Level 1 — RecognitionDoping & PN Junctions

Doping & PN Junctions

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Time limit: 20 minutes Total marks: 30

Section A — Multiple Choice (1 mark each)

Choose the single best answer.

Q1. Phosphorus added to silicon produces N-type material because phosphorus has:

(a) 3 valence electrons
(b) 4 valence electrons
(c) 5 valence electrons
(d) 6 valence electrons

Q2. Boron is used to create P-type silicon because it acts as a(n):

(a) donor atom
(b) acceptor atom
(c) intrinsic atom
(d) recombination center

Q3. In an energy-band diagram, the donor energy level lies:

(a) just above the valence band
(b) in the middle of the band gap
(c) just below the conduction band
(d) inside the conduction band

Q4. The depletion region of a PN junction is characterized by:

(a) a high concentration of free carriers
(b) a region depleted of mobile charge carriers
(c) a region of pure metal
(d) zero electric field

Q5. Under forward bias, the width of the depletion region:

(a) increases
(b) decreases
(c) stays the same
(d) becomes infinite

Q6. The Shockley diode equation is:

(a) $I = I_S\left(e^{V/nV_T} - 1\right)$
(b) $I = I_S\left(e^{nV_T/V} + 1\right)$
(c) $I = V/R$
(d) $I = I_S\, V^2$

Q7. At room temperature the thermal voltage $V_T = kT/q$ is approximately:

(a) 0.7 V
(b) 26 mV
(c) 1.1 V
(d) 300 mV

Q8. Zener breakdown (as distinct from avalanche) dominates in junctions that are:

(a) lightly doped, wide depletion region
(b) heavily doped, narrow depletion region
(c) undoped intrinsic
(d) reverse-current-free

Q9. Reverse saturation current $I_S$ increases when:

(a) temperature rises
(b) reverse voltage decreases
(c) doping increases
(d) the diode is forward biased only

Q10. In reverse bias, the small current that flows is due to:

(a) majority carriers only
(b) minority carriers (thermally generated)
(c) tunnelling only
(d) no current at all

Section B — Matching (1 mark each, Q11 total 5 marks)

Q11. Match each term (i–v) to its correct description (A–E).

Term	Description
(i) Arsenic	(A) Voltage across junction with no external bias
(ii) Built-in potential	(B) Capacitance dominant in forward bias
(iii) Diffusion capacitance	(C) Group-V donor dopant
(iv) Space charge	(D) Immobile ionised dopant charge in depletion region
(v) Avalanche	(E) Carrier multiplication by impact ionisation

Section C — True/False WITH justification (2 marks each: 1 T/F, 1 justification)

Q12. An N-type semiconductor carries a net negative electric charge. (T/F + justify)

Q13. Increasing the reverse voltage widens the depletion region. (T/F + justify)

Q14. The junction (depletion) capacitance increases as reverse bias increases. (T/F + justify)

Q15. A silicon diode conducts significantly as soon as any small forward voltage is applied. (T/F + justify)

Q16. Acceptor levels lie just above the valence band. (T/F + justify)

Q17. The built-in potential can be measured directly with a voltmeter across the terminals of an unbiased diode. (T/F + justify)

Q18. At the same magnitude of voltage, the "–1" term in the Shockley equation is negligible under strong forward bias. (T/F + justify)

Answer keyMark scheme & solutions

Section A (10 marks)

Q1 — (c). Phosphorus is Group V with 5 valence electrons; the fifth is weakly bound and becomes a free conduction electron → donor. (1)

Q2 — (b). Boron is Group III (3 valence electrons); it creates a hole and accepts an electron → acceptor. (1)

Q3 — (c). Donor levels sit just below the conduction band edge (small ionisation energy), so electrons are easily excited into the conduction band. (1)

Q4 — (b). Diffusion leaves the junction region stripped of mobile carriers, exposing fixed ionised dopants. (1)

Q5 — (b). Forward bias reduces the barrier and injects carriers, narrowing the depletion width. (1)

Q6 — (a). Correct Shockley form with ideality factor $n$ and thermal voltage $V_T$ . (1)

Q7 — (b). $V_T = kT/q \approx 0.0259$ V ≈ 26 mV at 300 K. (1)

Q8 — (b). Heavy doping → thin depletion region → high field at low voltage → tunnelling (Zener). (1)

Q9 — (a). $I_S$ depends on thermally generated minority carriers, which grow strongly with temperature. (1)

Q10 — (b). Reverse current is the minority-carrier (thermal generation) saturation current. (1)

Section B (5 marks)

Q11: (i)→C, (ii)→A, (iii)→B, (iv)→D, (v)→E. (1 each)

Section C (14 marks)

Q12 — FALSE. (1) Each donor ion's positive core is balanced by its electron; the crystal is overall electrically neutral. N-type refers to majority-carrier type, not net charge. (1)

Q13 — TRUE. (1) Reverse bias pulls carriers away from the junction, uncovering more ionised dopants, so the depletion width grows ( $W \propto \sqrt{V_{bi}+V_R}$ ). (1)

Q14 — FALSE. (1) Depletion capacitance $C_j = \varepsilon A / W$ ; since $W$ increases with reverse bias, $C_j$ decreases. (1)

Q15 — FALSE. (1) Current is negligible until forward voltage approaches the ~0.6–0.7 V threshold, where the exponential term rises sharply. (1)

Q16 — TRUE. (1) Acceptor levels lie just above the valence band, so electrons are easily excited from the valence band into them, leaving holes. (1)

Q17 — FALSE. (1) The built-in potential is balanced by contact potentials around the circuit; no net EMF appears externally, so a voltmeter reads ~0 V. (1)

Q18 — TRUE. (1) Under strong forward bias $e^{V/nV_T} \gg 1$ , so $I \approx I_S e^{V/nV_T}$ and the –1 is negligible. (1)

[
  {"claim":"Thermal voltage at 300 K is approx 26 mV","code":"k=1.380649e-23; q=1.602176634e-19; T=300; VT=k*T/q; result = abs(VT-0.026) < 0.001"},
  {"claim":"Shockley -1 term negligible for V=0.5V, n=1 forward bias","code":"import sympy as sp; VT=0.0259; V=0.5; ratio=sp.exp(V/VT); result = float(ratio) > 1000"},
  {"claim":"Depletion capacitance decreases as width increases (Cj = eps*A/W)","code":"eps=1.0; A=1.0; W1=1.0; W2=2.0; C1=eps*A/W1; C2=eps*A/W2; result = C2 < C1"},
  {"claim":"Depletion width grows with reverse voltage (W ~ sqrt(Vbi+VR))","code":"import sympy as sp; Vbi=0.7; W=lambda VR: sp.sqrt(Vbi+VR); result = float(W(5)) > float(W(0))"}
]