Question on Hspice: How to write the Source file?

[shartz]

I am a starter on HSPICE, and need to simulate a simple circuit by force the input, say V(INPUT), as a function of time:

V(input)=1*exp(-5)+2*exp(-6),

How can I achieve it? By defining
.PARAM V(INPUT)='1*exp(-5)+2*exp(-6)' ?

But, the error is: no DC path from node INPUT...

Is there anybody can help me out?

 

[Fahmy]

try to insert any dummy load on the node INPUT (for example a resistance in the order of 1meg ohm)

 

[shartz]

Dummy load?
I need a force, with function of time "t". Is dummy load work?

 

[Fahmy]

The duumy loads are just to fix the error of "no DC path to ground"

 

[hr_rezaee]

Hi
1*exp(-5)+2*exp(-6) is a constant value.
so you can calculate it and use a DC voltage source with this value.
regards

 

[shartz]

Sorry, I did not illustrate it clearly.
I want the


V(input)=1*exp(-t)+2*exp(-t), in which "t" is time, and I don't want to do it by PWL.
Any other suggestions?

And I don't know how the "floating" node works, can you explain it more clearly?

Best

 

[hr_rezaee]

Hi
I think hspice manual can help you.
I think you can make exponential source in it.
regards

 

[Fahmy]

Well I may have a solution but is not straight froward and it uses PWL source .
you can generate a ramp voltage that is proportional to time using a PWL source (0 0 Tend Vmax).
Then you can search your manual for a macromodel that generates the exponential of the input signal (the ramp signal) . Using a combination of the these blocks with some fixed gain blocks and a summer you can get the equation you want.

Added after 6 minutes:

And I don't know how the "floating" node works, can you explain it more clearly?

some times the tool refuses to have a node that is connected to one element and it reports that there is no DC path to ground. To force the tool to accept this circuit is to add a resistor with a very very huge value between this node and ground, This gives the tool a path to ground and will not impact the circuit behaviour considerably (provided that the value of the resistor is large enough)