bjt in active region : amplified output :confused:

[meowth08]

Three modes of operation of BJT

1. saturation: analogous to on switch
2. cut-off: analogous to off switch
3. active: produces amplification

For active region: NPN

E-B junction is forward biased.
B-C junction is reverse biased.

Given the above condition, the depletion region becomes wide.
And from concept of a pn junction (e.g. a diode) where the depletion region is wide, the resistance is high therefore not letting current to pass.

If there is a high resistance, then how come that there is AMPLIFICATION taking place?

 

[MrAl]

Hi,

Simple answer, because the resistance is not as high as you think it is. It is somewhere between cutoff and saturation, and as you know, saturation can look like a low resistance.

Aside from that, a somewhat higher resistance can still provide amplification if the external resistances are comparable.

 

[Ratchit]

meowth08,

Given the above condition, the depletion region becomes wide.

There are two depletion regions in a BJT, emitter-base and base-collector. To which one are you referring? The depletion width is less than the thickness of a hair. Is that wide?

And from concept of a pn junction (e.g. a diode) where the depletion region is wide, the resistance is high therefore not letting current to pass.

The depletion width is a symptom, not a cause of less current. Less current is caused by charge carrier starvation, not depletion width. In the emitter-base region, the charge carriers in the emitter diffuse into the base region and vice versa. This is caused by the difference in carrier concentration, i.e., the N-type material is full of free electrons and the P-type region is full of holes. That causes diffusion like a drop of dye into a glass of water. A positive charge leaving the P region leaves behind a negative charge and vice versa. This sets up a barrier voltage that eventually inhibits further diffusion. A proper voltage applied to the emitter-base will increase or decrease the barrier voltage thereby controlling the emitter current. Once the current from the emitter is in the base, it is swept across to the collector by the relatively high reverse bias of the collector. In the active region, the current in the collector is controlled by the emitter, because the emitter is the only place where its current mostly comes from. The diffusion mechanism also explains why the collector current and base current is exponentially related to the emitter-base voltage.

To answer your question, if the same current exists in the emitter as does in the collector, and the impedance of the collector is higher than the emitter, then a larger signal will result across the collector due to the resistance formula E = IR. Keep on asking questions.

Ratch

 

[meowth08]

Hello,

I could not still fully understand your explanations sir.
Could you present some of the parameters mathematically.
Hopefully,I would understand it better.

Thanks.

 

[Ratchit]

meowtho8,

I could not still fully understand your explanations sir.
Could you present some of the parameters mathematically.
Hopefully,I would understand it better.

Throwing some numbers at you will not help you much unless you understand the concepts of resistance, current and voltage. Perhaps, if you read some books on BJT operation, you can gain enough insight to ask more directed questions.

Ratch