I have the following doubts about a p-n junction diode:

1) If we short the diode ( connect it's terminals together without any power source) then how is the Kirchoff's law across the loop satisfied?

2) If the diode is forward bias, people say that the built in voltage Vin decreases by V. That is the new built in voltage is Vin-V. But what I don't get is shouldnt the voltage across the diode be V? How does a Kirchoff's voltage equation across the circuit look like?

3) Similarly, in reverse bias why does the applied voltage V add to the built in voltage? Shouldn't the voltage across the diode be V in the opposite direction? How does a Kirchoff's equation for this circuit look like?

Any help would be appreciated. Thanks :)

  • \$\begingroup\$ Your confusion comes from the idea the forward voltage is a constant. It isn't. This is a simplification for (relative) high currents. As soon you get into the µA range, the forward voltage drops to µV. At zero current, it is zero. \$\endgroup\$
    – Janka
    Aug 7, 2018 at 2:40
  • \$\begingroup\$ I still don't understand. \$\endgroup\$
    – pranav
    Aug 7, 2018 at 2:59
  • \$\begingroup\$ I think the questioner is asking about the "built-in voltage" which is a device physics thing, not the forward voltage. It's been years since I learned how diodes work so I can't give a great answer. It's something like "the built-in voltage is balanced by an equal and opposite voltage at the metal contacts" so that there is no current, and no way to measure the voltage. \$\endgroup\$
    – Matt
    Aug 7, 2018 at 3:21
  • \$\begingroup\$ @Matt You posted this comment just before I finished typing up my answer. Please note, I'm a different Matt. \$\endgroup\$
    – Matt
    Aug 7, 2018 at 3:25

1 Answer 1


I suspect your confusion comes from an incomplete picture of what a packaged diode is compared to the p-n junction diode you have learned about. You might notice that a discrete diode has metal leads coming out from the package, but your p-n junction diode model only has semiconductor. In order to interface with semiconductors we usually need to connect metal wires to them. Every introductory semiconductor book I have seen has a chapter on metal-semiconductor junctions which you can read about if you are interested. The short version is that the metal-semiconductor junctions also have a built in potential that cancels out the p-n junction built in potential. So to quickly address each of your specific questions:

1) If you short a diode (in the dark) you will have 0 current through everything, KCL is satisfied. The KVL equation looks something like \$0 = V_{bi,m1} + V_{bi,pn} + V_{bi,m2}\$ with one or two of those variables being negative which allows KVL to be satisfied.

2) If you forward bias a diode the built in voltage increases by the applied bias... sort of. This is technically going to be spread across all 3 junctions, but the vast majority should show up across the semiconductor p-n junction, so its fine to assume thats where it all is. KVL equation: \$-V_{ext} = V_{bi,m1} + (V_{bi,pn} - V_{ext}) + V_{bi,m2}\$

3) Basically the same as #2 but with a positive voltage: \$V_{ext} = V_{bi,m1} + (V_{bi,pn} + V_{ext}) + V_{bi,m2}\$

  • \$\begingroup\$ Thanks a lot. Is this valid even if other circuit elements like resistors and inductors are added to the circuit? Can we neglect the potential drop across the ohmic contacts in any case? \$\endgroup\$
    – pranav
    Aug 7, 2018 at 15:28
  • \$\begingroup\$ Yes, other circuit elements wont change this. If the contacts are truly ohmic, and they have a low resistance, then you can ignore them. Its possible to make rectifying contacts, or high resistance ohmic contacts, in which case you wouldnt be able to ignore them so easily. \$\endgroup\$
    – Matt
    Aug 7, 2018 at 15:31
  • \$\begingroup\$ @Matt So is there like a voltage gradient inside the diode itself. For example, if I apply an external 5V to a diode in a reverse-biased manner. The pn junction will expand and might have an internal potential across it of 0.9V (as opposed to 0.7V). The remaining 4.1V is dropped from the N contact to the edge of the depletion region, is that correct? \$\endgroup\$ Aug 23, 2021 at 22:53
  • \$\begingroup\$ @jaurunjljgrtutkwcy There is a voltage gradient if you have current, but its usually quite small compared to the voltage drop across the junctions and typically ignored. Applying an external 0 V means the metal-semiconductor junctions built-in voltage adds up to exactly cancel out the built-in voltage of the p-n junction. Applying any nonzero external voltage shows up entirely across the p-n junction (as a very good approximation). \$\endgroup\$
    – Matt
    Aug 24, 2021 at 2:04
  • \$\begingroup\$ @Matt That makes sense! Thank you. I didn't realise there were two extra junctions that I never considered. \$\endgroup\$ Aug 24, 2021 at 19:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.