If two identical diodes are in parallel and forward biased such that their anodes are at the same potential but their cathodes are at different potential how can we find the anode potential?

As an example I found the following circuit where diodes are identical and their forward voltage drop is 0.6V: enter image description here

I can simulate this and figure out what V6 is but I tried to find myself and puzzled.

First of all the diodes are identical and the diode on the left is reverse biased so it is out of the picture.

Now if only the diode with -6V cathode voltage existed V6 would be -5.4V. And if only the diode with -12V cathode voltage existed V6 would be -11.4V. My logic here is that the voltage drop must be 0.6V so one can find the anode terminal voltage by adding 0.6V to the cathode voltage. And both diodes are forward biased.

But when their anodes are tied together there must be only one voltage at that node V6. I have been thinking about it but probably missing a point. How should we approach this topology to figure out the voltage at V6?

  • \$\begingroup\$ Hint: they're not in parallel if they're not connected at both ends. \$\endgroup\$
    – The Photon
    May 18, 2019 at 2:05
  • \$\begingroup\$ Oh thats true I should have named not in parallel but naming it different doesn’t help me. \$\endgroup\$
    – pnatk
    May 18, 2019 at 2:07

2 Answers 2


if only the diode with -12V cathode voltage existed V6 would be -11.4V.

And if \$V_6\$ is at -11.4, then what operating regime is the other diode in?

But maybe it will be how can we know?

The general approach to these problems is,

  1. Guess which diodes are forward biased and which are reverse biased.
  2. Check whether your guess leads to any logical contradiction or non-physical result.
  3. If it does, go back to 1. and make a different guess.

It's possible (in a more complicated circuit) that more than one guess will produce no contradiction. In that case, the circuit may have multiple stable states, and the actual state will depend on the history of the input voltages over time, or may even develop randomly (through metastability, for example).

  • \$\begingroup\$ It would be reverse biased \$\endgroup\$
    – pnatk
    May 18, 2019 at 2:08
  • \$\begingroup\$ But maybe it will be how can we know? Or it cant? \$\endgroup\$
    – pnatk
    May 18, 2019 at 2:09
  • \$\begingroup\$ Is there any non-physical result if it's reverse biased at -11.4 V? What if \$V_6=-5.4\ V\$? Is there any non-physical result in that case? \$\endgroup\$
    – The Photon
    May 18, 2019 at 2:14
  • \$\begingroup\$ -5.4V yields contradiction. Because in that case both are forward biased. So my question arises. But if somehow -11.4V it doesn’t yield weirdness. One becomes reverse biased. But is there a way to analytically reach that point? Or this is the way? \$\endgroup\$
    – pnatk
    May 18, 2019 at 2:16

This circuit is a bit different, but I have always liked to think of this topology as a voltage selector.

enter image description here

If I’m not mistaken, this circuit is used in your phones and laptops, because if you connect a charger to your device it introduces a bit of a problem. With the charger connected, how does the device decide to take power from your battery or the charger? You would definitely want to use the higher of the two voltages, but how do you go about doing this?This circuit handles this by taking whatever voltage is higher, and allowing it to pass to the rest of the circuit.

In your case, it is simply the opposite, where instead of picking the highest voltage, the diode with the lowest voltage on its cathode will conduct.


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