How long does it take for a cap to reach steady state step response while current limited

Thread Starter

MikeJacobs

Joined Dec 7, 2019
201
ok stupid question. Just have not done this in a long time

Say you have a cap and some power supply. All you know is your cap is .01uF and you are going to current limit the power supply to 10 mA.

The supply is going to start at 0 volts and go to 10 volts.

How long does it take to have 10V across your cap?
 

wayneh

Joined Sep 9, 2010
17,152
A Farad is one Coulomb (C) per volt. Your cap is 0.01µF or 1x10^-8 Farad. At 10V, it contains 10^-7C Coulombs.

A Coulomb is an amp-second. Your constant current supply is delivering 10mA = 10mC per second = 10^-2C/s

10^-7C ÷ 10^-2C/s = 10^-5 seconds = 10µs
 

BobaMosfet

Joined Jul 1, 2009
1,776
ok stupid question. Just have not done this in a long time

Say you have a cap and some power supply. All you know is your cap is .01uF and you are going to current limit the power supply to 10 mA.

The supply is going to start at 0 volts and go to 10 volts.

How long does it take to have 10V across your cap?
Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3
 

Thread Starter

MikeJacobs

Joined Dec 7, 2019
201
So obviosly thing all comes from I = C*DV/DT

For some reason i didnt think you could treat the derivative like that and just algebraically re-arrange it.
Its been a long time since i have tried to exercise the rules of differential equations.

So apparently, you can just re-arrange this and solve for DT
ok then. thanks
 

wayneh

Joined Sep 9, 2010
17,152
So obviosly thing all comes from I = C*DV/DT

For some reason i didnt think you could treat the derivative like that and just algebraically re-arrange it.
Its been a long time since i have tried to exercise the rules of differential equations.

So apparently, you can just re-arrange this and solve for DT
ok then. thanks
That's doable because you specified constant current, dQ/dt.
 

djsfantasi

Joined Apr 11, 2010
7,693
\(T=\frac{CV}{I}=\frac{0.01uF*10V}{10mA}=10us\)
Be careful when presenting equations, that it is clear what the units are for each variable. C is in Farads. V is in Volts. And I is in amperes. In your case, the exponent of the units cancel out, but if capacitance was specified as mF instead of μF, the answer would have been wrong.
 
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