# Cap charging & discharging

Discussion in 'General Electronics Chat' started by aamirali, Apr 10, 2012.

1. ### aamirali Thread Starter Member

Feb 2, 2012
415
1
I have a voltage source connected to resistor, as in figure. Lets say i have 1.5V dc, 10K & 1nF.
1. Cap will charge >90% in 5*R*C.
2. Cap will also get 1.5v. Rough approx.
3. Now if my voltage source starts to fall & get 1V. So now cell=1V, cap=1.5v . What will happen now, how to analyse the ckt.

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2. ### #12 Expert

Nov 30, 2010
16,704
7,354
Vo = dV x (e^ -t/RC)
Volts out equals change in voltage times e to the negative time over (resistance times capacitance)

In this case, Volts out = (Volts on capacitor at the start) minus? change in volts times e to the -t/RC

unless I flopped a minus sign. If you calculate more than the battery voltage, switch a minus to a plus

3. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
The math works, but so does the graphical approach...

5TC = 99.3%, give or take.

This is from a thread I had about a year ago on RC time curves.

RC oscillators like the 555 live or die by this math.

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4. ### crutschow Expert

Mar 14, 2008
13,505
3,376
Just remember that the capacitor will always charge (or discharge) to the new voltage with a exponential change between the old and new voltages based upon the RC time-constant.

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5. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,439
1,627
One formula that continues to come in handy is to compute the time between two arbitrary voltages in an exponential charge. This formula always works for an RC network:

$T= RC ln\frac{(Vss - Vi)}{(Vss - Vf)}$

Where:

T = time to change between Vi and Vf
R = resistance
C = capacitance
ln = natural log
Vss = Steady State voltage (what the voltage goes to if you wait forever)
Vi = initial (starting) voltage
Vf = final voltage

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6. ### #12 Expert

Nov 30, 2010
16,704
7,354
Thanks. That form covers the scenario when Vi is not zero.