Semiconductor Troubleshooting MCQ Quiz — Part 1 of 5

Fundamentals — Diodes & PN Junction | 50 Practice Questions (Q1–Q50 of 250) | For ITI Electrician, Electronics Mechanic, Diploma ECE/EEE and industrial electronics maintenance trainees

Topic Focus: Fundamentals — Diodes & PN Junction
Covers intrinsic/extrinsic semiconductors, doping, the PN junction and depletion region, diode types (rectifier, Zener, Schottky, LED, photodiode, varactor), diode characteristics and multimeter diode testing.
This is Part 1 of a 5-part, 250-question Semiconductor Troubleshooting practice series covering real-world fault-diagnosis scenarios.

Question Palette (tap any number to jump)

Source basis: Standard semiconductor device physics and practical electronics troubleshooting principles as taught in ITI/Diploma electronics curricula, consistent with well-established component references (diode/transistor datasheets, IC families and standard test procedures). Prepared for practice and skill-building purposes; always follow your equipment manufacturer's manual and workplace safety procedures.

About this practice set

This page is Part 1 of 5 in the Semiconductor Troubleshooting MCQ Quiz series. It contains 50 multiple-choice questions (Q1 to Q50 of the 250-question set) on Fundamentals — Diodes & PN Junction. The questions are written for ITI, diploma and B.Tech trainees and for anyone preparing for a maintenance or fault-finding interview. Each question has four options; the quiz above checks your answer and shows an explanation, and the full answer key is printed below so you can also revise without running the quiz.

How to use this quiz

Attempt the questions in one sitting without searching for answers.
Read the explanation shown after each answer, even when you were right.
Write down the questions you missed and the reason in your own words.
Open the answer key below to revise those questions.
Retake the quiz after a day or two and compare your score.

Sample of what is covered

What is an intrinsic semiconductor?
What forms at the junction between p-type and n-type material when no external voltage is applied?
What primarily causes the small reverse saturation current in a reverse-biased diode?
The commonly used 1N4007 rectifier diode has what approximate PIV rating?
If a diode shows OL (open) in both directions during a multimeter diode test, what does this indicate?
A varactor (varicap) diode is used in circuits primarily because it acts as what kind of component?
What is 'thermal runaway' as it applies to a forward-biased diode?
What is 'recombination' in a semiconductor material?

Answer key with explanations (50 questions)

Q1. What is an intrinsic semiconductor?

A. A semiconductor doped with trivalent impurities

B. A semiconductor doped with pentavalent impurities

C. A semiconductor mixed with a metal conductor

D. A pure semiconductor with no added impurities ✓ Correct answer

Explanation: An intrinsic semiconductor is a pure crystal such as silicon or germanium in which electron and hole concentrations are equal because no doping atoms have been added.

Q2. Why is silicon used far more often than germanium in modern semiconductor devices?

A. Germanium cannot be doped with impurities

B. Silicon has a smaller bandgap than germanium

C. Silicon has a larger bandgap giving better thermal stability and lower leakage current ✓ Correct answer

D. Silicon is a better electrical conductor than germanium

Explanation: Silicon's larger bandgap (about 1.1 eV versus 0.66 eV for germanium) means it can operate at higher temperatures with much lower reverse leakage current, making it far more stable in practical circuits.

Q3. Which type of impurity is added to pure silicon to create an n-type semiconductor?

A. A trivalent (3-valence-electron) impurity such as boron or indium

B. An inert gas impurity

C. A tetravalent impurity identical to silicon

D. A pentavalent (5-valence-electron) impurity such as phosphorus or arsenic ✓ Correct answer

Explanation: Pentavalent atoms like phosphorus contribute one extra free electron beyond the four needed for covalent bonding, making electrons the majority carriers in n-type material.

Q4. Which type of impurity is added to pure silicon to create a p-type semiconductor?

A. A pentavalent impurity such as phosphorus

B. A hexavalent impurity such as sulphur

C. A trivalent impurity such as boron, indium, or gallium ✓ Correct answer

D. Pure carbon

Explanation: Trivalent atoms have only three valence electrons, so one covalent bond position is left empty, creating a hole; holes are the majority carriers in p-type material.

Q5. In an n-type semiconductor, what are the majority charge carriers?

A. Holes

B. Protons

C. Positive ions

D. Free electrons ✓ Correct answer

Explanation: The donor (pentavalent) atoms release loosely bound extra electrons into the conduction band, so free electrons vastly outnumber holes in n-type material.

Q6. In a p-type semiconductor, what are the majority charge carriers?

A. Free electrons

B. Donor ions

C. Neutrons

D. Holes ✓ Correct answer

Explanation: Acceptor (trivalent) atoms create vacant bonding positions called holes, which behave as mobile positive charge carriers and dominate conduction in p-type material.

Q7. What forms at the junction between p-type and n-type material when no external voltage is applied?

A. An open circuit with infinite resistance everywhere

B. A short circuit between the two regions

C. A depletion region with a built-in barrier potential ✓ Correct answer

D. A permanent magnetic field

Explanation: Diffusion of electrons and holes across the junction leaves behind fixed ions, creating a narrow depletion region and an internal electric field that opposes further diffusion, called the barrier potential.

Q8. What is the approximate barrier (junction) potential of a silicon PN junction diode at room temperature?

A. 0.3 V

B. 5 V

C. 0.7 V ✓ Correct answer

D. 1.5 V

Explanation: Silicon diodes typically need about 0.6-0.7 V of forward bias before significant forward current flows, a value used as a standard reference in circuit analysis.

Q9. What is the approximate barrier potential of a germanium PN junction diode?

A. 0.3 V ✓ Correct answer

B. 1.2 V

C. 0.7 V

D. 2.1 V

Explanation: Germanium has a smaller bandgap than silicon, so its junction turns on at a lower forward voltage, typically around 0.2-0.3 V.

Q10. What happens to the depletion region width when a PN junction diode is forward biased?

A. It remains exactly the same width regardless of bias

B. It converts into a conductor with zero resistance immediately

C. It narrows, allowing current to flow easily once the barrier potential is overcome ✓ Correct answer

D. It widens, blocking current flow completely

Explanation: Forward bias pushes majority carriers toward the junction, reducing the width of the depletion region and lowering the barrier so current can flow once the threshold voltage is exceeded.

Q11. What happens to the depletion region when a PN junction diode is reverse biased?

A. It disappears entirely

B. It narrows to zero width

C. It starts conducting heavily

D. It widens, increasing the barrier and allowing only a tiny leakage current ✓ Correct answer

Explanation: Reverse bias pulls majority carriers away from the junction, widening the depletion region and increasing its resistance so only a small reverse saturation (leakage) current flows.

Q12. What does the 'knee voltage' on a diode's forward V-I characteristic curve represent?

A. The voltage drop when the diode is fully reverse biased

B. The voltage at which the diode is destroyed

C. The maximum reverse voltage the diode can withstand

D. The forward voltage at which current begins to rise sharply ✓ Correct answer

Explanation: Below the knee voltage almost no forward current flows; once the applied voltage exceeds this point (about 0.7 V for silicon) current increases rapidly for small further increases in voltage.

Q13. What primarily causes the small reverse saturation current in a reverse-biased diode?

A. Leakage through the diode's plastic casing only

B. Majority carriers being pushed across the junction

C. Thermally generated minority carriers crossing the junction ✓ Correct answer

D. A short circuit inside the diode

Explanation: Heat continuously generates electron-hole pairs; the minority carriers on each side are swept across the junction by the reverse field, producing a small, temperature-dependent leakage current.

Q14. What is the avalanche breakdown mechanism in a reverse-biased diode?

A. Majority carriers tunnel directly through the depletion region

B. The diode's doping level spontaneously changes

C. The junction physically melts due to excess forward current

D. High reverse voltage accelerates minority carriers, which collide with atoms and knock loose more carriers in a chain reaction ✓ Correct answer

Explanation: At sufficiently high reverse voltage, minority carriers gain enough kinetic energy to knock additional electrons free on collision, multiplying the current rapidly; this dominates breakdown in lightly doped, higher-voltage diodes.

Q15. What distinguishes the Zener breakdown mechanism from avalanche breakdown?

A. Zener breakdown occurs when a very strong electric field in a heavily doped, narrow depletion region directly pulls electrons out of covalent bonds ✓ Correct answer

B. Zener breakdown only occurs under forward bias

C. Zener breakdown destroys the diode permanently

D. Zener breakdown requires no electric field at all

Explanation: Heavy doping makes the depletion region extremely thin, so a modest reverse voltage produces a field strong enough to directly rupture covalent bonds (a quantum tunneling effect), which dominates in diodes rated below about 5 V.

Q16. What is the primary practical application of a Zener diode?

A. Converting light energy into electrical energy

B. Voltage regulation by maintaining a nearly constant voltage when reverse biased in breakdown ✓ Correct answer

C. Amplifying weak signals

D. Rectifying AC to DC in power supplies

Explanation: Once in reverse breakdown, a Zener diode's voltage stays almost constant over a wide current range, so it is used as a simple voltage reference or shunt regulator.

Q17. Why must a series resistor always be used with a Zener diode in a voltage regulator circuit?

A. To increase the output voltage above the supply voltage

B. To reverse the polarity of the input signal

C. To convert the Zener diode into a rectifier

D. To limit current through the Zener diode and prevent it from being destroyed by excessive power dissipation ✓ Correct answer

Explanation: Without a current-limiting series resistor, the low dynamic resistance of a Zener in breakdown would let current rise uncontrollably, exceeding its power rating and destroying it.

Q18. What does PIV stand for when rating a rectifier diode?

A. Peak Input Voltage of the power supply

B. Positive Increment Voltage across the load

C. Peak Inverse Voltage, the maximum reverse voltage the diode can safely withstand ✓ Correct answer

D. Power Input Value in watts

Explanation: PIV specifies the highest reverse voltage the diode can block without breaking down; exceeding it in a circuit can permanently damage the diode.

Q19. The commonly used 1N4007 rectifier diode has what approximate PIV rating?

A. 100 V

B. 400 V

C. 50 V

D. 1000 V ✓ Correct answer

Explanation: The 1N4001-1N4007 series shares a 1 A forward current rating, but the PIV rating increases through the series, with the 1N4007 rated at 1000 V, the highest in that family.

Q20. What is the maximum average forward current rating of a 1N4007 diode?

A. 1 A ✓ Correct answer

B. 10 A

C. 100 mA

D. 5 A

Explanation: The 1N400x series of general-purpose rectifier diodes is rated for 1 A average forward current, sufficient for small power supplies but not for heavy industrial loads.

Q21. On the body of a typical axial-lead rectifier diode, what does the painted band indicate?

A. The cathode terminal ✓ Correct answer

B. The positive supply pin

C. The diode's wattage rating

D. The anode terminal

Explanation: The band corresponds to the bar in the diode's schematic symbol and marks the cathode; current flows conventionally from anode to cathode only when forward biased.

Q22. When a digital multimeter's diode-test mode is connected with the red probe on the anode and black probe on the cathode of a good silicon diode, what reading is expected?

A. A voltage drop reading of roughly 0.5 to 0.7 V ✓ Correct answer

B. A negative voltage reading

C. 0 V, indicating a dead short

D. OL (open loop / infinite resistance)

Explanation: In diode-test mode the meter forces a small current through the junction and displays the forward voltage drop, which for a healthy silicon diode falls around 0.5-0.7 V.

Q23. When the probes are reversed on a healthy diode in diode-test mode, what should the multimeter display?

A. The same forward voltage drop as before

B. A short beep with a low resistance reading

C. 0 V exactly

D. OL or 1, indicating the junction is blocking, effectively an open circuit ✓ Correct answer

Explanation: With the junction reverse biased, only a negligible leakage current flows, so a good diode shows an open-circuit (overload/OL) reading in the reverse direction.

Q24. If a diode shows a low voltage/resistance reading in BOTH directions when tested with a multimeter, what does this indicate?

A. The diode is a Zener diode operating correctly

B. The multimeter battery is weak

C. The diode has failed and is shorted ✓ Correct answer

D. The diode is functioning normally

Explanation: A healthy diode should block conduction in one direction; a low reading both ways means the junction has broken down internally and the diode must be replaced.

Q25. If a diode shows OL (open) in both directions during a multimeter diode test, what does this indicate?

A. The diode is reverse biased correctly

B. The diode is open-circuited (blown) internally ✓ Correct answer

C. The diode is functioning normally

D. The multimeter is set to the wrong function

Explanation: A functional diode must conduct in the forward direction; showing an open circuit both ways means the junction has burned open, usually from excessive current or a voltage transient.

Q26. What is the main advantage of a Schottky diode over a standard PN junction rectifier diode?

A. A much higher reverse breakdown voltage in all cases

B. It emits visible light when forward biased

C. A much lower forward voltage drop and very fast switching due to its metal-semiconductor junction ✓ Correct answer

D. It can amplify AC signals

Explanation: A Schottky diode uses a metal-to-semiconductor junction instead of a PN junction, giving a forward drop of only about 0.2-0.3 V and negligible charge storage, so it switches on and off very quickly with less power loss.

Q27. What is the typical forward voltage drop of a silicon Schottky diode?

A. About 3.3 V

B. About 1.5 V

C. About 0.7 V

D. About 0.2 to 0.3 V ✓ Correct answer

Explanation: Because it lacks the minority-carrier injection of a PN junction, a Schottky diode conducts at a much lower forward voltage than a standard silicon diode.

Q28. Why are Schottky diodes commonly used as rectifiers in switch-mode power supplies (SMPS)?

A. They can withstand extremely high reverse voltages above 5000 V

B. They only work with AC mains frequency

C. Their low forward drop and fast recovery reduce switching losses at high frequency ✓ Correct answer

D. They are cheaper than any other diode type regardless of application

Explanation: SMPS circuits switch at tens to hundreds of kHz, so a diode's recovery time and forward drop directly affect efficiency; Schottky diodes minimize both, cutting conduction and switching losses.

Q29. A photodiode used as a light sensor is normally operated in which mode?

A. AC bias that alternates polarity every cycle

B. Forward bias

C. Reverse bias ✓ Correct answer

D. Zero bias only, never any external voltage

Explanation: Reverse biasing widens the depletion region and lets the diode's reverse leakage current vary almost linearly with incident light intensity, giving a usable sensor signal.

Q30. How does increasing light intensity affect a reverse-biased photodiode?

A. It increases the reverse current roughly in proportion to the light intensity ✓ Correct answer

B. It decreases the reverse current to zero

C. It has no effect on the diode's current

D. It forces the diode into forward conduction

Explanation: Photons striking the depletion region generate additional electron-hole pairs, which are swept across by the reverse field, so more light produces more reverse (photo) current.

Q31. A varactor (varicap) diode is used in circuits primarily because it acts as what kind of component?

A. A light-emitting element

B. A voltage-controlled capacitor whose capacitance changes with reverse bias voltage ✓ Correct answer

C. A fixed-value resistor

D. A high-power rectifier for mains supplies

Explanation: The depletion region of a reverse-biased junction acts like the dielectric of a capacitor; as reverse voltage increases the depletion width increases and capacitance decreases, making it useful for tuning circuits like radio receivers.

Q32. What primarily determines the color of light an LED emits?

A. The bandgap energy of the semiconductor compound used ✓ Correct answer

B. The polarity of the applied voltage

C. The current-limiting resistor value in series with it

D. The physical size of the LED package

Explanation: When electrons recombine with holes, they release energy as a photon whose wavelength (and hence color) depends on the semiconductor material's bandgap energy, which is why different compounds like GaAs or GaN produce different colors.

Q33. Why must a current-limiting resistor be connected in series with an LED?

A. An LED has very low forward resistance, so without limiting the current it would draw excessive current and burn out ✓ Correct answer

B. To reverse the polarity of the LED automatically

C. To make the LED brighter than its rated maximum

D. To convert AC to DC before it reaches the LED

Explanation: Once an LED's forward voltage is reached, its current rises very steeply for small voltage increases, so an unregulated supply would quickly exceed the LED's rated current and destroy it.

Q34. Compared to a standard red LED, a blue or white LED typically has what forward voltage characteristic?

A. An identical forward voltage drop in all cases

B. No measurable forward voltage at all

C. A noticeably higher forward voltage drop, around 3 to 3.4 V versus about 1.8 to 2.2 V for red ✓ Correct answer

D. A much lower forward voltage drop

Explanation: Blue and white LEDs use wider-bandgap materials such as gallium nitride, which require a higher forward voltage to emit their shorter-wavelength, higher-energy photons than red LEDs made from lower-bandgap materials.

Q35. What happens if an LED is connected with reversed polarity across a low-voltage supply within its reverse voltage rating?

A. It lights up brighter than normal

B. It converts to a photodiode permanently

C. It simply does not light because it is reverse biased, and is normally undamaged ✓ Correct answer

D. It short-circuits the entire power supply

Explanation: An LED is a diode, so reverse bias blocks conduction and no light is produced; however, LEDs have a low reverse breakdown voltage, so excessive reverse voltage can still damage them.

Q36. What is the purpose of a freewheeling (flyback) diode connected across a relay coil driven by a transistor?

A. To increase the switching speed of the relay

B. To reverse the direction of current through the relay coil

C. To reduce the relay's holding current to zero

D. To provide a safe path for the collapsing magnetic field's induced current, protecting the transistor from a voltage spike ✓ Correct answer

Explanation: When the transistor switches off, the coil's collapsing magnetic field induces a high reverse voltage spike; the diode, connected in reverse across the coil, conducts this energy safely instead of letting it destroy the switching transistor.

Q37. What is 'thermal runaway' as it applies to a forward-biased diode?

A. A condition where the diode cools itself automatically as current increases

B. A self-reinforcing cycle where rising temperature lowers forward voltage drop, increasing current, which raises temperature further until failure ✓ Correct answer

C. A normal, harmless operating condition for all diodes

D. A phenomenon that only occurs in reverse-biased diodes

Explanation: Since a diode's forward voltage drop decreases with rising temperature, if current is not properly limited, more current flows as the diode heats up, generating even more heat in a destructive positive-feedback loop.

Q38. What does a diode clipper circuit do to an input waveform?

A. Removes or limits the portion of the waveform above or below a set voltage level ✓ Correct answer

B. Converts AC to a pure DC voltage with no ripple

C. Shifts the entire waveform up or down without changing its shape

D. Amplifies the waveform's peak amplitude

Explanation: By placing a diode, sometimes with a bias voltage, across or in series with the signal path, a clipper cuts off, or 'clips', the part of the waveform that exceeds the diode's conduction threshold.

Q39. What does a diode clamper circuit do to an input waveform?

A. Removes the peaks of the waveform entirely

B. Shifts the DC reference level of the waveform up or down while preserving its shape ✓ Correct answer

C. Converts a sine wave into a square wave

D. Rectifies the AC waveform into pure DC

Explanation: A clamper uses a diode and capacitor to add or subtract a DC offset so the waveform's shape is unchanged but its baseline is shifted to a new reference level, unlike a clipper which removes part of the signal.

Q40. How does a diode's reverse leakage current typically change as its temperature rises?

A. It becomes negative

B. It decreases to zero regardless of the diode type

C. It remains completely constant with temperature

D. It increases, roughly doubling for every 10 degree Celsius rise in silicon diodes ✓ Correct answer

Explanation: Higher temperature generates more thermally produced electron-hole pairs, which are swept across the junction under reverse bias, so leakage current rises sharply with temperature, a key reason for thermal derating.

Q41. How does the forward voltage drop of a silicon diode typically change as temperature increases?

A. It becomes negative above 50 degree Celsius

B. It increases sharply and unpredictably

C. It decreases, by roughly 2 mV per degree Celsius for a given forward current ✓ Correct answer

D. It stays exactly fixed at 0.7 V regardless of temperature

Explanation: As temperature rises, thermal energy makes it easier for carriers to cross the junction, so less applied voltage is needed to sustain a given forward current, a well-known negative temperature coefficient of about -2 mV/degree C.

Q42. How does increasing the doping concentration of a semiconductor generally affect its conductivity?

A. It has no effect on conductivity whatsoever

B. It decreases conductivity because doping atoms block current flow

C. It increases conductivity because more majority charge carriers become available ✓ Correct answer

D. It converts the semiconductor into a perfect insulator

Explanation: Adding more donor or acceptor atoms increases the number of free electrons or holes available to carry current, which lowers resistivity and raises conductivity.

Q43. What is 'recombination' in a semiconductor material?

A. The depletion region doubles in width

B. Two holes merge to form a free electron

C. A dopant atom splits into two separate atoms

D. A free electron falls into a hole, releasing energy and effectively neutralizing both carriers ✓ Correct answer

Explanation: When a conduction-band electron meets a valence-band hole, it drops into the vacant bond and releases energy as heat or, in LEDs, as light, removing one electron and one hole from the pool of free carriers.

Q44. What does a 'hole' represent as a charge carrier in semiconductor physics?

A. An extra electron trapped at a doping site

B. A vacancy left by a missing valence electron, which behaves as a mobile positive charge ✓ Correct answer

C. A free proton moving through the lattice

D. An actual physical hole drilled into the silicon crystal

Explanation: When a valence electron leaves its bond, the resulting vacancy can be filled by a neighboring electron, making the vacancy appear to move like a positive charge carrier; this apparent motion is called a hole.

Q45. What is the approximate energy bandgap of silicon?

A. About 0.1 eV

B. About 1.1 eV ✓ Correct answer

C. About 0.66 eV

D. About 5 eV

Explanation: Silicon's bandgap of roughly 1.1 electron-volts is wider than germanium's (about 0.66 eV), which is why silicon devices tolerate higher temperatures and have lower leakage current.

Q46. What typically happens to a rectifier diode if the reverse voltage applied to it exceeds its PIV rating?

A. Nothing happens; diodes have no reverse voltage limit

B. It undergoes reverse breakdown and is usually permanently damaged or destroyed ✓ Correct answer

C. Its forward voltage drop decreases

D. It automatically converts into a Zener diode

Explanation: Exceeding the peak inverse voltage rating drives the diode into breakdown, where the resulting current is normally uncontrolled and generates enough heat to permanently damage an ordinary rectifier diode not designed for that operation.

Q47. What is the most common cause of a rectifier diode failing as an open circuit?

A. Testing it with a multimeter in diode mode

B. Operating it at exactly its rated forward voltage

C. Storing it at room temperature

D. Sustained forward current exceeding its rated maximum, causing the junction to overheat and burn open ✓ Correct answer

Explanation: Excess current dissipates power as heat in the junction; if this exceeds the diode's thermal limits, the junction material can melt or fracture, most often leaving the diode open rather than shorted.

Q48. Zener diodes rated below approximately 5 V primarily break down by which mechanism?

A. Avalanche multiplication

B. Thermal runaway only

C. The Zener (field emission/tunneling) effect ✓ Correct answer

D. Photoelectric emission

Explanation: Low-voltage Zener diodes are heavily doped with a very thin depletion region, so the true quantum-tunneling Zener effect dominates below about 5 V, while avalanche multiplication takes over in higher-voltage diodes.

Q49. In a PN junction, what causes 'diffusion current' as opposed to 'drift current'?

A. Carriers cooling below freezing temperature

B. The natural movement of carriers from a region of higher concentration to lower concentration ✓ Correct answer

C. An externally applied electric field pushing carriers

D. Carriers being pulled by gravity

Explanation: Diffusion current arises simply from the random thermal motion of carriers moving from areas where they are plentiful to areas where they are scarce, whereas drift current is driven directly by an electric field.

Q50. Compared to a silicon diode, a germanium diode generally has what characteristic regarding reverse leakage current?

A. Zero reverse leakage current under all conditions

B. Identical reverse leakage current to silicon

C. Leakage current that decreases as temperature rises

D. Significantly higher reverse leakage current, especially at elevated temperature ✓ Correct answer

Explanation: Germanium's smaller bandgap makes it easier to thermally generate carriers, so germanium diodes exhibit much higher and more temperature-sensitive reverse leakage current than silicon diodes, one reason silicon dominates modern designs.

Frequently asked questions

Who is this Semiconductor Troubleshooting quiz for?

It is for ITI, diploma and B.Tech students, apprentices and working technicians who want to test fault-finding knowledge. The explanations are written to teach the reason, not just the answer.

How should I use the answer key?

Attempt the quiz first without looking. Then open the answer key, read the explanation for every question you missed, and retake the part after a day to see whether the explanation stuck.

Do I need to finish all five parts?

No, but the parts build on one another: fundamentals first, then components, then circuits and real-world scenarios. Part 5 is the best test of whether you can apply the earlier parts to a live fault.

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