Difficult Diode Exam Problem

Discussion in 'Homework Help' started by hitmen, Dec 5, 2008.

  1. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    This is my next exam paper which has left my cohort stumped.

    a) Initially, the diode D3 is not connected (replaced by open circuit). What is the max value of voltage Vo(t)? Is the diode D7 forward or reverse biased at this point? Explain whether this corresponds to the maximum power dissipated at the diode. Determine the max power dissipated at the diode.

    b) Both diodes are now connected as shown in Fig Q3. Each diode has a max power dissipation of 1/8 watt. What is the max allowable amplitude Y of Vi(t)?

    I dont understand the concept in (a). When D7 is reverse-biased, how can there be current conducted:confused: In that case, how can there be power? In that case, how can there be a max power???

    When both diodes are connected, how does the current flow? Does it go through one and then the other? Do give me a hint. Thanks

    Note : the diagram states that Vi(t) = 5 sin (2∏t)V, R2 = 100Ω, D3 forward voltage = 0.3 V and that D7 forward voltage = 0.7V. Vo(t) is the voltage across D7.
     
    Last edited: Dec 5, 2008
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    A)
    D7 will conduct in one direction only, every half cycle. The max voltage will be the peak voltage of the sinewave.
    Since you only have half a wave, you only have half the power times the normal calculation for a sine wave.

    B) Power will be calculated for D3 in the same manner when connected, but using the other half cycle.
    Compare the max power for each diode to obtain the max value allowable.
     
  3. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    ok. For part a, I take the negative half cycle, Imax = 4.8/100 = 48mA. I use P=IV to get 0.2304W. I dont understand how to "explain whether this corresponds to the maximum power dissipated at the diode".

    For part b, why are we only using 1 half cycle? I thought we can be taking both since the current can now flow both ways??
     
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Both diodes have the same wattage rating.
    power = volts x amps
    each diode has a different voltage drop
    therefore different power

    I did say compare these and hoped the outcome would then be obvious
     
  5. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    For +ve half cycle: D3 is on.

    Idiode = (1/8)/3 = 5/12A
    KVL: -Vi +0.3 +(100)(5/12) = 0
    Vi = -41.36V (does not make sense)

    For -ve half cycle D7 is on.
    Idiode = 1/8 /0.7 = 5/28A
    0.7 - Vi +(100)(5/28) = 0
    Vi = 18.577V in the opposite direction
    Doesnt make sense also since Vi is greater than 5.

    Thanks
     
  6. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Check your arithmetic

    Vi = +0.3 +(100)(5/12), which is positive.

    In any case what you have shown is that both diodes will not operate at maximum power, simultaneously. As we increase the supply voltage, one diode will reach max power first and this limits the max voltage to that diode. So you must work out separately for each half cycle the limiting voltage and choose the lowest.
     
    Last edited: Dec 6, 2008
  7. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    I see. Got it thanks.
     
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