Switch-cap Circuits! This is something most of the students are not very comfortable with. But they are very important for analog circuit design. You won't able to make ADC or charge-pump without using a switch cap. For the same reason, you won't able to crack an analog interview without a switch-cap.

In This tutorial, I will take one switch cap circuit and try to analyze it both quantitively and qualitatively.

Q) Find the steady-state output Voltage V2.

Where S1 and S2 are two non-overlapping pulses.

Solution:

Before you watch the solution, I would recommend you try solving it by yourself first.

S1 on for 1st time

At first take S1 is on (While S2 is off), then the equivalent schematic will look like bellow.

For the first S1 pulse VC1=VC2=0

$V_{1}=V_{C2}= 1\times \frac{C}{C+2C}= \frac {1}{3}V$

Also, $V_{C1}=\frac{2} {3}V$

S2 on for 1st time

Now, For S2 pulse, The equivalent circuit would look like this,

In this case, VC2 is = 1/3 V. Due to charge sharing the node V2 has to jump by

$V_{2}=V_{C2}\times \frac{2C}{2C+3C}= \frac{1}{3} \times \frac{2}{5}=\frac{2}{15}$V

Now, at the end of the First pulse of both S1 and S2, the voltage across capacitors are

$V_{C1}= \frac{2}{3}V,V_{C2}= \frac{2}{15}V,V_{C3}= \frac{2}{15}V$

S1 on for the 2nd time

Now, The voltage across C and 2C when S1 is on for the 2nd time are,

$V_{C1}= \frac{2}{3}+\{(1-\frac{2}{3}-\frac{2}{15})\times \frac{2}{3}\}=\frac{4}{5}V$

$V_{C2}= \frac{2}{15}+\{(1-\frac{2}{3}-\frac{2}{15})\times \frac{1}{3}\}=\frac{1}{5}V$

P.S: Wonder how I have written these equations? Well, I will suggest looking at this Basic Charge Conservation of Capacitor.

S2 on for the 2nd time

Now, The voltage across C and 2C when S2 is on for the 2nd time are,

$V_{2}=\frac{V_{C2}\times 2C+V_{C3}\times 3C}{2C+3C}=\frac{\frac{1}{5}\times 2C+\frac{2}{15}\times 3C}{2C+3C}=\frac{12}{75}V$

At the end of this, the voltage across different capacitors are,

$V_{C1}= \frac{4}{5}V,V_{C2}= \frac{12}{75}V,V_{C3}= \frac{12}{75}V$

S1 on for the 3rd time

Similarly, Capacitors voltage while S1 is on for the third time are,

$V_{C1}= \frac{4}{5}+\{(1-\frac{4}{5}-\frac{12}{75})\times \frac{2}{3}\}=\frac{62}{75}V$

$V_{C2}= \frac{12}{75}+\{(1-\frac{4}{5}-\frac{12}{75})\times \frac{2}{3}\}=\frac{13}{75}V$

S2 on for the 3rd time

$V_{2}=\frac{V_{C2}\times 2C+V_{C3}\times 3C}{2C+3C}=\frac{\frac{13}{75}\times 2C+\frac{12}{75}\times 3C}{2C+3C}=\frac{62}{375}V$

At the end of this,

$V_{C1}= \frac{62}{75}V,V_{C2}= \frac{62}{375}V,V_{C3}= \frac{62}{375}V$

Now, Observe the voltage across V2=VC2 after each cycle (in one cycle both S1 and S2 are on ).

1st Cycle, $V_{2}(T)=V_{C3}(T)=\frac{2}{15}$

2nd Cycle, $V_{2}(2T)=V_{C3}(2T)=\frac{12}{75}$

3rd Cycle, $V_{2}(3T)=V_{C3}(3T)=\frac{62}{375}$

Now, check if there is any relation between these voltages? Yes, it has!

$V_{2}(2T)-V_{2}(T)=\frac{1}{5} \times V_{2}(T)$

also

$V_{2}(3T)-V_{2}(2T)=\frac{1}{5} \times V_{2}(2T)$

Clearly, It's a GP series

$V_{2}(\infty)=\frac{2}{15} \times (1+\frac{1}{5}+\frac{1}{5^{2}}+\frac{1}{5^{3}}+.............)= \frac{\frac{2}{15}}{1-\frac{1}{5}}= \frac{1}{6}V$

Finally, we got the answer after so much hard work. You are very eager to hear the phase 'well done' from the interviewer, instead, you can hear the sound of snoring.

Well, don't get sad if you are able to solve by this method, it means your basic is very strong, That I can say for sure!

But you all know, This blog is only concerned about problem-solving by intuition only.

Intuitive Approach:

Well, first tell me what do you mean by steady-state?

Well, you might answer, steady-state is something that doesn't change with time (Voltage or current).

Correct!

At steady-state irrespective S2 being 'on' or 'off' $V_{C3}$ doesn't change. That means C3 can't take charge from C2 anymore. That concludes $V_{C2}=V_{C3}$, they are in parallel.

At steady-state equivalent circuit looks like,