# Capacitive Divider Confusion

Consider the following

The voltage Vin is on the top plates of all capacitors. The top half capacitors are charged to a voltage of Vin-Vref and the bottom half capacitors are charged to a voltage of Vin.

Next, we switch the circuit to the following:

The 2C capacitor on the top-half side now has GND on it's bottom plate and Vin has been removed from the common node.

The question is "What is the voltage on node Vdac?" now

My answer: After switching the circuit to the second state, the top plate of the 2C capacitor on top side now must go to Vin-Vref so as to maintain the same charge on it. If this capacitor top plate has Vin-Vref on it and all the other capacitors still have Vin on their top plates, then charge redistribution must occur.

My confusion is regarding the ratio of the charge redistribution. I'm trying to draw the capacitive divider but I'm having trouble now:

• Is Vin disconnected first or is the top left capacitor connected to GND first? If the latter applies: is Vin low impedant? Then the answer is Vin of course. Oct 27, 2019 at 19:58
• @Huisman Vin is disconnected first and then the top left capacitor is switched to GND. Oct 27, 2019 at 20:09
• Some (I hope) constructive criticism: Do use reference designators, as it is way easier and unique way to refer to an element. Next, The voltage Vin is on the top plates of all capacitors. There is a visual contradiction. The (visiual) top plate of the upper capacitors are connected to Vref in the first situation. Oct 27, 2019 at 20:27
• @Huisman Noted. Will remember that for future diagrams. Oct 27, 2019 at 20:50

For capacitors applies that the charge is conserved (as well as the energy), but the voltage will not be conserved, it will change.

Giving the element reference designators, assuming "top plate" means the positive terminal and using the same definition for the voltage across C1 being $$\V_{in}-V_{ref}\$$ as OP does I come to this picture.

In the first situation, the total charge is:

$$Q_{total} = (C_1+C_2+C_3)(V_{in}-V_{ref}) + (C_4+C_5+C_6)V_{in}$$ $$Q_{total} = (4C)(V_{in}-V_{ref}) + (4C)V_{in}$$

In the second situation the total charge is:

$$Q_{total} = (C_2+C_3)(V_{unknown}-V_{ref}) + (C_1+C_4+C_5+C_6)V_{unknown}$$ $$Q_{total} = (2C)(V_{unknown}-V_{ref}) + (6C)V_{unknown}$$

As charge is conserved, you can find the total charge from the first equation and therefore be able to solve $$\V_{unknown} = V_{DAC}\$$.

Not sure whether this is a homework assignment, below is the spoiler:

$$(2C)(V_{unknown}-V_{ref}) + (6C)V_{unknown} = (4C)(V_{in}-V_{ref}) + (4C)V_{in}$$ $$(8C)V_{unknown} = (8C)V_{in}-(2C)V_{ref}$$ $$V_{DAC} = V_{unknown} = V_{in}-\frac{1}{4}V_{ref}$$

• Very nice response. One thing though, in your third equation, how come C2 and C3 have Vin-Vref across them, yet C1, C4, C5, C6 have Vunknown. My thought process was that the positive plate of C1 would have to change to Vin - Vref in the 2nd picture, then that -Vref would be distributed across ALL the capacitors. Oct 27, 2019 at 20:44
• @AlfroJang80 You're right! I missed that one. Will update. Oct 27, 2019 at 20:47
• Amazing answer. Thank you for putting in the time to explain this to me. I think I'll abandon my capacitive divider method, was driving me insane. Oct 27, 2019 at 20:50