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In many documents and books, the collector base junction of a BJT transistor is usually represented as an inversely polarized diode, however there is no collector base voltage that we must overcome since as soon as we apply 0.7 volts to the base the transistor already conducts between emitter and collector without offering resistance

Is the collector base junction really misrepresented as a reverse biased diode?

Thanks

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  • \$\begingroup\$ Two diodes alone is not correct. Your text should at least reference the DC gummel poon model which includes a current source. \$\endgroup\$
    – sstobbe
    Jan 30 at 14:49
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however there is no collector base voltage that we must overcome since as soon as we apply 0.7 volts to the base the transistor already conducts between emitter and collector without offering resistance

That is correct.

For a full explanation, please consult a text on semiconductor theory. What I present here is a simplification which I hope is accurate enough.

Inside a diode there are three regions, an N-region, a P-region, and at the junction between them, a depletion layer. If the voltage between the N-region and the P-region is insufficient, very few electrons in the N-region cross the depletion layer, and very few holes in the P-region cross the depletion layer. When there is enough voltage between the N-region and P-region, electrons will cross from the N-region to the P-region, and holes will cross from the P-region to the N-region.

When electrons cross over from the N-region to the P-region, they do not immediately combine with holes, but continue travelling for some distance as "minority carriers". They are minority carriers because they are electrons in a P-regions. Similarly, when holes cross over from the P-region to the N-region, they do not immediately combine with electrons, but continue travelling for some distance as minority carriers.

If the N-region is more heavily doped than the P-region, then more electrons will cross over from the N-region into the P-region, than holes will cross over from the P-region to the N-region. Whether it is mostly electrons crossing the depletion region or mostly holes crossing the depletion region doesn't really matter in a diode. The diode is conducting current either way.

For the sake of simplicity, in what follows, when I refer to a "transistor", I will mean an NPN transistor. What happens in a PNP transistor is similar except that the roles of electrons and holes is swapped.

When the base-emitter junction of an (NPN) transistor is forward biased, electrons cross over from the N-region emitter to the P-region base. There, they are minority carriers. They don't recombine with holes in the base immediately. If the base is thin, and the average path of an electron before recombining is long enough, the electrons will reach the base-collector junction. Because they are minority carriers, they pass over the base-collector junction easily, and enter the collector. Once in the collector, they become majority carriers again, and travel freely to the metal collector terminal. That is why, when the the emitter-base junction is forward biased, current will flow from emitter, through base, into collector, even when the base-collector voltage differential is less than that of a typical diode junction or reverse biased.

The doping concentration of the collector is significantly lower than that of the base or emitter. So, for an NPN transistor, electrons easily flow in the direction just described. It is also possible for the roles of the emitter and collector to be reversed, but because the doping concentrations in the emitter and collector are significantly different, a transistor biased in "reverse active mode" doesn't have the same electrical characteristics as one in "forward active mode". Because the base is more heavily doped than the collector, it is much easier for holes to cross from the base into the collector, (when the base collector junction is forward biased) than for electrons to cross from the collector into the base. That is why the collector of a transistor does not make a good "emitter", hence why the reverse-active mode is not as widely used as the forward active mode.

The take away: When electrons cross from the emitter N-region emitter into the P-region-base, they do not recombine with holes immediately. They become minority carriers. As minority carriers, they may cross over the the base-collector junction even when or even though the collector may be at a lower voltage than the base.

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  • \$\begingroup\$ Thanks for the help, can we say then that the diode base collector is only reverse biased with the transistor in cutoff and forward biased with the transistor in its active region and in saturation? \$\endgroup\$
    – Mario
    Jan 30 at 14:36
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    \$\begingroup\$ No. The base-collector diode is forward biased if the transistor is in saturation, and reverse biased in the active region, as well as in cutoff. I may add some more to the explanation, but think of this. Minority carriers cross pn junctions easily. The reason diodes don't conduct much when reverse biased is that there aren't many minority carriers, because the semiconductor is heavily doped. However, there are lots of minority carriers in the base of a transistor that is conducting. The base collector junction is a very, very leaky reverse biased diode in active region. \$\endgroup\$ Jan 30 at 14:42
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In a way, the collector-base junction IS the transistor.
A bipolar junction transistor is basically a reverse polarized diode (the CB junction) whose inverse saturation current (you know, the asymptotic line of the seemingly flat portion that sits just under the negative voltage axis in the characteristic of a diode) is altered by the carriers injected by the base-emitter junction.

Consider the negative voltage portion of a diode and draw it for different values of the Isat current. You will see curves that resembles - albeit in reverse - the output characteristic of a transistor.
The missing part of the picture is a link between the collector and emitter current and that is also what allows a transistor to work as an amplifier: the diffusion current that takes place in the narrow base region allows for the decoupling of the current crossing the CB junction and the current crossing the BE junction from the respective voltages.
Note that you cannot have diffusion current in the conductor between two discrete diodes back-to-back. That's why you cannot build a transistor with two separate diodes, but need to use the same semiconductor in common with the two junctions.

So, in a common base configuration (the configuration usually employed to demonstrate transistor action) you can have a moderate input IE current associated with a small voltage, while on the output nearly the same current (IC is approx IE) is associated with a much bigger voltage. This decoupling also explains why you can use a transistor to match impedances.

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  • \$\begingroup\$ great explanation! \$\endgroup\$ Jan 30 at 16:11
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The base collector junction is a legitimate diode junction. If you have a transistor you can try it out. Force a small current through it (like 1mA) or probe it with a VOM in diode mode.

Also, in saturation, the base collector junction is forward biased. This is a common mode of operation (any time a BJT is used for switching).

FYI, you can swap the emitter and collector and still have a transistor with degraded properties (low beta, for example).

Note that the transistor is NOT just two diodes. Two separate diodes would not have gain and all the other properties of a transistor.

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  • \$\begingroup\$ In other words, as I understand, as the base current increases, the collector base junction becomes a diode, polarizing directly? \$\endgroup\$
    – Mario
    Jan 30 at 10:10
  • \$\begingroup\$ I guess I don't really understand your comment or question. \$\endgroup\$
    – mkeith
    Jan 30 at 10:24
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    \$\begingroup\$ Mario - simple statement (case npn): When the potential of the collector (Vc) is smaller than that of the base (Vb) the base-collector junction is forward biased. As an indication for this state: The base current is much larger than under normal operation (for example Ib is 10% of Ic or even larger). \$\endgroup\$
    – LvW
    Jan 30 at 10:37

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