Cap charging & discharging

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aamirali

Joined Feb 2, 2012
412
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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#12

Joined Nov 30, 2010
18,224
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
 

Wendy

Joined Mar 24, 2008
23,415
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.
 

crutschow

Joined Mar 14, 2008
34,285
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.
 

ErnieM

Joined Apr 24, 2011
8,377
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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