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  1. I'm wondering if below configuration forward biases D1.
    D2 is clearly reverse biased. Would D1 drop 0.7 V and D2 drop 10-0.7 = 9.3 V?

schematic

simulate this circuit – Schematic created using CircuitLab

  1. I know how leakage current flows in reverse bias with one diode: Due to thermal agitation electron-hole pairs get generated inside junction, electric field inside the junction pushes these away in opposite directions to P,N type regions. Battery then sucks them. But with two back to back diodes wouldn't the holes from both D1 and D2 get stuck in the wire joining them? So there shouldn't exist any leakage current?

enter image description here

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    \$\begingroup\$ I think the leaking current will flow as D2 is reverse biased and D1 is in "forward biased" but D1 diode voltage will be below the "knee voltage" but still forward biased. In simulation Vd1 = 35mV; and 9.965V across D2. \$\endgroup\$
    – G36
    Commented Jun 10, 2017 at 16:01
  • \$\begingroup\$ Ahh got it @G36 , Since D1 drop is just 35mV, am I allowed to replace D1 by a wire? Then leakage current flow makes perfect sense(problem simplfies to leakage current in a single reverse biased diode). I can't thank you enough, you've been so helpful on this site :) \$\endgroup\$
    – Hiiii
    Commented Jun 10, 2017 at 16:19

1 Answer 1

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Both diodes are opposing therefore they block current in either direction therefore neither diode is forward biased. You do actually need current to flow to get any meaningful forward bias situation.

Of course there might be leakage currents (25 nA for the 1N4148 at 25 volt reverse voltage) but these are so small that the relevance of considering if one or the other diode was forward biased is lost. For instance if you looked at the 1N4148 forward characteristic you'd see something like this: -

enter image description here

At 100 uA, the forward voltage is about 0.5 volts and the voltage will drop at a rate of about 0.1 volt for each decade reduction in current hence, with 0.4 volts applied, I would expect the current to be 10 uA. For 100 nA I would expact the terminal voltage to be only 0.2 volts and 0.1 volts at 10 nA.

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  • \$\begingroup\$ Thank you, that answers my first q. About the leakage current, I don't get how it can flow when holes from both the diodes are getting pushed to the wire joining P-type regions. I'll try to elaborate a bit more if my question is not clear... \$\endgroup\$
    – Hiiii
    Commented Jun 10, 2017 at 9:41
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    \$\begingroup\$ @Hiiii: Maybe that would work if your diodes were frozen at absolute zero, but at real temperatures you will always have minority charge carriers (electrons in the P-type regions and holes in the N-type regions). \$\endgroup\$ Commented Jun 10, 2017 at 11:20
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    \$\begingroup\$ @Hiiii, there essentially won't be holes pushed into the wire. Yes, at a quantum level, there'll be Brownian motion and charged particles moved away from the junctions because local conditions override the external. However, D2's junction reverse-bias effectively absorbs the electrical potential (voltage) from the battery... so there's no energy for those particles to move very far beyond the local conditions of the junctions. A charged particle moving is doing "work" from a physics perspective, so think about it in that reference and you'll have a clue to the answer to question 2. \$\endgroup\$ Commented Jun 10, 2017 at 12:36

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