# Voltage drop across each diode and current drawn from battery

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?

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?

• 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.
– G36
Jun 10, 2017 at 16:01
• 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 :) Jun 10, 2017 at 16:19