Electronic Devices - Bipolar Junction Transistors - Discussion
Discussion Forum : Bipolar Junction Transistors - General Questions (Q.No. 6)
6.
Refer to this figure. The value of VBC is:
Discussion:
16 comments Page 2 of 2.
Afsar Khan said:
9 years ago
Because base-collector is normally reverse biased.
Teezy said:
10 years ago
Why not positive?
Sagar said:
10 years ago
Not right because only one answer is required.
Adnan. said:
1 decade ago
Vbc = Vbe-Vce.
Vbe = 0.7.
Now, applying KVL between base and emitter,
-5+5000Ib+0.7 = 0 or, Ib = 8.6*10^-4 A.
We know, Ic = Beta*Ib or, Ic = 25*8.6*10^-4 = 0.0215 A.
Again, applying KVL between collector and emitter,
-20+470*Ic+Vce = 0 or, Vce = 9.895.
So, Vbc = 0.7-9.895 = -9.2 V.
Vbe = 0.7.
Now, applying KVL between base and emitter,
-5+5000Ib+0.7 = 0 or, Ib = 8.6*10^-4 A.
We know, Ic = Beta*Ib or, Ic = 25*8.6*10^-4 = 0.0215 A.
Again, applying KVL between collector and emitter,
-20+470*Ic+Vce = 0 or, Vce = 9.895.
So, Vbc = 0.7-9.895 = -9.2 V.
Vishal gupta said:
1 decade ago
Answer is 9.2v i.e. option A because.
Vce = Vbe+Vbc.
Vbc = Vce-Vbe.
Vce = Vbe+Vbc.
Vbc = Vce-Vbe.
Hariom Gupta said:
1 decade ago
Vbc=Vbe-Vce
So, first find out base current by kirchoff's loop law in input side .
Then collector current by Ic=b*Ib;
Now output kirchoff's current law at the output give value of Vce= 9.895;
Now Vcb = 0.7-9.895 = -9.195.
So, first find out base current by kirchoff's loop law in input side .
Then collector current by Ic=b*Ib;
Now output kirchoff's current law at the output give value of Vce= 9.895;
Now Vcb = 0.7-9.895 = -9.195.
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