Switch Capacitor Circuit-2

In the last two tutorials on switch capacitor circuits, we developed the basis of charge conservation. So, before you start reading this just go through them once.

 

Q) Find the steady-state voltage \(V_{out}\), where S1 and S2 get on periodically (Non-overlapping).

Solution: 

S1 is on for the first time

From the above picture, we get \(V_{C1}=V_{1}\) and \(V_{C2}=0\)

S2 is on for the 1st time

Here incremental  $ \Delta V= V_{2} - (-V_1)= V_{2} +V_{1}$

You can check Switch-Cap Basics (Charge Conservation) to have the basic understanding of charge conservation & distribution in a capacitor.

i) \(V_{C2}= 0 + \Delta V \times \frac{C_1}{C_1+C_2}\)

= \( \frac{V_{1}+V_{2}}{2}\)

      ii) \(V_{C1}= -V_{1} + \Delta V \times \frac{C_2}{C_1+C_2}= \frac{V_{2}-V_{1}}{2}\)

S1 is on for the 2nd time

i) \(V_{C1}= V_{1}\)

ii) \(V_{C2}=\frac {V_{1}+ V_{2}}{2}\)

S2 is on for 2nd time

here incremental \(\Delta V = V_{2} - (-V_{1})- \frac{V_{1}+V_{2}}{2}=\frac{V_{1}+V_{2}}{2}\)

i)   \( V_{C2}=\frac{V_{1}+V_{2}}{2} + \frac{V_{1}+V_{2}}{2} \times \frac{C_{1}}{C_{1}+C_{2}}\)

= \(\frac{V_{1}+V_{2}}{2} + \frac{V_{1}+V_{2}}{4} \)

ii) \( V_{C1}= (-V_{1}) + \frac{V_{1}+V_{2}}{2} \times \frac{C_{2}}{C_{1}+C_{2}}\)

S1 is on for the 3rd time

i) \(V_{C1}= V_{1}\)

ii) \(V_{C2}=\frac{V_{1}+V_{2}}{2} + \frac{V_{1}+V_{2}}{4}\)

S2 is on for 3rd time

here incremental \(\Delta V = V_{2} - (-V_{1})- \frac{V_{1}+V_{2}}{2}- \frac{V_{1}+V_{2}}{4}=\frac{V_{1}+V_{2}}{4}\)

i)   \(V_{C2}=\frac{V_{1}+V_{2}}{2} + \frac{V_{1}+V_{2}}{4} + \frac{V_{1}+V_{2}}{4} \times \frac{C_{1}}{C_{1}+C_{2}}\)

= \(\frac{V_{1}+V_{2}}{2} + \frac{V_{1}+V_{2}}{4} + \frac{V_{1}+V_{2}}{8}\)

ii) \( V_{C1}= (-V_{1}) + \frac{V_{1}+V_{2}}{4} \times \frac{C_{2}}{C_{1}+C_{2}}\)

Following this trend the steady-state voltage

 

\(V_{out}=V_{C2}= \frac{V_{1}+ V_{2}}{2} + \frac{V_{1}+ V_{2}}{4} +\frac{V_{1}+ V_{2}}{8}.......\) 

= \(\frac{V_{1}+V_{2}}{2} \times (1+ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....)\)

=\( \frac {V_{1}+V_{2}}{2} \times \frac {1}{(1-\frac{1}{2})}\)

=\( V_{1}+V_{2}\)

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Intuition:

Now, comes the best part (the cherry on the cake)! Obviously, you can't remember this formula, and also you shouldn't try. Let's try to understand it intuitively! 😀

At a steady-state, the potential of the different nodes shouldn't be changing with a change in the phase of the switches. Then one can easily conclude, the voltage across capacitor C1 remains unchanged. 

\(V_{C1}|_{S1,on}=V_{C2}|_{S2,on}=V_{1}\)

Applying KVL in the above diagram,

\(-V_{2}-V_{1}+V_{out}=0\)

\(=>V_{out}=V_{1}+V{2}\)

 

Well, to conclude this post, I would like to request everyone (who will visit this blog) to comment below if you find this content helpful and well explained. Also, if you have any suggestions to improve please let me know by commenting below. I would really appreciate this as you know, Negative feedback always makes a system stable. 😅

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Now you can try to attempt this Switch Cap Problem-1 to know how much you have understood this concept. For more content click here INDEX

Switch Capacitor Circuit-1

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. 

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Q) Find the steady-state output Voltage V2.

Where S1 and S2 are two non-overlapping pulses.

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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 

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

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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,

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

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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,

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

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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

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

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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. 

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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,  

Now, easily you can find the voltage V2 in steady-state using the capacitive voltage division rule.

\(V_{2}=1 \times \frac{C}{C+(2C||3C)}=1 \times \frac{C}{C+5C}=\frac{1}{6}V\)

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For More content click here INDEX

To have a hand on switch-cap, have a look at Basic of Charge conservation.

For More RC circuit questions for the interview.

For Interview question of Texas Instruments.