# BJT cut-off currents magnitudes and directions

I was reading the datasheet for the 2N3904 transistor (NPN BJT) recently and found out that I couldn't wrap my head around the cutoff current values.

For the given conditions, both junctions are reverse biased ($V_{CE} = 30V, V_{EB} = 3V$), the datasheet says that the base cutoff current is 50 nA, and the collector cutoff current is also 50 nA.

Consider the collector-base junction. Since it's reverse biased we have a reverse saturation current which goes into the collector and this is equal to 50 nA. Holes from the collector move to the base and then recombine with electrons there, thus generating some portion of the base current which is directed out of the base. So according to KCL, the emitter current must be zero or very close to it. On the other hand the emitter-base junction is also reverse biased so there also must be a reverse saturation current flowing into the emitter, so according to KCL, the base current must be a combination of the emitter and collector currents.

Either the emitter current is extremely low or I'm missing something. So the question is what really happens to carriers inside BJT when we use it in a cutoff mode.

• read about the Early Effect and diode leakage your assumptions are missing this info – Tony Stewart Sunnyskyguy EE75 May 23 '18 at 19:16