I was thinking about a question that was just up here, someone's interview I think. Anyway I'd never really thought about what happens when you have two diodes in series but reverse biased.


simulate this circuit – Schematic created using CircuitLab

Why is the voltage drop shared, or 10V across each of them? I mean I understand if it were resistors there'd be a current flow, I suppose here it's the very small reverse bias current flow? I just don't understand what's happening at a low level :)

Then having Vdrop be VCC/3 seems like a text book answer, like the real answer should be each diode's characteristics are a little bit different so depending on how much current each one is handling, however small somehow leads to the voltage drop across it.

So how does that work in real life?


3 Answers 3


You'd need to look at the I-V curve for each diode and find an operating point where the total voltage for the three makes up the source voltage (30 V in your example), and the current is equal through each one.

If the diodes are identical, then of course that leads to VCC/3 across each diode.

But if one diode is slightly warmer or colder than the others, or has more light shining on it, or just came out of the fab with a little bit more or less dopant in its junction, then they won't be truly identical, and you could have substantial variation in the voltages.

Putting high-value resistors in parallel with each diode (as in the earlier question) will reduce the effect of the diode variations on the resulting voltage drops and get you closer to achieving equal voltages across the diodes (if that's what you want).

  • 2
    \$\begingroup\$ also: Parasitic series resistance (aka pcb material) could even exceed the reverse leakage current of the diodes. \$\endgroup\$ Sep 22, 2015 at 2:05
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    \$\begingroup\$ @NilsPipenbrinck, good point. OP, typically this would be a problem mainly if there's some contamination on the pcb surface (like no-clean flux residue, for example). \$\endgroup\$
    – The Photon
    Sep 22, 2015 at 2:09

You seem to have the right idea. For diodes arranged in this way, it comes down to reverse bias leakage. Here's a screen shot from a 1N4148 datasheet made by Fairchild: enter image description here

As the voltage source turns on and rises from 0V to 30V in your schematic, the reverse voltage across the diodes will increase the same. So the reverse current will start at zero and start to climb up the curve in the figure above. Just as the reverse current reaches about 15nA through each diode, the reverse voltage drop on each diode will be just about 10V. At that point it'll stabilize and hold there indefinitely.

Obviously, the diodes won't be exactly the same, so they will stabilize with different voltages, at whatever value of curent that causes the three voltages to add up to 30V. Also, there are other causes of leakage current in the circuit that is entirely dependent on the circuit's topology, but that's beyond the scope of your question.


You state in your question, "I would understand, if the diodes were resistors." Well - they are! They are variable resistors. Let me explain, by looking at the I-V curve diagram (thanks to Dan Laks), you will see that at 10v, there is a reverse current of 15nA. This corresponds to a resistance of (10v/15nA =) 667 meg-ohms. Assuming that the diodes are "identical", then you have 3 equal resistors across a 30v supply. Therefore, the voltage across each diode is VCC/3. The sum of the three resistors is 2,000 meg-ohms, so the current across the three diodes is(30v/2,000mohms =) 15nA.
For the case that the diodes are not identical, then the voltage across each diode will be higher or lower than 10v, but their sum will be 30v.


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