In my understanding, in an NPN junction for example, an increase in the emitter-collector voltage should widen the depletion layer in the base-collector junction therefore increasing the electric field in that region. This should normally speed up any electrons that reach that region toward the collector and the negative terminal of the battery. This should logically increase the current since current is the number of charges (Coulombs) per second.

The only explanation I managed to find is that the number of free electrons in the emitter region is limited and therefore the number of charges per second can't increase because there simply are not enough charges. Unfortunately, this seems very counter-intuitive to me because in that case we should be facing the exact same problem when we increase the the emitter-base voltage.

Someone stated that the electrons in the emitter region are shielded by the base but the explanation he gave wasn't related to any kind of shielding at all and I don't understand what that shielding could be.

Why doesn't the current in a BJT transistor increase with an increase in emitter-collector voltage?

  • 1
    \$\begingroup\$ "Why doesn't the current in a BJT transistor increase with an increase in emitter-collector voltage?" In fact - it does (Early effect). \$\endgroup\$
    – LvW
    Jan 20, 2022 at 16:18
  • \$\begingroup\$ Well, maybe my question should be why does base-emitter voltage influence the collector current greater than the emitter-collector voltage ? \$\endgroup\$ Jan 20, 2022 at 17:17
  • \$\begingroup\$ Try these Qs... electronics.stackexchange.com/questions/590179/… and electronics.stackexchange.com/questions/101595/… \$\endgroup\$
    – user16324
    Jan 20, 2022 at 17:31

1 Answer 1


Effectively the Ic is controlled by Vbe as long as Vce is not in saturation, although there is a small load from Early Effect leakage.

In saturation, Vcb becomes forward biased and shunts the hFE significantly. such that standard ratings for hFE are done at Ic/Ib = 10, 20 or 50 for typical transistors rated with hFE ~100, 200 ,500 +/- tolerance. There is also a bulk series resistance with Rce such that at high currents, saturation region may well exceed 1V with Rce values typically <4 ohms = Vce(sat)/Ic.

In the forward active mode as a common emitter, the collector is a current source with an impedance from the Early Effect and Miller Capacitance, so neglecting those effects, it becomes insensitive to applied voltage changes.

  • \$\begingroup\$ Thank you for the clarification but I'm obviously interested in the active region mode. And my question is why a small increase in base-emitter voltage generates a huge change in collector current while an increase in base-collector voltages generates nearly no change in collector current ? According to my own logic and what I explained, they should have both the same effect. I am interested in the microscopic phenomenon. \$\endgroup\$ Jan 20, 2022 at 17:19
  • \$\begingroup\$ Let me try the following (short) explanation: The B-E voltage releases electrons from the emitter and allows a current Ie towards the pn junction (exponential relation). Most of the carriers swap through this (extremely small) region and are attracted by the collector voltage (larger than the base potential). Just a tiny portion of these carriers go to the base terminal (forming the base current). This process works - if the collector voltage Vc is 3 or 5 or 10 volts larger than the base potential. There is only a rather small dependence on Vc (Early effect, explained elsewhere). \$\endgroup\$
    – LvW
    Jan 21, 2022 at 8:21

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