Respuesta :
Answer:
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)
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)
