Differentiator

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gorgondrak

Joined Nov 17, 2014
61
http://www.allaboutcircuits.com/vol_3/chpt_8/11.html

To put some definite numbers to this formula, if the voltage across a 47 µF capacitor was changing at a linear rate of 3 volts per second, the current "through" the capacitor would be (47 µF)(3 V/s) = 141 µA.
I'm having difficulty grasping this new concept of capacitance value times voltage per second for total current. I would like to know how this relates to the method of calculating the current using the RC time constant and the universal time constant equation or calculating its peak current using its capacitive reactance in an ac circuit.
 

WBahn

Joined Mar 31, 2012
29,979
The defining relationship for a (linear) capacitor is

C = Q/V

Namely that the capacitance is, by definition, equal to the charge stored on the capacitor (+Q on one plate and -Q on the other) when a voltage V appears across it.

Current is, by definition, the rate at which charge is flowing past a given point (or through a given area).

Thus

I = ΔQ/Δt = dQ/dt (the latter being the derivative of charge with respect to time, the calculus way of expressing it).

If we change the charge on a capacitor by a ΔQ, then the voltage will change by an amount ΔV and these will be related by

ΔQ = C·ΔV

or, in calculus terms:

dQ/dt = C·(dV/dt)

If we use the Δ notations, then we have

I = (C·ΔV)/Δt = C·(ΔV/Δt)
 

wayneh

Joined Sep 9, 2010
17,496
In words, the rate of flow of charge - the current - is proportional to the rate of change in voltage across the capacitor, dV/dt, scaled by the capacitance of the capacitor C.

This makes sense since Q and V are directly proportional for a linear capacitor, scaled by the capacitance.
 
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