• Skip to main content
  • Skip to search
  • Skip to footer
Cadence Home
  • This search text may be transcribed, used, stored, or accessed by our third-party service providers per our Cookie Policy and Privacy Policy.

  1. Community Forums
  2. Custom IC Design
  3. Constant Power Source

Stats

  • Locked Locked
  • Replies 2
  • Subscribers 125
  • Views 16337
  • Members are here 0
This discussion has been locked.
You can no longer post new replies to this discussion. If you have a question you can start a new discussion

Constant Power Source

ssarkar
ssarkar over 10 years ago

Hi,

I have searched extensively for this problem but was unable to get a related post.

Suppose I have a diode-connected mosfet connected on the drain side by a constant power source (port cell in analogLib of Virtuoso); and on the source side to a capacitor and a current-source load, like a half-wave rectifier. The constant power source ("port" cell in analogLib) takes the value of source resistance, source type (sine), source frequency, and power amplitude of the source (in dBm).

I wanted to model the whole system, which is not achievable until I know how the voltage and current behaves at the output of the constant power source. The presence of a MOSFET makes it more non-linear. It would be of great help if someone can help me with this.

Thanks in advance,

Regards,

Soumik

  • Cancel
  • Andrew Beckett
    Andrew Beckett over 10 years ago

    Soumik,

    If you're asking how the port component works, then that's quite simple. It isn't a constant power source - if you look at the documentation for port (see "spectre -h port" in a terminal window, or look in cdnshelp) you'll see that it is implemented as an ideal voltage source in series with a resistor (the value of the resistor is the source resistance), and if the amplitude is specified in volts (using the ampl parameter), the internal source is double the value you specify - so that if the load matches the source resistance, it will be the specified value at the output (because you'll have a potential divider dividing the voltage by 2).

    If the source amplitude is specified in dBm, then the documentation says:

    The DC value given for the port voltage specifies the DC voltage across the port when it is terminated in its reference resistance (in other words, the DC voltage of the internal voltage source is double the user specified DC value, `dc'). The same is true for the values for the transient, and PAC signals of the port. However, the amplitude of the sine wave in the transient and PAC analyses can alternatively be specified as the power in dBm delivered by the port when terminated with the reference resistance.

    So, this means that if the load is matched, you'd get that power in the load. You can work out the voltage on the internal voltage source fairly simply then the source amplitude=

    Vsource=2*sqrt(10dbm/10*1e-3*Rsource*2)

    The 10^(dbm/10) inside the sqrt converts the dbm into the multiple of 1mW, and the 1e-3 converts it into Watts. All we're doing here is computing V=sqrt(power*R) - the factor of 2 inside the parentheses is to give the amplitude of the source rather than RMS value of the source, and the 2 outside the sqrt is to deal with the fact that the internal source is doubled so that if matched, you'll get sqrt(10dbm/10*1e-3*Rsource*2) at the output of the port.

    Does that help?

    Regards,

    Andrew.

    • Cancel
    • Vote Up 0 Vote Down
    • Cancel
  • ssarkar
    ssarkar over 10 years ago

    Thanks for the reply. This is really helpful, especially the "help" command. The misconceptions that got cleared:

    1) I was trying to think of the port as an antenna, and hence assumed that I am receiving the specified power at the input. According to your reply, the power that we specify in the "port" parameters only determines the voltage at the source depending on the source resistor.

    2) Initially, I thought the amplitude that I specify inside the "port" parameters is that of the internal source. According to your reply, this is the amplitude at the output during perfect matching. So I think I should calculate the source amplitude using the formula, and then consider it as fixed. Now the current should in turn depend on the load. (Earlier I was forcing the condition in my model that (voltage*current) should be equal to the input power)

    I haven't read the documentation yet, but this is what I understood from your reply.

    Thanks again,

    Regards,

    Soumik.

    • Cancel
    • Vote Up 0 Vote Down
    • Cancel

Community Guidelines

The Cadence Design Communities support Cadence users and technologists interacting to exchange ideas, news, technical information, and best practices to solve problems and get the most from Cadence technology. The community is open to everyone, and to provide the most value, we require participants to follow our Community Guidelines that facilitate a quality exchange of ideas and information. By accessing, contributing, using or downloading any materials from the site, you agree to be bound by the full Community Guidelines.

© 2025 Cadence Design Systems, Inc. All Rights Reserved.

  • Terms of Use
  • Privacy
  • Cookie Policy
  • US Trademarks
  • Do Not Sell or Share My Personal Information