Electronic – Voltage drop across each diode and current drawn from battery

diodesleakage-current

  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

Best Answer

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.