Electronic Devices - DC Biasing-BJTs - Discussion
Discussion Forum : DC Biasing-BJTs - General Questions (Q.No. 1)
1.
Calculate VCE.
Discussion:
5 comments Page 1 of 1.
Severina said:
2 years ago
ic = Bib.
1. kvl to find Ib.
100k ib + 0.7 - 9 = 0,
Ib = 83 micro A.
2. kvl on the right side to find Vce.
1.2k (Bib) + Vce -9 = 0,
Vce = 4.52 V=> answer.
1. kvl to find Ib.
100k ib + 0.7 - 9 = 0,
Ib = 83 micro A.
2. kvl on the right side to find Vce.
1.2k (Bib) + Vce -9 = 0,
Vce = 4.52 V=> answer.
(5)
PranavPriyatosh said:
1 decade ago
No need for VBB VEE serving role of VBB here. Just take KVL in BE loop to find IB.
FaKkU said:
1 decade ago
This is in saturation mode right?
Musharique said:
1 decade ago
How calculate vbb?
D.V.WAYKULE said:
1 decade ago
IB.RB+VBE=9V
IB=(9V-VBE)/RB
IB=8.3/100K=8.3x 10E-5
IC=BETA x IB= 45 x 8.3 X10 E-5
VCE+IC.RC=9V
VCE=9V-IC.RC=9V - 45 x 8.3 x 10 E-5 x 1.2K=9V-4.48V=4.52
IB=(9V-VBE)/RB
IB=8.3/100K=8.3x 10E-5
IC=BETA x IB= 45 x 8.3 X10 E-5
VCE+IC.RC=9V
VCE=9V-IC.RC=9V - 45 x 8.3 x 10 E-5 x 1.2K=9V-4.48V=4.52
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