interested in the gory details. <S> Most bulbs have a small switching supply in them, but this type is different. <A> The cause of this is the design of the electronics inside the bulb combined with stray capacitance between live and switched live. <S> In the UK it is most commonly seen on 2-way switching installations due to the physical conductor layout in the cable used for those. <S> Basically the circuitry inside the bulb contains a rectifier followed by a capacitor, followed by some electronics that has a minimum voltage threshold. <S> The capacitor charges (through the stray capacitance) until it reaches some threshold where the electronics in the bulb try to start up producing a flicker of light and discharging the capacitor. <S> This process repeats. <S> This forms a voltage divider with the stray capacitance preventing the voltage build-up. <S> (for those wondering why this is different from the capacitance inside the bulb it is because it is before the rectification while the capacitance in the bulb is after it) <S> In theory to calculate a minimum value for the capacitor you would need to know How low the voltage needs to be held to stop the LED bulb flickering. <S> What the stray capacitance is between live and switched live. <S> Unfortunately neither of these values is readilly available so it becomes a case of rules of thumb rather than an exact science. <S> The capacitor needs to be of a type suitable for connecting directly across the mains. <S> It's usually also good if it has some resistance in series to limit current surges at switch-on. <S> One forum thread I found suggested http://uk.rs-online.com/web/p/rc-network-capacitors/2067869/ which seems fairly reasonable to me. <S> The capacitor needs to be mounted inside a suitable electrical enclosure. <A> A lot of different electronic circuits will produce odd-ball behaviour as the power is lost. <S> Think about some music systems producing little pops when turned off (or on). <S> Microprocessor boards will produce odd behaviour when being forcibly powered down and a PC has to be powered down in a proper way or you might get hard disk failures (extreme but not unheard of) or possibly data corruptions. <S> TVs used to produce a white dot in the centre of the screen when switched off (it gradually faded of course) and <S> when you remove power from a motor it slowly comes to a halt. <S> These are all after-effects following the removal of electricity. <S> I had an old two stroke motor bike that got so hot even though I turned the ignition off <S> it was revving like mad - now that really was annoying. <S> A typical LED circuit might use a chip like this: - As the AC voltage is removed, the DC voltage across Cin remains high-enough for a few tens of milli-seconds to keep the LEDs powered <S> but then it will inevitably drop below a certain theshold and the chip will turn off the LEDs but, this removes the load from Cin and its voltage then falls more gradually and, who amongst us can say that at some low voltage the chip doesn't have some perculiar behaviour that flashes the LEDs for a couple of instants. <S> Yes it's annoying but will sometimes happen with some designs. <S> BTW I'm not saying the above chip will exhibit this behaviour, I'm just using it as an example of how strange things can appear to happen in some designs. <A> All the neutral wires carries the current back to the transformer neutral point that is grounded at the supplying transformer secondary side. <S> If it is not properly grounded, it can cause incidents such like yours. <S> Check <S> if the neighbors also having the same issue. <S> If it is, you better call the breakdown services.
| One fix is to add a a capacitor in parallel with the bulb.
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Correct way to measure a wall wart voltage I got a full drawer of second hand wall warts which I'm planning to use for powering arduino projects (mostly 5V "pro mini" or "nano"). I'm using a voltmeter to check that they are still working, however most of them seem to output a far greater voltage than what is printed on the case. For example the Arduino pro mini accept up to 12V so I've been looking at 9V wall warts, but they often give more than 12V. Example (but typical reading): written specification: 9 V 300 mA Measured voltage: 12.37 V Measured voltage while powering a led (through a 300 ohms resistor): 11.2 V I tried to add a led because I've read that measuring a voltage with no load may give a wrong result, but perhaps that's not enough to get a proper reading... My questions are: Is there a way to measure the correct voltage that will be issued in "normal usage conditions"? Can I safely use a "9V wall wart" to power a "12V max" device if I measure more than 12V? PS: So far I've use 6V wall warts which are giving about 10V, but I don't have any left so I'm trying to use 9V ones <Q> It sounds as though you've got some unregulated power supplies. <S> Generally these have a transformer, rectifier and smoothing capacitor. <S> With no load the capacitor will charge up to peak voltage. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> There is always some internal resistance in the transformer, represented by R1, which will cause a voltage drop which increases with current drawn. <S> In addition, C1, which maintains output voltage during troughs in the rectified AC (i.e., between pulses from the rectifier) will droop between cycles causing the average DC value to droop with increasing current. <S> e.g., for 100 mA load on a 9 V PSU you would use a 9 / 0.1 = 90 Ω test load. <S> Nearest standard values are 82 Ω or 100 Ω. Connect the test load across the terminals / connector and measure the voltage. <S> An alternative is a 12 V, 6 W car tail-light lamp. <S> At 12 V this should pull about 0.5 A. <S> Be aware that the resistance value will change dramatically with temperature which is voltage dependent. <S> To get a good feel for what's happening you would need to measure both the voltage and the current. <A> This is likely to continue until you draw the maximum stated current, or within a certain percentage of it. <S> Your best option is to use a pre-regulator. <S> With something small like a Nano or Pro Mini <S> you are better off dropping as much voltage as you can externally to save heat being dissipated on the Arduino board. <S> A 7805 is beefy enough to handle the currents typically required by an Arduino without a heatsink, even with a 12V input. <A> It's normal for <S> unregulated wall-warts to show higher voltage than the nominal one when tested without (or with small) loads. <S> It will show voltages closer to nominal as you increase the load, reaching nominal voltage when the current approaches the specified limit. <S> In your example, the LED isn't enough of a load to bring the voltage to nominal value, but it already shows that the wall-wart voltage decreases with higher loads. <S> You'll probably measure 9V if you apply a load that draws close to 300mA.
| If they're transformer supplies (rather than switched mode) then yes, you're going to get higher than the stated voltage. To test you should load them with something simulating the intended load.
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What is the purpose of a relay between a switch and motor? I was disassembling a belt disc sander today and noticed that there was a rather large relay between the switch and the motor. The relay consists mostly of a large copper coil. The motor doesn't have a capacitor and neither does the switch assembly. There are three wires coming out of the motor: red, white and black. All of the wires are of the same size about 14AWG. When I tried wiring the motor without the relay, I couldn't get it to start without turning the shaft by hand. This would mean that the relay was acting as a starter. I am perplexed because I was under the impression that the only way to start a motor was either by hand or a start capacitor. Can a relay be used to start a motor, if so, how? <Q> The relay is likely connected to an auxiliary motor winding that has a different number of turns, a different wire size and resistance compared to the main winding. <S> That would act to start the motor similarly to capacitor starting. <S> There are several possibilities for the motor design details. <S> It is not possible to completely and accurately explain the motor design without close examination of the construction. <S> Photographs might help, but that might still not be enough to determine exactly how the motor works. <S> Edit <S> Re Photos <S> Note that the relay has relatively few turns of heavy wire. <S> I believe it is a current relay, wired in series with the main winding. <S> When the motor is first turned on, a high current flows in the main winding through the relay coil causing the relay to connect the auxiliary winding. <S> Since the aux winding has a different inductance and resistance, its current is phase shifted compared to the main current similarly to the phase shift that a capacitor would cause. <S> That defines the direction of rotation and provides starting torque. <S> When the motor approaches full speed, the current drops and the relay disconnects the aux winding. <A> When you bypassed the relay, was the wire a sufficient size to handle the current needed by the motor? <S> The point of the relay is so that the large amount of current needed by the motor doesn't all have to go through the switch. <S> If you bypassed the relay and used the smaller switch wires, they may be too small to power the motor. <S> This may be especially true if it requires a large spike in power to get started (which is why they use start capacitors). <S> Did you notice the wires getting hot when it was running? <A> You might posted name plate details of this motor. <S> From there you may find what kind of motor is that. <S> From your information i guess its split-phase single phase induction motor. <S> In split phase induction motor, there are two windings 1. <S> Starting winding Fewer Turns with Lower Inductance and Higher Resistance. <S> 2. <S> Main (or) Running Winding <S> More no of turns with low Resistance and High Inductance <S> Here starting winding takes part while starting only, (up to 75-80% of rated speed) after attained this speed starting winding will be disconnected by centrifugal switch. <S> Here instead of centrifugal switch the relay might be used.
| A smaller, "signal" current can go through the switch, which actuates the relay which connects the much bigger wires that power the motor.
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Why is it safer to place a switch on the positive side of a voltage source? According to the research I've done, it is safer to place a switch on the positive side of a voltage source: http://www.allaboutcircuits.com/textbook/direct-current/chpt-3/safe-practices/ This only makes sense to me visualizing the circuit using conventional current (counter clockwise). When visualizing it in electron flow (clockwise), the opposite seems true. I think this stems from my use of the "water analogy." Why does a switch go on the positive side of a voltage source? <Q> In theory there is no difference, but practically there is because of how most engineers use "ground". <S> The problem is how sure are you that you truely are opening all paths for electricity to flow? <S> If "circuit ground" is tied to the actual ground, something which happens to be resting on the ground may form a closed loop. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This is common in "mains" electricity (i.e. the wall outlets), so if SW1 is closed, Joe is having a bad day. <S> SW2 doesn't matter, and it doesn't matter if V1 is positive or negative. <S> Another common example is when you connect "ground" to a metallic chassis, which is common in the automotive world. <S> The circuit for this looks similar to the one above. <S> If you want to use a low-side switch, you had better make sure you don't have any metal contact with the body of the car or anything electrically connected to the body. <A> In that case, the switch in the positive feed will remove all power from the load. <S> If the switch was in the negative side, the load would still have power applied, and an accidental ground somewhere could cause unpleasant surprises, as motors start, lights flash.... <S> A more general description of the preferred switch location is "in the ungrounded supply line, as close to the power source as practical". <S> This covers AC power and positive ground circuits as well as negative ground. <A> The circuit works the same no matter where you put the switch. <S> Part of the electrical code (for house wiring) says that the switch should always go between the hot conductor and the load. <S> This is for safety. <S> Since the neutral (white wire) is referenced to ground, it should not have any potential relative to someone touching it. <S> This is potentially dangerous because an unsuspecting person might touch a white wire thinking it is neutral. <S> In this case it would actually be hot. <S> It's an extra layer of safety <S> so you know that white wires are never hot, although you should still never touch live wiring.
| If the switch was placed to break the circuit on the neutral wire, it would mean that there is potential on a plugged in device and some length of neutral wire up to the switch even when it is turned off. Usually the negative terminal of the voltage source is considered Zero Volts, or Ground.
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Digital vs Analog signal for data transfer across boards I have 1 board that reads data from a sensor, while another board wants to get data from that sensor. Is it better to have an ADC on the first board, then send that data via i2c to the 2nd board? Or is it better to send the analog signal from the first board directly to the ADC of the 2nd board? My assumption is that sending the analog data across 2 boards is going to have a lot of noise. <Q> Sending the digital signal is the better option, as it will be less susceptible to interference, or distortion as it travels between the two points. <S> Along the way not only can the sensor pick up noise from various sources, but losses along the sense lines can distort the reading. <S> Sometimes sensors need to be placed far from the ADC due to environmental or other conditions (high heat, etc.), in which case those distortions are taken into account during the design of the system. <S> Place the ADC on the sensor board, and send the digital signal out from there. <A> Either approach might be alright: <S> analog point-to-point connection (down to, say, 10 or 12 bits of resolution), or I 2 C. CubeSats are small. <S> They operate far from external sources of EMI. <S> If the switch-mode power supplies in the EPS (electric power subsystem) are properly designed, they should not generate much EMI. <S> Is it likely that the payload will generate EMI? <S> From systems engineering standpoint, both point-to-point and bus have their pros and cons. <S> The pro of point-to-point is that it's simple (in the k.i.s.s. <S> way). <S> The con is that it doesn't scale as well as I 2 C. <S> The pro of the I 2 C is that it scales well, and it's convenient. <S> The con is that complexity increases. <S> The bus can also be a single point of failure: I 2 C bus can get stuck. <S> If you have more than one sensor on that board, and they lend themselves to an A/D connected via I 2 C. <S> Or, if you anticipate that there may be more sensors on that board in the future. <S> Then I 2 C might be convenient. <S> p.s. <S> In school, I have designed a C&DH board for the MAST CubeSat. <S> It had an extensive I 2 C bus, which worked without a hitch. <S> Later, I have designed and then had to grapple with an overgrown [because of lack of a better upfront judgement] I 2 C bus in an industrial instrument. <A> Digitize it. <S> What ground is your analog signal referenced at? <S> Chances are low <S> your intra-board harnessing has dedicated signal grounds vs. power grounds. <S> The long run will have a greater chance of susceptibility to noise and it is also additional harnessing which I can tell you from experience <S> is the last thing you want in a satellite. <S> Additionally, what reference voltage will it be sampled against, and where is that voltage located? <S> Now, if this signal has say a 1Hz filter on it (some kind of power rail), the discussion becomes different as its so slow as to be virtually DC <S> Finally, I'm not sure what your system bus is, but an I2C link can be extended to carry additional data with no extra penalty of wires.
| Digitizing the signal as early as possible is IMO in your best interest.
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Minimum and Maximum temperature for a copper wire to work? Is there a minimum and maximum temperature that a copper wire/s will still work on. Meaning that current will still be able to go through them. Let's say that I have battery that is heated or cooled depending on the environment, and that battery controls a light, that is also heated or cooled, and the wires connecting it are exposed to cold or hot environment. Is there a specific temperature, max / min, that will ensure that current still passes through the wires. I know this might depend on the size of the copper wire, and the specific gauge, but I am considering wires between 18 and 26 AWG. I also figured out that the melting point of cooper is about 1,085 °C, therefore that will in theory be the maximum temperature that a copper wire will be able to operate, I assume. And I am not sure about the minimum temperature. <Q> 0 K to 1358.15 K -273.15 <S> °C to 1085 <S> °C -459.67 <S> ° <S> F to 1984 °F <S> You can operate at higher temperatures if you can keep the wire-shaped puddle of molten copper where you need it, but it's usually easier to use platinum wire at that point. <S> (2041 K, 1786 C, 3215 F) and tungsten after you can't use platinum any more (3695 K, 3422 C, 6191 F) <A> Wire temperature limits sets first for safety reasons then to ensure circuit operation. <S> First priority is the insulation protection, not only in the particular cycle of operation (melt etc), but also in long term operation. <S> Insulation degrade each time that the temperature over the specs maximum temperature, shorten their life. <S> Second is to prevent surrounding materials ignition. <S> Hay for example ignites very easily. <S> IEC sets as maximum temperature increase the 35 degr Celsius above the environment temperature, which is can taken roughly to around 40 degr. <S> C. <S> The environment (clima etc, earth/space), the orientation, the encapsulation, the operation cycle (ON/OFF period), the installation place characteristics (in-house, open-air, buried etc) of the conductor plays important role. <S> As an example in normal operation and for copper magnet wire (formvar coated), the above temperature limit for AWG18 reached when 10A passing the conductor and for AWG26 is 2.8A. <S> However in adiabatic conditions (i.e short-circuit) current reduced to 3.5A for #18 and 0.9A for #26 (bare) suspended on air. <S> In case of coil temperature increase and new calculations should be done. <A> The resistance of the wire changes with temperature. <S> There is a given temperature coefficient of resistance for every material. <S> For copper, it is 0.0039/°C. <S> The new effective resistance of the wire is calculated by (deltaR)/(Ro)=(temperature coefficient of resistance)*(deltaT). <S> So, if you know the resistance of the wire at ambient temperature, you can calculate its change in resistance at various temperatures. <S> This will help you decide what size wires you need for a given temperature, or if the temperature is feasible at all. <A> So Copper melts at 1083°C. <S> But that has absolutely nothing to do with Minimum and Maximum temperature for a copper wire to work as a conductor. <S> Both are determined by the insulation and jacket materials. <S> Insulation breaks down if ambient temperature goes above 100°C. <S> The insulation goes through irreversible chemical changes and starts to burn typically causing fires. <S> Even if you have a jacket which is HOFR (Heat, Oil and Flame Retardant), XLPE and EPR, as insulation, act as wicks for fire and will spread a fire throughout the structure. <S> It is ultimately why residential wire insulation is colored white. <S> Discoloration means the wire has gotten to hot and the full wire must be replaced. <S> The portion that has burned is no longer an insulator, which means it is becoming a conductor. <S> Maximum PVC 75 <S> °C XLPE/EPR 85° <S> C Silicon 95 <S> ° <S> C <S> These maximums are one of the factors which must be taken into account when derating maximum current for a given wire size and insulation material. <S> Minimum PVC -55 <S> ° <S> C XLPE <S> -40 <S> °C EPR -60 <S> °C Silicon –80° <S> C Insulation becomes brittle below minimum. <S> Not a problem if you are dealing with buildings. <S> So copper will not melt at 200°C, but it is not much of a conductor if the insulation catches on fire. <S> Insulation and Jacket Materials
| Insulated wires are a bit of a problem at extreme temperatures as insulation tends to melt, burn, or become brittle and crack.
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Dealing with signal noise over 50 foot communication link Situation: Connecting a motor controller to an MCU. Motor controller is about 50 feet away from MCU. Connected via a single conduit (buried), containing the following cables: (There is no other electrical equipment in the area, and this is outdoors between two weather-proof boxes.) 24VAC power to motor controller (#14/2) (60Hz) Cat5 cable (24AWG copper) Signals being transmitted to MCU: Motor direction signal. This line is pulled high or low depending on the direction of the motor. No fast changes: motor runs one direction for at least 10 seconds. Motor position indicator. A TTL-level square wave (100Hz to about 1kHZ) outputting a number of pulses per revolution of the motor. The problem:Signals arriving at the MCU are very noisy - you can see the 60Hz line interference clearly - and you can see crosstalk on the two lines - the Direction signal is pulsing along with position indicator square wave. Any suggestions how to clean up these signals to be usable? I could possibly put a second MCU inside the motor controller box, but what protocol could I use to transmit across the existing Cat5 cable that would not be as susceptible to this type of interference? Would prefer I2C... suggestions? Hoping to avoid CAN bus as it adds too much complexity and extra hardware. Any suggestions appreciated. <Q> Single-ended signalling is the source of your troubles. <S> Differential signalling will drastically improve crosstalk and noise. <S> RS-422 would be an appropriate choice. <S> I2C is not ideal, as its bidirectionality (and, consequently, weakly-driven signals) does not lend itself well to differential signalling. <S> However, it seems you only need one direction. <S> And if you want bidirectional communication, you have extra pairs in your Cat5. <A> Building on uint128_t's answer: <S> This would not require the use of the RS-422 serial transmission protocol, so there is no need for an MCU at the motor end of the cable. <A> Start with the easy stuff - replace the cat5 UTP with cat 5, 6 or 7 STP - shielded twisted pair. <S> Ground the shield at one end. <S> If that solves it, you're done with little effort. <S> If not, whatever you do next (such as using proper differential line drivers) will work better. <S> I'm also wondering if you are making ground loop with motor power and signal ground. <S> Edit: as we elucidated in comments, both signals were running on a single pair, which is about as bad as they could get for cross-coupling the signals. <S> While differential transmission is what Cat5 is designed to do, simply moving each signal onto separate twisted pairs, with the other half of the pair grounded should make things much better from a cross-coupling point of view, and in fact, did, per further comments. <S> I'd say it still falls under the general heading of this answer, start with the easy stuff, (even easier if you don't the change cables, just how you use the wires) though I hadn't thought of cable-pair mis-use until after I wrote the first form of the answer, and asked the question in comments. <A> As you have 2 available conductors, plus GND, I would suggest using them for balanced transmission for the speed sensor signal. <S> YThis will reject the common mode interference quite nicely. <S> You also need to transmit direction, this could be done by a substantial common mode DC signal . <S> As it is known to be DC except when reversing, interference can be rejected by a low-pass filter.
| You could use RS-422 transceiver chips (single ended to differential converter and differential to single ended converter) at the MCU end and the far end of your cable to convert the existing signals from single ended to differential and back.
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Conditioning a voltage using constant current for an ADC range I wish to make a set of temperature measurements inside a relatively inaccessible box. The measurements can be made using Pt100 sensors - platinum Resistance Temperature Detectors (RTD) with a resistance of \$100\Omega\$ at \$0^{\circ}C\$. The change in resistance over \$80^{\circ}C\$ is around \$30\Omega\$. I want to digitize this result within the box. I'm currently looking at microcontrollers like this STM32F373 model, which includes (several) 16-bit ADC. To minimise the self-heating on the RTD, I want to minimise the current used. To make an accurate measurement, I would seem to need a constant current source. Wikipedia suggests several arrangements, at least one of which is given as an answer to this question - Zener Diode Current Source . How can I go about arranging a known constant current, with only a constant voltage source? Should I build a circuit similar to the 'Zener Diode' question, separate from any circuitry near the microcontroller, and then just pass the sense leads to the analogue input? Is there a simple trick to precondition the voltages to lie within a good range for the ADC? Seemingly a combination of gain to scale the range to fit \$V_{DD} = \frac{30\Omega}{I_{const}}\$ along with a constant offset from another circuit would be appropriate - but perhaps this is just because I'm stunningly ignorant. <Q> Making a constant-current source is, in principle, very easy if you have a precision voltage source. <S> The basis for conditioning your RTDs is simulate this circuit – <S> Schematic created using CircuitLab <S> In this case the current is set by R1, and is equal to V1/R1. <S> The output of the first op amp is equal to the current times the RTD value, but is negative. <S> So op amp 2 is used to invert the signal (assuming you want a positive voltage to the ADC), and this section may have some gain associated with it. <S> R5 is used to provide an offset to compensate for the fact that the RTD does not produce a zero resistance at one end of the useful range. <S> You'll need to be careful of your resistors. <S> If you're going to the lengths of using a platinum unit and a 16-bit ADC, this suggests that you're trying for very accurate and precise measurements. <S> Temperature coefficients in the resistors can adversely affect this accuracy, but you can get low-drift resistors with at tempco in the range of 25 ppm/deg C. <A> I use this constant current source for strain gauges and RTDs: - The strain gauge or RTD fits where the block called "current out" is. <S> You start with a decent constant voltage source (V+) and you set a voltage at the non-inverting input of the op-amp (called Vref). <S> This is set by R1 and R3. <S> The op-amp then tries to make the voltage on the inverting input equal the voltage on the non-inverting input. <S> To do so it drives the BJT until the current through R2 produces a voltage across R2 that equals the voltage across R1. <S> Current through the load is then regulated; if the load resistance rises, the op-amp drives the BJT a bit harder until the voltage across R2 = <S> the voltage across R1. <S> Ignoring the small current taken through the base/emitter to activate the BJT, the current flowing through R2 is the same as current in the load and this current is defined by the voltage across R1. <S> Try simulating it in LTSpice (or equivalent) <S> and you'll see that it is pretty effective and accurate. <S> If you find that the very small current drawn from the emitter and taken by the base is too much of an error (I'm talking about 0.2% from experience) then use a P channel MOSFET instead of a BJT. <S> Maybe if you don't quite follow the circuit, consider this NPN version with the load tied to the positive rail: - <A> As stated in the other answers you can build your own constant current source, but there are also designated ICs available for this purpose like the REF200 from TI. <S> You have to choose a low current to excite the RTD, or elsewise \$I^2R\$ losses will heat the sensor, thus affecting your measurements. <S> To achieve a good signal for your ADC you can use an instrumentation amplifier to amplify the voltage drop over the RTD. <S> Here is an reference design by TI which shows a expanded version of what I described. <S> It features a three wire connection and details filters on the input and output of the instrumentation amplifier.
| With a low excitation current you get a low voltage drop across the RTD.
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Is it OK to use a lower voltage to operate a relay coil than rated coil voltage? I want to know if I used a lower voltage than what was labeled on the relay was OK. The relay was a 24V relay.I check that the relay turns on when the voltage when the voltage was 13.5v.If I apply that voltage is it reliable?.EDIT: I do not have a datasheet but here is the part number: HTCA024V <Q> Electromechanical relays are really current operated devices. <S> The relay pulls in when the current is sufficient. <S> The current is determined by the operating voltage and the coil resistance. <S> But the plot thickens - the coil resistance is a function of temperature- <S> the coil temperature- increasing by about 0.4% for every degree C. <S> The coil is heated by the current through it, as well as by the ambient conditions. <S> It is also heated by resistive losses when the contacts are carrying current. <S> The temperature rise above ambient might be 35 C <S> so if the temperature outside your box is 45 C, and the temperature inside is 70 C the coil could easily be running above 100 C. <S> That means the pull in voltage will increase by 1/3. <S> In fact high temperature versions of relays have less tolerance than the usual ones. <S> Bottom line is that the relay engineers designed the relay to be reliable at the nominal voltage (with a bit of tolerance for unregulated supply voltage) over a reasonable temperature range, and with rated current flowing, and to account for manufacturing variations (friction, spring constant, magnetic properties). <S> If you know that those assumptions are somehow conservative in your situation (say you have a tightly regulated supply) you may be able to shave a bit off, but <S> you if you are wrong the relay operation may not be assured or contact life may be compromised due to more arcing from sluggish operation. <S> For example the relay may not pull back in on a hot day immediately after it has been on for a while. <S> Holding current is less than operating current (and contact pressure is not compromised in most relay designs) <S> so it's sometimes possible to give a higher voltage pulse to operate the relay then back off to a lower voltage, but that's seldom worth the trouble. <A> It depends on the application. <S> You will find that the relay switches much more quickly when powered at the rated voltage. <S> This results in clean switching with minimum arcing and contact bounce. <S> For a low current application it may be fine. <S> Note that as the relay armature pulls in that the magnetic circuit air-gap reduces and this increases the magnetic field strength even further, etc. <S> You should find that if you can pick from fully open at 13.5 V that it should hold on until the voltage drops to maybe 2/3 of that. <S> When the armature gap is closing it is working against the springiness of the contacts. <S> If you can get the cover off the relay you could energise the relay and then press the armature with your finger to confirm that it is fully closed. <S> If not then you might find that it stops when the first contact closes and the second contact doesn't make. <A> You must meet <S> at least the operate voltage of the relay as given in the datasheet. <S> If you do not then it is very possible that the relay will not switch. <A> The voltage value that is specified for a relay is the value at which the manufacturer guarantees the operation .From <S> experience, I know that usually the relay is activated with lower voltage on and off with a much less than the activation voltage. <A> Most relays have a range of voltage specified in their datasheet in which the relay can operate. <S> The minimum voltage of that range is only needed to reliably switch the relay to its new state, and is sometimes called the "must operate voltage". <S> After switching, a voltage significantly lower than the "must operate voltage" could be applied to keep the relay activated. <S> Take a look at a snippet from this this answer : 12 V relay at 5 V <S> Here we get to the Real Savings™. <S> First we'll have to explain the 5 V operation. <S> We already saw that we can operate the relay at 9 V, since the "must operate voltage" was 8.4 V. <S> But 5 V is considerably lower than that, so it won't activate the relay. <S> It appears, however, that the "must operate voltage" is only needed to activate the relay; once it's activated it will remain active even at much lower voltages. <S> You can easily try this. <S> Open the relay and place 5 V across the coil, and you'll see it doesn't activate. <S> Now close the contact with the tip of a pencil and you'll see that it remains closed. <S> Great.
| For a high-current application you will probably burn out the contacts fairly quickly. However, to ensure the functioning of the circuit, you must respect the nominal voltage, as specified by the manufacturer.
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Where does current go in these circuits? In both these schematics, the 9V source drives current through the circuit, but there doesn't seem to be any way in which the current can 'return' to the source to complete the circuit. In what ways would current flow through both these circuits from the DC source, and where would the current end up? Wouldn't all the current just go to ground? This is for an audio circuit, input is from a electric instrument and output goes to the speakers/recording equipment. <Q> It is implied that the 9 V source has its negative end connected to ground. <S> Current therefore flows out the positive end of the 9 V source, into the circuit, out the circuit's ground connection, and back into the negative end of the 9 V source. <S> Added <S> As Dave Tweed pointed out in a comment, and I didn't notice ealier, the second circuit uses a positive ground. <S> The common thread in both cases is that ground is implied as the 0 V reference for all other voltages that don't explicitly specify a reference. <S> Both the "+9v DC" point in the top circuit and the "-9v" point in the bottom circuit are therefore relative to ground. <S> Since they are labeled as supply points, it is assumed that a power supply is connected between these points and ground. <A> The actual 9 volt supply is not shown. <S> In these cases, the default assumption is that it is connected to ground, so current returns to the supply via circuit ground. <A> This is only a shift in notation to simplify circuit design especially in electronics. <S> One must be sure of the fact that fundamentals are not at all different. <S> There is current flow when there is potential difference. <S> Current flows from higher to lower potential. <S> This notation is basically extension to the concept of positive terminal used to provide voltage and negative is grounded with other wires. <S> In fact if you redraw the circuit using that concept you will be free from confusion.
| In that case, current flows out the positive end of the 9 V supply into the circuit ground, thru the circuit to the -9V connection, and back into the negative input of the 9 V source.
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Is it possible to boost current into a load without boosting voltage also? If a device "D" requires X watts power, and I have a power source Y that is less than the nominal voltage recommended by the device "D", can I obtain the same performance for the device by somehow boosting the current such that Y·(boosted current) = P? I know this is a very simplified question that doesn't take alot of factors and practical limitations into account, but rather than getting an answer to my specific problem I would like to see the idea generalized.If this can't be generalized, maybe some rules of thumb could be given along with some edge cases where it does not work could be helpful. Note that I won't have any issues with too high current burning out the wire(awg size can be adjusted). My main issue is that I do not have a high enough voltage as the device recommends and wanna know how much I can underpower and still get satisfactory performance. <Q> Along with P = <S> V <S> * <S> I you also need Ohm's Law which states that V = I * R. <S> In the case of a resistive load the voltage and current are proportional. <S> The answer, therefore, is no. <S> Note that since V and I are proportional that power changes with the square of the voltage or current. <S> So if you drop your voltage to 90% the power will go down to 81%. <S> (0.90 <S> * 0.90 = 0.81.) <S> Will you get satisfactory performance at 81% power? <S> It depends on the device. <S> A heater might work OK although it will take longer to heat the room. <S> An electronic device may not operate. <S> A motor may not have enough torque to start. <A> Device D needs X watts so providing source <S> Y can provide a 10% more power than device D needs you can use a boost regulator to step up the voltage to suit the voltage required by device D. <S> The 10% more power accounts for the losses in a boost switching regulator. <S> In other words if Y can supply a bit more power you can convert this (at 90% power efficiency) to give you the right voltage and the right current. <S> My main issue is that I do not have a high enough voltage as the device recommends and wanna know how much I can underpower and still get satisfactory performance. <S> Totally unanswerable but who cares - use a boost regulator and get the right voltage. <A> can I obtain the same performance for the device by somehow boosting the current <S> You cannot 'boost' the current. <S> The device will draw whatever current it decides to at the lower voltage. <S> My main issue is that I do not have a high enough voltage as the device recommends and wanna know how much I can underpower and still get satisfactory performance. <S> Totally depends on the device. <S> It may still work at a lower voltage, but with lower performance. <S> Or it may draw more current trying to compensate for the lower voltage, perhaps damaging itself. <S> Or it may not work at all. <S> For example, if the device is a DC electric motor then it will spin slower. <S> You might try to compensate by working it harder, increasing current draw above normal and causing it to overheat. <A> Yes, you can exchange voltage for current (and vice-versa), but there is always some power loss in the process, so the total input power will be somewhat higher than the power consumed by the end device. <S> With AC power, this is accomplished by using a transformer. <S> The turns ratio determines the voltage and current ratios from input to output. <S> Some extra "magnetizing" current is required on the input, and there are resistive losses in the coils and the core that generate waste heat. <S> With DC power, this is accomplished by using a DC-DC converter, which converts the incoming power to AC and then uses a reactive element (coil, capacitor and/or transformer) to make the transformation before converting the power back to DC at the new voltage and current. <S> There are losses associated mainly with the switching elements that generate waste heat. <A> Simply put, of you can't do what you are asking for. <S> For example, if a device requires a DC input voltage 10 volts and consumes 10 watts, then the input current will be 1 ampere. <S> If you only have a 5 volt supply, you can't force the device to draw 2 amperes so that the total power remains 10 watts. <S> Depending on the device, if you supply it <S> 5 volts it will either draw only 0.5 ampere (if it is a purely resistive load), or more or less current if it is some other type of load. <S> It probably won't work at all or not as well as with the specified input voltage. <S> If you don't have the required voltage, then you need to use some kind of conversion device between your supply and the device to get that voltage. <S> You can't change the characteristics of your device to suit your supply (at least in general).
| The only way to increase the current is to increase the voltage.
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Notation on schematic for circuit with n-channel switch I am designing a circuit with a load that will consume up to 2A @ 5V. I decided to use a slide switch rated at 100mAh to drive an N-channel MOSFET for the (relatively) high current, since a slide switch rated for 2A is significantly bulkier. I understand that when using N-channel as a switch it is connected on the low side between the load and GND. I'm just wondering about best practices for separating the circuit GND from the load GND that needs to go through the MOSFET. I'm guessing I should use different net names for those two GNDs. Which one of the grounds should I call GND? The external "raw" one? Or the one that goes through the MOSFET? I already have a dozen GND nets/symbols in my circuit that should be going through the MOSFET Should I use different symbols for the two grounds? Or same symbol with different net names? Does anyone have an example schematic with a similar circuit I can look at for reference? I am mainly asking about best practices and ensuring maximum readability. Thanks! <Q> If I understand correctly, you are asking what to call the net on the drain of your low-side N FET and what to call the net on the source of that transistor, correct? <S> That's very fine to do, but check my warning further down about the danger of accidentally bypassing the transistor. <S> I would call the GND that is connected to the source of your transistor as "the" power GND. <S> I would not give it its own GND symbol unless it is truly spanning multiple sheets. <S> If it is, I would have one schematic sheet with a large square drawn around the circuit element that connects it to GND through your FET. <S> As an engineer I generally follow GND on a schematic to its source, which is usually the power supply page that ends in a connector of some type. <S> If I saw some other GND symbol, I would follow it through the schematic until I find the tie / relationship to GND. <S> You should try to make it as clear as possible that any other connection between LOW_SIDE_LOAD and GND will effectively bypass your switch. <S> This is one of the dangers of low-side switching. <A> You can't separate the grounds when using an NMOS as a low side switch. <S> At least not without introducing some other component (like an optocoupler) between the control logic and the switch. <S> The control signal has to set the FET's gate-source voltage, and since the source is also connected to the load's ground, the control voltage has to be set relative to that also. <A> The low side of your load will not be GND. <S> It will be a switched line. <S> Depending on the characteristics of the Mosfet it may actually sit at a voltage slightly above GND potential. <S> Either allow the system to assign the node name, or create a new name, (eg. Switched_XX), if you need to link it to another point. <A> Here: simulate this circuit – Schematic created using CircuitLab R1 prevents possible damage to your MOSFET <S> (keeps G-S current low, just incase).R2 is a pull-down for your MOSFET's gate. <S> The voltage at the drain is opposite to your gate (gate goes high, drain goes low and load turns on).
| I would assign a net-name to the node on the drain as "LOW_SIDE_LOAD" or similar.
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Area under voltage vs time discharge curve Will the area under a voltage vs. time graph of a battery discharge curve (with a constant current load) give the amp-hour capacity rating of the battery? <Q> The area gives you volt-hours and not ampere hours. <S> A manufacturer uses a constant current load and whilever the battery is able to produce above a "certain" minimum terminal voltage the hours that count the amps will keep ticking. <S> This, however does not mean that it will suit a specific target design. <S> Here is an example of a discharge curve for a lightweight PP3 9V battery <S> : - Note that they define the end voltage and this means "end of test" because to drag more enrgy from the battery is going to either make it unsafe, unusable in 90% of target applications and render it impossible to recharge (if it were a rechargeable type). <S> You have to set an end limit that is meaningful <S> or you'd be conning the public. <S> Look at this AA battery: <S> - After just over 2 (and a bit) hours of 100mA discharge it's not a 1.5 volt battery any more. <S> Another thing to watch out for is how they have used a constant current source - calculating ampere hour rating at 1mA gives you a better figure than at 100mA. Buyer beware and read the spec. <S> Take a look at the Duracell MN1500 battery <S> and if you have any questions on this raise them. <S> It should give you a good idea what a good supplier does to test their batteries. <A> No, in your case it is not the area that tells you the capacity, but the length of time before the voltage is so low that the battery is considered discharged. <S> For example, if the battery started out a 4.0 V, then decreased over 1.9 hours to 2.7 V when you considered it empty, the capacity is 1.9 hours times the discharge current. <S> If the discharge current was 1.3 A, for example, then the capacity was (1.9 h)(1.3 A) = <S> 2.5 <S> Ah. <S> The area under the graph is proportional to the total energy the battery delivered. <S> Actually its merely the time integral of volts. <S> But you said that the current was constant, so volts is proportional to power, and the time integral of power is energy. <A> Will <S> the area under a voltage vs. time graph of a battery discharge curve (with a constant current load) give the amp-hour capacity rating of the battery <S> [emphasis mine, N.A.]? <S> No. <S> Not even close. <S> Even units of measurement would not match. <S> Here's how to get Ah rating for the battery from voltage vs. time chart. <S> One needs one more piece of information: discharge current. <S> It's usually written on the chart. <S> ( source ) <S> Let's say that the cutoff voltage for the battery in this chart is 6V. Under this voltage, we consider that the battery is depleted. <S> At 8A discharge current, it will take roughly 20 minutes to discharge the battery down to 6V level. <S> 8A <S> * 0.3h = <S> 2.4Ah <S> (notice that the supplier advertised 3Ah). <S> (By the way, here's a somewhat related thread with a different set of charts.)
| It's not the area under the curve but the partially discharged (final) level of the voltage that is important thing.
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How to break apart grooved PCB I recently got a set of 20 breakout PCBs. I didn't realize that they came in one piece- that they had not been cut apart yet. However, there are grooves between each board that are cut into the material. The grooves extend about 1/3 of the thickness through on each side, leaving about 1/3 left of circuit board in the middle. How do I break these circuit boards apart? Is a forceful slice from a box cutter appropriate? Can I snap them apart with my bare hands? Or should I use a saw or a bandsaw to cut them apart? <Q> Bare hands should be sufficient. <S> Grab the boards solidly and bend. <S> Do so in a consistent fluid motion, and hold them such that the boards themselves won't bend much. <S> You might have to work the seam in the opposite direction a few times to make them separate. <S> Doing it by hand is fine. <A> These are typical scored multi-up boards. <S> You should break these in whole lines from end to end, then snap off the individual pieces. <S> Don't try to break out a single board from the large piece. <S> One method is to place each score line over the edge of a straight table then apply hand pressure to the section hanging off the table. <S> That should give a clean break. <S> Note that any dust or scrap generated from separating the boards may contain glass fibers (a potential health hazard) <S> so vacuum up any mess and/or use a damp cloth to clean up well. <A> But given the small size of these boards, you may not get enough leverage. <S> In which case I would line the jaws of a vice with cardboard to protect the boards, and hold the boards with a score line right at the top of the jaws. <S> Then press close to the join with a bending motion, and repeat for each strip of boards. <S> Breaking boards off a strip should be no problem. <S> A fine file might be better than sandpaper for cleaning up the edges.
| I would probably break the boards into strips, then assemble a strip at a time (while there's enough material to hold!), then break the strip into individual boards. Bare hands would normally be fine. There are depanelling tools for these sorts of things, in the form of vertical wedge presses, or wheels similar to pizza cutters, but you only really need those for situations where you can't bend without damaging the boards, or when dealing with large quantities.
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Why isn't my LED slowly dimming and eventually turning off? I have a 22uf capacitors hooked up to a 5 volt power source to power a 3.2 LED I am using a 220 ohm resistor to bring down the voltage, why isn't the LED slowly dimming and then turning off? I had it running for a while. I also have it connected in parallel. I am pretty new to circuit boards and all. Also when would it be appropriate to add a capacitor? like in what situation on a small circuit board should I use one <Q> When the capacitor is powering the LED, the circuit is an RC circuit with an exponentially decaying voltage. <S> The LED is only going to stay lit for as long as the voltage remains above the forward drop voltage. <S> Since capacitor values are not very precise, and the LED doesn't produce much light at low currents, we can estimate the amount of time the capacitor can power the LED as the time constant of the RC circuit. <S> It's not the exact time, but it will give you a ballpark estimate of the decay time for the fade out. <S> The time constant for an RC circuit is the product of capacitance and resistance: $$\tau= <S> RC=(220\Omega)(22 <S> \mu \text{F}) \approx <S> 4\text{ms}$$ <S> The time to fade is only about 4 milliseconds! <S> That's why you aren't seeing the LED fade off, because it happens too quickly. <S> There are two things you can do. <S> The first thing you might try is increasing the size of the resistor. <S> This will make the time constant longer, but it will also limit the current, making the LED dimmer. <S> This will increase the time constant without effecting the brightness of the LED. <S> You can figure out the value for yourself, but I'm guessing you want a fade out time of around 1s, which means you need a capacitor roughly 1000 times larger than what you have currently. <A> Well basically what is occurring assuming a simple RC circuit is your tau is too small meaning your circuit's capacitor is discharging relatively quickly. <S> Can you provide a diagram and maybe we can see whats occuring more clearly? <S> Not sure how this will show up but the basic equations for charging and discharging RC circuits are this.tau is defined as $$tau = <S> Resistance <S> * Capacitance$$ Charging $$V(t)=V_{0}(1-e^{-t/\tau }) <S> $$ Discharging $$V(t)=V_{0}(e^{-t/\tau }) <S> $$ <S> Your current discharging EQ is \$5e^-((t)/(0.00484))\$ essentially this means your LED will only remain on for approximately 0.0095 seconds. <S> Which as you can see is not a long time. <S> If you increase tau you slow the discharge rate. <S> If tau is small. <S> Then the discharge rate is close to e^(-x) which is a rather fast discharge. <S> This is a lot of basic circuit stuff but let's get a diagram <S> and we can see whats causing it. <S> The simplest solution would be to use a larger capacitor <S> just so you know <S> If you calculate it and graph a curve in your favorite software. <S> i.e wolframalpha or MATLAB, and then see how long in seconds the voltage remains above the point where the LED stays on you will know what to expect. <S> I can go more indepth but at this point I need to know more about the circuit. <S> Also you connected what exactly in parallel? <A> Once the capacitor is charged to the LED forward voltage almost immediately after power is applied, the capacitor has no effect on the circuit until power is removed. <A> If your circuit is something like this: simulate this circuit – Schematic created using CircuitLab <S> then you have 2 problems: 1) 5*(22µF)*(220Ω) = <S> about 24mSec charge time , where your discharge time depends upon R2 (I recommend R2 = <S> R1 <S> so discharge and charge time are the same). <S> 2) Power supplied to the load (collector side of the transistor) & to the control circuit (R and C at the base of the transistor) is the same . <S> So, when your disconnect the power , the collector loses its positive voltage and can't pull current through the LED, regardless of the base voltage. <S> If this is the case, try something like this instead: simulate this circuit This way, when you disconnect the power, C1 (make this as large a cap as you have, probably) will continue to source current & discharge through the LED over a time of approximately: 5*(1000µF)*(1kΩ) = <S> 5sec ... <S> Hope this helps.
| With the circuit you have shown above, the capacitor will delay the LED from lighting for a very short time after you apply power, an may keep it lit briefly after you remove the power, but as shown, with no switching, the capacitor has no effect. The other thing you might try is using a larger capacitor.
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Charging a battery from both solar panel and wind generator I am a complete newbie to electronics. I have taken up a project that basically aims to generate power using solar and wind energy due to personal interest and have done research but I seem to have hit a dead-end. I have a 10 watt solar panel, 17.28 V (Vmp) that gives me current in the "AC" form DC. I have used a stepper motor with the following features: 2 Phase Rated Voltage : 12V Rated Current : 0.33A Winding resistance: 32.6 Ω Winding inductance: 48 mH UPDATE: I just went over the specifications again of the stepper motor, and it is a DC motor. The DC motor is 12 volts output, The solar panel as mentioned before is also the about the same. I know how to produce electricity using the motor as well after a bit of research, i used a propeller that turns the motor to produce power now. The question now is, i have two DC sources and I want to charge one battery with it that will store the energy produced. How do I do it safely and connect both to one battery ? Thanks for putting up with me, although i sound very stupid i am still trying to learn. Thanks and merry xmas !! <Q> simulate this circuit – Schematic created using CircuitLab <S> You're going to end up doing something like this. <S> The solar panel will generate a current and the charger circuit will regulate the current to the battery by regulating its output voltage. <S> The wind generator will generate a voltage output and its charger will regulate current to the battery in the same way as the solar charger. <S> A pair of diodes could be used to prevent each side back-feeding into the other when it's not supplying. <S> In practice the diodes may cause a little difficulty in sensing the true battery voltage - but you're a long way from this point. <S> My advice is to build each circuit separately and power a simple load such as a bulb or small DC motor. <S> Progress then to charging a battery with each one separately. <S> When you get each working you can then tackle the combination of the two. <A> Your question is too broad to permit a response here except to provide few general recommendations. <S> You have good specifications for the solar panel. <S> Learn how to interpret those specifications. <S> Vmp and Imp are the voltage and current at the maximum power point. <S> Vmp X Imp = Pmak, the maximum power that the panel will produce when pointed directly at the sun on the brightest day. <S> Do research on maximum power point tracking. <S> A stepper motor is probably not the best choice for use as a generator. <S> You should use a permanent-magnet DC motor (PMDC) with brushes and a commutator or a 3-phase, brushless DC motor (BLDC). <S> PMDC motors are used in many battery operated toys, tools, appliances and vehicle accessories. <S> BLDC motors are used in computer fans, printers, disc drives etc, but they often have electronic circuits built into them in a way that prevents them from being used as generators. <S> If you can connect directly to the phase coils, BLDC motors work as AC generators. <S> A micro grid is a system of solar panels, generators, batteries, charge controllers etc. <S> to supply a network of loads. <S> Do more research on micro grids and their components. <S> Some other possible components are buck converters, boost converters and inverters. <A> First of all the solar panel produces DC and the stepper motor AC. <S> You will need to convert AC to DC. <S> Also for solar-panel you need a controller that will step up the voltage of the solar panel to match that of the battery if needed <S> a silicon cell has an open circuit voltage of about 0.7v, and make sure that the voltage produced by the turbine matches that of the solar panel. <S> Then you can connect the two DC power sources in parallel. <S> Also you will need some protection circuitry for your generators. <S> For a beginner this is no easy task <S> so I will say do your hw and take time to set it up safely. <S> Also make sure you read the specs of your type of battery chemistry and do some research on it that way you don't damage it. <S> Ps sorry I couldn't find how to comment here I am a beginner.
| Learn all you can about your individual power production methods before considering how to combine their outputs.
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Can probing an EEPROM with a multimeter corrupt the data? Say there's an EEPROM on a board and you happen to try probing the pin one by one with a multimeter, is there a good chance that the sequence of probing the pins must have overwritten (thus corrupt) the data? I always thought it was fairly safe as for EEPROMS, from what I've skimmed through, the writing happens in hundreds of kilohertz or above. I was using the continuity tester in my multimeter that tests at around 0.4V. <Q> No, a multimeter putting out 400 mV shouldn't be able to erase or write a typical EEPROM. <S> You can easily get charged up to a few 100 V without even being aware of it. <S> Since your total capacitance to elsewhere is quite low, the total charge it takes is also low, and you may not even notice when you discharge yourself against something grounded. <S> Most ICs have diodes and other protection circuitry to be able to deal with some static discharges, but of course these only go so far. <S> Most current EEPROMs store data as tiny charges trapped on the gates of FETs by a insulating layer of basically glass. <S> These charges are very tiny, and a good zap to the chip can make a mess. <A> OK, another big dumb brain fart. <S> Thank you PeterJ for answering, though it was curious that you would delete your answer immediately. <S> A continuity tester is basically shorting the nodes to test for shorts... <S> It wasn't a problem with EEPROM, but with the fact of probing it while powered on (<- ok for most of the time) AND testing for shorts. <S> I've made too many mistakes tonight (including spilling solvent on my only working multimeter), so <S> I should probably hit the sack now before I do any more damage. <S> Though, if anybody have any suggestions as to what might've happened or any general tips, please feel free to post. <A> I use multi-meter to work out the pinouts for motor vehicle ECU's, which have EPROM and EEPROM chips. <S> Never had any issue, even with infineon tri-core. <S> The only way I know of that damages them, is using the incorrect voltage and current when trying to bdm flash or boot mode.
| If you observed corruption after poking at the EEPROM with a meter, it was probably due to static electricity on you or the meter.
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Given the current and voltage of a consumer electronic device, how to calculate its power draw? Given a device that uses electricity from the wall socket (220-240V AC@50Hz or 110V AC@60HZ), and it has its current draw specific (i.e 1A), how can I calculate the power (in watts) that it consumes? I know that for DC current, the calculation is very simple (Current multiplied by voltage) but I understood that for AC the calculation is rather more complicated. Also, its quite unreasonable that my charger, that uses 1.5A AC, would consume 345W and only output 19V * 4.74A = 90.06W (255W turning into heat) and it won't get extremely hot. Also, it is rated above 80% efficiency, which doesn't meet the previous calculation. How can I calculate an electric device's power usage given its current and AC voltage? <Q> Your DC formula is correct: P = V * I. <S> The formula applies in AC too, provided the load is resistive. <S> Things get a little more complicated if the 'power factor' is less than one as is the case with motors, for example. <S> The formula then becomes P = <S> V <S> * I <S> * PF. <S> Your laptop power supply should have a power factor close to one so we can stick with the basic formula for this. <S> The input current quoted will be worst case and may be heavily influenced by the in-rush current on first powering up. <S> On a switched mode power supply the maximum current will be drawn at lowest supply voltage. <S> If you check the ratings you will probably see that it's rated for 80 - 240 V AC. <S> As the voltage goes up less current will be required to provide the rated power. <S> Conversely highest current will be drawn at low voltages. <S> If you can borrow a plug-in watt-meter (wall adaptor type) you can read the actual watts being drawn by your load and compare with your calculated values. <S> Most will also give you the amps and power-factor too which can be quite instructive. <A> Also, its quite unreasonable that my charger, that uses 1.5A AC, would consume 345W and only output 19V <S> * 4.74A = <S> 90.06W <S> That would be unreasonable but that doesn't happen. <S> The 1.5 A of current is at the lowest AC supply voltage that your charger can tolerate and still work. <S> From what you are saying that sounds like 110V RMS. <S> So now your input power will be 165 watts <S> but that's still sounding a bit high <S> so maybe your charger can operate at the "generally accepted" "world-standard" of 85V to 265V. <S> At 85 volts RMS and 1.5 amps, the input power is 127.5 watts and this is a power efficiency of: - Power efficiency <S> = 100% x output power / inputted power = 71%. <S> Also, it is rated above 80% efficiency, which doesn't meet the previous calculation. <S> If it is rated at above 80% efficiency this may be at the 230V range with efficiency dropping to 71% at the lower end of the range. <S> Or you may have measured the value incorrectly. <S> Whether DC or AC, instantaneous power is instantaneous voltage x instantanous current. <S> You then find that if you average the instantanous power over time, it becomes real power of which your can be billed for. <S> Here are various scenarios: - How can I calculate an electric device's power usage given its current and AC voltage? <S> is a device that multiplies the waveforms of voltage and current. <S> These days power supplies like yours are usually what is known as "power factor corrected" but if yours is an older type supply, the current and voltage waveforms may both look very different and not be truly in-phase. <S> That makes simple power estimations problematic. <A> AC Power is calculated as RMS voltage times RMS current (provided the voltage is in phase with current which it typically is). <S> General formula is: \$ P_{AC} = <S> V_{AC-rms} \cdot <S> I_{AC-rms} \cdot <S> PF \$ PF is Power Factor. <S> For example, my laptop charger has 1.2 A max current draw but it works with voltages from 100 V to 240 V (all rms). <S> Output power is 20V*4.5A = <S> 90W. <S> Assuming 80% power conversion efficiency, the power draw on the AC side is 113 W, which is close to the minimum voltage of 100 V times the maximum current draw of 1.2 A. Charger cables/fuses are specified for maximum current. <S> Lower voltage <S> (as in the US) results in heavier current and higher voltage (in Europe) results in lower current but the components remain the same. <S> The current is specified to design the power distribution panels and power strips. <S> It's good to know how many chargers (worst case) can be connected to one, say, 110V 15 A outlet. <A> It is not just motors that have a low power factor. <S> Low power factor is also caused by electronic devices that distort the supply current waveform. <S> The current marked on a product is the maximum that could be drawn and may vary considerably with actual operating conditions. <S> Because of those factors, any calculation of power used is likely to be higher than the actual power. <S> Inexpensive plug-in wattmeters appear to be reasonably accurate. <S> The best way to accurately determine power usage is to measure it. <A> The problem of higher current ratings required on many power supplies is related to the fact that current is absorbed from distribution net in narrow pulses around the peak voltage. <S> That is due to the input filter capacitor that sinks current pulses of high peak value to keep the input voltage leveled. <S> The new PFC corrected supplies should avoid this problem by distributing current over the whole cycle of mains voltage. <S> This is different from reactive power factor of motors because power is absorbed in very short pulses while motors have only a delay between voltage and current. <S> If you have a 100 W charger working with 90 W load you can see peaks of about 2 to 3 A at 220 V with a duration of 1 or 2 millisecond over about 8 to 10 milliseconds of half wave cycle of mains voltage. <S> The problem is much greater with 110 V lines because current pulses are very high respect to medium value. <S> To do a quick computation in case of dc power supplies you should start backwards by multiplying output current by output voltage increasing the product about 5 to 10 % to consider conversion losses.
| If you want to do it accurately you have to use a wattmeter and this gives you true average power - at the heart of any wattmeter If you want to measure it, there are cheap digital power measurement instruments on the market at about 10 $ on chinese websites.
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Excessive temperature on cheap soldering iron - will series capacitor fix it? Soldering iron came in a $14 USD kit off eBay. Lots of neat stuff, but the iron became so hot as to badly damage a DIY Kit. After a 2nd chance w/ the same results, it was hacked, Heating element of 216 Ohms directly connected to two "hot" wires of VAC ! Has to be runaway current. After much research but before any purchases, it was decided to add a CURRENT LIMITING Xc Capacitor of X2 class to one of the wires, 6uF should do the trick to maintain 5oo mA and 120 volts @ 60Hz. Will this work? Did lots and lots of research, and is better than impedance of shaving off 1 to 10 watts to heat energy. The capacitive reactance becomes the current limithing "resistor", but cannot find any schematic or datasheet to show how it is properly done. Please advise. <Q> Tools like that just aren't safe. <S> They're that cheap for a reason. <S> I recommend ditching that iron and spending a little more on something that's built a little more properly. <S> Hakko and Weller have some simple entry level irons that cost anywhere from 25-50 dollars. <S> They're also a lot safer. <S> Hakko has a nice kit that is a full soldering station with a bunch of tips if you feel like expanding your tools a bit, but that starts around 100 dollars and might be more than you want to spend. <A> the problem is almost certainly not thermal runaway, nichrome wire (which is what almost all heating elements are made from) has a very small positive termperature coefficient of resistance, the resistance increases slightly, but not enough to bother with. <S> if you reduce the power to 15W it'll be usable, but still not as good as a thermostat controlled iron, and will be slow to recover whilst being used. <A> simulate this circuit – <S> Schematic created using CircuitLab Basic dimmer circuit. <S> Note that the 216Ω resistance value will increase as the iron heats up. <S> Your dimmer will not be calibrated and will have to be adjusted by trial and error. <S> You're working with mains voltage. <S> Be careful and check the earth connection has low resistance between the element case and the mains plug earth pin.
| If money is a problem and you can't afford a better iron then a lighting dimmer switch will reduce power to this iron (or a better iron). 216 ohms on 115V is about 60W, 60w is going to be too hot for use on electronics without some sort of thermostat to regulate it.
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Why should we use LED driver chips instead of much simpler structures? For example: I have two led driver chips: bcr321u and RT8474AGSP And think they both support to same output voltage and same output current etc. My question is about topological differences of them. BCR321U is definetely looks simpler and it is cheaper. So, why do I need to use RT8474AGSP? When you consider, they both need much voltage supply and they both need an adjusted resistor for proper current flow, but RT8474AGSP configuration takes more area and so more and more expensive. In addition to that, this configuration is also an option: If we were constrained by supply voltage, we would use boost led driver chips however, they all are buck structured so why we should chose RT8474AGSP over bcr321u and bcr321u over this simple opamp - npn structure? Where does the value of RT8474AGSP lies? <Q> Here's ahow the data sheet describes the chip: - With the internal 500kHz operating frequency, the size of the external PWM inductor and input/output capacitors can be minimized. <S> High efficiency is achieved by a 100mV current sensing control. <A> Just to add to this. <S> The main advantage of the RT8474AGSP , as already mentioned, is the efficiency gained due to a PWM controlled current source. <S> The BCR321U & the OPAMP solution drop the excess voltage in a pass transistor (inefficient) but as the original poster pointed out "it uses less parts" ... <S> that may be the case but that power that the BJT's are dissipating needs to go somewhere either into an additional heatsink or into over sized copper planes. <S> This also limits the operating ambient such linear solution can operate at. <S> There isn't really a right and wrong solution (just bad implementations). <S> The two methods both provide means to vary the intensity & are naturally tolerant of supply perturbations (to a certain degree) & a range of supply voltages. <S> As the RT8474AGSP is a switching topology you may find the EMC concerns on the supply side, if poorly managed, are a real concern. <S> Likewise the additional area for heatsinking and/or operating ambient is a concern which would rule out the linear method. <S> Or the input voltage may be quite wide that rules out the linear solution. <A> One good reason is listed on the schematic. <S> The dimming is achieved by pulse width modulation internally and results in low power losses.
| Another good reason is the driver's efficiency: - It's a current limited switching regulator there fore there is hardly any power lost in current limiting resistors or transistors unlike the other three circutis shown in the question.
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Help me identify this audio connector on a turntable I have recently purchased a West German NECKERMAN 830 record player (1960s), which requires a separate set of speakers for audio output. Could somebody tell me the technical name for this cable so that I can purchase the correct counterpart? <Q> It appears to be a 5pin DIN plug http://elektrotanya.com/philips_n4506.pdf/download.html http://www.ebay.com/bhp/bang-olufsen-din http://www.leadsdirect.co.uk/technical-library/pinouts-wiring-diagrams/din-midi-wiring/ <A> This will carry the audio signals from the pickup cartridge to a separate amplifier with a 'phono' input. <S> The signal levels are very low (because it's generated by the needle wiggling in the record groove) and the bass frequencies need to be boosted. <S> * Stereo amplifiers of the period were all fitted with phono pre-amplifiers to do this. <S> You may need to purchase a separate phono preamp to boost the signal to AUX IN levels on a modern hi-fi. <S> Pinout 1 <S> Left <S> +4 <S> Left - 2 Ground5 <S> Right <S> -3 <S> Right + <S> The DIN plugs were common on German equipment. <S> Phono plugs were more common on American and Japanese products. <S> You may need to cut the DIN plug off ane solder on some phone plugs. <S> Phone Plug LeftTip: <S> Left +Sleeve: <S> Left - and half the ground wire strands. <S> Phono Plug RightTip: <S> Right <S> +Sleeve: Right - and half the ground wire strands. <S> The signal levels and frequency response were defined by the RIAA standard. <S> You should see this referenced in any decent preamp. <A> It's a 5-pin DIN connector, probably a 180 degree version <S> (i.e. 45 degrees between each of pins)
| As others have pointed out, it's a 5-pin DIN connector. The corresponding socket was often found on cassette recorders of the same vintage, and the output levels will either be "phono" (very low) or less likely "line" (low) and insufficient to drive speakers without a separate amplifier.
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What standards or conventions are there for BOM line items for non-refdes parts? Some designs have parts that may lack a reference designator because they never appear on the schematic. Examples include: Battery holders PCB-mounted (soldered on) fasteners such as solder-nuts IC sockets Jumper wires on a single sided board (if they haven't been backannotated to the schematic, that is) Plugin modules such as keypads and LCD modules (if they haven't been given a schematic symbol) Heatsinks Are there any standards or conventions for how these are listed in a Bill of Materials for a circuit board? <Q> If you don't, you'll constantly be manually fixing the BOM that is automatically generated by your tools. <S> This applies to all the parts that belong to the immediate built board assembly. <S> At the least, these are the parts that get soldered or otherwise mounted to the board. <S> If the built board is part of a larger assembly, then the BOM for that assembly will show the built board as a single line item, and it can call out screws, nuts, standoffs, etc, for mounting the board into the larger unit. <S> And yes, this applies to sockets too. <S> For example, here is a snippet of a larger schematic for a real commercial product that exists in the company's formal manufacturing database: <S> P11 is a Phoenix Contact connector that accepts a plug with screw terminals. <S> Only P11 actually gets mounted to the board, but there always needs to be a plug with the board. <S> That's what PLG2 is. <S> It is there primarily for the purpose of automatically appearing on the BOM. <S> In this case I think showing it on the schematic is also a nice visual reminder that the board comes with a screw-terminal plug. <A> In general, what I have seen at slightly larger more organized companies is that they try to put all the BOM items on the schematic. <S> The reason is that the BOM is generated using the EE CAD. <S> So the BOM will be more complete if all those things are included. <S> The details of how you do it need to be tailored to your specific CAD tools and manufacturing data management software. <S> I would definitely try to capture anything that is installed on the circuit board in the schematic somehow. <S> If necessary, you can assign it a reference designator. <S> No harm. <S> For plugin modules, it might be a judgement call. <S> If you have a hierarchical BOM, the module could be a subassembly under the PCB BOM. <S> I am generally not dogmatic about these kind of things. <S> The key is to find a workflow that reduces the chance of errors (wrong part, failure to order parts, etc). <A> At my company, we have strong opinions on how to do this from both the mechanical and electrical teams. <S> The detenté we have settled on is that for primarily EE-centric assemblies (I.e. PCB assemblies), we will add mechanical components to our docs to centralize BOM generation. <S> As we use Altium, the simplest method is to toss parts on the schematic, and set them to 'Mechanical'. <S> They are then not included in electrical rule checking, nor are they present during layout -- they only appear on the BOM. <S> Most of these items are placed on the last page of the schematic (revision history) and include things like fasteners, the blank PCB itself, conformal coat, threadlock, heat sinks, etc. <S> The blank PCB gets a PCB? <S> Refdes, and we decided on MP? <S> for the others -- no conflict with traditional reference designators and all parties understand what they mean. <S> With this method, the EE team owns that assembly, and generates (automatically) BOMs that include everything we need, the MEs need and the systems team needs. <S> As far as symbols go, the PCB placeholder is a simple square and one of my artful colleagues made representations of various fasteners, Etc that we place there. <S> Finally, just to throw some fuel on the fire, we put our plated mounting holes on the schematic only (not on BOM) so that we can see what their electrical connectivity is.
| The best way to handle parts that don't appear on the schematic is to fix the schematic so that they appear.
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Is it possible to use a potentiometer as a DP3T style control? I'll try to be specific: I have an audio application that has 3 inputs and 1 output. The classic answer would be to use a DP3T to switch between the inputs. However, I would like to have a continuously variable output, one that blends as you sweep a potentiometer. Say for instance if the pot was fully clockwise it would give you the signal X, at the midway point, the signal Y, and counterclockwise the signal Z, but between noon and clockwise you would also get a blend of XY. X and Z would never blend, just meet with Y towards the middle of the pot throw. Is this something that can be done with one pot and VCAs, or a one pot mixer? I'd like to stay away from multiple gang pots, unless that is the only answer because my application requires vertical PCB mount and conformity of potentiometers. Thanks! <Q> This is certainly possible, and is fairly straightforward, but not terribly simple. <S> Let's assume your pots are in the 10k to 100k range, and you have VCAs with a 0 to 5 volt control input. <S> Then simulate this circuit – Schematic created using CircuitLab will do what you want with a single pot. <S> Note that TL081s are shown here, and will work fine, but <S> TL084 quads will also do, and only 2 packages are needed. <S> If you use something like this, the op amps need to be driven with +/- <S> 9 <S> to +/- <S> 15 volts. <A> It sounds like you mean that you will hook up you 3 signals to the 3 inputs of the pot, and take the output between one of those connections and ground. <S> No that won't work - <S> if you draw out the schematic you'll see what I mean. <S> In the full clockwise position you have the pot's full resistance between signals X and Z, so you'll probably get some odd combination of all 3 signals depending on where you tap the output. <S> Honestly I have never seen an application like the one you describe, most mixer designs like this use an active buffer, like a simple op-amp (741 or 301) with a pot on the feedback resistor. <S> the outputs are tied together. <A> if the pot was fully clockwise it would give you the signal X, at the midway point, the signal Y, and counterclockwise the signal Z <S> Yes this can be done and what springs to mind is a mixer where the three summing resistors were somehow varied to give the mix you want. <S> But they can't be simple summing resistors like this: <S> Basically, if you control the duty cycle of an analogue switch the audio signal you put in at one end gets multiplied by the average of the duty cycle and, providing you have a reasonable (but not excessive) low-pass filter to "clean-up" the "chopped" audio signal, you get a variable resistance ranging from 0 ohms to infinite ohms. <S> What next springs to mind is the Linear Technology LTC6993 pulse width modulator. <S> It can take a linear voltage demand and produce a pulse-width modulated signal like so: - <S> So now you can chop up all three audio signals with a linear control voltage using the above chip and three-quarters of a quad analogue switch. <S> You'll also need a bit of filtering on the op-amp summing amplifier (a capacitor across R1). <S> The switching can be at 1MHz and this is 50 times greater than audio <S> so the filtering is probably just a capacitor. <S> To finish the design you need to take the potentiometer and connect it across a stable voltage so that the pot wiper output is (say) 0V to 5V and, between 0V and 2.5V you mix X and Y with Y being dominant at 2.5V. Between 2.5V and 5V, Y loses dominance to Z so that at 5V on the wiper Z rules the roost. <S> You'll probably need a comparator here to differentiate between upper and lower parts of the pot control. <S> The LTC6992 PWM device you need to use is in the following table: - I'd choose the 0% to 100% device because at extremes it fully switches on or fully switches off the analogue signal it seeks to control. <A> The basic element of your system will be a voltage controlled amplifier (VCA).You would need six of these to control your three stereo channels (or three for mono). <S> simulate this circuit – <S> Schematic created using CircuitLab VCA based on AN-6603 application note . <S> Next you'll have to monitor the pot with a micro and use the signal to give three analog outputs to each of the AGC pairs. <S> simulate this circuit <S> The code for the micro would have to generate the pulse-width modulation (PWM) output signals as follows. <S> (Pseudo code assumes analog in goes from 0 to 100%, 100% being fully clockwise. <S> // <S> Output at CCW <S> Midway <S> CWPWM1 = <S> (50 - AIN) <S> * 2; // <S> 100 <S> 0 <S> -100if <S> (PWM1 < 0) PWM1 = 0; // 100 0 0PWM2 = 100 - ABS(AIN - 50) <S> * 2; // <S> 0 <S> 100 <S> 0PWM3 = <S> (AIN - 50) <S> * 2; // <S> -100 <S> 0 100if (PWM3 < 0) <S> PWM3 = 0; // <S> 0 <S> 0 <S> 100 <S> The PWM outputs are passed through low-pass filters to give a smooth DC control voltage to the VCA cells. <S> I have absolutely no idea <S> how much distortion or noise the VCAs would introduce and I haven't worked out weather <S> the VCA would give the correct logarithmic response to simulate a log-law potentiometer. <S> Any problems with the log-law could be sorted out in the software with a little work. <S> All analog solution <S> It's probably possible to create the VCA control signals from the potentiometer using analog electronics only. <S> simulate this circuit <S> The potentiometer wiper goes from -5 to +5 V and is (optionally) buffered by OA1. <S> OA2 and D1 form a basic precision rectifier which will pass any pot signal from mid to +5 end. <S> 0A3 inverts the pot signal and feeds it to the second precision rectifier built around OA4 which will pass any signal from mid to -5 end. <S> Finally OA5 is an inverting summing amplifer. <S> With the pot at mid position the voltage on the top end of R6 will drive OA5 output to +5 V. <S> If either of the rectifier outputs start to rise OA5 output will fall. <S> This satisfies the requirements for the midway channel. <S> Now tell us what on earth you are doing! <A> The simplest method I know of is to use a center tapped potentiometer. <S> Although center tapped potentiometers are very rare now days.
| It's possible but probably most easily with some digital magic. - You need to have electronically controlled resistors; consider what happens when you use an analogue switch that is turned on and off at a high frequency with variable duty cycle.
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Designing a gate voltage for N-channel MOSFETS for H-bridge controller We have a project where our 2nd year students build an H-bridge controller from scratch. They learn a lot about power dissipation. We have a simplified design using IRF540 for the N-channel, IRF9540 for the P channel. These are way over-spec for the application, tiny robot motors. WE use 6V batteries, so to turn on the gate of the N-channel transistors we can use the output of the Arduino, and to turn on the gate of the Pchannel we set gate to 0V, or leave in a high-Z state with a pullup resistor. This works. However, we were thinking of extending the exercise to show all N-channel. That's the way it's done at larger scale because the N-channel has such dramatically better on-resistance. I think the difference is 0.004 ohms vs. 0.113 ohms, so it's significant. This would require that we generate Vcc + 5V for the gate. For a teaching circuit, with a power side of 12V, is there a simple, clean way to generate a tiny current at a voltage 5V higher than Vcc? We would like to have them build the circuit, not just use a black box. <Q> 5V is barely enough to turn on an IRF540. <S> The specified 0.077 Ohms Rdson is achieved with 10V Gate drive. <S> You are probably getting away with the lower drive voltage because your motors are only drawing a small current. <S> The simplest way to get a voltage higher than the supply is to make a 'charge pump' using a couple of capacitors and diodes. <S> You also need a driving waveform with sufficient voltage and current, which could be produced by a 555 timer, CMOS gates, an op-amp etc. <S> Here's an example circuit:- simulate this circuit – <S> Schematic created using CircuitLab <S> This circuit should produce about 22V (10V of boost) at 5mA. Note that this voltage exceeds the FETs Gate voltage rating, so you must limit the Gate-Source voltage to less than 20V. <A> I believe your best bet is a dedicated MOSFET driver IC designed to feed high-side switches. <S> Yes, a discrete charge-pump solution can be used, but these devices integrate a charge-pump internally and are designed for applications such as yours where you wish to drive an enhancement-mode N-channel device in such a fashion. <S> They have logic-level inputs intended to take in PWM-type signals. <S> I am aware this is not exactly what you asked for, but I think it's the solution you are looking for unless I'm missing the point of your teaching lab. <S> Apologies if I'm missing the point of your question, but I think this type of IC is what you are really after. <S> I normally use the LM5113 in my application since I use WBG GaN transistors, but for conventional MOSFETs, something like the LTC1155 (from this page: Linear Tech MOSFET Drivers ). <S> It's particularly neat because they have on-die capacitors for the charge pump, saving you external parts. <S> If it does not meet your spec, Linear has others on that page that probably will, and Allegro, IRF, Infineon, etc. <S> certainly will as well -- a high-side MOSFET driver IC is a very common product and is found from many different manufacturers. <S> You can choose single, dual or quad channels versions of a driver IC based on what you need. <S> Some even pair a high-side and low-side in the same package in case your microcontroller (or other) input signal cannot deliver sufficient gate charge / turn-on / turn-off requirements, which become more important at higher switching frequencies. <S> You would use two of those ICs total in a H-bridge configuration. <A> At a 12V supply, you're not even exceeding the Vgs limit on your mosfets. <S> So you don't need to do anything fancy. <S> Pull up resistors and nmos driver are just fine. <S> The outputs on on the 2n7000. <S> This also has a lot of static dissipation in the mosfet drivers. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Also of note: If you use BJT push pulls and a voltage level shift (common source/emitter) you can dissipate a lot less power, but you need more transistors. <S> The whole "No Free Lunch" thing. <A> Easiest way is likely a charge pump voltage doubler. <S> As long as you only need a few mA, there are many that will do the job. <S> An example: the LTC1144 , which will supply 50mA and has an input voltage range of 2-18V. <S> If you need 17V for the gate drive, per the datasheet, you can feed it 12V and get ~22V out, which you can then regulate via a small LDO down to 17V. <S> (Note that, unfortunately, the common *660 charge pump inverter/doubler IC is not appropriate, as Vin is limited to 5.5V, giving only 11V out.)
| To be honest, you can pick just about any N-type transistor device.
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Circuit behaves erratically when using a long cable I've got a circuit, it is a dimmer that a friend of mine designed, it is providing energy to 15 meters of 3528 led strips. I'm using a 12v 5A power supply, and as long as I connect the PS directly to the circuit, it works just fine. However, when I connect the PS to a cable (1.5 meters long) and then I connect that cable to the circuit, it starts to behave erratically, sometimes it doesn't even start, other times it reach full power and then it shuts down. I have the circuit details if you need them, just wanted to quickly check out if there is something obvious that I'm not taking into account, since I have barely basic knowledge of electronics. My brother in law (who designed the circuit) is on vacations unfortunately, so I can't consult with him. Ant help is appreciated. EDIT: <Q> Circuit + LEDs = <S> fine Circuit + cable + LEDs = <S> bad <S> The logical conclusion is that the cable is causing your problems ! <S> No use looking at the circuits <S> just yet although it is common sense to include all information right from the beginning. <S> The power supply is 12V, 5A and you have a lot of LEDs. <S> So probably a lot of current is flowing. <S> Let's assume it almost 5 A, for 5 <S> A <S> you need a decent cable for it not to drop too much voltage. <S> What sort of cable are you using ? <S> Do you have a voltmeter which you can use to measure the 12V ath the other end of the cable (when the LEDs are on because only then current flows). <A> Probably the circuit has inadequate input capacitance on the supply. <S> You are adding to the supply output impedance by adding the cable. <S> A large-ish possibly low-ESR capacitor directly at the circuit input may calm things down. <S> Try at least 1000uF/16V. <S> If there is already one there, increase it significantly (like 5:1). <S> Where is your brother-in-law that he can't be contacted? <S> There are few places that Internet (perhaps via Iridium sat phone) cannot reach. <A> Wires have inductance. <S> When the current passing through an inductor changes it produces a voltage. <S> The faster the current changes, and the greater the inductance, the greater the voltage that will be produced. <S> What effect will this have on your circuit? <S> If the cable is short then its inductance will be low, so any voltage spikes produced probably won't be large enough to cause any harm. <S> At some point the spikes may become large enough to upset the controller. <S> Here's an LTSpice simulation that represents the essentials of your circuit. <S> L1 represents the inductance of the power cable. <S> R6 represents the regulator and its load (MCU). <S> With L1 = 0.1uH (representing a short power cable) <S> I got this:- <S> You can see a little spike each time the LEDs are turned on, and bigger spike when they turn off. <S> However the maximum and minimum voltages are well within a range that the regulator should be able to handle, so the MCU should still get a stable +5V. <S> A 1.5m length of 16/2 lamp cable has an inductance of about 1uH. <S> Here's what happened when I put that value into the simulation:- <S> Now the voltage spikes are extreme, going up to 24V and down to 1V. <S> The voltage regulator will not be able to create a stable +5V from this dirty power! <S> Finally I added a 47uF low ESR electrolytic capacitor across the supply, in parallel with C1. <S> Now the spikes have completely disappeared! <S> (the remaining square wave is just a small voltage drop caused by power supply and cable resistance):-
| A longer cable has more inductance, and therefore produces bigger spikes. Your circuit controls the brightness of the LEDs by rapidly switching them on and off, so the current is cycling between 0 and 5A.
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Switch PWM signal to two destinations Sadly i am not really experienced with electronics, i have recently obtained an Arduino and a motor driver and i am trying to hookup some lego motors.My motor driver outputs 2 signals, one for direction (high/low) and one for speed (PWM).My lego motor (Power functions medium) has 2 inputs for PWM signals, one forward and one reverse. So i was thinking i could connect these by sending my PWM signal from the motor driver to one of the two inputs based on the direction signal voltage from the motor driver.What kind of component could do this job? I have been looking into using a transistor as a switch but i am not sure if i will be able to make it work like i want. <Q> So upon looking at that motor, I believe you can not power it directly with Arduino PWM power, you need to buffer the signal because the PWM is directly powering the motor. <S> The problem here is that digital pins on an Arduino (Atmega 328p and all other microcontolers) are designed to only provide a small amount of current (typically < 20mA). <S> What you really want is an H Bridge system, this is a chip or device that typically takes a single PWM signal and a direction signal. <S> An example of one that would work for you is the L293, Sparkfun and Adafruit have both information and motor control boards based on these chips. <S> The good news is that they still are made in DIP packages so you can buy some and drop them in a bread board. <S> Here is an example using the L293 chip mentioned: <A> Assuming these PWM outputs are 5v TTL-like on both the Arduino and Lego sides, I'd go for something like a 4016 transmission gate with an inverter arranged so that with your direction line high, the PWM signal will be routed to this motor; with it low, it'll get routed to that motor. <S> This will leave the 'inactive' motor with a floating input, which may not be what it's expecting. <A> Hopefully I interpreted your question correctly. <S> It can be made with "logical gates" (shown at the top), or it can be built from scratch with N-channel MOSFET transistors (shown at the bottom). <S> Link to interactive circuit on Falstad
| I'd advise that you get the Arduino/Lego interface working in one direction to start with (using just one of the motors and ignoring the direction line) before trying to get a bidirectional circuit going.
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Best method/circuit for converting 240VAC to 5VDC (Low power) In short, I'd like to power Arduino-like devices (small PIC devices as well) run off of a mains socket at 240VAC - which obviously can't be done with a simple wire. I assume I'd need an output anywhere between 3 and 5 volts DC. I don't want to use a generic plug power supply as they're usually designed for higher current outputs (I'd probably need 500mA at most - the processor and a couple of LED indicators), can cost a reasonable amount and I'd like to up my electronics knowledge/DIY skills by creating something myself (even if it is just a couple of soldering joints). Also, when I say "best", I mean either the simplest to build or most efficient (or a combination of both). My apologies for the generic question, I've just never done this sort of thing and I'm not sure how to word it properly. <Q> For safety, you'll need to use a transformer. <S> You probably can't find one that is smaller than 500 mA @ <S> 5 V capable. <S> If you use a simple bridge rectifier and 7805 LDO, you'll have a system that is larger, less efficient and has worse wasted power than nearly any simple cellphone power adapter. <A> This is what is in every USB charger ever made... <S> So why design it again when you can just pick one up for 5$ somewhere, the design will be long, and if not done right, it could be deadly. <S> Keep in mind 5V @ <S> 500mA is not really 'low power' that was the normal USB supply for the last ~10 years. <S> That is 2.5W... <S> I'll also add they are fairly efficient. <S> A well designed USB charger with synchronous rectification should be >80% efficiency. <S> You will be hard pressed to get this level of efficiency from your first custom design... <A> The whole conversion would probably be best done with a simple SMPS topology like @MadHatter suggested. <S> Flyback is probably the simplest and integrates the transformer into the design for you. <S> You can get some off-the-shelf parts and build it yourself or just grab the parts out of a 5V USB charger. <S> Is fabbing a small PCB out of the question for you? <S> Would probably be a pretty quickly design.
| You can use a transformer for isolation and safety, and then rectify it with a full-bridge before doing DC-DC conversion. The way you would need to do it is design a flyback SMPS - Check out some datasheets online because you may just be able to get a few samples from digikey.
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How to power a circuit that needs +9VDC, GND, -9VDC? While I have an assortment of 5V, 9V, 12V wall-warts, but I don't remember ever seeing one that is 18VDC and trying to avoid a trip to specialized parts shop, since I have all other components of a circuit (of a microphone mixer) that I am building. Here is the circuit in question: (Source: http://www.aaroncake.net/circuits/mixer2.asp ) As one can see, it requires +9VDC, -9VDC and GND. I've never built such circuits that have negative potential (I know it is relative). Wondering if someone can help illustrate / explain, if it is possible to power this somehow using a 9VDC wall-wart ? <Q> You will probably find that two 9V wall warts will do the trick - connect the positive output from one to the negative output from another - this connection is the "new" 0V <S> and you'll have +9V and <S> -9V available - regard it as two 9V batteries in series with the middle (common) connection being called 0V. <S> Because wall warts produce isolated output voltages, they can used like this but always double check with a meter first to ensure everything looks and feels OK. <S> Looking at your op-amp circuit you should be able to get aways with a single 12V wall wart and a rail splitter made by two resistors across the supply. <S> The junction of those resistors will be 6V up from the negative rail and ALL <S> the points marked with an "earth" symbol on your circuit can connect to this point. <S> This is a much used technique but, as always, it won't suit every target circuit but looking at your circuit it should be OK. <A> If you have a transformer-based PSU available you can probably convert it from single to split-rail as shown below. <S> simulate this circuit – <S> Schematic created using CircuitLab Note that each rail is now only half-wave rectified <S> so I'd recommend some large caps or voltage regulators to eliminate hum. <S> If you want to keep a standard jack on the PSU then convert it to an AC PSU by removing the rectifier and capacitors and put the diodes and caps into your project case. <S> This configuration can also be used to make an 18 V PSU. <S> simulate this circuit <A> Such as these from Microchip: <S> http://uk.farnell.com/microchip/tc7660hcpa/dc-dc-volt-converter-7660-dip8/dp/9762698 <S> I tried these DC/DC converters in an audio project with single supply limitation (guitar effects pedal) and they worked a treat. <S> Here's an example of a simple audio project that uses the 7660, hopefully it will be of some use to you! <S> http://www.generalguitargadgets.com/effects-projects/distortion/kc/ <A> If you still want a dual supply after considering these other great ideas, you could build a simple voltage doubler, either by using a Cockroft-Walton type circuit, or a push-pull supply, with maybe a couple of analog voltage regulators thrown in. <S> The first approach is not a very stiff supply, but is probably adequate for your circuit's load. <S> To get to a split supply from the doubled supply, follow the previous responder's suggestions. <S> For the second approach, a transformer is needed but something that works is not hard to come up with. <S> Just get a big core (e.g. ferrite cup-core or toroid) and fill it up with trifilar windings (i.e. 3 wires wound together). <S> Pay attention to the phasing, make sure your switching transistors are driven hard enough to saturate, and use the clock frequency to control droop leading to ripple. <S> If you want, you could estimate each winding's inductance from the A(L) value for the core. <S> The output is square-wave a.c., so a split supply is generated by single rectifiers with capacitive filters. <S> Note that there are also IC's made to generate a negative supply directly from the positive supply, by capacitive switching. <S> Using this you would have a split supply right off the bat. <S> I suspect that this approach is also adequate for your circuit's load. <A> An option is to use a DC/DC converter. <S> You can use a single +/- <S> DC/DC converter such as XP Power here . <S> DC/DC converters will have switching frequency noise (10's of mV), so you can use a PI filter coupled with simple linear regulators (+/- <S> 9V), to reduce noise. <S> Select the DC/DC converter based on voltage, current load, switching noise frequency, and power-supply ripple rejection. <S> Set the DC/DC voltage higher than drop out of the linear actuators. <S> For high precision a good reference design and suitable (TI) chips can be found here .
| If you're wanting to produce a split-rail from a single supply, you might find DC-DC inverter ICs useful.
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Is the given voltage Peak Voltage or RMS Voltage In circuit diagram problems is the labelled voltage RMS voltage or Peak Voltage ? For example say in problems like this what is the usual convention? I recently started learning about circuits and i'm in high school.I don't know if the question is suited for this site or not.But I thought of giving it a try.Do let me know if it's off-topic. <Q> RMS should be assumed unless otherwise indicated. <S> In this problem, "effective current" should be assumed to mean RMS current and the supply voltage should be assumed to be 200V RMS. <A> In this problem, it doesn't matter. <S> As long as you consider all the voltages in the circuit to be the same (RMS or peak) (or even peak-to-peak), all your math will work out. <S> To my knowledge, there is not a well-established convention for this in basic circuits problems. <S> In "real-world" problems, it is dependent on context, but should be labelled clearly. <A> As uint128_t said, here it does not matter since values are going to be calculated with phasorial analysis and not with differential equations. <S> This eventually will matter when you need to calculate the exactly value of any variable in any specific time, for example how much electrical current is stored into the inductor, in time t. <S> So in that case you will need to use differential equations in the time with sinusoidal analysis and the source voltage should be previously clarified (peak or RMS). <S> I will highly recommend you the Sadiku's Book <S> , I think it is really good for self learning, go to chapter 9.4 Phasor Relationships for Circuit Elements: http://faculty.weber.edu/snaik/EE2260/alexander_sadiku_fundamentals_of_electric_circuits_4thed.pdf <A> In most cases it is rms and if it is peak it is mostly clarified separately to the student. <S> Also as previously stated the use of term "effective" corresponds to rms valueof parameters of AC circuit.
| Normally the "real-world" convention could be to use RMS voltage for this kind of electric circuits.
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Is it a good habit to route wire under chip resistor/capacitor? Is it good/safe to route under resistor/capacitor? As below, it's a 0.5mm wire under a 1206 chip resistor. <Q> It depends. <S> The problem might come from crosstalk. <S> If you have a sensitive analog signal, and you pass it between a 1206, and the 1206 is part of a digital circuit, <S> well you may get some coupling. <S> However, this is true regardless if placing a trace between a component. <S> Anytime you have any traces < 2-3 trace widths apart, will have cross talk. <S> How much so, depends on frequency, plane distance, geometry etc.. <S> Fortunately, from what in your picture, its a very short distance, and the coupled length is only the width of the pad. <S> This should not be a problem if its a digital signal. <S> If its analog, then it would depend on what your requirements are for that signal. <S> How much noise can you handle before the going out of spec. <S> So it safe to do so, from a manufacturers perspective, but will it affect your circuit would depend. <S> If you stick with the, keep analog and digital away rule, then you wouldn't have much of a problem. <A> Yes, you can do this as long as it passes the DRC (design rule check). <S> In fact, some designers use a 0-ohm resistor to "jump" over a trace, to avoid having to make have traces on the bottom of a PCB.. <A> I do not know about any habits, customs or commandments telling not to do so. <S> As long as your DRC are correct, there is no problem, I think. <S> I myself used this trick very often in production - exactly with 1206 - to save the vias (as they really wanted bit more space than I can save changing to smaller resistor. <A> Note that PCB manufacturers have some tolerance in the soldermask layer registration, so there could be maybe plus <S> /minus 0.010 inch offset (depending on PCB fab house). <S> If the trace is too close to the pad, and the PCB soldermask opening exposes the trace <S> , there's a possibility of a solder short between the trace and the pad. <S> For smaller components, like EIA size 0402, I think our company's CAD team adds a soldermask line to the part outline, to help separate the two pads. <S> This supposedly helps for those cases where the pads are so close together that there's one big soldermask opening instead of two separate openings.
| As long as there is enough clearance between the trace and the pads, it should be OK. If you have a high density board, you might have too, otherwise it means you need a via, or have to take some irregular path on your board.
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Safety when using ATX power supply as benchtop supply I found many pages where people convert an ATX power supply to a benchtop power supply to give 12, 5, 3.3V. However the current output can be very high (several amps). When using this and you NEED high current, that is great. However, when using this for typical benchtop hobbyist needs (e.g. supply 5V to a microcontroller) the current supply needs are small. I understand that it will draw as much current as it needs -- but my question is one of safety. Since the current "compliance" is essentially set to several amps, what happens if I accidentally touch + and -? Or what about if something in my circuit suddenly blows up and shorts + to -? Then I suddenly have several amps running thru my circuit. I believe then an internal fuse in the ATX unit will probably blow (right?) but at least for some short length of time there were several amps running thru my resistor right? or Wrong?Would anybody have a simple adjustable current limit circuit that I could build and add to my PSU? <Q> It's quite far from certain death (indeed, it's voltages "generally recognized as unlikely to be hazardous", but we just had a big long discussion of that starting from a more hazardous and less useful premise. ) <S> Summary - you're a lousy conductor, you won't carry several amps at those voltages. <S> Doesn't mean you can't make a big messy spark, or start a fire, but it won't electrocute you if you leave the box closed, unless you walk off into very marginally possible or probable scenarios. <S> The simple answer if you want a small-amperage supply is some in-line fuse holders with fast-blow fuses of whatever size floats your boat. <S> If a quarter-amp suits your needs, it does not matter if the supply can provide 50 amps. <S> You won't pull much more than 1/2 amp for a very short period before the 1/4 amp fuse vaporizes. <S> The short-term probably more expensive, possibly long-term cheaper option is DC circuit breakers. <A> A typical 600W ATX power supply is specified to supply up to 25A from its 3.3V and 5V rails (and 46A at 12V). <S> This is an immense amount of current — if shorted, it will probably melt your wiring. <S> When short circuited, the behavior of the power supply will be unpredictable. <S> A good power supply will shut off. <S> However, this behavior is not guaranteed; less expensive power supplies may attempt to continue supplying voltage, or may fail dramatically. <S> In short: do not short-circuit computer power supplies. <S> If you need a bench power supply for experimenting, I highly recommend that you purchase a "real" lab power supply — a 30V/5A single-rail model can be had for under $100. <S> Such a power supply will allow you to monitor current, and explicitly set a current limit beyond which the supply will either limit the voltage or shut down. <S> It's a worthwhile investment. <A> For safety reasons, I'd suggest getting a used bench top power supply on eBay that you can use current control on. <S> It's just the easiest solution. <S> If you're really set on using an ATX power supply you can build an external fuse box and then connect the input port of your project's network to the output port of the fuse box. <S> If you choose to do this, make sure you use wiring that won't melt when carrying current at or slightly above the fuse's rating. <S> This way you can use the same box setup for projects that require higher current..just make sure to change out the lower-rated fuses for ones with the correct rating. <S> I believe most Arduino's will use up to 500mA (USB limited.) <S> Fuses are relatively expensive so if you're looking for a cheap solution in the grand scheme of things and plan on doing a lot of projects, it might be worth it just to cave in and get a bench top supply, depending on how many fuses you plan on blowing :).
| If something shorts in your network then the fuse would blow before the supply can source enough current to cause serious problems. It's probably best to use higher rated (thicker gauge) wire coming from the power supply and going into the fuse box.
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LM358 as a comparator output problem I've got an LM358 based ground fault detector circuit that I've recently revised, and the revision broke it, so far as I can tell. To make a long story short(er), what it amounts to is using one of the LM358 stages as a comparator has whacky output levels. There's a 5 volt supply, and the inverting input has a 20k/100k voltage divider, for a fixed .833 volt comparison level. When the non-inverting input is 0 volts, I would expect the output to be 0. In fact, it's 4.32 volts. If I make the non-inverting input rise up above the threshold, it goes up to around 4.5 volts. Now, the old version of the circuit had a 1N4148 diode on the output with a pull-down resistor to ground after that. What is the diode and pull-down supposed to achieve? Why isn't an LM358 acting as I expect without it? <Q> Not every op amp is perfect, nor is an op amp a good substitute for a comparitor. <S> Please consider replacing it with an actual comparitor. <S> As for why you should, that op amp must be pulled down with a 2KOhm resistor or higher in order to meet TTL logic specs (Vcc-1.5 = 3.5V in this case). <S> It's in the datasheet. <S> The Diode is there to prevent the op amp from sinking current and even out the H->L and L->H transition time. <S> As for why your output is not working. <S> I'd have to see a diagram. <S> From what you describe it should be working. <A> We cannot guess what the function of the diode might have been without a schematic. <S> The LM358 output swings very close to the negative supply if it is only called upon to source current (perhaps 50mV). <A> I've been using the CA3140e IC with no probs. <S> You can also buy it with dual op <S> -ampNote: The o/p of the comparator does go to 0v - unlike others!Attached simple example...
| If the output is not connected to something else then there is something wrong with the LM358.
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How the pin2 of 555 timer triggers this simple circuit? First of all, I will go through my concept which make the doubt. When we connect positive side of LED to a positive terminal of a battery and negative of LED to floor (not to -ve of battery), current will not flow, as -ve of battery is not at a potential equal to that of floor (0V). But if I hold by standing on floor of phase line of commercial electric circuit, I get shock because neutral line is at same potential as ground. (Correct me if I'm wrong.) Now look to this circuit: (source: circuitsgallery.com ) Figure above shows a touch sensor using 555 timer. It work only if a low pulse applied to pin 2. But how when we touch the plate it get low pulse? Although we consider -ve terminal to be 0v, in practice, I think it is not. As like my first example, I expect the current never flow and pin 2 stays at high voltage. But it is not so. Then what is my mistake?Pls help me I am a beginner.Sorry for my English. EDIT Some answers tell me "how the 555 works" or "how sensitivity of circuit increase". But my question is different from that. May be this is due to unclear data from my question. My question is how the pin 2 trigger (or the voltage at pin 2 become less than \$\frac{1}{3}V_{cc}\$ ) when I touch the plate? <Q> Three things to note: <S> Your body can be modeled as a resistor. <S> The touchpad is likely made up of 2 pads; one connected to pin 2 of the 555 IC and the other connected to ground. <S> Pin 2 is being pulled up by R3 which is 1M ohms. <S> This means that it is extremely weak. <S> When your finger makes contact with both pads of the touchpad, the circuit sees essentially a resistor between pin 2 and ground. <S> Using the voltage divider equation:$$1/3 = <S> R_{finger}/(R_{finger}+R_3)$$$$\therefore R_{finger} = 500K$$ <S> The resistance provided by your finger needs to be less than 500K ohms, which is quite achievable if the 2 pads are close together. <A> I remember when I was in high school, I built a kit that works exactly like this. <S> I remember being disappointed that it didn't work for my intended purpose: I wanted a touch sensor hidden below a car's door handle for security purposes. <S> I digress... <S> The idea is fairly simply. <S> Pin 2 has a very high input impedance. <S> The 1M-ohm pull-up resistor thus pulls it up almost all the way to the supply rail, thus preventing the 555 from triggering. <S> The circuit relies on the ground of the circuit ACTUALLY being connected to real ground. <S> Now, when you touch the touch plate, your body (and the earth) creates a resistance that's much lower than 1M-ohm. <S> It pulls pin 2 down to below a third of the supply, which is what the 555 needs to trigger. <S> The 555 then pulls pin 3 high for the period as determined by R1 and C1. <S> I forgot how you do that <S> and I'm sure you know how that works. <S> For interest sake, a human body has a resistance of around 10k-20k. <S> I'm not sure about earth, might depend on how recently it rained :-) <A> You don't actually need a negative pulse but take the lvl of pulse less then 5/3 Volt due to comparator action. <S> The specific O/p of plate depends on type of h/W used. <S> If I'm correct in the diagram you are exploiting the monostable operation of the 555 timer. <S> Thus on negative pulse on 2 qbar of ff becomes 0, by effect on comparators and sr flip flop capacitor C1 charges to level of 2/3 Vcc and then quickly discharges. <S> This allows an internal transistor(pnp type) to be on "high" till charging occours and qbar to be on 0 thus allowing a 1 on O/Pport 3 which powers on LED. <A> Personally, i would add a resistor shown as Rx. <S> Its value will be chosen in conjunction with R3 such that the resting (untouched ) voltage will sit near the trigger level point, see pic...
| The 555 is triggered by taking pin 2 lower than 1/3 of the supply voltage.
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Why more, smaller transistors increase power efficiency? Transistors, bjt, MOSFETs OK got it.More transistors = better computing got it. But compressing the transistors closer to each in my mind only helps reduce the physical dimensions. So does a CPU or any electronic become more efficient because the transistors use less voltage? Does more computing reduce power usage, thus simply having more transistors the reason? I am asking because as a newb and soon to graduate engineer, I think basic stuff like this is important to understand. But I always learned this concept as a rule of thumb and not by "first principle" or actual theory of transistor efficiency. PS I did take a class in where the math of L and W was calculated and compared to new L' and W', reduced on a npn. The theoretical frequency increased but I don't think the math translation well in my head because I don't see how that helps power efficiency, only performance and/or area. <Q> Most of the power consumption in CMOS circuits is 'dynamic power' -- power from devices switching state. <S> This power is basically the power required to drive the capacitance of the other gates, as well as the capacitance of the wiring. <S> In modern CMOS, there is additional static leakage power because when the MOS transistor is 'off', it still allows a little leakage current to flow, and given the numbers of devices in a modern CPU, the total current becomes significant. <S> There are techniques to minimize this though. <S> As devices get smaller, you get a number of benefits: <S> Device capacitances decrease, so less power is required to drive this capacitance. <S> Devices are smaller and closer together, so parasitic capacitance of the wiring also decreases. <S> Devices can be made operate at lower supply voltages, so both leakage current, and power required to drive parasitic capacitances decreases. <S> Generally, the leakage currents increase as devices get smaller (not really because they are smaller, but because the lower threshold voltages allow higher leakages). <S> More complex power-switching techniques are used to keep this in check. <A> The dynamic power consumption is caused by gate capacitances charging and discharging. <S> The power is proportional to C * F <S> * V^2. <S> C is gate (or process) <S> capacitance, F is frequency and V is voltage. <S> Since the V term is squared, reducing voltage greatly reduces power consumption. <S> Smaller, lower-voltage transistors have thinner insulation layers at the gate, so they have more static current leakage ( <S> even though the material is a good insulator, the path is very short, so there is some leakage). <S> I remember at one time, maybe around 2000, there was talk that static and dynamic power would cross over, and the static would become more significant than dynamic for high-end processors (intel PC's and such). <S> I think they figured out tricks for reducing static power, though. <S> And I sort of stopped paying attention to that area of industry. <A> Up to a limit, smaller transistors helps to reduce voltage drive requirements because your gate oxide is thinner and therefore the gate control is stronger due to the gate being closer to the channel. <S> Both voltage and current being lower results in lower power requirements. <S> The limitation of this scaling is when your gate oxide thickness is so thin that quantum tunneling occurs and the channel length is so short that you have to battle short channel effects.
| Smaller transistors also helps reduce capacitance which results in lower dynamic drive current.
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Meaning of strong and weak drive in VHDL? What is the meaning and effect of "strong" and "weak" drive shown by (0,1) and (L,H) in VHDL's package ieee.std_logic_1164? <Q> The ieee.std_logic_1164 package allows two drive strengths. <S> Formally, ieee.std_logic_1164 distinguishes between the std_ulogic and std_logic types, even though they share the same set of values (U,L,H,0,1,X,Z etc). <S> The difference between them is this : std_ulogic is an unresolved type. <S> It is an error for a std_ulogic signal to be driven by two signal sources (like short circuiting 2 wires together - it's usually a mistake) - such errors will be caught by the compiler and fixed even before you get to simulation. <S> On the other hand, std_logic is a resolved type. <S> It is legal to have two or more drivers on a signal. <S> The result is determined by a "resolution function" which looks at all the driving values and combines them into the value you will see on the signal. <S> For example, 0 and '1' combine to produce X (the Unknown state) , and <S> X and anything else produces X so that once Unknowns happen, they propagate through the design to show there's a problem. <S> But '0' <S> and 'H' combine to produce '0' (because H is weak) and similarly, '1' and 'L' combine to produce '1'. <S> So a strong driver can legally overpower a weak one : no harm done. <S> Let's look at the 'Z' state : it signifies Undriven, or High Impedance (High-Z). <S> Now Z and H combine to produce H because H , though weak, is still stronger than no driver at all. <S> This is just a brief summary : for all details see a good VHDL book or even the ieee.std_logic_1164 package source code. <S> So one use of a weak H is in the so-called "wired-or" configuration, as used in I2C buses. <S> SDA <= <S> `H`; -- permanent pullup resistorSDA <= `0` when I2C_Master_SDA = '0' else `Z`;SDA <= `0` when I2C_Slave1_SDA = '0' else `Z`;SDA <= `0` when I2C_Slave2_SDA = '0' else <S> `Z`; We are safely combining 4 drivers on the same signal : one I2C master, two slaves, and a permanent weak pullup. <A> This is one of those cases where IEEE Std 1076-2008 gives a concise answer. <S> 16.8.2.2 <S> The STD_LOGIC_1164 values <S> The logical values '1', 'H', '0', and <S> The resolution function RESOLVED treats the values '0' and '1' as forcing values that override the weak values 'L' and 'H' when multiple sources drive the same signal. <S> The values 'U', 'X', 'W', and '–' are metalogical values; they define the behavior of the model itself rather than the behavior of the hardware being synthesized. <S> The value 'U' represents the value of an object before it is explicitly assigned a value during simulation; the values 'X' and 'W' represent forcing and weak values, respectively, for which the model is not able to distinguish between logic levels. <S> The value '–' is also called the don’t care value. <S> This standard treats it in the same way as the other metalogical values except when it is furnished as an actual parameter to the STD_MATCH functions in the IEEE.NUMERIC_STD package or as an operand to a predefined matching relational operator (see 9.2.3). <S> The STD_MATCH functions and the predefined matching relational operators use '–' to implement a “match all” or “wild card” matching. <S> The value 'Z' is called the high-impedance value, and represents the condition of a signal source when that source makes no effective contribution to the resolved value of the signal. <S> From 4.6 Resolution functions: A resolution function is a function that defines how the values of multiple sources of a given signal are to be resolved into a single value for that signal. <S> ... <S> You can read how synthesis treats std_ulogic values in 16.8.2.4 Interpretation of logic values. <S> 16.8.2.4.2 Interpretation of the forcing and weak values ('0', '1', 'L', 'H', FALSE, TRUE) <S> A synthesis tool shall interpret the following values as representing a logic value 0: <S> The BIT value '0' <S> The BOOLEAN value FALSE <S> The STD_ULOGIC values '0' and 'L' <S> It shall interpret the following values as representing a logic value 1: <S> The BIT value '1' The BOOLEAN value <S> TRUE <S> The STD_ULOGIC value '1' and 'H' <S> And this should tell us you don't short outputs together. <S> There's also provision for 'Z' inferring tristate buffers. <A> I don't know what that standard is, to which you refer , but 'strong' and 'weak' refer to how much current a logic gate output can provide to maintain the appropriate voltage level corresponding to what the logic level should be. <S> E.g. if a logic 1 (H) output is supposed to be 5V, that voltage will drop if you connect a heavy load to it, perhaps to the point where it will no longer be considered a valid logic level.
| (0,1) are normal drive strength and used for all normal purposes.(L,H) are weak drivers, they are normally used to model pull-up and pull-down resistors. 'L' of type STD_ULOGIC are interpreted as representing one of two logic levels, where each logic level represents one of two distinct voltage ranges in the circuit to be synthesized.
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How to "spread linearly" the force of a solenoid For a pet project of mine, I'm interested in finding a tubular solenoid which force is spread linearly over a longer length. The project is based on IA and uses a programmed controller (and sensors) to feed several solenoids (the question is for one solenoid) quickly in order to "grab" something falling vertically, from below, this silently (done) but also a bit more smoothly (the problem). Currently, while the programming seems to work, the test prototype with two solenoids lack control. Basically the solenoids force goes from 0 / 1 [didn't try to control current yet]. I bought and tried from the Net, something like but the force is basically 0 / 1... meaning the force grows exponentially over a short distance. Ideally, the force would be linearly spread over a length I decide. Thus my question Preferably with only one solenoid, how to choose/select the core and wires in order to spread the force over a longer length (i.e. not exponential growth) If 2+ solenoids are necessary please explain how you would arrange them to reach such result (1 is preferable though). Note : if that helps, the current applied to the solenoid can be precisely controlled from a chip. <Q> Solenoids don't work that way because magnetic fields don't work that way. <S> To get a constant linear force independent of position, a rack and pinion driven by a motor is a possibility. <S> Some voice coil mechanisms can be pretty constant in force with position too, especially if position feedback can be used to adjust the drive current. <A> The force of a solenoid is a function of both its position and the current. <S> If you have position feedback, you should be able to control the current to achieve nearly constant force over some portion of its travel. <S> That said, there are linear actuators that don't involve rotary motors with gears or screws. <S> They are basically linear BLDC motors, and they can be very quick. <S> The downside is that they are rather expensive and require a relatively complex controller. <A> The easiest way to find the ramping function is probably trial and error.
| It is possible to have some intermediate control of push/pull solenoids like this one by using very carefully controlled PWM signals.
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Blocking DC without blocking any frequencies? Is there any practical way of blocking DC without blocking any AC frequencies (or as few as possible)? Suppose I wanted to block DC and let through all frequencies between 2Hz and 10Ghz with minimum resistance. The first methods that spring to mind would be to either use a very large capacitor or many capacitors in parallel with different values adjusted for different frequencies. Is there something blatantly obvious I've missed? <Q> Because you asked for "practical" methods, I'll provide an answer which differs somewhat from the others. <S> Yes, a capacitor (followed by a resistor) will block the DC component of a signal. <S> And yes, if you want low-frequency rejection you'll need a big capacitor. <S> But, and this is a big but, large capacitors by their nature are physically large and have large inductive components. <S> This interferes with high-frequency performance, and I seriously doubt that you'll find a unit which will give you 2 Hz to 10 GHz. <S> About 3 orders of magnitude is a reasonable goal for low price. <S> See here, for instance for a line of RF DC blocks with a good unit reaching from 10 MHz to 40 GHz. <S> If you're willing to spend the bucks, there are specialty manufacturers such as Picosecond Pulse Labs (now part of Tektronix) which provide 7 kHz to 26 GHz . <S> A Google on DC block will provide other possibilities. <A> A single series capacitor isn't enough - you need a resistor on the output to prevent capacitor leakage putting DC on the output: - The 3dB point is defined in formula form as Fc = <S> \$\dfrac{1}{2\pi <S> RC}\$ <S> So, with R = 100 kohm and C = 100 uF, Fc is 0.0159 Hz <S> If you wanted to use those values and see what effect that might have on passing (say) <S> 1Hz use this formula: - <S> Where <S> \$\omega = <S> 2\pi f\$ <S> and for this formula f is the frequency you want to test i.e. 1 Hz as mentioned above. <A> Higher frequencies will pass with less impedance. <A> In theory an infinitely narrow slit in a transmission line will block DC and pass all AC... at least until the width of the split becomes significant on an atomic scale. <S> In practice, there is no transmission line which can pass all AC. <S> So the DC block is only half the problem. <S> The other half is finding a suitable transmission line to carry the signal from either side of the DC block.... <S> PCB traces, coax cables, and waveguides each have a frequency response. <S> Anyway, you are not missing anything simple. <S> Every low pass filter has a roll off, and so there will be significant low frequency attenuation. <S> The practical answer is to determine exactly the lowest and highest frequencies of interest and the acceptable insertion loss, then design / buy a low pass filter to suit.
| You only need one series capacitor, selected for the lowest pass frequency.
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Transistors hooked up to one another I stumbled upon a very badly drawn circuit on the interent which must be good . I have not been able to figure out how a series of transistors work in this circuit. It is depicted below. When this circuit powers, the leds turn on and stay on one by one until the last led is turned on, and then they all go off and the cycle starts over again. My question is in regards to if someone could explain to me why this behavior happen with the transistors. I am not sure how an astable pulse to this array of transistors would cause this cascading behavior. <Q> This circuit works by creating an exponential R/C ramp voltage to the base feed resistors of the stack of transistors. <S> The circuit design as shown however has some issues: <S> First off is that the sequencing of the five LEDs all happens within about a quarter of a second and the times are not uniform. <S> Another problem is that the higher LEDs get successively lower and lower current though them due to the stepping up of the saturated collector voltage on each successive transistor. <S> This will lead to LED brightness variation unless the values of the 330 ohm resistors are weighted accordingly. <S> Current variation as shown is from 14mA down to about 3mA. The LED current for upper stages in the stack has to flow through the base-emitter junctions of the transistors below it. <S> Some small signal transistors have a maximum specification of allowable base current and this could get to be a problem <S> if higher LED currents are attempted. <S> In this case the lowest transistor base current max is about 27mA and the BC547 is rated up to 200mA so should be OK. <A> Improved LED Sequencer Circuit This circuit has improvements over the posted circuit with the addition of a few more components. <S> Still all simple components!! <S> I did not include the astable oscillator in the simulation but that can easily be built with a 555 or a couple more transistors, resistors and capacitors. <S> A current source keeps the voltage ramp linear. <S> An emitter follower buffers the waveform to the threshold diode stack and all transistors switch the same current levels to each LED. <S> The net result of this is that timing steps of the LEDs turning on are much more even and the timing parameters are easily adjusted to get desired timing, whether this be for a turn signal sequencer or for a light sabre flash up. <A> pjc50 gave you an important clue, but the other thing you need to know is that the turn-on threshold for each transistor is one diode drop (V BE ) higher than the transistor below it. <S> The bottommost transistor turns on at about 1.3 V, which is its own V BE drop plus the V F drop of the diode. <S> That's a terrible schematic diagram, by the way. <S> It uses unconventional symbols for components; it has the output capacitor connected badly; and it lacks reference designators on the components, which would make it easier to talk about them. <S> See this article for some tips on drawing good schematics.
| Each transistor going up in the stack turns on at a higher voltage on the ramp due to the emitter of each being referenced to the base of the proceeding transistor.
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CJMCU-75 -- seting the I2C adress I just got embedded system CJMCU-75 which has LM75A I2C temperature sensor on top of it. I dig into the LM75A documentation and found out that it's I2C address consists of four fixed MSB and three LSB bits which can be set up externally by pulling pins A0/A1/A2 towards VCC or GND. So I looked again at my newly bought embedded system and found that designer already took care of this on the bottom of the board where he put three pads for setting A0/A1/A2 pins: Do you have any advice on how to solder A0/A1/A3 pads to VCC/GND in order to be able to change them easily afterwards. <Q> I just solder bridged the pads on mine. <S> Worked like a charm. <A> Pick a surface mount resistor that is just the right size to go from the center pad to either GND or VCC. <S> It should be low value resistor. <S> If you have 0 Ohm jumpers, that would be ideal. <S> But any low value will work fine, for example 10 or 22. <S> Solder it in the desired location. <S> Later it should be fairly easy to move or remove the resistor. <S> Make sure it is physically short so that it does not accidentally contact both VCC and GND. <S> That would not be good. <S> Good luck! <A> Apply plenty of flux to make a clean bridge. <S> I found a $5 flux pen works nicely for small jobs like this. <S> Roughly the size of a permanent marker. <S> Apply flux to each pad, a dab of solder on each, then a little more to join them into a bridge. <S> To remove the bridge later, add more flux and drag your soldering iron across a few times, or use a little solder wick between the pad and your iron to suck up any remaining solder.
| If you don't have any SMT resistors, cut off a small piece of copper wire or a small piece of a lead from a leaded component, and solder that in place as a jumper.
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Comparing current from two 9V batteries I have two 9V batteries, an Energizer and a very cheap Gettop that was bundled with some equipment. The Energizer is down to 7.2V, the Gettop is at 9V. If I connect each of these separately to an LED via a 560 ohm resistor, the Energizer's current through the circuit is 7.1mA, while the Gettop only manages 1.1mA. Can someone help me understand what is going on here. Is the chemistry of the Gettop limiting the current? Is this telling me to avoid the very cheap batteries, or only to know their limitations? Thanks for reading. <Q> As a battery goes flat, the voltage will slightly reduce. <S> But at the same time, the internal resistance rises substantially. <S> So even if the battery gives 9V with no load, as soon as you connect a load, the output voltage drops. <S> So if you're testing how good a battery is, you need to apply a load, and then measure the voltage. <A> You've just discovered that alkaline batteries are not carbon-zinc! <S> The unloaded voltage for a 9V alkaline batt will decrease as it's drained, while a (much cheaper) 9V zinc battery will show 9V even when "dead." <S> Also, a nearly-dead alkaline battery at 7.2V might still supply a hundred mA when shorted (zero volts,) but a carbon-zinc will only give a couple mA or even less. <S> To calculate the "dead" battery internal resistance, subtract 560 ohms from the VI load resistance, like so: R(int) <S> = <S> V(open) / <S> mA *1000 - 560ohm For the alkaline, internal R is: 7.2V / <S> 7.1mA <S> * 1000 - 560 = 454 ohms <S> For your zinc, internal R is: 9V / 1.1mA <S> * 1000 - 560 = 7.6K <S> (I figured out this stuff around 1992 while hooking 200 nearly-dead alkaline batteries in series, "bandolier" style. <S> 1.8 kilovolts! <S> Yes, I did manage to zap myself with it. <S> Also used a couple of nails to burn my initials into a 2X4 board in flaming carbonization.) <S> Also a secret: <S> You can recharge your expensive 9V alkalines, but only if the plates aren't yet destroyed. <S> (All batteries are rechargable. <S> But the plates of single-use batteries get destroyed near the end of their lives.) <S> If the batt voltage is still roughly 8V or higher, hook the battery up to a benchtop supply set for 10V output and 20mA current limit. <S> (Or use a 10V supply with a 100-ohm resistor in series.) <S> This will trickle-charge the battery. <S> If it works, then you'll see the battery voltage slooooowly rise to 10V over hours. <S> But if the alkaline is truly dead, it will go up to 10V in a few minutes, which demonstrates its damaged plate-area (and large internal resistance.) <S> Best is to not let your single-use batteries drop below 8.8V or so, which is the point where the battery plates are not just thinning, but being eaten away entirely. <S> It's possible to thicken the plates and partly restore a well-used alkaline battery, but we can't replace the missing plates of a truly dead battery. <A> The Gettop battery is what we call "dead". <S> Toss it and move on. <S> 7.2 V open circuit for a battery that is supposed to be 9 V is very low. <S> But again, this is open circuit. <S> The internal resistance has gone up dramatically too, so as soon as you try to load it, the voltage plummets. <S> This is a great illustration of how batteries should be tested under load, not open circuit. <A> Aged battery cells (and the 9V has N=6 series cells) , are as "strong" as the weakest cell, although the internal resistance adds and the capacitance divides by N for the C of each cell. <S> The electrical model usually drops to 10% of the capacitance and 100 to 100x the internal resistance, Ri = ESR. <S> However, memory capacitance an equivalent RC parallel network can restore the cell voltage over time but with much higher Ri = ESR. <S> But in reality there are multiple ESR's and multiple C values. <S> So voltage is a poor indicator of current* time capacity , only a load current. <S> However if a good rechargeable cell is declining in V due to a constant load, then and only then is cell voltage a good indicator of State of charge (SoC). <S> Read my answer here <S> How is it possible a dead old Lithium battery becomes charged after opened?
| Note that both the voltage and the current are indicating this battery is dead.
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Is it possible to use an optointerrupter as a optoisolator? I currently can't buy an optoisolator, so I was wondering if I could wrap a optointerrupter in tape to shield it from light and use that as an optoisolator.Is this possible? <Q> Don't have an opto-isolator? <S> Make one. <S> All transistors are photo-electric until put in a can or some opaque plastic. <S> If you have a small signal metal can transistor you can cut or file the top off and super-glue it to an LED. <S> LED superglued to 'photo-transistor'. <S> The finished article with heatshrink sleeving. <S> The transistor is a 35-year old BC108. <S> LED is a generic red - probably same vintage. <S> Watch out for light getting in the back of the LED. <S> Current transfer may not be very good so you may need to 'Darlington' the output. <A> Sure, it's possible, but opto-isolators are intended to provide isolation and usually have isolation levels and common-mode rejection levels specified. <S> An opto-interruptor has a different purpose and may not specify those things. <S> If isolation and/or common mode rejection are important in your application you may get into trouble if they are unknowns. <S> Also, your tape may not be opaque to all frequencies that the receiver is sensitive to, so you may get interference from some light/radiation sources. <S> Digi-Key has lots of opto-isolators and they ship almost anywhere so not sure why you can't buy an optoisolator, but depending on your application you may want to. <A> Yes. <S> Also keep in mind that they may be using IR, not visible, and <S> most black electrical tape is quite transparent in near IR. <S> Also, "can't buy optoisolator" doesn't make sense. <S> These things are readily available from the other end of the internet, and cheaper than opto-interruptors too. <S> It seems you are only saying this because you want the answer to be to use a interruptor.
| Most opto-interruptors have poor characteristics as opto-isolators, but it is possible.
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Piezo or vibration sensors? I see that there're vibration sensors and that piezo can detect vibrations. Which is better (more sensitive) for detecting knocks over a piece of wood? I'm looking for a small package, so I'm only looking at small piezo. I've tried tap-enabled accelerometers, but they don't seem to be very sensitive. Should piezo or vibration sensors have higher sensitivity than accelerometers? Thanks <Q> Piezo is very good for detecting vibrations, bias it to a non-zero voltage would also make it suitable for vibration in both ways. <S> You can check this post, where a few people helped me with a good circuit diagram. <S> Piezo into 0-5v ADC, with offset center voltage? <S> note : to protect the input of the opamps, use a 10k in series into the opamp input. <S> note2 : <S> with a bouncy material (like wood and rubber), it would have a lot of a small vibrations after the initial knocking, you would need to work it in the code to isolated the 1st knock to avoid multi triggers in a very short times. <A> A piezo can be used as a contact mic, and it sounds like that's pretty much what you want. <S> For good sensitivity, you may want to glue your piezo to the surface. <S> However, keep in mind that you need to protect the input of your preamp, because if a piezo receives a direct/large impact, it can generate a nasty voltage transient that will blow away sensitive op-amps and the like. <A> If you want to go digital - try ADXL312 <S> 3-axis accelerometer. <S> Cheap one, 3 kHz. <S> At ±1.5g range <S> and it's 12 bits <S> - it has to be really sensitive. <S> If you want to stay analog - you may give a try @ electrostatic microphone. <S> There is knock sensor definition in the automotive industry (engines). <S> It sounds like what you need. <S> I believe, that it's piezo based however.
| With the right amplifier, piezos can be very sensitive.
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Human touch pins on MCU (atmega328p) directly or other components needed? I would use atmega328p to read the pulses transferred in human's body. So I wonder if that's good to directly expose the pins of the chip. Or there should be some type of components for protection and noise filtering. e.g. maybe a 1n ceramic cap in series for filtering 50Hz house main noise? a 1K resistor in series for static protection? Thank you! <Q> There are two approaches that you can use that don't require much in the way of extra circuitry or processing. <S> The easiest is to use a pair of electrodes complete a circuit to provide base current to a transistor or darlington pair, and connect the emitter to the Atmel pin. <S> You could then configure the pin as an external interrupt (or maybe a comparator), or simply analog sample it, depending on what you want to do. <S> A good description of your options is at http://www.electronicshub.org/touch-sensors/ , the source of the images. <S> The second approach would be to build a capacitive sensor. <S> When the free end is touched, the noise floor will raise. <S> You would need to sample pretty fast and process it, as well as doing some analog conditioning to make the signal suitable for single-ended acquisition, and you might need some gain. <S> There are chips that do this for you that other people have pointed to that will do this job better than you and a microcontroller can achieve on your own. <A> Without acquisition circuitry, you won't read anything more significant than noise. <S> This could be useful or not, depending on your application. <S> But if you are trying to do somewhat serious signal processing and interpretation, you will definitely need amplifiers and filters. <S> As a rule of thumb, your total gain of the circuit should be enough so that the signal amplitude measured is slightly below the reference voltage of the ADC. <S> Filters to be used include notch main filters and band-limiting anti-alias filters. <S> I would like to ask you to kindly expand on your explanation of the desired application. <A> If you walk about on a synthetic carpet, you can easily discharge a short 4kV pulse into a chip and kill it. <S> So you need some sort of ESD protection. <S> But what are you actually trying to sense? <A> There are several ways to measure human touch, but since you are mentioning pulses you need to specify from where on the body in order to get a good answer. <S> If it is touch you want to sense <S> and you want a all digital solution you could look into using an extra chip AT42QT1070 ( http://www.atmel.com/images/atmel-9596-at42-qtouch-bsw-at42qt1070_datasheet.pdf ), but from the way you are writing your question i guess this is not what you want to do. <S> If you want to use the MCU <S> i suggest you connect a mosfet that can react on the potential from your body. <S> Since it's kinda hard to explain the connections <S> i suggest you look at the circuit proposed here ( http://www.aaroncake.net/circuits/Simple_Touch_Switch_Circuit.asp ) and modify it (only care about Q1, R1 and R2). <S> The drain could be connected to your input on the MCU.
| You could use a megaohm range resistor, and just connect one end to the pin and the other to the air, hanging free. If you are trying to make a (very) rough touch detector, add some static discharge protection to an ADC port, identify peaks of variation and go for it.
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Duplicate Eagle part (package) in same libraray How do I duplicate a package in my Eagle library (without grouping, and copying and pasting the contents or opening up the XML of the library in a text editor)? I know I can copy between libraries, but I can't see how to do it within the same library as there is no option to change the name. ThanksThomas <Q> Actually I just checked, and you can right-click on the package in the table of contents <S> (icon looks like a book) , then choose Duplicate which asks for the new name: <A> The lbr file is actually an XML file. <S> Packages are in elements (See example below.)Open the file in a text editor (e.g. Notepad++), make a duplicate of the package element (e.g. right under the original one), change the name and save. <S> <package name="RFM69HCW"><wire <S> x1="-8.09" y1="-7.95" x2="7.91" y2="-7.95" width="0.127" layer="21"/><wire <S> x1="7.91 <S> " y1="-7.95" x2="7.91" y2="8.05" width="0.127" layer="21"/><wire <S> x1="7.91" y1="8.05" x2="-8.09" y2="8.05" width="0.127" layer="21"/><wire <S> x1="-8.09" y1="8.05" x2="-8.09" y2="-7.95" width="0.127" <S> layer="21"/><wire <S> x1="-4.99" y1="5.95" x2="-0.09" y2="5.95" width="0.127" layer="21"/><wire <S> x1="-0.09" y1="5.95" x2="-0.09" y2="2.05" width="0.127" <S> layer="21"/><wire <S> x1="-0.09 <S> " y1="2.05" x2="-5.09" y2="2.05" width="0.127" <S> layer="21"/><wire x1="-5.09" y1="2.05" x2="-5.09" y2="5.95" width="0.127" layer="21"/><smd name="DIO2" x="6.91" y="7.05" dx="3" dy="1.2" layer="1"/><smd name="DIO1 <S> " x="6.91" y="5.05" dx="3" dy="1.2" layer="1"/><smd name="DIO0" x="6.91" y="3.05" dx="3" dy="1.2" <S> layer="1"/><smd name="3.3V" x="6.91" y="1.05" dx="3" dy="1.2" layer="1"/><smd name="DIO4" x="6.91" y="-0.95" dx="3" dy="1.2" layer="1"/><smd name="DIO3" x="6.91" y="-2.95" dx="3" dy="1.2" layer="1"/><smd name="GND_3" x="6.91" y="-4.95" dx="3" dy="1.2" <S> layer="1"/><smd name="ANT" x="6.91" y="-6.95" dx="3" dy="1.2" layer="1"/><smd name="GND_1" x="-7.09" y="7.05" dx="3" dy="1.2" layer="1"/><smd name="MISO" x="-7.09" y="5.05" dx="3" dy="1.2" <S> layer="1"/><smd name="MOSI" x="-7.09" y="3.05" dx="3" dy="1.2" layer="1"/><smd name="SCK" x="-7.09" y="1.05" dx="3" dy="1.2" layer="1"/><smd name="NSS" x="-7.09" y="-0.95" dx="3" dy="1.2" layer="1"/><smd name="RESET" x="-7.09" y="-2.95" dx="3" dy="1.2" layer="1"/><smd name="DIO5" x="-7.09" y="-4.95" dx="3" dy="1.2" layer="1"/><smd name="GND_2" x="-7.09" y="-6.95" dx="3" dy="1.2 <S> " layer="1"/><text <S> x="-0.09" y="0.05" size="1.27" layer="25" align="center">>NAME</text><text x="-0.09" y="-1.95" size="1.27" layer="27" align="center">>VALUE</text></package> <A> This command should be run from the library editor with the library you want to copy a package in to open. <S> The syntax of the command is: copy packagename.pac@libraryname newpackagename <S> Where: packagename is the name of the package to copy. <S> If you want to copy a device as well, you can put .dev <S> instead of <S> .pac . <S> libraryname is the name of the library to copy it from (including the currently open one). <S> If the library is not currently in Eagles library search paths, you can use a full file path as the library name AFAIK. <S> newpackagename is optional, and not used when you use the copy to library function in the Eagle control panel. <S> This is the argument you need to do what you want . <S> If supplied, this is the name that will be given to the copy in the library.
| This can be done using the Eagle copy command in the library editor in all version of Eagle (at least ones since 5.x).
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Using MOSFET’s For AC Rectification I recently saw this video that did inspire a little bit of interest on my part. https://www.youtube.com/watch?v=IrB-FPcv1Dc It appears that when the drain of a N channel mosfet is connect to the gate it can act as a diode. Also according to this video it can do that very efficiently. So I messed around a little bit on a very primitive inverter that I built a while back. I tried to rectify the output current of the inverter which was stepped down to 12 volts ac.So I tried to use the transistors as diodes and my results where the following.I tried an NPN and I got 8 volts of DC on my multi meter.I tried the MOSFET and got less than a volt even without the gate being connected. Simply put the Mosfet acted a lot like a wire and had no effect whatsoever.So I am curious how would you achieve rectification using transistors and is it even possible. <Q> It is possible to create a very efficient rectifier with a MOSFET - but not by just connecting the Gate to the Drain! <S> This diode is as good as normal silicon rectifier, so the FET will work quite well as a rectifier provided that it is not turned on at the wrong time (ie. <S> when the Drain voltage is positive relative to the Source, since that would nullify the blocking effect of the body diode). <S> If the Gate is left floating then it might pick up a static charge causing the FET to turn on continuously, which is obviously bad news. <S> However if the FET is only turned on when Source-Drain voltage is negative <S> then it will bypass the body diode and cut losses to almost zero. <S> The trick is to synchronize the Gate drive to the incoming AC voltage. <A> Not only is it possible, but there are specialized IC's to do just that. <S> For example the LT4320 : <A> This is not a full answer but I made a similar experiment in Falstad and here it is: Link to circuit <S> (The AC voltage sources are really only one, but I had to split it up to get the simulator to work.) <S> It is called Active Rectification .
| All discrete MOSFETs have an inherent reverse 'body' diode between Source and Drain.
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How to build Forth for embedded devices So far I have just found this paper which provide a guide to achieve that: MOVING FORTH. Part 1: Design Decisions in the Forth Kernel Is there a more recent guide? Which Forth is better for that purpose? I am planning to use it with PIC16F877A, and I know that PicForth exists, but it is just a compiler, not a system to communicate with a terminal window by a serial connection (that's why I am going to use Forth). <Q> The PIC16F877A has 14K of Flash, and only 368 bytes of RAM <S> so there is no way pForth is going to fit. <S> The PIC16F877A also has a 16-deep hardware stack used only for return addresses, and no parameter stack. <S> These restrictions would have probably ruled out using a PIC16 for pForth anyway, even if you could get one with enough Flash and RAM. <S> Based on gbarry's answer, it looks like you're going to need a much bigger microcontroller for pForth. <S> You might as well skip the PIC18 <S> (which has a maximum program size of 128K), and also the PIC24 (which also has a max of 128K) and go with a PIC32. <S> The PIC32MX170F256B has 256K of Flash and 64K of RAM, and costs under $4 . <S> It comes in a 28-pin DIP instead of a 40-pin one though. <S> You also might want to check out the following books: Programming A <S> Problem Oriented Language: <S> Forth - how the internals work by Charles H. Moore (the original creator of Forth) <S> This book is only available as a Kindle book, but you can get a free reader for the PC and most portable devices like the iPhone, iPad and Android equivalents. <S> Forth <S> Application Techniques: <S> Course Notebook , 5th Edition by Elizabeth D. Rather (who was a co-founder of Forth, <S> Inc.) <A> I downloaded pForth on @tcrosley's advice, compiled under OS X, and the standalone image is 180 K. <S> This is one of several stumbling points that has kept me from realizing a goal similar to the asker's. <S> Only I've been trying for 15 years :) <S> Most are too large <S> , some lack an interpreter, some are not in C, some wouldn't build, some expected a Windows environment, and so on. <S> The information provided by Brad Rodriguez is still applicable to building a number of FORTH systems. <S> I'm not aware of any updates, and since it's his site, that's where any updates would be. <S> I recently was referring to these papers and I am still attempting to create my own small FORTH written in C. <S> I've had a little success, but I can tell you, there's a lot in the details. <S> This is not the sort of thing you can do while simultaneously trying to make progress on an embedded project. <S> Note that FORTH originally expected to have a block of RAM where new dictionary words get compiled. <S> You either have to deal with not much RAM, or come up with a way of writing to the program flash. <S> I have also tried something called FlashForth (this one's written in assembler) and it did work, but I put it on a fairly large PIC18F6620. <S> I got distracted trying to figure out the dictionary internals and where stuff was stored, so I never got to where I could do development with it. <S> I suggest trying it if you can get your hands on something in the PIC18F family...even just to try it. <S> On that note, there are some 18F parts that are pin compatible with their 16F counterparts, so if cost is a concern, just use the bigger parts for development if FORTH isn't actually going into the final product. <A> Have a look at Mecrisp Forth . <S> I've been using it on STM32 microcontrollers. <S> It compiles to about 25K and is probably a lot of work (assembly language) to get it working on a PIC32. <S> But if you do, you'll have a fast small Forth implementation. <S> If you do, let me know. <S> It is worth a look as it is as cutting edge a Forth implementation as I've seen.
| You might want to check out pForth , which is a portable implementation in C.
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What are the terminals in this 6 pin potentiometer? I want to use my 6 pin potentiometer in circuits so I want to know what all the terminals are for. I know the basic principle in the 3 pin pot. I'm completely new so please answer it in simple terms. I have attached a circuit which I want to make which pins do I use for it. I have also checked readings with a multimeter here they are, I have numbered the pins from 1 to 6 starting from left here are the readings: Pins 1-6 shows a range of 0 ohms to 53k ohms when I turn the knob. Pins 1-2 show 0 ohms to 53k ohms. Pins 2-6 shows only 53k ohms no matter what position the knob is in. Pins 3-4 show 0 ohms to 53k ohms. Pins 3-5 show 0 ohms to 53k ohms. Pins 4-5 show 0 ohms to 53k ohms. I was thinking that 4-5 should only be at 53k ohms and not vary but it varies. Here's what my pot looks like: And here is the circuit I want to use it in: <Q> Edit: Check the datasheet for the TT Electronics P120KGE, it appears to be your exact potentiometer: <S> It contains 2 individual potentiometers, connected to a common shaft, so it has 2 wipers and 4 outer terminals. <S> In order to determine the pinout, first you will need to find the resistance of the potentiometer, which is usually marked on the back. <S> There should be 2 sets of 2 pins that show that resistance between them no matter what position the knob is in, and those are the outer terminals. <S> The remaining 2 pins are the wipers. <S> The resistance between them and the connected outer terminals will vary when the knob is turned. <A> Generally, these are dual pots used in audio circuits. <S> I would suggest going through with a multimeter and ohming it out. <A> Its simply a 2-in-1 potentiometer of the value indicated at the back. <S> The pins at the back, is just a replica of whats at the front
| This appears to be a standard dual potentiometer, commonly used as a volume control in stereo audio applications. It is not easy to say what this does without looking through the datasheet or probing. As in 3-pin potentiometer, the outer two pins are for fixed max resistance, then the middle pin is the wiper.
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Jumper wire for thin connector pin / holes I have a board with a Samtec SFM-115-L2-S-D-LC connector . This connector has 30 (15*2) pin holes, with a 0.050" / 1.27mm pitch, and accepts pins that are smaller than those used in breadboard / prototyping jumper wires. Therefore, I can't directly connect my prototyping jumper wires to the connector holes. In the datasheet for the male version of my connector ( TFM-something ) I can see that the pins have a width of 0.018" / 0.46mm, while my jumper wires have the standard thickness (0.0254" / 0.65mm). How can I connect individual pins/holes of my Samtec connector to, say, a breadboard? Are there jumper wires or headers with the appropriate thickness? <Q> You can try using standard 24 AWG solid wire. <S> This is the standard size used in solid-core CAT-5 Ethernet cable and is also one of the most common sizes used in the telephone industry. <S> The nominal diameter is 0.020". <S> If the conductor is just that tiny bit too thick, try squeezing the conductor slightly with a pair of smooth-jaw pliers. <S> It won't take much pressure to reduce the thickness by a couple of mils. <S> FWIW - I use 24 AWG solid-core wire for all of my breadboarding. <S> It fits standard breadboards perfectly and is also a perfect fit in machine-pin IC sockets. <S> You get huge amounts for free just by chatting with your local Telco guys. <S> One final tip: I normally cut the very end (tip) of the bare conductor at an angle using scissor-type cutters (Miller type 101 Wire strippers / cutters). <S> This angled cut gives the conductor a taper which can make it much easier to push into a connector. <A> Yes there are headers you can use... <S> like the Samtec TFM. <S> A better option for hand soldering a breakout wire would be the Samtec FTS or FTSH (low profile vs standard profile), a plain 0.05" male header, in smd or through hole. <S> It has some standard options and can be found through the typical distributors, Digikey, Mouser, etc. <S> If you want a standard or easily found part, your out of luck. <S> Either use a header, or grab some appropriately sized wire an make your own cables by splicing the thin wire to some thicker wire, each side going to the corresponding board. <S> Even a nail file would work. <S> Just rub it on it until the pin is thin enough to fit properly. <S> Shouldn't take more than a minute per pin at worst. <S> Alternatively slice some of the copper off with an exacto knife. <A> I think you may be out of luck here Luis. <S> I've never seen them ready made and as you've noted, most ready made jumpers are for the more standard size of .0254", which also conveniently corresponds to 22 gauge wire, at least making it easy to roll your own. <S> 0.018" corresponds most closely to 25 gauge (0.0179"), so you may have to search for a small roll of insulated wire in this thickness and make some of your own. <S> Even there I was coming up dry looking for this guage in insulated wire. <S> I did find some "Magnet Wire" with this guage on Amazon, at this link: TEMCo 25 AWG Copper Magnet Wire . <S> But of course its insulation is a coating you'd need to remove at the ends to make jumpers, and at that thickness they'll be a little floppy. <S> If you start with a single row header that will obviously be easier. <S> Good luck.
| The other possibility is to buy a few headers from the same company with matching pins, spend an afternoon soldering little wire loops to them (maybe with some of that more common insulated 22 guage wire), and later cut them apart. Another option would be to file the pin on your jumper wire down some with a piece of sand paper or a file.
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Power glitch when switching between voltage sources I have I device with USB. When connected to USB it is powered from the USB port (an LDO to adjust from 5V to 3.3V), and when not connected to USB it is powered from a CR2032 battery. Below is the schematic with a P-MOSFET to switch between battery and USB power. The output from the LDO is 3.6 V, since it will drop slightly over D1. simulate this circuit – Schematic created using CircuitLab The problem I'm having is that on USB disconnect, there is a rapid drop in voltage. See the oscilloscope image below. When connected to USB the voltage is 3.28 V, and on USB disconnect the voltage rapidly drops to 2.1 V, before stabilizing at 2.5 V for approx. 60 ms before rising to battery voltage of 3.08 V. The duration at 2.5 V can be anything from 1 ms to, as in this case about 60-70 ms. On a previous version of the device, the exact same components and schematic layout is used, just a different PCB/board layout. Below is an oscilloscope image for that version. The same behavior is observed, but the "in-between voltage" is 2.75 V instead of 2.5. The duration at 2.75 V can be anything from 1 ms to 60-70 ms in this case as well. But how come the difference (2.5 V vs. 2.75 V) with the exact same components, just different board layout? And how do I go about fixing this problem? There are components in the device that does not survive (loses settings) when forced to run on 2.5 V for as long as 60 ms, and maybe the rapid drop to 2.1 V is a problem as well. How do I fix this? UPDATE:The circuit load (across C1) is a low-power microcontroller, and some LEDs and buttons. What I refer to as "USB disconnect", i.e. removing the USB power source and switching to battery power is accomplished through removing the device from the computer USB port. I.e., the LDO input disappears, and hence the LDO output 3.6 V. ADDED:Voltage drop on R1 (10K) when USB is disconnected: Voltage drop on R1 (1K) when USB is disconnected: <Q> This answer was made before the OP corrected the MOSFET type!! <S> the IRF9530 - it can have a gate-source threshold voltage as high as -4V <S> and for a lot of circuits this is going to produce a vast spread of results. <S> In fact the spread would be so large that a few percent of units will not manage to turn the MOSFET on at all. <S> This could easily account for the differences seen between two PCBs. <S> As for the time taken to properly recover. <S> the 10k resistor has to charge up the gate capacitance and this might be in the order of 1nF. <S> This is a CR time of 10us <S> so it doesn't look likely that this will contribute more than maybe a millisecond maximum. <S> But there are other things to consider too: <S> - What do you have connected as a load across C1 (2.2uF)? <S> How did you remove the presence of the USB supply? <S> Did you remove the output from the LDO regulator or remove the input to the regulator? <S> Were you consistent on both occasions? <S> Bottom line is the MOSFET choice is poor. <A> Fix is easiest: Just use identical diodes. <S> The 3.6V source will win over the battery because the voltage drop should be identical. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> If you want to keep the fancy design (like if you're also reading the PMOS gate input to determine if you're running off your battery or not). <S> Try this solution. <S> simulate this circuit <S> We can always add more transistors to the design to get rid of L->H and H->L transition differences. <S> As for why certain physical orientations produce different results, without pictures I couldn't say for sure. <S> Most likely has to do with the mosfet discharging at different rates due to either parasitic resistance or parasitic capacitance. <A> If it were me, I'd try changing R1 to 1K. <S> That should make the switchover faster. <S> The 1K will draw 3.6 mA where the 10K drew 0.36mA, but since you have the USB bus to draw on it shouldn't be a problem. <S> R1 has no effect on current draw when running on battery.
| It's a poor choice of MOSFET
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What are QFN sockets for? I would like to use a QFN chip but soldering it is difficult as the required tools are unavailable. After searching online, I found that there are sockets available that provide pins for those chips, such as this one . However I saw on the title that they are called programmable adapters. My question now is, what are the main use of these sockets and can they be used to connect the QFN chip to a breadboard and interface with its pins? <Q> What they are referring to is programming programmable parts, like microcontrollers and EPROMs, before the chip is mounted to the board. <S> In these cases, the board doesn't include a means to programming the part or providing access to the programming lines. <S> You can use these sockets any way you like, but a over $100 each, it won't take many to make a hot air soldering station the cheaper alternative. <S> Then you can solder your own QFN and won't need sockets at all. <A> Since chips may die during development using such an adapter is often easier and cheaper than soldering the QFN. <S> Regarding soldering: In a pinch you can solder the chips using a bit of solder paste, a hot plate and some of practice. <S> It's easier if you get a solder stencil, but it works fine without it. <S> For prototyping these adapters may work. <S> Whether they work or not depend on the chips you use. <S> Some high-speed or high precision parts have very strict requirements on how far capacitors and other components are located with respect to the chip. <S> This is usually not a problem if you work with a PCB but the leads from the break-out board to the chip itself may already be longer then allowed. <A> That adaptor doesn't look suitable for breadboard. <S> Plugging it into one looks like it would short out pairs of pins. <S> You could probablly use it on stripboard though with careful track cutting.
| They are called programmable adapters because you'll often use them to upload firmware into the chip during development.
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Help selecting component to detect red laser light While I did find a little information in searching your siteI am still unsure what to use in my application.I want to detect the (red) laser light hitting a screen ortarget. My questing is what is a good device to use, I readthat photo diode is faster than a photo transistor. Can anyone give me any part numbers to try? ApplicationSimple laser target, when the training pistol laser hitsthe target I need a fast detection. I will experiment withthe final output later in this project, to start with I mightjust turn on a light. I also have realized that other indoorlighting may need to be filtered, I have knowledge of filters,I will later research this. <Q> I assume the 'training pistol' is hand held. <S> If it is, you likely only need detection in several milliseconds, or more. <S> So, as Scott Seidman says, fast detection is not likely an issue. <S> I'd recommend starting with a visible light phototransistor. <S> They are easy to use as detectors, and are cheap, i.e. under $1. <S> (e.g. Vishay TEPT4400 ). <S> It is over 90% sensitive at about 630nm. <S> These have a wide enough range of wavelengths that they will still pick up infrared light, so use an optical filter, to reduce the intensity of other light sources too. <S> To avoid confusing the detection of the laser with sunlight, or some other light source, the laser needs to be modulated. <S> This needs to be done in such a way that it is easy to discriminate the laser signal from other light sources. <S> Simple hand-held laser's can be modulated at about 5kHz, which is well away from mains-powered light modulation (50, 60, 100 or 120Hz), and Infrared remote control (36kHz or higher). <S> Depending on your level of skill, you might build the filter using 'pure electronics', or you might use a small microcontroller. <A> LEDs also work as photo diodes - just not characterized as such on the datasheet. <S> An LED with its spectral peak close to that of your RED laser will naturally ignore other broad spectrum light sources, and if the LED lens is RED (not clear) then it has a built in optical filter. <S> Finally, the LED viewing angle (field of view as a photodiode) can be reduced to match that of the maximum angle of incidence of the laser. <S> Here is one with a peak wavelength very close to the 635nm red laser and a 30 deg FOV. <S> https://www.digikey.com/product-detail/en/ALMD-EG3D-VX002/516-2162-1-ND/2389846 <S> As others have stated, modulating (psudo random pulsing of the laser) will also aid in detection. <S> With modulation and the appropriate signal processing you won't need an optical filter unless the red component of the ambient light is so bright that it approaches saturation (and the modulation is lost). <S> Finally I'll add, if used indoors, then be aware the white LEDs and florescent lights emit in spectral bands and often flicker at 120Hz. <S> You might want to choose a laser color that is different from their peak wavelengths. <A> 600nm - 680nm filter on photodiode(or phototransistor or thermistor) <S> what you need I suspect. <S> However, they are neither cheap nor as fast as you need it to be I guess. <S> Thats because of the filter of course. <S> They are in limited production usually, I remember that we had a quote for 10 photodiode with 880 nm filter from heimann 35€ for each. <S> And you are right, most of the photodiodes responds to unwanted lights unless they have filter on. <S> However, you can always specialize your application to satisfy your needs such as adding some modulation, arranging the receipt sensitivity and gain etc.. <S> Have a look at this page . <S> You will see some photodiodes with specific response ranges. <S> But they are limited to 10 - 100 MHz <S> ranges so I am guessing they do not have less than a few hundreds of nanoseconds response time. <S> edit: <S> particularly this one looks like what you need if you care for your budget. <S> I hope I could help.
| Almost all photodiodes and phototransistors are fast enough by a factor of 100x, or more.
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How can I wirelessly deliver ~30B of data to ~10 peripherals roughly every 50ms? I am thinking of a system where 10 or so peripherals need to receive 30-50B of data every 50-ish milliseconds (each peripheral gets unique data). The total throughput required might not be huge (6-10kB/s) but the problem lays in the number of devices which need to receive the data. I've looked into using BLE but even though the total throughput lies well within the maximum possible ~35kB/s, what I want is not possible due to the hard limit of 20ms Connection Intervals. Another solution I considered is using a simple RF transmitter and RF receivers which all listen to it simultaneously. I'd have to manually take care of error correction etc but I would not have to worry about switching between the receivers which takes precious time. A problem with that would also include the inability to have two of such systems deployed near one another without the risk of the transmitters talking over one another. So what other solutions are available to me? <Q> You could take a look at the IEEE 802.15.4 protocol. <S> There are lots of transceivers out there. <S> This would allow you to transmit all device data for all peripherals in one packet every 50ms (on air time for a packet like this will be <1ms). <S> Assuming that you don't need a mesh network, this could work nicely for you and minimize latency. <S> IEEE <S> 802.15.4 radios use CDMA, so they will not stomp on each other when transmitting in the same RF space. <S> Without a mesh network, you'd be running a star configuration with the "master" at the center of the hub and your peripherals acting as points on your star. <S> Good luck - sounds like a fun project. <A> Get a simple 433 MHz transmitter and a few of receivers, send a data preamble, then an address then <S> the data then a couple of byte CRC. <S> In other words develop a simple protocol and buy some cheap TX and RX <S> and you should be in business. <S> 433 MHz may not be allowed in your country but there will be a frequency that is. <A> I suspect that using nRF24L01+ transceivers would solve this for you. <S> Theoretically good for 1 or 2Mbps, less acknowledgements and some packet overhead and you should still be able to sustain over 100kbps of user data without any dramatic effort.
| If you need to send data to specific receivers then embed address and data into one long stream intersperced with CRCs so that the right receiver knows what it's looking for.
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Why can't you put a plane under a trace antenna? I'm working with RFID for the first time. All the app notes I've read show a separate region of PCB allocated for the trace antenna, with no planes above/below. I want to understand why this is, and if it's strictly necessary. I understand that the plane will shield the antenna from one direction. For my application, this is ok. I can fathom the following scenario:1) An electromagnetic wave is incident upon a trace antenna with a copper plane below it2) The distance between the plane and antenna is half the wavelength (not realistic for RFID, but this is a thought experiment)3) The field seen at the antenna is 0, because the incident wave and reflected wave are 180 deg out of phase. So, in this scenario, there's no changing field and no current induced in the trace antenna, so nothing is received, but this is only because the separation between the copper plane and antenna is lamba/2. When this is not the case, I would imagine there should be current induced in the antenna. I'm particularly interested in this because I want to save board space. Specifically, I want to put components and signals on top, and the antenna on the bottom, with a copper plane in between. If I cannot do this, can anyone explain, with precise language and in EM fundamentals (no handwaving armchair engineer stuff please!) why this will not work? Thanks <Q> An antenna converts a low impedance transmission line (from the chip or also via a coax or stripline) to a high impedance at the end of the antenna. <S> The high impedance is to suit the impedance of free space (377 or 120\$\pi\$ ohms). <S> An antenna does this gently and in doing so creates a standing wave along its length in order to allow the current to gently fall to zero at the end (such as in a quarter wave monopole) whilst the voltage reaches a maximum thus, the correct ratio of E field to H field is created (a ratio which defines the impedance of free space i.e. 377 ohms). <S> Generally, the near-field extends to one wavelength and anything in the vicinity can prevent the above happening effectively. <S> If you just have a track above an earth plane you are going to create a sudden discontinuity (at the end) and get a total reflection of power back to the transmitter with no power being emitted into space. <S> That's not an EM antenna. <S> I'm working with RFID for the first time. <S> Given this, you may actually be using an "antenna" where the dominant field is magnetic and there are plenty of things you can do that you cannot do if you were using an EM transmission to transmit power. <S> So, quit the generalized hand waving in your question and get specific - what operating frequency and what antenna design have you targeted? <A> The plane is in the near field , which means the antenna can "see" it. <S> It will cause the electrical properties of the antenna to be quite different than when there is no plane. <S> The plane will also act like a receiving antenna. <S> Depending on the size, coupling, shape, and orientation of the plane, it may re-radiate some of the power it takes from the antenna, but it will also certainly dissipate some as heat. <S> Even if it re-radiated all of the power, the pattern would be hard to know up front, and the currents running thru the plane and their associated offset voltages would not likely be good for the rest of the circuitry. <S> So leave the clear area around the antenna as specified. <S> It's better for the antenna, better for the rest of the circuitry, and the result will be more predictable. <A> Yes you can put a plane under a trace antenna. <S> There is nothing preventing you from doing it. <S> If you do it and your design "works" for your intended application, you are done. <S> However, if it does not perform to your satisfaction, perhaps following a proven design might do it, saving the time, expense, and effort to design your own.
| It's not that the plane will block transmission in one direction, but that it will absorb much of the power you put into the antenna, and also greatly change the antenna characteristics.
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Can a current source have a voltage across it? I'm reviewing material from a lower div circuits class in preparation for the upper div class by redoing problems, and this is one of them. Does it make sense for the branch with the dependent current source to have a voltage Va(t) across it? Wouldn't the voltage measured across it be zero assuming the current source is ideal and the wire has no resistance (since it's not modeled in the problem.) Can someone explain why these dependent sources make sense, please? <Q> Think of an extremely simple circuit: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> There is obviously a voltage of 1V across R1, so <S> by Kirchhoff's voltage law, there also must be a voltage of 1V across I1. <A> A current source can certainly have a voltage across it. <S> If the voltage across a current source is zero, then it is not delivering or absorbing any power. <S> However, if the voltage across the source is not zero, then it is either sourcing or sinking power into the rest of the circuit. <A> Yes. <S> An ideal current source is a device that always produces the given current regardless of what voltage is applied across it .
| A device that always has 0 V across it is called a 0 V voltage source, or, less formally, a short circuit.
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What is the difference between Ground and Low? What is the basic difference between sending a LOW signal vs GND to a circuit? What If I ground instead of sending LOW? Or similarly what is the difference between Vcc and HIGH? <Q> Take a look at this: - It's for a 5V logic system and it tells you what the voltage ranges that are acceptable for standard CMOS inputs. <S> So an input signal that is below 0.8 volts will be regarded as a low and that is a definite . <S> However, if the signal received is (say) 1V <S> it might still be regarded as a low <S> but there is no guarantee that it won't be regarded as a high. <S> Then you could take this further and look at what the specifications of a driver and receiver are: - You should be able to see that a logic driver output has a tighter spec than a logic input and this should make sense. <S> What If I ground instead of sending LOW? <S> You should be able to see that ground (aka 0V) is in the range of an acceptable "low". <S> Or similarly what is the difference between Vcc and HIGH? <S> You should be able to see that Vcc is in the range of an acceptable "high". <A> Logic Levels <S> A digital output will have a datasheet specification for its VOHmin <S> and VOLmax. <S> When driving a logic high output, the voltage could be as low as VOHmin. <S> When driving a logic low output, the voltage could be as high as VOLmax. <S> A digital input will have a datasheet specification for its VIHmin <S> and VILmax. <S> Input voltage between VCC and VIHmin is received as logic high. <S> Input voltage between VILmax and GND is received as logic low. <S> Input voltage between VIHmin and VILmax is not a valid signal, should never happen except during transitions. <S> Input voltage greater than VCC or less than GND is not valid, and can cause incorrect operation or even permanent damage. <S> Input voltage of GND is interpreted as logic low. <S> Input voltage of VCC is interpreted as logic high. <S> Analog circuits care about the signal level no matter how close it may be to ground. <S> Digital circuits only distinguish between "logic high" and "logic low", it doesn't matter what the exact voltage is as long as it is within one of these valid signal ranges. <A> low is less than something (check the documentation for what is considered as to be low) <S> ground is an absolute voltage, ground is almost always low, VCC is the positive supply voltage <S> , VCC is always considered high,high is above some other limit. <S> in digital circuits <S> ground is the lowest voltage, anything close enough to ground is considered low vcc is the highest voltage anything close enough to VCC is considered high high and low do not meet. <S> there is a range between them where it is not defined if the signal is high or low.
| Ground is the reference point against which voltages are measured: by definition, whatever voltage is on ground is defined as zero volts.
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Wireless Communication between microcontroller and PC I have a FRDM K64f board that is using the nRF24L01+ as it's wireless transceiver. I want to communicate between the K64f board and my PC locally. I am struggling to find the hardware needed on my PC's side. I had thought of using a USB to TTL adapter and connect a nRF24L01+ on the PC side and communicate this way. Would this be viable? Or is there a simpler way? <Q> If you have a lot of cash to spare, Nordic makes an evaluation kit for the nRF24L01+, part number nRF24L01P-EVKIT, that should probably fulfill your needs. <S> However, considering the sparkfun product Andy aka posted has the schematic, eagle, and code files available, it would probably be cheaper to have one of those made at oshpark or similar. <S> You could also use a Bus Pirate as a USB->SPI bridge, and hook that up to one of sparkfuns <S> nRF24L01+ modules , and write a little program to talk to it through the bus pirate. <S> The Bus pirate has an additional IO pin you can use to strobe the CE line on the Nordic chip. <A> I had thought of using a USB to TTL adapter and connect a nRF24L01+ on the PC side and communicate this way. <S> Would this be viable? <S> Your PC will need a driver installed (if there is one) and that driver has to match the Nordic chip's data protocol and take heed of any subtleties in the TTL to USB interface circuit. <S> Keep searching. <S> Or is there a simpler way? <S> Try and make the USB interface "con" the PC into thinking it is talking to a generic USB to RS485 converter ( <S> or RS232 if those things exist). <S> See this related question. <S> See also this sparkfun serial interface that has a Nordic nRF24L01 interface: - Apparently it's not a current design but you may be able to find one (I like to call this inspirational assistance). <A> The NRF24LU1+ is a variation of the NRF24L01 idea with an on-board microcontroller having a USB interface. <S> It is commonly found in wireless keyboards and mice, but also used for some custom projects. <S> You would have to find or create firmware for this. <S> Or if you feel in the mood for Arduino, those readily handle the NRF24L01+ too - basically, any MCU which can talk both SPI and USB will work if you are familiar with working with it.
| Another option would be to simply use another Kinetis Freedom board to bridge the radio to the PC over its USB interface, or use or create a simpler/even cheaper board based on another Kinetis M4 or even M0 part.
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Can I ground the negative DC bus of a variable frequency drive? Suppose I have a typical VFD: rectifier, DC bus caps, inverter, feeding a motor. In general it would not be advisable to tie the DC bus to earth ground, because the AC line may also be referenced to earth. Having two competing ground references in the same circuit can cause substantial issues. I suspect the rectifier diodes would fail rapidly, if a fuse didn't save them first. But suppose I wanted to power the drive through the DC bus terminals. In this configuration, the AC terminals are left open, meaning there is no competing ground reference. The VFD and motor chassis are tied to earth, which can lead to common mode issues, but I don't see that they'd be any worse than running from an earth-referenced AC line. In that configuration, would there be any problem with tying the negative DC bus to earth? Is it possible to make a general statement, or would it be installation- and equipment-dependent. <Q> It should be fine but be very careful where you make your earth bond. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You need to run the DC- all the way back to the DC supply ground point. <S> Don't take a shortcut and connect DC BUS - to the chassis and run that back. <S> If the DC- got disconnected the VSD chassis would go live. <S> Update <S> I forgot to explain that R1 provides inrush current limiting on power-up and RLY1 switches in when the voltage reaches half or two-thirds of normal DC bus voltage. <S> (Thanks, Charles.) <A> If your substation is solidly grounded (as opposed to HRT, or ungrounded), you’ll probably just hear a loud bang at some point. <S> If not, the bang may be in a seemingly unrelated piece of electrical equipment fed from the same sub. <S> Look at your rectifier stage at the front end of the drive and envision that the negative bus is always about 1 volt more positive than the most negative of the three incoming ac phases. <S> Is that “ground” potential? <S> Nyet!!! <S> (Must be talk like a Ruskie day) <A> You will probably find inductors in both the positive and negative sides of the DC filter. <S> That is done to provide impedance to limit any fast-rising fault current in case of a ground fault on the motor lines. <S> Make sure that external DC is connected to the line side of both DC inductors. <S> Some VFDs have AC inductors instead of DC inductors. <S> In that case, the manufacturer will likely say that external DC should not be connected without fault-limiting impedance. <S> You should always seek instruction from the manufacturer for this type of connection and probably not do it otherwise. <S> Note that another answer shows a charging current limiting circuit. <S> That is probably built into the VFD. <S> Make sure that is not bypassed when the external DC source is connected. <S> "Rapid failure" is an understatement of the consequence of getting this wrong. <S> Exploded components would be more accurate. <A> Usually when I wire the VFD, I strictly wire it from manual. <S> I never applied ground to DC bus in no circumstance. <S> As you self described the AC line may be referenced to ground - that's quite normal case (I never saw it floating) so your rectifier diodes would blow ASAP. <S> The only thing that is grounded on VFD is the casing and logic/analog inputs/outputs. <S> There are also DC bus terminals if you want to parallel some drive, but no way to connect the ground that would be your biggest mistake. <S> Edit: <S> But as you decided to power the inverter from different power source, battery bank like <S> , then I think <S> you should't be worried about. <S> But you should check if the electronic part of VFD is powered from DC link or it needs AC input. <S> I have seen special drivers that require DC/DC converter from DC link to supply the electronics in case they are supplied from DC link (or from regenerative voltage during breaking in case of power outage).
| Grounding the dc negative bus is a really bad idea. If the fault-limiting impedance is not compromised, there should be no problem using a grounded DC source.
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Positive and negative terminals on a battery Sorry, this may be a newbie question, but is the positive and negative terminals on a battery refer to which side has a higher potential? Or one side has positive charge and one side has a negative charge? <Q> The convention for current flow was established before the electron was discovered (by J.J.Thomson in 1897). <S> The convention is that current flows from + to -. <S> We now know that, in fact, the mobile charges are electrons and flow from - to <S> + <S> but the convention has endured and we all use + to - flow but keep the reality in the back of our minds. <S> As is universal practice in maths, physics and general engineering, + is higher than - <S> so we refer to the positive terminal of a battery or power supply as having higher potential. <S> (Conventional) current will flow from the higher potential to the lower, i.e., from + to -. simulate this circuit – <S> Schematic created using CircuitLab <S> The magnitude of the current will, for a resistance load, be proportional to the voltage or potential difference between the terminals and inversely proportional to the resistance between the terminals. <S> This is succinctly stated in Ohm's law. <S> $$ V = <S> I \cdot <S> R$$ which can be written as <S> $$ I = \frac {V}{R}$$ <A> derstrom8's comment is correct. <S> The way you need to look at it is there is a positive voltage potential at the <S> + terminal of the battery relative to its - terminal. <S> To expand on that, if you had two batteries, completely disconnected, and attached one probe of a volt meter to the + terminal of one battery, and the other probe to the - terminal of the other battery, you would get ~0 volts. <S> That is because the two batteries are not in reference to each other. <S> You could connect just the two + terminals together, or one + terminal of one battery to the - terminal of the other battery, and nothing would happen (no current would flow), assuming the terminal of each battery was left disconnected. <S> This is also why if you connect the positive probe of a volt meter to the + terminal of one battery, and the negative probe to the - terminal of that same battery, you will get a positive voltage reading. <S> If you switched the probes around, you would get a negative voltage, because you are now asking the meter to read the voltage potential of the battery as if the + terminal was ground (or 0 volts). <S> In other words, the - terminal has a negative voltage potential in reference to the + terminal. <S> As far as charge goes, Optionparty's comment hits on this. <S> The - terminal produces electrons (normally associated with a negative charge). <S> Current flow from negative to positive (- to +) is usually referred to as Electron Current flow. <S> However many circuits will refer to current flow as being from Positive to Negative, or Conventional Current flow. <S> There's a TON more theory and physics behind this question but for now, it's probably good enough to just remember that voltage is a potential between to points. <S> If you only have a battery by itself, those points are the + and - terminals. <A> is the positive and negative terminals on a battery refer to which side has a higher potential? <S> Battery positive has a higher potential (voltage) than the negative. <S> Or one side has positive charge and one side has a negative charge? <S> "So many" charges passing per second is current but, due to an anomaly in the thought processes of our forebears, current was believed to flow from positive to negative - we still "conventionally" regard it being the same today even though electrons flow/pass in the opposite direction when you get down to the real physics. <A> If you connect the red probe of Voltmeter to the negative terminal of the battery, and the black probe of Voltmeter to its positive terminal, the voltmeter will indicate -12 Volts. <S> Which means nothing but that the terminal of battery that is connected with the red probe is 12 volt less than that connected with the red one. <S> So if you see this way, 12 and -12 give the same information.
| Charge (in terms of simple conduction is an electron) flows from negative to positive terminals. When you have 12 Volts, this means that the positive terminal of the battery is at 12 Volt higher potential as compared to its negative terminal.
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Is there a device that can push down on a button with a wireless remote control? I'm looking for a device, perhaps one that releases a small piston, tiny object, or block, that can press down on a button via wireless remote control. The button is about the size of a dime, and is sensitive enough it requires minimal effort to push. The button is facing up, so I might not even have to mount the device on anything. Worse case scenario I'm hoping to duct tape the device over the button. An alternative over a push-button could be a timed button. A device that can press down on the button every ~5 minutes. Is there anything that's close to anything like this? I was hoping to find something from a marketplace website, but I've had no luck. <Q> Yes, this product. <S> it pushes what you want using WIFI and application control: https://www.switch-bot.com/?fbclid=IwAR2hdh84V8mSksl3F3-UPTUy3m94_0usT6hSwtfWXaob61n4ANCWSfBwnWo <A> There are some good answers here already that solve the problem directly as described, but I'm going to take a different approach to it. <S> the contacts across the button with some small wire Drive the relay coil from a cheap radio control car (replace the drive motor) <S> Look for a car that does not have speed control; just go/stop. <S> This approach would also leave the button uncovered and available for manual use. <A> Two ideas: Mechanical SLRs used a cable release to minimise camera shake. <S> Here's a hack (unfortunately on an external site) to make an remote controlled one. <S> There were also pneumatic remotes - a rubber bulb connected to 5 - 10 m of micro-tubing ending in a mini piston to screw into the shutter release button. <S> Perhaps one of these would be very easy to hack by just pressurising the tube with a small air pump - maybe using the bulb as the actuator. <S> Picture from sammys.com . <S> (I've never heard of them before.) <S> Lego Technics or Mindstorms must have had something with a remote? <A> I can see a number of options radio control servos (radio controlled model hardware) <S> this stuff is well esablised technology: <S> IIRC the IRA used to use these to detonate bombs. <S> a car alarm. <S> or garage door opener remote, and an auxilliary receiver driving a push solenoid modify a wireless doorbell (especially one that has real chimes) <A> this circuit will generate an approximately half second pulse <S> every approx 5 minutes, and can drive a solenoid to press the button simulate this circuit – Schematic created using CircuitLab <A> A small linear actuator solenoid allowing with a wireless relay module would work great. <S> http://www.trossenrobotics.com/p/small-solenoid.aspx <S> you would need to mount it so it can push the button, so you would need to build a small frame. <S> It could be made out of wood, and attached to the button with some hot glue for a non permanent solution. <A> There is a bluetooth controlled button that is held in place using double sided tape (or glue if you want it to be permanent) called the Microbot Push https://www.amazon.ca/MicroBot-Push-Wireless-Automation-Platinum/dp/B01FQN2T2W/ref=sr_1_1?ie=UTF8&qid=1493585609&sr=8-1&keywords=microbot+push
| If you have access to the electrical connections of the button you want to press, you could: Get a small relay Solder
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Will cutting the data lines of USB still let the power to go through? I am trying to Charge a USB connected device through a computer without transferring data. An USB 2.0 cable should have 4 wires: Vcc, Data+, Data- and Ground. If I only cut the Data wires and let the Vcc and Ground wires connected, will it still be able to transfer power or will the power cables be shut off because of the disconnected Data wires? <Q> It depends on the device. <S> And on the source. <S> Usb spec states that a device should only use 1 block of power until it enumerates with the host. <S> Then it can request up to 5 blocks. <S> Each block is 100 mA. <S> It should not pull more than 5 blocks. <S> In practice, very few, if any hosts limit a usb port to 100 mA at all. <S> Some do limit ports to 500 mA (Apple computers for one, though they do allow iDevices to request more in a Apple only standard, so they are physically capable just not normally allowed). <S> Some USB ports have very little power protection and may be hooked up straight to the 5V power source. <S> Now usb peripherals also have requirements for power. <S> Dumb devices may power directly from any power source, 5V or different. <S> These dumb devices don't even have the data lines connected. <S> Older cell phones also allowed dumb charging. <S> As time passed, more and more restrictions come into play. <S> Some devices monitor the power pins for regulation, and won't work on anything outside of usb spec power (4.75 to 5.25 volts). <S> Some devices won't work unless a resistor pairing setup is on the data pins. <S> Apple started this, and others followed, and there are competing standards. <S> You can't plug a dumb 5V power supply into an iPhone and get it to charge. <S> It will state unauthorized charger. <S> Some devices require enumeration before charging, but this is rare. <S> The PS4 Six axis controller does this. <S> And some devices will treat a dumb charger as unsafe, and limit charging to 500 mA or less. <S> Most modern cell phones do this. <S> If the data lines do not have a resistor indicating it is a high power charger. <S> As a side note, there are commercial usb cables that have a data or charging switch, allowing you to disconnect the data lines to prevent data communication. <S> They typically switch in the resistor setup, but it varies. <A> yes, but you are breaking the rules if you take more than 100mA the device may not like recognise the computer as a charging source with the lines cut, so charging may be much slower. <S> there are adaptors that make the compute pose as a 2A charger which will give a fast charge. <A> Cutting the data lines will let the device charge in most circumstances. <S> Some devices are too smart and want to negotiate with the host device to determine maximum current available, but even these I suspect will default to a low value if comms isn't available.
| In short, a usb cable plugged into a computer without the data lines will either not work, work slowly, or work fine, depending on your computer and your device.
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How fast can a transistor switch? An NPN transistor can be made a switch by applying some power to the base. Is there a limit to how fast I can switch the transistor on and off, or I can switch it on and off as fast as I can turn the signal to the base on and off? As switching, I mean completely off and completely on. Not in between (if there is even any.) <Q> The major limit to BJT switching time is related to the charge carriers and specifically how long it takes to move carriers into the base, and how long it takes to get them out. <S> The datasheet will include a few parameters that will give you the theoretical maximum switching frequency <S> *. <S> They are <S> Delay time (td) <S> - how long it takes to get out of cutoff Rise time (tr) <S> - how long is the transition from cutoff to saturation Fall time (tf) <S> - how long is the transition from saturation to cutoff Storage time (ts) - how long to get out of saturation Using the datasheet (these parameters are usually listed), you can figure out how fast a transistor can switch between the two states. <S> $$ f_{max} = <S> \frac{1}{t_d <S> + t_r + t_f + t_s} $$ <S> * <S> This is what transistor can theoretically do, but there are tricks that can be done to improve the switching speed. <S> Also, if you are switching a square wave, then in order to maintain a nice square waveform, the actual switching frequency will be much less. <A> When you try to switch a (bipolar -- NPN or PNP) <S> transistor fast, there are some effects that limit the rate it can respond. <S> 1) <S> The physical base of the transistor extends over some (small) area, and the base lead is only connected to one portion of this. <S> There is some internal resistance between the connection location and the most remote portions of the base. <S> When you switch fast, the time for the charges at the remote portions to be extracted is significant and limits how fast it can turn off. <S> This isn't as severe for turning on because you could get by (for a few ns) with just the local portion turning on. <S> 2) <S> There is a significant capacitance between the collector and base. <S> When turning off, the collector voltage rises (if it didn't there would be no need to turn the transistor off). <S> This couples via the base-collector capacitance and tends to counter the turn-off or -on voltage on the base, thereby making it harder to turn on or off. <S> 3) <S> It actually takes some time for the carriers to transition the base and collector regions. <S> However, bipolar transistors can be very fast -- they can be switched in picoseconds inside ICs. <S> Discrete devices can't be switched as fast because of greater parasitic capacitances, inductance of the wires that attach to the device, and because the fastest devices have breakdowns of just 1 or 2 V, and so are really only useful inside specialized circuits in ICs. <A> However, it is dependent on which transistor you use, as well as the characteristics of the driving signal, the power supply, and the load.
| Yes, there is a limit to how fast you can switch the transistor on and off.
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Effect of 0 Hz setting on VFD with continuously spinning fan As part of a HVAC system we have a supply fan and extract fan running on independent VFD 1 . With the current configuration we usually don't need to run the return fan as the supply fan pressurises the room adequately to force air out through the return duct and room is at the required pressure. So far, so good. The control system ramps down the return fan speed to control room pressure and so we're mostly running the return fan at 0 Hz. We are wondering what effect this will have on the drive. Variable Frequency Drive outline schematic thanks to Stephen Collings . We know that in normal VFD braking the normal commutation of the output transistors and generation from the motor result in a rise in DC bus voltage. If this exceeds a preset level the drive will bleed off power into a braking resistor (not shown on the schematic). Questions: Q1. What typically happens at 0 Hz? Will the drive hold the three motor phases at a particular DC levels (by PWM 2 )? Or will the output transistors turn off completely. Q2. Is this a bad thing? (I have experience of servo motor brake resistor burn-out. The resistor turned out to be more like a toaster element and I repaired the drive by connecting to an external high-wattage heater cartridge of the same resistance.) I'm going to check the DC bus voltage and motor current readings and will update the question. Is there anything else I should read from the drive? Update after inspection I visited the drives and recorded the following data. Both are ABB ACS550 series . ╔═════════╦════════════════╦════════════════╗║ ║ Supply ║ Extract ║╠═════════╬════════════════╬════════════════╣║ Model ║ ACS550-01-059A ║ ACS550-01-023A ║║ Rating ║ 30 kW ║ 11 kW ║║ DC bus ║ 555 V ║ 564 V ║║ Power ║ 12.4 kW ║ 0 kW ║║ Current ║ 30 A ║ 11 A <--- ! ║║ Torque ║ 44% ║ -4% to +10% ║║ Speed ║ 900 RPM ║ 0 RPM ║║ Temp ║ 30°C ║ 30°C ║╚═════════╩════════════════╩════════════════╝ The only thing that looks odd is the 11 A on the idle return fan. It could be that the voltage and current are 90° out of phase or that the motor voltage is zero. 1 VFD = variable frequency drive for AC induction motor. 2 PWM = pulse width modulation. <Q> VFDs used for HVAC applications are often special models designed for that industry. <S> You should look carefully at your manual and other documentation to determine what features are provided in your VFD. <S> Rather than braking, it may be equipped with a feature that only allows the load to decelerate as fast as it will decelerate without regenerating more energy than is absorbed by losses in the motor and VFD. <S> It may essentially "sleep" when the speed command is zero. <S> It may have a feature to find the motor speed and synchronize with it if it is turned on when the motor is already turning due to air flow produced by the other drive. <S> Various features can be enabled or disabled using drive set-up parameters. <S> It is essential to have full documentation for the VFD. <S> Edit re added information: <S> I believe the drive is configured as for sensorless vector control. <S> When zero speed is commanded it will energize the motor with magnetizing current, but try to drive the motor to zero speed if the shaft is turning. <S> It will not provide DC braking unless that is configured, and I suspect that it is not configured. <S> The drive has a braking chopper built in, but apparently no braking resistors. <S> There will be no dynamic braking unless optional external braking resistors are connected. <S> When deceleration is called for, the drive will decelerate at the slower of the rate called for by the deceleration ramp adjustment or the rate that prevents regeneration in excess of losses. <S> This drive has a group of preset adjustments that are loaded by selecting a "Macro. <S> " <S> The PFC macro is intended for use with fan and pump loads. <S> It sets the control method to scalar frequency control rather than vector speed control. <S> I am pretty sure that would cause the drive to completely de-energize <S> the motor when zero speed is called for. <S> I don't think this is configured by the PFC macro. <S> I have written technical literature under contract for ABB USA, but I have not done that for quite a while, so I am not completely familiar with this drive. <A> That is, if the commanded speed is 0 Hz, then the torque command will be such to force the velocity to 0 Hz. <S> Therefore, the current regulation loop is still active and the switches keep PWM'ing. <S> Since the bEMF is effectively zero, the control system will mostly command the zero space vector anyway. <S> However, most better VFDs have features that reduce the switch utilization in such circumstances. <S> For example, the switching frequency might be cut in half or a quarter to prolong the life of the gate drivers in case there is little to no current flowing through the motor windings. <S> The braking resistor engages if the drive absorbs energy. <S> The resistor should be designed for continuous operation up to certain power but such operation should not be prolonged because the drive was designed to drive motors and absorb the motor inertia energy only. <S> Here I'd suggest to contact the drive manufacturer directly. <S> They might be willing to share some info on how much power and for how long the drive can keep absorbing energy. <A> A1: <S> In vector control with feedback, setting the 0Hz will output magnetizing current to the motor (only d part - excitation) if there is no load applied. <S> The sensorless vector control does not work under certain frequency margin. <S> A2: <S> If you don't deal with servo loop, then this is unwanted mode of operation. <S> By the way, if you want to control fans, there is a special V/f curve that is used for. <S> As the torque of a fan vs. fan speed is rising quadratic, the ideal V/f curve is also quadratic, this reduces heating of the motor and matches the motor and fan torque. <S> You would normaly use a RUN command from PLC + analog input set speed, and you can set the minimum frequency (which indeed is needed to prevent motor overheating). <S> Therefore for your application, no 0Hz is ever needed.
| Since centrifugal fans and pumps produce zero or near zero flow at some speed above zero, I believe this drive has a configuration setting that would de-energize the motor if any speed below some set minimum speed is called for. It depends on VFD and type of operation. All VFDs that I have worked with or designed maintain closed-loop velocity control.
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Auto Switching 12 volt battery under load I have a 120 V / 60 Hz 2000 watt inverter which can draw up to 200 amps from a 12 V DC battery source. An AC oil heater drawing 6 amps max 15 amps is running from the inverter constantly. I want to automatically change batteries without interrupting the current to the inverter. The scenario is: several batteries are supplying the inverter. When a battery becomes depleted it is replaced with a charged battery. The invertor cannot go off line so a battery must be on line at all times. The depleted battery is switched to a charging circuit until it is fully charged then it is available to be switched to the inverter load. The battery being charged must be isolated from the inverter during charging. Having many batteries is not an issue. Having many battery chargers is not an issue. Currently this is a manual process and I'd like it to be automatic. I'm looking for: the best way to detect the depleted battery, swap the battery from the inveter to the charge circuit, detect the fully charged state then return the battery to the load circuit. This application is an off grid. Any advice will be helpful. <Q> There might be better ways to do this <S> but we need more details on the topic. <S> However one possible way around this is Voltage sensor + high current Mosfet switch. <S> Feel free to use a micro-controller of your choice. <S> Make sure it has adc capabilities(almost all have). <S> You job will be simple if it has as many adc pins as the number of batteries you are planning to switch between. <S> If that's not possible, you can always use multiplexers to do the job. <S> Voltage sensing: Make a voltage divider to bring the charged battery voltage within range that can be measured by your micro-controller (5V, 3.3V or whatever your adc pin allows). <S> Make sure your adc pin never sees a value which is out of range. <S> For ex - If battery has a max voltage of 15V and my adc can accept max voltage of 5V, I can use this circuit: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This will allow you to sense the battery voltages. <S> You will know which batteries are ready for getting fed into the inverter. <S> For switching, you need a high current mosfet - one for each battery. <S> Parametric search on digikey will give you some results. <S> If your max current can be 200A, better to select a value higher than that - probably 2-3 times. <S> Another solution (cheaper) is electromechanical relay but at that current, arcing might kill it very often. <S> If you are using mosfets, you will have heating issues so you will definitely need to attach massive heatsinks to your mosfets. <S> Having a common heatsink for your mosfets might help because only one mosfet will remain ON at any particular time. <S> Due to the big size, cooling will be more effective as compared to a small detached one. <S> Assume a power dissipation of 150-200 watts in your mosfet while carrying 200A. <S> If that's a concern, you probably shouldn't go this way. <A> As others have already noted in comments to your question, you have a fundamental problem in your setup in that you are using more power than you can supply. <S> Even assuming you start off with a bank of fully charged batteries all ready to be swapped in, no matter how many batteries you have you are only delaying the inevitable. <S> If you can't charge a battery from empty to full in the same or less time it takes to drain it from full to empty then you will only end up with more & more batteries waiting in the queue to be charged - eventually leading to a bank of empty batteries with one being charged and none available to be used. <S> No, this doesn't answer the actual generic question, but addresses the issue which really makes it unanswerable in your specific application. <A> There are so many problems here. <S> First, it is always better to have a single (or maybe two) large battery bank than many small ones. <S> Because then you will be discharging the individual cells more slowly, which means the discharge time will be longer, and the number of discharge cycles over cell lifetime will be greater. <S> This is probably the main reason people do not switch individual cells in and out in an engineered system. <S> As others have pointed out, if the charger cannot supply enough power to run the load, then eventually you are going to run out of batteries anyway. <S> But to delay that eventuality as long as possible, you want to use a large battery bank, not a bunch of small ones. <S> The max current rating should be higher than 200A and probably the highest you can find available, but I don't normally design power electronics. <S> Each battery should have a fuse in case something goes wrong with the switching scheme. <S> You can use a simple voltage threshold to detect depleted batteries. <S> Deciding on the best endpoint would require quite a bit of knowledge and work. <S> If you insist on your ill-advised plan, then you will be discharging a single battery at over 200 Amps, so you can probably consider the battery depleted at 10V. <S> If you create a large battery bank that can supply 200A for many hours, you could consider using a higher voltage (maybe 10.5?) as your depletion threshold. <S> Basically, the faster you discharge the battery (as a proportion of its capacity), the lower the voltage will be when the battery is at 80% (or whatever) discharge. <S> So the heavier the load, the lower the threshold. <S> Most 12V batteries are not designed to supply 200 Amps continuously. <S> I don't know what kind of batteries you are planning to use, but I am skeptical that they will last for very many charge/discharge cycles with this type of use.
| You will want to connect the new battery prior to disconnecting the old one, but the interval where both batteries are connected to the load should be kept as short as practical, because the "good" battery will be running the load AND charging the bad battery with a very high current. I imagine that you should be able to switch batteries using MOSFET's or IGBT's.
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How to compensate for weather changes in barometric pressure sensor (BMP180)? I am using BMP180 barometric pressure sensor on an elevator for extended periods (months together), below is the guidance from Sparkfun. Sparkfun guidance : “You should also remember that pressure changes due to weather will affect your altitude readings. The best accuracy will be obtained if you take a “fresh” p0 (baseline/reference pressure) when you need it and don’t rely on it to be accurate for extended periods due to changes in the weather” How can I obtain a fresh p0 (baseline/reference pressure) when the elevator is continuously moving ? <Q> Two options: Detect ground-floor and perform a reset on each arrival. <S> Install a second unit on a specific floor and update the reference pressure in the elevator micro - perhaps by wireless connection. <S> And ... Try to compensate in software. <S> If you can detect the difference between the rapid changes of pressure due to elevator motion and those due to atmospheric variation, you could adjust accordingly when the elevator stops (but see caution notes). <S> If there is reasonably frequent travel to upper and lower limits, you could recalibrate then. <S> i.e., if (p > pmax) { // <S> p is pressure reading. <S> pmax = p <S> ; // Must be at top floor. <S> pmin = p - bottom_to_top; // bottom_to_top is the pressure <S> span <S> } if (p < pmin) { pmin = p; // Must be at bottom floor. <S> pmax = <S> p + bottom_to_top; } You would have to manage power-up if the micro doesn't have non-volatile memory. <S> Caution <S> If this is an office building with air conditioning, you may have trouble with varying pressures on different floors. <S> This may be high enough to 'swamp' the readings between floors. <S> One way of avoiding this may be to read only when the elevator doors are closed and monitor lift-shaft pressure but this may vary also due to compression of the air during descent and vice versa. <A> Except for storm front events, the pressure will change quite slowly, but you can update as often as needed. <S> Cheers, Ross <A> Install a switch to detect a specific floor, it could be top, basement, or any in between. <S> Switch options are actual mechanical switch, magnet and Hall sensor, light and photodiode, the possibilities are legion. <S> Then take the pressure whenever you activate that switch. <S> How you use that reading to update your reference is another matter. <S> You could simply use the reading as the new reference, or you could average it with the last (say) ten readings to reduce the inevitable reading noise.
| Measure the ground level pressure and adjust your in-elevator reading accordingly.
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How can I turn off a 3.3V circuit with a 3.3V microcontroller I'm working with ATTiny85 that has low current (40 mA) GPIO. In the other side I have a power supply (battery with ~5V) connected to a LM317 and an ESP8266 that demands high current (more than the ATTiny85 can give, around 170 mA) and uses 3.3V. I need to switch the ESP8266 (on/off). I tried some Darlington transistors, MOSFET and optocoupler but nothing works. All the examples that I see in the web is to switch from a 3.3V signal to a 5V (or more) circuit. In my case I want to switch from a 3.3V signal (ATTiny85 low current) to a 3.3V circuit (ESP8266 high current). Is this possible? And is so, how can I do this? (I have a little background in electronic but I'm from computer science, so please try to be soft with your answers, thank you) <Q> You should show us what circuit you've tried, but your simplest solution is this case would be a single P-channel MOSFET. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> When your ATtiny's GPIO pin is high, the MOSFET will be switched off, and when the pin is low the MOSFET will be switched on. <S> Make sure you connect the Drain & Source pins the right way! <S> If you get it wrong <S> the MOSFET's body diode will conduct and you won't be able to turn it off. <S> Note that this simple circuit will only work when you're switching the same supply voltage that your ATtiny runs from, and requires a MOSFET with a suitably low gate threshold voltage spec. <A> Currently I don't have the ESP8266 module to test this, but I think that you can use CH_PD <S> pin to turn it on and off. <S> CH_PD <S> pin. <S> When you set CH_PD <S> pin high (3.3V), ESP8266 should turn on. <S> And when you set CH_PD pin low (GND), it should turn off. <S> Getting Started with ESP8266 - Basic wiring <A> You can use a Reed Relay. <S> Choose a relay with 5 volt coil. <S> (or if you can find 3.3v coil). <S> Your 3.3v output pin can drive an NPN transistor to energise the reed relay coil. <S> The relay contacts can connect however you desire to power the ESP8266. <S> Common Data Sheet for reed relay
| No additional components required, just connect ATTiny85 and ESP8266 directly to your power source (3.3V from LM317) and one of the MCU GPIO pins to the ESP8266
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Why float data type is not accurate on a STM32? I'm using STM32F103C8T6 and my compiler is KEIL 5. I have written a simple program to check something but I found out an interesting point. When I download the program to the MCU, aaa and bbb have to be 12.56 and 62.8 , but when I debug the program on the MCU, the initial values of these variables becomes these: aaa=12.5600004 and bbb=62.8000031 . the program: #include "stm32f10x.h"float aaa=12.56,bbb;int main(void){ bbb=aaa*5; while(1){}} Also keep in mind if I change float to double , it works without problem. <Q> Floating point numbers are represented as a sum of binary fractions times a power of two. <S> In single-precision IEEE-754 floating point numbers, the fractions range from \$1/{2 <S> ^0}\$ to \$1/2^{23}\$ and the powers of two range from \$2^{-126}\$ to \$2^{127}\$. <S> For example: $$12 = <S> \left(\frac <S> 1 1 + <S> \frac 1 2\right) <S> \cdot 2^3 <S> = 1.5 <S> \cdot 8 <S> = 12$$$$0.5 = <S> \left(\frac <S> 1 1\right) <S> \cdot 2^{-1} <S> = 1 <S> \cdot <S> \frac <S> 1 2 = 0.5$$$$12.5 = <S> \left(\frac <S> 1 1 + \frac <S> 1 2 + \frac 1 {16}\right) <S> \cdot 2^3 <S> = 1.5625 <S> \cdot 8 = 12.5$$$$62 = <S> \left(\frac <S> 1 1 + \frac <S> 1 2 + \frac <S> 1 4 + \frac <S> 1 8 + \frac 1 {16}\right) <S> \cdot 2^5 <S> = 1.9375 <S> \cdot 32 = <S> 62$$ <S> But 0.06 and <S> 0.8 cannot be evenly represented in this way, just as \$1/3\$ cannot be represented as a sum of decimal fractions: $$\frac 1 3 <S> \approx \left(\frac 3 {10} + \frac 3 {100} + \frac 3 {1000} + \frac 3 {10000} + \cdots <S> \right) \approx 0.33333\dots$$ $$0.06 <S> \approx <S> \left(\frac 1 1 + \frac <S> 1 2 + \frac <S> 1 4 + \frac <S> 1 8 + \frac 1 {32} + \frac 1 {128} + \frac 1 {256} + \frac 1 {512} + \frac 1 {16384} + \frac 1 {65536} + \frac 1 {1048576} + \frac 1 {2097152} + \frac 1 {4194304} + \frac 1 {8388608}\right) <S> \cdot 2^{-5} \approx 1.91999995708465576171875 <S> \cdot \frac <S> 1 {32} \approx <S> 0.0599999986588954925537109375$$ <S> The decimal representations may vary from system to system. <S> I'm not going to spend a lot of time writing out more examples, because hopefully you get the idea by now. <S> Floating point numbers are intended to be approximate. <S> If you need to represent exact values, use integers or another alternative . <S> If you don't know whether you need floating point, you probably don't. <S> For more information, I've found the Floating Point Guide to be easier to read than Goldberg's essay , but both are valuable. <A> The comments have covered the basics, but for a complete understanding of what's going on, I recommend <S> You're Going To Have To Think! <S> which is a series of articles. <S> In short, floats/doubles are approximations because they have to squeeze an infinite amount of values into a fixed size representation, so there will always be approximation errors. <S> To not get surprised by how and when those errors appear, you need to understand completely how floats work. <S> Not only do the articles do a good job of describing the principles, but it opens with a great intro: <S> The dragon of numerical error is not often roused from his slumber, but if incautiously approached he will occasionally inflict catastrophic damage upon the unwary programmer's calculations. <S> So much so that some programmers, having chanced upon him in the forests of IEEE 754 floating point arithmetic, advise their fellows against travelling in that fair land. <S> In this series of articles we shall explore the world of numerical computing, contrasting floating point arithmetic with some of the techniques that have been proposed as safer replacements for it. <S> We shall learn that the dragon's territory is far reaching indeed and that in general we must tread carefully if we fear his devastating attention. <A> There's an app note from STMicroelectronics, " AN4044 Application note Using floating-point unit (FPU) with STM32F405/07xx and STM32F415/417xx microcontrollers ". <S> I don't believe your STM32F103C8T6 has a hardware floating point unit (FPU), but its still interesting reading. <S> All ARM processors, as well as practically every other current microprocessor/microcontroller use a standard called IEEE 754-1985 to represent floating point numbers. <S> It was adopted in 1985 and its first use in a hardware FPU was the 8087 chip from Intel, which was a companion to the 8086. <S> (The original standard has actually been superseded by a newer one called IEEE-2008, which is backwards compatible.) <S> The two principal formats int IEEE standard are 32 bit (called float in the C language), and 64 bit (called double in C): <S> (This figure is adapted from the App note I referenced earlier.) <S> Prior to 1985, the representation of floating point numbers varied from one manufacturer to another, for example the IBM 360 mainframes introduced in the 1960's <S> had a similar format, but the exponent in the single precision floating point representation was 7 bits instead of 8, and the unsigned integer part was 24 bits instead of 23. <S> So it would of had one more bit of resolution, but only a exponent range of -62,+63 <S> instead of -126,+127. <S> The Digital Equipment VAX-11 format was really bizarre . <S> Others have included a bunch of math to show the imprecision of various floating point numbers, but just on an intuitive basis, if you look at just the 23 bit integer portion of a float, plus the 1 bit sign, and a corresponding 24 bit signed integer which has a maximum value of ±10\$^{23}\$, or ±8388608, and line up that value with one of your single precision floats, 8388608 <S> 12.5600004 <S> you will see that the precision is about the same. <S> I'm sure all the math pundits are groaning at this point, I'm just try to show how to look at this from a 50,000 foot level. <S> Anyway it helped me visualize it.
| The important thing is that floating point numbers are binary rational numbers , even though you put decimal rational numbers into your source code!
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Can I use V=IR in the analysis of AC circuits? My textbooks usually just assert this relation so I don't really know when it doesn't apply. EDIT: Reflecting on this question, I wanted to clarify that I wanted to know if Ohm's Law is valid for the analysis of time-varying current/time varying voltage/ time varying load i.e., V(t) = I(t)R(t). <Q> In an AC circuit <S> V= IR like this: - In other words I and V follow each other and have a constant ratio to each other. <S> That constant ratio (V/I) is still R. <S> For other types of components like capacitors the relationship between V and I is more complex: <S> - Now the current is "time displaced" and does not rise and fall in sync with the voltage. <S> Given that any resistor has a small amount of parasitic capacitance means that at high frequencies there begins a subtle change in the phase relationship between V and I. <S> This change gets bigger and bigger at really high frequencies and, the resistor begins to look more like a capacitor. <S> I've purposely not mentioned inductors just to keep the answer simple. <A> Simple answer: <S> Yes, Ohm's Law still applies in AC circuits. <S> The difference is that AC circuits involve complex sources and impedances which vary with either time or frequency, so your \$V, I, \&\ R\$ aren't always real numbers, but complex expressions. <S> Nevertheless, the relationships established in Ohms Law for DC circuits will ALWAYS apply. <A> Yes, generally you can, but you should also take into account electromotive force produced by C and L in your circuit at each time point. <A> It is used in DC analysis mostly. <S> For AC analysis, you will need to learn impedances too. <A> Of course, the Ohms's Law is still valid in AC circuits too. <S> However, you will have to look out for components like capacitors, inductors (commonly used in AC Circuits) which have a different behavior relative to resistors. <S> The mathematics dealing with them, is different as it climbs up into Complex Domains.
| \$V=IR\$ is always valid in any circuit at any frequency but there are some things to watch out for and be aware of when the frequencies get really high or when there are non-linearities such as diodes to consider.
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Control DC MOTOR from headphone I would to know if is possible with simply way ( i'm not electronic expert ) to drive one direction a DC MOTOR ( a little ) from headphone, with power supply of course. The difficulty is voltage of headphone, is very low 3mvw-6mw, is suppose is not enough to active logical input of L293. I suppose is possible with transistor with "biais voltage", but how ? I want to that because, I want to recycle my old android phone, into camera IP, rotate 380°, with eyefish. I see it exist the IOIO board, but I think it is possible with cheaper parts. Thanks by advance <Q> You want an audio level detector. <S> simulate this circuit – <S> Schematic created using CircuitLab C1 blocks any DC from this circuit getting back to your phone. <S> R2, R3 and C2 hold the '+' input at half-supply. <S> R1 and R4 set the gain of the op-amp circuit to 10. <S> (Actually it's -10 because it's inverted.) <S> C4 blocks the DC from getting through to Q1. <S> R5 slowly bleeds off C4 holding it at 0 V. D1 and C3 form a rectifier. <S> When the audio signal exceeds the forward bias voltage of D1, C3 starts to charge up. <S> If the audio level is high enough the voltage on C3 will rise enough to turn on Q1 pulling the 'on' output towards 0 V. <S> You'll need to choose an op-amp to suit your power supply. <S> Q1 will gradually turn off as when the audio stops. <S> You may need to add a Schmitt trigger after it to provide a clean signal to your motor controller. <A> How to drive a motor from LM3915 simulate this circuit – Schematic created using CircuitLab <A> how would the phone produce the sound that is needed by the circuit? <S> would you call it and <S> it would auto-answer <S> , then you would send it the sound? <S> it may be possible to attach light sensors onto the screen and control the motor by displaying areas of varying brightness on the screen <S> you could set the phone to display a webpage (with auto refresh) <S> that is served by your computer <S> the webpage would be a simple black and white picture then changing the webpage on your computer would control the phone
| Make an audio signal that goes from off to full on to full off to switch the motor.
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Does mutual inductance change individual inductances? If I have two inductors, with inductances \$L_1\$ and \$L_2\$. I put them really close to each other to make a transformer. Are their individual inductances still \$L_1\$ and \$L_2\$? Or has the mutual inductance changed their individual inductance? <Q> The classic way of characterising a transformer's performance (after establishing n, the turns ratio) is to first measure the inductance of the primary - with the secondary open circuit. <S> This measurement is the "primary inductance" - effectively unaffected by the other winding since no current flows in it. <S> And the primary inductance is an impedance connected across the power source - effectively wasted power, and as it is a low impedance at low frequency it determines the low frequency performance of the transformer. <S> Then re-measure the primary, but with the secondary short circuited. <S> This is the "leakage inductance" (technically it's the parallel combination of primary and leakage inductances, but the primary inductance is usually a large enough impedance that it can be regarded as infinite, and ignored). <S> Anyway the "leakage inductance" is essentially the coupling factor of the transformer into a short circuit - so in a good transformer it will be a very low impedance. <S> (The same pair of measurements can be made on the secondary, with the primary open/short circuit. <S> It should give you the same result, scaled by n^2). <S> So the leakage inductance doesn't change the winding inductances - it couples one winding to the other, allowing the load impedance (scaled by 1/n^2) to appear in parallel with the winding inductance. <S> And the series combination of source impedance and primary inductance determine the LF response, while the series combination of leakage inductance and (load impedance/n^2) determine the HF response. <A> Let me shed some light on the physics background here. <S> Inductance of each coil can be calculated as \$ L = \frac{N^2}{R} \$. <S> The reluctance term is calculates as: \$ R = \frac{core <S> -length}{\mu <S> A_c} \$. <S> Therefore, the inductance is proportional to the permeability \$ \mu \$, which is a product of absolute permeability \$ \mu_o \$ and the relative permeability \$ \mu_r \$, which is 1 for air and ~thousands for magnetic materials. <S> Therefore, if you have two inductors with closed-loop magnetic path close to each other, the leakage fluxes will affect each other very little, well below 1%, since all the flux is contained in a low reluctance magnetic circuit . <S> However, if you have two air coils or somehow connect the magnetic circuits of the two individual coils, you will get much stronger mutual coupling since the magnetic flux is not constrained to the magnetic core. <S> Air cores inductors will have their individual inductances unchanged because the reluctance will not change. <S> Magnetic core inductors will see some change or reluctance (decrease) and therefore the individual inductances will decrease as well. <S> This is similar to adding inductors in parallel but the coupling is not done electrically but magnetically. <A> Are their individual inductances still L1 and L2? <S> Think about measuring the inductance of L1 with L2 open circuit - you will measure exactly the same inductance for L1 if it were a million miles away from L2 because L2 has no current flowing. <S> Sure it is now producing an open circuit voltage (just like a transformer would) but there are no amps in the secondary that can affect the magnetic field produced by L1 when measuring it.
| It depends what is connected to the other winding, which is why "mutual inductance" is also called "coupling factor" (They are not identical, but closely related terms).
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Practical Approach for getting maximum Power from a DC motor Theoretically, a DC Machine achieves its maximum Power when back emf is equal to half the applied voltage. Why this condition is not realized in practice ? What is the Practical Approach for getting maximum Power from DC machines ? <Q> 50% back-emf occurs at maximum power output , at which point the motor is slightly less than 50% efficient. <S> Why? <S> Because at 50% rpm half the voltage going into the motor is being dropped across the resistance of the armature windings, so half the power is wasted. <S> This becomes obvious when you look at the equivalent circuit of a DC motor. <S> Here is an example:- simulate this circuit – Schematic created using CircuitLab <S> Back-emf is caused by the armature spinning and generating a voltage which opposes the supply voltage. <S> The rest of the supply voltage is dropped across Rm. <S> In this case back-emf is 3.5V, so Rm must have 7V-3.5V = 3.5V across it. <S> 3.5V / 0.21Ω = <S> 16.7A. 3.5V * 16.7A = 58.3W of 'copper' loss. <S> 0.75A is used up in 'iron' losses, leaving 15.95A to produce output torque and power. <S> 3.5V <S> * 15.95A = <S> 55.83W. <S> Efficiency = output power / input power. <S> 55.83W <S> / (7V * 16.7A) = 47.9% efficiency. <S> Maximum efficiency is achieved when copper loss equals iron loss. <S> This usually occurs at around 80-90% of no-load rpm. <S> If you know Io and Rm then calculating this point is simple. <S> Iron loss = Vm-(Io*Rm) <S> * Io. <S> In this case that works out to 5.13W. <S> At peak efficiency copper loss is the same, so <S> Im 2 <S> * Rm = <S> 5.13W. <S> Solving for I <S> m <S> we get 4.94A. <S> Now calculate the voltage drop across Rm at this current, 4.94A <S> * 0.21Ω = 1.04V. <S> That leaves 5.96V for back-emf. <S> Now calculate the no-load back-emf. <S> Voltage drop across Rm is 0.75A <S> * 0.21&ohm = 0.158V. <S> 7V - 0.158V = <S> 6.84V. <S> 5.96V/6.84V = <S> 87% of no-load rpm. <S> Here's an actual dyno test of the motor simulated above. <S> Peak efficiency occurred at 5A (very close to the 4.94A we calculated) at 7360rpm. <S> This is 86% of the no-load speed, 8550rpm. <S> You wouldn't want to run this motor at maximum power for long because it only weighs 50g. <S> With over 50W of heat to get rid of it would quickly burn up. <S> DC motors are generally operated somewhere between peak efficiency and maximum power output, as this provides the best compromise between power, efficiency, and size. <A> Commutator DC motors are not used when efficiency is a high priority. <S> They are used in situations where cost is a high priority and DC is the most readily available power source. <S> They are used for auxiliary equipment in vehicles, for portable tools and appliances, as traction motors in small inexpensive vehicles like wheel chairs and golf carts and for toys. <S> In many applications where commutator DC motors have been used, there is not much opportunity to operate the motors under the most efficient conditions because the applications require the motor to operate at a wide range of speeds and torques. <S> Where larger motors are required and it is economically attractive to optimize motor efficiency, more efficient types of motors are used. <S> With efficient electronic controls, most other types of motors are more efficient than commutator DC motors. <S> That includes brushless <S> DC / permanent magnet synchronous motors, induction motors, wound rotor synchronous motors and reluctance motors. <S> In other words, "the Practical Approach for getting maximum Efficiency from DC machines" is to use another type of machine. <A> A practical approach is the empirical one! <S> Couple <S> the shafts of two identical motors together. <S> Apply and input voltage to one (motor) and a resistor on the other (generator). <S> Measure the input dc voltage/current and the output dc <S> voltage/current. <S> Find the combo with the best Pin/Pout and that is maximum efficiency for that machine.
| Try different combinations of resistors and input voltages (which correlates to torque-speed, respectively).
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Wiring Transformers in Series With a 240V Plug From a Generator? How should I wire 2 transformers in series (either primaries or secondaries)(not sure which I should use) using a NEMA L14-30 plug from a generator socket? I'm using 2 Microwave Oven Transformers to make a homemade stick welder. I was thinking about using a shared neutral for both. Questions: 1)Is this a good configuration? (Generator/Shared Neutral) 2)If it is, should I wire the primaries or secondaries in series, and what advantages (or much more likely disadvantages) will be involved in either choice? 3) Can you show/describe the best (or at least a recommended) approach for doing this off of this type of plug? <Q> If you have microwave oven transformers that are rated 120 volts and a generator with 120/240 volts, then it makes sense to connect one primary winding to one of the two hot terminals and the other to the second hot terminal and both windings to the neutral. <S> That will load both sides of the transformer rather that putting all of the load on one side. <S> You will need to replace the high-voltage secondary winding of each transformer with a low-voltage winding that you wind according to instructions that you find on the internet. <S> You can connect the two secondary windings in parallel. <S> This scheme will provide twice the current you would have by using just one transformer. <S> That being said, asking this question here <S> indicates that you are not very well qualified to undertake this project. <S> You can not expect to learn enough here to do this safely. <S> You should study some more about basic electrical circuits, transformers etc. <S> You should also look at as many plans for microwave oven transformer projects as you can find and make sure that you understand them. <S> After looking at another answer that you have. <S> I agree that connecting the secondaries in parallel is better. <S> Double the size of the secondary wire compared to what is recommended for a welder using a single transformer and use half as many turns. <S> That will give you the current increase with series windings. <A> secondaries in parallel could cause circulating currents if the transformer ratios don't match. <A> hope I can be helpfull First observation: <S> either the transformers is wired in series or paralel, both pimary and secondary have to be wired! <S> Second observation: <S> if you shared neutral of the transformers, then the only configuration is paralel (seies don´t share neutrals nor fase) 1) <S> The best configuration depends on your needs. <S> 2) Series configuration: <S> Same current, double voltage. <S> Paralel configuration: <S> Same voltage, double current. <S> All this is assuming that both transformer are exactly the same. <S> If not, for paralel configuration you need to check that the primary and secondary voltages need to bee the same as well also the shortcircuit voltage. <S> If not, there will be current flowing between the transformers. <S> Warning:
| if they're 120v transformers (and seeing as you've got a NEMA connecter there they probably are) connect them in series, and connect the neutral to equalise the voltage if there's a mis-match between the two transformers, also wire the secondaries in series. I agree with Charles Cowie, it may be dangerous to work with this if you are not qualified.
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How to measure temperature of a syrofoam cutter hot wire? I would like to create a hot wire foam cutter device. The wire material is nichrome, diameter 0.9mm, resistance is about 1.7 Ohm/meter. For best cutting results, I would like to control the temperature electronically. This will be a 4 axis CNC system where the wire length between the mounting points changes while the axes move. Planned length of the wire is abount 1.5m. Another extra (non-heating) wire is added, that is able to change its length to ensure the tension of the wire while the axes move. So the length of the heated part of the wire would not change, but it would be hard to measure the contraction caused by the heat. The main question is this: how can I measure the temperature of the wire? I was first thinking about a thermistor, but then I realized that the hot wire itself could be acting as a thermistor. When temperature goes up, then probably the resistance will also go up. However, I have no idea how to create a circuit that measures the current flowing through (it can be 10A), and how to calculate the temperature from that. Another problem is that the hot wire will be heated with a 24V 200W transformator, unregulated AC power. My idea for power control was to use a diode bridge, a high power fet for switching (e.g. IRF44N or similar) and PWM signal to control the temperature. But this would make temperature measurement impossible - PWM controlled AC cannot be used together with the other idea. The workaround I have in mind is this: use regulated DC and a simple A/D converter to measure the resistance of the wire in every (say) 1 second. Heating would be turned off completely during resistance measurement. This would take just a few msec. Use the PWM controlled driver to heat the wire in the remaining time. Would that work? How can I separate the heating circuit from the measurement circuit? (E.g. to protect the A/D converter safely.) The secondary question is this: do any of you have an idea about how much power and voltage I need to heat the wire? I thought that 200W would be enough. That is about 8 amps at 24V. But don't have any experience. Also, even if I can measure the resistance of the wire, I have no idea how to convert it to a temperature value. (It would be nice to display the absolute temperature value, but it is not important - all I care is to adjust and maintain a constant temperature that has the best cutting results.) I have a 24V 210W transformator for heating, but the power may not be enough. Or maybe you have an easier solution in mind - please don't hesistate to share. Mechanical part: Heating goes through the green lines and the red wire. Tension is provided by the gray cable (which is not heated). <Q> Because your supply is unregulated, you can try something like simulate this circuit – <S> Schematic created using CircuitLab <S> Using a current sense resistor (in this case, 3 0.1 ohm power resistors in parallel), you can find the current through the wire. <S> Then, knowing the voltage across the wire you can divide the voltage by the current to get the resistance. <S> Nominal current voltage will be about 0.27 volts, which is a fairly convenient voltage level to measure, and a x10 amplifier using an op amp should be simple enough if you need it. <S> For the setup shown, 3 5-watt resistors could be used, since the nominal dissipation for each resistor is about 2 watts. <S> Heatsinking the resistors will help accuracy. <S> While this is a sort of roundabout way of measuring resistance, it is arguably simpler than interrupting the heating cycle to make more conventional resistance measurements. <A> High current AC control simulate this circuit – Schematic created using CircuitLab <S> Given that your specification requires variable length cutting wire it appears that current control may be better. <S> A quick web search shows up several devices. <S> Nu Wave current controller . <S> I have never heard of this company before but something along these lines could provide you with an off-the shelf solution. <S> A brief look at the data sheet indicates that you need a separate SSR to switch the load and a small transformer to power the module. <S> Note that since \$P = I^2R\$ the potentiometer will give linear control of current but exponential control of power. <S> (Confirm the linear current with the manufacturers' data sheets.) <A> One of the problems with hot-wire cutting is that the wire expands with temperature and a tensioning system is required. <S> You could use this to your advantage and control wire temperature by controlling wire length. <S> This in turn could allow very simple control from your AC transformer through a 'phase-angle firing module'. <S> These are glorified dimmer controllers with a DC control input. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Your challenge with this approach is to come up with the linear transducer and some means of calibration. <S> Setup would be by trial and error. <S> For my one attempt at this I used a guitar string, E, 1st, 0.010". <S> I had a 12 V, 5 A power supply and adjusted a contact along the wire until I got the required result.
| Referring to a table of resistance for your wire, you should be able to derive the resistance and hence the temperature.
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Changing the resistor to be in front or behind an LED doesn't affect brightness? Note: I am very new to the more advanced side of electricity. I do not have a firm grasp on the exact mechanics. Lets say I have an example circuit: 5V Power Supply -> LED -> Resistor -> Ground Now the LED is receiving 20ma and 5V.If I were to switch the LED and the resistor however, the LED should receive less voltage but the same current, so wouldn't it be dimmer?Or would it receive the same voltage? I've looked at these two threads and their answer doesn't help. Why doesn't it matter if a resistor is before or behind an LED wrt voltage drop? Why does a resistor need to be on the anode of an LED? <Q> The current through them would be exactly the same. <S> Thus the brightness of the LED would be the same. <S> The voltage drops across the resistor and the LED would be the same since the current through them <S> is the same. <S> This is a characteristic of all series circuits. <A> The led DOESN'T receive 5 V across its terminals <S> But it only DROPS <S> some amount of voltage across it <S> Any Diode has a property called Forward voltage <S> this is simply the amount of voltage that is needed to turn this diode on <S> (Let the current flows through it) <S> Lets assume your LED has a forward voltage = <S> 1.7 V <S> And this LED is connected in Series with a 250 Ω resistance <S> Two possible configurations could be achieved Source - <S> > <S> Resistance - <S> > <S> LED -> <S> GND Source - <S> > <S> LED ->Resistance <S> ->GND <S> For the 1st configuration If we measured the voltage drop across the resistance it will be 3.3 V <S> So your source provided 5 V , 3.3 V dropped across the resistance and 1.7 V dropped across the LED (Forward voltage) <S> The sum of the voltage drop across the circuit elements must equal to <S> 5 V (3.3 + 1.7) Not more or less than 5 <S> For the 2nd configuration Same this will happen ( 1.7 V dropped across the LED , then 3.3 V are dropped across the resistance <S> So it <S> DOESN'T <S> matter where is the resistance as long as its connected in series with the LED <S> So why we put the resistance ? <S> Well in many electronics books this resistance is called the "Current limiting resistance" its used to avoid short circuits and its value is used to control the amount of current flowing in your circuit without this resistance huge amount of current will flow and burn your LED <S> What does this resistance has to do with the brightness of the LED? <S> Well as mentioned , the MORE the value of this resistance the LESS the amount of current flows in your circuit , the LESS the LED brightness <S> How can i control the brightness of the LED ? <S> You will need to know how much current is following in your circuit , since your circuit contains only a resistance (250 ohms) and a diode , Neglecting the resistance of diode <S> The amount of current is (Source voltage / Resistance) <- ohm`s law <A> As explained, it would receive the same voltage either way, but an experienced engineer would say there is only one correct way to normally do this. <S> The LED should be connected to ground, with the resistor between it and 5 V. <S> The resistor has two functions. <S> The first is to set the amount of current into the LED in normal use, as you have imagined. <S> The second is to protect the circuit from accidental damage. <S> If you are working on this circuit, you might be using something like an oscilloscope that has one clip lead that is normally ground. <S> If you connect the LED to the 5V, and accidentally touch the other end with ground, you will have no current limiting and may burn out the LED. <S> If you connect the resistor to 5 V, and the LED to ground, then touching the connection between them with ground draws more current through the resistor, but the LED gets no voltage since it is now grounded on both ends, draws no current, and therefore no damage is done.
| Switching the positions of the LED and resistor will have no effect on the brightness of the LED because they would still be in series.
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Direction of flow of current in secondary circuit in AC transformer/rectifier In this transformer/rectifier if the current in the primary circuit is clockwise what will be the direction of current in the secondary circuit?Why? I'm feeling a bit confused about the direction of flow of current due to flux linkage of primary coil with secondary.Please help! <Q> Using <S> Ampere's right hand rule : <S> Point your thumb in the direction of the current in the primary side <S> and you can get the direction of the magnetic field from the direction of rotation of the other four fingers. <S> The opposite is true, rotating the 4 fingers in the direction of the magnetic field will point your thumb to the direction of current. <S> Try it on the secondary side. <S> Therefore <S> *: <S> * <S> Assuming the primary and secondary windings are wrapped in the same direction around the core of the transformer. <S> Read more about the dot notation: <A> The direction of current in the primary circuit is clockwise & anti-clockwise as it is an AC source. <S> The direction of current in the secondary is clockwise only (ignoring reverse leakage) <A> As the current increases the rate of change of flux generates an emf which opposes the applied voltage. <S> The transformed is designed so that only a small current builds up in each half cycle of the applied voltage. <S> Now if we add a load current then the secondary current induces opposing flux in the transformer core. <S> This requires the primary to supply increased flux in order to induce the sane emf. <S> This extra flux requires primary current to generate it. <S> As we increase load current then flux in the core is initially reduced (due to the opposing flux). <S> This reduces the opposing emf in the primary. <S> This causes more current to flow in the primary thereby restoring the flux level. <S> As secondary current only flows in the half cycle where the top of the winding is positive, then the additional primary current only flows in this half cycle too. <S> This is why the large current only flows clockwise in the primary. <S> There will still be a small anticlockwise primary current on the other half cycle, just enough to generate the opposing emf.
| If there is no secondary load, when the top of the primary is positive, a small current flows clockwise in the primary circuit.
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Why DRAM costs much more than flash memory? Comparing prices at a local store, I calculated prices 0.36 USD/GB for SSD NAND flash memory and 5.41 USD/GB for DRAM memory. The difference is 15 times. Why so big difference? Both are semiconductor devices. Memory cells of both types occupy approximately the same die area: 4 F^2 for NAND flash memory, 6 F^2 for DRAM. (I can't give a good source; this information is scattered over the internet.) MLC of flash memory gives 2 times improvement. (BTW, is the same technology possible for DRAM?) I can't blame market forces because both markets seem similar to me in terms of competition, maturity, and size. For example, DRAM revenue 35.74e9 USD, SSD revenue 10.9e9 USD in 2013. <Q> The key element in the difference is redundancy. <S> A DRAM has to be perfect (that is, it has no redundancy). <S> This is because, in order to get the speeds required of DRAM, there is no buffer level between the address inputs and the memory cells except for address decoding. <S> Flash, on the other hand, is not expected to run nearly as fast as DRAM, and this allows the insertion of various logic into the address logic. <S> Specifically, data is organized into blocks (equivalent to pages), and there are a fair number of spare blocks included on the chip. <S> When an access is attempted the address essentially goes into a lookup table which redirects access from a known bad block to a good (spare) one. <S> The result is that manufacturing can be much cheaper for flash than DRAM, since it is much harder to produce a bad (unusable) chip for flash. <A> Supply and demand. <S> Also, there are many manufacturers across many different process nodes. <S> There are only four (as of 2016) makers of DRAM on the latest process node. <S> 5$/GB is a per GB DRAM on a module. <S> 8 to 36 chips on a module depends on the configuration. <S> Cheap SSDs can have as little as two chips + passives. <S> (M.2 PCIe controller and NAND flash). <S> From the above you can see that on average, DRAM has a larger failure potential than an SSD, therefore, requires higher price. <A> From the IC designer perspective, FLASH is much more expensive to make than DRAMs because I use more masks, and they take more area. <S> I don't know where you got your area numbers, but I can believe that if you had a FLASH only process, I could tighten it up a bit. <S> Of course, that would be expensive. <S> You need to have a comparison that is uniform. <S> I get a single bit of FLASH in the space that I get 8-bits of DRAM trench caps on a modern FinFET process. <S> I've attached a picture of the DRAM layout for a 8-bit DRAM cell. <S> FLASH needs to not leak, and DRAM can be leaky <S> so I don't need to worry about my oxide thickness or quality so much. <S> Also, every time you program FLASH, you use hot electron injection that can cause oxide damage statically. <S> The greater the VDS, <S> the faster you program because of more hot-electrons, but <S> the more risky you get a chance of having a hot-hole make an intermediate state, so you want thicker oxide. <S> Growing ticker oxide costs time for deposition, and therefore money, but just on your question: If you have a comparable process, I would expect the cost between FLASH and DRAM to be 8x for bit equivalence just on area. <S> I expect that the discrepancy that you see between DRAM and FLASH is just process cost. <A> Flash can work and be sold with bad bits scattered all over the chip and the controller will just hide the faults. <S> DRAM must operate at much higher frequencies, and it must be perfect. <S> Any single bit error in a DRAM die could cause that die to be discarded (unless there is redundancy in its design), and the dies are not small. <S> See this previous question: Yields in DRAM and other Massively Redundant Processes
| I think the answer is that the different technologies have different yields. There is much bigger supply of NAND chips out there.
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How can I test a wire for small holes? I have stripped one wire, the black/hot/positive one, for a 120v appliance. I had to cut away the outer plastic that surrouneds all three wires, and now I'm afraid I may have nicked a small hole into one or more of the wires, and am deathly afraid of getting shocked. Is there a way to test for very small holes in wires? The ole put-water-on-something-inflatable-and-listen/feel-for-bubbles comes to mind, but this is electronics. I suppose I could closely inspect the wires, but am hoping there's a better way. <Q> Plastic insulation for the mains is quite thick so any damage has to be pretty macroscopic to cause danger. <S> Electrons won't come spilling out of a pinhole like water from a punctured tube (at least at mains potential). <S> Just inspect it thoroughly, bending it back and forth and look for any cracks or any visible copper. <S> If you find some, shorten the cord to eliminate it. <S> I could tell you to dunk it into saline solution and check the resistance <S> but I think that would be a terrible idea. <A> The problem with nicking the wire is that you have caused mechanical damage. <S> This isn't going to be a problem in the near term, but in the future as you wiggle the wire around\pull on it, it can cause a point of strain and break. <S> When manufacturing cables this can be a problem, and especially aerospace industries take great care in avoiding it. <S> Even inspecting pins after stripping with a microscope and having people sign off on it before they will put the pin in the housing. <S> You'll probably be fine if you haven't cut through to the copper or cut through more than 20% of the insulation. <S> If you have, try and find the point where you though you nicked it and wiggle it until you see the break, then electrical tape it tightly to give it strain relief. <S> If your worried about it for safety's sake, just replace the whole cable. <A> As someone else suggested, I would check for less than infinite resistance when the supposedly insulated part of the wire is immersed in saline solution. <S> Pinholes or cracks in the insulation will make conductive spots. <S> Keep both ends out of the solution when testing though. <A> Essentially you want to break the small holes out to a larger area. <S> Its not safe, but you are lookin for something along the lines of drape the cable through water (not the end). <S> Then measure the water for a voltage. <S> If you have a voltage, then you have a hole... <S> I don't think you should mess with this, as it is dangerous, but I just wanted to answer your question. <S> A Eugene Sh. had a much better suggestion of electrical tape.
| If you think you might have damaged the cord in some particular place, you may be able to just remove that end and shorten it a bit. Or you could connect it to a 25kV+ source and observe it in a dark room looking for a purple corona discharge (an even more terrible idea).
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Whats the standard for denoting the orientation of an LED on a PCB? Whats the standard for denoting the orientation of an LED (or diode in general) when printing a PCB? I'm drawing up a design for students to use in a course and want to make sure I'm not misleading them. <Q> Make a pad for the cathode square. <S> This works even if your PCBs don't have silk screen. <S> If you do have silk screen on the PCB, then draw a flat on the cathode side which would correspond to the flat on the actual part. <S> (mechanical drawing for a typical T-1 3/4 throughole LED from datasheet ) <S> Orient all diodes in the same direction. <S> It reduce errors during manual assembly. <S> This guideline applies to other types of polarized components too. <S> The PCB layouts above were sketched in ExpressPCB. <S> Typically, the footprints for diodes that come in the libraries with PCB layout packages have some way of showing polarity. <S> related: <S> How Should You Mark Your Diodes? <S> (write-up in a free industry journal) <A> I don't think there is a standard, but there definitely is some sort of convention. <S> If you see a design with a diode and some sort of marking, then your best bet is that the marked side is a Cathode. <S> Only in very tight layouts I will revert to a "dot" to mark the cathode. <A> Usually the cathode is marked with a bar similar to what Nick Alexeev showed. <S> But there is one simple advice: put a marking that matches the marking on the component! <S> If it's a round 5mm LED, draw the cut-off edge of the housing. <S> If the component has a marking on its body, try to copy it to the silk screen. <S> Nothing is more irritating than non-matching markings. <S> For example, I regularly use a 0402 LED with a "T"-shaped marking on its bottom. <S> Unfortunately the cross-bar of the T is on the anode side. <S> Always drives me crazy figuring out what the right orientation is... <A> Screaming Circuits weighing in here... <S> The closest thing to an actual "standard" is the cathode bar. <S> The problem is that a fair number of people mark their diodes with a minus sign, which usually indicates the cathode, but not always. <S> FLyback, Zeners, TVS and a few others generally have the anode on the negative side. <S> The biggest problem occurs with small surface mount LEDs. <S> Even worse, some part manufacturers have been known to use the same marking as both an anode and cathode indicator. <S> The diode symbol, if you can fit it, is about the only way to really get rid of this ambiguity. <S> I've since updated the document referenced to indicate that "K" is more commonly used to indicate Cathode than "C", to reduce possible confusion with a capacitor.
| I tend to follow the Diode Marking Guidelines from Screaming Circuits and place a little diode symbol next to my diodes. There is no standard for polarity marking on the part. Marking the Anode is not common, but some manufacturers do that, and some designers do when they've got common anode multi LEDs on the board.
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What does a capacitor capacity really mean? I am very new to electronics, and I am trying to use a capacitor to make my LED's stay on a little longer after I turn it off. I know that the voltage on the capacitor is the maximum limit of voltage it can take, and the fared unit of the capacitor is the capacitance, but that wasn't really clear to me, so I made a list of some stuff that I need to know to complete my goal. Does a higher capacitance affect the time the charge is held? If I had 5 100uf capacitors, and I connected them in parallel (which will give me 500uf), will it perform the same way as a 500uf capacitor (that was not combined with others)? Does different types of capacitors hold charges longer than others? Thank you all for feedback, help, and tips :) <Q> Does a higher capacitance effect the time the charge is held? <S> A higher capacitance can store more charge (\$Q = CV\$), which can then supply a current for a longer period (assuming a constant load). <S> If I had 5 100uf capacitors, and I connected them in parallel(which will give me 500uf), will it perform the same way as a 500uf capacitor(that was not combined with others)? <S> Yes. <S> Does different types of capacitors hold charges longer than others? <S> A 1uF ceramic cap and a 1uF electrolytic cap are both 1uF, and therefore hold the same amount of charge. <S> However, certain types of capacitors have different characteristics <S> (ceramic caps cannot really be used to make large caps, for instance). <S> If you want to know how long a capacitor which is not connected to anything will hold its charge, that is dependent on the leakage current of the capacitor, which in turn is highly dependent on capacitor type and construction. <A> If you compare a 100uF capacitor and a 500uF capacitor (by the way, that is huge) at a certain voltage, say 10V, the 500uF capacitor will contain 5x the number of electrons. <S> The equation that relates charge, voltage, and capacitance is$$Q = <S> C <S> \times V$$ <A> For easy understanding of capacitance of capacitor, think of similar notions that use a similar word. <S> The capacity of a milk carton is the volume of milk it can store <S> The heat capacity of an object is amount of energy an object <S> can store per unit of temperature difference " <S> IN SIMILAR MANNER" <S> The capacitance of the capacitor is the amount of charge that it can store per unit of potential difference.
| The "Capacitance" boils down to how much charge a capacitor holds at a particular voltage.
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Understanding Test Parameters on Datasheet (CL pF) On Page 7 of the datasheet for M74HC590 It uses a CL (pF) At the bottom of Page 8 is the test circuit which shows where the CL(pF) goes, in regards to a circuit diagram, but I don't understand that circuit diagram. Is the D.U.T. the chip they are testing? For context what I'd like to do when I get these chips is put 4 of them in series to count the amount of time it takes sound to go from one ultrasonic transducer to another with as high of accuracy as I can. So I can calculate the speed of air traveling through a pipe so I can know the volume of air traveling through it, In the volume of 1-200 Liter per minute hopefully with a resolution greater than .1 LPM through a 1" PVC pipe. I could probably go smaller but I am limited by the blower blowing the air to have at least 2 Kpa with a minimum of 200 LPM and I don't have any information on how the flow will drop as the pressure increases. So I'm worried if I go too small on the pipe I'll get to much of a pressure drop before I get to the outlet where I need that much pressure. 240 LPM @ 32k RPM 4.5A 0 Pressure vs At Zero flow 5Kpa of pressure 1.6A 36k RPM Is the blower output Side Question <Q> For the test conditions part of your question, CL is a specific test load applied to each output to measure parametric data, and a different load is used for different tests. <S> DUT is indeed the D evice U nder T est. <S> At the test circuit, there is a note: CL = 50pF/150pF or equivalent (includes jig and probe capacitance) . <S> Looking at the device parameters (page 7 in particular), under Test Conditions , you will see CL. <S> Note that the propagation delay and the high impedance to output valid timings are tested with load capacitances of both 50pF and 150pF, with (expected) longer timings with a higher capacitance load. <S> All other tests are measured with a 50pF total load. <S> An interesting parameter that you ignore at your peril is in the recommended operating conditions: Input Rise and Fall Time . <S> There are maximum times listed. <S> This is very well explained in a TI application note, and is a well known issue in CMOS devices. <S> [Updated for application use] <S> To use these together, I would operate the counter clocks synchronously (i.e. the counter clock for all the stages is the same clock) and use the application suggestion in the datasheet for gating using the RCO to CCKEN for each subsequent stage. <S> There will be some timing uncertainty in your measurement results due to the propagation delay from clock to output Q(n) changing state. <S> The maximum frequency you can run at in cascaded mode is determined by the propagation delay (max) from CCK to RCO (upstream output) and the input setup time (min) of CCKEN to CCK (downstream input) <S> Together, these have a worst case timing of 80 nanoseconds. <S> That yields Fmax of 12.5MHz <A> That's a digital chip, which means its outputs ideally transition instantaneously from one state to the other. <S> Capacitance on one of these outputs slows down the edges. <S> You should not add any deliberate capacitance. <S> Some unintentional capacitance will be unavoidable. <S> This is telling you how much of such capacitance is OK. <A> Yes DUT is 'Device Under Test' - <S> The circuit is just representative of the test circuit used to measure the timing waveforms shown on the next page, and CL is just the capacitance of the test probe - you don't have to add this in your application.
| They are telling you either the capacitance they did the tests with, or the maximum capacitance you can hang on a output and still have the chip perform according to the rest of the specifications.
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Using an opto-isolator in a car's circuit. How to ground? So I have a microcontroller that will read some of the car's sensor output (0-5v) and i plan to send the signals through an opto-isolator. The problem is that when i simulate this, it seems i cannot connect the car's ground with the MCU's ground. I later learn that this should not be done in an opto-isolated circuit. My question now is, how is this going to work if my MCU is powered off the car's power source? car's 12-14v & GND -> switching regulator-> 5v -> powers MCU car's sensor -> opto-isolator-> GND How can I get it to work without the two GND's coming into contact with each other? Would the 5V regulator isolate the two GNDs'? <Q> Ignoring the question of whether you actually need an optoisolator for the moment, the specific problem you're having in the circuit simulator is that the simulator itself requires a DC path among all nodes of the circuit being simulated — otherwise, the math does not converge on a solution. <S> The traditional fix for this is to connect a high-value resistor (e.g., 10MΩ or more) between your two "isolated" grounds only for purposes of simulation . <S> It will have negligible effect on the actual solution. <S> You would do this wherever you have an "island" of circuitry that is totally isolated from the simulator's ground (node 0) by capacitors, transformers, relays, optical couplers, etc. <A> Nothing in your description explains why you need isolation at all. <S> Connect the negative output of the 5 V power supply to the car's ground. <S> Car signals can now be connected directly to this circuit. <S> Of course the circuit has to take into account that the signals coming from the rest of the car can exceed the 0-5 V range of the micro, and can have occasional spikes on them. <S> This can be addressed by clamping, a transistor as a buffer, and the like. <S> A opto-isolator can be used as a buffer too, but since isolation isn't required, just a transistor is simpler and serves the same purpose. <A> It seems like you're mixing up things that you don't understand. <S> I think that in your case you can simply connect the grounds together. <S> "I later learn that this should not be done in an opto-isolated circuit. <S> "And <S> do you understand why that is ? <S> There are many cases where this is not required and where it is perfectly OK to just connect the grounds. <S> In other cases it is completely forbidden <S> (mains adapters and other mains isolated circuits). <S> Confused yet ? <S> My point is, such general "do this" and "don't do this" rules are useless if you do not understand the why . <S> "Would the 5V regulator isolate the two <S> GNDs'?"Only if you're using an isolated regulator which is not needed and too expensive for your application. <S> Note that almost all (non isolated) <S> regulators actually have both grounds shorted ! <S> You could use an LM2596 based switching regulator and it will have only one common ground connection. <S> So no, the regulator will not separate the ground . <S> And like I said, there is no need ! <S> So just get an LM2596 board, connect 12V to input, feed 5 V output to your MCU.
| Use the optoisolator to interface but connect all grounds together and that should work fine.
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