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From what I understand, for an NPN transistor the reverse saturation current Icbo flow can be shown as in the red current loop:

enter image description here

My conclusion was that for an NPN transistor the Icbo does not add up to the base current.

But if we look at to the following common emitter circuit:

enter image description here

In the above case(when the transistor in active forward region) does the Icbo adds up to the base current Ib? Can you draw/indicate the current path for the Icbo in this case?

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Icbo is leakage from collector to base. It is measurable from the base wire if in your 2nd drawing Rb is infinite (=removed) and the base is connected to the minus terminal of Vcc via an micro- or nanoampere meter.

If your second drawing has nothing else connected to base than Rb, Icbo continues down as a part of the base current. It can be considered as internal add-on to the base current. Think it's caused by invisible resistor (nonlinear one) between collector and base.

Proper circuit design in switching and amplifier applications tries to ensure that Icbo is sucked off from the base to prevent it causing unwanted Ic, which is difficult to predict due it's temperature dependence and variation between transistor individuals. That's why there's some conductive path between base and emitter in many applications. It can be omitted in cases where there's a big resistor in series with the collector which prevents exessive Ic.

Your first image has beautifully stayed alive in textbooks from the days when transistors were new and totally different than our normal bipolar junction transistors. Of course it's valid, but it does not give any idea how transistors are mostly used.

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