pss

Discussion in 'Cadence' started by fredo, Mar 22, 2007.

  1. fredo

    fredo Guest

    Hi,
    How is power defined in pss analog simulation?
    Thanks all.
    Fredo
     
    fredo, Mar 22, 2007
    #1
  2. Your question is a little non-specific! If you're asking about how to define
    input power, then this is not really anything to do with pss, but to do with how
    you set up your input sources. The "port" component (in analogLib, corresponding
    to the "port" component in spectre) can specify the input signal in dBm since it
    has a built in resistance.

    If you're talking about measuring output power, then ADE gives you a number of
    options - if you look on the direct plot form (Results->Direct Plot->Main Form),
    then there are several choices whenever you try to plot anything power related.
    For example, you can select a port - and it knows what the impedance of the port
    was, so can convert the voltage to power. Or you can specify the terminal of a
    device - in which case it will use the current and voltage to compute power (you
    need to save the current if you want to do that). Or there are several other
    options.

    Without knowing what you're actually asking, it's hard to answer your question
    more precisely!

    Regards,

    Andrew.
     
    Andrew Beckett, Apr 11, 2007
    #2
  3. fredo

    fredo Guest


    I found that the power obtained by integrating the product of voltage and
    current in one cycle in a transient simulation has large deviation to the
    power obtained from the pss plot option. So I wonder if it is a mistake in
    my calculation, a wrong assumption or that pss has a special definition to
    power.

    Sorry that I didn't make it more specific. I almost thought that it might be
    a naive question to post in here.

    fredo
     
    fredo, Apr 13, 2007
    #3
  4. Without knowing how you're measuring the power in both situations, it's hard to
    know what could be wrong. Are you summing up the power in all the harmonics? Are
    you measuring the power the same way?

    A bit more information might make this possible to answer... there's no
    fundamental reason why it should be wrong.

    Regards,

    Andrew.
     
    Andrew Beckett, Apr 19, 2007
    #4
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