Bipolar transistor inner workings

Thread Starter

Solenoid

Joined Mar 10, 2011
2
I've been reading some articles and transistor inner workings as my electronics course is pure math and I don't understand anything without atomic level explanation.

I found this nifty article on common-emitter amplifier from here and was curious about a line in it:
Remember that bipolar transistors are current-controlled devices: they regulate collector current based on the existence of base-to-emitter current, not base-to-emitter voltage.
Really? Another article says exactly the opposite:
Whenever voltage is applied between base and emitter, this insulating layer changes thickness. If (+)voltage is applied to the p-type (to the base wire,) while a (-) voltage polarity is applied to the n-type, (to the emitter wire,) then electrons in the n-type are pushed towards the holes in the p-type. The insulating layer becomes so thin that the clouds of electrons and holes start meeting and combining. A current therefore exists in the base/emitter circuit. But this current is not important to transistor action. What's important to notice is that the *VOLTAGE* across the base/emitter has caused the insulating Depletion Layer to become so thin that the charges can now flow across it.
(From: http://amasci.com/amateur/transis.html)

I'm kind of confused here, but the second article has a solid argument. Can anybody help me/correct something?
 

Wendy

Joined Mar 24, 2008
21,840
Welcome to AAC!

Feedback and Suggestions Forum for providing feedback and suggestions about All About Circuits, including corrections to the e-book. This forum is not for getting help with technical questions.
This really needs to be in the Electronic Chat area. A moderator will be along and move the thread shortly.

It is an old argument, and both sides have points. However, when you are modeling a transistor there is a very simple equation that comes extremely close to predicting what a transistor does.

Ic = β Ib

Where

Ic = Collector Emitter Current
β = Beta, the gain of the transistor found in the datasheet
Ib = Base Emitter Current

Because of this it is generally accepted that transistors are current controlled, and it works. There are other applications where the voltage theory help explain behavior, but they are very few and far between.

The voltage across a base emitter resembles a diode very closely. It starts around 0.6V for a silicon transistors (there are other materials that have different voltages, such as germanium). As the current goes up the voltage increases slightly, just like a diode. This voltage usually doesn't exceed 0.7V, but with power transistors that are handling many amps it can be up to 1.0V

One last point, an important one. While the Base Emitter is dropping voltage (0.6V), if the transistor is fully on (AKA, saturated) then the Collector Emitter is dropping considerably less. Some text books teach 0.2V, my personal experience is it is usually under 0.1V.

Hope this helps.
 

Adjuster

Joined Dec 26, 2010
2,147
Where transistors are driven from relatively large source impedances, it is often convenient and quite accurate to consider them as purely current-controlled. It is however important not to get stuck with the idea that the base-emitter voltage is invariant with signal conditions, so that voltage changes are not necessary to drive the transistor. This would have the implication that a single transistor amplifying stage has infinite voltage gain, which is certainly not the case.

There is a fairly strict relationship between a change VBE and the resulting change in emitter current. This can be used to model transistor amplification, particularly where voltage gain needs to be calculated. This may arise when the signal source has a low impedance, so that the base signal current is not independent of the transistor characteristics.
 

Wendy

Joined Mar 24, 2008
21,840
The idea of models is not that they are hyper accurate, though that is a worthy goal. It is that they can predict behaviors and be used to model with. The models need not be perfect to be useful.

There was a fierce debate a couple of years ago over just this issue. I had never heard of the voltage controlled model before, and have never had to use it, ever. It is good to know it exists, but it doesn't affect how you use the math.
 

Jaguarjoe

Joined Apr 7, 2010
767
One last point, an important one. While the Base Emitter is dropping voltage (0.6V), if the transistor is fully on (AKA, saturated) then the Collector Emitter is dropping considerably less. Some text books teach 0.2V, my personal experience is it is usually under 0.1V.

Hope this helps.
When the transistor is saturated and C-E voltage is very low, the B-C junction will be forward biased. This steals current from the B-E junction which causes the beta, or gain, to go to a very low value, typically 10.
 

Wendy

Joined Mar 24, 2008
21,840
Only for guaranteed saturation, and it is a rule of thumb, no more. I went decades not knowing this rule of thumb and having no problems designing and using BJTs. It is a useful rule to know, but not very pertinent to answering the OPs base question about voltage models vs current models.
 
Top