The Designer's Guide Community Forum https://designers-guide.org/forum/YaBB.pl Modeling >> Behavioral Models >> veriloga driver model https://designers-guide.org/forum/YaBB.pl?num=1561198183 Message started by michdundee78 on Jun 22nd, 2019, 3:09am

 Title: veriloga driver model Post by michdundee78 on Jun 22nd, 2019, 3:09am Hi all,can someone help me on writing a veriloga driver model (ron, roff, delay) please?thx in advanceBR,mich

 Title: Re: veriloga driver model Post by Ken Kundert on Jun 22nd, 2019, 4:36pm Doesn't really work like that here. You generally have to make an effort first by proposing something, and then we can help you by answering your questions.-Ken

 Title: Re: veriloga driver model Post by hafiz2431 on Jul 25th, 2019, 3:57am Hi,You can try the following code:`include "constants.vams"`include "disciplines.vams"module drv_ideal(A, Y, vcc);output Y;electrical Y;input A,vcc;electrical A,vcc;// INSTANCE PARAMETERS:parameter real vhi = 0.8;              parameter real vlo = 0  ; parameter real vth = 0.5*(vhi+vlo) ; parameter real tr = 1p; parameter real tf = 1p; parameter real tdel =1p; parameter real rpuon =35;parameter real rpdon =35;parameter real roff =10M;// LOCAL VARIABLES:real vy,rpu,rpd;       integer y;     analog begin           y =  (V(A) > vth);         vy = y ? vhi : vlo;            V(Y) <+ transition( vy, tdel, tr, tf);                rpu = transition(V(A)>vth ? rpuon : roff ,tdel, tr,tf);        I(vcc,Y) <+ V(vcc,Y)/rpu;        rpd = transition(V(A)

 Title: Re: veriloga driver model Post by Geoffrey_Coram on Jul 25th, 2019, 5:08am Hi, hafiz2431 -Your code probably not doing what you intended. You need to look at the rules of value retention in Verilog-A, because you are contributing to both the potential:    V(Y) <+ transition( vy, tdel, tr, tf);and flow:       I(Y) <+ V(Y)/rpd;of the same branch between Y and ground. The rules say that the V contribution will be discarded because of the subsequent I contribution.

 Title: Re: veriloga driver model Post by Ken Kundert on Jul 25th, 2019, 4:01pm Also you need a cross function to resolve the threshold crossings in the input voltage.

 Title: Re: veriloga driver model Post by hafiz2431 on Jul 25th, 2019, 8:09pm Hi Ken and Geoffrey,Thanks for your response. I'm not very familiar with Verilog-A. I can just write some simple codes. Actually, I tried to address the issue raised by michdundee78. As for the code, I could see the response at Y after a 'tdel' amount of delay from A. Moreover, I found pull-up and pull-down current at node Y equal to [V(vcc)-V(Y)]/rpuon and V(Y)/rpdon, respectively.  This served my purpose apparently but of course, I'd appreciate if you suggest some modifications.

 Title: Re: veriloga driver model Post by Ken Kundert on Jul 25th, 2019, 9:36pm You need a cross function to resolve the threshold crossings in the input voltage.

 Title: Re: veriloga driver model Post by hafiz2431 on Jul 29th, 2019, 3:23am Hi Ken,I added a cross function and it now looks like:`include "constants.vams"`include "disciplines.vams"module drv_ideal(A, Y, vcc);output Y;electrical Y;input A,vcc;electrical A,vcc;// INSTANCE PARAMETERS:parameter real vhi = 0.8;              parameter real vlo = 0  ; parameter real vth = 0.5*(vhi+vlo) ; parameter real tr = 1p; parameter real tf = 1p; parameter real tdel =30p; parameter real rpuon =35;parameter real rpdon =35;parameter real roff =10M;// LOCAL VARIABLES:real vy,rpu,rpd;       integer y;     analog begin    @ (cross(V(A) - vth, 0) )         y =  (V(A) > vth);         vy = y ? vhi : vlo;            V(Y) <+ transition( vy, tdel, tr, tf);                rpu = transition(V(A)>vth ? rpuon : roff ,tdel, tr,tf);        I(vcc,Y) <+ V(vcc,Y)/rpu;        rpd = transition(V(A) high, a current =(0.8-0.4)/35 =11.4 mA will flow at a delay of 30 ps from Vs1 to Vs2. When in=> low, Vs2 will source 11.4 mA to ground.

 Title: Re: veriloga driver model Post by Geoffrey_Coram on Jul 29th, 2019, 6:05am In small/short simulations, you may not notice a difference. If you ask the simulator to run a very long simulation, it may decide to take much larger timesteps, and it could step over a whole cycle and not notice that it was skipped.

 Title: Re: veriloga driver model Post by Geoffrey_Coram on Jul 29th, 2019, 6:06am If you comment out this line:  V(Y) <+ transition( vy, tdel, tr, tf);does the result change at all?

Title: Re: veriloga driver model
Post by Ken Kundert on Jul 29th, 2019, 10:45am

As Geoffrey says, the cross function controls the time point selection, so to see the difference you would need to look at the time point placement. Geoffrey mentioned that in long simulations you can miss entire output pulses, but if you control your time steps that is unlikely to happen, and if it does a cross function won't help. Instead the cross function helps by accurately resolving the time of the crossing. Without it the time of the output transition can be off by a large amount. An artificial jitter of 10-20% of the pulse width is not uncommon.

Your use of cross has the 'start up' problem. The start-up problem results in the output being wrong at the beginning of the simulation. It only corrects itself after the first input transition. Instead, I recommend that you put the if statement outside the cross function ...

Code:
 @(cross(V(A) - vth, 0))     ;if (V(A) > vth)     vy = vhi;else     vy = vlo;

Geoffrey is also trying to call attention to another very odd aspect of this model.  You are driving the output with a voltage source and also driving it with pull up and pull down resistors. The resistors would have no effect on the output voltage, instead they only affect the current in the supply and ground.

-Ken

 Title: Re: veriloga driver model Post by Ken Kundert on Jul 29th, 2019, 11:06am For simulation efficiency, it is better to set tdel to 0 and make tr and tf as large as practical. So unless there is some reason to use a nonzero tdel or small tr and tf, you should not.I mention this because people often try to make their models needlessly more accurate without realizing that they are paying a price in simulation speed.-Ken

 Title: Re: veriloga driver model Post by hafiz2431 on Jul 30th, 2019, 12:51am Thanks, Ken and Geoffrey for your valuable inputs. I appreciate that.How do you think about the code now:`include "constants.vams"`include "disciplines.vams"module drv_ideal(A, Y, vcc);output Y;electrical Y;input A,vcc;electrical A,vcc;// INSTANCE PARAMETERS:parameter real vhi = 0.8;              parameter real vlo = 0  ; parameter real vth = 0.5*(vhi+vlo) ; parameter real tr = 15p; parameter real tf = 15p; parameter real tdel =0; parameter real rpuon =35;parameter real rpdon =35;parameter real roff =10M;// LOCAL VARIABLES:real vy,rpu,rpd;       //integer y;     analog begin    @ (cross(V(A) - vth, 0) )         vy =  (V(A) > vth);        // vy = y ? vhi : vlo;             if (V(A) > vth)            vy = vhi;      else            vy = vlo;      // V(Y) <+ transition( vy, tdel, tr, tf);                //rpu = transition(V(A)>vth ? rpuon : roff ,tdel, tr,tf);        rpu = transition(vy ? rpuon : roff ,tdel, tr,tf);        I(vcc,Y) <+ V(vcc,Y)/rpu;        // rpd = transition(V(A)

 Title: Re: veriloga driver model Post by Ken Kundert on Jul 30th, 2019, 6:52am Have you tried it? How does it work?-Ken

 Title: Re: veriloga driver model Post by hafiz2431 on Jul 30th, 2019, 9:12am Yes, I tried to drive an extracted model of a transmission line. In=> input to the buffer, tx_pad<0> => output of the buffer driving the transmission line. The low level of the first 3 pulses are not zero but some other values. It's also due to initialization problem or else?

 Title: Re: veriloga driver model Post by Ken Kundert on Jul 30th, 2019, 3:48pm Yeah, your model has some relatively obvious mistakes. Perhaps they would be easier to see if you removed the dead code.  Also, you did not follow instructions very well.-Ken

 Title: Re: veriloga driver model Post by hafiz2431 on Jul 30th, 2019, 7:04pm OK, here it is:`include "constants.vams"`include "disciplines.vams"module drv_ideal(A, Y, vcc);output Y;electrical Y;input A,vcc;electrical A,vcc;// INSTANCE PARAMETERS:parameter real vhi = 0.8;              parameter real vlo = 0  ; parameter real vth = 0.5*(vhi+vlo) ; parameter real tr = 15p; parameter real tf = 15p; parameter real tdel =0; parameter real rpuon =35;parameter real rpdon =35;parameter real roff =10M;// LOCAL VARIABLES:real vy,rpu,rpd;       analog begin    @ (cross(V(A) - vth, 0) )         vy =  (V(A) > vth);                     if (V(A) > vth)            vy = vhi;      else            vy = vlo;                      rpu = transition(vy ? rpuon : roff ,tdel, tr,tf);        I(vcc,Y) <+ V(vcc,Y)/rpu;        rpd = transition(!vy ? rpdon : roff ,tdel, tr,tf);        I(Y) <+ V(Y)/rpd;      endendmodule

 Title: Re: veriloga driver model Post by Geoffrey_Coram on Jul 31st, 2019, 6:13am You allow vlo to take any value:parameter real vlo = 0  ; but your transition calls assume that vy == 0 means the driver is off:       rpu = transition(vy ? rpuon : roff ,tdel, tr,tf);What would happen if vlo = 1 (and vhi = 2)? Or if vlo = -1 and vhi = +1?

 Title: Re: veriloga driver model Post by hafiz2431 on Jul 31st, 2019, 8:57am Yes, this issue needs to be resolved as well, but before that, would you please assist to figure out why the first 3 pulses at drv-resp-2.jpg are not going to 0?

Title: Re: veriloga driver model
Post by Ken Kundert on Jul 31st, 2019, 11:51am

We cannot really tell you why your simulations are failing without having all the information. However, rather than providing us with the circuit and all the waveforms it would be better if you debugged the situation yourself. I recommend you thinking about the value of vy ? rpuon and !vy. Here you are treating vy, a real number, as a boolean. It is logically false if its value is 0.000000000000000000, and true otherwise. Is that really the way you want to write this model?

Given that you are clearly struggling, here are a few issues with your model:

Code:
 @(cross(V(A) - vth, 0) )    vy =  (V(A) > vth);

Why are you computing vy here. You immediately overwrite with the next statement.

The next problem is here:

Code:
 if (V(A) > vth)    vy = vhi;else    vy = vlo;

Why are you setting vy to vhi and vlo. vhi and vlo no longer have any meaning in this model, and the fact that you are using them leads to the problem Geoffrey is pointing out.

Perhaps you should replace the above code with the following:

Code:
 @(cross(V(A) - vth, 0) )    ;vy =  (V(A) > vth);

And while I am at it, why do you use vy as the name of your variable. The value is not a voltage, and y is meaningless. This poor choice of names is contributing to your problems. Perhaps you should use a name like on. It is very descriptive of both the meaning and the type of the value.

-Ken

 Title: Re: veriloga driver model Post by hafiz2431 on Jul 31st, 2019, 7:17pm Thanks for the hints. I'll try to debug the issues.