Here, R_C = 2 k$$\Omega $$ = 2 × 10³ $$\Omega $$

V₀ = 2 V, R_B = 1 k$$\Omega $$ = 1 × 10³ $$\Omega $$, $$\beta $$ = 100

The output voltage, across the load R_C

V₀ = I_CR_C = 2

The collector current (I_C)

I_C = $${2 \over {2 \times {{10}^3}}}$$ = 10^-3 A = 1 mA

Current gain($$\beta $$) = $${{{I_C}} \over {{I_B}}}$$ = 100

$$ \Rightarrow $$ I_B = $${{{I_C}} \over {100}}$$ = $${{{{10}^{ - 3}}} \over {100}}$$ = 10^-5 A

Input voltage, V_i

= I_BR_B = (10^–5 A) (1 × 10³ $$\Omega $$) = 10^–2 V = 10 mV

(n_i)² = n_e × n_h

(1.5 × 10¹⁶)² = n_e (4.5 × 10²²)

So n_e = 5 × 10⁹

Now n_h = 4.5 × 10²²

$$ \Rightarrow $$ n_h $$ >> $$ n_e

Hence, semiconductor is p-type

and n_e = 5 × 10⁹ m^–3

p-n junction is said to be forward biased when p side is at high potential than n side. It is for circuit (A) and (C).

Voltage across 250 Ω resistance = 20 V – 15 V = 5 V

Now current through 250 $$\Omega $$ resistance: 5/250 = 20 mA

If voltage across load resistance 1 k$$\Omega $$ is 15 V, then

current through 1 kΩ is 15/1000 = 15 mA

The current through the zener diode is

= Current through 250 $$\Omega $$ resistance – Current through 1 k$$\Omega $$ resistance.

= 20 - 15

= 5 mA