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.
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.
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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