I started studying the BJT Configurations and how they operate in active mode, saturation mode, and cutoff. And that every two neighboring junctions have either forward or reverse bias, But I wonder what happens when I only connect one battery between only collector and emitter (N.Terminal to the emitter and P.Terminal to the collector). what happens to the holes in the middle base. Does current flow? why? or why not?

  • \$\begingroup\$ Your battery "sees" two diodes back-to-back. There's a very small current flowing because one of the diodes is driven in reverse direction. The current is slighty higher if it's the BE diode (that one usually leaks more.) \$\endgroup\$ – Janka Dec 27 '17 at 10:52

The voltage \$V_{CE}\$ applied has to drop across the two junctions: collector junction and emitter junction. Now the sign of the potential will be such that one junction will be forward biased and the other will be reverse biased. Since base terminal is open, same current has to flow through these junctions. So a small current having a magnitude in the order of reverse saturation current of the reverse biased junction will be flowing.

Because of the bias conditions prevailing, say forward biased emitter junction and reverse biased collector junction, electrons from emitter will be injected to base (npn transistor assumed). These electrons will be transported to collector and contribute to current. The holes injected from base to emitter also contribute to current. But the applied voltage will be dropping mostly in the collector (reverse biased) junction and hence the current will be very small.

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