Best value resistor to discharge .45 F cap?

[Motolectric]

Hello,

I have a .45 farad 20V electrolytic cap that I need to discharge from time to time.

I need to know the best value resistor to use that will allow me to discharge the cap in 2-3 minutes safely.

By safely, I mean that the resistor will not overheat or even get very hot.

I am making a small plug to fit into a mating socket on the cap.

The plug will have an LED and the resistor so that when the discharge is complete my user will know.

Any tips/advice are appreciated.

M./

 

[Harald Kapp]

The discharge of a capacitor follows the law V = V0*exp(-t/tau) with tau = R*C.
For t=3*tau the capacitor is discharged by 95 %. If you consider this as still being to much charge on the capacitor, use t=5*tau for a discharge of 99 %.

From t = 3 min = 180 s = 5 * 0.45F * R it follows R = 80 Ω
The peak power is Pmax = V²/R = 400 V²/80 Ω= 5 W. The power dissipation will rapidly sink as the voltage on the capacitor falls due to being discharged. Therefore when you use a 5 W resistor you'll cover the worst case, the resistor willl not overheat (sufficient cooling assumed).
The resistor may, however, get hot. How hot you'll have to try. The only thing you can do is provide sufficient cooling. This is based on physics as you want to dissipate the energy stored within the capacitor within a limited time. Otherwise you'll have to oncrease the time for discharge and spread teh enery dissipation over a longer time, thus giving the resistor more time to get rid of the dissipated energy and stay cool.

 

[kellys_eye]

You'd need a fairly large (physically) resistor to NOT notice the heat too much. A 3W LED dissipates sufficient heat - in a bedside lamp environment - to be almost too hot to touch.

To make it

or even get very hot

you have to dissipate the heat over a large area.

 

[Motolectric]

The discharge of a capacitor follows the law V = V0*exp(-t/tau) with tau = R*C.
For t=3*tau the capacitor is discharged by 95 %. If you consider this as still being to much charge on the capacitor, use t=5*tau for a discharge of 99 %.

From t = 3 min = 180 s = 5 * 0.45F * R it follows R = 80 Ω
The peak power is Pmax = V²/R = 400 V²/80 Ω= 5 W. The power dissipation will rapidly sink as the voltage on the capacitor falls due to being discharged. Therefore when you use a 5 W resistor you'll cover the worst case, the resistor willl not overheat (sufficient cooling assumed).

I very much appreciate your thorough reply.

I am not able to understand the equations except in the most rudimentary fashion.

But if I summarize it I think you said I am okay with an 80 Ohm 5 watt resistor and just need to test a few times to make sure it does not run too hot.

So, my questions -

If I need it to discharge more quickly (say the above takes longer than I want), I need to lower the impedance, correct?

if the resistor gets too hot I should raise the impedance? or lower it? or increase the wattage capability?

Also, I am pursuing this because an acquaintance said that (based on his college courses 30 years ago) a cap will take a memory during use and then not be able to operate at 100% capability (speed, current flow, etc.).

This cap is just being used as an adjunct power source in combination with a 12V battery and my testing shows that it does not expend more than about 3 amps in use in the circuit.

Again, thanks much for the very useful advice.

M./

 

[Harald Kapp]

If I need it to discharge more quickly (say the above takes longer than I want), I need to lower the impedance, correct?

Yes. Less resistance -> more current -> faster discharge.

Also, I am pursuing this because an acquaintance said that (based on his college courses 30 years ago) a cap will take a memory during use and then not be able to operate at 100% capability (speed, current flow, etc.).

Capacitors do not exhibit a memory effect. Your acquaintance may mix this up with NiCd batteries.
Capacitors, specifically electrolytic capacitors, do wear out with time. One of the effects being loss of electrolyte. Discharging them will not help to improve lifetime. On the contrary: When you have an electrolytic capacitor sitting around withoud voltage being applied, it will fail to work after some time due to degradation of the insulating oxide barrier within the capacitor. Electrolytic capacitors require applied voltage at least from time to time to keep that insualtion in good working order. This process is called reforming.

if the resistor gets too hot I should raise the impedance? or lower it? or increase the wattage capability?

Raise the impedance. Increasing the wattage will not reduce the heating. It only serves to keep the resistor alive even under higher power dissipation.

In your application I see no need for a separate discharge resistor - unless you want to force the voltage on the capacitor to zero when the power supply is turned off. Usually the caapcitor will discharge slowly by itslef and by the attched compoennts.

 

[Ratch]

Hello,

I have a .45 farad 20V electrolytic cap that I need to discharge from time to time.

I need to know the best value resistor to use that will allow me to discharge the cap in 2-3 minutes safely.

By safely, I mean that the resistor will not overheat or even get very hot.

I am making a small plug to fit into a mating socket on the cap.

The plug will have an LED and the resistor so that when the discharge is complete my user will know.

Any tips/advice are appreciated.

M./

Here is the dissipation of an 80 ohm resistor plotted against time. Notice that after 10 seconds, it is still dissipating slightly less that 3 watts. Shown also is the dissipation equation.


Ratch

 

[Motolectric]

Hello,

So, I'm back with what I hope will be one of my last questions on the Cap.

I want to use a small "charging tool" with the cap.

It is an ATC fuse with the fuse portion cut out so that there are just the 2 isolated tangs.

Then I have an LED and a resistor soldered in series across the tangs. This allows me to insert the tool into an inline ATC fuse holder and the cap charges slowly and the LED dims as the cap charges up.

I would like to provide a tool that will provide a 2-3 minute charge time so that the cap does not suck down the battery too quickly. If it is connected directly to a 12V battery it will blow a 30 amp fuse instantly.

As for the best value for the resistor to provide the 2-3 minute timing I have been told everything from 100 ohms to 1Kohm.

Does anyone have any best value to suggest?

Again, I very much appreciate the excellent replies I've received, you guys really know your stuff.

Thanks,

M./

 

[Motolectric]

On the contrary: When you have an electrolytic capacitor sitting around withoud voltage being applied, it will fail to work after some time due to degradation of the insulating oxide barrier within the capacitor. Electrolytic capacitors require applied voltage at least from time to time to keep that insualtion in good working order. This process is called reforming.

Do you have any advice as to what this timing might be?

If I store the caps uncharged for 24 months is that going to cause a degradation if they were never charged up after manufacture?

Or are we talking longer times? Would the degradation be subtle as in 10% a year or something like that?

In your application I see no need for a separate discharge resistor - unless you want to force the voltage on the capacitor to zero when the power supply is turned off..

Yes, that is what I need to do. Safely bring is back down to zero for transportation/shipping.

Usually the caapcitor will discharge slowly by itslef and by the attched compoennts.

I have had this happen and was wondering how long it would take a .45 farad cap with a 12.8V charge on it to self-discharge if the battery was removed from the circuit.

I have been working on a spreadsheet that will show the self-discharge rate for lead-acid batteries across time and temperature and the cap discharge time is something I would try to add.

Thanks for any feedback or comments.

M./