[SI-LIST] Re: Question about hspice .probe command

  • From: "lgreen" <lgreen22@xxxxxxxxxxxxxx>
  • To: <jihong@xxxxxxxxx>
  • Date: Sun, 17 Apr 2005 19:39:43 -0700

Jihong,

In HSPICE, you can look at the current through any pin, such as a voltage
source or a resistor.  Some versions of SPICE are limited to looking at
currents through voltage sources.

Here are some things you could check:
* What other components are connected to that node?  In your equation, you
only have two voltage sources, which indicates there is at least one more
component connected to that node.
* What are the tolerance settings in .OPTIONS?  How large are your currents
relative to the allowed tolerance?
* If all else fails, double-check sign conventions for currents for each
component.

Best regards,
Lynne

-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of Craig Clewell
Sent: Sunday, April 17, 2005 6:19 PM
To: jihong@xxxxxxxxx
Cc: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: Question about hspice .probe command


Jihong, 

I don't know what node naming convention you are using, but your email seems
to show that you are trying to find the current in a voltage source.
Normally, you would look for the current through a resistance.  Try probing
the source resistance at V1 with I(Rx) [where Rx is the name of the source
resistance] instead of if I(V1).

Regards, 

Craig Clewell


Hi, I am trying to confirm that two currents in my circuit sum up to 0. I
tried the following two approaches. 1. ".probe tran Itotal = par('I(vin1) 
+ I(vin2)') " to view the total current from the two voltage sources. 2. 
use the expression tool from awaves to compute the total current. I expect
they should give similar result and the first one should give more accurate
result since the second approach reads data from .tr0 file which only
contains data points specified by the time step. However, these two gave me
completely different results, several orders of magnitude difference!  Did I
do something wrong? Thanks!

Jihong


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