Conceptual short circuit diode / zero-bias base collector junction BJT question

[Only a Mirage]

Why exactly is there zero current flowing through a diode when it is short circuited, given the presence of the barrier potential? My current understanding is that the drift current precisely balances the diffusion current. But if you take a diode and short it with a wire, wouldn't you then also have charges "diffusing" in the direction opposite to the junction (the normal direction of diffusion), at least giving rise to a transient current?

Related to this zero bias question, consider a simple NPN bipolar junction transistor circuit with the base-emitter junction forward biased and the base-collector junction zero-biased, with some load connected between the collector and emitter. The transport model equation under these conditions for Ic is Ic = Is(exp(Vbe/Vt) - 1). So in this case, there is current across a "shorted diode". But why? The best i can come up with is that the electrons flowing from the emitter into the base region eliminate the normal gradient of charge carriers in the "diode" so that there is no diffusion current to balance the drift current. (But Ic is a lot larger than the normal drift current because it depends on Vbe here?)

This was confusing to me because initially, I thought that since the base-collector junction was shorted, all current should simply "bypass" this junction, flowing through the short instead. My semiconductor devices teacher wasn't sure when I asked him, and any insight would be appreciated.

 

[cabraham]

Ic = alpha_n*Ies*(exp((Vbe/Vt)-1) - alpha_i*Ics*(exp((Vbc/Vt)-1).

The above is the complete Ebers-Moll representation of the bjt. In active or cutoff mode, the 2nd term is negligible. But as the device enters saturation, i.e. Vbc approaches or positively exceeds zero, then the 2nd term must be retained.

Claude

 

[cabraham]

Also, remember that the diode equation of Shockley, Id = Is*(exp((Vd/Vt)-1), cannot be directly applied to the b-c junction of a bjt because 2 things are happening. SDE (Shockley diode equation) applies for 2 terminals and anode current = cathode current. Having 3 terminals & 2 junctions complicates things.

THe collector current is not simply Ic = Ics*(exp((Vbc/Vt)-1), because the collector is collecting electrons from the emitter, for an npn device. The forward bias on the b-e junction results in lare numbers of electrons moving from emitter to base. But the base is so thin they don't stay in the base & recombine with holes, only a few do, but keep going to the collector. The E field from collector to base attracts these electrons.

That is why you need to look at both terms in the Ebers-Moll equations.

Claude