Rectifier Circuits MCQ Quiz - Objective Question with Answer for Rectifier Circuits - Download Free PDF

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.

a.) Full-wave Rectifier

No. of diodes required = 4

b.) Half-wave Rectifier

No. of diodes required = 1

c.) Centre-tapped full-wave rectifier circuit

No. of diodes required = 2

The number of diodes arranged in ascending order: b-c-a

Explanation:

DC Power Supply Using Single Phase Bridge Rectifier and Inductive Filter

Problem Statement: We aim to design a DC power supply capable of delivering 135 V and 20 A using a single-phase bridge rectifier with an inductive filter. The peak-to-peak current ripple in the output should be approximately 10%. The available AC source operates at 60 Hz. The task is to calculate the peak-to-peak current ripple in the system.

Key Formula and Concepts:

The peak-to-peak current ripple in the output of a DC power supply using an inductive filter can be calculated using the following equation:

ΔI_pp = (ΔV × T) / L

Where:

• ΔI_pp: Peak-to-peak current ripple
• ΔV: Ripple voltage across the filter
• T: Time period of the AC waveform = 1 / f
• L: Inductance of the filter

In this case, the ripple percentage is specified as 10%. Therefore, the peak-to-peak current ripple can be calculated as:

ΔI_pp = 0.1 × I

Where:

• I: DC output current, which is 20 A

Calculation:

Given data:

• DC output current (I): 20 A
• Ripple percentage: 10% (0.1 × I)

Substituting the values into the formula for peak-to-peak current ripple:

ΔI_pp = 0.1 × I

ΔI_pp = 0.1 × 20 A

ΔI_pp = 2 A

Hence, the peak-to-peak current ripple is 2 A.

Correct Option Analysis:

The correct option is:

Option 1: 19 A

This option is incorrect because it does not match the calculated peak-to-peak current ripple. The ripple percentage of 10% results in a value of 2 A, not 19 A.

Important Information:

To further understand the analysis, let us evaluate the other options:

Option 2: 20 A

This option is incorrect. The ripple cannot be equal to the full DC output current. A 10% ripple percentage implies a peak-to-peak ripple of 2 A, not 20 A.

Option 3: 2 A

This option is correct. A 10% ripple percentage of 20 A results in a peak-to-peak current ripple of 2 A, as calculated above.

Option 4: 1 A

This option is incorrect. A 10% ripple percentage does not correspond to a peak-to-peak current ripple of 1 A.

Conclusion:

Understanding the operation of a single-phase bridge rectifier with an inductive filter and the relationship between ripple percentage and current ripple is critical for accurate calculations. Based on the given data, the peak-to-peak current ripple is 2 A, which corresponds to the correct option 3. The other options are incorrect because they either misrepresent the ripple percentage or fail to align with the calculated values.

Explanation:

Calculation of Average DC Current in Each Diode

Problem Overview: The given AC source has an effective voltage of 120 V at a frequency of 60 Hz. The load connected to the circuit draws a DC current of 20 A. The task is to calculate the average DC current through each diode in the circuit.

Assumptions:

• The circuit is a full-wave rectifier configuration (common in such problems).
• The load connected to the rectifier is purely resistive, and the current drawn is constant DC.
• The diodes are ideal, meaning they have no forward voltage drop or reverse leakage current.

Key Concept: In a full-wave rectifier, the AC waveform is rectified such that during each half-cycle of the AC signal, only two diodes conduct (out of the four diodes in the bridge rectifier circuit). This ensures that the output DC current is distributed equally among the diodes over one complete cycle.

Step-by-Step Solution:

1. Full-Wave Rectifier Operation:

• In a full-wave rectifier, during the positive half cycle of the AC waveform, one pair of diodes conducts (e.g., D1 and D2).
• During the negative half cycle of the AC waveform, the other pair of diodes conducts (e.g., D3 and D4).
• As a result, each diode conducts for only half of the total AC cycle (i.e., for half the time).

2. Total DC Current:

• The total DC current drawn by the load is given as 20 A.
• This current is supplied alternately by two diodes during each half-cycle of the AC signal.

3. Average Current Through Each Diode:

• Since each diode conducts for half the time in a full-wave rectifier, the average DC current through each diode is half of the total DC current.
• Average DC current in each diode = Total DC current ÷ 4 (because two diodes share the current for half the time).
• Average DC current in each diode = 20 A ÷ 4 = 5 A.

4. Final Answer:

• The average DC current through each diode is 5 A.

Correct Option: Option 1 (5 A)

Important Information

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

Option 2 (10 A):

This option assumes that each diode conducts for the entire duration of the AC cycle, which is incorrect. In a full-wave rectifier, each diode conducts only during one-half of the AC cycle. Hence, the average DC current cannot be 10 A.

Option 3 (20 A):

This option assumes that each diode carries the full DC current of the load, which is incorrect. In a full-wave rectifier, the load current is shared between two diodes during each half-cycle, so no single diode carries the entire DC current.

Option 4 (0 A):

This option is entirely incorrect, as it contradicts the basic operation of the rectifier. Each diode conducts for half the AC cycle, so the average DC current in each diode cannot be zero.

Conclusion:

The correct answer is Option 1 (5 A). This value accurately represents the average DC current flowing through each diode in the full-wave rectifier circuit. The analysis of the other options highlights the importance of understanding the operation of rectifier circuits and the distribution of current among the diodes.

Concept:

In a full wave rectifier, the output frequency is twice the input frequency. This is because during each cycle of the input AC signal, the output signal completes two cycles.

Explanation: 

If the input frequency to a full wave rectifier is 50 Hz, the output frequency will be:

\( f_{out} = 2 \times f_{in} \)

Given that the input frequency \( f_{in} \) is 50 Hz,

\( f_{out} = 2 \times 50 \, \text{Hz} = 100 \, \text{Hz} \)

Therefore, the output frequency of the full wave rectifier will be 100 Hz.

The correct option is (2).

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%

Ripple factor:

The amount of AC present in the output of the signal is called as ripple.

The ripple factor indicates the number of ripples present in the DC output.

The output of the power supply is given by

\({\rm{Ripple\;factor}} = \frac{{{I_{rms}}\;of\;AC\;component}}{{{I_{DC}}\;component}}\)

It is given as:

\(\gamma = \sqrt {{{\left( {\frac{{{V_{rms}}}}{{{V_{DC}}}}} \right)}^2} - 1} \)

Thus if the ripple factor is less, the power supply has less AC components and power supply output is purer (i.e more DC without much fluctuations)

Thus ripple factor is an indication of the purity of output of the power supply. 

For full-wave rectifier: γ = 0.48

For half-wave rectifier: γ = 1.21

Transformer utilization factor

The transformer utilization factor (TUF) of a rectifier circuit is defined as the ratio of the DC power available at the load resistor to the kVA rating of the secondary coil of a transformer.

% TUF = \({V_{o(avg)}\times I_{o(avg)}\over V_{s(rms) \times I_{s(rms)}}}\)

For half-wave rectifier

The waveform of a half-wave rectifier is:

The parameters of half wave rectifier are

\(V_{o(avg)}={V_m\over \pi}\) and \(I_{o(avg)}={V_m\over \pi R}\)

\(V_{s(rms)}={V_m\over {2}}\) and \(I_{o(rms)}=I_{s(rms)}={V_m\over 2R}\)

% TUF = \({{V_m\over \pi}\times{V_m\over \pi R}\over {V_m\over \sqrt{2}}\times {V_m\over {2}R}}\)

% TUF = 28.6%

For full-wave rectifier

The waveform of a full-wave rectifier is:

The parameters of the full-wave rectifier are

\(V_{o(avg)}={2V_m\over \pi}\) and \(I_{o(avg)}={2V_m\over \pi R}\)

\(V_{s(rms)}={V_m\over \sqrt{2}}\) and \(I_{s(rms)}=V_m\)

% TUF = \({{2V_m\over \pi}\times{2V_m\over \pi R}\over {V_m\over \sqrt{2}}\times {V_m}}\)

% TUF = 57.2%

Mistake Points The rectification efficiency of the half-wave rectifier is 40.6%

The rectification efficiency of a full-wave rectifier is 81.2%

The correct answer is option 3): (100 Hz)

Concept:

The ripple frequency of the output and input is the same.

This is because one half-cycle of input is passed and another half-cycle is seized.

So, effectively the frequency is the same

so f_in  = f_r

the ripple frequency  = 100 Hz

The ripple frequency of a half-wave rectifier is f₀, where f₀ is the input line frequency. The ripple frequency of a center-tapped or a bridge-type full-wave rectifier is 2f₀.

Concept:

In a half-wave rectifier,

The average value of load current, \({I_{avg}} = \frac{{{V_m}}}{{\pi \left( {R + {R_L}} \right)}}\)

The RMS value of load current, \({I_{rms}} = \frac{{{V_m}}}{{2\left( {R + {R_L}} \right)}}\)

V_m is the peak value of supply voltage

R is the internal resistance

R_L is the load resistance

Calculation:

Given that, supply voltage = 50 sin ωt

Peal voltage (V_m) = 50 V

Internal resistance (R) = 20 Ω

Load resistance (R_L) = 800 Ω

Concept

The efficiency of the rectifier is defined as the ratio of the DC output power to the AC output power.

For half-wave rectifier 

The DC output power is given by:

\(P_{o(avg)}=​​V_{o(avg)}I_{o(avg)}\)

\(P_{o(avg)}=​​{V_m\over \pi}​​{I_o\over \pi R}\)

The AC output power is given by:

\(P_{o(RMS)}=​​V_{o(RMS)}I_{o(RMS)}\)

\(P_{o(RMS)}=​​{V_m\over 2}​​{I_o\over 2 R}\)

The efficiency is given by:

 \(η={V_m I_o\over \pi^2 R}\times {2^2 R\over V_m I_o}\)

η = 40.6%

For full-wave rectifier 

The DC output power is given by:

\(P_{o(avg)}=​​V_{o(avg)}I_{o(avg)}\)

\(P_{o(avg)}=​​{2V_m\over \pi}​​{2I_o\over \pi R}\)

The AC output power is given by:

\(P_{o(RMS)}=​​V_{o(RMS)}I_{o(RMS)}\)

\(P_{o(RMS)}=​​{V_m\over \sqrt{2}}​​{I_o\over \sqrt{2} R}\)

The efficiency is given by:

 \(η={4V_m I_o\over \pi^2 R}\times {2 R\over V_m I_o}\)

η = 81.2%

Full wave rectifier:

A bridge rectifier is of two types:

1) Bridge Type Full Wave Rectifier

2) Center-Tap Full Wave Rectifier

A Bridge type full wave rectifier contains 4 diodes as shown:

Ripple factor (RF):

The ripple factor indicates the number of ripples present in the DC output.

The output of the power supply is given by

\(RF= \sqrt {\frac{{{\rm{V}}_{{\rm{RMS}}}^2 }}{{{\rm{V}}_{{\rm{DC}}}^2}}-1} \)

\(RF= \sqrt{\dfrac{{({\frac{V_m}{\sqrt 2}})}^2}{({\frac{2V_m}{\pi })}^2}-1}=0.4834\)

Efficiency:

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

\(\eta = \dfrac{{{P_{dc}}}}{{{P_{ac}}}} = \frac{{\frac{{V_{dc}^2}}{{{R_L}}}}}{{\frac{{V_{rms}^2}}{{{R_L}}}}} = \frac{{{{\left( {\frac{{2{V_m}}}{\pi }} \right)}^2}}}{{{{\left( {\frac{{{V_m}}}{{\sqrt 2 }}} \right)}^2}}} = 81.2\;\% \)

Additional Information

Rectifier: A rectifier is a device that converts an alternating current into a direct current. A p-n junction can be used as a rectifier because it permits current in one direction only.

There are types of rectifiers i.e. half-wave rectifier and full-wave rectifier.

• In half-wave rectifier, there is only one diode, so during the positive half cycle diode conducts and gives the output similarly in the negative half cycle diode don’t conduct and gives no output.
• In a half-wave rectifier, the output frequency (ω) for the half-wave rectifier is the same as that of ac.

Full Wave rectifier:

• In a full-wave rectifier during a positive half cycle, one diode conducts and gives the output similarly in the negative half cycle another diode conducts and gives the output. Hence at a time, only one diode will be ON for a one-half cycle.
• In the case of the full-wave rectifier, the fundamental frequency = 2 × main frequency.

Half wave rectifier

Case 1: During +ve half cycle

Diode is forward biased, hence short-circuited

V_o = V_s

Case 2: During -ve half cycle

Diode is reverse biased, hence open-circuited

Vo = 0 

The output waveform is shown below:

The average value of a half-rectified sine wave is:

\(V_{o(avg)}={1\over 2\pi}\int_{0}^{\pi}V_m\space sin\omega t\space d\omega t\)

\(V_{o(avg)}={V_m\over 2\pi}(1-cos\pi)\)

\(V_{o(avg)}={V_m\over \pi}\)

Hence, the correct answer is option 3.

Mistake Points Full wave rectifier

The average value of a full-rectified sine wave is:

\(V_{o(avg)}={2V_m\over \pi}\)