Finding the parallel resistance across CL capacitors 01 PUBLIC

If CL applies the same sort of techniques as SPICE then CL probably places a very high value resistor in parallel with a capacitor as part of the model.

This simulation presents three ways to find the value of that resistance.

If CL applies the same sort of techniques as SPICE then CL probably places a very high value resistor in parallel with a capacitor as part of the model.

This simulation presents three ways to find the value of that resistance. Let's call this resistance Rpar.

In the Time Domain simulation, SW1 is in series with a very large resistance (1T) so that when the switch is off, the minimum open circuit resistance across C1 is guaranteed equal to be at least that resistance. This minimises the effect of any switch of resistance appearing in parallel with Rpar.

V(cap) = 1V when SW1 opens at T = 1ps.

V(cap) falls to 999mV at T = 1000s.

From CV = IT the value of the effective parallel resistance is given by:

Rpar = 1000s/((1V - 0.999V)*1pF)

This is the calculation that is then used to set the value of Rparcalc.

Rparcalc is then set up in series with C2. The voltage across R2 in series with C2 is 1V. Since V(dccap) = 0.5V, this demonstrates that Rpar = Rparcalc.

The DC Sweep simulation sweeps the value of Rparcalc and plots V(dccap).

When V(dccap) = 0.5V then Rparcalc = Rpar.

Simulate > Time Domain > Run Time-Domain Simulation

Simulate > DC Sweep > Run DC Sweep

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