Question

If, for a particular junction, the acceptor concentration is 1017/cm3 and the donor concentration is 1016/cm3, find the junction built-in voltage. Assume ni = 1.5 × 1010/cm3. Also, find the width of the depletion region (W) and its extent in each of the p and n regions when the junction terminals are left open. Calculate the magnitude of the charge stored on either side of the junction. Assume that the junction area is 10 µm2.

Answer 1

A) V_o = 0.7543 V

B) W = 328.42 nm

C) X_n = 298.56 nm and X_p = 29.86 nm

D) Q_j = 47.767 × 10^(-15)

Step-by-step explanation:

We are given;

Acceptor concentration; N_A = 10^(17) /cm³

donor concentration; N_D = 10^(16) /cm³

ni = 1.5 × 10^(10) /cm3

A) The formula for the built - in - voltage at the junction is given by;

V_o = V_T(In (N_A × N_D/ni²))

Where V_T is thermal voltage at room temperature with a value from online sources as 25.9 mV

Thus;

V_o = 25.9(In (10^(17) × 10^(16))/(1.5 × 10^(10))²

V_o = 0.7543 V

B) Now, formula for the width of the depletion region (W)is given as;

W = √(2ε_s/q[(1/N_A) + (1/N_D)]V_o)

Where;

ε_s is the permittivity in the semiconductor with a constant value of 1.04 × 10^(-12) F/cm

q is the electron charge = 1.6 × 10^(-19) C

Thus;

W = √(2 × 1.04 × 10^(-12)/(1.6 × 10^(-19)) [(1/10^(17)) + (1/10^(16)]0.7543)

W = 32.842 × 10^(-6) cm

Converting to m gives;

W = 328.42 nm

C) Formula for extent of the n region is;

X_n = W × (N_A/(N_A + N_D))

X_n = 32.842 × 10^(-6) × (10^(17)/(10^(17) + 10^(16))

X_n = 29.856 × 10^(-6) cm

Converting to m gives;

X_n = 298.56 nm

Formula for extent of the p region is;

X_p = W - X_n

X_p = 328.42 nm - 298.56 nm

X_p = 29.86 nm

D) The charge stored on either side of the junction is given by the formula;

Q_j = Aq[(N_A × N_D)/(N_A + N_D)]W

Where A is junction Area and we are given junction area as 10 µm² = 100 × 10^(-8) cm

Thus;

Q_j = (100 × 10^(-8) × 1.6 × 10^(-19))[(10^(17) × 10^(16))/(10^(17) + 10^(16)] × 32.842 × 10^(-6)

Q_j = 47.767 × 10^(-15)