This circuit contains forward and reverse bias. The question is to find Vd and Id. In this case, my understanding is that the current from the leftmost mesh will not flow to the middle and rightmost mesh due to reverse bias. In another case, the current from the middle and rightmost mesh will not will into the leftmost mesh because of the reverse-biasing condition (negative terminal of diode connected to the positive terminal of external voltage).

First, in the leftmost mesh, I solve using Ohm's law to find the respective current and voltage.

Second, the middle and rightmost mesh, I apply Thevenin's theorem to find the Vth and Rth.

But then, I am stuck with the subsequent steps. Any guidance/suggestions are welcomed. Thank you. enter image description here


1 Answer 1


Try assuming the diode is not conducting (reverse biased) and calculate the voltage on either side to see if that assumption is correct. If it is, the current is zero and the diode voltage is the difference of the two voltages.

If it turns out the diode is conducting, you can calculate the Thevenin equivalent resistance on each side (and you already have the Thevenin equivalent voltages from the first step) so you can calculate the diode current.

A diode can only be forward biased or reverse biased at any given time. The issue with the problem circuit is that it's not immediately obvious which it is.

  • \$\begingroup\$ meaning that, the diode is open circuited in both case and we see the voltage(left or right whichever is higher), right? \$\endgroup\$ Commented Mar 31, 2022 at 4:24
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    \$\begingroup\$ In the second case (diode is forward biased) the diode can be considered a short (if your simple diode model includes a forward voltage then it must be forward biased by more than the fixed forward voltage and you replace the diode with a voltage source of eg. 0.7V). So you have two voltage sources (maybe 3) and two Thevenin equivalent resistors in series, find the voltage difference and the current is Vdiff/Rtotal. \$\endgroup\$ Commented Mar 31, 2022 at 4:27
  • \$\begingroup\$ does it mean that if the diode is forward, then only I apply Thevenin theorem, whereas if it is reverse bias, we are no longer needing the Thevenin but instead we can directly find the potential difference across the diode, am I understanding it correctly? \$\endgroup\$ Commented Mar 31, 2022 at 4:37
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    \$\begingroup\$ Yes, that is correct. If no current is flowing through the diode, then the voltage dividers don't affect each other. \$\endgroup\$ Commented Mar 31, 2022 at 4:38
  • \$\begingroup\$ assume the node at left of diode (-ve side) is Va and right of diode is Vb (+ve side), I got Va>Vb, in which voltage at negative terminal of diode is greater than its positive terminal, then the current diode is zero and V diff is -1.2V. \$\endgroup\$ Commented Mar 31, 2022 at 4:39

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