Do transistors *only* work with small current in base?

[cmkluza]

Quick basic question on how transistors work. Do they only work when a small current goes through the base, or would they function the same with a large current as well?

 

[LvW]

Small? Large? µA or mA ?
In the linear operating range the base current always is a certain portion of the collector current.
That`s all I can say.

 

[anorlunda]

Transistors come in many sizes. When I worked at ASEA in Sweden, we made some the size of large pizzas for use in power electronics.

 

[SherLOCKed]

If you generalize the word transistor then there are two classifications of transistor depending on the power ranges they operate: signal level transistor and power transistors.
For power transistor I( base) ranges can be in Amperes when compared to signal level with milli Ampere ratings.

'Function the same'- does it imply type of operations? While we could use signal level transistors for switching and amplification application, we don't use power transistor for amplification purposes because we already have a large signal which can be easily studied. Therefore we use it only for switching purposes basically high frequency switching(cut off and saturation)

Hope it helped.

 

[Tom.G]

Transistors have a limit on the current they can handle that is based on their temperature. Get them too hot and they melt internally. Their temperature is determined by the amount of current, their size, and how well the heat is removed.

If you want to use them as an amplifier then you must limit the base current to the range of linear operation. That is, the operating region where the collector current is proportional to the base current.

If the base current is greater than that, then the collector current increases by a very small amount. If you want to operate a transistor as a switch (as is done in a switching power supply), increasing the base current, to, say, 1/10 of the collector current, often reduces the power dissipation in the transistor; thereby allowing it to switch a higher current before overheating.

 

[meBigGuy]

Not sure what you are trying to understand. Do you have ANY idea how a transistor works? From the question, I would guess not.
Perhaps some basic study is in order.

Once the transistor turns on completely (saturates) increasing the base current has little effect. IS it still operating? Depends on your definition of operating.

 

[cmkluza]

Not sure what you are trying to understand. Do you have ANY idea how a transistor works? From the question, I would guess not.
Perhaps some basic study is in order.

Once the transistor turns on completely (saturates) increasing the base current has little effect. IS it still operating? Depends on your definition of operating.

Thanks for the response!

I definitely do need to spend more time studying the basics of transistors (as well as a bunch of other components for that matter). My question stems from my seeing almost every description of a transistor specifically reference that (when working as a switch) it requires a small input current on the base to allow current through the collector and emitter, so I became curious as to how it might work with a larger current. In retrospect I realize the descriptions "small" and "large" are pretty subjective, and my question is most likely worded incorrectly and/or based on a faulty idea of transistor functionality.

 

[meBigGuy]

A transistor has the concept of beta, which the the ratio of base current to collector current. For example, a transistor might have a Beta of 100 at Ic=10ma. That means that 100uA of base current will cause 10ma of collector current. Beta is not constant though, and at Ic=100ma it may only be a beta of 50. So, 2ma of base current is required to get 100ma of collector current. As you drive the base harder, you reach a point there the transistor is saturated. I'll let you read about saturation.

You can look at a few transistor data sheets and see this specified. Two interesting common transistors are the 2n3904 (or MPS3904) and the 2N2222 (or MPS2222).
You can look at the Beta values and see they are optimized differently.

 

[Jeff Rosenbury]

If the base current times the beta is high enough the transistor will act a lot like a diode. The collector will basically be shorted, but depending on the circuit that may not be a problem. (Though it is in every actual application I've seen, so it's a pedantic point.) However unlike a diode, the emitter base junction is usually not built to support much current.

So one could theoretically do this, but I've never seen it done. Transistors are usually used as amplifiers or switches, not diodes.

 

[rbelli1]

Also please realize that a bipolar junction transistor is a voltage controlled device. Current transfer is a simplification.

I only recently learned this so I can't be any more helpful on this concept. The more knowledgeable will be a of greater assistance.

BoB

 

[meBigGuy]

The simplified Ebers-Moll model explains the active region.
https://en.wikipedia.org/wiki/Bipolar_junction_transistor#Ebers.E2.80.93Moll_model

Ie = Ies (e^(Vbe/Vt) -1) where Ies and Vt are internal characteristics. So, basically, the emitter current is an exponential function of the base voltage. Base current is a price you pay.

But it still works out that Ic = Beta* Ib

 

[rbelli1]

I agree the simplification is useful. Many (most?) transistor designs are based on current. It makes the design process easier. However understanding the details leads to a better understanding of the materials you are working with.

BoB

 

[LvW]

I agree the simplification is useful. Many (most?) transistor designs are based on current. It makes the design process easier. However understanding the details leads to a better understanding of the materials you are working with.
BoB

Sorry, but I disagree.
May I ask you: Which step within the design process is "easier"? For my opinion, this is simply a "misbelief".
Using the relation IB=IC/beta is not a "simplification", because it is a fact that a base current does exist. However, some people think that they are using a "current-control" model during calculation of transistor based gain stages. But that`s not true. We are all using the same set of formulas - and, of course, also the "voltage-control" party takes into account that there is a finite IB. However, this does not automatically mean that IB controls IC. Each calculation starts with a suitable value for VBE - and it is the purpose of DC feeedback to reduce the influence of this VBE uncertainty upon the desired DC operational point.
This is even true for bias point calculkation in case of "current injection" into the base (using a single resistor RB instead of a voltage divider): The resistor value for RB is calculated using again, of course, a value of VBE=0.65...0.7 volts. We are using the term "current injection" - however, in fact we are designing a voltage divider between RB and the B-E branch, which produces the desired VBE value.
Finally: There are many circuits and observable effects which can be explained with voltage-control only!.

 

[rbelli1]

After some more reading I seem to have a flaw in my understanding here. I will have to do more research and make some smoke.

BoB

 

[meBigGuy]

We had a long ranting thread about this a while back. Basically, for the most part small signal analog ASIC designers think in terms of transconductance (gm). When I look at an LED driver, I think in terms of the base current to drive it into saturation and assume a constant Vbe. So, both methods are useful depending on your task and required accuracy.

Once upon a time I posted: https://www.physicsforums.com/threa...tor-amplify-thread.709791/page-2#post-4502095
" I've been talking to the analog IC guys here and the consensus view is that, for small signal, gm rules, and for large signal, consider Vbe constant and Beta rules. (which isn't so different from what is being said)."

Also:
One of the guys I talked to made the observation that if you think about it in terms of current, you have to deal with charging depletion regions, etc until the voltage "appears", where as if you just "put" the voltage at the junction, its done.

 

[LvW]

Yes, I should add that my comment applies primarily to small signal linear operation