Diodes

32 questions

Question 1Question →

A technician is troubleshooting a power supply circuit featuring a full-wave bridge rectifier composed of four standard silicon diodes. During each single half-cycle of the alternating current (AC) input signal, how many of the four diodes conduct current simultaneously through the load?

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Answer: Two diodes conduct while the remaining two diodes are reverse-biased.

Answer

Two diodes conduct while the remaining two diodes are reverse-biased.
In a standard four-diode bridge rectifier, alternating current (AC) is converted to pulsating direct current (DC) by routing current through pairs of diodes. During any given half-cycle, two diagonally opposite diodes are forward-biased and conduct current through the load in a single direction, while the other two diodes are reverse-biased and block conduction.

Step-by-Step Solution

1
Analyze the bridge rectifier circuit topology under an AC input half-cycle
One AC input terminal is positive with respect to the other terminal.
AC voltage alternates polarity every half-cycle, determining the bias conditions of each diode in the bridge.
2
Trace the current flow through the forward-biased diodes and the load
Current enters the bridge, passes through one forward-biased diode, travels through the load resistor, and returns through a second forward-biased diode to the opposite AC line.
A complete circuit path requires a forward and return path through the bridge network.
3
Determine the state of the remaining two diodes
The opposite pair of diodes experiences reverse bias and blocks current flow.
Diodes only conduct when their anode is more positive than their cathode.

Key Concept

Full-Wave Bridge Rectification and Diode Conduction Pairs
Estimated Time:1m 0s
Question 2Question →

In electronic systems, standard PN-junction and specialized diodes exhibit distinct electrical behaviors depending on bias polarity and operating conditions. Match each diode operating condition with its characteristic electrical behavior.

Click a left item, then click its matching right item

Items

Standard silicon diode in forward bias
Standard silicon diode in reverse bias (below breakdown)
Zener diode operating in reverse breakdown
Light-emitting diode (LED) in forward bias

Matches

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Answer

Standard silicon diode in forward bias matches conducting current readily with an approximate 0.7 V0.7\text{ V} potential drop; Standard silicon diode in reverse bias matches blocking current flow with high resistance and negligible leakage; Zener diode in reverse breakdown matches maintaining a constant reference voltage across a range of currents; Light-emitting diode in forward bias matches releasing optical energy through electron-hole recombination.
Each diode state corresponds directly to its semiconductor junction physics: a forward-biased silicon diode exhibits a characteristic 0.7 V0.7\text{ V} barrier drop during conduction; a reverse-biased PN junction widens its depletion region to block current; a Zener diode exploits controlled reverse breakdown for voltage stabilization; and an LED emits light via radiative recombination under forward bias.

Step-by-Step Solution

1
Analyze standard forward bias characteristics of silicon diodes.
Identified that overcoming the silicon barrier potential (0.7 V0.7\text{ V}) permits large forward current conduction with a steady 0.7 V0.7\text{ V} drop.
Forward bias narrows the depletion region once the threshold voltage is reached.
2
Analyze standard reverse bias behavior below breakdown limits.
Identified that reverse bias widens the depletion layer, preventing majority carrier flow and resulting in high resistance.
A reverse-biased PN junction acts essentially as an open circuit.
3
Analyze specialized Zener diode reverse breakdown operation.
Identified that reverse-biased Zener diodes hold a stable breakdown voltage across varying currents.
Controlled avalanche/Zener breakdown enables voltage regulation applications.
4
Analyze light-emitting diode (LED) forward conduction mechanics.
Identified that forward current injection causes radiative recombination emitting photons.
Electrons dropping across the direct bandgap release energy as light.

Key Concept

Diode Operating States and Biasing Characteristics
Estimated Time:1m 0s
Question 3Question →

A 12 V12\text{ V} DC power source is connected in series with a 100 Ω100\ \Omega load resistor and a parallel pair of silicon diodes (D1D_1 and D2D_2). In the parallel branch, the anode of D1D_1 is connected toward the positive supply terminal, while the cathode of D2D_2 is connected toward the positive supply terminal. Assuming each silicon diode has a standard forward voltage drop of 0.7 V0.7\text{ V} and zero reverse leakage current, what is the voltage drop across the 100 Ω100\ \Omega load resistor?

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Answer: 11.3 V11.3\text{ V}

Answer

11.3 V11.3\text{ V}
Because Diode D1D_1 has its anode connected toward the positive terminal of the DC supply, it is forward-biased and conducts current, dropping its forward barrier potential of 0.7 V0.7\text{ V}. Diode D2D_2 has its cathode toward the positive supply, making it reverse-biased, so it acts as an open branch and conducts no current. According to Kirchhoff's Voltage Law, the total source voltage equals the sum of the series voltage drops. Subtracting the 0.7 V0.7\text{ V} diode drop from the 12 V12\text{ V} source gives a remaining 11.3 V11.3\text{ V} across the 100 Ω100\ \Omega resistor.

Step-by-Step Solution

1
Analyze the biasing condition of each diode in the parallel branch
Diode D1D_1 is forward-biased (anode toward positive terminal) and Diode D2D_2 is reverse-biased (cathode toward positive terminal)
Diodes conduct conventional current only when the anode is positive relative to the cathode (forward bias).
2
Determine the effective voltage drop across the parallel diode combination
The parallel branch voltage is clamped at Vdiodes=0.7 VV_{\text{diodes}} = 0.7\text{ V}
Since D1D_1 is forward-biased, current flows through D1D_1, establishing its characteristic 0.7 V0.7\text{ V} forward barrier drop. D2D_2 is reverse-biased and behaves as an open circuit.
3
Apply Kirchhoff's Voltage Law (KVL) around the single series loop
Vresistor=Vsource−Vdiodes=12 V−0.7 V=11.3 VV_{\text{resistor}} = V_{\text{source}} - V_{\text{diodes}} = 12\text{ V} - 0.7\text{ V} = 11.3\text{ V}
The sum of the voltage drops across the series elements must equal the source voltage.

Key Concept

Diode biasing state determination and junction voltage drop in series-parallel DC circuits
Estimated Time:1m 30s
Question 4Question →

Specialized semiconductor diodes are engineered with distinct physical and electrical properties to perform specific functions in electronic circuits. Match each specialized diode and operating condition on the left with its corresponding functional characteristic or application on the right.

Click a left item, then click its matching right item

Items

Zener diode operated beyond its reverse breakdown voltage (VZV_Z)
Varactor (Varicap) diode operated under variable reverse bias
Schottky barrier diode in forward conduction
Light-Emitting Diode (LED) in forward bias

Matches

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Answer

The Zener diode in reverse breakdown matches stable voltage regulation; the Varactor diode under reverse bias matches voltage-variable capacitance for RF tuning; the Schottky diode matches fast switching with low forward voltage drop; and the Light-Emitting Diode matches electroluminescent photon emission.
Each diode type corresponds to its unique semiconductor junction physics: Zener diodes maintain stable reverse breakdown voltages for regulation; varactor diodes provide voltage-dependent junction capacitance; Schottky diodes utilize metal-semiconductor junctions for high-speed switching and low forward drop; and LEDs release photons through forward-biased electron-hole recombination.

Step-by-Step Solution

1
Analyze the operational principle of the Zener diode in reverse bias.
Identified as maintaining a constant voltage (VZV_Z) across terminals.
Zener diodes are specifically doped to operate safely in the avalanche or Zener breakdown region to provide voltage regulation.
2
Analyze the operational principle of the Varactor diode under reverse bias.
Identified as a voltage-controlled variable capacitor.
The reverse bias modulates the width of the dielectric depletion region, changing junction capacitance proportionally to the applied reverse voltage.
3
Analyze the construction and forward characteristics of the Schottky barrier diode.
Identified by its metal-semiconductor junction providing low forward drop (0.2–0.3 V0.2\text{--}0.3\text{ V}) and negligible reverse recovery time.
Because majority carriers dominate conduction, there is no minority carrier recombination delay, enabling high-frequency switching.
4
Analyze the forward bias behavior of the Light-Emitting Diode (LED).
Identified as emitting light via electroluminescence.
Direct bandgap semiconductor recombination converts injected forward current into light.

Key Concept

Operating characteristics and specialized semiconductor applications of Zener, Varactor, Schottky, and Light-Emitting Diodes.
Estimated Time:1m 30s
Question 5Question →

An electronics technician is testing a power supply circuit featuring a four-diode full-wave bridge rectifier driven by a 60 Hz60\text{ Hz} AC secondary winding. During testing under load, one of the four rectifier diodes fails open. Which of the following best describes the electrical output observed across the load resistor?

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Answer: The circuit delivers pulsating DC with a ripple frequency of 60 Hz60\text{ Hz} because only one half-cycle path remains functional.

Answer

The circuit operates as a half-wave rectifier, supplying pulsating DC with a ripple frequency of 60 Hz60\text{ Hz} because only one alternation pair conducts.
A standard bridge rectifier uses two diodes per half-cycle to route current through the load in the same direction. When one diode opens, the series loop for that half-cycle is broken, producing zero output during that alternation. The other pair continues to conduct normally during the opposite half-cycle, resulting in half-wave rectification where pulses occur once per AC period at 60 Hz60\text{ Hz}.

Step-by-Step Solution

1
Analyze normal full-wave bridge rectifier conduction paths.
In normal operation, two diodes conduct in series on the positive half-cycle, and the opposite two diodes conduct in series on the negative half-cycle, yielding full-wave pulsating DC at twice the input line frequency (2×60 Hz=120 Hz2 \times 60\text{ Hz} = 120\text{ Hz}).
Understanding normal operation establishes the baseline for analyzing component failure.
2
Determine the impact of one diode failing open on both AC half-cycles.
The half-cycle requiring the open diode cannot complete its series loop, resulting in zero conduction during that alternation. The opposite half-cycle still has both diodes forward-biased and fully operational.
Diodes operate as one-way directional switches in specific series pairs per AC polarity.
3
Identify the resulting waveform characteristics across the load.
The circuit delivers unidirectional current pulses during only one half-cycle of the input AC sine wave, which defines half-wave rectification with a fundamental ripple frequency equal to the input line frequency (60 Hz60\text{ Hz}).
A single pulse per full AC cycle corresponds directly to the source frequency of 60 Hz.

Key Concept

Diode bridge rectification mechanics and single-diode open fault behavior
Estimated Time:1m 30s
Question 6Question →

When a freewheeling (flyback) diode is installed across an inductive DC relay coil to suppress transient voltage spikes, its cathode must be connected to the positive supply rail so that the diode remains reverse-biased during normal energized operation and becomes forward-biased only when the collapsing magnetic field inverts the coil's terminal polarity.

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Answer: True

Answer

True: In a flyback protection configuration, connecting the diode's cathode to the positive supply rail keeps it reverse-biased during steady-state energization, while allowing inductive collapse back-EMF to forward-bias the diode and clamp harmful voltage transients.
The statement is true because connecting the cathode of a freewheeling diode to the positive DC rail holds the diode in reverse bias during normal circuit operation, preventing unwanted conduction. When the control switch opens, the inductor's collapsing magnetic field inverts the terminal voltage polarity (back-EMF), which makes the anode positive relative to the cathode, placing the diode in forward bias to clamp voltage spikes and dissipate stored inductive energy.

Step-by-Step Solution

1
Analyze the biasing condition of the diode during normal steady-state energization of the inductive relay.
With the cathode tied to the positive DC supply rail and the anode connected to the switched low side, the diode is reverse-biased and draws only negligible leakage current, allowing full operating voltage across the coil.
A semiconductor diode conducts only when its anode is positive relative to its cathode; applying a higher potential to the cathode establishes reverse bias.
2
Determine the electrical response of the relay coil when the control switch opens to de-energize the circuit.
In accordance with Lenz's Law, the collapsing magnetic field induces a high-voltage back-EMF across the coil that reverses the terminal polarity, driving the switched terminal significantly positive relative to the supply rail.
An inductor opposes instantaneous changes in current by reversing its induced terminal voltage.
3
Evaluate the diode's conduction state during the induced back-EMF transient.
The reversed polarity places a higher positive voltage at the diode's anode than at its cathode, forward-biasing the diode to create a circulating discharge path that clamps the voltage spike to approximately 0.7 V0.7\text{ V} above the supply rail.
Forward conduction through the diode safely dissipates the stored magnetic energy and protects switching components from overvoltage breakdown.

Key Concept

Flyback diode polarity and inductive transient voltage clamping in DC circuits
Estimated Time:1m 30s
Question 7Question →

Two standard silicon rectifier diodes are arranged in a dual-input diode OR logic configuration. The anode of Diode 1 is connected to a +5.0 V+5.0\text{ V} DC supply, and the anode of Diode 2 is connected to a +2.0 V+2.0\text{ V} DC supply. Both cathodes connect to a common junction that ties to ground through a 2.2 kΩ2.2\text{ k}\Omega load resistor. Assuming each forward-biased silicon diode exhibits a nominal barrier potential drop of 0.7 V0.7\text{ V}, what is the resulting output voltage measured at the common cathode junction?

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Answer: 4.3 V4.3\text{ V}

Answer

The voltage measured at the common output cathode junction is 4.3 V4.3\text{ V}.
The output potential is determined by the diode with the highest anode voltage. Diode 1 has +5.0 V+5.0\text{ V} applied to its anode, which forward-biases its PN junction and establishes a common cathode potential of 5.0 V−0.7 V=4.3 V5.0\text{ V} - 0.7\text{ V} = 4.3\text{ V}. Because Diode 2 has only +2.0 V+2.0\text{ V} at its anode while its cathode is held at +4.3 V+4.3\text{ V}, it experiences a net reverse bias of −2.3 V-2.3\text{ V} and remains completely non-conductive (OFF).

Step-by-Step Solution

1
Determine which diode enters forward conduction first by comparing anode potentials.
Diode 1 has an anode voltage of +5.0 V+5.0\text{ V}, which is higher than Diode 2's anode voltage of +2.0 V+2.0\text{ V}.
In a common-cathode configuration, the diode with the most positive anode potential forward-biases first and sets the output voltage.
2
Calculate the output node voltage established by the conducting diode.
Vout=5.0 V−0.7 V=4.3 VV_{\text{out}} = 5.0\text{ V} - 0.7\text{ V} = 4.3\text{ V}.
A conducting silicon diode drops approximately 0.7 V0.7\text{ V} across its PN junction.
3
Evaluate the biasing state of Diode 2 with the cathode node established at 4.3 V4.3\text{ V}.
Vanode2−Vcathode=2.0 V−4.3 V=−2.3 VV_{\text{anode2}} - V_{\text{cathode}} = 2.0\text{ V} - 4.3\text{ V} = -2.3\text{ V} (Reverse Biased / OFF).
Because the cathode is more positive than the anode, Diode 2 is reverse-biased, acts as an open circuit, and contributes no current to the load.

Key Concept

Diode conduction states, forward barrier potential drop, and reverse-bias isolation in multi-diode circuits
Estimated Time:1m 30s
Question 8Question →

A circuit is constructed with a 10 V10\text{ V} DC power supply connected in series with a 220 Ω220\ \Omega current-limiting resistor. After the resistor, the circuit splits into two parallel branches that eventually reconnect to the negative terminal of the power supply. The first parallel branch contains a Zener diode (rated at VZ=6.2 VV_Z = 6.2\text{ V}) installed in a reverse-biased orientation. The second parallel branch contains a standard silicon diode installed in a forward-biased orientation. Based on this configuration, what is the approximate voltage measured across the parallel diode combination, and which component dictates this voltage?

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Answer: 0.7 V0.7\text{ V}, dictated by the forward-biased standard silicon diode clamping the voltage.

Answer

The correct answer is 0.7 V0.7\text{ V}, dictated by the forward-biased standard silicon diode clamping the voltage.
The correct answer accurately identifies that in a parallel circuit, the component with the lowest forward voltage drop or breakdown voltage dictates the voltage across the entire parallel block. Since the standard silicon diode is forward-biased, it drops approximately 0.7 V. This prevents the voltage across the parallel branches from ever reaching the 6.2 V required to cause the Zener diode to break down and conduct. The Zener diode remains effectively an open circuit, and the standard diode clamps the branch voltage at 0.7 V.

Step-by-Step Solution

1
Determine the biasing state of each diode in the parallel branches based on the power supply orientation.
The Zener diode is reverse-biased (requires 6.2 V to reach breakdown), and the standard silicon diode is forward-biased (requires ~0.7 V to conduct).
A diode's behavior depends fundamentally on whether it is forward or reverse-biased.
2
Analyze the voltage rules for parallel circuits.
The voltage across all components connected in parallel must be identical.
This fundamental circuit law means both the Zener diode branch and the standard diode branch must share the exact same voltage drop.
3
Determine the clamping voltage of the parallel combination.
The forward-biased silicon diode begins conducting heavily at 0.7 V, clamping the voltage across both branches at this level.
Because the silicon diode provides a low-resistance path at 0.7 V, the voltage cannot rise high enough to reach the Zener diode's 6.2 V breakdown threshold. The Zener diode remains effectively off (open circuit).

Key Concept

Diode Biasing and Parallel Circuit Voltage Clamping
Estimated Time:1m 30s
Question 9Question →

A semiconductor diode is a two-terminal electronic component that acts as a one-way valve for electric current. For current to conduct freely through a standard diode in a direct current (DC) circuit, how must the diode be connected relative to the power source?

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Answer: The anode must be connected toward the positive voltage terminal and the cathode toward the negative terminal.

Answer

The anode must be connected toward the positive voltage terminal and the cathode toward the negative terminal.
A diode conducts electric current when it is placed in forward bias. In a direct current circuit, forward bias is established by connecting the anode toward the positive terminal of the power supply and the cathode toward the negative terminal.

Step-by-Step Solution

1
Identify the two terminals of a standard diode and the condition needed for conduction.
A diode has an anode (p-type material) and a cathode (n-type material), and it conducts current only when forward-biased.
Understanding diode terminal polarity is necessary to determine proper circuit orientation.
2
Determine the voltage orientation required to forward-bias the diode.
The anode must be at a higher electrical potential than the cathode.
Applying a positive voltage to the anode and a negative voltage to the cathode overcomes the internal barrier potential, narrowing the depletion region and permitting current flow.

Key Concept

A semiconductor diode allows current to flow when it is forward-biased, which occurs when the anode is connected toward the positive supply voltage and the cathode toward the negative side.
Estimated Time:45s
Question 10Question →

In order for a standard PN junction diode to conduct conventional electric current, its cathode must be connected to a higher positive electrical potential than its anode.

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Answer: False

Answer

False
The statement is false because forward biasing a PN junction diode requires the anode (P-type material) to be at a higher positive potential than the cathode (N-type material). Applying a higher positive potential to the cathode places the diode in reverse bias, which widens the depletion layer and prevents current flow.

Step-by-Step Solution

1
Identify the relationship between diode terminals and semiconductor materials.
The anode is connected to the P-type semiconductor region, and the cathode is connected to the N-type semiconductor region.
Understanding terminal orientation is essential to determining how external voltage affects the internal PN junction barrier.
2
Evaluate the biasing condition described in the statement.
Connecting the cathode to a more positive potential than the anode creates a reverse-bias condition, widening the depletion region and preventing conventional current conduction.
For conduction to occur, the diode must be forward-biased by applying a more positive potential to the anode relative to the cathode.

Key Concept

Diode Biasing and Polarity
Estimated Time:45s
Question 11Question →

In an electronic power supply circuit, what is the primary function of a rectifier diode?

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Answer: To convert alternating current (AC) into direct current (DC) by allowing current to flow in only one direction

Answer

To convert alternating current (AC) into direct current (DC) by allowing current to flow in only one direction
A semiconductor diode permits electrical current to flow easily in only one direction (when forward-biased) and blocks current flow in the reverse direction (when reverse-biased). In power supply circuits, this unidirectional property is utilized in rectification to convert alternating current (AC) into direct current (DC).

Step-by-Step Solution

1
Identify the fundamental operational characteristic of a semiconductor diode.
A diode acts as a one-way electrical valve, exhibiting very low resistance in the forward direction and extremely high resistance in the reverse direction.
Understanding the unidirectional conduction property of the PN junction is essential for determining its circuit application.
2
Connect this one-way conduction property to its purpose in a power supply circuit.
When alternating current (AC) is applied, the diode conducts only during the positive half-cycles, producing a pulsating direct current (DC) output.
Rectification is the process of converting bidirectional AC into unidirectional DC.

Key Concept

Unidirectional current conduction and AC-to-DC rectification by semiconductor diodes
Estimated Time:45s
Question 12Question →

A semiconductor diode is a fundamental component that controls current flow in electronic circuits. Match each diode term on the left with its corresponding characteristic or operational definition on the right.

Click a left item, then click its matching right item

Items

Anode
Cathode
Forward bias
Reverse bias

Matches

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Answer

The Anode pairs with the P-type terminal represented by the triangle base, the Cathode pairs with the N-type terminal identified by the painted stripe, Forward bias pairs with the state allowing conduction when the P-type region is positive, and Reverse bias pairs with the state blocking conduction when the N-type region is positive.
A semiconductor diode consists of an anode (P-type side, represented by the base of the triangle symbol) and a cathode (N-type side, indicated by a band on the physical package and a vertical bar on the symbol). Connecting the positive supply to the anode creates forward bias, collapsing the depletion layer and allowing current conduction. Connecting the positive supply to the cathode creates reverse bias, expanding the depletion layer and insulating the circuit.

Step-by-Step Solution

1
Identify the two terminals of a PN junction diode
The anode represents the P-type material (triangle base on schematics), while the cathode represents the N-type material (vertical line on schematics and painted ring on physical components).
Correctly identifying component leads is essential for analyzing circuit orientation and polarity.
2
Determine the operating conditions for current conduction and cutoff
Applying positive voltage to the anode creates forward bias and allows conduction; applying positive voltage to the cathode creates reverse bias and halts conduction.
Forward bias shrinks the internal depletion layer, whereas reverse bias widens it to prevent electron movement.
3
Match each term to its exact operational and physical descriptor
All four terms are unambiguously paired with their respective terminal characteristics and biasing behaviors.
Verifies complete conceptual understanding of diode fundamentals.

Key Concept

Diode terminal identification (anode vs. cathode) and junction biasing states (forward vs. reverse bias)
Estimated Time:45s
Question 13Question →

On the cylindrical casing of a standard axial-lead semiconductor diode, what feature is identified by the painted stripe or band near one of the ends?

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Answer: The cathode terminal

Answer

The cathode terminal
On standard through-hole axial diodes, the painted stripe marks the cathode terminal. In schematic diagrams, this corresponds to the vertical bar toward which the triangle points.

Step-by-Step Solution

1
Identify the standard packaging markings used on axial-lead semiconductor diodes.
Standard cylindrical diodes (such as the 1N4001 series) use a single painted band at one end to distinguish polarity.
Because diodes are polarized components that conduct current primarily in one direction, physical orientation must be identifiable.
2
Correlate the physical stripe with the corresponding electronic terminal.
The stripe aligns with the cathode (negative lead during forward conduction), matching the vertical bar in the standard schematic symbol.
Correct alignment ensures the diode is wired in forward bias rather than accidental reverse bias.

Key Concept

Physical diode lead identification and polarity marking
Estimated Time:45s
Question 14Question →

A light-emitting diode (LED) produces light when it is connected in a reverse-bias orientation within a direct current (DC) circuit.

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Answer: False

Answer

False. A light-emitting diode (LED) produces light only when forward-biased; in a reverse-bias configuration, it blocks current flow and stays dark.
The statement is false because a light-emitting diode (LED) must be forward-biased (anode connected to the positive potential and cathode connected to the negative potential) to conduct electrical current and release energy in the form of light. In a reverse-bias orientation, the diode blocks current and produces no light.

Step-by-Step Solution

1
Identify the operating requirement of a semiconductor diode and LED.
Diodes conduct electric current in only one direction (forward bias) and block current in the opposite direction (reverse bias).
An LED functions as a standard PN-junction diode with the specialized property of emitting light when electric current conducts through it.
2
Determine the state of the LED in a reverse-bias orientation.
Under reverse bias, the internal depletion region widens, preventing current flow across the junction.
Because no charge carriers cross the junction under reverse bias, no electron-hole recombination occurs, preventing any light emission.

Key Concept

Light-emitting diodes (LEDs) require forward biasing (anode connected to a higher positive potential than cathode) to conduct current and emit light.
Estimated Time:45s
Question 15Question →

In a standard electronic schematic diagram, what does the triangular arrowhead of a semiconductor diode symbol indicate?

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Answer: The permitted direction of conventional current flow

Answer

The permitted direction of conventional current flow
In standard electronic schematic symbols, the triangular arrowhead points from the anode toward the cathode, designating the direction that conventional electric current (from positive to negative) is allowed to flow during forward bias.

Step-by-Step Solution

1
Analyze the anatomy of a standard semiconductor diode schematic symbol.
The symbol features a triangle pointing directly toward a vertical barrier line.
Recognizing the individual elements of schematic drafting symbols is necessary to read circuit diagrams.
2
Correlate the triangular symbol element with electrical flow conventions.
The triangle represents an arrowhead pointing from the anode (positive terminal) to the cathode (negative terminal), matching conventional current direction.
Standard electronics schematics use conventional current (positive-to-negative flow) to define component polarity arrows.

Key Concept

A semiconductor diode schematic symbol uses a triangular arrowhead pointing from anode to cathode to indicate the permitted direction of conventional current flow.
Estimated Time:45s
Question 16Question →

Different semiconductor diode designs are tailored for specific roles in electronic systems. Match each specialized diode type on the left with its primary operating function on the right.

Click a left item, then click its matching right item

Items

Zener Diode
Light-Emitting Diode (LED)
Photodiode
Schottky Diode

Matches

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Answer

Zener Diode matches maintaining a steady reference voltage in reverse breakdown; Light-Emitting Diode matches emitting light when forward biased; Photodiode matches acting as an optical sensor in reverse bias; Schottky Diode matches providing high switching speeds and low forward voltage drop.
Each diode type utilizes semiconductor properties for a targeted function: Zener diodes maintain a stable clamping voltage in reverse breakdown, LEDs emit photons when forward-biased, photodiodes sense light through light-induced reverse current, and Schottky diodes allow high-speed rectification with a lower forward voltage drop.

Step-by-Step Solution

1
Identify the primary electrical behavior of a Zener diode
Zener diodes uniquely conduct in reverse breakdown to regulate voltage across loads
They provide constant reference voltages in DC power supplies
2
Differentiate optoelectronic diodes (LED vs. Photodiode)
The LED emits light when forward-biased, while the photodiode detects light by conducting reverse current
One transforms electrical energy to light, whereas the other transforms light into an electrical signal
3
Identify the metal-semiconductor barrier characteristic of the Schottky diode
Schottky diodes feature low barrier voltage (~0.2 V to 0.3 V) and rapid switching capability
They are selected for high-frequency clamping and power rectification

Key Concept

Specialized Diode Types and Functional Applications
Estimated Time:45s
Question 17Question →

A technician installs a standard semiconductor diode in series with a direct current (DC) motor circuit to provide reverse polarity protection. Which of the following best describes the operating characteristic of the diode that enables this protection?

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Answer: It allows current to flow easily in only one direction while blocking current flow in the reverse direction.

Answer

The diode allows current to flow easily in only one direction while blocking current flow in the reverse direction.
A semiconductor diode functions as an electrical one-way check valve. When forward-biased (anode more positive than cathode), it allows current to pass with minimal resistance. When reverse-biased (voltage polarity reversed), the diode exhibits extremely high resistance and blocks current, safeguarding downstream components from incorrect power connections.

Step-by-Step Solution

1
Identify the primary electrical function of a semiconductor diode
A diode is designed to allow electric current to pass easily in one direction (forward bias) while presenting very high resistance to current flowing in the opposite direction (reverse bias).
Understanding the fundamental one-way conduction property explains how it functions as a circuit safeguard.
2
Apply the diode characteristic to the reverse polarity protection scenario
When power is connected with normal polarity, the diode is forward-biased and allows normal circuit operation. If the battery is connected backward, the diode becomes reverse-biased and blocks the reverse current, preventing component damage.
This directly demonstrates why unidirectional conductivity provides reverse polarity protection.

Key Concept

Unidirectional current flow and biasing in semiconductor diodes
Estimated Time:45s
Question 18Question →

A standard silicon diode requires a forward-bias voltage of approximately 0.7 V0.7\text{ V} across its terminals to overcome its internal barrier potential and conduct current.

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Answer: True

Answer

The statement is true because standard silicon semiconductor diodes require approximately 0.7 V0.7\text{ V} of forward-bias voltage to overcome the internal barrier potential and allow current conduction.
Silicon semiconductor diodes have an internal PN junction barrier potential of approximately 0.7 V0.7\text{ V}. In a forward-biased circuit, current cannot flow freely until the applied external voltage meets or exceeds this 0.7 V0.7\text{ V} threshold.

Step-by-Step Solution

1
Identify the semiconductor material and its characteristic barrier potential.
Silicon diodes possess an intrinsic barrier potential of approximately 0.7 V0.7\text{ V}.
The material properties of the silicon PN junction determine the electric field strength across the depletion layer.
2
Evaluate the condition required for current conduction.
Applying a forward voltage of at least 0.7 V0.7\text{ V} neutralizes the depletion region and allows charge carriers to cross the junction.
Voltages below 0.7 V0.7\text{ V} leave the diode in a non-conducting state, while voltages at or above 0.7 V0.7\text{ V} permit continuous current flow.

Key Concept

Silicon Diode Forward Voltage Drop and Barrier Potential
Question 19Question →

To maintain a steady output voltage in a voltage regulator circuit, a Zener diode must be connected in a forward-bias orientation.

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Answer: False

Answer

False
The statement is false because Zener diodes must be reverse-biased to exploit their precise breakdown voltage characteristic for stabilizing and regulating DC voltages.

Step-by-Step Solution

1
Identify the primary operating mode utilized by a Zener diode for voltage regulation.
Zener diodes are engineered to operate in their reverse breakdown region (the Zener region).
When reverse voltage reaches the breakdown threshold (VZV_Z), voltage across the diode remains virtually constant despite significant changes in reverse current.
2
Evaluate the electrical behavior of a Zener diode when placed in forward bias.
In forward bias, the Zener diode conducts like a standard silicon diode with a typical forward voltage drop of approximately 0.7 V0.7\text{ V}.
Forward biasing does not activate the controlled avalanche or Zener breakdown mechanism necessary for regulating higher voltages.
3
Evaluate the truth value of the given statement.
The statement asserts that forward bias is required for voltage regulation, which is incorrect.
Reverse biasing is required for a Zener diode to act as a voltage regulator.

Key Concept

Zener Diode Reverse-Bias Voltage Regulation
Estimated Time:45s
Question 20Question →

An electronics technician uses an ohmmeter to test an unmarked semiconductor diode. When the positive lead is placed on Terminal 1 and the negative lead on Terminal 2, the meter indicates a low resistance. When the meter leads are reversed, the meter indicates infinite resistance (an open circuit). Based on these test results, which of the following conclusions is correct?

Show answer & explanation

Answer: Terminal 1 is the anode, Terminal 2 is the cathode, and the diode is operational.

Answer

Terminal 1 is the anode, Terminal 2 is the cathode, and the diode is operational.
A properly functioning semiconductor diode conducts current when forward-biased (positive potential on the anode, negative potential on the cathode) and blocks current when reverse-biased (positive potential on the cathode, negative potential on the anode). Because the ohmmeter indicates low resistance when the positive lead connects to Terminal 1 and the negative lead connects to Terminal 2, Terminal 1 is the anode and Terminal 2 is the cathode. The infinite resistance displayed upon reversing the leads verifies that the PN junction is intact and the diode is operational.

Step-by-Step Solution

1
Analyze the first meter reading for biasing conditions
Low resistance occurs when the positive lead is on Terminal 1 and the negative lead is on Terminal 2, which corresponds to forward bias.
A semiconductor diode only allows significant current to flow when its anode is made positive relative to its cathode.
2
Analyze the second meter reading for reverse-bias integrity
Infinite resistance when the leads are reversed corresponds to reverse bias.
A functioning PN junction establishes a depletion region that blocks current when the cathode is positive relative to the anode.
3
Identify terminal polarities and verify component health
Terminal 1 is the anode, Terminal 2 is the cathode, and the high front-to-back resistance ratio confirms normal diode operation.
A healthy diode exhibits low forward resistance and very high/infinite reverse resistance.

Key Concept

Diode terminal identification and continuity verification using front-to-back resistance measurements
Estimated Time:1m 0s
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