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Electronics and Communication Engineering - Analog Electronics - Discussion

Discussion Forum : Analog Electronics - Section 17 (Q.No. 8)
Analog Electronics
8.
In figure the coordinates of Q point on the load line are (Neglect VBE)
IC = 10 mA, VCE = 15 V
IC = 5 mA, VCE = 22.5 V
IC = 5 mA, VCE = 15 V
IC = 10 mA, VCE = 22.5 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
5 comments Page 1 of 1.
Sachin said:   9 years ago
Anyone provide a clear solution. Please.
Ajit said:   9 years ago
Since beta is not given. Question is incomplete I think. So, Here beta is taken as 100.
T Ravi Kumar Reddy said:   7 years ago
Ib=Vcc-Vbe/Rb.

Ib=30-0/300*10^3=10mA,
Vce=Vcc0-IcRc,
Vce=30-10*1.5=15mA.
GOPALA R said:   7 years ago
Please explain the clear solution.
(1)
Tindelaraa said:   6 years ago
Vcc-IbRb-Vbe=0.
30V-Ib(300k)=0.
*here neglect VBE*.
Ib=1x10^-4.

Ic=Beta(Ib).
*take beta as 100,
c=100(1x10^-4),
Ic=10mA.

Vcc-IcRc-Vce=0.
30V-10mA(1.5k)-Vce=0.
Vce= 15V.
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