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Since someone edited my question and asked me to post a new question instead with the edits I did, so here is the new schematic for this question:

Can someone help me find the voltage and charge across each of the capacitors? This is not just a simple/ parallel circuit so I am having a problem in formulating the right equation on how to find the voltage and charge across each capacitor.This is actually one branch of a charge pump wherein capacitors pump charges to the load capacitor, CL. I also want to know how charge transfer occurs between capacitors. enter image description here

Below is the original circuit: enter image description here

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  • \$\begingroup\$ Please label all the nodes in your schematic so we can discuss them and you will know what we are talking about. \$\endgroup\$ – The Photon Dec 1 '14 at 2:23
  • \$\begingroup\$ Hint: The node connecting R2, C2, and Cx has no dc path to ground, so there is no unique solution to your problem. (You see why it would have been easier if you would have labeled that node "node 2" or something?) \$\endgroup\$ – The Photon Dec 1 '14 at 2:24
  • \$\begingroup\$ Finally, have you tried entering the circuit into a SPICE-like simulator? What result did you get? What about the result don't you understand? \$\endgroup\$ – The Photon Dec 1 '14 at 2:25
  • \$\begingroup\$ Charge-pump? Then it should have some switches in the circuit? Right? \$\endgroup\$ – diverger Dec 1 '14 at 2:30
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    \$\begingroup\$ Is Vd meant to be an ac source? You need to add that information to the question if it is. \$\endgroup\$ – The Photon Dec 1 '14 at 2:56
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Considering the top circuit :

Applying KCL at the nodes :

A : (Vd - Va) / R1 = ( Va / sC1) + ( (Va - Vb) / sCx)

B : ( (Va - Vb) / sCx ) + ( (Vd - Vb) / sC2) = ( Vb - VL) / R2

L : ( (Vb - VL) / R2 ) = VL / sCL

3 equations for 3 unknowns.

( Don't forget to put Laplace transform of input ie of Vd).

(Did I miss something ?)

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  • \$\begingroup\$ I've tried this solution which is very lengthy but at the end evaluating s to infinity for steady state I got an answer of zero. So, might as well try another solution. \$\endgroup\$ – djambalong Dec 1 '14 at 5:16
  • \$\begingroup\$ I've reviewed it together with my instructor and we seemed to find no error in the process of solving it (evaluating algebra, nodal analysis, etc). Somebody told me to try the charge transfer but I had a problem analyzing it. \$\endgroup\$ – djambalong Dec 1 '14 at 5:25
  • \$\begingroup\$ @djambalong . Yes the solution will be lengthy cuz you asked for voltage across all capacitors. And btw, why take limit s-> infinity ? For steady state, you use final value theorem for laplace transform acc to which you take lim s->0 ( s*F(s) ). \$\endgroup\$ – Plutonium smuggler Dec 1 '14 at 8:36
  • \$\begingroup\$ @djambalong . Also, if you evaluate s to infinity, you will get initial values, as per initial value theorem. And since you are getting zero, this suggests that indeed my solution should be correct, since all voltages are zero initially. \$\endgroup\$ – Plutonium smuggler Dec 1 '14 at 8:40
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Looking at the botton circuit, it can be calculated as folows. As you know Kirchoff applies in each node all the time. Instead of writing Kirchoff using currents, we do it using charge (current integral over time):

a) $$Q_{R1}=Q_{C1}+Q_{Cx}$$

b) $$Q_{Cx}=Q_{C2}+Q_{CL}$$

The total charge in each capacitor, at the end, is its initial charge plus the charge that has circulated afterwards, i.e. QC1total=QC1initial + QC1 (in previous equation).

Therefore the initial charges (before current starts to flow) must be known to solve the problem. We will assume the initial charge of all capacitors is 0, and therefore QC1total=0+QC1=QC1, QC2total=QC2, etc.

Using second law of Kirchoff we can now write the final voltages, knowing that final currents will be 0, and therefore there is no voltage drop in resistors.

I) $$-Vd+V1=0 \rightarrow -Vd+\frac {Q_{C1}} {C1}=0$$

II) $$-V1+Vx+V2+Vd=0 \rightarrow \frac{-Q_{C1}} {C1}+\frac {Q_{Cx}} {Cx}+\frac {Q_{C2}} {C2}+Vd=0$$

III) $$-Vd-V2+Vout=0 \rightarrow -Vd-\frac {Q_{C2}} {C2}+\frac {Q_{CL}} {CL}=0$$

With those 3 equations plus the equation in b) you get 4 equations with 4 unknowns.

The equation in b) is saying in fact that \$\Delta Q_{Cx}=\Delta Q_{C2}+\Delta Q_{CL}\$, i.e. $$Q_{Cx}-Q_{C2}-Q_{CL}=Q_{previousCx}-Q_{previousC2}-Q_{previousCL}$$

Taking equation II, III and the last one, you can write in matrix form $$ \left( \begin{array}{c} 0 \\ Vd\\ Q_{previousCx}-Q_{previousC2}-Q_{previousCL} \end{array} \right) = \left( \begin{array}{ccc} \frac 1 {Cx} & \frac 1 {C2} & 0 \\ 0 & -\frac 1 {C2} & \frac 1 {CL} \\ 1 & -1 & -1 \end{array} \right) \cdot \left( \begin{array}{c} Qx \\ Q2\\ QL \end{array} \right)$$

This can be solved by inverting the matrix and multiplying by the vector in the left where You must use the previous values of capacitor's charge (at the beginning they will be 0).

After doing this, you should get the equations for the other value of CLK. You will see that the matrix doesn't change, only the vector changes (Vd swaps position with 0) and you will use as Qprevious the charges obtained in the previous step. After multiplying you will get the new values of Q and you can iterate again with other value of CLK.

You will see that the output voltage has a ripple (oscillates between 2 values). For your simulations choose a slow clock and you'll see that same ripple. If the clock is fast, the output voltage won't have ripple, but its final value will be the average between the 2 values obtained from equations.

Example using Vd=5V, C1=1uF, Cx=0.1 uF, C2=1uF, CL=1uF and a slow CLK of 10 Hz, here is the simulation: Simulation 10 Hz

You can see it oscillates between 2.619 V and 238.1 mV.

We get those same results solving, in Matlab, the equations previously stated: Equations solved

If you increase clock, the output voltage becomes smooth as you can see in this simulation (CLK changed to 100 kHz): enter image description here

The final voltage is 1.43 V, which you can recognize as the average between 2.619 V and 238.1 mV.

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  • \$\begingroup\$ thanks @Roger C. but why is it that the resistor isn't included in performing the KVL? \$\endgroup\$ – djambalong Dec 6 '14 at 15:59
  • \$\begingroup\$ Because when the capacitors are charged the current I stops to flow, i.e. I=0 A. Therefore the voltage drop in the resistors V=I·R=0·R=0 V. Of course this is assuming that input voltage (CLK in your diagram) is steady enough time to allow the capacitors to be fully charged. \$\endgroup\$ – Roger C. Dec 6 '14 at 16:09
  • \$\begingroup\$ That means there is an initial charge across the capacitors?how could i get the initial charge @Roger C. ? I've tried solving it and I assume that the value in pF of capacitor Cx should affect the Vout based on my simulation results but when I subsituted values on it the output is still Vd in just one branch. Adding the two branches would result to twice Vd. But Cx should affect the Vout, like, when Cx=1/2 C1; Vout=1.5xVd. \$\endgroup\$ – djambalong Dec 6 '14 at 17:26
  • \$\begingroup\$ @djambalong, I've updated the answer. \$\endgroup\$ – Roger C. Dec 7 '14 at 12:29
  • \$\begingroup\$ @djambalong, I followed this sign convention for voltages and charges in vertical components: + top and - bottom and for horizontal components + left and - right. Note: always considering the bottom circuit in your question. \$\endgroup\$ – Roger C. Dec 7 '14 at 17:30

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