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In synopsis, little increase in the base terminal will result in a large change in the collector terminal. If you input uer "small waveform" to the base terminal, u will have a "big waveform" at the output!

We kan bias the BJT using a DC voltage supply and resistors(most commonly and practically as potential divider bias). Note that this biasing should be done in order to achieve the active region of the BJT(B-E junction forward bias and B-C junction reverse bias such that VCE>VBE). In the active region, IC= hFE.IBwhere hFE (assume it as 100)denotes DC current gain or Beta, IC denotes the current towards the common terminal(C) and IB denotes the current towards the base terminal(B). Therefore, IC= 100IB. Whenever you change IB, IC will be changed in 100 times! That's how a BJT transistor has been constructed by modifying the p-n junction doping. Also that's simply how it amplifies....

What if we input an a.c. waveform(mid-frequency about 10kHz) which is smaller in magnitude through the base terminal(i.e. with the help of a coupling capacitor) and take the output across C and the ground (i.e. with the help of a coupling capacitor)?

Applying an alternating input across B-Ground results in chaniging the base voltage regularly(accordingly base current will also be changed). As explained earlier, change in the base current, will change the emitter current enormously(accordingly emitter voltage, VC). Hence voltage across C-Ground will be change largely. So uer input has been amplified!! As I mentioned earlier this output should be separated away across a coupling capacitor.

Note that, this amplification is happened with the help of the supply voltage to the transistor. Supply voltage is the one who provides energy to increase the amplitude of your small signal.

However, to analyse a BJT amplifier, we may have to perform both D.C analysis and a.c. analysis seperately[which will generally be teached in the Bachelors level].

Regards,

-R-

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Q: What is the theory of operation of a BJT amplifier?
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