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

Given:

• Input voltage: V(t) = sin²(t) + cos²(t) V
• Load resistor: 1 kΩ
• Ideal diode (no voltage drop when conducting)

Key Analysis:

1. Input Voltage Simplification:

   Using the trigonometric identity:

    sin2(t) + cos2(t)  = 1 

   Therefore, V(t) = 1 V (constant DC voltage at all times)

2. Diode Behavior:

   Since the input is always positive (1V):

   • The diode remains continuously forward-biased
   • Acts as a short circuit (0V drop)
3. Output Voltage:

   The entire input appears across the resistor:

   \( V_{out}(t) = V(t) = 1 \text{ V} \)

Average Voltage Calculation:

For a constant DC voltage:

\(V_{avg} = \frac{1}{T}\int_0^T 1\ dt = 1 \text{ V} \)

Final Answer:

4) +1 V

The Correct Answer is:  4) The transformer Utilisation Factor (TUF) is better for a bridge rectifier than for a centre-tapped rectifier.

Explanation:
Transformer Utilisation Factor (TUF):
TUF indicates how efficiently the transformer is used in a rectifier circuit.

Bridge Rectifier TUF ≈ 0.812

Centre-Tapped Full-Wave Rectifier TUF ≈ 0.693

Therefore, a bridge rectifier uses the transformer more efficiently, making option 4 correct.

Additional Information 
1) The PIV of both rectifiers is the same
False — In a bridge rectifier, each diode withstands only V_m (peak voltage),
while in a centre-tapped rectifier, each diode must withstand 2V_m ⇒ PIV is higher.

2) The transformer utilisation factor is the same for both
 False — TUF is better in a bridge rectifier, not the same.

3) A bridge rectifier has double the PIV compared to the centre-tapped rectifier.
False — It's the centre-tapped rectifier that has higher PIV, not the bridge.

Final Answer:
4) The transformer utilisation factor (TUF) is better for a bridge rectifier than for a centre-tapped rectifier.

Concept:

A half-wave rectifier converts AC to DC by allowing only one half-cycle of the input waveform to pass. The DC output voltage is the average value of the rectified waveform.

The average DC voltage for a half-wave rectifier is given by:

\(V_{dc} = \frac{V_m}{\pi} \)

where V_m is the peak voltage of the input AC signal.

Given:

• Primary voltage: \(V_p = 240 \sin(\omega t) \; V\)
• Turns ratio \( \frac{N_1}{N_2} = 1:1 \)
• Rectifier type: Half-wave

Calculation:

1. Determine Secondary Voltage:

   Since the turns ratio is 1:1, the secondary voltage equals the primary voltage:

   \(V_s = V_p = 240 \sin(\omega t) \text{ V (peak)}\)

2. Calculate DC Output Voltage:

   For a half-wave rectifier:
   \( V_{dc} = \frac{V_m}{\pi} = \frac{240}{\pi} \text{ V} \)

Thus the correct answer is option 2

The correct answer is: 1) The transformer utilization factor (TUF) is equal for both the bridge and center-tapped rectifier.

Explanation:

Transformer Utilization Factor (TUF)

TUF measures how effectively a transformer's secondary winding is used in a rectifier.

Concept:

The DC resistance of a reverse-bias diode is calculated using Ohm's Law:

\( R_{DC} = \frac{V_{reverse}}{I_R} \)

Where:
V_reverse  = Reverse voltage across diode
I_R  = Reverse leakage current

Given:

• Anode (P-terminal) connected to Ground (0V)
• Cathode (N-terminal) connected to +10V
• Cut-in voltage = 0.7V (not relevant for reverse bias)
• Reverse current \( I_R = 1 \mu A \)

Calculation:

1. Determine Reverse Voltage:

   \(V_{reverse} = V_{cathode} - V_{anode} = 10V - 0V = 10V\)

      Calculate DC Resistance:

      \( R_{DC} = \frac{10V}{1 \times 10^{-6}A} = 10 \times 10^6 \Omega = 10 M\Omega\)

Key Notes:
In reverse bias, the diode's resistance is very high (leakage current is small).

The cut-in voltage (0.7 V) is only relevant for forward bias and does not affect reverse-bias calculations.​

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:

\(\eta = \frac{{{P_{dc}}}}{{{P_{ac}}}}\)

\(\eta= \frac{{\frac{{V_{dc}^2}}{{{R_L}}}}}{{\frac{{V_{rms}^2}}{{{R_L}}}}} \)

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:

\(V_{DC}=\frac{V_m}{\pi}\)

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

\(V_{rms}=\frac{V_m}{2}\)

Calculation:

The efficiency for a half-wave rectifier will be:

\(\eta= \frac{{{{\left( {\frac{{{V_m}}}{\pi }} \right)}^2}}}{{{{\left( {\frac{{{V_m}}}{{2 }}} \right)}^2}}} = 40.6\;\% \)

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 < V_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

​           \(C=\frac{A\epsilon}{W}\)

• 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.

Concept

The value of Zener current is given by:

\(I_Z=I_S-I_L\)

\(I_Z={V_S-V_Z\over R_S}-{V_Z\over R_L}\)

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

\(I_Z={30-20\over 0.1}-{20\over 1}\)

I_Z = 100 - 20 mA = 0.08 A

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.

Symbols of diodes:

Zener Diode
: 
Varactor Diode

Tunnel Diode

Photo Diode

• 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