# Formula for calculating capacitor capacity?

Discussion in 'General Electronics Chat' started by MikeA, Apr 3, 2014.

1. ### MikeA Thread Starter Member

Jan 20, 2013
125
17
I have a large capacitor that I'd like to check, but it's above the 10,000uF my multimeter can do.

I think that there should be a formula that would take (a) time (b) voltage A (c) voltage B (d) resistance, and calculate capacity.

That is, say I charge the capacitor to 30V, put a 10000ohm resistor across the leads, and in say 30 seconds it reads 1V. According to my understanding of Ohm's law those 4 variables should be enough to calculate capacity. But I'm not smart enough to come up with a formula on the fly.

2. ### AnalogKid Distinguished Member

Aug 1, 2013
4,691
1,297
You will need a calculator that can do natural logarithms, the ln function. That's LN but both letters are lower case. Can be confusing in this font.

V2 = 1V
V1 = 30V
R = 10K

V2 / V1 = 1 - (e ^ -(t/RC)

V2 over V1 equals 1 minus e to the minus t over R times C power.

Measure T in seconds, solve for C in farads.

Hint: the natural log of e raised to an exponent is the exponent. ln and e cancel each other out. So rearrange things then take the ln of both sides of the equation, and you'll be left with high school algebra to solve.

As a test, solving for 5 seconds should give 15,000 uF.

Note, this method is only as accurate as your time and voltage measurements. By the time the cap discharges to less than 5% of its initial voltage, the voltage will be changing very slowly.

ak

3. ### MrChips Moderator

Oct 2, 2009
12,633
3,453
Here is the simple way. Use the formula for time constant = R x C.
The time constant is the time it takes for the voltage to fall to 37% of the initial voltage.

It works the other way too.
If you are charging the capacitor through a resistor R, time constant is the time it takes to reach 63% of the charging voltage.

So, assume the capacitor is charged to 10V. Short the capacitor with the resistance R and measure the time it takes to reach 3.7V.

Or the other way around. Discharge the capacitor. Now charge the capacitor from a 10V supply using a series resistor. Measure the time it takes to reach 6.3V.

The capacitance C (in farads) = time in seconds divided by resistance in ohms.

You can use any charging voltage. Simply multiply the voltage by 0.63 for charging or 0.37 for discharging to determine your target voltage.

4. ### wayneh Expert

Sep 9, 2010
12,382
3,240
V/V0 = e^(-t/RC)

ln(V/V0) = -t/RC

C = t/R/ln(V0/V)

 Take the natural log of the voltage ratio you observed at the two time points. Change the sign to positive.
 Divide this into the ∆t in seconds
 Divide that by R to obtain C

5. ### THE_RB AAC Fanatic!

Feb 11, 2008
5,435
1,305
We were taught the cap charge time is 5 Tc.

Charge the cap through a resistor, divide the total time by 5. Easy.

6. ### AnalogKid Distinguished Member

Aug 1, 2013
4,691
1,297
For your application, the product of R x C is called the time constant, usually written RC. Timing circuits frequently are discussed in terms of the number of time constants rather than dealing with specific component values. RC is a number of seconds.

MrChips' shortcut comes from rearranging the standard equation to start with the RC product (which gives you the C since you choose the R) and solve for the V2/V1 ratio. If you work with these kinds of circuits for a while, some standard shortcut values stick in your head:

1RC (1 x R x C) = 63.2%
2RC = 86.5%
3RC = 95.0%

Another handy one is this:
0.693RC = 50% (often shortened to .7RC)

.7RC is the basis for many timing circuits that compare an RC charge or discharge curve to a voltage divider that is 50% of the circuit power supply voltage, a very common trip point. In your case with a 30V power source, catching 15.0 volts on you DVM might be easier than catching 18.96V (1RC). OTOH, at 3RC the capacitor voltage is discharging very slowly. 1.500 volts should be pretty easy to catch, and the math is do-it-in-your-head easy. Example:

30V charge on the cap, apply 1.0K resistor, start timer.
Voltmeter hits 1.500V, stop timer.
For this example, timer reads 180 seconds.
3RC = 180, RC = 60, R = 1000, C = .06 F, or 60,000 uF

The 1K resistor makes it easy - timer value divided by 3 equals the capacitor value in thousands of microfarads.

ak

7. ### MikeA Thread Starter Member

Jan 20, 2013
125
17
The cap I have is rated 35v, and I connected it up to a 24.27v DC power supply through a 10K resistor.

At exactly 180 seconds the voltage reached 15.34v (63.2% of 24.27v).

So the cap is roughly 18,000uF if my math is right?

8. ### crutschow Expert

Mar 14, 2008
13,496
3,373
Easy but not precise. There is very little change in voltage with time at that point, reducing the accuracy. For example 4 time-constants is 98.2% of the final value, 5 time-constants is 99.3% of the final, and 6 time-constants 99.8% of the final value , so a small error in the voltage measurement gives a large error in the calculated time-constant. Much better to measure the time to the 50% point (.693RC) or to 1 time-constant (63.2%) where the voltage is changing much more rapidly.

The handy thing about using the 50% point is that you don't need to be concerned about whether it is charging or discharging to do the measurement.

THE_RB likes this.
9. ### crutschow Expert

Mar 14, 2008
13,496
3,373
Seems correct.

10. ### MikeA Thread Starter Member

Jan 20, 2013
125
17
Awesome! Thanks.

On subject, why do most multimeters limit themselves to 10,000uF if capacity is purely a math function? Seems like there is no reason for a meter that can do 10,000uF not to be able to do 20,000uF.

11. ### k7elp60 Senior Member

Nov 4, 2008
479
69
If you just want to check is value(uF's) put a smaller value in series with it
and let your multimeter tell you the resultant value. Then with total value and the known value you can get the actual value. I do it all the time with
my meter.

12. ### AnalogKid Distinguished Member

Aug 1, 2013
4,691
1,297
k7 - NICE. fizix is your friend.

ak