Capacitor discharge... - Page 1
Forum section: electronics / beginners

author: banedon
author_rank: Regular Contributor
timestamp: March 16, 2015, 08:11:44 pm
content: Hi guysI don't know a huge amount about capacitors (hence this post) and I have a question regarding them:When you short out a capacitor to discharge it, I assume that it all it discharges at the stored voltage and dumps the current in one hit?I ask this as I put together a test unregulated 12V PSU made of the following:Transformer (230V primary, 2x 12v secondary)Full bridge rectifier1000uF 16V electrolytic capacitor(see attached diagram)I put the circuit together, played around with it and then decided to disassemble it.As I'm cautious around mains (although the cap wasn't storing mains... I know - just being careful), I decided to short the capacitor against my solder real instead of just picking it up.A nice spark and crackle later I examined the solder real to find pits in the solder.I suspect that part of this is the energy from the cap leaping the spark gap just before making contact with the solder real, thus making the pits, but I'm not sure. And I do wander what 1000uF @ 12V (14V unloaded) would actually give when discharged at once. I.e. What happens with the current? Could you get a bad shock from a large capacitor but rated low voltage-wise? I assume yes as I've always been told that current kills, not volts.

author: TimFox
author_rank: Super Contributor
timestamp: March 16, 2015, 08:22:10 pm
content: As the capacitor discharges, the voltage falls. The charge Q = C x V, so the voltage V = Q/C falls as the charge flows out of the capacitor. This is true for any value of the discharge-circuit resistance: lower resistance makes the discharge current higher and therefore the time required to remove the charge faster.This is the difference between discharging a capacitor and discharging a battery. The total energy stored in the capacitor is E = (1/2)QV, while the equivalent energy stored in a battery is E = QV (for an ideal battery).When discharging a capacitor through your body, the current is limited by the resistance of your skin, which is higher than that of your body's interior. If your skin is wet, then a relatively low voltage could deliver a lethal current, usually considered to be between 100 mA and 200 mA.

author: picandmix
author_rank: Frequent Contributor
timestamp: March 16, 2015, 08:26:12 pm
content: You can indeed get very bad shock or burns from un-discharged capacitors.The ones on a basic low voltage circuit are bad enough but when working on a switch mode power supplies where the input capacitors are at very high voltage then its can be real trouble.Its always best to have a 100 ohm or so, 3 to 7 watt resistor so you can carefully use that across the capacitor to drain it off.A led and suitable resistor would give you a visual indication as well.Many will say just use a pair of pliers, but that risks getting a shock and burning tracks or components.

author: Zero999
author_rank: Super Contributor
timestamp: March 16, 2015, 09:56:45 pm
content: A 16V capacitor on a 12V transformer is not enough. 14V does sound a little low. The peak voltage is 12*sqrt(2) = 17V minus the rectifier losses but it can be much higher off load and even more if the mains is on the higher end of its tolerance. The capacitor should be rated to 25V or more.There's a lot of misunderstanding about electrical shock.No you won't get a shock from a capacitor charged to such a low voltage. It may be current which kills but your skin has such a high resistance that it will not allow a lethal current to flow at only 16V. If the voltage was quadrupled (64V) then you may feel a slight tingle if your skin was moist but again it would be unlikely to kill and you may not even feel it. In fact capacitors charged to less than 100V pose little electric shock risk. DC is less likely to disrupt the heart's rhythm than AC and there's a finite amount of energy stored: unless the capacitor is massive, at very low voltage the current will decay to a safe level before any real damage is done, so may feel a shock but nothing more.It's also the energy stored inside a capacitor which kills. A 100pF capacitor charged to 10 000V will give you a small shock but it won't kill you because the current will decay to a safe level before it has a chance to cause any damage. It will feel like a shock from static electricity because that's what it is.It's capacitors from photo flash units and switched mode power supplies which are very dangerous. Not only does the voltage exceed a couple of hundred volts but there's a lot of energy stored in there. The large current pulse can easily stop your heart.The current path is also important. If you hold one terminal of the capacitor in one hand and the other in the other hand, the current can flow through your heart and cause it to stop if it's high enough and lasts for along enough but if you just bridge the terminals with your finger, it'll hurt like hell and if it's a large capacitor charged to a high voltage, suffer severe burns but it won't kill you.The capacitor made such a large spark because a huge current flowed through the solder which also vaporises a bit, making it appear more dramatic. This should be avoided because it's not good for the capacitor and if it were in a circuit, the large current could damage other components.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 17, 2015, 04:00:20 am
content: Sparks and short circuits are surprisingly destructive, even when small!When you short out a capacitor, you are placing a low resistance across it. Exactly how this happens varies with voltage. At low voltages (under 30V or so), nothing happens until the metals very nearly make contact (within nanometers) at some point. When that happens, the voltage across the gap suddenly equalizes, drawing current and launching an electromagnetic wavefront (aka a pulse). The risetime of this transient is very small: under a nanosecond. (For a long time -- until the 1960s or so -- the fastest pulse generators were, in fact, mechanically switched transmission lines. Mercury-wetted reed relays yield a risetime in the 300ps range!)Risetimes this short necessarily contain wide ranges of frequencies, well into the GHz -- hence why sparking causes widespread radio interference, even against advanced, modern, high frequency (cellphone, Wifi, etc.) channels!As the wave propagates, it communicates the current flow along the wires connecting the short to the rest of the circuit. When the wave bounces into something with a low impedance (like a capacitor), it bounces off, and ricochets back and forth. Each time, inducing more and more current flow. Eventually (after a few bounces or so), the sharpness of the wave decays, and it blurs into a straight ramping current flow.For a "long" time, not much happens: the wave decays in nanoseconds (depending on the physical dimensions of the structure, and the speed of light, or something near it), and current continues to ramp up over microseconds.Eventually, the current rises so high that the voltage drop across circuit resistances and capacitances becomes considerable. Now the voltage begins to drop, while the current continues to rise (but more slowly, as the voltage drops). If there isn't much resistance in the loop, the voltage will go down to zero (and current ceases to increase), but the current is still huge, so the voltage is pulled negative. After it goes negative a while, current reverses, then undershoots, and so on -- it rocks back and forth between voltage and current, resonating.If the resistance is relatively large, the inductive component will only matter for the rising edge; once there, current kind of saturates (it levels off along an inverse exponential curve), then voltage and current decay, proportionally, over a longer time period.We can model the discharge event with an RLC series circuit. Suppose the capacitor is 1000uF, and initially charged to 17V. The resistor might be 50 milliohms (typical of the ESR of such a capacitor), and the inductance, probably less than 20nH (more if you're shorting it from the opposite end of some wires, etc.). We can calculate which situation will result, based on the value of R: if R < sqrt(L/C), it will resonate and polarity will reverse; if R = sqrt(L/C), it will be critically damped, and kind of go "thud" as fast as possible; if R > sqrt(L/C), it will rise with an L/R time constant, then decay with an R*C time constant.Since 0.05 ohm >> sqrt(0.02uH / 1000uF), it will follow the resistive decay pattern.It will take on the order of t1 = (0.02uH)/(0.05 ohm) = 0.4us for current to ramp up (the wave-bouncing-around, then dampening, then ramping, then leveling off, phase), and t2 = (0.05 ohm)*(1000uF) = 50us for current and voltage to decay together. At the peak, current will be about (17V) / (0.05 ohm) = 340A, for a peak power dissipation of 5.78kW! Most of this will be in the capacitor's internal ESR, but the same current flows through everything, even the nanoscopic contact point, which gets heated into a microscopic melt zone, then vaporized into a crater as ionized air and metal forms a plasma ball -- a spark.At higher voltages, the metal surfaces don't quite make contact before a spark jumps through the air. This is most likely to occur at a sharp point, edge, scratch or other feature on the surfaces, that causes the electric field to be stronger than elsewhere. Polished, round surfaces have higher breakdown voltage than sharp needles. The RLC equivalent circuit remains relevant, but at higher voltages, the capacitance is probably smaller, and you're more likely to see resonant type behavior.At very high voltages, the sheer physical size required to attain high capacitances introduces so much inductance that a resonant discharge is very common, limited only by resistance instead.Take ESD (electrostatic discharge) for example. The human body has a capacitance (against free space) of about 100-200pF. Skin resistance varies dramatically with voltage, frequency and hydration, but at the voltages and frequencies we're talking here (kV+ and MHz+), the equivalent resistance is in the 100-2000 ohm range. The inductance depends on what part of your body makes the zap, but it'll be in the 0.1 to 2uH range.Let's say it's a good dry winter's day and you've been scuffling around on carpet, and you get charged to 10kV. Then you sit down at the computer and plug in a thumbdr*ZAP*OW!If your ESR is 1kohm, your peak current is about 10A, with a time constant of about sqrt(L*C) ~= 14ns! That's a sharp pulse, and no current to laugh at -- considering 20mA is enough to kill, at least if it's applied for quite a bit longer than 14ns.Suppose that zap was applied directly to the thumbdrive's (or computer's) USB D+/- pin: that's 10A in 14ns, ripping through a circuit designed for only 5V and 20mA! This much current is more than enough to vaporize microscopic bits of semiconductor and destroy ports. This is why ESD protection is so important to get right!Tim

author: banedon
author_rank: Regular Contributor
timestamp: March 17, 2015, 07:07:16 pm
content: Thanks for the replies, guys - very thorough and interesting (although I don't pretend to understand it all - yet!).I'll definitely get some higher voltage 1000uF capacitors.With regard to the transformer I have this one:http://uk.farnell.com/pro-power/ctfcs50-12/transformer-50va-2-x-12v/dp/1780891It has 2x12V output. I assume this is two secondary windings and not one winding which is centre tapped? I wanted to produce +12V 0V -12V but this doesn't work - I end up with two +12V rails. I did this by joining the two grounds together.Should I have joined the first winding's ground to the second winding +12V, taken my 'common' off of that and use the second winding ground as -12V?Something like the diagram attached? Also, do I need to anchor the common to earth or can I just leave that floating? I suspect the latter, but I have seen diagrams which show it joined to earth.

author: Electro Fan
author_rank: Super Contributor
timestamp: May 11, 2015, 11:24:57 pm
content: Quote from: T3sl4co1l on March 17, 2015, 04:00:20 amSo in this video how much current is flowing over the test leads between the capacitor and the resistor?To make similar discharge tool, what gauge wire or what current rating for the test leads would be needed to safely discharge say a 400 Volt 10,000 microfarad capacitor with a with a 50 ohm 50 watt resistor?

author: mariush
author_rank: Super Contributor
timestamp: May 12, 2015, 12:34:44 am
content: Quote from: banedon on March 17, 2015, 07:07:16 pmAs the datasheet for that transformer says, there are two secondary windings. For plus and minus voltages, you connect the two windings in the middle and that middle becomes the common, like in your picture. But, note that in your picture there's a capacitor missing for your negative voltage section.In fact, here's a more detailed schematic for your +/- 12v power supply :You can ignore the F1 (fuse, optional) and just assume your secondary windings are 12v, not 15v as the picture says.When AC voltage from the two secondary windings is converted to DC using the bridge rectifier ( the four 1n4007 diodes), the voltage coming out will be a bunch of DC pulses that have a peak voltage equal to 1.414 x V ac , minus the losses in the two diodes that are always working ... so, you have peak voltages in the range of 1.414 x 24v = 34v.Now if you look at the datasheet of the transformer, you will notice there in a chart a column sayingReg%, and for your 50 VA transformer, that value is equal to9. In a simplified way, this means that when you don't use a lot of power from the transformer, when it's not "loaded", the output voltage of the transformer may be up to 9% higher than normal. Rounding it to a 10%, it means our peak DC voltage of 34v could actually be about 38-39v, or about 17-18v on each side of the common wire.This is before we take in account the losses in the diodes. Depending on which diodes, the losses are anywhere from 0.3v per diode to 1-1.2v per diode, for 1n4007 you can safely work with 0.7v per diode.So anyway, the point is that after all's said and done, right after the diodes, you could have DC voltages that have peaks which could be above 16v, so you should use capacitors rated for at least 25v.Like I said, after diodes, you'll have a series of DC pulses, 2 x fAc ... if you're in US where you have 60Hz mains frequency, then you'll have 120 pulses of DC on each side of the common wire. In Europe, you'll have 100 pulses.That's where capacitors come into play, they charge up with energy when the pulses reach their peaks and fill in the gaps when the transformer doesn't provide the power needed.Depending on how big the capacitor is, you can make sure the voltage on each side never drops below a particular amount of voltage. There's a formula which can tell you how much capacitance you need to keep the voltage above a particular level, and that's done with this:C = Current / ( 2 x fAC x Vripple) where Vripple is how low you're willing to let the voltage go down from that peak voltage, down to a minimum voltage.The current amount is how much current you expect your devices to consume. But basically, you have a 50 VA transformer and 24v output, so your transformer can provide 2.08 A of current, but after rectification and all that you're looking at about Idc = 0.62 x I ac = 0.62 x 2.08 = 1.29 A. So each side of the common realistically will only be able to provide up to about 0.7A of current.In the picture above, a 7812 and a 7912 linear regulator are used to smooth out what comes from the capacitors and to provide clear +12v and -12v.If you look at the datasheet for such linear regulators ( seehereorhereyou will notice a line in the charts saying Dropout Voltage and a typical value is 2v at 1A for these linear regulators. This means that if a device wants 12v and pulls 1A of current from the linear regulator, the regulator must have at least 12v + 2v at the input to work properly, otherwise you get less than 12v at its output.There are regulators out there that don't need as much voltage above the output, for example 1117 linear regulators only need about 1v, but let's go along with these classic 7812 and 7912 linear regulators.Since we know from the math above that our maximum current on each side will be about 0.7A, and let's assume you're in US where you have 60 Hz mains frequency and let's assume you'll have peak voltages of about 16v then let's figure capacitance to always have at least 14v :C = I / ( 2 x fAC x Vripple) = 0.7A / [ 2 x 60 Hz x (16v-14v)] = 0.7 / 240 = 0.002916 Farads or 2916 uF. Of course, there's no such thing, it's not a standard value, so you can go with the next standard value which is 3300 uF or maybe you don't really need to design this for 0.7 A on each +12v and -12v, maybe you're satisfied with only 0.5A on each side, or even less, in which case you can do the math with the new current value.In the picture above, C3 and C4 are decoupling capacitors, and those must be ceramic and close to the input and ground pins of the regulators. C7 and C8 are a bit "overkill", they're not really required, and C5 and C6 can be any value you want within reason ( 10uF - 220uF, won't make much of a difference, and the voltage obviously has to be higher than 12v, so use capacitors rated for 16v or more)

author: IanB
author_rank: Super Contributor
timestamp: May 12, 2015, 01:10:02 am
content: Quote from: Electro Fan on May 11, 2015, 11:24:57 pmJust do a simple direct calculation.If you have 400 volts feeding 50 ohms, then you have an initial current of 400/50 = 8 amps. So any wire that can carry 8 amps will be fine. Pretty much any ordinary wire.Now if you have 8 amps flowing at 400 volts, then the power will be 400 x 8 = 3200 watts. Your 50 watt resistor might not be sufficient if the capacitor is too big.Take a worst case estimate:400 V x 10,000 µF = 4,000,000 µA-seconds = 4 amp-secondsSo the capacitor will take ~2 seconds to discharge at 8 amps. Can a 50 watt resistor withstand 3200 watts for a few seconds? Possibly, though it might be borderline. On the plus side is the exponential discharge curve so the power will fall off rapidly. On the minus side is vastly exceeding the power rating of the resistor. The wires inside might melt before the power can escape.(By the way, a 10,000 µF capacitor charged up to 400 V isscaryI would treat it a bit like a loaded gun.)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 12, 2015, 01:12:19 am
content: Quote from: Electro Fan on May 11, 2015, 11:24:57 pmThe peak current will be no more than 400V / 50 ohms, since we can assume the RLC circuit is dominant RC (R >> sqrt(L/C)). The average current, over time scales that wires care about, will be quite teensy; you'll spend more on insulation ("600V" wire perhaps?) than copper when building such a tool.The resistor needs to handle a peak rating of (400V)^2 / (50 ohm) or 3.2kW. This is well over the momentary overload rating of most power resistors (usually something like 5x for 5 seconds), but the duration isn't very long, either (10mF * 50 ohm = 500ms, where the majority of the energy is dissipated; the capacitor is "fully" discharged in a couple seconds).Very loosely speaking, power rating usually scales as sqrt(t), so for a 0.5s pulse (instead of 5), we should expect *maybe* 3.2 times the power rating, or 15 x rated, or 750W. Still not really enough, so, I'd suggest a "high energy" bulk type (they are available, though poorly stocked and expensive), or a bigger vitreous wirewound type that's more likely to handle the overload.The metal-case resistors are notoriously cheap -- well, perhaps not "notorious" enough if few people know their construction. The black stuff on the ends is indeed an epoxy potting compound, which retains a wirewound ceramic former inside the aluminum heatsink. So guess what: the heat transfer is from wire to epoxy to aluminum. Even thermal epoxy is not a terrific conductor, so this is a pretty crappy method, and it's why they're only rated to modest temperatures (150C or thereabouts) and overloads. Note also, the continuous rating is ONLY possible with a full heatsink, including mounting hardware and grease. They can't be used willy-nilly -- the ambient (non-heatsinked) rating is something like 10% of the full rating.Whereas a vitreous resistor is what it says it is; metal wire on a ceramic body, bonded with glass. You have to hit them with a punishing amount of energy to destroy them, and they will operate at red hot for momentary overloads (your connections will desolder or burn themselves off before the resistor itself is compromised!).Not that continuous ratings are key here (they aren't!), but taking note of the heat transfer materials in the construction is.The metal case resistor might not survive that kind of treatment forever (probably dying due to fatigue, fracture and burning of the plastic, then overheating at a spot and burning out; possibly failing shorted in the process, which will of course discharge capacitors quite a lot faster afterwards!), and we're probably talking in the 1000s of cycles range here. So on a human scale, hardly a problem. I'd still feel better with the vitreous or bulk resistor though; one advantage might be having one in your junk box already (they were popular in old TVs and radio gear, back in the days when voltage was dropped with resistors rather than converting it with a transformer, let alone a switching supply or anything!).Tim

author: IanB
author_rank: Super Contributor
timestamp: May 12, 2015, 01:28:57 am
content: Quote from: T3sl4co1l on May 12, 2015, 01:12:19 amWhich produced that nostalgic burning smell when turning on a TV that had been stored unused for some time. Dust would settle on the hot running power resistors, which would proceed to burn off the first time the TV was switched on again.

author: Electro Fan
author_rank: Super Contributor
timestamp: May 12, 2015, 02:11:10 am
content: Quote from: T3sl4co1l on May 12, 2015, 01:12:19 amThanks - the 400V, 10,000 microfarad is somewhat of a stretch spec. Likely, 100V and 4,700 microfarad would be the upper limit of use and even that is probably on the high side - but I appreciate the description of how the math works. Given the resistor in the attached photo, do you think 100V and 4,700 microfarad is safely doable, or would these values be borderline?- I'm just looking for a decent safety margin. Most likely my use would be with caps in the range of 35V 4700 microfarad to 25V 470 microfarad, or smaller caps. And the cycle count might eventually be in the dozens (not a thousand). The plan is to build a tool that safely discharges caps and then limit the use of caps to those that can be safely managed with the tool. It just seems to me that if someone is going to learn to use capacitors it would be good to be able to safely discharge them.- I think the youtube configuration above was a 10 ohm 50 watt resistor with a 150 Volt 3600 microfarad capacitor.

author: Electro Fan
author_rank: Super Contributor
timestamp: May 12, 2015, 02:52:57 am
content: Quote from: IanB on May 12, 2015, 01:10:02 amRoger all that. I have no plans for charging up a 10k uF cap to 400V. This is just a theoretical set of numbers used to make sure I understand the math, and good practice - and to build some safety margin. As you point out, 8 amps doesn't seem too difficult to address with respect to wire gauge but 3200 watts through a 50 watt resistor might be something else. In fact, when I first saw the youtube video above I couldn't quite get my head around the wattage aspect of the tool's design. Then I figured out the dissipation curves and it started to look doable - but knowing how well (or if) a 50 watt resistor can operate at such a spec even for a short period of time is different than knowing that a cap is supposed to dissipate within a couple or a few seconds.The alternative seems to be the "screwdriver method" which doesn't look too good itself. If there is a better way to discharge a cap, I'm open to suggestions.To be square, this all started when someone pointed out that if you measure a charged cap with an ESR meter it's possible you could damage the ESR meter - which wouldn't be good. In researching cap charges and cap discharging it became clear that beyond damage to a meter there could be some hurt or damage to the user(Someone posted a thread a while back where he was surprised how much charge was still held by a cap that had been removed for some considerable period of time from a working device.)So, in an effort to safely work with caps I'm just looking for a tool to safely discharge caps.All recommendations welcome! Thx

author: IanB
author_rank: Super Contributor
timestamp: May 12, 2015, 02:56:59 am
content: High wattage resistors are readily available. Space heaters and electric kettles come immediately to mind. Or just use a higher value resistor and let the discharge take longer. Instead of 50 ohms, use 500 ohms, say.

author: Electro Fan
author_rank: Super Contributor
timestamp: May 12, 2015, 03:12:49 am
content: Quote from: IanB on May 12, 2015, 02:56:59 amRight, that's why after watching the youtube example I decided to go from 10 ohms (as used in the video) to 50 ohms, but perhaps 500 ohms would be better, or possibly 1k ohms.... so I'm looking for a way to make the selection. My main idea is to build the tool with whatever size resistor and then stay well south of what that resistor should handle by limiting the size (volts and microfarads) of the caps used. I don't need any big caps, I just want something practical to discharge modestly small caps. I guess you could use a 5k ohm resistor but at some point the discharge process will require the cap to be set aside for a discharge waiting period. I'm thinking that it would be better to design a tool that will allow for a few seconds (less than 10 seconds and preferably less than 5 seconds) discharge and then stay with cap sizes below whatever that safely allows. This way the "workflow" always says "discharge this cap now" rather than "I wonder if this cap was previously discharged?" This is all simple-silly for the experienced practitioner but that's why I posted this in the beginner forum.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 12, 2015, 04:19:29 am
content: Quote from: IanB on May 12, 2015, 01:10:02 amhttp://en.wikipedia.org/wiki/Muzzle_energy.357 Magnum ~= 873 Jhttps://www.google.com/search?q=.5+*+(400V)%5E2+*+10+millifarad800 JThe bullet is probably more destructive (it's already carrying its energy in mass), but they're both about the same speed (a general purpose, computer grade electrolytic would probably be capable of discharging in under 100us; a bullet going 300 m/s is going 0.3 mm/us, or for a 10mm long bullet, it can stop in no less than 33us).(So by the way, if you see people dicking around with coil guns, and showing off how they can dent pop cans and shoot darts into cardboard, with capacitors this big... yeah, they have a lot of work to go. The efficiency is just zilch.)Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 12, 2015, 04:27:21 am
content: Quote from: Electro Fan on May 12, 2015, 03:12:49 amYeah, I'd go with 10-1000 ohm, vitreous or "high energy", over 20W (or equivalent energy rating).Large resistances (1k range and up) may be inconveniently slow, which makes it even more of a nag, and less likely to be used. Undesirable for a safety tool.Could put a high-efficiency LED across it, with a nice large resistor (100k 2W?), so you can see the discharge progress. If it blinks, you shouldn't have touched it, and it needed the discharge. If it doesn't blink, you were being overly cautious (which is to say, better safe than sorry!).You'd want a high efficiency LED just so you can see the glow down to very low voltages (with a 100k resistor, it should still be visible under indoor lighting conditions with all of 10V applied).If you wanted to get fancier, you could even put a meter on it.Tim

author: Electro Fan
author_rank: Super Contributor
timestamp: May 13, 2015, 11:33:24 pm
content: Quote from: T3sl4co1l on May 12, 2015, 04:27:21 amTim, how about this vitreous resistor (50 Ohm, 225 Watt)... does this one look like what you had in mind?http://www.digikey.com/product-search/en?mpart=FVT20020E50R00JE&vendor=2741.125" Dia x 10.500" L (28.58mm x 266.70mm)(Could also do 100 ohm, 225 watt for the same price but I’m inclined to keep the testing time down to encourage regular use and keep the cap size down as needed to be safe.)http://www.digikey.com/product-search/en?mpart=FVT20020E100R0JE&vendor=2741.125" Dia x 10.500" L (28.58mm x 266.70mm)On the other hand, when looking at the dimensions of these resistors, this 50 ohm 50 watt resistor looks like it will take up less bench spacehttp://www.digikey.com/product-search/en?mpart=FVT05006E50R00JE&vendor=2740.563" Dia x 4.000" L (14.30mm x 101.60mm)I'm kind of leaning toward a portable resistor vs a bench resistor- might have to go with the 50 ohm, 50 watt unit and just keep the cap volts and microfarads down accordingly.Thoughts? Thx, EF

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 14, 2015, 01:32:15 am
content: Quote from: Electro Fan on May 13, 2015, 11:33:24 pmExactly

author: Electro Fan
author_rank: Super Contributor
timestamp: November 08, 2015, 06:34:35 pm
content: resurfacing this thread.... to go slightly adjacently off topic....after having discharged a capacitor, it would be good to be able to measure a capacitor and understand the reading(s)...In the following thread was a discussion about charts.https://www.eevblog.com/forum/beginners/esr-values-for-electrolytic-caps/This thread had some insightful discussion. I'm not sure if the conclusion was that charts are of no value, limited value, or potentially high value. It would seem that such a chart could be of some value if it fit appropriately with some (TBD) attributes of the capacitor being tested:Is there a recommended or preferred chart (or multiple charts for various common use cases) that provides reasonable parameters for determining a capacitor's condition (and that indicate the attributes of a capacitor for which the chart is applicable)?