If \$V_{ext}\$ is applied to a RC circuit, then voltage across capacitor is given by: $$v(t) = V_{ext} + (V_i -V_{ext})e^{-t/RC} $$

Using above to find the ripple bounds \$v_l\$ and \$v_h\$, get two equations:
$$v_h=A + (v_l - A)e^{-aT/RC}$$ $$v_l = 0 + (v_h-0)e^{-(T-aT)/RC}$$

Where \$A,T, a\$ are the amplitude, period, duty cycle of the input square wave.
Eliminating \$v_l\$ gives: $$\color{blue}{A(1-e^{-aT/RC})} = \color{purple}{v_h(1-e^{-T/RC})}$$

If I'm looking at it correctly, the left side represents:
\$\color{blue}{\text{voltage charged by the capacitor in time $aT$ when external dc voltage is $A$. }}\$
the right side represents:
\$\color{purple}{\text{voltage charged by the capacitor in time $T$ when external dc voltage is $v_h$. }}\$

The equation above says these two voltages are EQUAL!
Is this a coincidence or something interesting going on here? I'm not able to see further why they are equal... Love to hear your insights!


simulate this circuit – Schematic created using CircuitLab

enter image description here

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    \$\begingroup\$ Your reduction equivalence is incorrect as the Vavg steadystate rise to the duty cycle ratio of A with T time constant. \$\endgroup\$ Jul 26, 2021 at 13:23
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    \$\begingroup\$ Get the average (vh+vl)/2= d*A… or aA \$\endgroup\$ Jul 26, 2021 at 13:33
  • \$\begingroup\$ @TonyStewartEE75 ty working on it... wil get back 10 min \$\endgroup\$
    – across
    Jul 26, 2021 at 13:38
  • \$\begingroup\$ @TonyStewartEE75 I'm getting $$\frac{v_h+v_l}{2} = \frac{A}{2}(1-e^{-aT/RC})$$ not sure where im doing wrong. still working... ty:) \$\endgroup\$
    – across
    Jul 26, 2021 at 13:45
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    \$\begingroup\$ Exactly! feels something wrong in my work. have to think more clearly... thank you for suggesting to find the average value.. you're awesome:) @TonyStewartEE75 \$\endgroup\$
    – across
    Jul 26, 2021 at 14:18

1 Answer 1


The exponential terms on the two sides of the equations are not the same. You only have equality if the duty factor is 1, which means that you have a step input rather than a rectangular wave.

This makes sense. If the input is a step that stays high then the capacitor voltage will eventually be equal to \$A\$, the input voltage.

  • \$\begingroup\$ Right, \$A, v_h\$ are thought of as dc voltage sources.. but in that equation \$T\$ is constant, not the variable time \$t\$. I think that equation is an identity, not sure with terminology... there are no variables in that equation... \$\endgroup\$
    – across
    Jul 26, 2021 at 13:34
  • \$\begingroup\$ Yes, there is the variable \$a\$, the duty factor of the waveform. \$\endgroup\$ Jul 26, 2021 at 13:41

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