Doping & PN Junctions

Q1 [10]

(a) [4] P is Group V (5 valence e⁻), Si Group IV (4 valence e⁻) [1]. P forms 4 covalent bonds with neighbouring Si atoms using 4 of its electrons [1]; the 5th electron is not needed for bonding [1] and is weakly bound (donor binding energy ~0.045 eV), so at room temperature it is thermally ionised into the conduction band as a nearly-free electron → n-type [1].

(b) [3] B is Group III (3 valence e⁻) [1]. It forms only 3 bonds, leaving one bond incomplete — an acceptor state [1]. An electron from a neighbouring bond hops in to complete it, leaving behind a mobile hole; B becomes a fixed negative ion → p-type [1].

(c) [3] Donor level $E_D$ lies just below $E_c$ (≈0.045 eV below) [1]; acceptor level $E_A$ lies just above $E_v$ [1]. "Shallow" because the ionisation energy ($\sim$tens of meV) is comparable to $k_BT\approx26$ meV at 300 K, so nearly all dopants are ionised at room temperature [1].

Q2 [12]

(a) [7]

• At equilibrium, net electron current = 0: drift balances diffusion [1]: $J_n = q\mu_n n E + qD_n\frac{dn}{dx} = 0$
• Use Einstein relation $D_n/\mu_n = V_T$ and $E=-d\psi/dx$ [1]: $\frac{d\psi}{dx} = V_T\frac{1}{n}\frac{dn}{dx}$
• Integrate across junction from p-side to n-side [1]: $V_{bi} = \psi_n-\psi_p = V_T\ln\frac{n_n}{n_p}$
• On n-side $n_n \approx N_D$; on p-side (minority) $n_p = n_i^2/N_A$ by mass-action law [2]:
• Substitute [2]: $V_{bi} = V_T\ln\frac{N_D N_A}{n_i^2}$

(b) [5] $V_T = kT/q = 0.02585$ V [1]. Argument $= \dfrac{10^{17}\cdot10^{16}}{(1.5\times10^{10})^2} = \dfrac{10^{33}}{2.25\times10^{20}} = 4.44\times10^{12}$ [2]. $\ln(4.44\times10^{12}) = 29.12$ [1]. $V_{bi} = 0.02585\times29.12 = 0.753\ \mathrm{V}$ [1].

Q3 [10]

(a) [3] Near the metallurgical junction, electrons diffuse from n→p and holes from p→n, recombining and leaving behind uncovered fixed ionised dopants [1]. This region has no mobile carriers → depleted [1]. n-side: positive space charge (ionised donors); p-side: negative space charge (ionised acceptors) [1].

(b) [4] Overall charge neutrality requires total positive = total negative charge [1]: $q N_D x_n = q N_A x_p \Rightarrow N_D x_n = N_A x_p$ [2] If $N_A \gg N_D$ then $x_n \gg x_p$: the depletion region extends mostly into the lightly-doped n-side [1].

(c) [3] $\left(\frac1{N_A}+\frac1{N_D}\right) = 10^{-17}+10^{-16} = 1.1\times10^{-16}\ \mathrm{cm^3}$ [1]. $W = \sqrt{\frac{2(1.04\times10^{-12})}{1.602\times10^{-19}}(1.1\times10^{-16})(0.753)}$ Inner $= 1.298\times10^7 \times 1.1\times10^{-16}\times0.753 = 1.075\times10^{-9}$ [1]. $W = \sqrt{1.075\times10^{-9}} = 3.28\times10^{-5}\ \mathrm{cm} = 0.328\ \mu\mathrm{m}$ [1].

Q4 [12]

(a) [3] $I = I_S\!\left(e^{V/(nV_T)}-1\right)$. $I_S$ = reverse saturation current, $V$ = applied voltage, $n$ = ideality factor, $V_T = k_BT/q$ thermal voltage. [3] (1 for eqn, 2 for symbols)

(b) [4] $V_T=0.02585$ V. $V/V_T = 0.6/0.02585 = 23.21$ [1]. $e^{23.21}=1.20\times10^{10}$ [1]. $I = 10^{-14}(1.20\times10^{10}-1) \approx 1.20\times10^{-4}\ \mathrm{A} = 0.12\ \mathrm{mA}$ [2].

(c) [5]

import numpy as np
def diode_current(V, Is, n, T):
    kB = 1.381e-23; q = 1.602e-19
    VT = kB*T/q
    return Is*(np.exp(V/(n*VT)) - 1.0)

[3] for correct function. Forward: exponential rise, current ~0 below ~0.6 V then steep knee [1]. Reverse: current saturates at $-I_S$ (tiny, ~constant) until breakdown [1].

Q5 [10]

(a) [4] $I_S$ is carried by minority carriers (thermally generated electrons in p-region, holes in n-region) swept across the junction by the field [2]. Nearly voltage-independent because it's limited by the rate of thermal generation, not the field, once carriers are collected [1]. Strongly T-dependent because generation $\propto n_i^2 \propto e^{-E_g/kT}$ — roughly doubles every ~10 °C [1].

(b) [6]

Q6 [6]

(a) [3] Depletion (junction) capacitance and diffusion capacitance [1]. Depletion capacitance dominates in reverse bias [1]; diffusion capacitance dominates in forward bias (stored minority charge) [1].

(b) [3] $C_j = \dfrac{\varepsilon_s A}{W} = \dfrac{1.04\times10^{-12}\times10^{-4}}{3.28\times10^{-5}}$ [1] $= \dfrac{1.04\times10^{-16}}{3.28\times10^{-5}} = 3.17\times10^{-12}\ \mathrm{F}$ [1] $= 3.17\ \mathrm{pF}$ [1].

[
  {"claim":"V_bi = 0.753 V for NA=1e17,ND=1e16,ni=1.5e10 at 300K",
   "code":"kB=1.381e-23;q=1.602e-19;T=300;VT=kB*T/q;NA=1e17;ND=1e16;ni=1.5e10;Vbi=VT*log(NA*ND/ni**2);result=abs(float(Vbi)-0.753)<0.01"},
  {"claim":"Depletion width W approx 3.28e-5 cm",
   "code":"eps=1.04e-12;q=1.602e-19;NA=1e17;ND=1e16;Vbi=0.753;W=sqrt(2*eps/q*(1/NA+1/ND)*Vbi);result=abs(float(W)-3.28e-5)<0.1e-5"},
  {"claim":"Diode current at 0.6V, Is=1e-14, n=1, T=300 approx 1.2e-4 A",
   "code":"kB=1.381e-23;q=1.602e-19;T=300;VT=kB*T/q;Is=1e-14;I=Is*(exp(0.6/VT)-1);result=abs(float(I)-1.2e-4)/1.2e-4<0.05"},
  {"claim":"Junction capacitance approx 3.17 pF",
   "code":"eps=1.04e-12;A=1e-4;W=3.28e-5;Cj=eps*A/W;result=abs(float(Cj)*1e12-3.17)<0.1"}
]