Electronic Devices - DC Biasing-BJTs - Discussion
Discussion Forum : DC Biasing-BJTs - General Questions (Q.No. 32)
32.
Calculate VCE.
Discussion:
4 comments Page 1 of 1.
MemmEd Hacili said:
9 years ago
9 - U(be) - 10^5 * I(b) = 0 ( U(be) = 0.7 V ).
So,
I(b) = (9 - 0.7)/ 10^5.
I(b) = 0.083 mA.
0 - 9 - U(ce) - I(c) * R(c) = 0 ( I(c) = I(b) * 45 = 3.735 mA.
U(ce) = 3.735 * 1.2 - 9.
U(ce) = 4.482 - 9.
U(ce) = - 4.518 ~= -4.52V.
So,
I(b) = (9 - 0.7)/ 10^5.
I(b) = 0.083 mA.
0 - 9 - U(ce) - I(c) * R(c) = 0 ( I(c) = I(b) * 45 = 3.735 mA.
U(ce) = 3.735 * 1.2 - 9.
U(ce) = 4.482 - 9.
U(ce) = - 4.518 ~= -4.52V.
(2)
Swati singh said:
1 decade ago
VEE-VEB-100kohm*IB = 0.
9-0.7 = 100kohm IB.
IB = 0.083mA.
VEE-VEC-1.2kohmIC = 0.
IC = IB*45.
IC = 0.083mA*45 = 3.735mA.
9-1.2kohm*3.735mA = VEC.
9-4.482 = VEC.
4.518V = VEC.
VCE = -4.518.
VCE = -4.52V.
9-0.7 = 100kohm IB.
IB = 0.083mA.
VEE-VEC-1.2kohmIC = 0.
IC = IB*45.
IC = 0.083mA*45 = 3.735mA.
9-1.2kohm*3.735mA = VEC.
9-4.482 = VEC.
4.518V = VEC.
VCE = -4.518.
VCE = -4.52V.
(1)
Kimi said:
4 years ago
Why VCE is negative?
I solved it but my answer is in positive. Please explain me.
I solved it but my answer is in positive. Please explain me.
Glenn said:
8 years ago
Good explanation, Thank you @Memmed.
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