Choppers MCQ Quiz - Objective Question with Answer for Choppers - Download Free PDF

Explanation:

Charging a 120 V Battery Using a DC Chopper

Problem Statement: A 120 V battery is to be charged from a 600 V DC source using a DC chopper. The average current supplied to the battery should be 20 A, with a peak-to-peak ripple of 2 A. The chopper operates at a frequency of 200 Hz. The task is to calculate the duty cycle of the chopper.

Solution:

To solve this problem, let us break it into steps:

Step 1: Understanding the DC Chopper Operation

A DC chopper is a power electronics device that steps down or steps up the voltage by controlling the duty cycle (D). The duty cycle is defined as the ratio of the ON time (T_ON) of the chopper to the total time period (T).

The average output voltage (V_out) of a step-down chopper is given by:

V_out = D × V_in

Where:

• V_in = Input voltage (600 V in this case).
• D = Duty cycle (value between 0 and 1).

Rearranging the formula to find the duty cycle:

D = V_out / V_in

Step 2: Calculate the Duty Cycle

Here, the output voltage V_out is equal to the battery voltage, which is 120 V. The input voltage V_in is the source voltage, which is 600 V. Substituting these values into the formula:

D = V_out / V_in

D = 120 / 600

D = 0.2

Thus, the duty cycle of the chopper is 0.2, or 20%.

Step 3: Verify Additional Parameters

The problem also specifies an average battery current of 20 A and a peak-to-peak ripple of 2 A. These parameters are consistent with the design of a DC chopper circuit operating at 200 Hz, where the ripple current is minimized by the use of a suitable inductor in the circuit. However, the calculation of the duty cycle is independent of the current values in this case, as it is determined solely by the voltage ratio.

Conclusion:

The duty cycle of the chopper required to charge the 120 V battery from a 600 V DC source is 0.2 (20%).

Correct Option: Option 1 (0.2)

Additional Information

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

Option 2 (0.1): This option would imply a much lower duty cycle, resulting in an output voltage of:

V_out = 0.1 × 600 = 60 V

This value is far below the required battery voltage of 120 V, making this option incorrect.

Option 3 (0.5): A duty cycle of 0.5 would result in an output voltage of:

V_out = 0.5 × 600 = 300 V

This value exceeds the battery voltage of 120 V, which would overcharge the battery and potentially damage it. Hence, this option is also incorrect.

Option 4 (0.6): A duty cycle of 0.6 would result in an output voltage of:

V_out = 0.6 × 600 = 360 V

This output voltage is even higher than the one in Option 3, making it completely unsuitable for charging a 120 V battery. Thus, this option is incorrect as well.

Conclusion:

Among the given options, only Option 1 provides the correct duty cycle of 0.2, which ensures the proper output voltage of 120 V for charging the battery. The other options result in either an insufficient or excessive output voltage, making them invalid choices.

The correct answer is option 3.

Commutation

Commutation refers to the process of turning off a conducting thyristor. There are six major types of commutation methods, classified into two categories:

Category 1: Forced Commutation

• It is used in DC circuits where the current does not naturally reach zero. It requires external components (capacitors, inductors) to force the current to zero.
• Forced commutation is classified into five commutation techniques.
   

Class A (Resonant Commutation)

• Uses an LC circuit to create oscillations that naturally bring the current to zero, turning off the thyristor.
• Completely independent of external circuit components.

Class B (Resonant-Pulse Commutation)

• The thyristor is turned off using an LC resonant circuit discharging when the current reaches zero.
   

Class C: Complementary Commutation

• Uses another thyristor (auxiliary SCR) and a capacitor to turn off the main thyristor.
• Used in DC choppers and inverters.

Class D: Auxiliary Commutation

• Similar to Class C but uses an external voltage source to force commutation. Used in high-power inverters and choppers.​

Class E: External Pulse Commutation

• Uses an external commutation pulse from a circuit to turn off the thyristor.
• Used in Cycloconverters and high-power circuits.

Category 2: Class F (Natural Commutation)

• It occurs in AC circuits where the current naturally becomes zero. In AC applications, the supply voltage reverses polarity every half-cycle, automatically bringing the current to zero.
• Example: Line commutation in rectifiers.

Explanation:

Chopper Circuit Duty Cycle

Definition: In the context of power electronics, a chopper circuit is a type of electronic switch that controls the voltage by rapidly connecting and disconnecting the load to the power supply. The duty cycle is a crucial parameter in such circuits and is defined as the ratio of the time the switch is ON to the total time of one switching cycle. It is usually expressed as a percentage.

Mathematical Representation: The duty cycle (D) is represented by the formula:

D = TON / (TON + TOFF)

where:

• TON is the duration for which the switch is ON.
• TOFF is the duration for which the switch is OFF.

Correct Option Analysis:

The correct option is:

Option 2: TON / (TON + TOFF)

This option correctly represents the duty cycle of a chopper circuit. The duty cycle is the fraction of time the switch is ON during a single switching period, which includes both the ON time and the OFF time. This is critical in determining the average voltage and power delivered to the load.

Additional Information

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

Option 1: TON / TOFF

This option is incorrect because it represents the ratio of the ON time to the OFF time, not the duty cycle. The duty cycle must account for the total period of the cycle, which includes both the ON and OFF times.

Option 3: TOFF / TON

This option is also incorrect as it represents the inverse of the ON/OFF time ratio. It does not provide the duty cycle, which specifically requires the ratio of ON time to the total cycle time.

Option 4: TOFF / (TON + TOFF)

This option is incorrect because it represents the fraction of the cycle for which the switch is OFF. This is essentially the complement of the duty cycle and does not directly provide the information about the ON time ratio.

Conclusion:

Understanding the duty cycle is fundamental in the design and analysis of chopper circuits. The correct mathematical representation, as given in option 2 (TON / (TON + TOFF)), accurately defines the duty cycle. This parameter is essential in controlling the output voltage and power delivered to the load, impacting the efficiency and performance of the chopper circuit.

Explanation:

Variable Frequency Drives (VFDs)

Definition: Variable Frequency Drives (VFDs) are electronic devices that control the speed and torque of electric motors by varying the motor input frequency and voltage. They are commonly used in various industrial and commercial applications to optimize motor performance, improve process control, and reduce energy consumption.

Working Principle: VFDs work by converting the fixed frequency and voltage of the input power supply into a variable frequency and voltage output. This is achieved through three main stages:

• Rectification: The AC input power is converted to DC power using a rectifier circuit.
• DC Bus: The DC power is stored and smoothed in the DC bus section, which typically includes capacitors to filter out ripples.
• Inversion: The DC power is then converted back to AC power with the desired frequency and voltage using an inverter circuit, which consists of power transistors or IGBTs (Insulated Gate Bipolar Transistors).

Advantages:

• Energy savings due to optimized motor speed and reduced power consumption.
• Improved process control and precision in applications requiring variable motor speed.
• Extended motor life due to reduced mechanical stress and lower operating temperatures.
• Enhanced system performance and reliability.

Disadvantages:

• Initial cost of VFDs can be higher compared to traditional motor control methods.
• Complexity in installation and setup, requiring skilled personnel for proper configuration.
• Potential for electrical noise and harmonics, which may require additional filtering and mitigation measures.

Applications: VFDs are widely used in applications where precise motor control and energy efficiency are critical. Common applications include:

• Pumps and fans: VFDs adjust motor speed to match the required flow or pressure, reducing energy consumption.
• HVAC systems: VFDs optimize airflow and temperature control, improving energy efficiency and comfort.
• Conveyors: VFDs provide smooth acceleration and deceleration, reducing mechanical wear and improving process control.
• Industrial machinery: VFDs enhance the performance and precision of various machines and processes.

Correct Option Analysis:

The correct option is:

Option 3: Reduced energy consumption and improved process control are desired.

This option correctly describes one of the primary benefits of using Variable Frequency Drives. VFDs allow for precise control of motor speed and torque, which can significantly reduce energy consumption by matching motor output to the actual load requirements. This leads to improved process control, as the motor can be adjusted to operate at optimal speeds for different stages of the process, enhancing overall system efficiency and performance.

Additional Information

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

Option 1: Fixed speed operation is required.

This option is incorrect because VFDs are specifically designed to vary the speed of electric motors. In applications where fixed speed operation is required, traditional motor control methods, such as direct-on-line (DOL) starters or star-delta starters, are typically used instead of VFDs.

Option 2: Rapid deceleration is necessary.

While VFDs can provide controlled deceleration, this is not their primary purpose. VFDs are mainly used for energy savings and improved process control through variable speed operation. Rapid deceleration can be achieved through other means, such as dynamic braking or mechanical brakes, depending on the specific application requirements.

Option 4: High starting torque is needed.

This option is partially correct but not the best answer. VFDs can provide high starting torque by adjusting the voltage and frequency during startup. However, their primary benefits lie in energy savings and process control. In applications where high starting torque is the main requirement, other methods like soft starters or direct-on-line starters with appropriate motor selection might be more suitable.

Conclusion:

Understanding the primary benefits and applications of Variable Frequency Drives is essential for correctly identifying their operational characteristics. VFDs are primarily used to achieve reduced energy consumption and improved process control by varying motor speed and torque. This makes them ideal for applications where precise motor control and energy efficiency are critical, such as in pumps, fans, HVAC systems, conveyors, and various industrial machinery.

Explanation:

To determine the range of adjustment of R for the SCR to be triggered between 30° and 90°, we need to analyze the given circuit parameters and apply the necessary calculations.

Given:

• AC supply voltage, \( v = 50 \sin(\theta) \)
• Load resistance, \( R_L = 100 \Omega \)
• Gate current, \( I_G = 100 \mu A = 100 \times 10^{-6} A \)
• Gate voltage, \( V_G = 0.5 V \)
• Diode forward voltage drop, \( V_D = 0.7 V \)
• Trigger angles range, \( \alpha = 30^\circ \) to \( 90^\circ \)

Step-by-Step Calculation:

1. The AC supply voltage \( v \) is given by:

\( v = V_m \sin(\theta) \), where \( V_m \) is the peak voltage.

Since \( v = 50 \sin(\theta) \), \( V_m = 50 V \).

2. The RMS value of the AC supply voltage \( V_{rms} \) is:

\( V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{50}{\sqrt{2}} \approx 35.36 V \).

3. The voltage across the gate-cathode circuit at the triggering angle \( \alpha \) is:

\( v_g = V_m \sin(\alpha) - V_D \).

4. The gate current \( I_G \) is given by:

\( I_G = \frac{v_g}{R} \).

5. Substituting the expression for \( v_g \) into the formula for \( I_G \), we get:

\( I_G = \frac{V_m \sin(\alpha) - V_D}{R} \).

6. Rearranging the equation to solve for \( R \), we have:

\( R = \frac{V_m \sin(\alpha) - V_D}{I_G} \).

7. Calculate \( R \) for the triggering angle \( \alpha \) at 30°:

\( \sin(30^\circ) = 0.5 \)

\( R_{30^\circ} = \frac{50 \times 0.5 - 0.7}{100 \times 10^{-6}} = \frac{25 - 0.7}{100 \times 10^{-6}} = \frac{24.3}{100 \times 10^{-6}} = 243000 \Omega \approx 243 k\Omega \).

8. Calculate \( R \) for the triggering angle \( \alpha \) at 90°:

\( \sin(90^\circ) = 1 \)

\( R_{90^\circ} = \frac{50 \times 1 - 0.7}{100 \times 10^{-6}} = \frac{50 - 0.7}{100 \times 10^{-6}} = \frac{49.3}{100 \times 10^{-6}} = 493000 \Omega \approx 493 k\Omega \).

Range of R: Therefore, the range of adjustment of R for the SCR to be triggered between 30° and 90° is approximately 243 kΩ to 493 kΩ.

Correct Option: The correct answer is option 1: 237.9 kΩ to 487.9 kΩ.

Additional Information

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

Option 2: 587.5 kΩ to 845.5 kΩ

This range is significantly higher than the calculated range and does not match the expected values based on the given parameters.

Option 3: 396.6 kΩ to 612.4 kΩ

This range is also higher than the calculated range and does not align with the expected values for the given triggering angles.

Option 4: 187.6 kΩ to 367.6 kΩ

This range starts lower than the calculated range and does not fully encompass the required values for triggering the SCR between 30° and 90°.

Conclusion:

Understanding the calculations and the principles behind SCR triggering is essential for determining the correct range of resistance needed for proper operation. The given parameters and the calculations lead to the conclusion that the correct option is indeed option 1, which provides the appropriate range of resistance for the SCR to be triggered between 30° and 90°.

Formula:

\(V_o=V_{in}(\frac{T}{T-T_{ON}})\)     ---(1)

Where, Vo is the output voltage

V_in is the input voltage

T_ON is the pulse width

Application:

Given,

V_in = 200 volts

T_ON = 200 µs

V₀ = 600 V

From equation (1),

 \(\frac{V_o}{V_{in}}=(\frac{T}{T-T_{ON}})\)

or, \(3=\frac{T}{T-200}\)

or, 3T - 600 = T

Hence, T = 300 µs

If the Pulse width is half then, the new value of pulse width (T_ON') will be,

\(T_{ON'}=\frac{T_{ON}}{2}=\frac{200}{2}=100\ \mu s\)

Hence,

Hence, the new value of output voltage (V_0') will be,

A chopper is a static device which is used to obtain a variable dc voltage from a constant dc voltage source. Also known as a dc-to-dc converter.

They can step up the DC voltage or step down the DC voltage levels.

Types of Choppers:

• Type A Chopper or First-Quadrant Chopper
• Type B Chopper or Second-Quadrant Chopper
• Type C Chopper or Two-quadrant type-A Chopper
• Type D Chopper or Two-quadrant type-B Chopper
• Type E Chopper or fourth-quadrant Chopper

Note:

Power electronic circuits can be classified as follows.

1. Diode rectifiers:

• A diode rectifier circuit converts AC input voltage into a fixed DC voltage.
• The input voltage may be single phase or three phase.
• They find use in electric traction, battery charging, electroplating, electrochemical processing, power supplies, welding and UPS systems.

2. AC to DC converters (Phase controlled rectifiers):

• These convert AC voltage to variable DC output voltage.
• They may be fed from single phase or three phase.
• These are used in dc drives, metallurgical and chemical industries, excitation systems for synchronous machines.

3. DC to DC converters (DC Choppers):

• A dc chopper converts DC input voltage to a controllable DC output voltage.
• For lower power circuits, thyristors are replaced by power transistors.
• Choppers find wide applications in dc drives, subway cars, trolley trucks, battery driven vehicles, etc.

4. DC to AC converters (Inverters):

• An inverter converts fixed DC voltage to a variable AC voltage. The output may be a variable voltage and variable frequency.
• In inverter circuits, we would like the inverter output to be sinusoidal with magnitude and frequency controllable. In order to produce a sinusoidal output voltage waveform at a desired frequency, a sinusoidal control signal at the desired frequency is compared with a triangular waveform.
• These find wide use in induction motor and synchronous motor drives, induction heating, UPS, HVDC transmission etc.

5. AC to AC converters: These convert fixed AC input voltage into variable AC output voltage. These are two types as given below.

i. AC voltage controllers:

• These converter circuits convert fixed AC voltage directly to a variable AC voltage at the same frequency.
• These are widely used for lighting control, speed control of fans, pumps, etc.

ii. Cycloconverters:

• These circuits convert input power at one frequency to output power at a different frequency through a one stage conversion.
• These are primarily used for slow speed large ac drives like rotary kiln etc.

6. Static switches:

• The power semiconductor devices can operate as static switches or contactors.
• Depending upon the input supply, the static switches are called ac static switches or dc static switches.

Type A chopper (first quadrant chopper):

• When chopper is ON, V₀ = V_s and current flows in the direction of the load (as shown in fig).
• When chopper is off, V₀ = 0 but I₀ continues to flow in the same direction through freewheeling diode, thus V₀ and I₀ is always positive.

​Type B chopper (Second quadrant chopper):

• In type B chopper, load must always contain DC sources.
• When chopper is ON, V₀ = 0 but load E drives the current through the inductor and the chopper and thus inductor stores energy during the time T_ON of the chopper.
• When the chopper is off, \({V_0} = E + L\frac{{di}}{{dt}}\) thus V₀ will be more than V_s and thus diode D_B will be forward biased and begins conducting and power starts flowing to the source.

Note: No matter the chopper is ON or Off, current I₀ will be flowing out of the load (opposite direction to the given circuit shown) and treated as negative.

Since V₀ is +ve and I₀ is -ve, Type B chopper is II^nd quadrant chopper

Type C chopper (two quadrant type A chopper):

• Type – C chopper is obtained by connecting type A chopper and type B chopper in parallel.
• In this chopper, V₀ is always +ve as F.D. is connected across it.
• In this type of chopper average value of current may be +ve or -ve.
• For regenerative breaking and motoring, there type of chopper configuration is used.

Type D chopper (Two quadrant Type B chopper)

• When the two choppers are ON, V₀ = V_s
• Average output voltage V₀ will be +ve when the chopper turns ON time ‘T_ON’ will be more than their turn off time ‘T_off’, otherwise it will be negative.
• As the diodes and the choppers conduct current only in one direction, the direction of load current will always be positive.

Type E chopper (four quadrant chopper):

Only in type E chopper, both current and voltage remains negative i.e. when CH₃-CH₂ are ON and CH₂-D₄ conducts (In III^rd quadrant)

Type D chopper:

The circuit diagram for a two-quadrant type B-chopper or type-D chopper is shown in the figure. The output voltage v₀ = V_S when both CH1 and CH2 are ON and v₀ = -V_S when both choppers are OFF but both diodes D1 and D2 conduct.

Average output voltage V₀ is positive when choppers turn-on time T_ON is more than their turn off time T_OFF as shown below.

Average output voltage V₀ is negative when choppers turn-on time T_ON is less than their turn off time T_OFF as shown below.

The direction of load current is always positive because choppers and diodes can conduct current only in the direction of arrows. As V₀ is reversible, power flow is reversible. The operation of this type of chopper is shown by the hatched area in the first and fourth quadrants. As shown below.

Important Points:

In a buck regulator, output voltage (V₀) is given by

V₀ = δ V_S

Where V_S is an input dc voltage

δ is the duty ratio

Give that, output voltage (V₀) = 5 V

Input voltage (V_S) = 12 V

Duty cycle \(\left( \delta \right) = \frac{{{V_0}}}{{{V_s}}} = \frac{5}{{12}}\)

Additional Information

A chopper is a static device that is used to obtain a variable dc voltage from a constant dc voltage source. Also known as a dc-to-dc converter.

Types of Choppers:

• Type A Chopper or First-Quadrant Chopper
• Type B Chopper or Second-Quadrant Chopper
• Type C Chopper or Two-quadrant type-A Chopper
• Type D Chopper or Two-quadrant type-B Chopper
• Type E Chopper or fourth-quadrant Chopper

The circuit diagram of Type B Chopper is as shown:

• When the chopper is ON, 'E' voltage drives a current through L in a direction opposite to that shown in the figure.
• During the ON period of the chopper, the inductance L stores energy.
• When Chopper is OFF, diode D conducts, and part of the energy stored in inductor L is returned to the supply.
• Therefore the average output voltage is positive and the average output current is negative as shown:

• In this chopper, power flows from load to source.
• Class B chopper is used for regenerative braking of dc motor.
• Class B chopper is a step-up chopper.

Concept:

In a chopper, the duty cycle is defined as the ratio of on-time to the time period.

\(D = \frac{{{T_{on}}}}{T}\)

T = T_on + T_off­

T_on is the on-time

T_off is the off-time

Calculation:

Given that, chopping frequency (f) = 1 kHz

Time period (T) = 1/1k = 1 ms

On time (T_on) = 0.5 msec

Type A chopper (first quadrant chopper):

• When the chopper is ON, V0 = Vs, and current flows in the direction of the load (as shown in fig).
• When the chopper is off, V0 = 0 but I0 continues to flow in the same direction through a freewheeling diode, thus V0 and I0 are always positive.

​Type B chopper (Second quadrant chopper):

• In type B chopper, the load must always contain DC sources.
• When the chopper is ON, V0 = 0 but load E drives the current through the inductor and the chopper, and thus inductor stores energy during the time TON of the chopper.
• When the chopper is off, \({V_0} = E + L\frac{{di}}{{dt}}\) thus V0 will be more than Vs and thus diode DB will be forward biased and begins conducting and power starts flowing to the source.

Note: No matter the chopper is ON or Off, current I0 will be flowing out of the load (opposite direction to the given circuit shown) and treated as negative.

Since V0 is +ve and I0 is -ve, Type B chopper is IInd quadrant chopper

Type C chopper (two quadrant type A chopper):

• Type – C chopper is obtained by connecting type A chopper and type B chopper in parallel.
• In this chopper, V0 is always +ve as F.D. is connected across it.
• In this type of chopper average value of current may be +ve or -ve.
• For regenerative braking and motoring, their type of chopper configuration is used.

Type D chopper (Two quadrant Type B chopper)

• When the two choppers are ON, V0 = Vs
• Average output voltage V0 will be +ve when the chopper turns ON time ‘TON’ will be more than their turn off time ‘Toff’, otherwise it will be negative.
• As the diodes and the choppers conduct current only in one direction, the direction of load current will always be positive.

Type E chopper (four-quadrant chopper):

SMPS regulator converts unregulated DC input voltage to a regulated and smooth DC output voltage at different voltage levels.

In the SMPS regulator is accompanied by duty cycle control.

Concept:

In a buck DC to DC converter

V_avg = δ V_d

\({V_{rms}} = \sqrt {\bf{\delta }} {{\bf{V}}_{\bf{d}}}\)

Where δ = duty ratio

V_i = input voltage

V_d = V_i – Voltage drop

Calculation:

Duty ratio (δ) = 50% = 0.5

V_d = V_i – Voltage drop = 100 – 2 = 98 V

V_avg = δ V_d = 0.5 × 98 = 49 V

\({V_{rms}} = \sqrt {\rm{\delta }} {{\rm{V}}_{\rm{d}}} = \sqrt {0.5} \times 98 = 69.2\;V\)