In bipolar transistor, what's the real reason for base to become narrow?

[ThinkerCorny]

I'm kind of new in this.
Now, I'm especially talking about npn common emmiter configuration.
I've read somewhere that base is made narrower to stop the recombination of minority charge with holes in base. And I assume that if base would broad then electron combine with holes creating strong electric field which repel further electron to flowing from base to collector while this field is weak in narrow base.
But if narrow base results in weak electric field, then is only connection of battery between emitter and collector result in flow of current.
So what's the real reason behind making of narrow base?
I know I'm messing it all but I really need help with this.
Thanks in advance!

 

[LvW]

Here you`ll find a very good desription how a BJT works as a voltage controlled current source Ic=f(Vbe):.:
http://www.eecs.berkeley.edu/~hu/Chenming-Hu_ch8.pdf

Regarding your question - read page 298.

 

[ThinkerCorny]

That's too complicated. There's lot of mathematics in there. I want simple intuitive explanation.

 

[cnh1995]

That's too complicated. There's lot of mathematics in there. I want simple explanation here.

Most of the electrons injected by the emitter are 'collected' by the collector. To make this happen, recombination should be minimized in the base. This is why the base is thin and lightly doped.

 

[ThinkerCorny]

Most of the electrons injected by the emitter are 'collected' by the collector. To make this happen, recombination should be minimized in the base. This is why the base is thin and lightly doped.

I'm trying to say that if less recombination occur in base so there would be narrow depletion region between base emitter. As base collector is reverse biased though has very narrow depletion region so the direct connection of battery with collector and emitter should result in flow of current. But it's not the case. Why?

 

[LvW]

Forget all the recombination stuff (for a simple explanation).
* Start with the principle function of a pn junction (diode):
* A base-emitter voltage of app. Vbe=0.65...07 volts across the pn junction releases electrons in the n-doped emitter.
* This would cause a corresponding current between E and B - unless there is a larger positive voltage at the collector which causes the majority of electrons to cross the base region (electrons are attracted by the pos. collector voltage).
* This works best if the base region (the depletion region) is as small as possible.

 

[ThinkerCorny]

Forget all the recombination stuff (for a simple explanation).
* Start with the principle function of a pn junction (diode):
* A base-emitter voltage of app. Vbe=0.65...07 volts across the pn junction releases electrons in the n-doped emitter.
* This would cause a corresponding current between E and B - unless there is a larger positive voltage at the collector which causes the majority of electrons to cross the base region (electrons are attracted by the pos. collector voltage).
* This works best if the base region (the depletion region) is as small as possible.

I can totally understand that base is made narrow to make narrow depletion region but if base emitter battery is removed then can't due to strong electric field by emitter collector battery and narrow depletion region in reverse biased base collector diode(I'm breaking transistor into two diodes) should conduct electric current but this doesn't happen. Why?

 

[cnh1995]

so the direct connection of battery with collector and emitter should result in flow of current.

The depletion region between collector and base is bigger compared to that between base and emitter. So when you remove the BE junction battery, emitter can't inject electrons because of the reverse biased B-C junction.

To make the emitter inject electrons in the base, BE junction diode should be forward biased. As LvW said earlier, it is a voltage controlled current source.

You can see how collector current varies with base current (which is a function of B-E voltage).

 

[cabraham]

The bjt is current controlled. The 0.65 volt b-e junction forward DROP does NOT "release" electrons. The emitter is n type silicon (for an npn device), while the base is p type. The p type base has an abundance of free holes, as they are the majority carrier. The n type emitter has an abundance of free electrons, as they are the majority carrier. The 0.65 volt b-e voltage drop is a consequence after the emitter current has transited from emitter, through the base, then arrive at collector due to attraction via collector base E field.

An external current or voltage source, can forward bias the b-e junction. When power is switched on, the b-e barrier voltage is around 25.7 mV at temp of 25 C. As soon as electrons transit through the conductors & arrive at the base & emitter terminals, they easily conduct through the base & emitter silicon material. Although the Vbe is a measly 25.7mV, holes easily move through the p type base, while electrons easily move through the n type emitter.

As the electrons emitted approach the collector, the electric field from base to collector attracts these electrons, as a result, nearly all electrons emitted transit through the base & reach the collector/ One or two electrons out of 10,000 emitted, end up recombining in the base with a hole.

The base emits holes as well. These holes cross the b-e junction & recombine with electrons headed towards the base. If 10,100 electrons enter the emitter terminal, around 99 of those electrons recombine at the edge of the b-e depletion region on the emitter side. The remaining 10,001 electrons continue & enter the base region, where 1 of them recombines in the base region with a hole, & the other 10,000 reach the collector.

The objective in fabricating a bjt is to obtain as much current gain as possible, without incurring undesirable consequences. The device described above has a forward current gain, beta, or "hfe", of 100 in value. For every 10,100 electrons entering the emitter terminal, 100 electrons exit the base terminal, & 10,000 electrons exit the collector terminal. Thus collector current is 100 times base current, for a beta of 100. Collector current is 100/101 of emitter current. This ratio is "alpha". It is desirable that alpha be as close to unity as possible.

If the base is wide, 2 undesirable things happen. First, the number of holes emitted by the base is large. An external power source (microphone, antenna, battery, photodiode, etc.) biases the b-e junction plus connected devices (resistors). The b-e junction electric field imparts motion to free electrons in emitter, & free holes in base. A large volume of base material puts more holes into conduction. The volume equals the base length times width times height. By making the width very narrow, we get a small volume of free holes being yanked into conduction by the external source E field. Also, light doping is used in the base. Heavy emitter doping puts lots of electrons in conduction towards the base & collector. Light base doping puts few holes in conduction towards the emitter.

Also, the emitted electrons from the emitter incur fewer recombinations with a narrow base region. If the electrons must travel further due to a thicker base, more recombinations occur. Both of these reduce current gain. Current from base to emitter in the form of holes is called "injection current". Current in the form of electrons recombining in the base is called "transport current". These are 2 components of the overall "base current", the injection component being much larger. To maximize current gain, we must lightly dope the base, & make it thin.

For a bjt in the common emitter or common base configuration, this base current does not power the load (load is in the collector side), so it is often regarded as "lost", a quantity that needs to be minimized. So a bjt is fabricated with base current minimization in mind, to preserve high current gain. But it is also desirable to prevent "punch through". The base-collector voltage breakdown rating needs to be sufficient for the device in the application used. If the supply can reach 30 volts, the bjt must not break down at or below that value. Some devices must withstand hundreds of volts Vce. In order to insure no break down, we need a thicker base, and/or heavier doping of acceptor atoms.

But both of these decrease current gain, which is the price paid. Examining data sheets will reveal that the higher the collector-emitter breakdown voltage, the lower the current gain. It is a classic tradeoff.

One more thing, with emitter follower configuration, the load is on the emitter side. So base current actually powers the load, & is not "lost". But, if the base is driven by a signal source with very limited current capability, it is still desirable to have high current gain. In some cases, that may not be the case. A linear voltage regulator powers the base & collector terminals from the input supply, so that high base current du to low beta is not an issue. Again, if the base is driven by a high impedance source with small current capability, high beta is needed.

Feedback/comments/questions welcome. Best regards.

Claude Abraham