Solutions to Homework #4

( ~ 02/20/04)

3.43

N_A = 10¹⁵ / cm³ N_D = 10²⁰ / cm³ n_i = 10¹⁰ / cm³ and V_T = 0.025 V at 300 K

Since we are require to calculate capacitance/unit area we use

For V_R = 5 V and A= 0.01cm²

3.56 Applying KVL to the circuit loop we get

5 = 10⁴ * I_D + V_D ------- Eqn 1

This equation represents a straight line that can be plotted using the following two points

When V_D = 0V, I_D = 0.5 mA

When I_D = 0 mA, V_D = 5V

From the diode characteristics, for Q point

V_D = 0.5 V

From Eqn 1 we get

I_D = 0.45 mA

3.83

(a)

V_Z = 15 V

Therefore,

I₁ = (50 – 15)/150k = 0.2333 mA

I₂ = 15 / 75k = 0.2 mA

Iz = I3 = I₁ – I₂ = 0.2333 – 0.2 = 0.0333 mA

Power Dissipation in the Zener Diode is given by

P_Z = V_Z*I_Z = 15 * 0.0333

P_Z = 0.5 W

(b)

R₂ à infinity

Hence, it is equivalent to an open circuit in place of R₂

In this case,

I_Z = I₃ = I₁ = 0.2333 mA

P_Z = V_Z * I_Z = 15 * 0.2333

P_Z = 3.5 W

3.91

V_RMS = 6.3V

V_DC = - (V_p - V_ON) = -(8.91 – 1)

V_DC = - 7.91V

The PIV rating is given by

V_PIV = 2*V_P = 2 * 8.91

V_PIV = 17.82 V

The surge current is given by

I_S = w*C*V_P = 2 * 3.14 * 60 * 8.91

I_S = 3526 A

To find the repetitive current,

I_P = 838 A