Transistor amplification

[Idoubt]

Sorry this seems to be the wrong forum. I've posted this in the electrical engineering forum.Im studying basic electronics and transistor amplification is confusing me.

Now I understand the basics of a npn\pnp transistor but here's what I don't get

What I've read, says that the amplification occurs because the emitter base jn is forward biased and hence low resistance leading to a low potential drop across it and the collector-base jn is reverse biased and has a high resistance and hence a higher pd across it.

Suppose we are looking at an npn transistor,
The electrons from the emitter enter the base with little resistance and hence only a low voltage need be applied.

Now since electrons are minority carriers in the base they can pass through the reverse biased collector-base jn and there is a high current through the collector

some electrons recombine and is lost a base current

My question is how can the reverse biased c-b jn offer high resistance to electrons which are minority carriers in the base?
If the electrons can move from the base to collector with ease wouldn't that mean that there would be less of an energy loss and hence consequently only a small pd?

 

[Valerie Park]

I don't understand how the reverse biased c-b jn is offering high resistance. The answer to your question lies in the concept of minority carrier injection. In a reverse biased diode, the electric field at the junction points to the anode side (in this case, the collector side of the transistor). This field pushes the minority carriers away from the junction and thus creates a depletion region as a result. As the electrons pass through this region, they experience a potential barrier which results in a higher resistance, leading to a higher voltage drop across the junction. So, in summary, the reverse biased junction offers high resistance to the electrons due to the increasing potential barrier that it creates.