Does current lead voltage during discharge in capacitors?

Question

I can see that how current leads voltage while capacitor is charging. Looking at any capacitor charging diagram will explain this: (e.g figures in http://en.wikipedia.org/wiki/RC_circuit)

However, I don't see how the current leads voltage while discharging? Because for discharging, both current and voltage look alike in the same descending format. And, it doesn't seem that there is a phase difference between current and voltage curves during the discharge! Can someone please explain what is happening?

Answer 1

Talk about "current leading voltage" or "phase difference" only applies to AC analysis. In the more general case, one could say what a capacitor really does is differentiate voltage, according to:

$$ i = C\frac{dv}{dt} $$

From this, you can derive all sorts of well-known things about capacitors. Such as, if you want a linearly changing voltage across a capacitor, you must apply a constant-current source to it. As an example, consider a 1 ampere current source connected to a 1 farad capacitor:

$$ \require{cancel} \begin{align} 1A &= 1F \frac{dv}{dt} \\ 1A &= \frac{1 A \cdot s}{V} \frac{dv}{dt} \\ \frac{1\cancel{A}\cdot V}{1\cancel{A}\cdot s} &= \frac{dv}{dt} \\ \frac{1V}{s} &= \frac{dv}{dt} \end{align} $$

If you consider the case where the applied voltage is sinusoidal, then so too is the current:

$$ \begin{align} i &= C\frac{dv}{dt} \\ i &= C\frac{d\sin(t)}{dt} \\ i &= C\cos(t) \end{align} $$

because \$\cos\$ is the derivative of \$\sin\$.

You will also see if you graph these functions, that \$\cos\$ (current) leads \$\sin\$ (voltage) by 90 degrees, as an electrical engineer would put it:

Answer 2

You can only realistically talk about phase angles when sine waves are applied and if you apply a sinewave voltage, the cap current will lead voltage by 90 degrees all the time.

Current = \$ C\dfrac{dV}{dt}\$ and the differential of a voltage sinewave is a cosinewave of magnitude C. Cosine lead sine by 90 degrees.

Answer 3

Generally speaking, we can only meaningfully speak of a relative phase difference between waveforms if the two waveforms have the same form but are displaced in time.

Now, as others have pointed out, the current through a capacitor is proportional to the rate of change of the voltage across so, in general, the current and voltage associated with a capacitor do not have the same form.

For example, if the capacitor voltage is a ramp, the capacitor current is a constant. If the capacitor voltage is parabolic, the capacitor current is a ramp.

How can we meaningfully talk about the relative phase between a parabolic voltage and a current ramp?

Thus, for it to be possible to meaningfully speak of a phase difference, we need a very special type of waveform; a waveform that has the same form as its rate of change.

An example of such a waveform is

$$v_C(t) = \sin( \omega t)$$

The rate of change (the time derivative) of this is

$$\dot v_C(t) = \omega \cdot \cos (\omega t) = \omega \cdot \sin(\omega t + 90^\circ)$$

So

$$i_C(t) = C \,\dot v_C(t) = \omega C \cdot \sin(\omega t + 90^\circ)$$

Now, it's easy to see that, in this case, the voltage across and current through a capacitor have the same form and that there is a relative phase of \$90^\circ\$.

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In the case of the RC circuit charge and discharge waveforms, note that the solutions are, for DC excitation:

$$v_C(t) = V_{DC}(1 - e^{-t/RC}) + v_C(0)\cdot e^{-t/RC}$$

$$i_C(t) = \dfrac{V_{DC} - v_C(0)}{R}e^{-t/RC}$$

For zero initial condition (the capacitor is charging), these are:

$$v_C(t) = V_{DC}(1 - e^{-t/RC})$$

$$i_C(t) = \dfrac{V_{DC}}{R}e^{-t/RC}$$

For zero DC excitation (the capacitor is discharging), these are:

$$v_C(t) = v_C(0)\cdot e^{-t/RC}$$

$$i_C(t) = - \dfrac{v_C(0)}{R}e^{-t/RC}$$

As you can see, in either case, there isn't any apparent relative phase parameter we can identify in the above voltage and current waveforms.

There is a subtle reason for this. In the case of a sinusoidal waveform, we can add a constant to the argument which has the effect of displacing the waveform in time; adding this constant changes the phase of the sine waveform:

$$\sin(\omega t + \phi)$$

is a sine waveform shifted in time by \$\frac{\phi}{\omega}\$ seconds.

However, if we add a constant to the argument for the exponential, the result is not a displacement in time but a scaling (multiplication by a constant).

$$e^{-t/RC + \phi} = e^{-t/RC}e^\phi = Ke^{-t/RC}$$