I know how we can amplify a given signal using BJT transistors by biasing them. But I would like to know what is that crux property which enables BJT transistor to act like an amplifier. Is it the constant nature of reverse saturation current or is it the definite relationship between the base and the collector current or anything else?

I am specifically talking about the BJT.

  • 6
    \$\begingroup\$ Have you read the wikipedia article on transistors? \$\endgroup\$
    – HandyHowie
    May 1 '19 at 11:12
  • 2
    \$\begingroup\$ Are you specifically thinking of the BJT as an amplifier or transistors in general (e.g. FETs)? \$\endgroup\$
    – edmz
    May 1 '19 at 11:23
  • 1
    \$\begingroup\$ @edmz i am specifically talking about the BJT \$\endgroup\$ May 1 '19 at 11:26
  • \$\begingroup\$ Here you can find a different view electronics.stackexchange.com/questions/355899/… \$\endgroup\$
    – G36
    May 1 '19 at 18:05

A transistor on its own does not make it an amplifier.

The transistor needs a circuit around it to do the actual (signal) amplification.

Depending on the circuit a transistor can amplify current changes and/or voltage changes and that means power amplification. Power amplification means that you need a smaller power to control or output a larger power.

In my opinion, the most basic property of a transistor which results in (power) amplification is the current relation between base current \$I_B\$ and collector current \$I_C\$. Their ratio is often referred to as \$\beta\$:

$$\beta = \frac{I_C}{I_B} $$

This \$\beta\$ is also quite "visible" in the actual transistor as it is linked to the ratio between the doping levels of the emitter and the base. The emitter will have the highest doping level, the base has a lower doping level (it could be \$\beta\$ times lower) and the collector will have the lowest doping level.

So if we increase the doping level of the base region, \$\beta\$ will increase and "amplification" goes up.

Does that mean I will always get a higher amplification if I use a transistor with a higher \$\beta\$?

No, it depends on the circuit you're using.

In some circuits indeed a higher \$\beta\$ will give you more amplification.

For example, a transistor controlling a relay. When \$\beta\$ is increased, we could use a smaller base current.

In others it will not give you more amplification.

For example, a Common Emitter amplifier, assuming we do not change the DC current \$I_C\$. In a CE amplifier, the voltage gain is \$gm*R_{load}\$. To get more gain we would need to increase \$gm\$ or \$R_{load}\$. Both can be done irrespective of \$\beta\$.


The crux property that allows anything to act as an amplifier is that it can control a high power signal, using a low power input.

In the case of a transistor, it's the fact that a low power base current or gate voltage can change a large collector or drain current.

There's a whole host of other devices that can be used as amplifiers. One of the earliest audio amplifiers used a diaphram microphone to modulate the tension of one end of a string wound round a rotating drum. The several turns of slipping string could control a large output tension which pulled a loud-speaking diaphram. Also look up fluidic amplifiers, magnetic amplifiers, Travelling Wave Tube Amplifiers (TWTAs).

  • 5
    \$\begingroup\$ Interesting history of amplifiers you've mentioned. Please consider adding to your answer with links to more information, as some may (myself included) want to read more about it.. \$\endgroup\$ May 1 '19 at 14:14
  • \$\begingroup\$ Interestingly, all the examples are transducers - which means: It is NOT a small quantity that controls a larger quantity OF THE SAME KIND. And - the same must apply also to the bipolar transistor: It is not (and cannot be) the small base current which controls the much larger collector current (how could 2 additional charged carriers in the base region enable the release of 500 additional carriers in the emitter, assuming a beta value of 250 ?). No - the BJT is a voltage-to-current transducer and it is the voltage Vbe that determines the collector current. \$\endgroup\$
    – LvW
    May 1 '19 at 15:49
  • 4
    \$\begingroup\$ @LvW With your assertion that base current doesn't control collector current, I am minded of Douglas Adams' philosophers who argue that black is white, and get killed on the next zebra crossing. I would be more sympathetic if you were to argue that gate current doesn't control drain current. I'd like to see you change base voltage without changing base current, and then argue which one is really controlling the collector current. \$\endgroup\$
    – Neil_UK
    May 1 '19 at 18:31
  • \$\begingroup\$ @Neil_UK, do you really still believe that the BJT is current-controlled? And - it is just a believe, because there is not a single proof for current-control. Can you explain WHY and HOW such a current control could work? In contrary, there are many proofs and indication for voltage control (and many good (!!) textbooks and university papers explain this fact (it is not my "assertion"). The famous Barrie Gilbert says that the unwanted base current would be "best viewed as a defect"! Do you need examples/indications for voltage-control? \$\endgroup\$
    – LvW
    May 2 '19 at 9:03
  • \$\begingroup\$ @LvW I'm with Gilbert that base current is unwanted, but it always tends to be there in BJTs. You can choose a gm (voltage controlled) or beta (current controlled) model for a transistor, both work well within their own range. When it comes to reality, what is really happening, then explanations that this is, or what your POV seems to be isn't controlling it don't make much sense. Physics doesn't deal with reality, even down in QM it's still equations. So I won't tell you it is current, if you don't tell me it isn't, and we'll let people still design with hybrid parameters. \$\endgroup\$
    – Neil_UK
    May 2 '19 at 10:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.