# Cap charging & discharging

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

1. ### aamirali Thread Starter Member

Feb 2, 2012
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2
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
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9,678
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
21,528
2,973
The math works, but so does the graphical approach...

5TC = 99.3%, give or take.

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

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4. ### crutschow Expert

Mar 14, 2008
21,375
6,124
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
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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
18,076
9,678
Thanks. That form covers the scenario when Vi is not zero.