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Why is D1 considered cutoff and D2 considered conducting ? Note : I am working with constant voltage model where the diode voltage is 0.7v and I need to figure out I and V. Please explain why D1 is considered cut off and D2 conducting, this is what chegg solutions say. -Thanks for your time.


1 Answer 1


The voltage across the diode is only 0.7V (in this model) if the diode is conducting. You can (mentally) remove the diode and see if the voltage across where it was would be more or less than 0.7V. If it would be less than 0.7V, then the diode is not conducting (cut off).

In your example:-

Remove both diodes and the voltage goes to -3V (so voltage across where D1 would be is +4V and where D2 would be is +6.0). So we know that either diode by itself will be conducting.

Replace D2, and remove D1. Since we know D2 will be conducting, the voltage goes to +2.3V (3V - 0.7V) (so voltage across where the diode D1 would be is -1.3V).

Replace D1, and remove D2. Since we know D1 will be conducting, the voltage goes to +0.3V (1V - 0.7V) (so voltage across where the diode D2 would be is +2.7V).

Thus you can see that D1 will be cut off and D2 will be conducting.

  • \$\begingroup\$ Hey Spehro I was analyzing through your statements. So when we are removing D1 the voltage across where D1 was is -1.3 v (agreed) and here we are basically saying that D2 is conducting and V= 2.3 v as a result. On the other case when we remove D2 we have V=0.3v and the voltage across where D2 was is 2.7 V. So here we initially are saying that D1 is conducting. So my conclusion is 1) When we consider D2 conducts D1 is non conducting 2) whereas when we consider initially D1 is conducting D2 is also conducting.So why are we finally choosing conclusion 1) over conclusion 2) \$\endgroup\$
    – Timmy
    Sep 12, 2014 at 3:36
  • \$\begingroup\$ When diode D2 is present the the voltage across where D1 would be is negative, so it will never conduct when D2 is there. The opposite is true if you consider D1 present and D2 optional. \$\endgroup\$ Sep 12, 2014 at 3:43
  • \$\begingroup\$ So essentially we will have two cases where V=0.3v and V=2.3v and I would have to choose conclusion 1 because in conclusion 1 the diode D1 is reverse biased and D2 is Conducting. and we don't choose the second case as the voltage drop across the D2 when D1 conducts become 2.7V which is not consistent with the constant voltage model that requires the diode voltage to be 0.7v for conduction. -Thanks \$\endgroup\$
    – Timmy
    Sep 12, 2014 at 4:00

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