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The voltage at the top most node of this circuit should be the average of the input voltages.

Vtop*CT = C1*V1 +C2*V2 +C3*V3 where CT = C1+C2+C3

I get 0V instead.

enter image description here


UPDATE


Thanks for the help. I managed to make it work by adding resistances in parallel with the capacitors. enter image description here

CT = (20uF+10uF+10uF) = 40uF

4Vtot = 15V*2 +3V + 4V

Vtot=9.25V as seen in the graph

Here's what i was trying to simulate: http://sci-hub.tw/https://www.sciencedirect.com/science/article/pii/S1434841116308524

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  • \$\begingroup\$ Replace C with equal R’s \$\endgroup\$ – Sunnyskyguy EE75 Jul 13 '18 at 18:03
  • \$\begingroup\$ Use uic, or startup in the simulation settings. I'd wager Multisim has the same solution. \$\endgroup\$ – a concerned citizen Jul 13 '18 at 18:09
  • \$\begingroup\$ What did you want to learn or now about this stupid and unrealistic example? Enable "skip initial operating point solution" in transient options \$\endgroup\$ – G36 Jul 13 '18 at 18:11
  • \$\begingroup\$ I wanted to add DC voltages without using an amplifier to reduce power consumption. I thought this should work, I worked out the math. I added the the large resistance to add a path to ground for simulation purposes. \$\endgroup\$ – ElecNoob Jul 14 '18 at 4:27
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    \$\begingroup\$ Thanks for the help. I'm not trying to blame the tool. I was wrong to say it was a simulation problem. I just couldn't find the right way to describe it. I should really read the tutorials first. I dove right in without knowing anything. \$\endgroup\$ – ElecNoob Jul 14 '18 at 6:58
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If you are performing a dc or operating point analysis then the capacitors are treated as an open circuit, as if they were not there at all. A transient analysis with only dc sources is essentially the same as dc simulation. R1 then provides the only conducting path to ground, so the voltage is 0V.

Change your voltage sources to ac sources, all at the same frequency. Compare the peak-to-peak voltage at the top node to the source voltages.

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The RC timeconstant is \$ \tau = R C\$

which is going to be 10uF*1GΩ= 10e-6*1e9=10000s

which means that the two seconds of time will not be near enough to see the changes in the voltage with such a large time constant.

The other problem is you have no starting condition, If you specify a voltage the simulator solves the operating point for those values, all the voltages will be settled. You need a step input so it starts the inital conditions at zero and then step the voltages.

To fix this, increase the simulation time and use a step input as shown below (shown for one source) notice the time axis is in kiloseconds.

enter image description here

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