Electronic Devices - DC Biasing-BJTs - Discussion
Discussion Forum : DC Biasing-BJTs - General Questions (Q.No. 9)
9.
Determine the change in IC from 25ºC to 175ºC for the transistor defined in this table for fixed-bias with RB = 240 k
and 
= 100 due to the S(VBE) stability factor.
and = 100 due to the S(VBE) stability factor. 
145.8 
A
A145.8 nA
–145.8 A
A–145.8 nA
Discussion:
5 comments Page 1 of 1.
Christiangalvez said:
3 years ago
=> 100/240x10^3 ( 0.3-0.6) = 145.8uA.
Bon said:
7 years ago
Why the answer is not -145. 8μA?
The current changes from 25 to 175μdegrees so it should be a negative change.
The current changes from 25 to 175μdegrees so it should be a negative change.
PONZO said:
7 years ago
First get the value of Ib @25°C and @175°C.
Use beta=100 in all equations;
@25°C Vbe=0.65V
Ib1=.65/240x10^3=2.71x10^-6A.
@175°C Vbe=0.3V
Ib2=.3/240x10^3=1.25x10^-6.
I=B*Ib1-(B*Ib2)=(100*2.71x10^-6)-(100*1.25x10^-6)
=146*10^-6A.
Use beta=100 in all equations;
@25°C Vbe=0.65V
Ib1=.65/240x10^3=2.71x10^-6A.
@175°C Vbe=0.3V
Ib2=.3/240x10^3=1.25x10^-6.
I=B*Ib1-(B*Ib2)=(100*2.71x10^-6)-(100*1.25x10^-6)
=146*10^-6A.
Etits said:
8 years ago
Can someone give a detailed explanation for this?
Beke said:
9 years ago
Can anyone please explain this for me?
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