Capacitor voltage divider network simplification

[Dino1400]

Hi, I’m try to calculate the voltage Vs with a function of VG or (VG function VS) in this capacitor network
Any idea, suggestion
Thank you all in advance

 

[BradtheRad]

In order for this quiz to be solvable, we must assume that (a) all capacitors start at zero charge, and (b) there is an unchanging resistance in series with VG.

Since there are no other components besides capacitors, we can analyze this network similarly to a resistor network... although we invert certain formulae. For one thing, the smaller Farad values will have the greater voltage on them (as a general rule).

The two upper capacitors are in parallel. Their charge will always be identical.
The formula for capacitors in parallel says the total value is the sum of their individual Farad values.

Then you continue analyzing the network as you would do with a resistor network. Figure out the series and parallel branches, etc.

Eventually you'll have the net Farad value of all except the bottom capacitor.
You can then calculate the proportion between the bottom capacitor and the rest of the network. This will give you the charge on each.
From there you can calculate the volt levels on all the capacitors.

 

[Dino1400]

Actually the upper capacitor are not in parallel, and I don’t have the values I only need an approximation expression, this network doesn’t have a series or parallel branches and we don’t know charges on each one. It must be some rules to divide the capacitance like we for Miller capacitance. I still don’t have the full idea how to simplify this.

 

[albbg]

Why you say that the two upper capacitors (Cox and 2Cox-vr) are not in parallel ? According to your diagram, they are as BradtheRad said (however their charge in general will be different; q=CV, then since the voltage is the same but the capacitance is different, then the charges have to be different).
It's not clear to me what you exactly want to calculate:

the voltage on each capacitor after they charge due to a DC input voltage (Vg) or the behaviour of the network when Vg is an AC voltage ?

Are you sure the resistor (due for instance to non-ideality) on each branch are negligible ? Is there no load at Vs node ?

 

[Dino1400]

VG is the only DC voltage applied to the circuit normally around 1V Vs is the output of the circuit.

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Thanks for your help and I did not notice that both upper capacitors are in Series
But I still not getting the answer VG=f(Vs) or Vs=f(VG)

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here is another view

 

[prestonee]

Ok So you might want to go back and look at your application or homework because if Vg is your only supply and is a DC supply, then you have a lot of open circuits and Vs = floating. Remember DC current current cant pass through a cap.
if Vg was ac source. simply convert the caps into resistors of 1/(s*Cap_value), s = j*w or s=j*2*pi*f.typically esr in a capacitive voltage divider is negligible. so just combine the parallel top caps, place in series with the 1/cox2 for // with 1/cbox all in series with csi and you have your total current. then determine current in each leg, once you get the current in the leg of interest use a simple I = V/R to get your Vs value.
-Pb

 

[albbg]

Let's call

C1=Cox+2Cox-vr
C2=Cox2
C3=Csi
C4=2Cbox

First of all we can find the equivalent capacitance. If Co=C1*C2/(C1+C2) then Ceq=(Co+C4)*C3/(Co+C4+C3)

The total charge will be q=Vg*Ceq.

Since C1 will share the same charge of C2 and C3 will share the same charge of C1 and C2 plus the charge of C4, that is the total charge (you can treat the charge as a current) and since q=C*V, then if Vi is the voltage across Ci:

q=C1*V1+C4*V4
q=C2*V2+C4*V4
q=C3*V3

Furthermore, since C1 and C2 share the same charge ==> C1*V1=C2*V2 ==> V1=C2/C1*V2
and, from Kirchoff V1+V2=V4

We want Vs=V3+V2

q*Vg*Ceq and q=C3*V3 then Vg*Ceq=C3*V3 ==> V3=Vg*Ceq/C3

Using now: V1=C2/C1*V2 and V1+V2=V4 ==> V4=V2*(1*C2/C1)
substituting in q=C2*V2+C4*V4 ==> q=[C2+C4*(1*C2/C1)]*V2
equating this last to q=Vg*Ceq we will obtain V2=Vg*Ceq/[C2+C4*(1*C2/C1)]

Then Vs=Vg*Ceq*{1/C3+1/[C2+C4*(1*C2/C1)]}

in the similar way you can solve all other voltages.