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

The correct answer is: 2) Forced commutation

Explanation:

In a chopper circuit, the thyristor (SCR) is used to control the output voltage by rapidly switching ON and OFF. However, unlike transistors or MOSFETs, thyristors cannot turn OFF on their own once triggered; they must be commutated (i.e., brought back to the OFF state).

• In DC circuits like choppers, there is no natural zero crossing of current (as in AC circuits) to help turn OFF the thyristor.

• Therefore, natural commutation (which relies on the current going to zero naturally) cannot be used.

• Instead, a forced commutation technique is used, where an auxiliary circuit forces the current through the thyristor to zero, turning it OFF.

The correct answer is: 4) All of the above

Explanation:

DC/DC converter topologies are used to convert one DC voltage level to another. The major types include:

1. Buck Converter

   • Steps down the voltage

   • The output voltage is less than the input voltage

2. Boost Converter

   • Steps up the voltage

   • The output voltage is greater than input voltage

3. Flyback Converter

   • An isolated converter

   • Can step up or step down voltage

   • Often used in low-power applications and provides galvanic isolation using a transformer

These are all commonly used topologies in DC/DC converter designs.

Hence, Option 4: All of the above is correct.

Explanation:

Buck-Boost Converter Analysis:

The given problem involves analyzing the performance of a buck-boost converter under specific operating conditions. The parameters provided are:

• Input voltage (V_in) = 24 V
• Switching frequency (f_s) = 100 kHz
• Duty ratio (D) = 0.4
• Load resistance (R) = 5 Ω
• Inductance (L) = 20 μH
• Capacitance (C) = 80 μF

The two main quantities to determine are:

1. The minimum value of the inductor current (I_L,min).
2. The percentage output voltage ripple.

Step 1: Output Voltage Calculation

The output voltage (V_out) of a buck-boost converter is determined using the formula:

V_out = V_in × D / (1 - D)

Substituting the given values:

V_out = 24 × 0.4 / (1 - 0.4)

V_out = 24 × 0.4 / 0.6

V_out = 16 V

Therefore, the output voltage of the buck-boost converter is 16 V.

Step 2: Average Load Current (I_load)

The average load current is given by:

I_load = V_out / R

Substituting the values:

I_load = 16 / 5

I_load = 3.2 A

Thus, the average load current is 3.2 A.

Step 3: Ripple Current (ΔI_L)

The peak-to-peak ripple current through the inductor is calculated using:

ΔI_L = V_in × D × (1 - D) / (f_s × L)

Substituting the values:

ΔI_L = 24 × 0.4 × (1 - 0.4) / (100 × 10³ × 20 × 10^-6)

ΔI_L = 24 × 0.4 × 0.6 / (100 × 10³ × 20 × 10^-6)

ΔI_L = 24 × 0.24 / (2 × 10^-3)

ΔI_L = 5.76 A

The peak-to-peak ripple current (ΔI_L) is 5.76 A.

Step 4: Minimum Inductor Current (I_L,min)

The minimum value of the inductor current is given by:

I_L,min = I_load - ΔI_L / 2

Substituting the values:

I_L,min = 3.2 - 5.76 / 2

I_L,min = 3.2 - 2.88

I_L,min = 2.93 A

Therefore, the minimum inductor current is 2.93 A.

Step 5: Output Voltage Ripple (ΔV_out)

The output voltage ripple is calculated using:

ΔV_out = ΔI_L / (8 × f_s × C)

Substituting the values:

ΔV_out = 5.76 / (8 × 100 × 10³ × 80 × 10^-6)

ΔV_out = 5.76 / (8 × 8 × 10^-3)

ΔV_out = 5.76 / 0.064

ΔV_out = 0.09 V

The percentage output voltage ripple is calculated as:

Percentage Ripple = (ΔV_out / V_out) × 100

Percentage Ripple = (0.09 / 16) × 100

Percentage Ripple = 0.5625%

Approximating, the percentage ripple is approximately 1%.

Final Results:

• Minimum inductor current: 2.93 A
• Percentage output voltage ripple: 1%

Correct Option: Option 3

Additional Information

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

Option 1: 7.33 A and 1% respectively

The inductor current calculation in this option is incorrect. The ripple current and load current do not lead to a minimum inductor current of 7.33 A under the given conditions. Additionally, while the percentage voltage ripple is close to 1%, this option does not provide the correct inductor current.

Option 2: 7.33 A and 10% respectively

This option incorrectly calculates both the inductor current and the percentage voltage ripple. The ripple is approximately 1%, not 10%, and the minimum inductor current cannot be 7.33 A based on the provided formulae and values.

Option 4: 2.93 A and 10% respectively

While the minimum inductor current is correctly calculated as 2.93 A, the percentage voltage ripple is not 10%. The actual ripple is approximately 1%, making this option partially correct but ultimately inaccurate.

Conclusion:

Through detailed analysis, it is evident that Option 3 provides the correct values for both the minimum inductor current (2.93 A) and the percentage output voltage ripple (1%). This highlights the importance of precise calculations and understanding the behavior of buck-boost converters under various operating conditions.

Concept:

In a buck converter, the output voltage ripple (ΔV_o) is due to the charging and discharging of the output capacitor. The ripple is affected by the inductor, capacitor, load, switching frequency, and duty cycle.

The approximate formula for ripple voltage (in continuous conduction mode) is:

\[ \frac{\Delta V_o}{V_o} = \frac{1 - D}{8 L C f^2} \]

Where:

• D = Duty cycle
• L = Inductance
• C = Capacitance
• f = Switching frequency

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.

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 chopper, chopping or switching frequency is given by

\(f=\frac{1}{T}\)

Average output voltage (V₀) is given by

\({{V}_{0}}={{V}_{s}}\cdot \frac{\left( {{T}_{ON}} \right)}{\left( {{T}_{ON}}+{{T}_{Off}} \right)~}\)

\({{V}_{0}}={{V}_{s}}\frac{{{T}_{ON}}}{T}\)

V₀ = Vs⋅ f ⋅ T_ON

Where,

V_S is the supply voltage

T = Total time period (T_ON + T_OFF).

T_ON = ON-time period

T_OFF = OFF-time period

Calculation:

Given that,

T_ON = 25 μs, T_OFF = 5 μs  

T = 25 + 5 = 30 μs

∴ Chopper frequency can be calculated as 

\(f=\frac{1}{30\times 10^{-6}}\)

f = 33.33 kHz