I was trying some simulations and tried the below circuit.

enter image description here

My question is, how is the below marked voltage negative (-3 V)?

I thought like, since there is a diode, there will be approximately 4.3 V across the capacitor and 0.7 V across the diode. Why is this not the case in the above circuit?

  • \$\begingroup\$ Try to add a small series resistor 1Ω and restart the simulation. \$\endgroup\$
    – G36
    Aug 17, 2022 at 18:23
  • \$\begingroup\$ Thank you. But why do you say that? What's the reason behind that logic? \$\endgroup\$
    – user220456
    Aug 17, 2022 at 18:36
  • \$\begingroup\$ I don't know. I suspect a bug. Because changing the diode to a "real model" (1N4148) also fix the circuit. tinyurl.com/2m8sg3yw \$\endgroup\$
    – G36
    Aug 17, 2022 at 18:47
  • \$\begingroup\$ Try resetting the simulation \$\endgroup\$
    – user253751
    Aug 17, 2022 at 19:17
  • 1
    \$\begingroup\$ the capacitor is pre-charged to 8.09 V ... give it a way to discharge \$\endgroup\$
    – jsotola
    Aug 17, 2022 at 19:19

1 Answer 1


Simulators can't model everything, they simulate only within their limits and according to the models of components they have, and using discrete finite time steps.

Your circuit has ideal voltage source, ideal wires, ideal capacitor and ideal diode, simulated with discrete time steps. And it is not known which capacitor simulation you have selected.

As such an ideal circuit like you have drawn can't exist in real life, simulating it will end up in non-real life result in simulator.


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