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Below is a circuit build in LTSpice. It is capacitor network simulation.

What is of interest to me is the voltage at node N005. A voltage controlled switch rises the voltage of node N003 to 29.42 Volt as expected from voltage source V1.

LTSpice Circuit Siumlation capacitor network

In a series capacitor network the voltage is highest at the least capacitance value according to formular:

Vc1 = Q / C1
Vc4 = Q / C4

Vtotal = Vc1 + Vc4.

However, node N005 stays at zero (2 mV) all the time. I would expect a higher voltage at N0005. Even using non ideal capacitors does not change this behaviour. Can you give me a hint why this occurs? Measuring at a real lab experiment shows 26.5 volt at N0005.

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  • \$\begingroup\$ I would try bypassing C4 with something large (10 Meg) as LTSpice sometimes has issues with DC biasing. Your ground node looks attached, but is it? \$\endgroup\$ Dec 4, 2018 at 14:35
  • \$\begingroup\$ Are you sure N003 produced 29.42 volts? \$\endgroup\$
    – Andy aka
    Dec 4, 2018 at 14:35
  • \$\begingroup\$ Ground node is attached, I put it now some distance of the lower line and made a new connection to it but behaviour stays the same. Putting 10meg parallel to c4 doesnt change the voltage level but produces ripples on voltage of c4. N003 produces 29.42 Volts. \$\endgroup\$
    – hendrik2k1
    Dec 4, 2018 at 14:38
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    \$\begingroup\$ You might try adding a ground connection to the V2 circuit. In an ideal world it wouldn't matter, but SPICE really likes to have a path to ground for everything. \$\endgroup\$ Dec 4, 2018 at 14:40
  • \$\begingroup\$ Changes nothing in circuit behaviour. \$\endgroup\$
    – hendrik2k1
    Dec 4, 2018 at 14:42

2 Answers 2

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The net N005 is a floating net, that means it does not have a connection to the rest of the circuit. Imagine that a charge was present on that net, it would be trapped and cannot leave!

In a simulator capacitors can be ideal meaning that the have an infinitely good isolation. For the circuit simulator it will then be a challenge to determine which capacitor is charged, C1, C4 or both have a charge?

To prevent this the circuit simulator can (temporarily) add a very high value resistor (like 1000 Mohm) to the circuit to help it find a solution. That can explain why you see the 2 mV, net N005 is pulled down by this resistor which the simulator added.

On your bench things are very different. Capacitors are not ideal and have leakage. Obviously the 100 uF capacitor leaks a lot more than the 10 uF one and that pulls the voltage at N005 up and you measure 26.5 V.

If you measure with a normal voltmeter, realize that it has a 10 Mohm input impedance, that can cause a current to flow in the same order of magnitude as the leakage currents of the capacitors. That means your meter influences the voltages you're trying to measure.

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  • \$\begingroup\$ But charge doesn't need to leave N005, does it? We just expect charge to move back and forth between the bottom plate of C1 and the top plate of C4 as a voltage is applied. \$\endgroup\$ Dec 4, 2018 at 14:46
  • \$\begingroup\$ Yes I see that putting the voltmeter across C4 shows 26.5 for an instance and the voltage starts declining. So in other words only leakage current in non ideal capacitor are the source for the appearing of 26.5 volts an N0005? \$\endgroup\$
    – hendrik2k1
    Dec 4, 2018 at 14:56
  • \$\begingroup\$ @ElliotAlderson What you write is correct for the simulated situation, however a circuit simulator does not know what charge it should start with. Circuit simulators generally start with solving the circuit for DC so leaving out all capacitors. That results in N005 floating so the simulator cannot determine the voltage, so it adds a "gmin" which is a very high value resistor to ground. \$\endgroup\$ Dec 4, 2018 at 14:57
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    \$\begingroup\$ only leakage current in non ideal capacitor are the source for the appearing of 26.5 volts an N0005 Yes, that is correct. And that's why experienced EEs never use capacitors in series without resistors in parallel to define a proper DC voltage. Now the voltages across the caps depend on leakage which is very unpredictable. \$\endgroup\$ Dec 4, 2018 at 15:00
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The issue is that the switch has a finite off resistance. When you start the simulation, V1 is already live, and has charged the caps up, in some fashion that doesn't seem to be well defined. Looking at the current through R1 shows this - the transfer of charge that would result in the divider effect of the different capacitances doesn't occur.

If you add the startup condition, which sets everything to zero until the simulation starts, you get the expected behavior. enter image description here

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