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I have been studying about time response of RC circuits. I have analyzed the below circuit. Circuit

The input signal is a pulse signal of width 1.5 sec and period 3 sec. Alternating between 0 V and 5 V. So here is transient response i have obtained.

Transient response

The first figure is voltage across the capacitor C1 and the second is the voltage across C0. Now what i understood is, when the input switches from 0 v to 5 v, the capacitors acts as a short circuit and the resistor does not have any action at this point. Hence both capacitors get charged instantly to 2.5 v as in the figure. However once the input voltage stabilize at 5 V, the resistance comes into effect.

So my doubt is, why is the C1 capacitor discharging from 2.5 v to 0 v while c0 capacitor charges from 2.5 v to 5 v?

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Here is another way to look at the problem:

The capacitors are initially uncharged and they get charged instantly to same voltage 2.5V (since both have same capacitance) when the pulse is applied. Now that there is 2.5V across C1 (and hence across R), the resistor demands a current of 2.5mA at the instant.

Now the only path for the current is through C0. This charges up the capacitor C0. This is to say that capacitor C1 is discharging since the voltage across C0 and C1 must add up to 5V.

You can also use Thevenin's theorem to gain proper insight into the circuits and obtain quicker solutions.

It is however easier to obtain the steady state values of voltages and currents by simplifying the circuit using complex impedance. If the circuit is of first-order (which usually is the case when problems of this kind are given), you would know that the step response would be exponential Here is how you do it:

The effective impedance of parallel combination of resistor R and 1/sC1 (impedance of capacitor in laplace domain) is:

$$ Z1 = \frac{R*(\frac{1}{sC1})}{R+\frac{1}{sC1}} = \frac{R}{1+sRC1} $$ Z2 = 1/sC0

The steady state voltage across C1 would be:

$$ Vc1 =\frac{ VinZ1}{Z1+Z2} $$ Since s=0 for DC voltages, steady state impedance Z2 is infinity and Z1=R.

$$ Vc1 =0 $$

$$ Vc0 = \frac{ VinZ2}{Z1+Z2} = \frac{Vin}{\frac{Z1}{Z2}+1} = 5V $$ Once you know the steady state value, you can plot the required circuit parameters if you know the initial values.

Instead of using vpulse for plotting circuit transients, you can use a DC source(vdc) and set the initial voltage of capacitor (or initial current through inductor) to zero and run the simulation. If you don't, the simulator will show only the steady state value.

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A simple way to look at it is as follows:

Hence both capacitors get charged instantly to 2.5 v as in the figure.

Correct. From this point on C1 is discharged by R0 so its voltage drops. Since the voltage across C1 and C0 must sum to 5 V then C0 must be charging up.

Alternative way to look at it:

If C1 was not there then C0 would charge up to 5 V via R0. Since C1 is there it gives a kick to 5/2 V and the RC charge continues from there.

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  • \$\begingroup\$ Why is that C0 not getting discharged through R0? \$\endgroup\$ – Anjana Verma Mar 5 '16 at 12:01
  • \$\begingroup\$ It does when the input voltage drops to zero. My answer addresses the charge cycle (0 to 5 V) which is what you asked about. During that period the top of C0 is connected to +5 V through R0 so it must charge up. \$\endgroup\$ – Transistor Mar 5 '16 at 12:13
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    \$\begingroup\$ So what actually happens if the voltage source gets replaced by a current source of say pulse alternating between 0 A and 5 A? \$\endgroup\$ – Anjana Verma Mar 6 '16 at 10:19
  • \$\begingroup\$ If you have a current source of 5A, the voltage across capacitor C0 would go on increasing linearly (ramp) since the current source is pushing a definite amount of charges through C0. But since there is a resistor in parallel with C1, the voltage across it follows an exponentially decaying charging curve finally reaching a steady state value of 5A * 1K = 5KV! But what happens when the current from source is 0? There is no path for C0 to discharge. It will retain its voltage. But C1 has a discharging path through R. C1 keeps discharging through R until the next high pulse arrives. :) \$\endgroup\$ – Aditya Patil Mar 10 '16 at 4:09

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