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in Semiconductor Electronics by (48.5k points)
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In the circuit, the value of β is 100. Find IB , VCE , VBE and VBC , when IC = 1.5 mA. The transistor is in active, cut off or saturation state?

1 Answer

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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.

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