Geoffrey_Coram
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Francesco - I wasn't intending for you to use ddt and integral; those two are with respect to time, but we're interested in the derivative and integral with respect to voltage.
I'm a little puzzled by two things in your model: 1) you have R(Vd) where Vd is the voltage across the whole device, not just across the resistive part 2) you have 1.01 * Vd, rather than just re-scaling vr0 and vr1 (and R1) such that R =R0 + R1*tanh((Vd-vr0)/vr1))/Nf;
If Vd were the voltage across the resistor, then when you wrote: Rfit=(R0+R1*tanh((1.01*Vd-vr0)/vr1))/Nf; V(res) <+ Rfit*I(res); you'd have the equation V = (R0 + R1*tanh((V-vr0)/vr1)) * I (let's remove the 1.01 and Nf), or equivalently, I = V/(R0 + R1*tanh((V-vr0)/vr1)))
Now, the small-signal conductance would be the partial derivative of I with respect to V, G = 1/(R0 + R1*tanh((V-vr0)/vr1))) + V/(R0 + R1*tanh((V-vr0)/vr1)))^2 * R1*sech^2((V-vr0)/vr1) / vr1 where the first term is what you wanted.
I suggested that, if the nonlinearity is small -- that is, R1/R0 is small -- then this second term might be small enough that you don't notice it.
I don't have time right now to think about it, but the fact that this is a varactor, and in fact, there is no large-signal current, might mean that your solution is not as bad an approximation as I thought. The resistance/conductance you measure isn't the derivative of the large-signal current, since (except for leakage), there is no large-signal current.
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