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