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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 , γ . The emitter injection efficiency is just the ratio of the electron current flowing in the emitter to the totalcurrent across the emitter base junction:

γ I e I Ee I 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

γ 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, α 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

α T 1 1 2 W B L e 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 D e τ r

Clearly, if the base is very narrow compared to the diffusion length, and since the electron concentration is falling off like 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 α 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 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" α ." Thus we say

I C α I E

Since the electron current into the base is just γ I E and α T of that current reaches the collector, we can write:

I C α I E α T γ 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 I C α I C

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

I C α 1 α I B β 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 α is a number close to (but still slightly less than) unity, then β which is just given by

β α 1 α
will be a fairly large number. Typical values for a will be on the order of 0.99 or greater, which puts β , 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.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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cm
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Introduction to physical electronics. OpenStax CNX. Sep 17, 2007 Download for free at http://cnx.org/content/col10114/1.4
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