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 1 of 2.
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.
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.
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.
Mrs NAIK said:
8 years ago
IC is 3.2mA so 3.2/30=106 is Ans.
Faami said:
8 years ago
I too agree @Wilson.
Wilson Komla said:
8 years ago
I think the exact value for Ic can't be read from the graph unless approximations are done.
Mini said:
9 years ago
But If we compare both Vce = 15V and Ib = 30uA combined then the value of Ic should be greater than 3mA.
(1)
Piyush said:
9 years ago
Why we are taking the value of turning curve i.e. Ic?
Pasha said:
1 decade ago
Ic it come from the graph we can see the graph where the two axis there one is Ic(mA) and other Vce.
Look the graph where the curve turn to the Ib 30uA.
And Ic measured in (mA)3.
Look the graph where the curve turn to the Ib 30uA.
And Ic measured in (mA)3.
Khan said:
1 decade ago
As it seem from graph that at Vce=15v, Ie>3ma so beta must be greater than 100, ie 106.
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