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pnp- junction ( transistor ) question ?

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rabso

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how transistor amplifies the current when I(emitter) = I (base)+ I(collector) , do the mean by amplifying is controlling the current like the current in circ. is 10 A and then reduce it to 5 and then amplify it to 10 again ?
 

Amplifying in BJT transistors is by Beta factor (B)
Ic= B * IB
IB the base current
Ic the collector current
B Beta factor

GOOD LECTURE NOTES AVAILABLE WITH EXAMPLES
**broken link removed**
 

Amplifying in BJT transistors is by Beta factor (B)
Ic= B * IB
IB the base current
Ic the collector current
B Beta factor

GOOD LECTURE NOTES AVAILABLE WITH EXAMPLES
**broken link removed**
thanks very much , i am going to read your lectures and then reply to you
 

Amplifying in BJT transistors is by Beta factor (B)
Ic= B * IB
IB the base current
Ic the collector current
B Beta factor

GOOD LECTURE NOTES AVAILABLE WITH EXAMPLES
**broken link removed**

Hi Mahmoud, congratulation to the good lecture notes.
Contrary to many books and internet publications (and also in contrast to your own wording) the transistor is treated in this note as a voltage-controlled device (which is the physical truth).
The relation Ic=B*beta is NOT used to verify amplification but only to calculate the (unavoidable) base current IB=Ic/beta.
(see the discussion regarding current or voltage control of the BJT we had in this forum some days ago)
 

the transistor is only control element it will provide the amplification by scaling the input current from the power supply. it mean that the amplification is converting the D.C power in to an A.C power with a certain efficiency.
when you use the bipolar as a voltage amplifier then the input is base and the output is the collector so you get difference in current of factor beta

how transistor amplifies the current when I(emitter) = I (base)+ I(collector) , do the mean by amplifying is controlling the current like the current in circ. is 10 A and then reduce it to 5 and then amplify it to 10 again ?
 

the transistor is only control element it will provide the amplification by scaling the input current from the power supply. it mean that the amplification is converting the D.C power in to an A.C power with a certain efficiency.
when you use the bipolar as a voltage amplifier then the input is base and the output is the collector so you get difference in current of factor beta
beta is never bigger than one so ib when multiplied be beta it will decrease .

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but i found that is used to separate two circuits so the current begin one the first circ on low r and goes to another circ with big r ( r = resistor )
 

I dont know who told you that B=1 ??????????? it is in order of tens

beta is never bigger than one so ib when multiplied be beta it will decrease .

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but i found that is used to separate two circuits so the current begin one the first circ on low r and goes to another circ with big r ( r = resistor )
 

Rabso,
You are confusing Beta (Common Emitter Current Gain) with Alpha (Common Base Current Gain). Alpha is always less than one for a junction transistor. The relationsip between the two is Beta = Alpha/(1-Alpha).
 

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