Semiconductors and Electronic Devices Lesson Test

Answer: C

Correct answer: N-type silicon with free electrons as majority carriers. Silicon has four valence electrons. A dopant with five supplies one extra, loosely bound electron, so it acts as a donor and creates n-type material in which electrons are the majority carriers.

• A: Doping raises conductivity by adding carriers; it does not turn silicon into an insulator.
• B: This is what a three-electron acceptor such as boron produces, not a five-electron donor.
• C (correct): Antimony donates a loosely held fifth electron, so electrons become the main mobile carriers and the material is n-type.
• D: The type is right, but in n-type material holes are the minority carriers, not the majority.

Answer: B

Correct answer: Wider than a conductor's but much narrower than an insulator's. Conductors have overlapping bands, insulators have a wide gap that electrons almost never cross, and semiconductors have a small gap that some electrons can cross with heat or light energy.

• A: A gap that wide would almost never be crossed, which describes an insulator, not a semiconductor.
• B (correct): Conductors have overlapping bands, insulators have a wide gap, and semiconductors fall in between with a small gap.
• C: Overlapping bands describe a conductor; pure silicon conducts poorly because it has a real, though small, gap.
• D: Overlapping bands with no gap describe a conductor such as copper.

Answer: B

Correct answer: Light-emitting diode. Matching the function to the device: light output comes from an LED. When forward biased, electrons drop into holes at its junction and the released energy leaves as light.

• A: An IC combines many parts on one chip but is not the device that produces indicator light.
• B (correct): An LED is an output device that turns electrical energy into light when forward biased.
• C: A Zener diode holds a steady reference voltage in reverse breakdown; it does not give off useful light.
• D: A photodiode senses light that falls on it; it is an input device, not a light source.

Answer: A

Correct answer: A photodiode. A photodiode is the sensing mirror image of an LED: light in, electrical response out. Light striking its reverse-biased junction frees electrons and holes, producing a current that grows with brightness.

• A (correct): A photodiode, usually reverse biased, lets light free carriers at its junction, so its current rises with the light level.
• B: A Zener is used for a voltage reference, not for detecting light.
• C: A BJT switches or amplifies current controlled at its base; it is not described as a light sensor here.
• D: An LED gives off light; it is an output device rather than a light sensor.

Answer: C

Correct answer: The cathode, or n side. A diode's p side is the anode and its n side is the cathode. The printed band on the part, like the bar in the symbol, marks the cathode.

• A: The anode is the unmarked end; the triangle in the symbol sits on that side.
• B: Ratings are listed on the datasheet; the band shows polarity.
• C (correct): The band marks the cathode end, which matches the bar in the schematic symbol.
• D: A gate belongs to a field-effect transistor; a diode has only an anode and a cathode.

Answer: D

Correct answer: Emitter, base, and collector. A BJT is built from three layers of doped silicon. Its terminals are the emitter, the base (the control terminal), and the collector, which usually connects toward the load.

• A: This mixes a MOSFET terminal name into the BJT list; the control terminal of a BJT is the base.
• B: Anode and cathode name the two diode terminals; a BJT uses different names.
• C: These are the terminals of a field-effect transistor such as a MOSFET.
• D (correct): A BJT has three doped layers whose terminals are named emitter, base, and collector.

Answer: D

Correct answer: Four. Silicon sits in group 14 of the periodic table, the same column as carbon, so it has four electrons in its outer shell. In a crystal each atom shares one electron with each of four neighbors, so every atom ends up surrounded by eight shared electrons, a stable arrangement. That shared total of eight is the most tempting wrong answer. Remember that doping works by swapping in atoms with three or five valence electrons, creating holes or extra free electrons.

• A: Three valence electrons describes acceptor dopants such as boron, not silicon.
• B: Eight is the number of electrons an atom shares around itself when all four bonds are filled, not its own valence count.
• C: Five valence electrons describes donor dopants such as phosphorus, not silicon.
• D (correct): Each silicon atom has four outer electrons and forms four covalent bonds with its neighbors.

Answer: B

Correct answer: A Zener diode. A Zener diode is built to operate safely in reverse breakdown. At its rated voltage, such as 5.1 V, it conducts while keeping the voltage across it nearly constant, so it works as a voltage reference.

• A: A photodiode responds to light; it is not used to set a fixed reference voltage.
• B (correct): A Zener is designed to work in reverse breakdown and keep the voltage across it nearly constant.
• C: An ordinary rectifier diode can be damaged if pushed into breakdown with too much power.
• D: An LED is used to give off light, and it is operated forward biased.

Answer: D

Correct answer: An integrated circuit. An integrated circuit puts many transistors, diodes, resistors, and sometimes capacitors on one piece of silicon. This saves space, lowers cost, and improves reliability.

• A: A relay is a mechanical switch with a coil and moving contacts, not a silicon chip.
• B: A bridge is an arrangement of four diodes for full-wave rectification, not a chip of many parts.
• C: A P-N junction is a single boundary between p and n regions, the basis of one diode.
• D (correct): An IC, or chip, combines many devices on one piece of silicon.

Answer: C

Correct answer: The gate. A MOSFET has gate, drain, and source terminals. The gate-to-source voltage controls the channel between the drain and source, and almost no steady current flows into the insulated gate.

• A: The base is the control terminal of a BJT, not a MOSFET.
• B: The drain is one end of the channel that carries the load current, not the control terminal.
• C (correct): The voltage between the gate and the source controls the channel, and the gate is insulated from it by a thin oxide layer.
• D: The collector is a BJT terminal that usually connects toward the load.

Answer: A

Correct answer: 460 Ω. Subtract the LED drop from the supply, then divide by the current in amperes.

• A (correct): The resistor drops 9 V minus 2.1 V, or 6.9 V, and 6.9 V divided by 0.015 A is 460 Ω.
• B: This divides the LED's 2.1 V drop by the current instead of the voltage left for the resistor.
• C: This divides 6.9 V by 15 instead of 0.015 A, mixing milliamps and amperes.
• D: This divides the full 9 V by the current and forgets to subtract the LED's forward drop.

Answer: D

Correct answer: 150. Divide collector current by base current, using the same unit for both.

Beta is a ratio, so it has no units.

• A: This subtracts the base current from the collector current instead of dividing.
• B: This divides base current by collector current, which flips the ratio upside down.
• C: This multiplies the two currents, which does not give a gain ratio.
• D (correct): Beta is collector current divided by base current: 45 mA divided by 0.3 mA is 150.

Answer: B

Correct answer: Forward biased, so it conducts. Compare the terminals: -3 V is higher than -9 V, so the anode is 6 V more positive than the cathode. An anode more positive than the cathode means forward bias, and 6 V is well above the 0.7 V needed for a silicon diode to conduct.

• A: Bias depends on the difference between the terminals, not on whether the values are negative.
• B (correct): The anode at -3 V is 6 V more positive than the cathode at -9 V, which is forward bias.
• C: This judges by the minus signs instead of comparing the two voltages; the anode is the more positive terminal.
• D: Breakdown requires the cathode to be well above the anode, but here the anode is higher.

Answer: D

Correct answer: 5.14 mA. Subtract the model diode drop, then apply Ohm's law to the resistor.

• A: This adds the diode drop to the supply instead of subtracting it.
• B: This uses the full 12 V and ignores the 0.7 V the diode drops.
• C: This divides only the diode's 0.7 V by the resistance.
• D (correct): The resistor gets 12 V minus 0.7 V, or 11.3 V, and 11.3 V divided by 2,200 Ω is about 5.14 mA.

Answer: A

Correct answer: 20 mA. The Zener holds its node at 6.2 V, so the resistor has the rest of the supply across it.

With no load, all of that current flows through the Zener.

• A (correct): The resistor carries 15 V minus 6.2 V, or 8.8 V, giving 8.8 V divided by 440 Ω, which is 20 mA, all of it through the Zener.
• B: This divides the full 15 V by the resistor and ignores the voltage held by the Zener.
• C: This adds the Zener voltage to the supply instead of subtracting it.
• D: This divides the Zener voltage by the resistor instead of the voltage left across the resistor.

Answer: C

Correct answer: 2.4 W. Diode power loss equals its forward drop times its current: 0.8 V times 3 A equals 2.4 W. Volts times amperes gives watts.

• A: This adds the volts and amperes; power comes from multiplying them.
• B: This divides the current by the voltage instead of multiplying.
• C (correct): Power equals voltage times current: 0.8 V times 3 A is 2.4 W.
• D: This divides the voltage by the current, which does not give power.

Answer: A

Correct answer: Humps of one polarity during half of each cycle, with zero between them. This is a half-wave rectifier. During the half-cycle that forward biases the diode, the load sees a positive hump; during the other half, the diode is reverse biased and the load voltage is zero.

• A (correct): The diode conducts only during the half-cycle when its anode is positive and blocks during the other half.
• B: The diode blocks one polarity, so the load does not see a full sine wave of any size.
• C: A rectifier alone gives pulses; a filter and regulator are needed for smooth DC.
• D: Gap-free humps come from a full-wave bridge, which uses both half-cycles; a single diode uses only one.

Answer: B

Correct answer: 120 Hz. A bridge rectifier uses both half-cycles and sends each through the load in the same direction. That gives two humps for every input cycle, so the output pulses at twice 60 Hz, or 120 Hz.

• A: This would be true of a half-wave rectifier, which gives one pulse per input cycle.
• B (correct): A bridge turns both half-cycles into same-polarity humps, giving two pulses per input cycle, or 120 Hz.
• C: This halves the frequency; a bridge produces more pulses per cycle, not fewer.
• D: A rectifier does not turn the input into a constant 0 Hz signal; its output still pulses.

Answer: D

Correct answer: Neutral, because each hole is balanced by a fixed negative acceptor ion. Both p-type and n-type silicon are electrically neutral overall. In p-type material, every acceptor atom that captures an electron becomes a fixed negative ion that balances the hole it created.

• A: Each captured electron is balanced by the hole it leaves, so there is no net negative charge.
• B: Boron atoms in the lattice are balanced by their own electrons; doping does not create a net charge.
• C: The p label names the majority carrier, not the net charge; each hole is balanced by a fixed negative acceptor ion.
• D (correct): Doping adds no net charge: the acceptor ions and the holes cancel, so the block is neutral.

Answer: B

Correct answer: It widens, raising the barrier. With the cathode more positive than the anode, majority carriers are pulled away from the junction. The depletion region widens, the barrier rises, and only a tiny leakage current flows until breakdown.

• A: Removing the barrier is closer to forward bias; reverse bias makes the barrier larger.
• B (correct): Reverse bias pulls majority carriers away from the junction, so the depletion region grows and only tiny leakage flows.
• C: Narrowing and a lower barrier happen under forward bias.
• D: The applied voltage changes the region; reverse bias pulls carriers away from the junction.

Answer: D

Correct answer: A PNP bipolar transistor. The emitter arrow shows the direction of conventional current through the base-emitter junction. Arrow out of the base means NPN; arrow into the base means PNP.

• A: A Zener has two terminals and a cathode bar with bent ends, not an emitter arrow.
• B: A MOSFET has gate, drain, and source terminals, not an emitter lead.
• C: An NPN symbol has its emitter arrow pointing out, away from the base.
• D (correct): An emitter arrow pointing in, toward the base, identifies a PNP transistor.

Answer: A

Correct answer: Just beginning to conduct a small current. The threshold voltage is only where conduction begins, at a very small current. To switch a large load with low resistance, the gate must be driven well above threshold to the value the datasheet gives for its rated on-resistance.

• A (correct): Threshold is the point where a channel starts to form, measured at a very small test current.
• B: Reaching threshold is only the start of conduction; full on requires driving the gate well above threshold.
• C: 2 V is the threshold for this part, not a voltage that harms the gate.
• D: Breakdown is a diode concept; the MOSFET channel is controlled by the gate voltage.

Answer: C

Correct answer: Cathode to the 12 V supply end, anode to the collector end. While the coil is energized, its top end is at +12 V, so the diode must be reverse biased then: cathode to the supply end and anode to the collector end. When the transistor turns off, the coil's voltage reverses, the diode conducts, and the stored current circulates safely through it.

• A: A flyback diode goes across the coil, not in series with it.
• B: A diode in the base path does not give the coil current a path when the transistor turns off.
• C (correct): This keeps the diode reverse biased during normal operation, and the turn-off spike then forward biases it so coil current can die away safely.
• D: This puts the diode forward biased while the coil is energized, so it would short the coil and carry heavy current.

Answer: A

Correct answer: The diode is reverse biased and blocks current, protecting the radio. With correct polarity, the anode is more positive than the cathode, so the diode conducts with a small drop. Reversed, the cathode becomes more positive, the diode is reverse biased, and it blocks current so the circuit is protected.

• A (correct): With reversed polarity, the cathode is more positive than the anode, so the diode blocks.
• B: The 0.7 V drop applies with correct polarity; reversed, the diode does not conduct.
• C: Reversing the battery makes the cathode more positive, which is reverse bias, not forward bias.
• D: Battery voltages in a small radio are far below the breakdown level of a rectifier diode.

Answer: C

Correct answer: The diode is shorted. A good silicon diode reads about 0.5 to 0.7 V forward and open in reverse. Near zero in both directions means current flows freely either way, which indicates a shorted diode.

• A: An open diode would show OL, or open, in both directions.
• B: A Zener still shows a forward reading and blocks low reverse test voltages; it does not read zero both ways.
• C (correct): A good diode blocks in one direction; reading near zero both ways means it conducts both ways, a short.
• D: A good silicon diode reads about 0.5 to 0.7 V forward and open in reverse.

Answer: B

Correct answer: 60.4 mA. The emitter carries both the collector and base currents:

= 60 mA + 0.4 mA = 60.4 mA.

• A: This multiplies the two currents, which has no meaning here.
• B (correct): Emitter current equals collector current plus base current: 60 mA plus 0.4 mA is 60.4 mA.
• C: 150 is the beta ratio, 60 divided by 0.4, not a current.
• D: This subtracts the base current; the emitter carries the sum of collector and base currents.

Answer: B

Correct answer: 0.25 W. The resistor has 12 V minus 2 V, or 10 V, across it and carries 0.025 A. Its power is 10 V times 0.025 A, which is 0.25 W. Checking another way, the resistor is 10 V divided by 0.025 A, or 400 Ω, and

• A: This uses the full 12 V; the resistor only has 10 V across it after the LED drop.
• B (correct): The resistor drops 12 V minus 2 V, or 10 V, and 10 V times 0.025 A is 0.25 W.
• C: This multiplies 10 V by 25 instead of 0.025 A, a milliamp slip.
• D: This is the LED's own power, 2 V times 25 mA, not the resistor's.

Answer: D

Correct answer: NAND. An AND gate gives 1 only when both inputs are 1. Inverting that output gives a gate that gives 0 only when both inputs are 1, which is the NAND gate.

• A: An OR gate outputs 0 only when both inputs are 0.
• B: An AND gate outputs 1 only when both inputs are 1, which is the opposite of this behavior.
• C: A NOR gate outputs 1 only when both inputs are 0.
• D (correct): A NAND gate is an AND followed by an inverter, so it outputs 0 only when both inputs are 1.

Answer: C

Correct answer: A large filter capacitor. Each power-supply stage has its own job. The rectifier makes one-direction pulses, a filter (usually a large capacitor) smooths them with some ripple left over, and a regulator then holds the output at a set voltage.

• A: A photodiode senses light and has no role in smoothing a supply.
• B: A switching transistor turns a load on and off; it does not store charge to fill in the gaps between pulses.
• C (correct): A filter capacitor smooths the rectifier's pulses into a fairly steady voltage with some remaining ripple.
• D: The bridge is the rectifier itself; it creates the pulses rather than smoothing them.

Answer: C

Correct answer: Carriers still cross both ways, but the flows cancel, so net current is zero. At equilibrium, carriers keep diffusing across the boundary while the built-in field pushes some back. The two flows are equal and opposite, so there is lots of motion but zero net current.

• A: Carriers keep diffusing and drifting at equilibrium; only the net current is zero.
• B: The depletion region and built-in voltage form on their own and exist even with no outside voltage applied.
• C (correct): Diffusion one way is balanced by carriers pushed back by the built-in field, so there is motion but no net current.
• D: Without an external source there is no steady net current; that would create energy from nothing.

Answer: A

Correct answer: 30 mA. Beta predicts 150 times 0.4 mA, or 60 mA. But the outside circuit allows at most 12 V divided by 400 Ω, which is 30 mA. The transistor saturates and the collector current is limited by the load to about 30 mA.

• A (correct): Beta predicts 60 mA, but the load allows at most 12 V divided by 400 Ω, or 30 mA, so the transistor saturates.
• B: This uses the base current as the collector current and ignores the gain entirely.
• C: This adds base current to the beta prediction, which is an emitter-current idea and still ignores the load limit.
• D: This multiplies beta by the base current but ignores that the load cannot pass that much current.

Answer: D

Correct answer: 18 mA. The resistor carries the supply voltage minus the Zener voltage.

That current splits between the load and the Zener: $I_Z$ = 30 mA - 12 mA = 18 mA.

• A: This divides the full 24 V by 300 Ω, ignoring the Zener voltage, and then subtracts the load.
• B: This is the total resistor current; part of it goes to the load and must be subtracted.
• C: This adds the load current to the resistor current instead of subtracting it.
• D (correct): The resistor carries 9 V divided by 300 Ω, or 30 mA, and the Zener gets what the 12 mA load does not take, 18 mA.

Answer: A

Correct answer: 400 Ω. In a series string, the drops add and the current is the same everywhere. The LEDs take 2 V + 2 V = 4 V, leaving 8 V for the resistor.

• A (correct): The two LEDs drop 4 V in total, leaving 8 V for the resistor, and 8 V divided by 0.02 A is 400 Ω.
• B: This divides the full 12 V by the current and ignores both LED drops.
• C: This subtracts only one LED's 2 V drop, though both LEDs are in the series path.
• D: This doubles the current for two LEDs, but in a series string the same 20 mA passes through every part.

Answer: C

Correct answer: 10.6 V. In a bridge, two diodes conduct during each half-cycle, so each path has two drops. The load peak is 12 V minus 2 times 0.7 V, which is 12 V minus 1.4 V, or 10.6 V.

• A: This subtracts only one diode drop, but every conducting path in a bridge passes through two diodes.
• B: This ignores the diode drops completely, which is the ideal-diode model, not the 0.7 V model given.
• C (correct): Each half-cycle passes through two diodes, so the load peak is 12 V minus 1.4 V, or 10.6 V.
• D: This subtracts all four diode drops, but only two diodes conduct at a time.

Answer: B

Correct answer: 600 J. First find power, then multiply by time in seconds.

• A: This adds the volts and amperes to get 3.3 before multiplying by 300 s; power requires multiplying volts by amperes.
• B (correct): Power is 0.8 V times 2.5 A, or 2 W, and 2 W times 300 s is 600 J.
• C: This uses only 60 seconds, one minute, instead of the full 5 minutes.
• D: This multiplies the 2 W power by 5 without converting minutes to seconds.

Answer: B

Correct answer: 5.3 V correctly installed and about 0 V backward. With correct polarity, the diode is forward biased and takes its 0.7 V model drop, leaving 6 V minus 0.7 V, or 5.3 V, for the load. With the battery reversed, the diode is reverse biased and blocks, so the load voltage is essentially 0 V.

• A: Reversed, the diode is reverse biased and blocks, so the load does not see a reversed voltage.
• B (correct): Correctly installed, the diode conducts and drops 0.7 V, leaving 5.3 V; reversed, it blocks, so the load gets essentially nothing.
• C: This adds the diode drop to the battery voltage instead of subtracting it.
• D: This ignores the diode's forward drop when the battery is installed correctly.

Answer: D

Correct answer: The BJT needs steady base current, while the charged MOSFET gate needs almost none. A BJT is current controlled, so base current must keep flowing to hold it on. A MOSFET is voltage controlled: its gate is insulated by a thin oxide layer, so once the gate is charged, almost no current is needed to hold it on, apart from brief pulses when switching.

• A: This reverses the two devices: the BJT is current controlled and the MOSFET is voltage controlled.
• B: A BJT turns off when its base current is removed, and a MOSFET turns off when its gate is discharged.
• C: Only the BJT needs continuous control current; the MOSFET gate is insulated by oxide.
• D (correct): A BJT is current controlled, but a MOSFET's insulated gate draws almost no current once charged.

Answer: A

Correct answer: 0.5 mA. Full coil current is 24 V divided by 480 Ω, which is 50 mA. Since

the base current must be at least 50 mA divided by 100, or 0.5 mA. In practice designers supply more to ensure saturation.

• A (correct): The coil needs 24 V divided by 480 Ω, or 50 mA, and 50 mA divided by a beta of 100 is 0.5 mA.
• B: This is the coil current itself; the base needs only a fraction of it set by beta.
• C: This divides the supply voltage by beta instead of dividing the collector current by beta.
• D: This is ten times too large, a decimal slip when dividing 50 mA by 100.

Answer: C

Correct answer: 0.7 V. A Zener holds its rated voltage only in reverse breakdown, with its cathode toward the positive side. Installed backward, its anode is the more positive terminal, so it is forward biased and behaves like an ordinary silicon diode, dropping about 0.7 V.

• A: The diode is not blocking here, so the full supply does not appear across it.
• B: The Zener only holds its rated voltage when reverse biased; installed backward, it is forward biased.
• C (correct): Installed backward, the Zener is forward biased and acts like an ordinary silicon diode with about a 0.7 V drop.
• D: This is the voltage across the resistor in the correct circuit, 12 V minus 5.1 V, not the voltage across the Zener.

Answer: A

Correct answer: The result is inconclusive because 2 V is below the LED's forward drop. Blue and white LEDs drop about 3 V or more. A meter that supplies only about 2 V cannot forward bias the LED, so it shows open both ways even if the LED is good. A test with a higher voltage, such as a supply and a series resistor, is needed.

• A (correct): Blue LEDs drop about 3 V or more, so a 2 V test cannot forward bias the LED, and OL both ways is expected even from a good part.
• B: The meter showed OL in both directions, so lead orientation does not explain the result.
• C: A blue LED needs about 3 V or more to conduct, so an open reading from a 2 V test is not proof of failure.
• D: A shorted part would read near zero, not OL, in both directions.

Answer: D

Correct answer: P-type silicon with holes as majority carriers. A three-valence-electron dopant can form only three complete bonds in the silicon crystal. Each missing electron is a hole, so the dopant is an acceptor and the result is p-type silicon with holes as the majority carriers. The material is still electrically neutral overall.

• A: N-type doping needs a dopant with five valence electrons that donates an extra electron; gallium has only three.
• B: Even a few dopant atoms per million raise silicon's conductivity greatly; that is the purpose of doping.
• C: The holes act as positive carriers, but each is balanced by a fixed negative acceptor ion, so the material stays neutral.
• D (correct): Gallium completes only three of silicon's four bonds, leaving holes, so it is an acceptor and makes p-type material.

Answer: B

Correct answer: Germanium. Semiconductors such as silicon and germanium sit between conductors and insulators. Their conductivity can be raised precisely by doping, heat or light, which is what makes them useful for electronic devices.

• A: Copper has many free electrons and carries current easily, so it is a conductor.
• B (correct): Germanium, like silicon, conducts poorly when pure but can have its conductivity raised in a controlled way, which makes it a semiconductor.
• C: Rubber has almost no free charge carriers and blocks current, so it is an insulator.
• D: Aluminum is a metal with plenty of free electrons, so it is a conductor.

Answer: D

Correct answer: Conventional current flows from anode to cathode, while electrons move from cathode to anode. In a forward-biased diode, conventional current flows in the direction the symbol's triangle points, from anode to cathode. Electrons, being negative, travel the opposite way, from cathode to anode.

• A: Conventional current goes from anode to cathode; it is electrons that move from cathode to anode.
• B: Electrons carry negative charge, so they move opposite to conventional current.
• C: This reverses both directions; conventional current follows the symbol's triangle toward the bar.
• D (correct): Conventional current follows the triangle from the p-side anode to the n-side cathode, and electrons move the opposite way.

Answer: A

Correct answer: OR. The output is 1 whenever at least one input is 1 and is 0 only when every input is 0. That is the definition of an OR gate.

• A (correct): An OR gate outputs 1 when any input is 1, so either open window, or both, sounds the alarm.
• B: An AND gate outputs 1 only when all inputs are 1, so it would need both windows open before sounding.
• C: A NOT gate has a single input and simply flips it; this alarm combines two inputs.
• D: A NOR gate outputs 1 only when all inputs are 0, so it would sound only when both windows are closed.

Answer: B

Correct answer: 17.3 V and 60 humps per second. In a half-wave rectifier, current passes through one diode, so the load peak is

The diode conducts during only one half of each cycle, so a 60 Hz source gives 60 humps per second, with flat gaps between them.

• A: This treats the circuit as a bridge, subtracting two diode drops and counting two humps per cycle.
• B (correct): One diode drops 0.7 V, leaving 18 − 0.7 = 17.3 V, and a half-wave rectifier passes one hump per input cycle, 60 per second.
• C: The voltage is right, but a half-wave rectifier blocks every other half-cycle, so it gives only one hump per cycle.
• D: The hump count is right, but this ignores the 0.7 V drop across the conducting silicon diode.

Answer: B

Correct answer: It switches very fast and has no moving contacts to wear out. A relay moves physical contacts, which limits its speed and causes wear. A transistor switches electronically with no moving parts, so it can turn a load on and off thousands or even millions of times per second without wearing out.

• A: Complete isolation is an advantage of a relay, whose coil and contacts are separate circuits.
• B (correct): Transistors have no moving parts, so they can switch millions of times per second without contact wear or arcing.
• C: Relays are generally more tolerant of brief overloads than semiconductor switches.
• D: Closed relay contacts have very low resistance; this is a relay strength, not a transistor one.

Answer: D

Correct answer: A regulator. Match the job to the device. Holding the output at a fixed voltage while the input or load changes is what a voltage regulator does.

• A: A rectifier changes AC into one-direction pulses; it does not hold a voltage steady.
• B: An amplifier makes a weak signal larger; it does not hold a supply at a set voltage.
• C: An inverter changes DC into AC, which is not this stage's job.
• D (correct): A regulator, such as a Zener circuit or regulator IC, holds the output at a set voltage despite changes in input or load.

Answer: A

Correct answer: It lets current flow mainly in one direction and blocks it in the other. A diode is an asymmetric, one-way device: it conducts when its anode is more positive than its cathode and blocks current when the polarity is reversed.

• A (correct): A diode conducts well when forward biased and blocks current when reverse biased, making it a one-way device.
• B: Storing charge is the job of a capacitor or battery, not a diode.
• C: Amplifying a signal is the job of a transistor amplifier, not a diode.
• D: Changing AC voltage levels is the job of a transformer.

Answer: C

Correct answer: 200. The emitter carries both currents,

mA.

• A: This divides the emitter current by the base current; beta uses the collector current, which is the emitter current minus the base current.
• B: This divides the base current by the collector current, which turns the ratio upside down.
• C (correct): The collector current is 50.25 − 0.25 = 50 mA, and beta = 50 mA ÷ 0.25 mA = 200.
• D: This divides the emitter current by the collector current, which is not the definition of beta.

Answer: C

Correct answer: No series resistor limited the current, so the LED drew far too much. Once an LED's forward voltage is reached, a small increase in voltage causes a large jump in current. A 9 V battery is far above a red LED's roughly 2 V drop, so without a series resistor the current is excessive and the LED fails.

• A: A reverse-biased LED simply blocks and stays dark; this LED flashed, so it was conducting in the forward direction.
• B: A battery supplies DC; the problem is the size of the current, not the type of supply.
• C (correct): Above its forward voltage an LED's current rises steeply, so without a current-limiting resistor it overheats and burns out.
• D: Outward arrows mark a light emitter; inward arrows would mark a sensor such as a photodiode.