Level 2 — RecallDoping & PN Junctions

Doping & PN Junctions

⏱ 30 minutes50 marksprintable — key stays hidden on paper

Topic: Semiconductor Doping and PN Junctions Level: 2 — Recall & Standard Problems Time Limit: 30 minutes Total Marks: 50

Use $k = 1.38\times10^{-23}\ \text{J/K}$ , $q = 1.6\times10^{-19}\ \text{C}$ , thermal voltage $V_T = kT/q \approx 25.85\ \text{mV}$ at $T=300\ \text{K}$ unless told otherwise. Take $n_i = 1.5\times10^{10}\ \text{cm}^{-3}$ for silicon.

Q1. (4 marks) (a) Define N-type and P-type doping. (2) (b) Give one common donor atom and one common acceptor atom used in silicon, and state how many valence electrons each has. (2)

Q2. (4 marks) Explain, with reference to the band gap, where the donor energy level and acceptor energy level lie. State which level is close to the conduction band and which is close to the valence band, and why.

Q3. (5 marks) Describe the formation of the depletion region when a PN junction is created. In your answer explain (a) what happens to majority carriers near the junction, and (b) what "space charge" refers to.

Q4. (5 marks) The built-in potential of a step junction is $V_{bi} = V_T \ln\!\left(\frac{N_A N_D}{n_i^2}\right).$ A silicon junction has $N_A = 1\times10^{17}\ \text{cm}^{-3}$ and $N_D = 1\times10^{16}\ \text{cm}^{-3}$ at $300\ \text{K}$ . Calculate $V_{bi}$ .

Q5. (6 marks) Compare forward bias and reverse bias for a PN junction. For each, state: (a) the polarity of the applied voltage relative to the P side, (b) the effect on the depletion region width, (c) the resulting current behaviour. (2 marks each row)

Q6. (6 marks) The Shockley diode equation is $I = I_S\left(e^{V/(nV_T)} - 1\right).$ (a) Define each symbol. (3) (b) For a diode with $I_S = 1\times10^{-14}\ \text{A}$ , $n = 1$ , $V_T = 25.85\ \text{mV}$ , calculate the current at a forward voltage of $V = 0.6\ \text{V}$ . (3)

Q7. (5 marks) (a) What is the reverse saturation current $I_S$ ? (2) (b) State two physical factors that increase $I_S$ . (2) (c) Why does raising temperature strongly increase $I_S$ ? (1)

Q8. (6 marks) Distinguish between avalanche breakdown and Zener breakdown in a reverse-biased junction. For each, describe the physical mechanism and state whether it dominates in lightly-doped or heavily-doped junctions.

Q9. (5 marks) (a) Name the two contributions to a diode's junction capacitance. (2) (b) State which one dominates under reverse bias and which dominates under forward bias. (2) (c) State how the depletion (junction) capacitance changes as reverse voltage increases. (1)

Q10. (4 marks) Sketch (or describe in words) the diode I–V characteristic curve. Label: the forward "knee"/turn-on voltage, the small reverse saturation current region, and the reverse breakdown region.

Answer keyMark scheme & solutions

Q1. (4 marks) (a) N-type: doping with donor atoms that donate free electrons; majority carriers are electrons (1). P-type: doping with acceptor atoms that create holes; majority carriers are holes (1). (b) Donor: phosphorus or arsenic — 5 valence electrons (1). Acceptor: boron — 3 valence electrons (1).

Q2. (4 marks) Donor level lies just below the conduction band (~small ionization energy) so electrons are easily excited into the conduction band (2). Acceptor level lies just above the valence band, so valence electrons are easily promoted into it, leaving holes (2). Why: small energy gaps mean near-full ionization at room temperature.

Q3. (5 marks) On contact, electrons from the N side diffuse into the P side and holes from the P side diffuse into the N side (1). Near the junction majority carriers recombine and are removed, leaving the region depleted of mobile carriers (2). The fixed ionized dopant ions left behind (positive donors on N side, negative acceptors on P side) form the space charge (1). This space charge sets up an internal electric field opposing further diffusion (1).

Q4. (5 marks) $V_{bi} = 0.02585 \ln\!\left(\frac{(10^{17})(10^{16})}{(1.5\times10^{10})^2}\right)$ Numerator $= 10^{33}$ ; denominator $=2.25\times10^{20}$ (1). Ratio $=4.44\times10^{12}$ (1). $\ln(4.44\times10^{12}) = 29.12$ (1). $V_{bi} = 0.02585 \times 29.12 = 0.753\ \text{V}$ (2). Answer ≈ 0.75 V.

Q5. (6 marks)

	Forward bias	Reverse bias
(a) Polarity on P	Positive (+)	Negative (−)
(b) Depletion width	Narrows	Widens
(c) Current	Large, rises exponentially	Very small (~ $I_S$ ) until breakdown

2 marks per correct row.

Q6. (6 marks) (a) $I$ = diode current; $I_S$ = reverse saturation current; $V$ = applied voltage; $n$ = ideality factor; $V_T=kT/q$ thermal voltage (3, any 3 correct ×1). (b) Exponent $=0.6/0.02585 = 23.21$ (1). $e^{23.21}=1.20\times10^{10}$ (1). $I = 10^{-14}(1.20\times10^{10}-1) \approx 1.2\times10^{-4}\ \text{A} = 0.12\ \text{mA}$ (1).

Q7. (5 marks) (a) The small, nearly voltage-independent current that flows under reverse bias, due to minority carriers drifting across the junction (2). (b) Any two: higher temperature; smaller band gap material; larger junction area; higher minority carrier concentration/lower doping (2). (c) $I_S \propto n_i^2 \propto e^{-E_g/kT}$ ; raising T sharply increases $n_i^2$ , roughly doubling $I_S$ every ~10 °C (1).

Q8. (6 marks) Avalanche: high reverse field accelerates carriers which impact-ionize, creating a multiplication cascade of electron–hole pairs; dominates in lightly-doped junctions (wide depletion region), higher breakdown voltages (3). Zener: very strong field in a narrow depletion region causes direct quantum tunnelling of electrons across the gap; dominates in heavily-doped junctions, low breakdown voltages (≲5 V) (3).

Q9. (5 marks) (a) Depletion (junction) capacitance and diffusion capacitance (2). (b) Depletion capacitance dominates under reverse bias; diffusion capacitance dominates under forward bias (2). (c) Depletion capacitance decreases as reverse voltage increases (depletion width grows, $C_j \propto 1/\sqrt{V_{bi}+V_R}$ ) (1).

Q10. (4 marks) Curve description:

Forward region: current ≈ 0 until knee/turn-on voltage (~0.7 V Si, 0.3 V Ge), then rises steeply (1.5).
Reverse region: tiny constant reverse saturation current $-I_S$ (1.5).
At large negative voltage, sharp breakdown (current increases rapidly, near-vertical) (1).

[
  {"claim":"Vbi for NA=1e17, ND=1e16, ni=1.5e10 is approx 0.753 V","code":"VT=0.02585; NA=1e17; ND=1e16; ni=1.5e10; Vbi=VT*log((NA*ND)/(ni**2)); result = abs(float(Vbi)-0.753)<0.005"},
  {"claim":"Diode current at V=0.6 with Is=1e-14, n=1, VT=0.02585 is approx 1.2e-4 A","code":"Is=1e-14; VT=0.02585; V=0.6; I=Is*(exp(V/VT)-1); result = abs(float(I)-1.2e-4)<0.2e-4"},
  {"claim":"ln argument ratio equals 4.44e12","code":"ni=1.5e10; ratio=(1e17*1e16)/(ni**2); result = abs(float(ratio)-4.444e12)<1e10"}
]