VerilogA for Varactor

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desperado0729

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varactor veriloga

Hi,

I tried to build a veriloga model for varactor and the code is listed below.
I used a math equation to fit the CV curve from the foundry (from -1.8V to 1.8V).
But when I did the ac simulation to double check the value in the equation, I got two different CV curve.

Could someone help me?

Thanks!

=====================
module CVAR_SMIC( S, G, T);
inout S,G,T;
electrical S,G,T;

real c, v, q;

parameter real A1 = 259.33f;
parameter real A2 = 730.28f;
parameter real x0 = -0.09478;
parameter real dx = 0.19275;

analog begin

v = V(G,S);
c = A2 + (A1-A2)/(1 + exp((v-x0)/dx));
q = c * v;
I(G,S) <+ ddt(q);
V(T) <+ c;

end

endmodule
=====================

V
 

u cann't use Q=CV as this is for linear capacitor
so Q=int(CdV) and u cann't get C out of integration if it is function of V
try calculating the Q from the equation then use the result
 
I think there is one statement that specifies the equations for ac analysis and Transient analysis separately. Specifying that you use the same equation for both analysis may help.
 

To safwatonline,

I just did what you said using Mathematica to integrate C(V) and then these 2 CV curve are a good match!

Thanks again.

V
 

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