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

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).

Explanation:

Average Value of Full-Wave Rectified Voltage

Definition: The average value of a full-wave rectified voltage is the mean of all the instantaneous values of the rectified voltage over one complete cycle of the input AC signal. It is a measure of the DC component of the rectified output.

Working Principle: In a full-wave rectifier, both the positive and negative halves of the AC signal are converted to positive voltage. This means that the output voltage is always positive, regardless of the input voltage polarity. The average value of this rectified voltage can be determined by integrating the rectified signal over one complete cycle and then dividing by the period of the cycle.

Formula: The average value (V_avg) of a full-wave rectified voltage with a peak value (V_m) can be calculated using the following formula:

V_avg = (2 * V_m) / π

Where:

• V_m is the peak value of the input AC voltage.
• π is a mathematical constant approximately equal to 3.14159.

Calculation:

Given that the peak value (V_m) is 66 V, we can substitute this value into the formula to find the average value:

V_avg = (2 * 66 V) / π

V_avg = 132 V / 3.14159

V_avg ≈ 42 V

Correct Option Analysis:

The correct option is:

Option 2: 42 V

This option correctly represents the average value of the full-wave rectified voltage with a peak value of 66 V, as calculated using the formula for the average value of a full-wave rectified signal.

Important Information

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

Option 1: 66 V

This option incorrectly suggests that the average value is equal to the peak value of the input voltage. However, the average value of a full-wave rectified signal is always less than the peak value. The peak value represents the maximum instantaneous voltage, not the average.

Option 3: 33 V

This option may confuse the average value with half the peak value. However, the correct average value is derived from the formula (2 * V_m) / π, which does not result in half the peak value.

Option 4: 88 V

This option may stem from a misunderstanding of the rectification process. The value 88 V is not related to the average calculation and is not derived from the provided peak value using the correct formula.

Conclusion:

Understanding the calculation of the average value of a full-wave rectified voltage is crucial in analyzing rectifier circuits. The correct average value for a full-wave rectified voltage with a peak value of 66 V is approximately 42 V, as determined by the formula (2 * V_m) / π. This understanding helps in designing and analyzing power supplies and other electronic circuits that utilize rectification to convert AC to DC.

Concept:

• The efficiency of a half-wave rectifier is the ratio of the DC power delivered to the load to the AC power supplied to the rectifier.
• The formula for rectifier efficiency (η) is:
• η = P_DC / P_AC
• The current equations for a half-wave rectifier are as follows:
  • Peak current: I_m = V_m / (R_L + R_F)
  • DC current: I_DC = I_m / π
  • RMS current: I_rms = I_m / √2
• DC power: P_DC = (I_DC)² × R_L
• AC power: P_AC = (I_rms)² × (R_L + R_F)

Calculation:

Amplitude of the voltage (Vm) = 25 V

Frequency = 50 Hz

Load resistance (RL) = 1000 Ω

Forward resistance of diode (RF) = 100 Ω

Peak current:

⇒ I_m = V_m / (R_L + R_F)

⇒ I_m = 25 / (1000 + 100)

⇒ I_m = 25 / 1100

⇒ I_m = 24.75 mA

DC current:

⇒ I_DC = I_m / π

⇒ I_DC = 24.75 / 3.14

⇒ I_DC ≈ 7.87 mA

RMS current:

⇒ I_rms = I_m / √2

⇒ I_rms = 24.75 / 2

⇒ I_rms = 12.37 mA

DC power:

⇒ P_DC = (I_DC)² × R_L

⇒ P_DC = (7.87 × 10^-3)² × 10³

⇒ P_DC ≈ 61.9 mW

AC power:

⇒ P_AC = (I_rms)² × (R_L + R_F)

⇒ P_AC = (12.37 × 10^-3)² × (10 + 1000)

⇒ P_AC ≈ 154.54 mW

Efficiency:

⇒ η = P_DC / P_AC

⇒ η = (61.9 / 154.54) × 100

⇒ η ≈ 40.05%

∴ The rectifier efficiency is 40.05%.

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:

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