How Does Reverse Bias in Transistors Amplify Signals Despite High Resistance?

[Idoubt]

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?

 

[cabraham]

Gain occurs because charges emitted from emitter, transit through the base & are collected by the collector. The bjt works because of extreme proximity between the 2 junctions. If the base region weas very large, the device is just 2 diodes & the gain is so low it is not useful for amplifying.

Emission, proximity, collection, are the 3 events that describe bjt action.

 

[Idoubt]

Can you elaborate a bit? I've heard this but I don't understand the relevance of the proximity of the 2 ends

 

[Naty1]

It's analogous to traditional vacuum tube amplifiers...which may be easier to visualize. There a heated cathode emits electrons towards a distant cathode...to control the flow of electrons, an electrically charged screen is placed between the two, much closer to the cathode...so a SMALL change in screen voltage produces a significant and much larger change in flow of electrons from cathode to anode...hence small changes in screen voltage are AMPLIFIED by the large effect on cathode to anode electron flow...

Some background here: http://en.wikipedia.org/wiki/Vacuum_tube_amplifier

 

[dlgoff]

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?

No. The potential difference is high.

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html#c4"

 

[Idoubt]

Ok so tell me if I've got the right idea now.

Because of the geometric structure of the BJT transistor, only a very small current will flow through the base compared to the collector.

But since the EB jn is forward biased the conductance is very high, so a very small increase in voltage across the EB jn will be enough to increase the base current proportionally and the collector current by a much higher degree.

Since the base is so small, the base resistance is also very small, so the pd across the base (base current * base resistance ) remains small, while the collector current is high and the collector also has a higher resistance, so the pd across the collector ( collector current * collector resistance ) is high

To sum up, doubling the base current is like increasing the collector current by 20 ( ratios unimportant ) and this high current increase across the collector automatically means a higher voltage across it.

 

[Naty1]

excellent additional diagrams at the above reference:
http://hyperphysics.phy-astr.gsu.edu...trans2.html#c4

Check out "More on Transistor Regions"...good details
Both doping and relative physical size play important roles.

 

[sophiecentaur]

It's analogous to traditional vacuum tube amplifiers...which may be easier to visualize. There a heated cathode emits electrons towards a distant cathode...to control the flow of electrons, an electrically charged screen is placed between the two, much closer to the cathode...so a SMALL change in screen voltage produces a significant and much larger change in flow of electrons from cathode to anode...hence small changes in screen voltage are AMPLIFIED by the large effect on cathode to anode electron flow...

Some background here: http://en.wikipedia.org/wiki/Vacuum_tube_amplifier

That, of course, is a good 'way in' to understanding what's going on. The effect of a small voltage over a small gap (high field in V/m) on the current that can pass through the grid has instant appeal as an idea - compared with the 'cleverer' Physics of what goes on in the two semiconductor junctions. (Which took many more years to discover and the first transistors had virtually no gain at all! )
But you need to remember that the transistor is,essentially, a (base) current controlled device whereas the vacuum tube is very much voltage (field) controlled device.

Having said that, the very high gain transistors which are produced today can often be regarded as (near enough) voltage controlled whilst high power valves may well pass current through the grid. Nothin's easy.