Figure out the Vout wave form of the following circuit: ( assume the name of the 9u capacitor to be C2 and not C1)

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Is there any way to intuitively plot the waveform by understanding the behavior of the circuit? Assume that initially all the capacitors are discharged.

My attempt at the problem:

C1d(Vin - Vout)/dt = C2 dVout/dt + Vout/R

For t<0,the output voltage remains zero.

For t>0, the input voltage remains constant resulting in an equation -C1 dVout/dt = C2 dVout/dt + Vout/R

or (C1+C2) dVout/dt = - Vout/R

Solving this equation,

ln (Vout) = - t/[R(C1+C2)] with initial condition of Vout being zero which poses a problem with the logarithmic function. Am I making any mistake in my analysis?

  • \$\begingroup\$ What do you understand so far? What have you tried that didn't work? \$\endgroup\$
    – David
    Jan 28 '16 at 13:02
  • \$\begingroup\$ Why do both capacitors have the same reference designator? That makes them hard to talk about. \$\endgroup\$
    – Dave Tweed
    Jan 28 '16 at 13:12
  • \$\begingroup\$ I have included my attempt and please assume the 9u capacitor to be C2 rather than C1. \$\endgroup\$ Jan 28 '16 at 13:21
  • \$\begingroup\$ On initial rise, the caps will share charge to output will rise straight away to Vin * C1 / (C1 + C2). Then normal CR discharge with time constant (C1 + C2) * R. \$\endgroup\$ Jan 28 '16 at 13:26

If by intuitively you also mean qualitative, yes.

Considering both capacitors are ideal and initially discharged, you have a mathematical discontinuity in front of you if the input impedance coming at Vin is zero, and that should ideally be tackled with Laplace transformation.

Now considering the expected capacitors behavior, both capacitors will charge (the 1u cap faster than the 9u) and stabilize at certain in-between voltage.

If you want actual waveforms, you can solve the maths using Laplace transform (which is the correct way, but I will not be troubled to doing it), or take a wild guess as to what is the voltage after the discontinuity in each cap is and draw some exponential assymptotes - with a discontinuity peak at t=0 and exponential decay/growth to final voltage.

Edit: correct answer, thanks to user1582568

The voltage Vout after the discontinuity can be calculated by the divisor rule [C1/(C1 + C2)]*Vin = 1V. The final voltage is zero, and the resulting waveform is an exponential decay with time constant T = (C1 + C2)*R.

  • \$\begingroup\$ Wouldn't Vout final voltage be zero? I would expect the 1 uF charge to be absorbed by the 9 uF capacitor (rising to 1/(9 + 1) V_in?) and discharging through R to end up at zero. \$\endgroup\$
    – Transistor
    Jan 28 '16 at 13:39
  • \$\begingroup\$ I am not sure about this, as discharge from the 9u cap through R causes a voltage difference through the 1u cap, which allows for "recharge" current from Vin. One thing is certain, as the circuit could not present an oscillatory behavior, it definitely stabilizes in a constant voltage. If it is zero or not, I personally can't tell without simulations or maths. \$\endgroup\$ Jan 28 '16 at 13:42
  • \$\begingroup\$ When he says final value, he means final initial value, not final final value :-) \$\endgroup\$ Jan 28 '16 at 13:43
  • \$\begingroup\$ There will be an exponential decay with a time constant (C1 + C2) * R towards zero (but never getting there of course). \$\endgroup\$ Jan 28 '16 at 13:44
  • \$\begingroup\$ I really meant final final value @user1582568. Seeing your comment to OP's post, I believe you are correct at the initial cap voltage division, yet I cannot immediatly assess your time constant. Any reasoning I'm missing? \$\endgroup\$ Jan 28 '16 at 13:46

The turn on will create a fast spike, a differential voltage, so we must look at impedances. at this dV/dT the resistor plays little part.

The two caps form a voltage divider to the dV/dT change, so the amplitude will be 1/(9+1) * Vin. The rise time due to this will essentially follow the rise time of VIn isnce there is no series resistor shown.

Immediately the 9uF cap starts charging, it will start discharging through the resistor, slowing the charge rate on 9u at effectively RC time constant.

As soon as the capacitive voltage divider has 'stabilised' @zC9/(zC1+zc9) the 1uF the cap divider has a high Z and no current can flow. At the values shown about Vin/10.

The charge in 9u exponentially decays to zero through 1K resistor @RC constant. dV/dT is small, so there is high Z in 1u and very little recharging through the caps. What little there is simply slows the process.

The circuit stabilises at Vout=0.

At each stage, each component is either contributing to or inhibiting a fast rise/decay time.


As Vicente Cunha remarked, the best way to deal with this problem is using a Laplace transform, as this allows you to deal with the discontinuity at switch-on. As no-one has done that yet, I'll try to give the solution here.

For clarity, I'll call the 9uF capacitor \$C_2\$, and say that current \$I_1\$ flows through \$C_1\$, \$I_2\$ through \$C_2\$, and \$I_R\$ through the resistor. Applying Kirchoff's law to the circuit I obtain three equations:

\$V_{out} = I_R R = Q_2/C_2\$, and

\$V_{in} = Q_1/C_1 + V_{out}\$.

Let us differentiate the final equation, to convert \$Q_1\$ to \$I_1\$:

\$d V_{in}/dt = I_1/C_1 + d V_{out}/dt \$.

We know from conservation of current that \$I_1 = I_2 + I_R\$, so we can replace \$I_1\$ in the above equation to obtain:

\$d V_{in}/dt = I_2/C_1 + I_R/C_1 + d V_{out}/dt \$.

Since \$d V_{out}/dt = I_2/C_2\$ (from differentiating the first equation), and \$I_R = V_{out}/R\$, we can now obtain the differential equation for the circuit:

\$d V_{in}/dt = (C_2 / C_1) d V_{out}/dt + V_{out}/(R C_1) + d V_{out}/dt \$.

We can now solve this equation by going to the Laplace domain, noting that the derivatives will be replaced by factors of "s" (the Laplace coordinate):

\$s {\tilde V}_{in} = (C_2/C_1) s {\tilde V}_{out} + {\tilde V}_{out}/R C_1 + s {\tilde V}_{out} \$.

The step discontinuity in \$V_{in}\$ means that its Laplace transform is \${\tilde V}_{in} = 1/s\$. Substituting this and rearranging the terms gives the result:

\${\tilde V}_{out} = 1/(a s + b)\$, where \$a = (C_1 + C_2)/C_1\$ and \$b = 1/R C_1\$. We can now transform back to the time domain to obtain the final answer:

\$V_{out}(t) = V_0 \ C_1/(C_1 + C_2) \ e^{-t/(\tau_1 + \tau_2)}\$,

where the time constants are \$\tau_1 = R C_1\$ and \$\tau_2 = R C_2\$. As expected, the voltage is split by the divisor rule, and the result is an exponential decay with a time constant of \$\tau = R (C_1 + C_2)\$. We can clearly see there is no oscillatory behavior, and the circuit indeed finally stabilises at a voltage of zero.


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