Measuring the C of the RC parallel circuit

[rikie_rizza]

Guys, I need help.

How can I precisely measure the C in RC parallel circuit?

I force 1mA current from a DC source.
One way that I did was to fully charge the cap so it will become a high impedance. After that I measure the R. The voltage is 998mV so this is close enough.

But then I realize that the voltage will never reach 1V..but its okay thou.

Now how to measure the C precisely?
What is the equation correlate V/I with RC?
Discrete please, since s or jω is just a fantasy :lol:

 

[enjunear]

When you connect the DC source to the parallel system, once the voltages reach steady state, you essentially have all of the current flowing through the resistor and the capacitor effectively "disappears". When you disconnect the DC source, instantly you have a cap charged to 1V, discharging through a fixed resistor. This sounds like a time-constant problem to me, where (tau) τ = R*C. You could connect an oscilloscope to the common node and wait for it to reach {e}^{-1} V (or 0.368 V).

See also:
RC circuit - Time-domain considerations
and
RC time constant

 

[rikie_rizza]

Woah dude,
Thanks.
You mean that I can only measure the discharge? And then just use the V(t)=V0+V0*e^-(t/(tau))
I will try to look through the scope and see the actual step response of the system.

Yeah I've searched wiki too but since the equation still have this s and j stuff, I cannot measure it in the real world.

 

[enjunear]

Woah dude,
Thanks.
You mean that I can only measure the discharge? And then just use the V(t)=V0+V0*e^-(t/(tau))
I will try to look through the scope and see the actual step response of the system.

Yeah I've searched wiki too but since the equation still have this s and j stuff, I cannot measure it in the real world.

I believe that you can only do the scope measurement on discharge, since it looks like a series RC when the DC source is disconnected. Tau = RC doesn't apply for a parallel RC network, only series.