Capacitor discharge questions

[Anndy]

I've googled this extensively, but couldn't find a decent answer.
What resistors are ideal for discharging motherboard and/or power supply capacitors? I found all kind of different recommendations, like 5W 2K, 5W 100K, 5W 20K and the list goes on and on, but nothing definitive!
All help will be greatly appreciated.

[volkswagenvan]

Re: Capacitor discharge questions

100ohm 5w or similar would do the job for most caps. Just don`t use your finger, ring, watch etc.

[mariushm]

Re: Capacitor discharge questions

The power dissipated in a resistor can be calculated with the formulas below:

P = VxV / R or P = IxIxR

So the higher the resistance, the less power is dissipated in the resistor therefore the capacitors discharge slower.

A 2k resistor would discharge a 200v capacitor dissipating 200x200/2000 = 20 watts, you would blow up the 5w resistor. They probably recommended the 2k for smaller capacitors, like maximum 50v capacitors.

A 20k resistor would discharge at 200x200/20000 = 2 watts, so a 5w resistor would be suitable.

A 100k would discharge the capacitors slowly, a 200v capacitor discharges at a rate of 200x200/100000 = 0.4 watts ... so even a 1w resistor would do.

Now this is all assuming the capacitor would remain at 200v, but the capacitor won't stay at 200v... the voltage will slowly (well, "slowly" is a relative term) decrease as the energy stored is dissipated in the resistor.
So as the voltage on capacitor decreases, so will the power dissipated in the resistor, unless you decrease the resistor value along with the capacitor voltage (remember, power dissipated is V x V / R).


My 2 cents... for small capacitors, under 63v, I would use a 1-3k 7w resistor. For more, I would use a 33-47k 7w resistor (as these are with big leads and wirewound, in ceramic package, so I can safely hold them in hand)

I personally don't know of the top of my head how to determine the relationship between the resistor value and the discharge time, like I said the power dissipated varies with voltage so I'm sure there's a formula somewhere.

-- note, not sure if this is correct below (except the Q,E and P formulas), take everything with a grain of salt (or more) --

Just at a quick google ...

The total charge in a capacitor is Q = Capacitance x Voltage,
the Energy is (1/2) x C x V x V ,
Power in capacitor is P = dW / dt (potential power / dissipation time)

So let's say we have a 560 uF 200v capacitor.

Q = 1/2 * 560 x 10^-6 x 200 = 11.2 Joules

So power is dW / dt ... if I understand correctly, if we want to discharge the capacitor within 1 second, we would need to dissipate 11.2 watts.

P = VxV/R = > R = VxV/P = 200x200/11.2 = 3571 ohm?

[mariushm]

Re: Capacitor discharge questions

Oh, it turns out Digikey has a nifty Discharge calculator made in an Excel spreadsheet:

http://www.digikey.com/Web%20Export/...s?redirected=1



and looks like I have to correct myself, seems that the discharge time matters quite more than I estimated.

Doing the math for that 560uF 200v capacitor, turns out a 10k resistor will discharge the capacitor to 10v within 17 seconds.
A 4.7k ohm one would do it in eight seconds but the power would be 8.5w so a 7w resistor might not like it that much (but the resistor may see more than 7w for half a second or so)

[kaboom]

Re: Capacitor discharge questions

The total charge in a capacitor is Q = Capacitance x Voltage,
the Energy is (1/2) x C x V x V ,
Power in capacitor is P = dW / dt (potential power / dissipation time)

So let's say we have a 560 uF 200v capacitor.

Q = 1/2 * 560 x 10^-6 x 200 = 11.2 Joules

So power is dW / dt ... if I understand correctly, if we want to discharge the capacitor within 1 second, we would need to dissipate 11.2 watts.

P = VxV/R = > R = VxV/P = 200x200/11.2 = 3571 ohm?

Only for that second. Thermal mass of the resistor will keep its temp reasonable. Most ceramic wirewound resistors can take 5-10x their rated power for several seconds, easily. The bigger they are, the greater the intermittent "overload" can be.

Take a resistor whose temperature rise, at rated power is "X." Fully discharging a cap, storing 10 Joules, into a 5W resistor, in one second, will not result in a temp rise of "2X." In fact, it won't go much above room temperature.

[mariushm]

Re: Capacitor discharge questions

Thank you for the clarification.

I didn't really stop to think about this issue before this thread appeared, hence why my initial post was that.. how to say it.. very basic in reasoning.

Once I saw that calculator, I noticed how steep the discharge curve is and I can see how the resistor would tolerate such initial power dissipation.

[kaboom]

Re: Capacitor discharge questions

Thank you for the clarification.

You're welcome.

I didn't really stop to think about this issue before this thread appeared, hence why my initial post was that.. how to say it.. very basic in reasoning.

Once I saw that calculator, I noticed how steep the discharge curve is and I can see how the resistor would tolerate such initial power dissipation.

There's usually not much energy to speak of, even though initial power may be a bit high, at least for 'normal' size caps. The caps in bigger stuff, like large VFDs/multi-kW power inverters, can supply current into a discharge resistor for quite some time. We're talking about several (or more) large caps, able to supply maybe tens of milliamps into a resistor of a few k ohms.

Smaller caps, such as those in computer/monitor power supplies, only store a few Joules. At least in the case of old-fashioned "rectifier/doubler" cap banks... Those with PFC boost-supplies can store more. A 560u cap charged to 425V stores 50.575 joules!

For example, I have an Ohmite 10 ohm, 25 watt tubular ceramic resistor, of the type seen here:


This would be more than adequate for discharging caps, but we use a highish resistance to reduce sparking and peak currents while clipping it across the caps. One of these, 500 to 2500 ohms would be perfect. An easy way to determine discharge currents is with Ohm's law. (I=V/R)

Caps are usually considered charged or discharged after 5 RC time constants. Simple! T= R*C, with C being in Farads, so make sure you divide properly. For instance, a 470u cap, charged to 200 volts, takes about 2.5 seconds to discharge through a 1k resistor. Total energy is 9.4 joules, average resistor power is 3.76W. Thermal mass, and only thermal mass of the resistor, will determine how hot it gets while discharging the cap. In other words, it is possible to use a 1W, or smaller, resistor. And that would only work where repeat discharges do not occur until the resistor has a chance to cool off!

In the case of large HV film caps or CRT envelope capacitance, put a jumper lead across it after discharging. Dielectric absorption can "surprise" you.

[Anndy]

Re: Capacitor discharge questions

Thanks for all the help guys!
I've certainly learned a lot.
I've just been busy with work lately so couldn't find the time to reply til now.