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Consider the following situation

enter image description here

We have two capacitors, C1 and C2, both with a value of 10μF. The capacitor C1 through some switching action now has a voltage of -Ain + Vref across it whereas C2 has just -Ain across it.

My understanding is that, current will flow from C1 to C2, redistributing the charges and the circuit will settle to a steady-state in which both capacitors have a voltage of -Ain + Vref/2 across them as shown below:

enter image description here

My question is, what would happen if C1 had a value of 10μF but C2 had a value of 50μF? I understand that V = Q/C, so in order to equalise the voltage, it would take a lot more charge on the plates of C2. With the conservation of charge in mind, does that mean the final voltage across both capacitors would be -Ain +Vref/10? In that case, we are missing (8/10)*Vref - where did that voltage go?

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  • \$\begingroup\$ The "voltage" doesn't "go anywhere". On a charged capacitor a certain voltage means a certain charge. If you change the capacitor value (C) while the charge Q remains the same, then V will change. Maybe related: the two capacitor paradox: en.wikipedia.org/wiki/Two_capacitor_paradox \$\endgroup\$ Commented Sep 25, 2019 at 14:11
  • \$\begingroup\$ @Bimpelrekkie So, what would the final voltage be? \$\endgroup\$ Commented Sep 25, 2019 at 14:17
  • \$\begingroup\$ So basically, charge is still conserved always, so in this case, the final voltage across both capacitors would be -Ain +Vref/10. The missing 8/10*Vref is due to ideal circuit theory failing? \$\endgroup\$ Commented Sep 25, 2019 at 14:27

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Your first circuit is invalid and can not be analyzed by the usual means. You have two ideal capacitors in parallel, which means that the voltage across them must be the same. However, you specify that they have different voltages. This is not possible if you are talking about ideal wires and ideal capacitors.

To make the circuit analyzable you must add a resistance between the capacitors. Then you need to analyze the current flow and determine how much energy is lost in the resistor before you can see the final states of the capacitors.

For a more detailed discussion, see Capacitor Paradox

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  • \$\begingroup\$ That's the equivalent circuit from a SAR DAC. see electronics.stackexchange.com/questions/460100/… \$\endgroup\$ Commented Sep 25, 2019 at 14:14
  • \$\begingroup\$ No, I think the situation in the linked question is different. In that question it looks like the capacitors in parallel are connected to a voltage source and that they all have the same voltage, or they are all disconnected from any voltage source but all have the same voltage. \$\endgroup\$ Commented Sep 25, 2019 at 14:18
  • \$\begingroup\$ Initially yes, but then the MSB capacitor is charged to -Ain + Vref as the switch S1 changes to Vref from ground and thus now the MSB capacitor has a voltage of -Ain + Vref across it, yet all the other capacitors still only have -Ain, hence leading us to the problem I have shown here. \$\endgroup\$ Commented Sep 25, 2019 at 14:26
  • \$\begingroup\$ @AlfroJang80, that is the simplified equivalent model of a SAR DAC, in particular it ignore the resistance (and inductance) that is always present in real electronics because for the purpose of demonstrating what they wished to show, it didn't matter. The (introductory) circuit theory notion of a wire as being something of zero resistance, zero inductance and having zero capacitance to other wires (or even to other points on itself) is a nice simplification that more or less works for many teaching problems (because they are designed that way), but it gets you into trouble in the limits. \$\endgroup\$
    – Dan Mills
    Commented Sep 25, 2019 at 15:22
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    \$\begingroup\$ There's a two spring "paradox" just like the tow capacitor one. The answer to where the energy went is "into the resistor" for two capacitors, and "into the dashpot" for two springs. \$\endgroup\$
    – john
    Commented Jun 6, 2021 at 1:09
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Since you know V=Q/C, you know voltage is inversely proportional to C. Whatever difference there is in the two voltages C1-C2 will then be multiplied by C1/(C1+C2) or 1/6 in your second example.

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  • \$\begingroup\$ Ah yes! So steady-state voltage at the end will be -Ain + (1/6)Vref \$\endgroup\$ Commented Sep 25, 2019 at 15:52
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The capacitor C1 through some switching action now has a voltage of -Ain + Vref across it whereas C2 has just -Ain across it.

To make the "switching action" apparent, let's redraw your schematic to show a switch

schematic

simulate this circuit – Schematic created using CircuitLab

In each capacitor, the voltage, charge and capacitance are related by

$$C_nV_n = Q_n$$

Where \$V_n\$ is the voltage across capacitor \$C_n\$ and \$Q_n\$ is that capacitor's charge.

When the switch is closed, the voltages must become equal, but the total charge must be conserved. We will use primes to indicate values after the switch is closed, and the absence of primes to indicate values before the switch is closed.

So.

$$C_1V_1 = Q_1$$ $$C_2V_2 = Q_2$$

$$Q_{total} = Q_1 + Q_2 = Q'_1 + Q'_2$$ $$V' = V'_1 = V'_2 = C_1Q'_1 = C_2Q'_2$$

with a bit of re-arranging.

$$Q_{total} = V_1C_1 + V_2C_2 = V'C_1 + V'C_2 = V'(C_1 + C_2)$$

$$V' = \frac{V_1C_1 + V_2C_2}{C_1 + C_2}$$

My understanding is that, current will flow from C1 to C2, redistributing the charges and the circuit will settle to a steady-state in which both capacitors have a voltage of -Ain + Vref/2 across them

If \$C_1 = C_2\$, yes.

what would happen if C1 had a value of 10μF but C2 had a value of 50μF?

Then the final voltage will be

$$V' = \frac{10V_1 + 50V_2}{10 + 50} = \frac{V_1 + 5V_2}{6} = \frac{-A_{in} + V_{ref} -5A_{in}}{6} = \frac{-6A_{in} + V_{ref}}{6} = -A_{in} + \frac{1}{6}V_{ref}$$

So, the answer is \$-A_{in} + \frac{1}{6}V_{ref}\$ not \$-A_{in} + \frac{1}{10}V_{ref}\$

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