Doping & PN Junctions

Question 1

(a) Phosphorus = donor (n-side); boron = acceptor (p-side). (1) n-side majority carriers = electrons; p-side majority carriers = holes. (2)

(b) $V_{bi} = V_T \ln\!\dfrac{N_A N_D}{n_i^2}$. (1) $\dfrac{N_A N_D}{n_i^2} = \dfrac{(2\times10^{17})(5\times10^{15})}{(1.5\times10^{10})^2} = \dfrac{1\times10^{33}}{2.25\times10^{20}} = 4.44\times10^{12}$. (1) $\ln(4.44\times10^{12}) = 29.12$. (1) $V_{bi} = 0.02585 \times 29.12 = 0.753\,\text{V}$. (1)

(c) $\Delta V_{bi} = V_T\ln(100) = 0.02585 \times 4.605 = 0.119\,\text{V}$ increase (~+119 mV). (2) Increasing $N_D$ raises the doping product, so $V_{bi}$ rises logarithmically. (1)

(d) Diagram: $E_C$ at top, $E_V$ at bottom, donor level $E_D$ just below $E_C$; $E_F$ close to $E_C$ (above mid-gap). (2)

Question 2

(a) $W = \sqrt{\dfrac{2\varepsilon_{Si}}{q}\left(\dfrac{1}{N_A}+\dfrac{1}{N_D}\right)V_{bi}}$. (1) $\varepsilon_{Si}=11.7\times8.854\times10^{-12}=1.036\times10^{-10}\,\text{F/m}$. Convert doping to m$^{-3}$: $N_A=2\times10^{23}$, $N_D=5\times10^{21}$. $\dfrac{1}{N_A}+\dfrac{1}{N_D}=5\times10^{-24}+2\times10^{-22}=2.05\times10^{-22}\,\text{m}^3$. (1) $\dfrac{2\varepsilon}{q}=\dfrac{2\times1.036\times10^{-10}}{1.602\times10^{-19}}=1.293\times10^{9}$. $W=\sqrt{1.293\times10^9 \times 2.05\times10^{-22}\times0.753}$ (1) $=\sqrt{1.996\times10^{-13}}=4.47\times10^{-7}\,\text{m}=0.447\,\mu\text{m}$. (1)

(b) Charge neutrality: $N_A x_p = N_D x_n$, so $\dfrac{x_n}{W}=\dfrac{N_A}{N_A+N_D}=\dfrac{2\times10^{17}}{2.05\times10^{17}}=0.976$. (2) The depletion region extends mostly into the lightly-doped n-side because fewer dopant atoms per volume there require a wider region to expose equal charge. (1)

(c) $C_j = \varepsilon_{Si}A/W$. $A=10^{-4}\,\text{cm}^2=10^{-8}\,\text{m}^2$. (1) $C_j=\dfrac{1.036\times10^{-10}\times10^{-8}}{4.47\times10^{-7}}=2.32\times10^{-12}\,\text{F}=2.32\,\text{pF}$. (2)

(d) $W\propto\sqrt{V_{bi}+V_R}$, so $W_5=W_0\sqrt{(0.753+5)/0.753}=W_0\sqrt{7.64}=2.76\,W_0=1.24\,\mu\text{m}$. (1) $C_5=C_0/2.76=0.84\,\text{pF}$; ratio $C_5/C_0=1/2.76=0.362$. (2) As reverse voltage increases the depletion width grows, reducing capacitance; thus $C$ is controlled by applied voltage (varactor behaviour). (1)

Question 3

(a) $I=I_S\big(e^{V/nV_T}-1\big)$. $V/V_T=0.60/0.02585=23.21$. (1) $e^{23.21}=1.20\times10^{10}$. (1) $I=2\times10^{-14}\times1.20\times10^{10}=2.4\times10^{-4}\,\text{A}\approx0.24\,\text{mA}$. (1)

(b) For $\times10$: $\Delta V=nV_T\ln10=0.02585\times2.303=0.0595\,\text{V}\approx59.5\,\text{mV}$. (3)

(c) At $V=-2\,\text{V}$: $e^{-2/0.02585}=e^{-77.4}\approx0$, so $I\approx-I_S=-2\times10^{-14}\,\text{A}$. (2) Real diodes show much larger reverse current due to generation in the depletion region and surface leakage. (1)

(d) $118\,\text{mV}=nV_T\ln10 \Rightarrow n=\dfrac{0.118}{0.0595}=1.98\approx2$. (2) $n>1$ arises from carrier recombination-generation within the depletion region (recombination current). (1)

(e) The Shockley equation assumes no breakdown; it does not model avalanche multiplication / Zener tunnelling, which cause the large reverse current at breakdown. (2)

Question 4

(a) Diode A (4.3 V, low breakdown voltage) → Zener: high doping ⇒ thin depletion region ⇒ very high field ⇒ direct band-to-band tunnelling. Diode B (12 V) → avalanche: lighter doping, wider depletion region, carriers gain enough energy for impact ionisation. (Crossover ~5–6 V in silicon.) (4)

(b) Zener (Diode A): negative temperature coefficient — higher T reduces band gap, making tunnelling easier so breakdown voltage falls. (1.5) Avalanche (Diode B): positive coefficient — higher T increases phonon scattering, shortening carrier mean free path, so a higher field/voltage is needed for ionisation. (1.5)

(c) Combining a negative-TC Zener diode in series with a positive-TC avalanche diode (or using a ~5.6 V device where both mechanisms coexist) lets the two temperature coefficients cancel, giving a near-zero net TC voltage reference. (3)

Question 5

(a) $I_S\propto n_i^2\propto T^3 e^{-E_g/kT}$. $\dfrac{I_S(350)}{I_S(300)}=\left(\dfrac{350}{300}\right)^3 \exp\!\Big[-\dfrac{E_g}{k}\big(\tfrac1{350}-\tfrac1{300}\big)\Big]$. (1) $(350/300)^3=1.588$. (1) $E_g/k=1.12\times1.602\times10^{-19}/1.381\times10^{-23}=1.299\times10^4\,\text{K}$. $\tfrac1{300}-\tfrac1{350}=4.762\times10^{-4}$; product $=1.299\times10^4\times4.762\times10^{-4}=6.186$. (1) $e^{6.186}=486.5$. (1) Ratio $=1.588\times486.5=772.6$; $I_S(350)\approx7.7\times10^{-11}\,\text{A}$. (1)

(b) Over 50 K the current rose ~773×; per 10 K factor $=773^{1/5}=3.76$. (2) This is somewhat above but consistent in spirit with the common "leakage roughly doubles per ~10 °C" engineering rule. (1)

(c) $I_S\propto1/N_A$ (p-side term), so tenfold $N_A$ reduces $I_S$ by ~10×; at fixed forward current $V_F=nV_T\ln(I/I_S)$ increases by $V_T\ln10\approx60\,\text{mV}$. (2)

[
  {"claim":"Vbi = 0.753 V for given doping","code":"VT=0.02585; NA=2e17; ND=5e15; ni=1.5e10; Vbi=VT*log(NA*ND/ni**2); result = abs(float(Vbi)-0.753)<0.005"},
  {"claim":"Delta Vbi for 100x ND is ~119 mV","code":"VT=0.02585; d=VT*log(100); result = abs(float(d)-0.119)<0.002"},
  {"claim":"Depletion width ~0.447 um","code":"eps=11.7*8.854e-12; q=1.602e-19; NA=2e23; ND=5e21; Vbi=0.753; W=sqrt(2*eps/q*(1/NA+1/ND)*Vbi); result = abs(float(W)-4.47e-7)<0.1e-7"},
  {"claim":"n-side fraction of W is ~0.976","code":"NA=2e17; ND=5e15; f=NA/(NA+ND); result = abs(float(f)-0.976)<0.002"},
  {"claim":"Forward current at 0.6V ~0.24 mA","code":"IS=2e-14; VT=0.02585; I=IS*(exp(0.6/VT)-1); result = abs(float(I)-2.4e-4)<0.3e-4"},
  {"claim":"Decade voltage step ~59.5 mV for n=1","code":"VT=0.02585; dV=VT*log(10); result = abs(float(dV)-0.0595)<0.001"},
  {"claim":"Effective n from 118mV decade is ~2","code":"VT=0.02585; n=0.118/(VT*log(10)); result = abs(float(n)-1.98)<0.05"},
  {"claim":"IS ratio 300->350K is ~773","code":"k=1.381e-23; q=1.602e-19; Eg=1.12*q; r=(350/300)**3*exp(-Eg/k*(1/350-1/300)); result = abs(float(r)-772.6)<10"},
  {"claim":"Per-10K factor ~3.76","code":"f=772.6**(1/5); result = abs(float(f)-3.76)<0.05"}
]