# How do you determine voltage output with time graphically? simulate this circuit – Schematic created using CircuitLab

If the step input is to 2V (0 to 2V at t=0, step function) what will be the plot of output voltage be?

I'm not able to proceed because of the Capacitor in series with the input.

• Can you give us a bit more background? Where is this circuit used. What is it you're actually looking for? Or is this your homework? Jul 28 '16 at 11:10
• Isn't homework, I am preparing for a technical interview for a few electronic companies, and this was one of the questions that came up while I was solving RC circuit related questions online. I'm looking for an approach, or a starting point. Do you straight away write the KCL equations, assuming C as an impedance of (1/sC) and find the Vout/Vin and then do reverse Laplace to get Vout? That was the only solution I could think of, but it seemed too lengthy, and calculation oriented. Jul 28 '16 at 11:16
• The input cap will turn the step into a spike. At the end of the transient it will be like an open circuit, so the final value of Vout will be Vdd/2 (due to the voltage divider R-R). The time evolution will be exponential with time constant RC (edit: not so fast---). You are only left with finding the initial value of Vout. Jul 28 '16 at 11:51
• To begin with the (steady state) voltage at Vout will be half Vdd by inspection. Jul 28 '16 at 11:52
• So initial value of Vout will be Vdd/2 + 2V (because capacitor acts like a short circuit at t = 0) ? Jul 28 '16 at 11:56

As long as intuition is concerned,

• Capacitor cannot change its voltage instantly. Hence, after switching, cap will still hold the previous value for an instant.

• As cap act as open circuit at steady state, the steady state vout here is 2.5V which is by res division.

• as soon as switch is closed, cap will hold that 2.5V across it. And now it has got a kick of 2V hence, output should go to 4.5V. BUT, as there are two caps, this 2V is divided by them and kick at the output will be of 1V. Hence, output will go to 2.5+1=3.5V. This 3.5V is there for very short time.

• Then, as previously stated, cap is open at steady state, final voltage is going to be 2.5V. hence, voltage will come down from 3.5V to 2.5V. this will take certain time constant.

• Time constant here by thevenin is R||R and C||C which is RC. Hence, the discharge from 3.5v to 2.5v will take 5 time constant to settle which is 5RC.

This is proved, with simulation as seen as attachment. I found a rather elegant solution to my own problem.

The circuit given is a basic first order RC circuit (if we remove the Resistors, or open circuit them, we are left with 2 Capacitors in series, which can thus be reduced to one- C/2)

Solution to an RC circuit to step response is an exponential which saturates to a value at steady state. Mathematically :

V(t) = Vs + ( V(0) - Vs )( exp (-t/ Tau) )

V(t) = voltage of capacitor at time t.
V(0) = Capactior voltage at t = 0+.
Tau = time constant, Tau = Reffective * Ceffective

Proceeding in a step wise manner we need to do the following
1)Find tau
2) Find Vs
3) find V(0+)

1) For tau, set Voltages to zero, and find Ceff, Reff Here it'll be 2 R's in parallel, 2 C's in parallel. Therefore R/2 * 2C = RC
2) For Vs, steady state voltage, open circuit the capacitors and find Vout Here it'll be Vdd/2 ( Voltage division across resistors)
3) V(0+) is the tricky part When you open circuit the Resistors and analyse at t=0, there seems to be a voltage source- Capacitor loop. This will cause a jump in Capacitor voltage at t=0 (To satisfy KVL, energy conservation)

So when you find the voltage across the 2nd Cap( i.e Vout) it'll be

Voltage drop across cap -> (2V-Vdd) * C/(C + C) = (2V- Vdd)/2) = 1V - Vdd/2

So Vout = 2V - (V - Vdd/2) = 1V + Vdd/2
This is V(0+)

So the final answer is - The voltage starts off at 1V + Vdd/2 and exponentially falls to Vdd/2 with a time constant of RC

This corresponds with Omibuddy's answer (substitute Vdd = 5, you get initial value as 2.5V + 1V = 3.5 V, decaying to 2.5 V at infinity or steady state)

Reference : http://www.ee.iitm.ac.in/videolectures/doku.php?id=ec1010_2014nk:start Refer to lectures 24,25,26 - RC first order circuits