Основные вопросы об усилении транзисторов

17

Кто-нибудь может объяснить, как транзистор может усиливать напряжение или ток? По моему мнению, усиление означает: «Вы отправляете что-то маленькое, оно получается большим. Скажем, например, я хочу усилить звуковую волну. Я шепчу на усилитель звука, и получается, в 5 раз больше (в зависимости от коэффициента усиления)

Но когда я читаю о действии усиления транзистора , все учебники говорят, что, поскольку небольшое изменение базового тока ΔIb, но соответствующее большое изменение тока эмиттера ΔIe, происходит усиление. Но где усиление? Что усиливается, как я это определил? Мое понимание термина амплификации неверно? И как ток передается из зоны низкого сопротивления в зону высокого сопротивления?

Думаю, я понял, как построен транзистор и как протекают токи. Так может кто-нибудь объяснить объяснение действия усиления транзистора и связать его с тем, что я понимаю об усилении.

Зеленый нуб
источник
@ChrisStratton Вот вопрос о заземлении electronics.stackexchange.com/q/24598/7364
Грин Нуб
Вы спрашиваете себя, почему в книгах говорится об изменении базового тока вместо просто «тока»?
0x6d64
@ 0x6d64 Можете ли вы быть более сложным?
Зеленый Noob
Здесь есть несколько довольно плохих ответов. Много путаницы, усиливают ли транзисторы ток, усиливают ли транзисторы напряжение и т. Д.
Роди

Ответы:

18

Я начну сначала с определения усиления. В наиболее общем смысле усиление - это просто соотношение между двумя значениями. Это не означает, что выходное значение больше, чем входное значение (хотя это наиболее часто используемый способ). Также не важно, является ли текущее изменение большим или маленьким.

Теперь давайте перейдем к некоторым общим используемым значениям усиления:

Наиболее важным (и тот, о котором говорит ваш вопрос) является . Определяется как β = I cβ , гдеIc- ток, поступающий в коллектор, аIb- ток в базу. Если мы немного изменим формулу, получимIc=βIbβ=IcIbIcIbIc=βIb которая является наиболее часто используемой формулой. Из-за этой формулы некоторые люди говорят, что транзистор «усиливает» базовый ток.

Как это соотносится с током эмиттера? Ну, у нас также есть формула Когда мы объединяем эту формулу со второй формулой, мы получаем β I b + I b + I e = 0 . Из этого мы можем получить ток эмиттера как - I e = β I b + I b = I b ( β + 1 ) (обратите внимание, что I eIc+Ib+Ie=0βIb+Ib+Ie=0Ie=βIb+Ib=Ib(β+1)Ie ток идет в эмиттер, так что это отрицательно).

Из этого вы можете видеть, что, используя в качестве удобного инструмента в вычислениях, мы можем видеть взаимосвязь между базовым током транзистора и током эмиттера транзистора. Поскольку на практике β находится в диапазоне от сотен до тысяч, мы можем сказать, что «маленький» базовый ток «усиливается» в «большой» ток коллектора (который, в свою очередь, создает «большой» ток эмиттера). Обратите внимание, что я не говорил ни о каких дельтах до сих пор. Это потому, что транзистор как элемент не требует изменения тока. Вы можете просто подключить базу к постоянному постоянному току, и транзистор будет работать нормально. Если требуется изменение тока,ββ

Также используется другое значение, и его имя . Вот что это такое: α = I cα . Когда мы это переставляем, мы видим, чтоIc=αIe. Таким образом,α- это значение, на которое усиливается ток эмиттера, чтобы произвести ток коллектора. В этом случае усиление фактически дает нам меньшую выходную мощность (хотя на практикеαблизко к 1, что-то вроде 0,98 или выше), потому что, как мы знаем, ток эмиттера, выходящий из транзистора, является суммой тока базы и ток коллектора, который идет в транзистор.α=IcIeIc=αIeαα

Теперь я немного расскажу о том, как транзистор усиливает напряжение и ток. Секрет в том, что это не так. Усилитель напряжения или тока делает! Сам усилитель представляет собой более сложную схему, которая использует свойства транзистора. Он также имеет входной узел и выходной узел. Усиление напряжения представляет собой отношение напряжения между этими узлами . Усиление тока - это соотношение токов между этими двумя узлами:Ai=IoutAv=VoutVinAi=IoutIin. We also have power amplification which is the product of current and voltage amplification. Do note that the amplification can change depending on the nodes we chose to be input node and output node!

There are few more interesting values related to transistors which you can find here

Итак, подведем итог: у нас есть транзистор, который что-то делает. Чтобы безопасно использовать транзистор, мы должны быть в состоянии представить, что делает транзистор. Одним из способов представления процессов, происходящих в транзисторе, является использование термина «усиление». Таким образом, используя усиление, мы можем избежать реального понимания того, что происходит в транзисторе (если у вас есть какие-либо классы физики полупроводников, вы узнаете это там), и просто есть несколько уравнений, которые будут полезны для большого числа практических задач.

AndrejaKo
источник
Thanks a lot for answering my earlier questions. But can you tell me why the author has introduced a 5 Kohm resistance in series while explaining voltage 'amplification'?? & where did he get the 20 ohm input resistance? Link
Green Noob
Does't really answer where the amplification comes from.
rhody
@rhody Back when I was looking at the question, I determined that the main issue is the use of terminology and therefore provided a terminological answer. Since OP already had a referece about transistors, there was no need to go into details explaining what actually happens.
AndrejaKo
It is my understanding that amplification is when you increase the signal strength,which is based on the energy carried by the signal and measured in terms of Power(Watts). So an amplifier increases the power. A "voltage" amplifier boosts the signal voltage without lowering the current and this in turn boosts the output power. A "power" amplifier boosts both the AC voltage and the AC current so there is a significant power gain(more than a voltage amplifier).
Mr X
@Mr X I explicitly disagree with your understanding. Namely, we have the "amplification" as an abstraction tool in general, and then we have practical uses of this tool. I explicitly decided not to try to muddy the water in this answer by referring to the practical uses, because I believe that it's very useful to first understand the abstraction tool on its own.
AndrejaKo
7

Transistor does not amplify. Imagine sound waves hitting a microphone: what happens actually is that the sound signal does not pass into the microphone, but the microphone produces a signal corresponding to the sound signal; It is not the actual signal.

Remember that the actual signals in real world cannot be amplified or attenuated. Can you catch a sound or any other real world signal? No. They are as they are, we can only make a system which can work on the effect of the real world signal; sound waves hit on a microphone, light hits on a camera lens etc.

But when it comes to the case of a transistor, you apply an input signal to the base and you obtain a new signal corresponding to the input signal with greater amplitude in the collector. Keep in mind that this happens because a small change in the input side will correspond to a large change in the output side, due to the variation in the resistance. It is only an effect one to one. The output signal is totally a new signal of a grater amplitude, not the actual signal.

gopal
источник
This doesn't answer the question at all.
rhody
Actually the electrical wave and a considerable part of the charge carriers do pass from base to emmiter, hence, we could say the new signal is in part composed by the previous one. But this is quite phylosophical, when for us, signals are voltage levels, measurable, repeatable...
Brethlosze
5

The signal is being amplified. Depending on the design of the transistor amplifier the actual base current may or may not be part of the output current. Don't get hung up on a definition of amplification that requires every input electron to get larger and then pass to the output...

Ben Jackson
источник
Please explain?
Green Noob
@GreenNoob - most transistor amplifiers have bias currents that ensure the circuit is operating linearly. With just the bias currents present, it will be true that emitter current is greater than base current, but this isn't very interesting since these currents are just constants. The books speak about changes in current b/c the signals we ordinarily think of amplifying are imposed as fluctuations on top of the bias currents.
JustJeff
4

The working principle of a BJT (Bipolar Junction Transistor), which makes it a useful thing, is that it amplifies current. Throw a small current in, get a larger current out. The amplification factor is an important parameter of the transistor, and is called hFE. A general purpose transistor may have an hFE of 100, for instance, sometimes higher. Power transistors have to do it with less, like 20 to 30.
So if I inject a 1 mA current in the base of my general purpose NPN transistor I'll get 100 mA of collector current. That's amplification, right? Current amplification.

How about voltage amplification? Well, let's add a couple of resistors. Resistors are cheap, but if you want to make money you can try to sell them expensive by calling them "voltage-to-current converters" :-).

enter image description here

We've added a base resistor, which will cause a base current of

IB=VB0.7VRB

And we know that the collector current IC is a factor hFE higher, so

IC=hFE(VB0.7V)RB

Resistors are really great things, because next to "voltage-to-current converters" you an also use them as "current-to-voltage converters"! (we can charge even more for them!) Due to Ohm's Law:

VRL=RLIC

and since VC=VCCVRL

we get

VC=VCCRLhFE(VB0.7V)RB

or

VC=hFERLRBVB+(hFERLRB0.7V+VCC)

The term between the brackets is a constant which we're not interested in at the moment. The first term shows that VC is VB multiplied by some factor depending on three constants. Let's use concrete values: 100 for hFE, 10 kΩ for RB and 1 kΩ for RC. Then (again ignoring the constant factor)

VC=hFERLRBVB=1001kΩ10kΩVB=10VB

So the output voltage is 10 times the input voltage plus a constant bias. Looks like we can use the transistor for voltage amplification as well.

stevenvh
источник
1
In the strict physics sense, transistors do not amplify current, since even the bipolar transistor is controlled using the base emitter voltage, but I agree it's a convenient shorthand. amasci.com/amateur/transis.html
Mister Mystère
@MisterMystère: a bipolar transistor in common emitter is controlled by base current, not voltage. It's the base current that causes an X times larger collector current. You're wrong.
Joris Groosman
@JorisGroosman Ever heard of the textbook "Art of Electronics?" They teach bipolar transistors with voltage-input design philosophy, not current input. Author Win Hill specifically points out all the flaws in the hfe-based, current-input viewpoint, and shows how they're solved by seeing BJTs as voltage driven; ruled by the Ebers-Moll equation. He points out that current-input doesn't apply to diff amp, current mirror or cascode. Check out one of his forum responses about BJT voltage input versus current input: cr4.globalspec.com/comment/720374/Re-Voltage-vs-Current
wbeaty
@wbeaty: Yes, I know AoE. Odd thing: since the 1950s engineers have calculated collector current as a function of base current is a gazillion of practical applications, and they all work! Current as a function of base voltage probably doesn't go beyond the blackboard.
Joris Groosman
No,you don't know AOE, since they show why hfe DOESN'T work for analog design. Amps based in hfe will fail if temp drifts a couple of degrees. The authors push the voltage-based BJT design philosophy. As Win Hill points out, hfe doesn't explain voltage-input stages such as emitter-followers or diff amps. Op amps and their voltage inputs are hardly a blackboard-only concept. They worK, and are immune to vast changes in the hfe of the transistors involved. Yes, hfe is a useful concept, but without voltage-based signals and Ebers-Moll, a large part of modern analog design would fail.
wbeaty
3

Amplifiy sound, and you're amplifying the energy-flow: the input watts of sound become larger output watts.

Note that an electrical transformer doesn't amplify. It can step up voltage, but it cant increase the watts.

Transistors (and any sort of valve or switch) can amplify. They do it by using a tiny wattage to control a power supply which can output a huge wattage. The large output comes from the power supply, while the input signal is valving the transistpr on and off.

If you have a giant hydraulic press, you can crush cars by touching a valve switch with your little finger. The valve amplified your finger motion to mash Chevys. But actually it was the hundreds-HP haudraulic supply which provided the increased wattage. With NPNs, same idea. Transistors are valves for flowing charge instead of flowing haudraulic fluid.

wbeaty
источник
Nice explanation... To move it to the electrical domain, we can simply say the transistor is an "electrically-controlled resistor" inserted in series ("rheostat") or in parallel ("shunt") to the load. Thus it forms a voltage or current divider. To be more precise, we can only add that this "resistor" is non-linear, and it is controlled both by the side of the input source and the load. And also, the transistor is a passive, not active device (regarding the power).
Circuit fantasist
From this "energy viewpoint", the transistor does not amplify; contrary, it attenuates the power of the source... it does not produce energy; it consumes energy.
Circuit fantasist
reading all the answers of yours really helps me a lot, especially thanks to @wbeaty, your explaination is realllly nice!
Your car crush analogy is soooo much easier to understand than a water valve. Thanks!
dval
0

What is my understanding is that for a transistor to amplify you need to bias it properly. Forward biasing of BE junction makes it a conducting diode so input resistance is less. Reverse biasing CE junction makes it non conducting diode so output resistance is high. And if Ic is almost equal to Ie then the current causes a low voltage drop at input and large one at output. This is why its called an Amplifier.

Aditya
источник
0

With a transistor, you can achieve this: Give a small signal(ac) at input, and get a larger valued(higher amplitude) signal at output. But this is not all. You have to give DC supply at collector and base; emitter if required. This is called biasing the dc point. The rms power you get at the output will be less than the dc power you have supplied.

If you want to do analysis, there are two steps involved for any circuit.

  1. DC analysis: don't consider any ac signal. Find out the values of all diode currents based on dc voltage at various nodes(Collector, base , emitter). This is done by using KVL along various loops.

  2. AC model: Image has been taken from the book "Electronic Devices and Circuit Theory
    This makes very clear: what we draw as a circuit v/s what elements are actually present inside. Going further, the diode has forward resistance. So the actual model will be like this:

From DC analysis, you must have found the value of Ie. According to diode theory, Re = (26mV/Ie). Our aim is to find Vout/Vin.
1. Vout will depend on Ic.
2. Ic will depend on Ib.
3. Ib will depend on Vin and Re.
4. Re we have found from DC analysis.

enter image description here In AC analysis, we make all the DC supply to 0V. By looking at this, you can make out that the output signal will be an amplified one, right?

Note: This was just to give you an intuitive idea that amplification does take place. But whether you will get amplification or not depends on whether the transistor is in linear(amplifier), saturation or cut off(switch). Again, what will be amplified(current or voltage) depends on type of configuration. So that all comprises of 3-4 chapters of any standard book on analog theory.

explorer
источник