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PN junction with different doping concentrations

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SanjKrish

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I have difficulty in understanding the BJT working
It led me to the doubt

Suppose if there is a PN junction diode with N heavily doped and P lightly doped and thiner than the N region say..
and we supply a forward bias voltage to it..

What will happen:

Will the heavily doped N inject more amount of electrons in the P region in exchange for the fewer holes from the P side..
OR
Will the N region inject exactly the amount of electrons equal to the holes from the P region..

My intuition says the second one is right, but while understanding the BJT all authors state that since since the base is lightly doped it is very easy for the emitter to flood it with electrons and the excess electrons will pass through the collector...
 

If there is a PN junction diode with N heavily doped and P lightly doped and thiner than the N region, then more eletrons will inject on p side and the transition region will spread deeper towards the p region which is called as "punch through effect".
Now in case of bjts the emitter is heavily doped, base is lightly doped, & collector is moderately doped. Hence there will be more recombination will take place on the emitter region compare to collector region and the width of space charge will expand more towards base & collector region as forward bias increased which result in base width modulation.
 
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If there is a PN junction diode with N heavily doped and P lightly doped and thiner than the N region, then more holes will inject on n side and the juction will spread deeper towards the n side region which is called as "punch through effect".
Now in case of bjts the emitter is heavily doped, base is lightly doped, & collector is moderately doped. Hence there will be more recombination will take place on the emitter region compare to collector region and the width of base will expand more towards emitter region as forward bias increased which result in base width modulation.
Click to expand...

How can more holes inject on the N side, when there is deficit of holes in the P region..
What do you mean by "The junction will spread deeper towards the N side region "
How can the width of the base expand physically??
 

Width of the base doesn't expand physically- i mean the space charge(region of uncovered charge both side of junction) expands more dipper towords the less doped side as per the biasing level which is known as early effect(vb=qn(w)2/2"ep"), where w is the space charge width,q is the charge of electron, n is the doping concentration, ep:8.854x10-19.
 

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Thanks sam , very helpful.. Now I understand BJT good...
 

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