Exercises
(14.1) In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer
(c) Holes are minority carriers and pentavalent atoms are the dopants.
n-type silicon is made by doping with a pentavalent element (such as P, As or Sb). Four of its five valence electrons form covalent bonds and the fifth is donated as a free electron, so electrons become the majority carriers and holes the minority carriers.
(14.2) Which of the statements given in Exercise 14.1 is true for p-type semiconductors.
Answer
(d) Holes are majority carriers and trivalent atoms are the dopants.
p-type silicon is doped with a trivalent element (such as B, Al or In). Each dopant atom has one electron fewer than needed to complete four bonds, creating a hole, so holes are the majority carriers and electrons the minority carriers.
(14.3) Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (E_g)C, (E_g)Si and (E_g)Ge. Which of the following statements is true?
(a) (E_g)Si < (E_g)Ge < (E_g)C
(b) (E_g)C < (E_g)Ge > (E_g)Si
(c) (E_g)C > (E_g)Si > (E_g)Ge
(d) (E_g)C = (E_g)Si = (E_g)Ge
Answer
(c) (E_g)C > (E_g)Si > (E_g)Ge
Carbon (diamond) has a band gap of about 5.4 eV, silicon about 1.1 eV and germanium about 0.7 eV. The band gap decreases going down the group, because the valence electrons of larger atoms are less tightly bound.
(14.4) In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
Answer
(c) hole concentration in p-region is more as compared to n-region.
Diffusion always takes place from a region of higher concentration to one of lower concentration. The p-side has many holes and the n-side very few, so holes diffuse into the n-region. The barrier potential actually opposes this flow, so (b) is wrong.
Q5. (14.5) When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above.
Ans. (c) lowers the potential barrier.
In forward bias the applied voltage opposes the built-in barrier potential, so the effective barrier height becomes (V₀ − V). The depletion layer narrows and majority carriers cross the junction easily, giving a large forward current.
(14.6) In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.
Answer
Half-wave rectifier: only one half of each input cycle passes, so there is one output pulse per input cycle
⇒ output frequency = 50 Hz.
Full-wave rectifier: both halves of each input cycle give an output pulse in the same direction, so there are two pulses per input cycle
⇒ output frequency = 2 × 50 = 100 Hz.
Important Extra Questions
Distinguish between conductors, semiconductors and insulators on the basis of energy bands.
Answer
In conductors the valence and conduction bands overlap (or the conduction band is partly filled), so electrons move freely. In insulators the band gap is large (E_g > 3 eV) and no electrons reach the conduction band at room temperature. In semiconductors the gap is small (E_g < 3 eV), so some electrons are thermally excited into the conduction band.
What is the depletion region and how is it formed?
Answer
When a p-n junction forms, electrons diffuse from n to p and holes from p to n, leaving behind immobile positive donor ions on the n-side and negative acceptor ions on the p-side near the junction. This thin layer, emptied of free charge carriers, is the depletion region; its ions set up the barrier potential.
Why does the reverse current of a p-n junction diode stay almost constant before breakdown?
Answer
Reverse current is due to minority carriers drifting across the junction. Their number depends on temperature, not on the applied voltage, so the current stays small and nearly constant (saturation current) until breakdown.
Draw the I–V characteristic shape of a silicon diode and state its threshold voltage.
Answer
In forward bias the current is negligible until the threshold (cut-in) voltage, about 0.7 V for silicon (0.2 V for germanium), after which it rises steeply (exponentially). In reverse bias a tiny, almost constant current flows (microamperes), until a sharp rise at the breakdown voltage.
What is the intrinsic carrier concentration relation, and how does doping change it?
Answer
In an intrinsic semiconductor nₑ = n_h = nᵢ. In a doped (extrinsic) semiconductor at thermal equilibrium nₑ × n_h = nᵢ², so increasing one type of carrier by doping decreases the other.