exponential delay on/off transition?

[NiC]

Hi everyone,
I would like to apply an exponential time delay to an current source that i model using verilog-a using Cadence.
I would like something like the transition operator but with emponential delay not a linear.
One more thing is that I want the on time delay to be different than the off time delay.

I have tried to implement this using RC network, but the network messes up the function of the circuit in time slots it supposed not to.

Any other idea?

 

[BradtheRad]

There are op amp circuits that achieve logarithmic response. Never constructed one.

One of those might produce output to fit your exponential curve.

Or else it might charge a capacitor to a level based on your exponential curve.

 

[JoannesPaulus]

In verilog-a you can use your own smoothing function that creates the behavior you need. For example:

analog function real my_fun;
input x; real x;
my_fun=(x<0)?0:(x>=1)?1:(1-exp(x/T);
endfunction

and you can use it in the analog block as:

output=my_fun(input);

 

[NiC]

In verilog-a you can use your own smoothing function that creates the behavior you need. For example:

analog function real my_fun;
input x; real x;
my_fun=(x<0)?0:(x>=1)?1:(1-exp(x/T);
endfunction

and you can use it in the analog block as:

output=my_fun(input);

where x is what? the simulation time?

---------- Post added at 19:54 ---------- Previous post was at 19:51 ----------

There are op amp circuits that achieve logarithmic response. Never constructed one.

One of those might produce output to fit your exponential curve.

Or else it might charge a capacitor to a level based on your exponential curve.

my exponential curve is just the response curve to a final value which is depended on specific voltages which are given from a formula..

---------- Post added at 20:58 ---------- Previous post was at 19:54 ----------

where x is what? the simulation time?

---------- Post added at 19:54 ---------- Previous post was at 19:51 ----------

ok I think I got it. x is the input formula. I just have to use the correct smothing function and apply the correct parameters.

 

[JoannesPaulus]

ok I think I got it. x is the input formula. I just have to use the correct smothing function and apply the correct parameters.

Not exactly: x is the variable you want to shape (voltage in your case).

 

[NiC]

actually i want to shape the output current. so I do this:

I(V) <+ my_sine(transition(Id,0,2e-3,1e-3));

where Id is a pulse in time and my_sine:

analog function real sinfn;
input x; real x;
sinefn = (x<=0) ? 0 : (x>=1) ? 1 : x-sin(`twopi*x)/`twopi;
endfunction

---------- Post added 10-11-11 at 00:04 ---------- Previous post was 09-11-11 at 23:41 ----------

I now notice that the problem with this kind of implementation is that that my upper limit is not constant..

 

[JoannesPaulus]

I do not understand what you mean: only when the signal is between 0 and 1 the transition will be affected.
I believe your current Id takes 2 discrete values and you might need to modify the function to take care of that appropriately scaling x in your formula.

 

[NiC]

I decoupled from the load the cap and it is working perfect. Thank you all