Switch-Cap Basics (Charge Conservation)

Let's go through the basics of Charge Conservation. To make it clearer, I will take a problem through which I will try to explain the concept. This a very very important concept and if you are clear with this concept, you can easily tackle any kind of capacitor related problem asked by the interview. Let's start! 

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Q) Find the voltage across each capacitors after the switch S is closed at t=t0. Assume the initial voltages across C1 and C2 are VC1 and VC2 respectively.

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Solution:

Initially when the switch S was open, \(V_{C1}(0)=V_{C1}(t_0-)=V_{C1}\) and \(V_{C2}(0)=V_{C2}(t_0-)=V_{C2}\)

After the switch is closed at \(t=t_0\), there will be an impulse current \(\left [\frac{C_1C_2}{C_1+C_2}V_s \delta (t-t_0)  \right ]\) flowing through both the capacitors and hence capacitor voltages will rise instantaneously.

\(V_{C1}(t_0+)=V_{C1}+x\) and \(V_{C2}(t_0+)=V_{C2}+y\)

As the capacitors are in series, so \(i_{C1}=i_{C2}\Rightarrow C_{1}\frac{\mathrm{d} V_{C1}}{\mathrm{d} t}=C_{2}\frac{\mathrm{d} V_{C2}}{\mathrm{d} t}\Rightarrow C_{1}\times x=C_{2}\times y\)     (1)

Applying KVL, \(V_{s}=V_{C1}+x+V_{C2}+y\)     (2)

Solving (1) and (2), 

\(x=\left ( V_{s}-V_{C1}-V_{C2} \right )\times \frac{C_2}{C_1+C_2}\)

\(y=\left ( V_{s}-V_{C1}-V_{C2} \right )\times \frac{C_1}{C_1+C_2}\)

Hence the final voltage across C1,  \({\color{Red} {V_{C1}(t\geq t_0)=V_{C1}+\left ( V_{s}-V_{C1}-V_{C2} \right )\times \frac{C_2}{C_1+C_2}}}\)

And the final voltage across C2,  \({\color{Red} {V_{C2}(t\geq t_0)=V_{C2}+\left ( V_{s}-V_{C1}-V_{C2} \right )\times \frac{C_1}{C_1+C_2}}}\)

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Conclusion / Take away:

So the net increment in the capacitor voltages will be \(\Delta V=\left (V_{s}-V_{C1}-V_{C2}   \right )\), which will be eventually divided according to the capacitive voltage division rule.

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Intuition:

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

You can follow 3 simple steps:

Step1: Find the net increment in the capacitor voltages, i.e.

\(\Delta V=\left (V_{s}-V_{C1}-V_{C2}   \right )\)

Step2: Apply the capacitive voltage division rule, i.e.

\(\Delta V_{C1}=\Delta V\times \frac{C_2}{C_1+C_2}\)

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

Step3: Add the increment voltage with the initial voltage to find the final voltage.

\(V_{C1}(t\geq t_0)=V_{C1}+ \Delta V_{C1}\)

\(V_{C2}(t\geq t_0)=V_{C2}+ \Delta V_{C2}\)

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 suggestion 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 Capacitor Circuit-1 and Switch Capacitor Circuit-2 to know how much you have understood this concept. For more content click here INDEX