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In a p-type semiconductor, the majority charge carriers are:

  1. A.Electrons
  2. B.Holes
  3. C.Protons
  4. D.Positive ions

Correct answer

B. Holes

Explanation

The correct answer is B, holes. A trivalent dopant such as boron, aluminium, gallium or indium forms only three covalent bonds with the surrounding silicon atoms, leaving one bond incomplete; that vacancy behaves as a mobile positive carrier called a hole, and holes therefore become the majority carriers while thermally generated electrons remain the minority carriers. Option A is wrong because electrons are the majority carriers in an n-type crystal and only the minority carriers here. Option C is wrong because protons are bound inside nuclei and never move through a crystal lattice. Option D is wrong because the dopant ions are fixed in the lattice and cannot carry current, even though they are charged. Remember also that a p-type crystal is electrically neutral overall, exactly like an n-type crystal.

Read the full article: Semiconductors and the Electronics Mission: Notes

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Q1.Science & TechnologyMedium

The forbidden energy gap of silicon at room temperature is approximately:

  1. A.0.3 eV
  2. B.0.7 eV
  3. C.1.1 eV
  4. D.5.5 eV
Show answer

Correct answer: C. 1.1 eV

Explanation

The correct answer is C, about 1.1 eV. The energy gap between the valence band and the conduction band decides whether a solid conducts: in silicon the gap is roughly 1.1 electron volts, small enough for heat or light to lift some electrons across it, which is what makes silicon a semiconductor. Option A is wrong because 0.3 volt is the forward barrier voltage of a germanium diode, a different quantity altogether. Option B is wrong because about 0.7 eV is the band gap of germanium, the other elemental semiconductor, and it is also the forward voltage drop of a silicon diode, which is why candidates confuse the two numbers. Option D is wrong because about 5.5 eV is the gap in diamond, an insulator. Learn the trio together: germanium 0.7 eV, silicon 1.1 eV, gallium arsenide about 1.43 eV.

Q2.Science & TechnologyAsked in: Delhi · 22 June 2022, Shift 1Medium

If np and ne is the number of holes and electrons respectively in an intrinsic semiconductor then:

  1. A.npne = 1
  2. B.np > ne
  3. C.np = ne
  4. D.np < ne
Show answer

Correct answer: C. np = ne

Explanation

The correct answer is C, the number of holes equals the number of electrons. In an intrinsic, that is a chemically pure, semiconductor every free electron is produced by breaking a covalent bond, and breaking a bond leaves behind exactly one hole, so electrons and holes are created strictly in pairs and their concentrations must be equal. Option A is wrong because a product of two concentrations cannot equal a bare number one; the product is fixed by temperature and the band gap, not by unity. Option B is wrong because holes can outnumber electrons only in a p-type crystal, which needs a trivalent dopant. Option D is wrong because electrons outnumber holes only in an n-type crystal, which needs a pentavalent dopant. Equal numbers are therefore the signature of the pure crystal, and any inequality means doping has taken place.

Q3.Science & TechnologyEasy

Which of the following impurities is added to silicon to make an n-type semiconductor?

  1. A.Boron
  2. B.Aluminium
  3. C.Phosphorus
  4. D.Indium
Show answer

Correct answer: C. Phosphorus

Explanation

The correct answer is C, phosphorus. Silicon has four valence electrons, so adding a pentavalent atom such as phosphorus, arsenic, antimony or bismuth leaves one electron over after four bonds are formed; that spare electron is loosely held and becomes a free carrier, making electrons the majority carriers and the crystal n-type. Option A is wrong because boron is trivalent and creates a vacancy, giving a p-type crystal. Option B is wrong because aluminium is also trivalent and is another p-type dopant. Option D is wrong because indium is trivalent as well. Note that an n-type crystal is not negatively charged: the donor atom keeps its positive nucleus, so the crystal as a whole stays electrically neutral, and examiners test exactly this point in statement-based questions.

Q4.Science & TechnologyMedium

The transistor was invented in 1947 at Bell Laboratories by which team of scientists?

  1. A.Bardeen, Brattain and Shockley
  2. B.Kilby, Noyce and Moore
  3. C.Faraday, Maxwell and Hertz
  4. D.Townes, Basov and Prokhorov
Show answer

Correct answer: A. Bardeen, Brattain and Shockley

Explanation

The correct answer is A, Bardeen, Brattain and Shockley. John Bardeen, Walter Brattain and William Shockley built the first working transistor at Bell Laboratories in the United States in 1947, and the three shared the Nobel Prize in Physics in 1956 for the discovery of the transistor effect; the device replaced the bulky vacuum tube and made all later electronics possible. Option B is wrong because Jack Kilby and Robert Noyce independently developed the integrated circuit in 1958 and 1959, and Gordon Moore gave the industry its famous doubling rule. Option C is wrong because Faraday, Maxwell and Hertz belong to the history of electromagnetism. Option D is wrong because Townes, Basov and Prokhorov won the Nobel Prize of 1964 for work on the maser and the laser.

Q5.Science & TechnologyAsked in: RRB ALP · 21 Jan 2019, Shift 3Medium

In a bipolar transistor, alpha is the ratio of:

  1. A.collector current to emitter current
  2. B.emitter current to collector current
  3. C.base current to collector current
  4. D.collector current to base current
Show answer

Correct answer: A. collector current to emitter current

Explanation

The correct answer is A, collector current to emitter current. In a bipolar junction transistor the emitter injects carriers, the thin lightly doped base lets most of them through, and the collector gathers them, so the emitter current divides into a large collector current and a small base current; alpha is defined as the collector current divided by the emitter current and is therefore always a little less than one, typically between 0.95 and 0.99. Option B is wrong because it inverts the ratio and would give a value greater than one. Option C is wrong because the base to collector ratio is a small fraction with no standard name. Option D is wrong because the collector current divided by the base current is beta, the current gain, which runs into the tens or hundreds. Keep alpha below one and beta well above one.