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 1 of 2.
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.
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.
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.
Sagar said:
10 years ago
Not right because only one answer is required.
Teezy said:
10 years ago
Why not positive?
Afsar Khan said:
9 years ago
Because base-collector is normally reverse biased.
Isha said:
9 years ago
For active region, Vbe = 0.7v get ib by using kcl in input loop ib = (Vbb - Vbe) / Rb and ic = β Ib and use Kcl in outer loop and get Vce and then Vbc = Vbe - Vce = -9.2.
Shivam said:
9 years ago
The equation is Vce = Vcb + Vbe.
Vbe = 0.7V is forward bias voltage of pn junction diode which is formed by base and emitter.
And Vce = 9.895 V as we calculating.
So, Vcb = Vce - Vbe = 9.195V~ = 9.2V.
So, Vbc = -Vcb = -9.2v is correct option.
Only the sign of magnitute of voltage measured changed.
Vbe = 0.7V is forward bias voltage of pn junction diode which is formed by base and emitter.
And Vce = 9.895 V as we calculating.
So, Vcb = Vce - Vbe = 9.195V~ = 9.2V.
So, Vbc = -Vcb = -9.2v is correct option.
Only the sign of magnitute of voltage measured changed.
(1)
Sabar said:
9 years ago
How Vce is 9.895?
Manoj said:
9 years ago
Apply kvl to base emitter loop.
-Vb + (Ib * 5k) + Vbe = 0. Vbe = 0.7v
Thus, Ib= 8.6*10^-4A.
B=Ic/Ib.
Thus, Ic = 0.0215A.
Appliy KVL to collector to base loop.
-20+Ic*470 +Vcb - Ib * 5k + 5 =0.
Thus, Vcb = 9.195v
Thus, Vbc = -9.195~ -9.2v.
-Vb + (Ib * 5k) + Vbe = 0. Vbe = 0.7v
Thus, Ib= 8.6*10^-4A.
B=Ic/Ib.
Thus, Ic = 0.0215A.
Appliy KVL to collector to base loop.
-20+Ic*470 +Vcb - Ib * 5k + 5 =0.
Thus, Vcb = 9.195v
Thus, Vbc = -9.195~ -9.2v.
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