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.