Electronic Devices - BJT Devices - Discussion
Discussion Forum : BJT Devices - General Questions (Q.No. 13)
13.
Use this table of collector characteristics to calculate 
ac at VCE = 15 V and IB = 30 
A.
ac at VCE = 15 V and IB = 30 A. 
Discussion:
15 comments Page 2 of 2.
Faami said:
8 years ago
I too agree @Wilson.
Mrs NAIK said:
8 years ago
IC is 3.2mA so 3.2/30=106 is Ans.
Rajashekar said:
6 years ago
(IC3-IC2)÷5 = 30uA where ic2 =150uA.
(Ic2-ic1)÷5 = 20uA,
Ic1-0)÷5 = 10uA so ic1=50uA,
Ic3 = 3uA.
(Ic2-ic1)÷5 = 20uA,
Ic1-0)÷5 = 10uA so ic1=50uA,
Ic3 = 3uA.
Ayaz satti said:
5 years ago
From graph.
Ic = 3mA and Ib given is 30uA.
So, B= Ic/Ib = 3mA/30uA = 100A.
Ic = 3mA and Ib given is 30uA.
So, B= Ic/Ib = 3mA/30uA = 100A.
Bertoroyto said:
5 years ago
@ Bac,
Vce is constant, that's why Ic(sat) = 3 mA
Bac= Ic/Ib.
Bac= 3mA/30uA,
Bac= 100.
Vce is constant, that's why Ic(sat) = 3 mA
Bac= Ic/Ib.
Bac= 3mA/30uA,
Bac= 100.
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