I am trying to understand how current flows in a BJT NPN and how amplification happens at the collector current.

Here is the link I was reading http://www.learningaboutelectronics.com/Articles/How-to-connect-a-transistor-in-a-circuit-for-amplification

This is the way I understand it: a Vcc supply to NPN Ic flows from Vcc to the collector but they also say amplification happens when minority and majority carries replace and diffuse from base into collector and a current flow comes which is amplified.

I cant understand how the current is amplified as per circuit laws. Vcc which gives the collector current which flows into the NPN collector. How is this current amplified? as it depends on the load resistance.

  • \$\begingroup\$ Welcome to the EE stack. Please rework your question and fix your spelling. Ensure the meaning of "ic" ("collector current" / "integrated circuit"). Please make the phrase "... as it depends on the load resistance really absurd" more clear. \$\endgroup\$ Commented Jan 31, 2017 at 20:34
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    \$\begingroup\$ don give bookish answers What does that mean ? There are plenty of questions and answers about how a BJT works on this site. Have you even looked at any of those ? You're also trying to understand too much all at the same time. First step for you to understand is why Ic is beta times larger than Ib. \$\endgroup\$ Commented Jan 31, 2017 at 20:35
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    \$\begingroup\$ It is also not clear if you're trying to understand how a BJT works (electrons, holes, PN junctions etc.) or simply how to use a BJT. You can use a BJT without understanding how it works as long as you just replicate the circuit others use. \$\endgroup\$ Commented Jan 31, 2017 at 20:38
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    \$\begingroup\$ The current flow through the BJT (Ic) is controlled by the size of the base current (Ib). Every BJT has a particular ratio between the size of this 'collector' current and the 'base' current which is called current gain (beta). Unless you really want to know the physics of the BJT forget about terms such as majority carriers, minority carriers, depletion layers etc. and work with terms such as gain, current, voltage and resistance. \$\endgroup\$ Commented Jan 31, 2017 at 20:59
  • \$\begingroup\$ Welcome to Electrical Engineering :-) It would help us to help you if you can tidy up your question as TCF asked and give some background about your situation (school/uni?). \$\endgroup\$
    – TonyM
    Commented Jan 31, 2017 at 21:02

3 Answers 3


It's not clear whether you're asking how a transistor works down at the physics level, or how it works in a circuit. From the apparent level of your question, you should start out with the latter. You don't actually need to know how a transistor does what it does on the inside to learn what it does from the outside world point of and to start using transistors in circuits.

Once you understand that, it is useful to have a basic grasp of the device physics, but there is a lot you can do with a simple circuit model. Also, this is the electrical engineering site. The internal workings of a transistor are better asked on the physics site.

Here is what a NPN BJT (bipolar junction transistor) does:

You can think of it as a adjustable valve that allows up to a certain amount of current to flow into the collector and out the emitter. That valve is controlled by the base current, which also comes out the emitter. The emitter current is therefore the base current plus the collector current.

So what's the big deal? The main useful property is that the collector current the valve allows is proportional to the base current, but much larger than it. The transistor has gain, which in this case is the ratio of collector current it will allow to the base current it takes to allow that collector current.

Small signal transistors easily have a gain of 50 or more. Large power transistors maybe 15-30 typically. There are some signal transistors that have guaranteed gains of several 100.

For example, let's say you have a transistor with a gain of 50. If you ground the emitter and inject 1 mA into the base, then the collector can now sink up to 50 mA.

A couple more wrinkles, and you can actually design useful circuits with these things:

  1. B-E the transistor looks like a diode to the circuit. That means for a normal silicon transistor, the B-E voltage will be in the 500-700 mV range when you're putting useful control currents thru the base. Just like a diode, the voltage is reasonably insensitive to the current, but there will always be some variation.

  2. When the valve is adjusted to allow for more current than is being supplied into the collector, you would expect the C-E voltage to go to 0. It doesn't quite. Figure about 200 mV for a small signal transistor in saturation (when it would allow more collector current than the circuit is giving it). Transistors can make decent switches for many applications, but some minimum on-voltage across the switch will always be there.

  • \$\begingroup\$ how do u take the collector current out of collector when it flows into the collector as per the circuit daigram in this link :electronicshub.org/npn-transistor \$\endgroup\$ Commented Jan 31, 2017 at 22:04
  • \$\begingroup\$ i don want the physics too ..i want it from a eee perspective as am a eee student \$\endgroup\$ Commented Jan 31, 2017 at 22:04
  • \$\begingroup\$ Can u explain how amplification is done and from where u tap the collector curent ....as per diagram when we connect vcc to the collector ic goes into the collector ...but if collector current is amplified it should be taken as output and not fed as input from vcc... \$\endgroup\$ Commented Feb 1, 2017 at 0:23
  • \$\begingroup\$ How will use npn as amplifier \$\endgroup\$ Commented Feb 1, 2017 at 0:26
  • \$\begingroup\$ @user: The collector current goes in the collector and out the emitter. See the diagram. Your other issues are getting too far afield for this question. Some of them might be appropriate for new questions here. But, you have to show some research and the answer can't be a book. "Explain the device physics of a transistor" is way too broad, for example. \$\endgroup\$ Commented Feb 1, 2017 at 11:39

I read that you are a EE and don't want to know about the physics. But knowing the physics is often the key to understanding the external behavior! In a bipolar junction transistor you can imagine the central base region as a hill that charge must climb to get from the emitter to the collector. Even though the collector-to-emitter voltage is positive, the hill prevents the electrons in the emitter from simply moving into the collector. That is why a NPN transistor with no base connection draws no appreciable current. A voltage between the base and emitter raises or lowers this hill. A positive Vbe lowers the hill in an NPN. The electrons in the emitter can now swarm over the hill. The effect is exponential - A linear change in base-emitter bias causes an exponential increase in emitter-collector current. Now, where does current gain come from? How come all of the electrons that enter the base region dont get caught by the positive bias on the base? Aren't electrons attracted to the positive base battery terminal? Answer: As the swarm of electrons move (diffuse) through the base region (move over the hill) some of them do get caught in the base and are collected by the base battery. But if the base region is manufactured to be very thin then most of the swarm gets over the hill and through the base without getting caught. This is the curious aspect of transistors. The base bias allows the electrons to diffuse, but the electrons diffuse so far before being caught that they are well into the collector before most of them could be caught (This distance is called the recombination length, and the base region is made much thinner than this length). So current gain is the ratio of the rate of electrons that make it across the base versus the rate that electrons are caught.


"Can u explain how amplification is done"

@user3252171 - let me try to answer your question. However, at first I must direct your attention to the fact that in the answers given up to now you will find two different explanations for the collector current control mechanism.

Hence, we need the answer to the question: Is the collector current Ic controlled by Ib or by Vbe ? Only one answer is possible. And the answer is: Ic=f(Vbe).

The collector current is controlled by Vbe and NOT by Ib. Ib is a result of the applied base-emitter voltage but it is neither the cause of Vbe nor Ic. On the other hand, we have a - more or less - fixed ratio between Ic and Ib (called current gain B resp. beta), but this is a simple correlation and has nothing to do with a control mechanism (causation).

Don`t get confused - I know that these two different (contradictory) explanations can be found also in textbooks. The reason may be that during calculation of transistor-based circuits these physical properties are sometimes overlooked. It is, however, surprising that all people make use, of course, from a bias voltage Vbe (in the range 0.65...0.7 volts). Nevertheless, some of them still think that this voltage would have no controlling function. A severe misconception - because there are many effects, circuit properties and working principles which can only be explained based on voltage control.

Now - the correct knowledge of the BJT`s control principle is the key to understand how the transistor can amplify: There is a transfer characteristics which relates the transistors input (Vbe) to the transistors output (Ic):

The classical equation formulated by the great W. Shockley: Ic=Io*exp[(Vbe/Vt)-1]. Now - asking for the small-signal gain we need the SLOPE of this function which is called transconductance gm=d(Ic)/d(Vbe).

This parameter gm is the gain determining parameter because it says how much the collector current will change its value as a result of an input voltage change d(Vbe)=d(Vin). If we use Ohms law (and a collector resistance Rc) to transfer the output current change d(Ic) into an output voltage change d(Vout)=-d(Ic)*Rc we have an amplifying stage with the gain A=d(Vout)/d(Vin)=-[d(Ic)*Rc]/d(Vin)=-gmRc.

This simple formula gives the gain for a non-stabilized stage. Because of several reasons (stability against temperature variations, parts tolerances,..) we normally make use of emitter degeneration (negative feedback) in form of an emitter resistance RE. In this case, the gain is A=-gm*Rc/(1+gmRE).

(The minus sign results from the fact that the ground referenced collector voltage is Vc=Vout=Vcc-Ic*Rc and d(Vc)=d(Vout)=-d(Ic)*Rc because d(Vcc)=0. Hence, we have amplification with phase inversion.)

  • \$\begingroup\$ There must be a "nice" forum member who is not able to contribute to the question - however, he always is able to confuse the questioner when downvoting a correct answer - without any attempt to give explanations/comments. Perhaps a case for FORUM ADMINISTRATION? Regarding BJT-related problems this happens not for the first time !! \$\endgroup\$
    – LvW
    Commented Feb 2, 2017 at 12:16
  • \$\begingroup\$ @user3252171, don`t be confused because of downvoting (-1). There is somebody who has some problems and is downvoting every BJT-related answer from my side. Just ignore it! \$\endgroup\$
    – LvW
    Commented Feb 2, 2017 at 12:20
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    \$\begingroup\$ I didn't downvote because your answer is technically correct, but I do think it confuses much more than helps, especially considering the apparent level the OP is at. You have a bug up your butt about BJTs being voltage-controlled devices. That may be so deep down at the semiconductor physics level, but is not a practical way to look at BTJs in most cases when designing circuits with them, especially for beginners that are still shaky with basic circuit design concepts. Take it to the physics site, or reserve your sermons for advanced discussions away from confusable newbies. \$\endgroup\$ Commented Feb 2, 2017 at 12:48
  • \$\begingroup\$ @Olin Lathrop, so you think that a correct answer "confuses much more than helps". The questioner was asking "how amplification is done" - and my explanation (in your view: "sermons", thank you) ended with a gain formula. Did YOU in your contribution explain how and why a BJT together with some resistors is able to perform voltage amplification? I wonder, if YOU are able to arrive at a gain expression (with explanation) WITHOUT using the transconductance gm. And the transconductance is an indication for voltage control only! That`s the logic behind it. I am very surprised about your view. \$\endgroup\$
    – LvW
    Commented Feb 2, 2017 at 12:59
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    \$\begingroup\$ That's all under the hood stuff largely not relevant to designing with BJT's. This is the electrical engineering site, not the physics site. Here we deal with circuit-level issues. Even if a BJT were strictly voltage-controlled, the low impedance and diode-like I-V characteristics of the B-E junction make actually controlling the voltage directly impractical. Then there is the fact that current control actually works over a wide range of operating conditions. Beta or hFE is a lot simpler than your gain equation, and controlling base current much simpler than controlling base voltage. \$\endgroup\$ Commented Feb 2, 2017 at 13:24

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