Electronic Devices - BJT Devices - Discussion
Discussion Forum : BJT Devices - General Questions (Q.No. 30)
30.
For a properly biased pnp transistor, let IC = 10 mA and IE = 10.2 mA. What is the level of IB?
0.2 A
200 mA
200 
A
A20.2 mA
Discussion:
7 comments Page 1 of 1.
Anithamanoharan said:
3 years ago
Very nice. Thanks.
Adrian said:
7 years ago
For pnp, IE(pnp) is opposite that of IE(npn).
-IE = -(IC+IB) [negative both sides].
-IE = -IC-IB.
IC = IE-IB.
10mA = 10.2mA - IB,
IB = 200uA.
-IE = -(IC+IB) [negative both sides].
-IE = -IC-IB.
IC = IE-IB.
10mA = 10.2mA - IB,
IB = 200uA.
Aruna said:
8 years ago
Thank you for your simple explanation @Dulan.
Dulan said:
1 decade ago
Ic=Ie-Ib
Ib=Ie-Ic
=.2mA=200uA
Ib=Ie-Ic
=.2mA=200uA
Ravichandranq said:
1 decade ago
Answer:
alpha=Ic/Ie where 10/10.2=0.98
and
beta= Ic/Ib and it is related to alpha is alpha=(beta/(beta+1))
So using alpha value, solving we get beta= 49
and substitute these values in the above said formula
Ib=Ic/beta which is 10/49= 0.02mA= 200micro Ampere.
alpha=Ic/Ie where 10/10.2=0.98
and
beta= Ic/Ib and it is related to alpha is alpha=(beta/(beta+1))
So using alpha value, solving we get beta= 49
and substitute these values in the above said formula
Ib=Ic/beta which is 10/49= 0.02mA= 200micro Ampere.
Komali said:
1 decade ago
Ie=Ic+Ib no chance of getting in micros even in practicle case.
Subhransu said:
1 decade ago
Ie=Ic+Ib
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