Safe discharging of capacitors - Page 1
Forum section: electronics / beginners

author: FenderBender
author_rank: Super Contributor
timestamp: December 17, 2012, 01:07:45 pm
content: If I have a 330uF capacitor charge to 1000V and I discharge it across a 50ohm resistor, what is the total power applied to that resistor?This is not a homework problem. I'm just trying to figure it out.T = R x CE = (V² x C) / 2Time constant is 0.0165. So 0.0165 x 5 = 0.0825seconds to discharge it.Energy = 165J.But how do I calculate power? 165/5? 33W?Thanks!

author: AndyC_772
author_rank: Super Contributor
timestamp: December 17, 2012, 01:13:46 pm
content: The power isn't constant, so it's not really a meaningful question. The initial power is V^2/R, but that falls off sharply as the cap discharges.The total energy stored in the cap is (1/2)*C*V^2, and in theory it takes forever for every last bit of it to be turned into heat in the resistor - so the average power in the resistor will depend entirely on the point at which you choose to call the capacitor "discharged".

author: FenderBender
author_rank: Super Contributor
timestamp: December 17, 2012, 01:47:28 pm
content: After 5 time constants then?

author: Bored@Work
author_rank: Super Contributor
timestamp: December 17, 2012, 02:30:09 pm
content: P = dE / dt

author: grumpydoc
author_rank: Super Contributor
timestamp: December 17, 2012, 04:27:21 pm
content: QuoteWhich I think, neatly, comes to 2kW as the "average" power during 5RC seconds.It might be more interesting to work out the temperature rise because the above doesn't mean you need a 2kW resistor - the cap has 165J which will raise 3.9ml of water by 10oC or 23g of carbon by the same amount.Edit: Can't say I've ever thought about it particularly but if you re-arrange the equations the "average" power over 5 time constants is V2/10R, the C's cancel out - which I found surprising until I thought about it for a moment.

author: Smokey
author_rank: Super Contributor
timestamp: December 17, 2012, 06:04:19 pm
content: Here is a spreadsheet that calculates cap discharge stuff. Recently found it on digikey and thought it was useful for basic stuff.It does initial power, time constant, and total energy discharged.http://www.digikey.com/Web%20Export/Supplier%20Content/tt-electronics-welwyn-985/docs/tt-electronics-capacitor-discharge-calculator.xls?redirected=1

author: FenderBender
author_rank: Super Contributor
timestamp: December 17, 2012, 10:12:16 pm
content: So if I gave a 50W wirewound resistor 2kW for say 0.001 seconds, does that mean that it would blow up? The 50W rating is a for a sustained 50W, right? Different chemistries might have different peak power ratings?I'm going to build an ESR meter and I want to build in a function that allows you to discharge the capacitor before you try to test it.Before I go on, is there an accepted threshold at which it is safe to measure caps? 1V? Less? More? Depends on the meter? Any rule of thumb you can think of? Reason I say that is because I want to have a simple op-amp circuit that measures the voltage across the capacitor and lights an LED when it drops below the threshold to indicate that it is safe to test.

author: Zero999
author_rank: Super Contributor
timestamp: December 17, 2012, 10:29:48 pm
content: The question should be what is the total energy delivered to the resistor? The answer is all of the energy stored in the capacitor. E =1/2CV2

author: FenderBender
author_rank: Super Contributor
timestamp: December 17, 2012, 10:45:06 pm
content: I think typically the charge on the capacitor would be <500J.Let's see if I can remember some thermochemistry and work out the numbers.

author: IanB
author_rank: Super Contributor
timestamp: December 17, 2012, 10:57:07 pm
content: I don't think you are going to get close to 100,000 kJLet's take the example you started with, a 330 uF capacitor charged to 1000 V. That's pretty extreme actually unless you deal in exotic capacitors.As mentioned, the energy stored in that example would be 165 J.To put that into perspective, that could be generated by a load of 165 W for 1 second, or a load of 16.5 W for 10 seconds, or a load of 1.65 W for 100 seconds. Basically, any 20 W wire wound resistor is likely to shrug that off without a murmur.What you want to avoid is the power being released too explosively before the resistor can absorb it. For the 1000 V charge, we can look at the instantaneous power dissipation which for a 50 ohm resistor was 20 kW. That seems like a bit much for a little resistor, so maybe it would be better to use a 1000 ohm resistor or so. That would bring the instantaneous power down to 1000 W and that should be OK for a fraction of a second. You basically want to avoid blowing up the wire inside the resistor before the heat can spread out. I might even go for a 5 k or 10 k resistor to be really safe.

author: FenderBender
author_rank: Super Contributor
timestamp: December 17, 2012, 11:12:57 pm
content: Quote from: IanB on December 17, 2012, 10:57:07 pmHaha I messed up. I didn't convert uF to F. I calculated that 100,000kJ is enough to turn about 500kg of water into steam. Yeah not quite right there...Thanks a whole lot though.I've been reading a few datasheets:http://www.vishay.com/docs/30208/30208.pdfVishay says that this particular resistor can handle 10x the rated power for up to 5 seconds. And as you said, the instantaneous power is there only shortly and it quickly degrades. So a 100W resistor could handle 1000W for 5 seconds if needed.

author: FenderBender
author_rank: Super Contributor
timestamp: December 18, 2012, 03:12:11 am
content: One additional question...I wanted to incorporate an indicator that would tell me when the cap is "okay" to test, that is, it has been discharged fully.I was thinking about using an op-amp configured as a comparator with a reference voltage of say "1V".But there is such a wide range of voltages that the capacitor could be charged to...and many of them out of the safe operating range of a op-amp's inputs. Is there any simple feasible way to light an LED at that threshold voltage regardless of the input voltage? Still struggling a bit with analog design. Any techniques that could be used?

author: IanB
author_rank: Super Contributor
timestamp: December 18, 2012, 03:56:01 am
content: Well, why not use an actual comparator (like the LM139) rather than an op amp? It's best to use something designed for the task at hand rather than adapting something else.For protecting against excess voltages you could use a zener diode. Place the zener diode in series with a current limiting resistor across the capacitor being discharged. Measure the voltage across the zener. This will limit the maximum voltage to the zener voltage and protect the comparator input.

author: westfw
author_rank: Super Contributor
timestamp: December 18, 2012, 06:08:50 am
content: You might want to take a look at the "safe capacitor discharge circuit" from Sam Goldwasser's RepairFAQ on Strobes (which are in the voltage/capacitance range that you're talking about):http://www.repairfaq.org/sam/strbfaq.htm#strbcapIt recommends "a high wattage resistor of 5 to 50 ohms/V of the working voltage of the capacitor."... And it has an indicator.Sam's FAQs are pretty neat stuff; predating the WWW...

author: FenderBender
author_rank: Super Contributor
timestamp: December 18, 2012, 09:29:09 pm
content: Quote from: IanB on December 18, 2012, 03:56:01 amFair enough.How would the voltage across a zener ever change unless the voltage was less than the breakdown voltage, in which the voltage across it would be...0?

author: IanB
author_rank: Super Contributor
timestamp: December 18, 2012, 10:00:19 pm
content: Quote from: FenderBender on December 18, 2012, 09:29:09 pmIf you had a 5.6 V diode, the voltage across the diode would always be less than or equal to 5.6 V. If your capacitor was charged up to 3 V you would measure 3 V across the diode. However, if your capacitor was charged up to 50 V you would measure 5.6 V across the diode. So you can safely feed the diode voltage to the comparator input and set the detection threshold at 1 V.

author: FenderBender
author_rank: Super Contributor
timestamp: December 18, 2012, 10:19:19 pm
content: Quote from: IanB on December 18, 2012, 10:00:19 pmOhh...so it's not necessarily conducting. Just there's a voltage across it. This is a reverse biased zener I presume. What series resistance would work? 100K? This would be in parallel with the main load resistor?And then when the voltage across the zener goes less than the reference voltage on the comparator, the output is triggered and the LED lights.If I understand this correctly, this has been very informative. Thanks.

author: SeanB
author_rank: Super Contributor
timestamp: December 19, 2012, 04:02:35 am
content: Zener just limits the max voltage to something the comparator can handle, it normally will only have a small leakage current and will effectively not be in circuit. That way you do not smoke the comparator if the capacitor is charged to 100V, it will just discharge through the resistor/zener combo until the voltage drops to the zener voltage. If your input can handle say 5V Dc on a capacitor then you use a 4v7 zener and have the comparator switch the terminal to the measuring circuit when the capacitor is below 5V.

author: FenderBender
author_rank: Super Contributor
timestamp: December 19, 2012, 01:13:53 pm
content: Can there be a voltage across a zener if it is not conducting? Yes right?If its a 1V zener, and only 0.5V is across it, that's okay?

author: SeanB
author_rank: Super Contributor
timestamp: December 19, 2012, 03:28:55 pm
content: Below the knee voltage it is just a normal diode. Nothing special otherwise about them, just they are designed to have a well defined low voltage breakdown. They can be used as regular diodes, which is what I am busy doing with the 10k I bought on auction for $2. 5k of 12V 0.4W diodes and 5k of 24V 0.4W diodes. I use them as general small signal diodes in low voltage circuits.

author: FenderBender
author_rank: Super Contributor
timestamp: December 20, 2012, 05:13:55 am
content: I drew up a little schematic in LTSpice. I'm a bit confused by the results though. When V1 (linear sweep from 100V) hits 1V, the comparator sets the output high (red trace). But if you look at the green trace, current through R3, the current went low when V1 hit 1V...Why? I would expect if the output voltage went high, the current would also go high, not vice versa.What am I missing here?Thanks.

author: IanB
author_rank: Super Contributor
timestamp: December 20, 2012, 05:35:16 am
content: Quote from: FenderBender on December 20, 2012, 05:13:55 amYes. That's what the circuit is configured to do. I know the answer, but I don't want to just give it to you. Look at the circuit for a second or third time, consider it carefully, trace the logic and think about what it will do under your test scenario. (Don't fall into the trap of making it do what you expected it to do, put your expectations to one side and figure out what it willactuallydo.)

author: FenderBender
author_rank: Super Contributor
timestamp: December 20, 2012, 10:20:40 pm
content: Quote from: IanB on December 20, 2012, 05:35:16 amI actually figured it out when I was laying in bed last night. If I set the output high, you have no potential difference across across the resistor and LED because both the power supply and the output are at 5V...so deltaV is 0, so no current flows. If you set the output low, then current flows from the power supply through the comparator to ground.I had the inputs mixed up. I had it so that the output was normally low and then went high (which then reduced the V to 0 across the LED + resistor). I needed it to be normally high.Thanks for not telling me. Feels better when I figure it out. Silly mistake. But I have almost no experience in circuit design. I can follow a schematic but rarely do I make my own.

author: IanB
author_rank: Super Contributor
timestamp: December 20, 2012, 10:25:56 pm
content: Good. I'm glad you figured it out.

author: FenderBender
author_rank: Super Contributor
timestamp: December 21, 2012, 01:06:41 pm
content: Now another question. This design is polarity specific. What's a good way I can do reverse polarity protection? Without a diode (voltage drop)?

author: FenderBender
author_rank: Super Contributor
timestamp: December 29, 2012, 10:48:56 pm
content: I'm a bit stumped again.Decided to play around with this discharger circuit thing again.The circuit I came up with is polarity-specific. Applying the capacitor the wrong way will lead to the possible destruction of the circuit.So I want to add a circuit which analyzes the polarity of the capacitor and if it is properly connected, switches a relay (or some sort of switch) and connects the capacitor to the discharge circuit.I have an idea of how to do it. It would be similar to the circuit I already posted. I would measure the voltage between R1 and R2. If the voltage is greater than 0, then the capacitor is properly oriented. If it is connected in reverse, no current flows and thus, there is no voltage at that point...However, I think in actuality, there may be some leakage current, albeit very small, through the diode. So there would technically always be a voltage > 0 at that point, just a very very very small voltage in this case.Would this be an issue? Any ideas? I was thinking of using an LM393, just for reference.

author: FenderBender
author_rank: Super Contributor
timestamp: December 29, 2012, 11:11:27 pm
content: Wait actually now that I think of it, that leakage current isreversecurrent, so therefore the voltage between R1 and R2 is actually negative with respect to ground.Therefore, I think I'm in the clear. Let's simulate some more.