Question

Q1

the built-in voltage Vo, the equilibrium electric field at the Junction ε0, and the depletion layer width D. Sketch the electric field and the potential as a function of position across the junction.

Q2

Q3

Answer 1

Q.1) Given:

• Doping on p-Side: 10¹⁷/ cm³ Boron (Valence electrons in Boron- 3)
• Doping on n-side: 10¹⁷/ cm³ Phosphorus (Valence electrons in Phosphorus- 5)
• Net p-type Dopant: N_a = 3 x 10¹⁷/ cm³
• Net n-type Dopant: N_d = 5 x 10¹⁷/ cm³

To find out: Built-in voltage: V₀ =?

V₀ = V_t log {(N_dN_a)/n_i²}

Where,

V_{t =} Thermal Voltage = 26 mV = 0.026 V (At room temp.)

n_i = Intrinsic Carrier concentration = 1.5 x 10¹⁰ in Silicon at room temp.( T = 300K)

N_a = 3 x 10¹⁷/ cm³

N_d = 5 x 10¹⁷/ cm³

Thus substituting the values in the equation above we get:

V₀ = 0.026 log {(5 x 10¹⁷ x 3 x 10¹⁷)/(1.5 x 10¹⁰)²}

= 0.026 log (6.6667 x 10¹⁴)

  = 0.026 x 14.8239

V₀ = 0.3854 V = Built in voltage.