# Weird result when finding ripple bounds of RC circuit with square wave input

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

• Your reduction equivalence is incorrect as the Vavg steadystate rise to the duty cycle ratio of A with T time constant. Jul 26, 2021 at 13:23
• Get the average (vh+vl)/2= d*A… or aA Jul 26, 2021 at 13:33
• @TonyStewartEE75 ty working on it... wil get back 10 min Jul 26, 2021 at 13:38
• @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:) Jul 26, 2021 at 13:45
• 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 Jul 26, 2021 at 14:18

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.
• 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... Jul 26, 2021 at 13:34
• Yes, there is the variable $a$, the duty factor of the waveform. Jul 26, 2021 at 13:41