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constant current charging capacitor (Read 2233 times)
A.D.J
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constant current charging capacitor
Aug 31st, 2016, 11:34am
 
I was trying to use constant I to charge one capacitor.
The charging process is based on the equation: I*t=CV.
If C=1pF, t=8us, I=10nA, so V=I*t/C=80mV. The schematic simulation is very closer to 80mV.
However, if changing the C=100fF, the expected V=8us*10nA/100fF=800mV. The simulation result is less than 800mV.
I have no idea what results in the voltage reduction.
Thank you.

--------------------------
Is it related to capacitor leakage current issue?
I read a article talked about the leakage current I is proportional to the  charges on the capacitor plate. For smaller capacitance, we got higher charge, so the leakage current is bigger. As a result, the final charged voltage could be smaller than expected.
Is that reasonable?
Thank you.
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davidshw
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Re: constant current charging capacitor
Reply #1 - Aug 31st, 2016, 6:25pm
 
to aid convergence, the simulator will add a resistor parallel with capacitor, which may cause leakage. you can try to set the gmin a smaller value to reduce the leakage.

.option gmin=1e-15
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Geoffrey_Coram
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Re: constant current charging capacitor
Reply #2 - Sep 16th, 2016, 10:14am
 
Or, if you don't want to play with gmin (which could affect other parts of your circuit), you could put a very large resistor in yourself.  Or, some simulators have a "switch" that you can have short to ground for the time=0 solution and then open up for t > 0.
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