Difficult Diode Exam Problem

hitmen

Joined Sep 21, 2008
161
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 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.

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studiot

Joined Nov 9, 2007
4,998
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.

hitmen

Joined Sep 21, 2008
161
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??

studiot

Joined Nov 9, 2007
4,998
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

hitmen

Joined Sep 21, 2008
161
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

studiot

Joined Nov 9, 2007
4,998
Idiode = (1/8)/3 = 5/12A
KVL: -Vi +0.3 +(100)(5/12) = 0
Vi = -41.36V (does not make sense)

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.

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