# i = C dv/dt for start capacitor

Discussion in 'Math' started by Volttrekkie, Aug 4, 2018.

1. ### Volttrekkie Thread Starter Member

Jul 27, 2017
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0
I am trying to use Ohm's law I = C dv/dt to figure out the capacitance of a start capacitor. My meter will not measure it directly because of the resistor between the terminals makes it go OL. Voltage measured 370 across the terminals. How do I calculate dv/dt?

2. ### crutschow Expert

Mar 14, 2008
22,483
6,581
I = C dv/dt is not Ohm's law.

What resistor?
You can't readily measure the capacitance with a resistor across the terminals.

3. ### Volttrekkie Thread Starter Member

Jul 27, 2017
60
0
I read it was Ohm's law for capacitors. And start capacitors in hard start kits have a resistor across the terminals. Therefore, my meter will not read the capacitance. It says OL. An electrical engineer told me I could measure the current and voltage and calculate the capacitance. I = C dv/dt must be the formula to use. But how do you calculate dv/dt? How do you take the derivative of the voltage?

4. ### crutschow Expert

Mar 14, 2008
22,483
6,581
An approximation is to connect a DC power supply to the capacitor through a resistor and measure the time for the voltage to reach one RC time-constant (0.632) of the final value.
For this you select a resistor large enough so that you can accurately measure this time, T (would you use a meter or an oscilloscope to measure the time to voltage?).
The capacitance value is then C = T/Req.

Note that the value of Req is the value of the added resistor in parallel with the resistance across the capacitor.
This resistance across the capacitor will also reduce the "final value" voltage used in the measurement that the cap is charging towards.

Do you know the approximate value of the capacitor and the resistance across it?

Last edited: Aug 4, 2018
5. ### Alec_t AAC Fanatic!

Sep 17, 2013
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2,397
Ohm's Law works for resistors.
You could charge the capacitor with a constant current. The voltage would then increase linearly with respect to time (dv/dt = constant). BUT, for any practical charge current, the cap would charge pretty quickly, so an ordinary meter probably wouldn't be able to sense the rate of voltage rise. You would need an oscilloscope for that (or for most other methods of measuring the capacitance).

6. ### LesJones Well-Known Member

Jan 8, 2017
2,195
651
As you mention that it is a start capacitor it should have an AC voltage rating that would allow it to me connected to the mains supply in your country. Also I would guess it has a value between about a few uF and a few hundred uF. (If you had given some more background information like the power rating of the motor it is used to start we would have a better idea of it's approximate capacitance.) If you connect it in series with say a 40 watt (tungsten filament.) bulb and connect it across the mains and measure the current through it and the voltage across it you can work out it's reactance. with the reactance and the mains frequency in your country you can calculate the capacitance. If the voltage across it is too small or too large to get an accurate reading you would need to select a more suitable rating bulb. The internal resistor on the capacitor is only there to discharge it so it will be a high enough value not to effect the readings.

Les.

7. ### Volttrekkie Thread Starter Member

Jul 27, 2017
60
0
I see. So, dv/dt is measured using an oscilloscope by feeding DC current slowly through a resistor? The capacitor is 187 to 225uF. I don't know what the resistor across it is.

8. ### crutschow Expert

Mar 14, 2008
22,483
6,581
No.
You are measuring the time-constant of the circuit, not dv/dt.

To readily measure dv/dt you need a constant-current source.
Using a resistor avoids the need for that source.
Can you measure it?
If you ohmmeter doesn't work for that, you can use a series resistor with an applied DC voltage.
Select a resistor that gives a steady voltage across the capacitor equal to 1/2 the applied voltage.
The capacitor resistor is then equal to the series resistor value.

9. ### Volttrekkie Thread Starter Member

Jul 27, 2017
60
0
Oh I see. T stand for Tau. Tau = RC. Oh, OK. I have to go study this.
I also noticed that some multimeters can measure this capacitance. Others cannot. I just returned mine. I have to order another one that costs \$50 more.

Mar 14, 2008
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11. ### MrAl AAC Fanatic!

Jun 17, 2014
6,226
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Hello there,

I am not entirely sure what you are trying to do, but to measure dv/dt you could approximate by taking two instantaneous measurements over a small time interval 'dt' and then calculate it directly as:
dv/dt=(v2-v1)/dt

so if you measure v1=10v and v2=11v and that is over a time interval of 0.1 seconds, then the slope is approximately:
dv/dt=(11-10)/0.1=1/0.1=10
in units of volts per second.

The thing is, in order to make the two voltage measurements over a small interval of time you almost always use an oscilloscope, but you can also use a modern ADC (analog to digital converter) that you can find in most modern microcontrollers.

You dont really need a constant current source though, as long as the slope does not change significantly during the time interval over which the two voltage measurements are made. That means over a relatively small time interval. You also need to know the current at the time of measurement if you want to calculate the capacitance.

The equation:
i=C*dv/dt

is a definition for the capacitor, it's not really Ohm's Law.

Sometimes Ohm's Law for capacitors is stated in the frequency domain as something like:
I(s)=s*V(s)*C

and that makes more sense because it is algebraic, but that's not really Ohm's Law either even though stated as such sometimes. It may be acceptable in some forums though.

Last edited: Aug 8, 2018