How does the 2nd junction in a BJT work?

Discussion in 'General Electronics Chat' started by bryancostanich, Feb 27, 2018.

1. bryancostanich Thread Starter New Member

Feb 10, 2018
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Hi folks!

I have a question about BJTs. I understand how P-N junctions work in and out now, but in modeling a PNP in my head (and on paper), something doesn't quite make sense. In a P-N Junction applying a forward bias reduces the junction (more or less), and you can overcome the junction potential and you get electron drift:

However, consider the following PNP junction:

The PN junction (Collector to Base) is just a diode, so as long as you overcome the junction field, all is good. But the I don't understand how the NP junction on the right works. I recognize that it's not _quite_ reverse biased, since the N layer is hooked to ground (good), but the P is also hooked to ground. Why would that cause electron drift through the NP junction and hole flow out the Emitter?

2. WBahn Moderator

Mar 31, 2012
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3. sjgallagher2 Active Member

Feb 6, 2013
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Currents can either be hole currents or electron currents - equivalent, but opposite in direction. In an n-type semiconductor, electrons are in abundance, and are called the majority charge carrier. Charge carrier meaning a hole, electron, ion, or any other electrically charged particle that can be current. In p-type semiconductor, holes are the majority charge carrier. Conversely, holes are the minority charge carrier in n-type semiconductor and electrons are the minority charge carrier in p-type semiconductor. This distinction is important because, though n-type regions carry primarily electron currents, they also carry hole currents, or minority charge carrier currents.

For example, let's say a chunk of n-type semiconductor has a million electrons and 100 holes. Holes are the minority, but applying an electric field will still cause those 100 holes to flow. It just won't be noticeable when millions of electrons are flowing at the same time.

Bipolar junction transistors are what's called minority charge carrier devices. They actually rely on minority charge carriers being driven into the base from the emitter, and then drifting into the collector over the reverse biased junction.

So let's take it through the PNP transistor.

1. The emitter is P-type. Its majority charge carrier is holes. Holes drift into the emitter, and drift across the BE junction into the base.
2. The base is N-type. Its majority charge carrier is electrons. However, holes are flowing in from the emitter. Holes from the emitter are now minority charge carriers.
3. The collector is P-type, its majority charge carrier is holes. The NP junction from base to collector is reverse biased, and no electrons can flow across the junction.

That is why it's important that the minority charge carriers are the main current in the base. In a normal PN junction, the minority carrier current is a leakage current, due to the fact that there will always be both electrons and holes in a region. A large depletion region will not allow the majority charge carrier to flow from N to P, but the minority carrier can flow just fine.

Note: With many electrons in the base by nature of it being N-type, a current will also flow from base to emitter. We want to minimize that, so the base is made quite thin. A recombination current (a current of electrons and holes going towards each other, effectively a current in one direction) will also occur in the base, as electrons recombine with holes from the emitter.

In the image you give, the transistor is not properly biased for "transistor action", it's connected in a diode configuration. At least, it is assuming the left N region is the emitter and the right N region is the collector. The principle is still the same.

4. bryancostanich Thread Starter New Member

Feb 10, 2018
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@sjgallagher2; why do you say that the transistor in the image above isn't properly biased for transistor action? are you saying that I've got the C * E swapped? in the illustration, I was using C on left, E on right (though unlabeled). Is that not an ok thing to do?

5. sjgallagher2 Active Member

Feb 6, 2013
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If you intended the left side as the collector and the right as the emitter then the transistor certainly is going to work poorly. A PNP transistor should have the emitter at a higher voltage (by 0.6V) than the base, and the base at a higher voltage (or lower by a max of 0.6V) than the collector. You have this backwards if you intend to have the collector on the left. In your drawing it makes no difference, because all regions are the same size, and doped roughly equally. This would however be a very poor transistor, due to excess base drive current among other factors. A real transistor is properly proportioned for forward activity, though you can still drive it backwards (as shown). Itll just work poorly.

Even still, the base and emitter are shorted together, preventing the EB junction from being biased.

Id suggest doing some research on transistor biasing!

6. bryancostanich Thread Starter New Member

Feb 10, 2018
13
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Ah, right. I just forgot to put the second voltage source. I was focused on trying to figure out the internals, which made no sense as they had been explained to me.

In any case, I read the link that @WBahn posted and it explained it well.