Test: Electronic Devices - 4 - Electronics and Communication Engineering (ECE) MCQ

∴ Voltage gain can be greater than unity but current gain is always less than unity.

The common emitter current gain is given by:

α = γ ∗ β ∗ δ

= 0.99 × 0.995 × 0.998 = 0.983

Collector current

I_c = αI_E = 0.983 mA

Base current

I_B = I_E – I_C

= 17 μA 

The Base transit time

⇒ 0.5832 × 10^-10

Applying KVL in collector to emitter loop

V_CC = R_LI_C + V_CE

 

Now applying KVL in collector to base loop

I_BR_B + V_BE = V_CC

Given: V_EC = 5V^-, this means pnp transistor is operating in the active node

I_CEO = 410 μA, I_CBO = 5 μA, I_B = 30 μA

I_C­ = βI_B + (1 + β) I_CBO

I_CEO = (1 + β) I_CBO

410 = (1 + β)5

β = 81

I_C = 81 × 30 + 82 × 5

I_C = 2.84 mA

Apply kVL in base emitter loop

at saturation V_BE(sat) = 0.8

substituting the value of V_BE

I_B = 0.0525 mA

Apply Kvl , Form the collector to emitter

5 × 10³ × I_C + V_CE = 12 V

= 2.36 mA

At the edge of saturation

Hence for β value greater than 45 the transistor will be in saturation

= 5.96 × 10¹⁴ cm^-3

Assume Transistor in Active region

20 – 2 (I_B + I_C) – 0.7 – 250 I_B – 10 = 0

20 – 2 × 101 I_B – 0.7 – 250 I_B – 10 = 0

10 – 452 I_B – 0.7 = 0

20 – 2 (I_B + I_C) - V_EC – 2I_C = 0

20 – 202 I_B - V_EC = 0

V_EC = 11.68 V

V_CB = V_CE + V_EB = -V_EC + V_EB = - 11.68 + .7

V_CB = -10.98

∴ Collector base junction is Reverse Biased

∴ Transistor is operating in forward active region

Second Method

Assume transistor in saturation region

I_B min ≤ I_B

i) V_{EB (sat)} = 0.8 V

ii) V_{EC (sat)} = 0.2 V

I_C ≠ βI_B ----- Because it is valid only active region

20 – 2 (I_B + I_C) – 0.8 - 250 I_B – 10 = 0

20 – 2 I_B – 2I_C – 0.8 – 250 I_B – 10 = 0

10 – 252 I_B – 2I_C – 0.8 = 0

252 I_B + 2I_C = 9.2      -------- (1)

20 – 2 (I_B + I_C) – V_EC(sat) – 2I_C = 0

20 – 2I_B – 2I_C – 0.2 – 2I_C = 19.8      -------- (2)

By solving equation (1) and (2)

I_B = 0.00278 mA

I_C = 4.95 mA

∴ I_B(min) > I_B

∴ Transistor is operating in forward active region

Given μ_n = 1300 cm²/V-s and μ_p = 450 cm²/V-s

D_p = 450 × 0.0259

= 11.655 cm²/s

Base diffusion length  = 1.08 mm

For injected holes from emitter to reach the collector they should not be recombined in base. Hence the base width should be very less compared to base diffusion length.

W ≪ L_P

In all options the base width is significant compared to base diffusion length.