Electronic – Analyzing resistive circuit with ideal diodes

circuit analysisdiodes

The circuit is this:


The diodes are ideal. The solution for this question is D1 OFF and D2 ON, and hence current \$I = 3mA\$.

But when I assume that both D1 and D2 are OFF, I am not able to prove that my assumption is wrong.

How can I analyze the circuit if I assume that both diodes are OFF?

Best Answer

Just for the benefit of the larger audience: when you deal with purely resistive circuits containing n ideal diodes you can analyze them by guessing the state of every diode (either ON or OFF), do the calculations and then check if the results are coherent with the assumption you have made.

If you guessed right, i.e. the calculations confirm the assumption, then it can be shown that you've found the solution, which is unique. Otherwise you have to redo all the calculations changing your assumption about the state of the diodes. At worst you have to check \$2^n\$ circuit configurations (all the possible combinations of states of the n diodes), but with a bit of common sense and experience you can make the right guess with many fewer attempts.

Ok, then, but how do you check whether your assumptions are correct? Just remember what's the behavior of an ideal diode in either of its two states:

  • ON state: 0 voltage drop across the diode (it's like a short circuit) and current flowing from anode to cathode.
  • OFF state: 0 current flowing (it's like an open circuit) and diode is reverse biased, i.e. the voltage on the cathode is more positive than the voltage on the anode.

So, let's consider your specific circuit. If you assume both diodes are OFF, you replace them with open circuits. Then think what is the voltage at the upper leg of the resistor: since no current flows in the resistor there is no voltage drop across it... you should be able to continue from here and see that the results contradict the assumption. Hence the diodes cannot be both off.