Why is the current so high in the saturation region of Common-Base output characteristics of a NPN transistor?
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The saturation region, in which both the collector and emitter junctions are forward-biased, lies in the portions of the curves wherein Vcb is just very slightly above 0 V and Ie > 0. When both are forward biased then how can so much current be flowing through the circuit.
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Answer:
This is a touchy-feely kind of question that you usually only see in fourth-rate textbooks. There does NOT have to be a "lot" of current to saturate a transistor, for instance if the collector resistor is 1 megohm fed from 1 volt and the transistor beta is 1,000, the collector and emitter currents are 1 microamp and the base current is 1 picoamp in saturation. None of those are "high" currents. In general when a transistor is saturated there is only a minimum, like 0.3 volts collector to emitter so the emitter and collector currents are as high as they can go given the collector resistance.
George Gonzalez at Quora Visit the source
Other answers
In that case, when both the junctions are in FB, you can mentally replace the transistor with a short circuit and hence the current is controlled by the external series resistance only. The current can not exceed that amount.
Manjunath Pai H
In a Common-Base circuit, 99% of the current is through the emitter. Indeed, in one way of calculating bias, most of the current in any transistor is through the emitter lead, if I recall correctly. If I'm wrong, mea culpa.
Ron Garrett
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