How to solve this? problem 3 and 4 3. Figure below shows the common emitter npn
ID: 1362071 • Letter: H
Question
How to solve this? problem 3 and 4
Explanation / Answer
3.a) Ib = 8 uA
Ic = 1.2 mA
as Ic = beta*Ib
beta = Ic/Ib = 150
also , Ic/Ie = alpha , where Ie = (1+beta)Ib
alpha = Ic/Ie = beta / (1+beta) = 150/151 = 0.99337
Ie = (1+beta)Ib = 151*8 uA = 1.208 mA
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b) Ie = 2.42 mA, Ic = 2.4 mA
as, alpha = Ic/Ie = beta / (1+beta)
Ic/Ie = 2.4/2.42 = 0.992
beta / (1+beta) = 0.992
beta = 0.992/0.008 = 124
Ib = Ic/beta = 2.4 mA / 124 = 0.0194 mA
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c) Ib = 16 uA , beta = 80 ,
Ic = beta *Ib = 1.28 mA
Ie = (1+beta)Ib = 1.296 mA
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4. ) Ib = 250 uA , Ie = 15mA
as Ie = (1+beta)*Ib
15 x 10^-3 = (1+beta)250 x 10^-6
1+beta = 60
beta =59
collector current gain = alpha = beta / (1+beta) = 59/60 = 0.9833
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