2
\$\begingroup\$

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!

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

\$\endgroup\$
9
  • 1
    \$\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
  • 1
    \$\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
  • 1
    \$\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

1
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.