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It is known that semiconductor transistor can be used as a switch as well as amplifier. How does the linear amplifying characteristic and non-linear switching characteristic come in a single system?

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    $\begingroup$ A transistor all by itself is not a linear amplifier, not even when it's being used as a current amplifier, even though the linearity in that case isn't all that bad. In reality it's the feedback that we add to the transistor in amplifier circuits that linearizes the non-linear circuit element. The most simple feedback is a simple emitter resistor. It lowers the gain, but at the same time leads to a much more linear response. $\endgroup$
    – CuriousOne
    Jul 11, 2015 at 8:50
  • $\begingroup$ Would Electrical Engineering be a better home for this question? $\endgroup$
    – Qmechanic
    Jul 11, 2015 at 15:21
  • $\begingroup$ @Qmechanic I too felt the same as above comment but I wanted to compare this non linear behaviour with that in a spin based transistor. Hence I posted in physics stackexchange $\endgroup$ Jul 12, 2015 at 11:56
  • $\begingroup$ @CuriousOne So is it safe to say that the exponential dependence of the carrier concentration on the chemical potential is the source for the intrinsic non-linear behaviour in a semiconductor transistor and that the linearity is forced onto it by feedback mechanism? $\endgroup$ Jul 12, 2015 at 11:58
  • $\begingroup$ Pretty much. One can prove in general that feedback linearizes non-linear elements (at least as long as their transfer function is monotonous). Most ultra-linear circuits (like op-amps) use very high gain in combination with very large feedback to achieve their performance. If you look at the open-loop small signal gain, it's usually atrocious. $\endgroup$
    – CuriousOne
    Jul 12, 2015 at 18:48

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I will restrict my answer to bipolar junction transistors. Let us take an npn transistor. Lots of electrons in the emitter, and if we forward bias the emitter-base junction they will head into the base. However, in the base they are minority carriers and are quickly gobbled up through recombination as the base attempts to keep $np = n_{i}^{2}$. None of the electrons make it across the base to be sucked into the collector to roam free in the circuit. The minority carrier lifetime is just too short.

Now however lets start injecting some electrons into the base. By providing more electrons you increase the minority carrier lifetime (you can delve deep into Shockly-Read-Hall recombination theory if you wish for more detail). With enough base current, the electrons injected from the emitter start having a chance of making it across the base and the transistor turns on. More base current, lifetime increases more, and more current flows.

This is gain - a little base current modifies carrier lifetimes and lots of current can make it across the base, avoiding being eaten by majority carriers. Near a given base current, small changes in the base current result in ~proportional changes in the collector current, and the proportionality is the small signal gain.

Now, as you keep adding more base current, the minority carrier lifetime keeps increasing, and the probability of an injected carrier making it across the base becomes ~1. More base current does nothing more - you have reached saturation - as much current as wants to makes it across.

So - for gain you need to be in an operating region where modulating the base current affects the minority carrier lifetime in the base. To switch, you drive the base current hard enough to go into saturation (in a Darlington pair one transistor is used to drive the second, making sure it will go into saturation with very little input current).

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