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

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.

1ϕ diode rectifier:

The supply power factor is given by:

$cos\varphi =\frac{V_o{I}_o}{V_{in(RMS)}{I}_{s(RMS)}}$

where, cosϕ = Power factor

V_o = Average output voltage

I_o = Average output current

V_in(RMS) = RMS value of supply voltage

I_in(RMS) = RMS value of supply current

Calculation:

Let the input voltage be: V_in = V_m sin(ωt)

Vin(RMS) = $\frac{V_m}{\sqrt{2}}$

Vo = $\frac{2V_m}{\pi }$

Iin(RMS) = Load current = I_o

$cos\varphi =\frac{2V_m\times I_o\times \sqrt{2}}{\pi \times V_m\times I_o}$

cos ϕ = 0.9

Concept:

A diode bridge is an arrangement of four diodes in a bridge circuit configuration that provides the same polarity of output for either polarity of input.

Three-Phase Full Wave Diode Rectifier:

• The full-wave three-phase uncontrolled bridge rectifier circuit uses six diodes, two per phase.
• A 3-phase full-wave rectifier is obtained by using two half-wave rectifier circuits. The advantage here is that the circuit produces a lower ripple output than the previous half-wave 3-phase rectifier as it has a frequency of six times the input AC waveform.

• As before, assuming a phase rotation of Red-Yellow-Blue (VA – VB – VC) and the red phase (VA) starts at 0o. Each phase connects between a pair of diodes as shown. One diode of the conducting pair powers the positive (+) side of the load, while the other diode powers the negative (-) side of the load.
• Diodes D1 D3 D2 and D4 form a bridge rectifier network between phases A and B, similarly diodes D3 D5 D4 and D6 between phases B and C and D5 D1 D6, and D2 between phases C and A.
• If we start the pattern of conduction at 30o, this gives us a conduction pattern for the load current of D1-4 D1-6 D3-6 D3-2 D5-2 D5-4, and return again to D1-4 and D1-6 for the next phase sequence as shown.
• So the average DC value of the output voltage waveform from a 3-phase full-wave rectifier is given as:
   

$V_0=\frac{1}{\pi /3}{\int }_{\pi /3}^{2\pi /3}{V}_{mline}sin\omega t.d(\omega t)$

$V_0=\frac{3}{\pi }{V}_{mline}$

$V_{mline}=\surd 3\times V_{mphase}$

$V_0=\frac{3\surd 3}{\pi }.V_m$

The correct answer is option 3):(Idc T)

Concept:

• Shunt Capacitor filter As the name suggests, a capacitor is used as the filter and this high-value capacitor is shunted or placed across the load impedance.
• The ‘T’ is the total non-conducting time of the capacitor. The charge per unit of time will give the current flow.
• The discharge current is given by the rate at which charge is lost by the capacitor 
• Q = $I_{dc}\times T$

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:

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

For Half wave rectifier maximum efficiency = 40.6%

Note: For Full wave rectifier maximum efficiency = 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

$RF=\sqrt{\frac{{\mathrm{V}}_{\mathrm{R}\mathrm{M}\mathrm{S}}^2}{{\mathrm{V}}_{\mathrm{D}\mathrm{C}}^2}-1}$

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

Efficiency:

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

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

Additional Information

Peak Inverse Voltage(PIV):

The maximum voltage across a reverse bias diode is known as Peak Inverse Voltage.

PIV for different rectifiers is shown below:

• Half Wave rectifier: Vm
• Full Wave centre tap rectifier: 2Vm
• Full Wave Bridge rectifier: Vm

​

Therefore, the PIV of a conducting and a non-conducting diode in a bridge rectifier is V_m i.e. Peak value of a.c. input.

Peak Inverse Voltage is the maximum voltage that appears across a reverse-biased diode:

If E is the maximum voltage of the sinusoidal input signal 

Then

For Half Wave rectifier $\mathrm{P}\mathrm{I}\mathrm{V}={\mathrm{E}}_{}$

For Full Wave rectifier $\mathrm{P}\mathrm{I}\mathrm{V}=2{\mathrm{E}}_{}$

For Bridge Wave rectifier $\mathrm{P}\mathrm{I}\mathrm{V}={\mathrm{E}}_{}$

Comparison between Half-Wave and Full-Wave Rectifier:

Concept:

Rectification efficiency:

It is the measure of the presence of the DC content in the output of a rectifier.

The rectification efficiency for a bridge rectifier is given by:

$\eta =\frac{V_{o(avg)}\times I_{o(avg)}}{V_{o(rms)}\times I_{o(rms)}}$

Calculation:

For bridge rectifier:

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

$I_{o(avg)}=\frac{2V_m}{\pi R}$

$V_{o(rms)}=\frac{V_m}{\sqrt{2}}$

$I_{o(rms)}=\frac{V_m}{\sqrt{2}R}$

$\eta =\frac{\frac{2V_m}{\pi }\times \frac{2V_m}{\pi R}}{\frac{V_m}{\sqrt{}2}\times \frac{V_m}{\sqrt{2}R}}$ × 100

η = 0.812 × 100

η = 81.2%

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:

Working:

For a positive half cycle, two diodes will act as short-circuit as shown

Similarly, for a negative half cycle, the other two diodes will be short-circuited resulting in a full-wave rectified output.

Additional Information

Center-Tap Full Wave Rectifier: 

It consists of two diodes as shown:

1ϕ diode rectifier:

The supply power factor is given by:

$cos\varphi =\frac{V_o{I}_o}{V_{in(RMS)}{I}_{s(RMS)}}$

where, cosϕ = Power factor

V_o = Average output voltage

I_o = Average output current

V_in(RMS) = RMS value of supply voltage

I_in(RMS) = RMS value of supply current

Calculation:

Let the input voltage be: V_in = V_m sin(ωt)

Vin(RMS) = $\frac{V_m}{\sqrt{2}}$

Vo = $\frac{2V_m}{\pi }$

Iin(RMS) = Load current = I_o

$cos\varphi =\frac{2V_m\times I_o\times \sqrt{2}}{\pi \times V_m\times I_o}$

cos ϕ = 0.9

The correct answer is option 3):(Idc T)

Concept:

• Shunt Capacitor filter As the name suggests, a capacitor is used as the filter and this high-value capacitor is shunted or placed across the load impedance.
• The ‘T’ is the total non-conducting time of the capacitor. The charge per unit of time will give the current flow.
• The discharge current is given by the rate at which charge is lost by the capacitor 
• Q = $I_{dc}\times T$

Concept:

Bridge rectifier:

It consists of four diodes and two didoes operate in each half-cycle. The average output voltage of a bridge rectifier is,

$V_0=\frac{2V_m}{\pi }$

The ripple frequency = f

Half wave rectifier:

The average output voltage of a half-wave rectifier or bridge rectifier is,

$V_0=\frac{V_m}{\pi }$

The ripple frequency = f

Application:

In a bridge rectifier, if one of the diodes is defective, the rectifier will work as a half-wave rectifier.

Therefore, the dc load voltage and ripple frequency will be lower than its expected values.

Peak surge forward current is the maximum permissible surge current in a forward direction having a specified waveform with a short time interval. It is largely because the filter capacitor must charge to peak voltage during the first few cycles. When one of the diodes is defective in the bridge rectifier, the surge current increases.