The common emitter forward current gain of the transistor shown is &be...
Assume Transistor in Active region
20 – 2 (IB + IC) – 0.7 – 250 IB – 10 = 0
20 – 2 × 101 IB – 0.7 – 250 IB – 10 = 0
10 – 452 I
B – 0.7 = 0
20 – 2 (IB + IC) - VEC – 2IC = 0
20 – 202 IB - VEC = 0
V
EC = 11.68 V
VCB = VCE + VEB = -VEC + VEB = - 11.68 + .7
VCB = -10.98
∴ Collector base junction is Reverse Biased
∴ Transistor is operating in forward active region
Second Method
Assume transistor in saturation region
IB min ≤ IB
i) VEB (sat) = 0.8 V
ii) VEC (sat) = 0.2 V
I
C ≠ βI
B ----- Because it is valid only active region
20 – 2 (IB + IC) – 0.8 - 250 IB – 10 = 0
20 – 2 IB – 2IC – 0.8 – 250 IB – 10 = 0
10 – 252 IB – 2IC – 0.8 = 0
252 IB + 2IC = 9.2 -------- (1)
20 – 2 (IB + IC) – VEC(sat) – 2IC = 0
20 – 2IB – 2IC – 0.2 – 2IC = 19.8 -------- (2)
By solving equation (1) and (2)
IB = 0.00278 mA
I
C = 4.95 mA
∴ I
B(min) > I
B∴ Transistor is operating in forward active region