We are learning about the different modes of BJT transistors currently. Something that is a bit odd to me is that our textbook states that:

"... an npn transistor whose EBJ is forward biased (usually, VBE ≃ 0.7 V) will operate in the active mode as long as the collector voltage does not fall below that of the base by more than approximately 0.4 V. Otherwise, the transistor leaves the active mode and enters the saturation region of operation."

Why does the emitter-base junction need the 0.7 V we're all familiar with to become forward biased, but the base-collector junction (\$V_{BC}\$) need to only be 0.4 V to become forward biased?

My line of thinking is that the collector is less doped than the emitter, so there is a smaller barrier voltage to overcome.

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    \$\begingroup\$ ah! the thing is, the EBJ is already biased to at least 0.7 V, when the BCJ needs to be 0.4 V. Think about it: that PN junction can't be quite as the other, since due to the EB-bias you've already got a different charge carrier concentration in the base. \$\endgroup\$ Apr 1, 2020 at 17:05
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    \$\begingroup\$ There is nothing in the quoted passage that says that the collector-base junction is "forward biased" at 0.4V. The passage only says that the transistor will enter saturation. \$\endgroup\$ Apr 1, 2020 at 17:24
  • \$\begingroup\$ Elliot, it is implied that the BCJ will become forward biased in saturation mode. I figured that didn't need to be explained. Marcus, I believe you've answered my question. If I understand correctly, the barrier voltage across the BCJ is decreased because of the increase in minority charge carriers in the base? \$\endgroup\$
    – ModularMan
    Apr 1, 2020 at 18:43
  • \$\begingroup\$ It is not 0.7V to be forward-biased - the junction will start to conduct at around 0.3-0.4V then the voltage will rise logarithmically with the current. \$\endgroup\$ Jul 8, 2021 at 21:11
  • \$\begingroup\$ It is not only "implied" that the BCJ is forward biased for saturation - this condition is the DEFINITION for saturation. The large base current is only the result of this operational mode (some people think that a base current which is larger than given by beta would be the definition for saturation, but this is not the case). \$\endgroup\$
    – LvW
    Aug 14, 2022 at 14:05

2 Answers 2


Majority carriers (electrons for N type semiconductors and holes for P type semiconductors) do not significantly cross a PN junction until a characteristic forward voltage (dependent upon the material and temperature) is reached. For silicon diodes, this is typically around 0.6-0.7 volts. In contrast, minority carriers (electrons in a P type region, or holes in N type region) cross a PN junction with relative ease.

When majority carriers cross a PN junction, they become minority carriers, i.e. electrons in a P type region, or holes in an N type region. Likewise, when minority carriers cross a PN junction, they become majority carriers. So, if a majority carrier crosses a PN junction, and becomes a minority carrier, and if it arrives at another PN junction (which will, logically need to be oriented in the opposite way to the first), then that minority carrier will easily cross the second PN junction (and become a majority carrier again.) The fact that it will easily cross the second junction is the basic principle behind bipolar junction transistors.

At this point, I would like to make a tangent point that is not strictly necessary for an answer to your question, but is interesting nonetheless. I wrote above that if a majority carrier crosses a PN junction, and becomes a minority carrier, and if it then reaches a second (reversely oriented) PN junction, then it will cross that second junction easily. However, if a majority carrier crosses a PN junction and becomes a minority carrier, and if that minority carrier now finds itself in a vast sea of majority carriers, the chances are good that the minority carrier and a majority carrier will meet and recombine. Obviously, once a minority carrier recombines, it will not continue its journey to a second (reversely oriented) PN junction. Thus, BJT transistor designers typically 1) dope the emitter much more heavily than the base, so that the "sea" of majority carriers in the base is not all that concentrated with majority carriers, and 2) make the distance between the emitter-base junction and the base-collector junction fairly small, so that the "sea" of majority carriers is not that "vast". These two design factors increase the percentage of majority carriers from the emitter that cross over the emitter-base junction and become minority carriers in the base, and then cross over the base-collector junction and become majority carriers in the collector. 99% efficiency in this regard is not atypical.

In a normal diode, even when there is not a sufficiently forward voltage for a significant number of majority carriers to cross the PN junction to generate a concentration of minority carriers on the other side, there is nevertheless always a certain number of minority carriers that are formed "spontaneously" on each side of the junction due to thermal effects. As mentioned, these minority carriers can pass easily across the PN junction, even when the PN junction is only slightly forward biased, but importantly, even when the PN junction is reverse biased. The flow of minority carriers across the PN junction when the junction is reverse biased accounts for most of the leakage current of a diode. The leakage current is small because the concentration of minority carriers is small. However, in a transistor, when a great deal of minority carriers are injected into the base from the emitter because the emitter-base voltage is sufficiently forward biased, the current across the base-collector junction can be large.

There are a number of ways saturation of a BJT has been defined. One commonly used definition is that a BJT is in saturation if both the base-emitter and base-collector junctions are forward biased. For an NPN transistor, this will be true when 0 < Vce < Vbe.

Now, why does the base-collector junction conduct even when the voltage across it is less than the typical 0.6-0.7 volts required for a diode to conduct?

The reason is that, just like the case where the transistor is the forward active region and the base-collector is reverse biased, when the base-emitter junction is sufficiently forward biased to conduct significantly, the base gains lots of minority carriers. Whether or not the base-collector junction is forward biased, or reverse biased, these minority carriers (or at least a large fraction of them) diffuse to the base-collector junction and cross it. When the base-collector junction is forward biased, but not sufficiently to cause significant majority carrier current, this base-collector current corresponds to the very small current that exists through a normal diode that is forward biased, but not sufficiently for significant majority carrier current. However, because there is more minority carriers in the base of a BJT in which the emitter-base junction is forward biased, than in a simple diode, there is correspondingly more current into the collector. The injection of minority carriers into the base from the emitter, permits a large current to flow across the base-collector junction which is forward biased, but not sufficiently to cause majority carrier current.

[Note that if the base-collector junction is sufficiently forward biased, majority carriers in the base and collector will also cross over the base-collector junction and contribute to the collector current. However, if the base-emitter and base-collector junctions are both sufficiently forward biased for significant majority carrier current, then \$V_{CE}\$ is unlikely to be more than 0.2V and will probably be less.]


When transistor is in active state the Base-Collector junction needs to be zero or reverse biased. When \$V_{BE}=0.7V\$ and \$V_{CE}=0.3V\$ you get \$V_{BE}-V_{CE}=V_{BC}=0.4V<0.7V\$.

It's not important for Base-Collector junction to be forward biased hence your confusion.\$V_{CE}\$ can be as low as \$0.2V\$ (\$V_{BC}=0.5V\$), and transistor would still be in forward active region. You just need to make sure Base-Collector voltage doesn't go over \$0.6-0.7V\$. That would mean you have entered saturation region, or in other words both of your PN juntions are forward biased.

  • \$\begingroup\$ I am not concerned about the modes, I already understand that if both are forward biased then it is in saturation mode. My question is why the BCJ will be forward biased with a smaller forward voltage than the EBJ. I believe Marcus answered my question above. Thank you for trying though, I appreciate it! \$\endgroup\$
    – ModularMan
    Apr 1, 2020 at 18:44

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