# Capacitor discharge current and time.

Discussion in 'General Electronics Chat' started by authorjim, Sep 30, 2014.

1. ### authorjim Thread Starter New Member

Jul 15, 2013
3
0
First off I'm not an electronics guy so here goes.

I was trying to calculate the discharge time for a capacitor and found the following
CV=IT so if I have a 5f capacitor and its charged to 10 volts.

So the discharge time for 250 ma draw is 200 seconds. Right?

5x10/.25=200 seconds

Here is where I get confused unless I messed up the first part.

If I have 4 capacitors. All 5 volt all 5 farad caps and i put them in a series parallel arrangement
Two together in series which gives me 10 volt 2.5 farad equivalent. Then I take the other two and do the same.
Then I put both these series loops in parallel I get 5 farad and 10 volts. right?

So these 4 capacitors together only give me the same 250 ma current draw for 200 seconds?
I know this is a stupid question for you guys, but it does not seem right.

Thanks
Jim

2. ### crutschow Expert

Mar 14, 2008
16,524
4,453
The difference is that the 4 caps are charged to only 5V so each cap contains less energy then the single cap charged to 10V. In either case you have 1F of capacitance charged to 10V, so why would you expect a difference?

3. ### authorjim Thread Starter New Member

Jul 15, 2013
3
0
As I said, I'm a newbie and I expected that the four caps would provide more time at the same discharge rate. After thinking about it what I forgot is that the 5 farad 10 volt cap will be alot bigger(i think then the 5 volt 5 farad caps so that makes sense.
What I don't understand your last comment that in either case I have 1 farad charged to 10 volts. Why 1 farad?

Jim

4. ### MrChips Moderator

Oct 2, 2009
14,509
4,277
Just a typo. He meant 5F at 10V.

5. ### MrCarlos Active Member

Jan 2, 2010
400
135
Hello authorjim

Since you're a newbie must learn something.
These Are laws:
If you connect two capacitors in parallel, the total capacitance is the result of adding the value of those two capacitances.
If you connect two capacitors in series, the total capacitance is the result of multiplying the value of the first capacitance by the value of the second capacitance divided by the sum of both capacitances.

Easier way:
(A) Parallel: RT = C1 + C2
(B) Series: RT = (C1 x C2) / (C1 + C2).

Now:
If you have two 5 F. capacitors charged to 5 volts and plug them in series they'll have at the ends of the series 10 Volts. but the capacitance is exactly half because the formula (B).
But you have another circuit as here above, you will have a 5 F. capacitor @ 10 volts, again, because the formula (A).

6. ### crutschow Expert

Mar 14, 2008
16,524
4,453
Yup, you beat me to it.

7. ### ScottWang Moderator

Aug 23, 2012
5,478
864
In practical, whatever you want to charging +5V or +10V, you may always need to in series with two or four caps, as two 1F/2.7V in series to 5.4V for charging +5V or four 1F/2.7V in series to 10.8V for charging +10V.

If you need to charge 5F/10V then you will need 20 caps.

8. ### Sensacell Senior Member

Jun 19, 2012
1,511
403
Does one need to worry about charge balancing with series connected super caps?

9. ### ScottWang Moderator

Aug 23, 2012
5,478
864
If you care about that then you can in parallel with resistors or zener diodes for each cap to do the job.

For a higher voltage as over 20V or more current then what your concerned will make sense and should be care about that.

10. ### ScottWang Moderator

Aug 23, 2012
5,478
864
The attached picture was what I tested for two 1F/2.7V supercaps in series and charged for 5V about 3 seconds, it can be keeping brightness about 30 seconds for ten 3V/20mA leds, and 5 minutes for one 3V/20mA led.

Last edited: Oct 1, 2014
11. ### authorjim Thread Starter New Member

Jul 15, 2013
3
0
Thanks to all, I did know the capacitance rules that is why I said I had two 5f 5 volt caps in series for 10 volts and 2.5 f
and two more the same way for 10 volts and 2.5 farads. THen I put bought of those loops in parallel for 10 volts and 5 farads.

Thanks for all the help.

Jim