Connect with us

Calculating Dynamic Power Dissipation- VLSI Design

Discussion in 'Electronics Homework Help' started by aditya639, Jun 7, 2021.

Scroll to continue with content
  1. aditya639

    aditya639

    4
    0
    Jun 7, 2021
    Hi!

    I am trying to compute the dynamic (switching) power dissipation of a few digital sytems including full adders and multipliers.
    Is the following expression correct to calculate the average power dissipation?:

    P.D.(average)= 1/T*integral{(input supply voltage)*(output current)}

    Can someone please verify if the expression is correct.
    Thanks and regards
    Aditya
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    11,417
    2,619
    Nov 17, 2011
    1. Why do you use input voltage but output current? What about power dissipated by current from the input power supply and returned to the power supply without being an output current?
    2. Average power can be computed by different means:
      - mean power - this is what your equation does
      - RMS power - is this possibly what you want instead?
     
  3. aditya639

    aditya639

    4
    0
    Jun 7, 2021
    Hi Harald,
    I need to calculate the average power dissipated due to switching (when the output switches from high to low or vice-versa).
    I am using output current because I need to calculate the power which is dissipated when the output switches states. But I am using input supply voltage as it is this voltage which is responsible for the output states.
    Do you think this method is correct?
    Thanks
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    11,417
    2,619
    Nov 17, 2011
    No.
    The output current is only partly related to the switching losses of the output river MOSFETs. Switching losses in the driver MOSFETs occur mainly due to two effects:
    1. Gate current to charge and discharge the Gate-Source and Gate-Drain capacitances when turning the MOSFET on and off.
    2. Power dissipated during the transition of the MOSFET from on to off and vice versa. During the transition the Drain-Source voltage rises (turn-off) of falls (turn-on) while current falls or rises. This gives rise to increased power dissipation compared to the on-state or off-state.
      The current during switching can have two main components (depending on how the timing of the MOSFETs drive signals (Gate-Source voltages):
      - output current due to output voltage across load resistance
      - transverse current between the MOSFETs of the output stage if both MOSFETs are at the same time (gate drive signal overlap
    More info e.g. here.
    So when measuring the power dissipation use the current drawn from the power supply times the power supply voltage and average that value to get the total power consumption.
     
    aditya639 likes this.
  5. aditya639

    aditya639

    4
    0
    Jun 7, 2021
    Thank you for the reply. This made things somewhat less murky.

    But wouldn't this give us the power which is supplied to the circuit, and not the power which is actually dissipated by the circuit, a simple CMOS inverter for example.
     
  6. Harald Kapp

    Harald Kapp Moderator Moderator

    11,417
    2,619
    Nov 17, 2011
    What would be the difference? If the power supplied to the circuit is not dissipated by the circuit, where would it go?
     
  7. aditya639

    aditya639

    4
    0
    Jun 7, 2021
    Only a fracction of the power supplied to the circuit from the source is dissipated in the circuit, the rest is used to drive the further stages of the circuit.
    Like in CMOS inverter, C(load)*V(dd)^2 is taken from the source, half of it is dissipated in the circuit during switching, while the other half is available for the next stage i.e. the load capacitance.
    Here's a link for inverter power dissipation: http://people.ece.umn.edu/~kia/Courses/EE5323/Slides/Lect_04_Inverter2.pdf
    Please take a look at the first 2 slides.
    I am unable to determine this dissipated power for more complex systems like adders.
    Thanks.
     
  8. Harald Kapp

    Harald Kapp Moderator Moderator

    11,417
    2,619
    Nov 17, 2011
    Then subtract the power that goes into other devices from the power supplied to the device you're analyzing.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-