0
\$\begingroup\$

enter image description here

I have confusion in this circuit. I know at t=0 both capacitor charges instantly. and voltage divided between both capacitors equally. but after some time output loop start active and capacitor C1 discharges through Resistor (R1) value. so when the voltage starts decreasing across C1, why not again charge through C3?

enter image description here

and if you consider time in answer that helps me a lot. for example at t=0 when input started then capacitor shorted/open,after some time --- happen with capacitor like intuitive way.

\$\endgroup\$
2
  • 2
    \$\begingroup\$ Where would current come from to keep C1 charged? Charging current would certainly have to flow through C3. Since V1 is constant (after t=0) this is impossible. \$\endgroup\$
    – glen_geek
    May 1 at 12:12
  • 2
    \$\begingroup\$ Considering the Thevenin equivalent might help. \$\endgroup\$ May 1 at 12:47

3 Answers 3

3
\$\begingroup\$

The reality is that the charging current in C3 and the discharging current of C1 sum together at the junction to provide the current flowing to ground in R1. The current in each capacitor is equal to C*dv/dt but in opposite directions to each other. Summing the two currents gives a current equal to V/R at any instant where V is the voltage across the resistor.

For C3 to discharge (voltage across it decreases) and C1 to charge (voltage across it increases), positive charge would have to be added to the right hand side of C3 and to the top of C1, R1 would have to source conventional current instead of sinking it that is to say electron flow would have to be from the positive terminal of the power supply to the negative terminal. No can do!

If you were to suggest taking the positive charge from the top of C1 and adding it to the right hand side of C3 then the voltage across both capacitors would have to reduce. Also not possible because they are connected together and must share the same voltage at the connection point.

When the input step voltage is applied, C1 will charge to a lower voltage than C3. This is because the charging current of C3 must be equal to the charging current of C1 plus the current through R1. Therefore C1 will have a lower charging current than C3 resulting in it having a lower voltage across it than C3 immediately after the step input. I=C*dv/dt So if the charging current is less, the dv/dt is less and so the voltage across C1 is less than the voltage across C3 immediately after the step input.

It's basically Kirchoff's current law at the junction between the three components when the step input is applied and the magnitude of the current supplied by the supply must be equal to the magnitude of the current returned.

\$\endgroup\$
3
\$\begingroup\$

Using Laplace and nodal analysis.

Let \$\small V(s)\$ be the voltage at the C1/C3/R1 node. The input voltage step is \$\large \frac{V_1}{s}\$

The node equation is: $$ \small \left (V(s)-\frac{V_1}{s}\right )sC + V(s)sC+ \frac{V(s)}{R}=0$$

$$ \small V(s)\left (2sC+\frac{1}{R}\right )=V_1 C $$ $$\therefore \small V(s)=\frac{V_1 RC}{1+2RCs}=\frac{V_1 /2}{s+\frac{1}{2RC}} $$ Inverse Laplace transform: $$v(t)\small =\frac{V_1}{2}e^{-t/2RC} $$

Therefore, the node voltage starts (\$\small t=0)\$ at \$\large \frac{V_1}{2}\$ and decays to zero exponentially with a time constant, \$\small \tau=2RC\$.

Thus, at steady state, C3 has \$\small V_1\$ across it, i.e. there is 0 V on its right hand plate. Consequently, there is 0 V across C1, and 0 V across R, so no need to 'top-up' C1.

Conceptually, at t=0 the capacitors look like short-circuits so a large instantaneous current (infinite, in theory) flows into both capacitors, charging each of them to \$ \frac{V_1}{2}\$. The series combination of these two voltages balances the applied voltage step.

\$\endgroup\$
1
\$\begingroup\$

... so when the voltage starts decreasing across C1, why not again charge through C3?

My answer is about a little different "real" case for helping.
In fact, it can't charge ... because it is "discharging" ... See the C1 current.
Made a simulation where I added some (low value) serial resistors to the capacitors.
(NB: if not, the currents are "infinite" ... and the @Chu answer apply)

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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