β = 100; IC = 1.5 mA= 1.5 x 10-3 A, VCC = 24 V
To calculate VCE , We apply Kirchhoff’s rule to loop CEFDC, therefore
VCC = IC x 4.7 kΩ + VCE
24 = 1.5 x 10-3 x 4.7 10-3 + VCE
VCE = 24 – 7.05 = 16.95 V
Again, applying Kirchhoff’s rule to loop
ABEFDCA, We get,
VCC = IB x 220 kΩ + VBE
VCC = 15 x 10-6 x 220 x 10-3 + VBE
VBE = 24 -3.3
VBE = 20.7 V
Going along loop ABCA, we get
IB x 220 kΩ + VBC = IC x 4.7 kΩ
15 x 10-6 x 220 x 103 + VBC = 1.5 x 10-3 x 4.7 x 103
VBC = 7.05 -3.3 = 3.75 V
As VCE < VBE , both the junctions are forward biased. So, the transistor is in the saturation state.