- #1

ShreyasR

- 88

- 2

## Homework Statement

Find the voltage V

_{o}for the following circuit. Consider the diodes to be ideal.

## Homework Equations

When a diode is forward biased, Diode resistance = 0Ω (Short circuited)

When a diode is reverse biased, Diode resistance is infinite (Open circuit)

Ohm's Law: V = IR

Voltage divider rule.

Basic Nodal analysis.

Kirchoff's voltage law.

## The Attempt at a Solution

1) Undoubtedly, the diode D1 is forward biased. Hence it is replaced by a short circuit.

2) Now there are two possibilities for D2.

Case (i) If D2 is assumed to be forward biased (replaced by a short circuit):

We directly obtain V

_{o}= 10 V

Current flows in the branch containing D2. (short circuit path)

By applying Kirchoff's voltage law, we get

15V - I*(10 kΩ) -10V = 0

I = 0.5 mA

We infer that 0.5 mA flows through diode D2 from the P side to N side (Anode to cathode), without any potential drop across the diode. Hence, the assumption that the diode D2 is forward biased is CORRECT.

Case (ii) If D2 is assumed to be reverse biased (replaced by an open circuit):

We have an open circuited branch in parallel with the 10 kΩ resistance.

Hence, current flows through the two 10 kΩ resistors. (effectively in series)

By applying voltage divider rule, we obtain

V

_{o}= 15(10k/20k) = 7.5 V.

The anode voltage of the diode V

_{A}= V

_{o}= 7.5 V.

The cathode voltage of the diode V

_{k}= 10 V.

Hence, V

_{A}< V

_{k}...

As V

_{A}< V

_{k}, and no current is flowing though the diode, the assumption that the diode D2 is reverse biased is CORRECT.

These are the two different solutions that I have obtained for the same problem, and both seem to be valid, and are not violating any laws. But getting two solutions of voltages for a well defined circuit is not possible isn't it? Where have I gone wrong?