Let imagine a RC circuit that requires 20 uAmp to charge in 0.2 sec. But the electric source can supply only 1uAmp. Voltage is not a problem. We have enough to pass through the resistor.
The question is what will happen with this limited current???
I think it will take more time to charge the capacitor compared with the theoretical RC value. Is this correct????
E
Purposefully limit the supply current, and you have just invented the linear ramp generator. One of the textbook definitions of capacitance, after all, is I =3D C dV/dt.
I don't think the RC value means what you think it means, unless the charging cure is at least mostly exponential. For example, look at tiny portions of the exponential curve but not the whole thing: you'd come up with entirely different dV/dt's. In fact limiting the application to a tiny portion of the curve is a time-proven method for waveform generation.
Tim.
I thought that the time a capacitor take to charge and discharge dipends on the RC constant??? Half charge time 0.693RC.
Although i don not know what is a linear ramp generator (google did not help much) i have not understood what will happen if the supply current is limited.
E
Yeahbut that only matters if the R is limiting the current. In the example you gave, it doesn't. So R becomes irrelevant.
Again, a textbook definition of the relationship between charge rate (dV/dt), current, and capacitance is:
I =3D C dV/dt
You'll see that as long as the current remains constant, dV/dt will remain constant and that's the definiton of a linear ramp.
Now in real life, keeping I constant, or for that matter keeping C constant, can become a challenge :-).
Tim.
yeah, it'll take much longer
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