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So, As I am studying I keeping finding confusions in my concept of Diodes. Sorry if the question is too noobish. So in plotting the Vout with respect to Iin, my manual says that using the constant voltage model, Vout will be undefined in Iin<0. I don't understand why. I believe calculating using the first branch(with diode D1 and resistor) gives us something for Vout, but D2 dictates that Vout should be equal to VDon(.8 v).Is it a paradox? I don't understand.


simulate this circuit – Schematic created using CircuitLab

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up vote 5 down vote accepted

Maybe because for Iin < 0 both the diodes are surely off, because the current is forced to flow in a direction opposite to that permitted by the diode. In that case D1 and D2 would be open circuits and the current could not flow nor in the first branch (with R1 and D1), neither in the second (with D2): Vout would be then the voltage across the Iin generator and if Iin is ideal, voltage is not defined across it (only the current is defined).

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For ideal diodes (with no reverse leakage current), you have an infinite resistance. So for any current flowing, you have an infinite voltage.

$$ V = I * R $$ $$ V = I * \infty $$

If current is not zero, voltage is immediately infinity. In the real world, you will have a certain finite, extremely high resistance. Also, a real world constant current source can only reach a certain voltage, rather than infinity. Once you reach the breakdown voltage, D2 will fail. D1's failure would take more voltage, depending on the value of R1. If R1 is small with reference to the reverse bias resistance of D1, then it not greatly alter the breakdown voltage.

enter image description here

If you take a given diode with a known leakage current profile, you CAN calculate the voltage for a given reverse bias current, up until reverse breakdown of the diode. However, the current will be very small before breakdown occurs.

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R1 will not materially affect the reverse breakdown of D1, unless it has such a large value that the reverse leakage current through D1 produces a significant voltage drop across R1. – Dave Tweed Apr 9 '14 at 15:15
Right. I should reword my fairly small. I mean this is reference to the resistance of a reverse biased diode. So it could meant a few megaohms being "fairly small". – Joe Apr 9 '14 at 15:25

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