# 2.1 Transistor equations

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Transistor Equations

There are several "figures of merit" for the operation of the transistor. The first of these is called the emitter injection efficiency , $\gamma$ . The emitter injection efficiency is just the ratio of the electron current flowing in the emitter to the totalcurrent across the emitter base junction:

$\gamma =\frac{{I}_{e}}{{I}_{\mathrm{Ee}}+{I}_{\mathrm{Eh}}}$

If you go back and look at the diode equation you will note that the electron forward current across a junction is proportional to ${N}_{d}$ the doping on the n-side of the junction. Clearly the hole current will be proportional to ${N}_{a}$ , the acceptor doping on the p-side of the junction. Thus, atleast to first order

$\gamma =\frac{{N}_{{d}_{E}}}{{N}_{{d}_{E}}+{N}_{{a}_{B}}}$

(There are some other considerations which we are ignoring in obtaining this expression, but to first order, and for most"real" transistors, is a very good approximation.)

The second "figure of merit" is the base transport factor, ${\alpha }_{T}$ . The base transport factor tells us what fraction of the electroncurrent which is injected into the base actually makes it to collector junction. This turns out to be given, to a very goodapproximation, by the expression

${\alpha }_{T}=1-\frac{1}{2}\left(\frac{{W}_{B}}{{L}_{e}}\right)^{2}$

Where ${W}_{B}$ is the physical width of the base region, and ${L}_{e}$ is the electron diffusion length, defined in the electron diffusion length equation .

${L}_{e}=\sqrt{{D}_{e}{\tau }_{r}}$

Clearly, if the base is very narrow compared to the diffusion length, and since the electron concentration is falling off like $e^{\frac{-x}{{L}_{e}}}$ the shorter the base is compared to ${L}_{e}$ the greater the fraction of electrons who will actually make it across. We saw before that a typical value for ${L}_{e}$ might be on the order of 0.005 cm or 50 μm. In a typical bipolar transistor, the base width, ${W}_{B}$ is usually only a few μm and so $\alpha$ can be quite close to unity as well.

Looking back at this figure , it should be clear that, so long as the collector-base junction remains reverse-biased, the collectorcurrent ${I}_{\mathrm{Ce}}$ , will only depend on how much of the total emitter current actually gets collected by thereverse-biased base-collector junction. That is, the collector current IC is just some fraction of the total emitter current ${I}_{E}$ . We introduce yet one more constant which reflects the ratio between these two currents, and call it simply" $\alpha$ ." Thus we say

${I}_{C}=\alpha {I}_{E}$

Since the electron current into the base is just $\gamma {I}_{E}$ and ${\alpha }_{T}$ of that current reaches the collector, we can write:

${I}_{C}=\alpha {I}_{E}={\alpha }_{T}\gamma {I}_{E}$

Looking back at the structure of an npn bipolar transistor , we can use Kirchoff's current law for the transistor and say:

${I}_{C}+{I}_{B}={I}_{E}$
or
${I}_{B}={I}_{E}-{I}_{C}=\frac{{I}_{C}}{\alpha }-{I}_{C}$

This can be re-written to express ${I}_{C}$ in terms of ${I}_{B}$ as:

${I}_{C}=\frac{\alpha }{1-\alpha }{I}_{B}\equiv \beta {I}_{B}$

This is the fundamental operational equation for the bipolar equation. It says that the collector current is dependent onlyon the base current. Note that if $\alpha$ is a number close to (but still slightly less than) unity, then $\beta$ which is just given by

$\beta =\frac{\alpha }{1-\alpha }$
will be a fairly large number. Typical values for a will be on the order of 0.99 or greater, which puts $\beta$ , the current gain, at around 100 or more! This means that we can control, or amplifythe current going into the collector of the transistor with a current 100 times smaller going into the base. This all occursbecause the ratio of the collector current to the base current is fixed by the conditions across the emitter-base junction, andthe ratio of the two, ${I}_{C}$ to ${I}_{B}$ is always the same.

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