# PSPICE simulation of capacitor voltage and current disagrees with math

I'm testing a simple enough circuit on PSPICE. The zener has a BV = 220V, so the capacitor has a voltage of 220V until the switch closes. • Using the time equation for voltage of a capacitor: Vc(t) = Vo * e^(-t/tau)

Vc(1ns) = 220V * e^(-1ns/50*500pF) = 211.37V

Vc(25ns) = 220V * e^(-25ns/50*500pF) = 80.93V

Vc(50ns) = 220V * e^(-50ns/50*500pF) = 29.77V

Vc(75ns) = 220V * e^(-50ns/50*500pF) = 10.95V

Vc(100ns) = 220V * e^(-50ns/50*500pF) = 4.02V

• Calculate dV/dt:

1ns --> dV/dt = (220V - 211.37V)/1ns = 8.63E9

25ns --> dV/dt = (220V - 80.93V)/25ns = 5.56628E9

50ns --> dV/dt = (220V - 29.77V)/50ns = 3.8046E9

75ns --> dV/dt = (220V - 10.95V)/75ns = 2.787E9

100ns --> dV/dt = (220V - 4.02V)/100ns = 2.1598E9

• And using the current equation for capacitors: Ic(t) = C*dV/dt

Ic(1ns) = 500pF*8.63E9 = 4.315A

Ic(25ns) = 500pF*5.56628E9 = 2.78A

Ic(50ns) = 500pF*3.8046E9 = 1.90A

Ic(75ns) = 500pF*2.787E9 = 1.3935A

Ic(100ns) = 500pF*2.1598E9 = 1.079A

Seems great. Unfortunately, PSPICE disagrees.

The peak voltage node R2 reaches after the switch closes is only around 28V. It's clear to see that based on the topology this is just one big voltage divider. There shouldn't be any voltage at all on the top node after the switch closes because the overwhelming majority of voltage is dropped across R1. This doesn't explain why the capacitor releases such a wimpy peak in voltage though. So I simply decide to test out 500nF, and poof! I get an output that I was expecting for the 500pF capacitor. But, I still don't understand why neither of these simulations discharges the capacitor within the expected time constant of 25ns(4) = 100ns. It should be almost completely discharged at 4tau.

Obviously PSPICE cannot be wrong. What am I missing here? • "Obviously PSPICE cannot be wrong." Hah... – pipe Nov 21 '18 at 21:29
• Your calculations for dV/dt and hence Ic are incorrect. dV/dt can be calculated directly by differentiating the equation for Vc and then substituting the desired values of t. Your calculations assume that Vc is falling linearly when it is actually falling exponentially. – Barry Nov 21 '18 at 21:50