Diodes and Its Applications MCQ Quiz - Objective Question with Answer for Diodes and Its Applications - Download Free PDF

Explanation:

The efficiency of a voltage regulator is defined as the ratio of output power to input power, expressed as a percentage:

Efficiency (%) = (Output Power / Input Power) × 100

Rearranging this formula, we can calculate the input power:

Input Power = Output Power / Efficiency

Next, we use the relationship between power, voltage, and current to determine the input current:

Power = Voltage × Current

Rearranging this, the input current can be calculated as:

Input Current = Input Power / Input Voltage

Solution:

The output power is the product of the output voltage and the load current:

Output Power = Output Voltage × Load Current

Here, the output voltage is 5 V, and the load current is 0.75 A.

Output Power = 5 × 0.75 = 3.75 W

Using the efficiency formula, we calculate the input power. The efficiency is given as 75%, or 0.75 in decimal form:

Input Power = Output Power / Efficiency == 3.75 / 0.75 = 5 W.

Using the relationship between power, voltage, and current, we find the input current. The input voltage is 10 V:

Input Current = Input Power / Input Voltage = 5/10 = 0.5 A

Concept:

We are given a square wave input V_S of ±10 V, 100 Hz frequency, and 50% duty cycle. The diode is ideal.

We need to find the average voltage across V_L, which is the voltage across the load resistor R_L.

Step-by-step Analysis:

When V_S = +10 V (50% of the time):

• The diode is forward-biased.
• Current flows through the upper 100 Ω resistor, diode, and lower 100 Ω + R_L.
• It acts like a voltage divider between two 100 Ω resistors.
• So,

When V_S = -10 V (other 50% time):

• The diode is reverse-biased, so it acts as an open circuit.
• No current flows; hence,

Average value over full cycle:

Concept:

A Zener diode regulator maintains a constant output voltage (V_Z) across the load resistor (R_L) as long as the Zener diode current is greater than or equal to its knee current (minimum operating current).

To maintain regulation, the Zener must conduct at least the knee current (I_ZK = 5 mA).

Given:

Input Voltage (V_in) = 14.5 V

Zener Voltage (V_Z) = 7 V

Series Resistance R_s = 100 Ω

Zener Knee Current I_ZK = 5 mA

Calculation:

Total current through R_s:

Minimum current required by Zener = 5 mA

So, maximum load current:

Using Ohm’s law to find minimum R_L:

Final Answer:

Option 3) 100 Ω 

Explanation:

Ideal Half-Wave Rectifier

Definition: A half-wave rectifier is an electronic circuit that converts an alternating current (AC) input signal into a pulsating direct current (DC) output signal. It achieves this by allowing current to flow through the load only during one half-cycle of the input AC waveform, while blocking the current during the other half-cycle. This operation is typically achieved using a single diode in the circuit.

Working Principle: In a half-wave rectifier, the diode is forward-biased during one half of the AC cycle (positive half-cycle for a typical configuration), allowing current to pass through. During the other half of the AC cycle (negative half-cycle), the diode becomes reverse-biased and blocks the current. As a result, the output voltage is non-zero only during one half of the AC input cycle and zero during the other half.

Correct Option Analysis:

The correct option is:

Option 1: Non-zero during only one half cycle of input.

This option is correct because the primary characteristic of an ideal half-wave rectifier is that it allows current to flow through the load only during one half of the AC input cycle. For a zero mean sine wave input, this means the rectifier will produce a pulsating output that corresponds to the positive or negative half-cycles of the input waveform, depending on the orientation of the diode. The output will remain zero during the other half-cycle when the diode is reverse-biased.

Mathematical Analysis:

Let the input AC signal be represented as:

v_in(t) = V_msin(ωt), where:

• V_m = Peak amplitude of the sine wave.
• ω = Angular frequency of the AC signal.

During the positive half-cycle of the input (0 ≤ ωt ≤ π), the diode is forward-biased, and the output voltage is:

v_out(t) = V_msin(ωt)

During the negative half-cycle of the input (π ≤ ωt ≤ 2π), the diode is reverse-biased, and the output voltage is:

v_out(t) = 0

Thus, the output of an ideal half-wave rectifier is non-zero only during one half-cycle of the input AC signal

Explanation:

Full-Wave Rectifier

Definition: A full-wave rectifier is an electrical device used to convert alternating current (AC) into direct current (DC). Unlike a half-wave rectifier, which only utilizes one half-cycle of the input AC signal, a full-wave rectifier makes use of both the positive and negative half-cycles of the AC input signal, thereby providing a more efficient conversion and a smoother DC output.

Working Principle:

The full-wave rectifier operates by using two diodes (in a center-tap transformer configuration) or four diodes (in a bridge rectifier configuration) to rectify both the positive and negative half-cycles of the AC input. The resulting output is a pulsating DC signal, which can be further smoothed using filters.

• Center-Tap Full-Wave Rectifier: In this configuration, a center-tap transformer is used, and two diodes are connected to each half of the secondary winding. During the positive half-cycle, one diode conducts, while during the negative half-cycle, the other diode conducts.
• Bridge Full-Wave Rectifier: In this configuration, four diodes are arranged in a bridge circuit. During each half-cycle of the AC input, two diodes conduct to allow current to flow in the same direction through the load.

Advantages of Full-Wave Rectifier:

• Utilizes both positive and negative half-cycles of the AC input, leading to higher efficiency in conversion.
• Provides a smoother DC output compared to a half-wave rectifier.
• Higher average output voltage and current.
• Reduced ripple factor, which results in less need for filtering.

Correct Option Analysis:

The correct option is:

Option 3: Higher efficiency and utilisation of both half-cycles of the AC input.

This option accurately highlights the primary advantage of a full-wave rectifier over a half-wave rectifier. By utilizing both the positive and negative half-cycles of the AC input, the full-wave rectifier achieves higher efficiency and provides a more stable and higher average DC output. This makes it more suitable for applications requiring consistent and reliable DC power.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Lower efficiency and utilisation of only one half-cycle of the AC input.

This option describes the characteristics of a half-wave rectifier, not a full-wave rectifier. A half-wave rectifier only uses one half of the AC input signal (either the positive or the negative half-cycle), leading to lower efficiency and a higher ripple factor.

Option 2: Lower cost.

While a half-wave rectifier is generally less costly due to its simpler design and fewer components (only one diode is required), this is not an inherent advantage of a full-wave rectifier. Full-wave rectifiers are relatively more expensive because they require additional components such as diodes or a center-tap transformer.

Option 4: Lower output voltage.

Full-wave rectifiers provide a higher average output voltage compared to half-wave rectifiers because they utilize both half-cycles of the AC input. This statement is incorrect as it contradicts the fundamental advantage of full-wave rectifiers.

Option 5: Lower output voltage.

This is a repetition of Option 4 and is also incorrect for the same reason. Full-wave rectifiers offer a higher output voltage and current, making them more efficient and effective for DC power supply applications.

Conclusion:

By utilizing both half-cycles of the AC input signal, the full-wave rectifier achieves higher efficiency, higher output voltage, and a smoother DC signal compared to the half-wave rectifier. These advantages make it the preferred choice in most applications where a stable and efficient DC power supply is required. The correct answer, therefore, is Option 3: "Higher efficiency and utilisation of both half-cycles of the AC input."

Concept:

The efficiency of a rectifier is defined as the ratio of dc output power to input power.

The efficiency of a half-wave rectifier will be:

V_DC = DC or average output voltage

R_L = Load Resistance

For a half-wave rectifier, the output DC voltage or the average voltage is given by:

Also, the RMS voltage for a half-wave rectifier is given by:

Calculation:

The efficiency for a half-wave rectifier will be:

For Half wave rectifier maximum efficiency = 40.6%

Note: For Full wave rectifier maximum efficiency = 81.2%

• A diode is an electronic device allowing current to move through it only in one direction.
• Current flow is permitted when the diode is forwaforward-biased
• Current flow is prohibited when the diode is reversed-biased.
• The direction of the arrow represents the direction of conventional current flow when the diode is forward biased
• In the figure given above, the symbol represents the circuit symbol of a semiconductor junction diode.
• The ‘P’ side of the diode is always positive terminal and is designated as anode for forward bias.
• Another side that is negative is designated as cathode and is the ‘N’ side of diode.

Concept: 

The working of the Zener diode is explained in the below figures.

Calculation:

Given,

Zener voltage Vz = 10 V

E_in = 6 V ⇒ E_in z

Hence zener will be reverse biased and get open-circuited.

Output voltage E₀ = 0 V

Varactor diode:

• It is represented by a symbol of diode terminated in the variable capacitor as shown below:

• Varactor diode refers to the variable Capacitor diode, which means the capacitance of the diode varies linearly with the applied voltage when it is reversed biased.
• The junction capacitance across a reverse bias pn junction is given by

​           

• As the reverse bias voltage increases, the depletion region width increases resulting in a decrease in the junction capacitance.
•  Varactor diodes are used in electronic tuning systems to eliminate the need for moving parts
• Varactor [also called voltacap, varicap, voltage-variable capacitor diode, variable reactance diode, or tuning diode] diodes are the semiconductor, voltage-dependent, variable capacitors
• Varactors are used as voltage-controlled capacitors and it operated in a reverse-biased state

• A limiter circuit is also known as a clipper circuit.
• A clipper is a device that removes either the positive half (top half) or negative half (bottom half), or both positive and negative halves of the input AC signal.
• The clipping (removal) of the input AC signal is done in such a way that the remaining part of the input AC signal will not be distorted
• In the below circuit diagram, the positive half cycles are removed by using the series positive clipper.

​Note: A Clamper circuit can be defined as the circuit that consists of a diode, a resistor, and a capacitor that shifts the waveform to the desired DC level without changing the actual appearance of the applied signal.

• A tunnel diode is a highly doped semiconductor diode.
• The p-type and n-type semiconductor is heavily doped in a tunnel diode due to a greater number of impurities. Heavy doping results in a narrow depletion region.
• When compared to a normal p-n junction diode, tunnel diode has a narrow depletion width.
• The Fermi level moves in the conduction band on the n-side and inside the valence band on the p-side.
• Below the Fermi level, all states are filled and above the Fermi level all states are empty

 

The tunnel diode is represented by the symbol

 

The symbols of different diodes are given below.

Diode	Symbol
Tunnel diode
Varactor diode
Zener diode
Schottky diode
Photo diode

Concept

The value of Zener current is given by:

where, Iz = Zener current

V_z = Zener voltage

V_s = Source voltage

R_s = Source resistance

R_L = Load resistance

Calculation

Given, Vz = 20 V

Vs = 30 V

Rs = 100 Ω = 0.1 kΩ 

RL = 1 kΩ

I_Z = 100 - 20 mA = 0.08 A

Symbols of diodes:

Zener Diode
: 
Varactor Diode

Tunnel Diode

Photo Diode

Light Emitting Diode (LED)

• A light-emitting diode (LED) is a semiconductor device that emits light when an electric current flows through it.
• When current passes through an LED, the electrons recombine with holes emitting light in the process.
• LEDs allow the current to flow in the forward direction and blocks the current in the reverse direction.
• Light-emitting diodes are heavily doped p-n junctions made of a semiconductor material such as gallium and arsenide.
• Based on the semiconductor material used and the amount of doping, an LED will emit colored light at a particular spectral wavelength when forward-biased.

• The maximum value of the reverse voltage that a PN junction or diode can withstand without damaging itself is known as its Peak Inverse Voltage
• This rating of Peak Inverse Voltage (PIV) is given and described in the datasheet provided by the manufacturer
• PIV rating of Crystal diode is lower than that of the equivalent vacuum diode