duty cycle watts

Discussion in 'Homework Help' started by richbrune, Apr 29, 2008.

  1. richbrune

    Thread Starter Senior Member

    Oct 28, 2005
    106
    0
    If you have a load that is "on" for 20 uS, off for 3.5 mS, which is 2.3 volts when on, zero volts off, 10mA on and zero volts off, and the current and voltage are in phase, how many watts is the load?
    Thanks again, Rich
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    I guess 2.3 is a dc voltage, am i correct?

    what is your load?
     
  3. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Calculate for DC conditions and divide by the duty cycle.
     
  4. richbrune

    Thread Starter Senior Member

    Oct 28, 2005
    106
    0
    So I don't need to use RMS?
     
  5. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Only if the applied voltage is AC. It's somewhat unusual to specify a duty cycle for AC.
     
  6. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
    5,699
    907
    It's a square wave, so: [(amplitude)^2]^1/2 = amplitude. That is, the RMS of a square wave is the amplitude. John
     
  7. Caveman

    Active Member

    Apr 15, 2008
    471
    0
    The power, which is Joules/sec when on is 2.3*10mA = 23mW.
    So the energy when on is 23mW*20us = 460nJ
    The energy when off is 0nJ.

    The average power for the load is the total Joules divided by the total time.
    460nJ / 10020us = 45.9e-6W or about 46uW.
     
  8. richbrune

    Thread Starter Senior Member

    Oct 28, 2005
    106
    0
    How do I arrive at "10020us"?
     
  9. richbrune

    Thread Starter Senior Member

    Oct 28, 2005
    106
    0
    I thought the RMS voltage of a rectangular wave was:
    SQRT(duty cycle) * Peak voltage
    As in:
    SQRT (.005714) * 2.3 = .0756 * 2.3 = .17388 volts
    and the RMS current of the .57% duty cycle was:
    SQRT (.005714) * .01 = .000756
    so the power would be .000756*.17388 = .0001314 watts or 131uW
     
  10. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
    5,699
    907
    Where did your formula come from, particularly the 0.005714 component?

    I refer you to Horowitz and Hill, page 18, Figure 1.23. Clearly duty cycle enters into the calculation of the power, but the mean amplitude of a square wave pulse is its amplitude and the square root of the amplitude squared is also just the amplitude. John
     
  11. Caveman

    Active Member

    Apr 15, 2008
    471
    0
    Sorry, I screwed that up.
    On power is 2.3V*10mA = 23mW. Off power is 0mW.
    For one cycle, the total energy consumed is 23mW*20us = 460nJ
    This is averaged over one cycle, which is 20us + 3.5ms = 3520us.
    So average power is 460nJ/3520us = 0.13mW

    Sorry about that.
     
  12. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,164
    You can use I * E * DutyCycle.

    Yes, duty cycle is part of the solution.

    20 uS on and 3.5 mS off is 20 uS/3.52 mS or 0.005681818

    2.3 V * 10 mA is 0.023W

    W * Duty Cycle is 130.681 uW
     
Loading...