Diodes

32 questions

All types32 Multiple choice17 True / False8 Matching7

In electronics design, technicians must evaluate semiconductor ratings to ensure circuit reliability and prevent component failure. Match each diode rating or parameter on the left with its correct operating definition on the right.

Click a left item, then click its matching right item

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Answer

Peak Inverse Voltage matches the maximum reverse-bias voltage before breakdown; Forward Voltage Drop matches the potential difference needed to overcome the internal barrier potential; Reverse Recovery Time matches the time required for stored charge to clear during switching; Maximum Forward Current matches the upper limit of continuous current before thermal damage occurs.

Each diode specification defines a critical operating limit or behavioral parameter: Peak Inverse Voltage (PIV) defines reverse breakdown tolerance, Forward Voltage Drop defines barrier potential overcoming voltage, Reverse Recovery Time defines switching speed between conduction and cutoff, and Maximum Forward Current defines the continuous forward conduction limit governed by thermal dissipation.

Step-by-Step Solution

Identify voltage ratings associated with diode polarity states.

Peak Inverse Voltage (PIV) pertains to reverse-bias breakdown limits, while Forward Voltage Drop (

V_F

) pertains to the forward-bias junction barrier potential.

Diodes behave as one-way valves with distinct electrical characteristics depending on whether bias polarity is positive (forward) or negative (reverse).

Distinguish between dynamic switching characteristics and steady-state current limits.

Reverse Recovery Time (

t_{rr}

) is a high-frequency switching parameter measuring carrier clearance time, whereas Maximum Forward Current (

I_{F(\max)}

) is a steady-state thermal safety boundary.

Proper component selection requires evaluating both high-speed switching performance and continuous power/current handling capability.

Pair each parameter with its corresponding functional definition.

All four parameters are mapped uniquely to their technical specifications.

Each rating addresses a specific operational constraint in semiconductor rectifiers and signal diodes.

Key Concept

Semiconductor Diode Ratings and Electrical Specifications

Estimated Time:1m 15s

Question 22Question →

In a basic electronic power supply circuit, an alternating current (AC) voltage source is connected in series with a standard semiconductor diode and a load resistor. Which of the following describes the electrical behavior of this circuit across a complete AC cycle?

It permits current to flow freely in both directions while stepping down the alternating voltage to a lower amplitude.

It stores electrical energy in an electrostatic field during the positive half-cycle and discharges it into the load during the negative half-cycle.

It conducts current only during the positive half-cycle when forward-biased, blocking current during the negative half-cycle to produce pulsating direct current.

It inverts the polarity of the negative half-cycle to convert the alternating waveform into continuous, perfectly smooth direct current.

Show answer & explanation

Answer: It conducts current only during the positive half-cycle when forward-biased, blocking current during the negative half-cycle to produce pulsating direct current.

Answer

The diode conducts current only during the positive half-cycle when forward-biased, blocking current during the negative half-cycle to produce pulsating direct current.

A semiconductor diode operates as a one-way electrical valve. During the positive half-cycle of an applied AC signal, the diode is forward-biased and conducts current through the load resistor. During the negative half-cycle, the diode is reverse-biased and blocks current flow. This action converts bidirectional AC into unidirectional pulsating DC through half-wave rectification.

Step-by-Step Solution

Analyze the diode biasing state during the positive half-cycle of the AC input voltage.

The anode becomes positive with respect to the cathode, placing the PN junction in forward bias.

Forward bias narrows the depletion region and allows current to pass through the diode and into the load resistor.

Analyze the diode biasing state during the negative half-cycle of the AC input voltage.

The anode becomes negative with respect to the cathode, placing the PN junction in reverse bias.

Reverse bias widens the depletion region, preventing current flow through the circuit during this half-cycle.

Determine the net output current and voltage waveform across the load resistor.

Current flows through the load only during the positive half-cycles in a single direction, resulting in pulsating direct current (DC).

This conversion of alternating current to pulsating direct current using a single diode is the foundation of half-wave rectification.

Key Concept

Half-wave rectification and unidirectional conduction of semiconductor diodes

Estimated Time:1m 0s

Question 23Question →

In a semiconductor PN junction diode, applying an external reverse-bias voltage causes the depletion region at the junction to widen, which prevents the conduction of majority charge carriers.

True

False

Show answer & explanation

Answer: True

Answer

True

Applying reverse bias attracts majority carriers away from the PN interface, widening the non-conductive depletion zone and preventing current conduction through the component.

Step-by-Step Solution

Identify the polarity of the applied external voltage across the diode terminals in reverse bias.

The positive terminal of the power source connects to the n-type cathode, and the negative terminal connects to the p-type anode.

Establishing the reverse-bias electrical configuration determines the direction of the electrostatic forces acting on internal charge carriers.

Determine how internal majority carriers respond to the applied electrostatic field.

Holes in the p-type region are pulled toward the negative terminal, and electrons in the n-type region are pulled toward the positive terminal.

Opposite electrical charges attract, pulling majority carriers away from the central PN boundary.

Evaluate the effect of charge carrier movement on the depletion layer and overall current flow.

The removal of mobile carriers from the junction widens the depletion layer of uncovered ionized atoms, preventing majority carrier flow and allowing only negligible minority leakage current.

The widened depletion layer acts as a high-resistance barrier to electric current, confirming the statement as true.

Key Concept

PN Junction Biasing and Depletion Layer Dynamics

Estimated Time:45s

Question 24Question →

In electronic circuit assembly and analysis, understanding diode terminology and physical structure is fundamental for proper circuit operation. Match each diode term or terminal to its correct operational description or physical characteristic.

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Answer

Anode pairs with the P-type semiconductor terminal requiring positive potential; Cathode pairs with the N-type terminal marked by a physical stripe; Depletion Layer pairs with the internal boundary zone depleted of mobile charge carriers; Forward Voltage Drop pairs with the fixed potential difference across the conducting PN junction.

Each diode term accurately matches its functional definition: the anode is the P-type terminal requiring higher positive voltage, the cathode is the N-type terminal identified by a package stripe, the depletion layer is the carrier-free junction zone, and the forward voltage drop is the barrier potential (approximately

0.7\text{ V}

for silicon).

Step-by-Step Solution

Identify the polarity and terminal designations of a semiconductor diode.

The anode corresponds to the P-type section (requiring positive bias), and the cathode corresponds to the N-type section (marked with a band on axial packages).

Correct terminal identification prevents reverse installation and ensures proper forward bias orientation in circuits.

Distinguish between junction phenomena and operating electrical characteristics.

The depletion layer is the carrier-free barrier region formed at the interface, while the forward voltage drop is the typical

0.7\text{ V}

threshold maintained during conduction.

Differentiating physical barrier structure from electrical conduction metrics clarifies semiconductor physics.

Key Concept

Semiconductor Diode Structure, Polarity Identification, and Electrical Characteristics

Estimated Time:1m 0s

Question 25Question →

A direct-current (DC) solar charging circuit contains a standard silicon diode connected in series between the solar panel and a 12-volt battery. Which of the following best explains the primary function of this diode when the solar panel is not producing voltage at night?

It converts the direct current from the battery into alternating current to power the solar panel.

It prevents the battery from discharging backward through the dark solar panel.

It allows bidirectional current flow so the panel and battery maintain identical voltage levels.

It steps up the battery voltage to continuously supply power to the solar cells.

Show answer & explanation

Answer: It prevents the battery from discharging backward through the dark solar panel.

Answer

It prevents the battery from discharging backward through the dark solar panel.

A semiconductor diode permits electrical current to flow in only one direction (from anode to cathode under forward bias). When darkness falls, the solar panel's output drops to zero while the storage battery remains charged at 12 V. This higher potential at the cathode places the diode in reverse bias, creating an open circuit condition that prevents the battery from discharging through the solar panel.

Step-by-Step Solution

Analyze the circuit state under nighttime conditions.

The unilluminated solar panel produces 0 V, while the connected storage battery maintains approximately 12 V.

Determining the relative electric potentials at the diode terminals identifies its biasing condition.

Determine the bias state of the series diode.

The cathode connected to the battery positive terminal is at a higher potential than the anode connected to the panel, placing the diode in reverse bias.

A semiconductor diode conducts current only when its anode is more positive than its cathode by at least the forward barrier potential.

Identify the resulting circuit behavior.

The reverse-biased diode exhibits extremely high resistance, preventing current from leaving the battery.

This blocking action prevents the battery from draining energy into the internal resistance of the dark solar panel.

Key Concept

Unidirectional Current Flow and Reverse-Bias Blocking

Estimated Time:1m 0s

Question 26Question →

The Peak Inverse Voltage (PIV) rating of a semiconductor diode specifies the maximum reverse-bias voltage that can be applied across its terminals before the PN junction breaks down and begins conducting in reverse.

True

False

Show answer & explanation

Answer: True

Answer

True: The Peak Inverse Voltage (PIV) rating defines the maximum reverse-bias voltage a diode is designed to withstand without entering junction breakdown.

The statement is true because the Peak Inverse Voltage (PIV) rating is the manufacturer's maximum allowable reverse-bias voltage. Operating beyond this limit causes avalanche breakdown, allowing destructive reverse current to surge through the component.

Step-by-Step Solution

Identify the standard definition of the Peak Inverse Voltage (PIV) specification for semiconductor diodes.

PIV represents the peak voltage experienced by the diode when it is reverse-biased (cathode positive relative to anode).

Understanding diode ratings is critical for selecting components capable of handling reverse voltage without failing.

Determine the physical consequence of exceeding the diode's PIV rating.

Exceeding the PIV rating forces the PN junction into avalanche or Zener breakdown, leading to sharp reverse conduction and potential permanent destruction.

Standard rectifier and switching diodes are not designed to sustain continuous operation in the reverse breakdown region.

Evaluate the accuracy of the stem statement against semiconductor theory.

The statement accurately reflects the definition and operational limit designated by the PIV parameter.

The PIV specification exists precisely to prevent unintentional reverse breakdown.

Key Concept

Diode Peak Inverse Voltage (PIV) Rating and Reverse Breakdown

Estimated Time:45s

Question 27Question →

In a direct-current (DC) control circuit, a silicon rectifier diode is connected in parallel across the terminals of an electromechanical relay coil, with its cathode connected to the positive supply rail. What is the primary function of this diode in the circuit?

To convert the direct-current supply into alternating current needed to keep the relay energized

To conduct continuous forward-bias current around the coil while the circuit is actively powered

To safely dissipate high-voltage transient spikes generated by the collapsing magnetic field when the coil is de-energized

To step up the supply voltage across the coil terminals to increase contact closing speed

Show answer & explanation

Answer: To safely dissipate high-voltage transient spikes generated by the collapsing magnetic field when the coil is de-energized

Answer

To safely dissipate high-voltage transient spikes generated by the collapsing magnetic field when the coil is de-energized

An electromechanical relay coil is an inductor that stores energy in a magnetic field. When the control switch opens to turn off the relay, the sudden interruption of current causes the magnetic field to collapse rapidly. In accordance with Lenz's Law, this collapse generates a high-voltage back-EMF spike of reverse polarity. Because the diode is connected in anti-parallel (cathode to positive rail, anode to switched side), this reverse voltage places the diode into forward bias. The diode conducts, clamping the voltage spike to its forward drop (approximately

0.7\text{ V}

) and circulating the current through the coil until the energy safely dissipates as heat, protecting upstream transistors or switch contacts from electrical breakdown.

Step-by-Step Solution

Analyze the biasing condition of the diode during normal steady-state operation

The diode's cathode is connected to the positive supply rail and its anode to the switched ground side, placing it in reverse bias with negligible leakage current.

Understanding normal circuit operation confirms that the diode does not interfere with energizing the relay coil.

Analyze what happens when the circuit switch opens to de-energize the relay coil

The current suddenly drops to zero, causing the magnetic field around the inductive coil to collapse rapidly.

Inductors resist instantaneous changes in current by generating an induced electromotive force (back-EMF).

Determine the polarity of the induced voltage and the resulting state of the diode

The collapsing magnetic field reverses the polarity across the coil terminals, making the anode positive relative to the cathode and placing the diode into forward bias.

The forward-biased diode clamps the inductive spike to approximately

0.7\text{ V}

and circulates the decay current safely until the stored magnetic energy dissipates.

Key Concept

Flyback (freewheeling) diode protection across inductive loads

Estimated Time:1m 0s

Question 28Question →

In electronic systems, specialized diodes are engineered to leverage distinct semiconductor properties for control, sensing, and signaling. Match each specialized semiconductor diode with its primary operating characteristic or function.

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Answer

Zener diode matches with maintaining constant voltage in reverse breakdown; Light-emitting diode matches with emitting radiation under forward current; Photodiode matches with generating reverse current proportional to illumination; Varactor diode matches with functioning as a voltage-controlled variable capacitor.

Each diode type is engineered for a specific physical phenomenon: the Zener diode exploits controlled reverse breakdown for voltage stabilization; the LED converts electrical energy to light via forward-biased carrier recombination; the photodiode converts optical energy to current via reverse-biased photon absorption; and the varactor diode leverages the voltage-dependent width of the depletion layer to act as an adjustable capacitor in tuning circuits.

Step-by-Step Solution

Analyze the functional mechanism of a Zener diode

Identified as a device designed for stable reverse breakdown regulation

Zener diodes maintain a steady voltage rating when reverse biased beyond their breakdown threshold.

Analyze the functional mechanism of a Light-Emitting Diode (LED)

Identified as a forward-biased electroluminescent semiconductor

Forward current drives electron-hole recombination across the bandgap, releasing visible or infrared light.

Analyze the functional mechanism of a Photodiode

Identified as a reverse-biased light sensor

Incident light striking the depletion region creates free charge carriers, increasing reverse leakage current.

Analyze the functional mechanism of a Varactor diode

Identified as a voltage-tuned capacitive component

Varying the reverse-bias voltage changes the depletion layer thickness, which inversely varies junction capacitance.

Key Concept

Specialized semiconductor diode operational modes and functional characteristics

Estimated Time:1m 0s

Question 29Question →

A dual-power electronic circuit uses two silicon diodes to connect a primary $12\text{ V}$ DC power supply and a secondary $9\text{ V}$ backup battery to a single shared load. The anode of each diode is connected to the positive terminal of its respective source, and both cathodes are connected together at the positive terminal of the load. When the $12\text{ V}$ primary supply is operating normally, which of the following best describes the condition of the diode connected to the $9\text{ V}$ battery?

Forward-biased, because current from the higher-voltage supply flows backward into the battery

Reverse-biased, because the cathode voltage established by the primary supply is higher than the battery anode voltage

Forward-biased, because both voltage sources continuously force current simultaneously through the load

Reverse-biased, because the diode converts the direct-current input into alternating current to block it

Show answer & explanation

Answer: Reverse-biased, because the cathode voltage established by the primary supply is higher than the battery anode voltage

Answer

Reverse-biased, because the cathode voltage established by the primary supply is higher than the battery anode voltage

A standard silicon diode becomes forward-biased only when its anode is roughly

0.7\text{ V}

more positive than its cathode. When the primary

12\text{ V}

DC supply conducts, the common cathode bus is held at approximately

12\text{ V} - 0.7\text{ V} = 11.3\text{ V}

. Because the backup battery diode has only

9\text{ V}

at its anode while its cathode is at

11.3\text{ V}

, the cathode is more positive than the anode, placing the backup diode into a reverse-biased (cutoff) state.

Step-by-Step Solution

Determine the voltage at the shared cathode connection using the active primary source.

The primary

12\text{ V}

supply forward-biases its diode, dropping approximately

0.7\text{ V}

and establishing a common cathode bus potential of approximately

11.3\text{ V}

Silicon diodes in the conducting state exhibit a standard forward voltage drop of roughly

0.7\text{ V}

Compare the voltage at the anode of the backup battery diode to the shared cathode voltage.

The backup diode anode is held at

+9\text{ V}

by the battery, while its cathode is held at

+11.3\text{ V}

A diode requires its anode to be more positive than its cathode by at least

0.7\text{ V}

to conduct.

Determine the biasing state of the backup diode.

Because the cathode potential (

11.3\text{ V}

) exceeds the anode potential (

9\text{ V}

), the backup diode is reverse-biased (in cutoff) and conducts negligible current.

When the cathode is at a higher electrical potential than the anode, the semiconductor PN junction depletion region widens, blocking current flow.

Key Concept

Diode Biasing and Power-OR Steering

Estimated Time:1m 0s

Question 30Question →

In a direct-current circuit, applying a voltage of $+5.0\text{ V}$ to the anode and $+4.5\text{ V}$ to the cathode of a standard silicon PN junction diode is sufficient to overcome its barrier potential and place the diode into full forward conduction.

True

False

Show answer & explanation

Answer: False

Answer

False. The diode does not enter full forward conduction because the potential difference across its terminals (

0.5\text{ V}

) is below the

0.7\text{ V}

barrier potential threshold required for silicon.

For a standard silicon PN junction diode to conduct significant forward current, the anode potential must exceed the cathode potential by at least the internal barrier potential of approximately

0.7\text{ V}

. With

+5.0\text{ V}

on the anode and

+4.5\text{ V}

on the cathode, the net forward voltage is only

0.5\text{ V}

, leaving the diode in a non-conducting cutoff state.

Step-by-Step Solution

Calculate the forward voltage drop (

V_D

) across the diode from anode to cathode.

V_D = V_{\text{anode}} - V_{\text{cathode}} = 5.0\text{ V} - 4.5\text{ V} = +0.5\text{ V}

Diode biasing depends on the relative potential difference between the anode and cathode terminals.

Compare the applied forward voltage to the internal barrier potential of a silicon diode.

The applied voltage of

+0.5\text{ V}

is less than the required silicon barrier potential threshold of approximately

+0.7\text{ V}

A silicon PN junction requires approximately

0.7\text{ V}

of forward bias to overcome the depletion region barrier and permit significant forward current flow.

Determine the operating state of the diode.

The diode remains in the cutoff/non-conducting region with only negligible leakage current.

Since the threshold voltage is not reached, the statement claiming full forward conduction is false.

Key Concept

Silicon Diode Barrier Potential and Forward Conduction Threshold

Estimated Time:45s

Question 31Question →

A technician connects a standard silicon diode in series with a $220\,\Omega$ resistor and a $12\text{ V}$ DC power supply. The cathode terminal of the diode is connected to the positive supply terminal, and the anode terminal is connected through the resistor to the negative supply terminal. What is the expected voltage drop across the $220\,\Omega$ resistor?

11.3\text{ V}

0\text{ V}

0.7\text{ V}

12.0\text{ V}

Show answer & explanation

Answer: $0\text{ V}$

Answer

The voltage drop across the resistor is

0\text{ V}

because the diode is reverse-biased and prevents current from flowing through the circuit.

A diode is a polarized semiconductor component that permits current flow in only one direction. When the cathode terminal is connected to the positive supply side and the anode is connected toward the negative side, the diode is reverse-biased. In this state, the diode acts as an open circuit with extremely high resistance, preventing current from circulating (

I = 0\text{ A}

). By Ohm's Law (

V = I \times R

), with zero current flowing through the

220\,\Omega

resistor, the voltage drop across the resistor is

0\text{ V}

Step-by-Step Solution

Determine the biasing state of the diode based on the terminal connections.

The cathode is connected to the positive potential (

+12\text{ V}

) and the anode is connected toward the negative potential (

0\text{ V}

), meaning

V_{\text{anode}} < V_{\text{cathode}}

, which places the diode in reverse bias.

A PN junction diode only conducts when the anode is positive relative to the cathode by at least its forward threshold voltage.

Determine the circuit current under reverse-bias conditions.

The reverse-biased diode acts like an open switch, restricting current flow to negligible leakage levels (

I \approx 0\text{ A}

The reverse voltage widens the internal depletion region, blocking majority carrier flow.

Apply Ohm's Law (

V = I \times R

) to calculate the voltage drop across the

220\,\Omega

resistor.

V_R = 0\text{ A} \times 220\,\Omega = 0\text{ V}

Without current flowing through the resistor, no electromotive force is dropped across its resistance, and the full

12\text{ V}

source potential appears across the open diode.

Key Concept

Diode Biasing and One-Way Current Flow

Estimated Time:1m 0s

Question 32Question →

In a basic power supply circuit, an alternating current (AC) source is connected in series with a single PN-junction diode and a resistive load. Which of the following best describes the waveform and polarity of the voltage developed across the load resistor?

Continuous direct current (DC) that maintains a constant, ripple-free voltage level across both halves of the input cycle

Alternating current (AC) that flows in both directions with reduced amplitude across both halves of the cycle

Pulsating direct current (DC) that appears only during the half-cycles when the diode is forward-biased

Alternating voltage that is amplified into high-frequency oscillations by converting current into magnetic energy

Show answer & explanation

Answer: Pulsating direct current (DC) that appears only during the half-cycles when the diode is forward-biased

Answer

Pulsating direct current (DC) that appears only during the half-cycles when the diode is forward-biased

A standard PN-junction diode conducts current only when it is forward-biased (anode more positive than cathode) and prevents current flow when reverse-biased. In an AC circuit, this unidirectional conduction allows current to pass through the load during only one half of each alternating cycle, converting the bidirectional AC input into a pulsating direct current (DC) output.

Step-by-Step Solution

Analyze the one-way conduction property of the PN-junction diode.

The diode allows current to pass only when its anode is positive relative to its cathode (forward bias) and blocks current when polarity reverses (reverse bias).

Semiconductor PN junctions function as directional electrical check valves.

Evaluate the circuit response across both halves of the sinusoidal AC cycle.

During the positive half-cycle, the diode conducts and produces a voltage across the load resistor; during the negative half-cycle, the diode blocks conduction, dropping current to zero.

This process constitutes half-wave rectification.

Classify the resulting load voltage waveform.

The voltage across the load is unidirectional and periodic, known as pulsating DC.

Because current flows in only one direction in pulses without filter smoothing, the output is pulsating DC.

Key Concept

Diode rectification and unidirectional current conduction

Estimated Time:1m 0s

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Diodes

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