# Where's the RC timing formula?

#### StephenDJ

Joined May 31, 2008
58
There used to be a formula I knew of for RC circuits where you could plug in the starting voltage or current, the ending voltage or current, and the amount of time in between, and project exactly what the voltage or current will be after exactly x amount time has passed. I know it rises/drops to 63% within the first time constant. But where's the formula for the other times? Also would like to have the vise versa: i.e. plug in the voltage or current and find the time.

#### mik3

Joined Feb 4, 2008
4,843
For an RC network with initial voltage on the capacitor Vo its voltage with respect to time is given by:

Vc(t)=V(1-exp(-t/RC)+Vo*exp(-t/RC)

where

V=the steady state voltage across the capacitor when fully charged
Vo=initial voltage (if exists)

#### StephenDJ

Joined May 31, 2008
58
For an RC network with initial voltage on the capacitor Vo its voltage with respect to time is given by:

Vc(t)=V(1-exp(-t/RC)+Vo*exp(-t/RC)

where

V=the steady state voltage across the capacitor when fully charged
Vo=initial voltage (if exists)
Mik3,Thanks so much. I have been needing a formula like this for so long!

Ratch,I believe the plain V (steady state) in mik3's formula is what the voltage is only AFTER the time has passed and it is then resting comfortably at its new level of satisfaction, or "charged" state.

Through it all, I've failed to mention one very important point:I'm assuming that this formula of course implies that the RC's supply voltage changed completely and suddenly (like a square wave) at the BEGINNING of the time, and stayed that way during the time the capacitor was catching up. If the supply voltage were to go wiggleing all up and down during this time only to wait untill the very end to reach its final steady state, V, then it's very difficult to establish any kind of formula as this would affect the way in which the capacitor charges. But I believe this formula implies steady state of the supply while capacitor is catching up. That is what I'm looking for.

#### StephenDJ

Joined May 31, 2008
58
There was a message from Ratch in here a while ago... looks like he may have came back and deleted it.

#### Wendy

Joined Mar 24, 2008
22,155
Naw, it got moved to here by a moderator.

#### Ron H

Joined Apr 14, 2005
7,014
One formula covers both charging and discharging:

V=Vf+(Vi-Vf)*e^(-t/(R*C))
Where
V=instantaneous voltage
Vi=initial voltage
Vf=final voltage

Which is another way of expressing mik3's equation.
(I posted this before thoroughly reading his post.)

Last edited:

#### mik3

Joined Feb 4, 2008
4,843
The formula posted is valid for both charging and discharging.

#### KL7AJ

Joined Nov 4, 2008
2,225
Mik3,Thanks so much. I have been needing a formula like this for so long!

Ratch,I believe the plain V (steady state) in mik3's formula is what the voltage is only AFTER the time has passed and it is then resting comfortably at its new level of satisfaction, or "charged" state.

Through it all, I've failed to mention one very important point:I'm assuming that this formula of course implies that the RC's supply voltage changed completely and suddenly (like a square wave) at the BEGINNING of the time, and stayed that way during the time the capacitor was catching up. If the supply voltage were to go wiggleing all up and down during this time only to wait untill the very end to reach its final steady state, V, then it's very difficult to establish any kind of formula as this would affect the way in which the capacitor charges. But I believe this formula implies steady state of the supply while capacitor is catching up. That is what I'm looking for.

The natural logarithm.....shows up EVERYWHERE in nature. Learn to love it.

eric

#### mik3

Joined Feb 4, 2008
4,843
The exponential function is the solution to many differential equations and because many natural effects are described by differential equations is logic to see the exponential function a lot.