BJT NPN current flow

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

"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.)