Hey all,

So I know how a basic NPN transistor works and how it amplifies, but the one thing I don't understand is WHY a base current is needed. I've seen all over the internet and in videos that a small change in the much smaller base current results in a greater change in the bigger collector current, and that you need 0.7V across the emitter-base junction (for silicon) to activate flow from emitter to collector, but, WHY?

If, for example, you already have 12V across the emitter/collector, why do you need an extra 0.7V to push the electrons across the base? Shouldn't they be able to just jump across the base since there is already way over 0.7V being supplied? I get that there is the forward bias to cross and I understand that the base/collector junction is reverse biased but I just don't understand why that extra base/emitter voltage is needed. Shouldn't the emitter/collector voltage also apply to the base automatically? As in, if there is 12V across the emitter/collector, shouldn't there therefore be 12V also across the base already?

A detailed answer would be appreciated... Thank you kindly!

 

In a transistor with zero volts across the base-emitter junction there is a potential energy barrier that prevents charge carriers from crossing the junction. Actually it minimizes the probability that a charge carrier will have enough kinetic energy to cross the potential energy barrier. The bias voltage on the base lowers that potential energy barrier so that charge carriers can move freely across the junction. Actually it raises the probability that almost any charge carrier will have enough kinetic energy to cross the potential energy barrier.

 

In a transistor with zero volts across the base-emitter junction there is a potential energy barrier that prevents charge carriers from crossing the junction. The bias voltage on the base lowers that potential energy barrier so that charge carriers can move freely across the junction.

Yes I understand that, you are referring to the forward bias voltage of the junction. But why doesn't the voltage across the emitter/collector apply to the base as well?

 

Yes I understand that, you are referring to the forward bias voltage of the junction. But why doesn't the voltage across the emitter/collector apply to the base as well?

There are actually two junctions in the transistor base-emitter and collector-base. The base region is very narrow and if you look at the node voltages of the collector, base, and emitter; you will note that a transistor in saturation has approximately 0.2V across the collector emitter junction. What is happening is that the charge carriers have enough kinetic energy to approach the first junction, move quickly through the base and emerge on the other side with energy to spare.

 

There are actually two junctions in the transistor base-emitter and collector-base. The base region is very narrow and if you look at the node voltages of the collector, base, and emitter; you will note that a transistor in saturation has approximately 0.2V across the collector emitter junction. What is happening is that the charge carriers have enough kinetic energy to approach the first junction, move quickly through the base and emerge on the other side with energy to spare.

So then if the electrons have enough energy, why do they need extra across the emitter/base junction to push them through?

(Thank you for your persistent replies haha)

 

So then if the electrons have enough energy, why do they need extra across the emitter/base junction to push them through?

(Thank you for your persistent replies haha)

Without the potential of 0.7V on the base the charge carriers have a low probability of having enough kinetic energy to cross the potential energy barrier. They don't get any "extra energy" and the don't need it because the potential energy barrier is lowered. Once the barrier is lowered, they have enough kinetic energy to approach the barrier, shoot through the base, and come out on the other side with something left over. No extra energy here.

 

Without the potential of 0.7V on the base the charge carriers have a low probability of having enough kinetic energy to cross the potential energy barrier. They don't get any "extra energy" and the don't need it because the potential energy barrier is lowered. Once the barrier is lowered, they have enough kinetic energy to approach the barrier, shoot through the base, and come out on the other side with something left over. No extra energy here.

Hmmm I think I'm just going to have to trust you. I just don't quite comprehend how (for example) a supplied voltage of 12V across the emitter/collector doesn't lower the barrier already. Wouldn't the kinetic energy already be enough to shoot through the potential energy barrier?

Thank you for your help

 

It might be better to stop thinking about current and look at charge concentration, electron and hole pairs.

An N-type semiconductor has impurity atoms that donate extra electrons, while a P-type semiconductor lacks electrons (supplies holes).

When you put a P-type and N-type semiconductor together, the electrons and holes recombine at the junction and thus creates a shortage of charge, i.e. a deletion region is formed.

There are two depletion regions, one at the collector-base, N-P junction, and another at the base-emitter, P-N junction.

When you apply a positive potential difference across the collector-emitter terminals of an N-P-N transistor, you pull the electrons towards the collector and push the holes towards the base. Hence you increase the depletion region at the collector-base junction.

The base-emitter P-N junction behaves like an ordinary P-N diode. By applying a positive potential difference across the P-N junction, you forward bias the junction and increase charge mobility.

The base region is very thin. Electrons being pulled from the emitter towards the base are swept across the thin base region and cross the base-collector depletion region. Thus collector-emitter current is created. Without the base-emitter bias, the base-emitter depletion region prevents flow of charge across the collector and emitter.

For further reading, check out this AAC tutorial on transistor theory:
https://www.allaboutcircuits.com/textbook/semiconductors/chpt-2/bipolar-junction-transistors/

 

Hey all,

So I know how a basic NPN transistor works and how it amplifies, but the one thing I don't understand is WHY a base current is needed. I've seen all over the internet and in videos that a small change in the much smaller base current results in a greater change in the bigger collector current, and that you need 0.7V across the emitter-base junction (for silicon) to activate flow from emitter to collector, but, WHY?

If, for example, you already have 12V across the emitter/collector, why do you need an extra 0.7V to push the electrons across the base? Shouldn't they be able to just jump across the base since there is already way over 0.7V being supplied? I get that there is the forward bias to cross and I understand that the base/collector junction is reverse biased but I just don't understand why that extra base/emitter voltage is needed. Shouldn't the emitter/collector voltage also apply to the base automatically? As in, if there is 12V across the emitter/collector, shouldn't there therefore be 12V also across the base already?

A detailed answer would be appreciated... Thank you kindly!

The typical drawing of the transistors as three stacked block is misleading.

 

A transistor will conduct with no base current if you exceed the collector-emitter breakdown voltage.

 

Transistors don't need a base voltage; they can lay in a box without a base voltage for years and still be OK. You need to apply the correct base voltage to a transistor to get the function you want.

 

I just don't quite comprehend how (for example) a supplied voltage of 12V across the emitter/collector doesn't lower the barrier already. Wouldn't the kinetic energy already be enough to shoot through the potential energy barrier?

Only if you apply enough voltage to break down the reverse-biased collector-base junction barrier.
Otherwise there is only potential energy, no kinetic energy.

When collector-emitter voltage is applied, there is only a tiny leakage current through the reverse-biased collector-base junction, which is not enough to forward-bias the base-emitter junction.
You have to supply voltage/current to the base-emitter junction so that the carriers can cross this reverse-biased junction.

If the transistor didn't need any base voltage or current to operate then I don't understand how you think it would work to provide gain?

 

With no connection to the base, a bipolar junction transistor is like two diodes back to back. Does current flow when you connect the cathode of one diode to 12V, the anode of that diode to the anode of another diode and the cathode of the second diode to 0V? That is basically what you have when you put 12V at the collector and 0V at the emitter of an NPN transistor. Putting current the the base emitter junction modifies things because it shares a layer (the base) with the collector base junction. The current modifies the behaviour in such a way that allows a much bigger current to flow from the collector to the emitter.

Bob