Solutions to Homework # 2

(1/26/04- 1/30/04)

2. 24 N_D = 10¹⁶ /cm³ N_A = 4 * 10¹⁶ /cm³ n_i = 10¹¹ /cm³

N_A > N_D since p – type

N_A – N_D = 4 * 10¹⁶ – 10¹⁶ = 3*10¹⁶/cm³

N_A – N_D >> 2*n_i

p N_A – N_D = 3 * 10¹⁶ / cm³

n = n_i²/p = 10²²/(3 * 10⁶) = 3.33 * 10⁵ / cm³

2. 25 N_D = 2 * 10¹⁷/cm³ N_A = 10¹⁷ /cm³ n_i = 10¹⁷ /cm³

N_D > N_A

N_D – N_A = 2 * 10¹⁷ – 10¹⁷ = 10¹⁷ /cm³

2 * n_i = 2 * 20¹⁷ /cm³

N_D – N_A and 2n_i are comparable

n =

n = 1.62 * 10¹⁷ /cm3

p = n_i²/n = (10¹⁷)²/(1.62 * 10¹⁷)

p = 6.18 * 10¹⁶ /cm³

2.34 N_A = 5.5 * 10¹⁶ / cm³ N_D = 4.5 * 10¹⁶ / cm³

At 300K

Since N_A > N_D it is a p-type material

p = 10¹⁶ / cm³

n = n_i² / p = 10²⁰ / 10¹⁶

n = 10⁴ / cm³

From the figure 2. 8 in the text we get

cm² / V-s

cm² / V-s

Now to find resistivity, we use

ohm – cm

2.35 ohm – cm

since it is a p-type material (n = 0)

Therefore,

Using Iterations on the graph in fig 2. 8 we find

N_A = 7 * 10¹⁷ / cm³

2.45

Now,

At 300K V_T = 0.025V

=

J_diff = 1.6 * 10^-19 C * 350 cm²/V-s * 0.025 V * 10²²/cm⁴

J_diff = 14 kA/cm²