Let us discuss, how does a transistor work
Before going to the actual topic, let us review the concept of forward, and reverse biased
PN junction. Here, we connect the positive terminal of a DC source with the P-type side of the junction
We connect the negative terminal of that DC source with the N-type side of the junction
Adjacent to this junction edge, free electrons in the N-type material will recombine with
holes in the P-type material. After absorption of free electrons in N-type side of in depletion region, uncovered positive
donor ions will remain there. After absorption of holes in P-type side of in depletion region, uncovered negative of
acceptor ions will remain there. The current through the junction can only start flowing in the forward direction, when
we apply the DC voltage, more than forward barrier potential. We know that, forward barrier potential of a silicon PN junction is 0.7 volt
So, we have to apply forward voltage more than 0.7 volts in the case of silicon PN junction
This condition we call, the forward biased condition of the PN junction
Now, we come to the reverse biased condition. Here, we connect the positive terminal of a DC source with the N-type side of the junction
We connect the negative terminal of that DC source with the P-type side of the junction
Here, due to the attraction of positive potential of the source, the free electrons in the N-type
side will shift away from the junction, leaving positive ions there in the junction
In the same way, due to the attraction of negative potential of the source, the holes
in the P-type side will shift away from the junction leaving negative ions there in the junction
As a result, the width of the depletion region will increase significantly
Ideally, there is no more current flowing through the junction in the direction of applied voltage
This condition we call, reverse biased condition of the PN junction. But here, one interesting thing happens
We know that, there always some holes, present in the N-type side of the junction as minority
charge carriers. Also, there always few free electrons, present in the P-type side of the junction as minority
charge carriers. Due to electrostatic force of applied reverse voltage, the holes in the N-type side, and
free electrons in the P-type side will cross the junction easily. This is because, the barrier potential of reverse biased depletion region cannot abstract
the flow of free electrons from P-type side to N-type side, and holes from N-type side
to P-type side. This causes, a tiny current to flow through the PN junction even in reverse biased condition
We have discussed here this phenomenon, of reverse biased PN junction, because it is
quite essential for explaining, basic working principle of a bipolar junction transistor
Now, we will try to construct a semiconductor transistor. There are two types of semiconductor transistor
These are NPN transistor, and PNP transistor. For sake of understanding, we will draw an NPN transistor
There are two N-type regions, and one P-type region in between them
One N-type region is highly doped. The size of this region is moderate
We call this region as emitter. The middle P-type region is lightly doped
We make this region very narrow. We call this region as base
Other N-type region is moderately doped. The size of this region is largest among all regions
We call this region as collector. Now, let us first connect the positive terminal of a DC voltage source to the collector, and
negative terminal of same voltage source to the emitter. In this situation, the PN junction between collector, and base becomes reverse biased
The PN junction between base, and emitter is forward biased. Let us consider, the voltage applied between collector and emitter is VCE, and the voltage
drop between reverse biased collector base junction is VCB. The voltage drop across the forward biased base emitter junction is VBE
In this situation, the voltage VBE is fixed, and it equals to forward biased voltage of
PN junction. If, we consider it, a silicon transistor, the forward biased voltage of PN junction
would be 0.7 volt. Thus, VCE equals to VCB plus VBE which implies VCE equals to VCB plus 0.7 volt
That equation indicates the entire applied voltage between collector, and emitter except
0.7 volt drops across the reverse biased collector base junction. So, whatever may be the applied voltage between collector and emitter, there would always
be a potential barrier of 0.7 volt between emitter and base. Because of that, no free electron can cross the base emitter junction for any applied
collector emitter voltage. So no current can flow through the transistor. Now, we connect negative terminal of the collector emitter voltage source to the negative terminal
of another smaller DC voltage source. We connect the positive terminal of this second voltage source to the base region
After that the total scenario will change. If, we increase the voltage between base, and emitter from 0, after reaching the base
emitter voltage just above 0.7 volt, the free electrons start crossing the junction from
emitter to base region. As we told earlier, the emitter is highly doped
The number of free electrons available in this zone is quite huge
So a large number such free electrons cross the junction and appear in base region
But, the base is very lightly doped, so the number holes available in this region is quite low
So, maximum free electrons come to this region do not find hole to recombine
If you, about 2% to 5% of the free electrons recombine with available holes there
These electron hole recombinations contribute the base current, which circulates through base, emitter, and voltage source
Rest, 98% to 95% of free electrons cross the collector base junction due to their kinetic
energy, gained from electrostatic force of base emitter voltage. And the excess free electrons in the base region can easily cross the reverse biased
collector base junction. Because, these free electrons appear as minority carrier in the base region, and the reverse
biased collector base junction does not oppose the flow of these electrons to collector region
Now, if we increase the base emitter voltage slightly, the number of free electrons cross
the junction is also proportionately increased. The base current is still 2% to 5% of the total current
Here, we see that total electrons enter into emitter from source then base from emitter
are diverted into two paths. Very few of them are combined with holes in the base region and contribute base current
Maximum of them cross the base collector junction, and contribute collector current
So, emitter current is the sum of base current and collector current
The ratio of collector current to base current is known as current gain in common emitter
mode and it is denoted as beta. Therefore, B2 equals 2 I C divided by I B. This was the basic concept of a transistor
Hope you got the idea. Thank you