I read many books but none of them explains why the collector current is independent on collector base voltage in common base transistor in its active region.

Please explain it with clear mechanism.


2 Answers 2


I believe that you're reading books on analog electronics - these books are not aiming to provide you with an understanding of semiconductor devices. Take any book on semiconductor physics and you'll find the complete derivation there.

In short (the below applies to NPN BJT):

It is because Collector-to-Base junction is reverse biased in the active mode of operation.

This reverse biased junction just "passes through" the carriers injected from Emitter into Base and then diffused through Base to Collector. The amount of these carriers is dependent upon BE voltage, and this whole amount is swept across reverse biased CB junction, regardless of CB voltage (as long as it is reverse biased). Therefore the current depends strongly on \$V_{BE}\$, but is independent of \$V_{CB}\$.

It is the depletion region charge which creates the electric field which "sweeps" the carriers from Base to Collector side of the junction.

Please note that there are second order effects ("weaker" effects) which contain \$V_{CB}\$ dependency (see for example Early effect).


First, that's how BJTs largely work. Take a look at a set of collector current as a function of C-E voltage curves, and you will see they are fairly flat once you get past a volt or so.

Further, in the common base configuration, the emitter current is governed completely by the input circuit (as long as there is at least some minimum collector voltage), which is connected to the base and emitter only. Once the emitter current is set, the collector current is then only a function of the gain, which again is reasonably constant accross a wide range of collector voltages.


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