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NanoMilton New Member Offline Posts: 3	Bsim VerilogA model and AC analysis Jan 24^th, 2007, 2:23pm Hi, Is there a Verilog A Version of BSIM model? If so, where can I find it? Also, it is not possible to run analysis with model implemented in Verilog A,isn't it? Thanks a lot.
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Geoffrey_Coram

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Re: Bsim VerilogA model and AC analysis
Reply #1 - Jan 25^th, 2007, 4:48am

Silvaco (now simucad) has "not for commercial use" bsim3 and bsim4 verilog-a models, available (last I checked) after registration from their web site. (I think there's a link if you click Verilog-AMS from the nav bar of this site.) Others have done the translation, also, but I don't think they've put them out for the public.

If your simulator supports Verilog-A, it should support Verilog-A for any analysis type, so I'm not sure I understand your second question.

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NanoMilton

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Re: Bsim VerilogA model and AC analysis
Reply #2 - Jan 26^th, 2007, 11:48am

Thanks a lot. I got the BSIM models from the web.

With regard to my second question, I remember that there is a post somewhere in this forum saying that Verilog A doesn't support complex number calculation. Actually, what I want is to run a frequency domain analysis on my model so as to characterize it's ac behavior. Is it possible to do so?

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Geoffrey_Coram Senior Fellow Offline Posts: 1999 Massachusetts, USA	Re: Bsim VerilogA model and AC analysis Reply #3 - Jan 26^th, 2007, 12:39pm You can run an ac analysis with a Verilog-A model. The complex numbers are handled by the simulator -- it linearizes the equations in the Verilog-A model, and capacitive terms get multiplied by j ω. The Verilog-A model itself does not have any complex numbers in its equations.
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NanoMilton New Member Offline Posts: 3	Re: Bsim VerilogA model and AC analysis Reply #4 - Jan 26^th, 2007, 12:53pm So, to describe a capacitor, I use the statement: I(Drain,Source) <+ ddt(Cds*V(Drain,Source)); Does the simulator automatically know that it's a capacitor?
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Geoffrey_Coram

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Re: Bsim VerilogA model and AC analysis
Reply #5 - Jan 29^th, 2007, 4:32am

NanoMilton wrote on Jan 26^th, 2007, 12:53pm:

Does the simulator automatically know that it's a capacitor?

Yes.

Actually, probably the better way to describe what happens is: the simulator sees the "ddt" operator, which has a Fourier transform involving multiplication by j ω, which is used to compute the frequency response of the linearized system.

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