NPN transistor and how it operates....

[fog37]

Hello Everyone,
I would to ask you for a couple of clarifications if possible:

• I understand that a NPN transistor has 3 legs (emitter, base and collector) and that if a very small current goes from the base to the emitter (current $I_{be}$, the transistor will be turned ON and a larger current, called collector current, will go from the collector to the emitter. The transistor is essentially an electronic switch. That said, I would think that a NPN transistor is a current-controlled device since a nonzero, small (not too small) current must pass from the base to the emitter. The presence of a voltage $V_{be}$ between the base and the emitter is not enough to turn the transistor on. A current must exist On the other hand, FET transistors, instead, seem to only require the presence of a voltage and don't draw a current to turn the transistor on. Is that correct?

• A transistor, whatever it may be, can be used as a switch (completely ON or OFF) and also as an "amplifier". When used as an amplifier, it creates an magnified replica of a smaller input signal that is applied to the base. The transistor essentially modulates a large DC signal to give it the same shape as the smaller input signal, correct? The input signal, and corresponding amplified replica, can be either digital or analog, correct? Or do different types of a transistors deal with digital and analog signals? Also, when talking about signal amplification, are we talking about current signals or voltage signals? Or does that distinction not matter since current and voltage can always be related by Ohm's law $V=IR$?

Thanks,
fog37

 

[Averagesupernova]

There has been considerable discussion and disagreement on this forum about whether a bipolar junction transistor is a current or voltage controlled device.

 

[fog37]

Ok, thanks. I will dig out some information.

What is your personal opinion? Based on what I mention above, a BJT seems to be a current-controlled devices since no current means that the transistor will not be operating as an open switch...

 

[jim hardy]

Also, when talking about signal amplification, are we talking about current signals or voltage signals?

Voltage amplifiers are more common than current amplifiers

That question has generated volumes of discussion.
Since current and voltage are dependent on one another but not necessarily related by a linear proportion , arguments usually boil down to a chicken or egg conundrum.

The simple current controlled switch you described is a mental model that works well for everyday circuit analysis and is a good one to start with.
You should learn about biasing, current gain(Beta) , and study transistor amplifier circuits. A transistor switch circuit is just a transistor amplifier circuit that's overdriven and no longer in its linear range of operation. That range is set by values of the resistors and voltages the designer chose to use.
Start someplace like this. From a search on 'Transistor Basics'
http://www.pitt.edu/~qiw4/Academic/ME2082/Transistor Basics.pdf
or
http://www.ti.com/lit/an/sloa026a/sloa026a.pdf
Google all the terms you're uncertain.about because good understanding starts with good vocabulary.

Or does that distinction not matter since current and voltage can always be related by Ohm's law V=IR ?

You will with practice learn where you must use a more precise relation.
For example, the current through a forward biased junction such as the Base to Emitter of a small transistor is i = e^qV/kt
where
e is the math constant 2.718etcetera
q is charge of the electron
V is voltage across the junction
k is Boltzmann's constant
t is absolute temperature of the junction .

In everyday work that'll usually work out close enough to a constant that you can just use 0.6 for silicon and 0.3 for germanium, or even ignore it.

Not dodging your question, just trying to forewarn you that seemingly hard & fast rules always have exceptions.
In your basic transistor voltage amplifier the distinction usually doesn't matter and we analyse them assuming V=IR .

Here's an example of why you mustn't lock in on one analytical aproach:
Observe that since in a forward biased jubction i = e^qV/kt
ln(i) = qV/kt,
V= kt/q X ln(i)
since k and q are natural constants and t can be kept constant
we could write V = ln(i) X K[a constant] .
So, voltage across a diode tells you the logarithm of current through it. There are places where that's real handy.
Here's two tutorials on such logarithmic amplifuers.
http://www.ti.com/lit/an/slyt088/slyt088.pdf
http://www.ti.com/lit/an/snoa575b/snoa575b.pdf
A log current amp similar to those is at the heart of many reactor power measuring instruments. My nuke plant used them.

So dive in, work lots of examples , and soon enough schematic diagrams will become your second language.

old jim

 

[LvW]

Since current and voltage are dependent on one another but not necessarily related by a linear proportion , arguments usually boil down to a chicken or egg conundrum.
old jim

Hi Jim - I certainly do not intend to start the discussion again - however, one thing is clear: The answer to the question is NOT a "chicken or egg conundrum."
The background is simple: Each current is the RESULT of a voltage. And this applies, of course, also to the current-voltage relation across the base-emitter diode.
The discreapency has been solved - and it is really not too difficult to show that it is the VOLTAGE across the B-E junction that matters alone.

 

[dlgoff]

The discreapency has been solved - and it is really not too difficult to show that it is the VOLTAGE across the B-E junction that matters alone.

I don't think it's that simple. Please show.
You may find this interesting.
http://ecee.colorado.edu/~bart/book/book/contents.htm

 

[jim hardy]

The discreapency has been solved - and it is really not too difficult to show that it is the VOLTAGE across the B-E junction that matters alone.

Well,

Maybe it's less 'Chicken vs Egg' than ' Experimenter's choice of Dependent vs Independent variable'.
One can control either voltage or current and measure its effect on the other.
When we did this experiment in high school on a vacuum tube diode we adjusted current(independent variable) and measured resulting voltage(dependent variable) ..
old jim
.

 

[NascentOxygen]

That said, I would think that a NPN transistor is a current-controlled device since a nonzero, small (not too small) current must pass from the base to the emitter.

The current-controlled device is a popular model, and neatly matches the equation i_c = β.i_b

The presence of a voltage Vbe between the base and the emitter is not enough to turn the transistor on. A current must exist

I'm not sure why you would make this erroneous statement. If V_BE is controlled precisely along its characteristic logarithmic curve, the collector current is controlled by the base voltage (whilever operation is kept clear of the BJT's saturation region).

On the matter of amplification, it is the circuit it is built into that determines whether you see voltage amplification or current amplification, or both. The same device can be used in either circuit. Though if a transistor has properties which make it especially attractive to fast switching at low power, it will be marketed as a "switching transistor" but there is nothing stopping you using it in any circuit where it can comfortably operate within its voltage and power limitations.

 

[jim hardy]

You may find this interesting.
http://ecee.colorado.edu/~bart/book/book/contents.htm

OMG i remember that course, Fall of '65. The math has grown to the point i doubt i'd pass today..

 

[LvW]

I don't think it's that simple. Please show.
You may find this interesting.
http://ecee.colorado.edu/~bart/book/book/contents.htm

Thank you for the reference, which - however - was not new to me.
With respect to the subject of our discussion I like to refer to Fig. 5.3.3 of the Colorado-Univ. paper, which shows Shockleys well known exponential relationship between Vbe and the forward current If. That means: The model as dicussed in this paper is based on voltage-control.

 

[jim hardy]

Hi Jim - I certainly do not intend to start the discussion again - however, one thing is clear: The answer to the question is NOT a "chicken or egg conundrum."
The background is simple: Each current is the RESULT of a voltage. And this applies, of course, also to the current-voltage relation across the base-emitter diode.

One needs to be careful with such generalizations.
A beginning student will think you mean current is always the result of an externally applied voltage.

When we did this experiment in high school on a vacuum tube diode we adjusted current(independent variable) and measured resulting voltage(dependent variable) .

I assure you when we ran that experiment on the vacuum tube diode, current flowed in the absence of external voltage when we connected a microammeter anode to cathode. We had to apply external voltage to stop it.
Of course in a vacuum tube there's an external energy source, the cathode heater, to drive that current so no conservation laws are violated.

A PN junction lacks the energy source but still has contact potential , a depletion region, and when connected to an ammeter will reach zero current equilibrium . .
Voltage and current follow the well known equations but there's a lot more going on inside the device than simple volts = ohms X amps .
At the atomic level it's local electric fields that cause charge carriers to drift and they're not all of external origin.
So we must not lead beginners to think the semiconductor life forms he's studying are as simple as Ohm's Law.I maintain we're arguing semantics. If an experimenter forces current and measures volts, then from his point of view amps caused volts to appear. And vice versa.

So we must be careful to not plant any concepts that a student will have to un-learn when he studies semiconductor physics or encounters his first thermocouple.

That's all.

Anycomments from professional educators ?

old jim

 

[NascentOxygen]

Thanks to all the regular contributors. I think the thread can now be closed.