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

Last updated on Mar 21, 2025

Latest Step Up Choppers MCQ Objective Questions

Step Up Choppers Question 1:

Find the relation between output voltage (Vo) and input voltage (Vin) for a Boost converter, where d is the duty ratio. 

  1. \(\rm v_o=v_{in}*\frac{d}{1-d}\)
  2. \(\rm v_o=v_{in}*\frac{1}{d-1}\)
  3. \(\rm v_o=v_{in}*\frac{d}{d-1}\)
  4. \(\rm v_o=v_{in}*\frac{1}{1-d}\)

Answer (Detailed Solution Below)

Option 4 : \(\rm v_o=v_{in}*\frac{1}{1-d}\)

Step Up Choppers Question 1 Detailed Solution

Explanation:

Boost Converter

Definition: A Boost Converter is a DC-DC power converter that steps up the input voltage to a higher output voltage. It is widely used in applications where the voltage needs to be increased from the input supply to the load.

Working Principle: The Boost Converter operates by storing energy in an inductor during the switch-on period and releasing it to the load during the switch-off period. The operation is controlled by a switch (typically a transistor) and a diode.

During the switch-on period, the input voltage is applied across the inductor, causing current through the inductor to rise linearly. When the switch is turned off, the energy stored in the inductor is transferred to the output capacitor and load via the diode, resulting in a higher output voltage than the input.

Key Equations: The relationship between the output voltage (\(V_o\)) and the input voltage (\(V_{in}\)) for a Boost Converter is derived from the principle of inductor volt-second balance and capacitor charge balance.

The duty ratio (d) is defined as the fraction of time the switch is on during one switching cycle. The key equation for the Boost Converter is:

Correct Option Analysis:

The correct option is:

Option 4: \(\rm v_o=v_{in}*\frac{1}{1-d}\)

This equation correctly describes the relationship between the output voltage (\(V_o\)) and the input voltage (\(V_{in}\)) in a Boost Converter. The derivation of this equation involves analyzing the voltage and current during the switch-on and switch-off periods.

**Derivation:** 1. **Switch On Period (dT):** When the switch is on, the input voltage (\(V_{in}\)) is applied across the inductor (L). The current through the inductor increases linearly. \[ V_{in} = L \frac{dI_L}{dt} \] 2. **Switch Off Period ((1-d)T):** When the switch is off, the inductor releases its stored energy to the output capacitor and load through the diode. The inductor voltage is \(V_{in} - V_o\). \[ V_{in} - V_o = L \frac{dI_L}{dt} \] 3. **Volt-Second Balance:** Over one complete switching cycle, the average voltage across the inductor is zero. Therefore, \[ V_{in} \cdot dT = (V_o - V_{in}) \cdot (1-d)T \] 4. **Solving for \(V_o\):** \[ V_{in} \cdot d = V_o (1-d) - V_{in}(1-d) \] \[ V_{in} \cdot d = V_o - V_o \cdot d - V_{in} + V_{in} \cdot d \] \[ V_o (1 - d) = V_{in} \] \[ V_o = \frac{V_{in}}{1-d} \]

This derivation confirms that the output voltage \(V_o\) is equal to the input voltage \(V_{in}\) divided by (1-d), where d is the duty ratio.

Additional Information

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

Option 1: \(\rm v_o=v_{in}*\frac{d}{1-d}\)

This equation is incorrect. It suggests that the output voltage is directly proportional to the duty ratio (d) and inversely proportional to (1-d). However, this does not align with the fundamental principles of the Boost Converter, as derived above.

Option 2: \(\rm v_o=v_{in}*\frac{1}{d-1}\)

This equation is also incorrect. It implies that the output voltage is inversely proportional to (d-1), which is not a valid representation of the Boost Converter’s operation. The denominator (d-1) would lead to negative or undefined values for certain duty ratios, which is not physically meaningful.

Option 3: \(\rm v_o=v_{in}*\frac{d}{d-1}\)

This option is incorrect as well. It suggests that the output voltage is proportional to the duty ratio (d) and inversely proportional to (d-1). This formulation does not correspond to the correct relationship derived from the Boost Converter's operation.

Conclusion:

Understanding the correct relationship between the output voltage and input voltage in a Boost Converter is essential for its design and application. The correct equation, \(V_o = V_{in} \cdot \frac{1}{1-d}\), accurately represents the step-up voltage conversion, where the output voltage is higher than the input voltage, depending on the duty ratio (d). This relationship is crucial for ensuring the proper functioning and efficiency of the Boost Converter in various electronic and power supply applications.

Step Up Choppers Question 2:

A step - up chopper is used to deliver load voltage of 400 V from a 200 V d.c. source, if the blocking period of the thyristor is 80 μs, compute the required pulse width.

  1. 80 μs
  2. 160 μs
  3. 500 μs
  4. 200 μs

Answer (Detailed Solution Below)

Option 1 : 80 μs

Step Up Choppers Question 2 Detailed Solution

The correct answer is 80 µs.

Key Points

  • A **step-up chopper** is also known as a **boost converter**. It converts a lower input DC voltage to a higher output DC voltage.
  • In a step-up chopper, the **output voltage (Vo)** is related to the **input voltage (Vi)** and the **duty cycle (D)** of the chopper. The formula is given by:

    Vo = Vi / (1 - D)

  • Given the **load voltage (Vo)** is 400 V and the **input voltage (Vi)** is 200 V, we can rearrange the formula to find the duty cycle:

    D = 1 - (Vi / Vo)

  • Substituting the given values:

    D = 1 - (200 / 400)

    D = 1 - 0.5

    D = 0.5

  • The **duty cycle (D)** is 0.5, which means the pulse width (Ton) is equal to the blocking period (Toff).
  • Given the blocking period (Toff) is 80 µs, the pulse width (Ton) is also:

    Ton = 80 µs

 Additional Information

  • Step-Up Chopper (Boost Converter)
    • A **step-up chopper** is used in various applications such as **renewable energy systems** (e.g., solar panels), **electric vehicles**, and **power supply units**.
    • It operates by storing energy in an **inductor** during the on-state and releasing it to the load during the off-state.
    • Key components include a **switching device (like a MOSFET or IGBT)**, **diode**, **inductor**, and **capacitor**.
  • Duty Cycle (D)
    • The **duty cycle** is the ratio of the **on-time (Ton)** to the total time period (T) of the chopper.
    • It is expressed as:

      D = Ton / (Ton + Toff)

    • A higher duty cycle means a higher output voltage in a step-up chopper.
  • Blocking Period (Toff)
    • The **blocking period** refers to the time during which the switch (thyristor) is off, and no current flows through it.
    • In a boost converter, this period is crucial for **energy transfer** from the inductor to the load.

Step Up Choppers Question 3:

What is the average value of output of a chopper with duty ratio 0.5 and source voltage 50 V?

  1. 50 V
  2. 100 V
  3. 200 V
  4. 25 V

Answer (Detailed Solution Below)

Option 4 : 25 V

Step Up Choppers Question 3 Detailed Solution

The correct answer is 25 V.
Concept:


To find the average output voltage of a chopper with a duty ratio of 0.5 and a source voltage of 50 V,
we can use the formula for the average output voltage of a chopper:

V out(avg)=V source ×Duty Ratio

Where: out(avg)
V out(avg) is the average output voltage,

source V source is the source voltage,
Duty Ratio is the ratio of the ON time of the chopper to the total period.
Given that the duty ratio is 0.5 and the source voltage is 50 V:
out(avg)=50×0.5=25V
V out(avg)=50×0.5=25 V

So, the average output voltage of the chopper is 25 V.
Therefore, the correct option is:25 V

Step Up Choppers Question 4:

The chopper circuit shown in figure (i) feeds power to a 5 A DC constant current source. The switching frequency of the chopper is 100 kHz. All the components can be assumed to be ideal. The gate signals of switches S1 and S2 are shown in figure (ii). Average voltage across the 5 A current source is 

F1 Savita ENG 16-11-23 D80

  1. 10 V
  2. 6 V
  3. 12 V
  4. 20 V

Answer (Detailed Solution Below)

Option 2 : 6 V

Step Up Choppers Question 4 Detailed Solution

 

F1 Savita ENG 16-11-23 D81

when switch S1 ON → ν0 = νs = 20V

D2 ON → ν0 = 0 volt

S2 ON → ν0 = 0 volt (no energy stored)

F1 Savita ENG 16-11-23 D82

\(V_{0(\text { ang })}=\frac{20 \times 3}{10}=6 \mathrm{Volt}\)

Hence, the correct option is (B).

Step Up Choppers Question 5:

A step up chopper delivers an average output voltage of 100 V from an input supply of 60 V when operating with a continuous source current. What is the operating duty ratio for the switch?

  1. 1/3
  2. 0.6
  3. 0.4
  4. 2/3

Answer (Detailed Solution Below)

Option 3 : 0.4

Step Up Choppers Question 5 Detailed Solution

Concept:

The circuit diagram of a boost converter is shown below.

F1 U.B Deepak 24.10.2019 D 10

A step-up or Boost converter is used to obtain the output voltage greater than the input voltage.

The relation between the output voltage and the input voltage is given by

\(Vo=\frac{{{V}_{s}}}{1-D}\)

Where D is the duty cycle of the chopper

Calculation:

Input voltage (VS) = 100 V

Average output voltage (Vo) = 60 V

\(100 = \frac{60}{{1 - D}}\)

\( \Rightarrow D = \frac{2}{5}\)= 0.4

Additional Information Duty Cycle:

The duty cycle for an alternating quantity describes the fraction of time that the pulse is ON in one pulse period.

Mathematically, this is defined as:

Duty Cycle = \(\frac{T_{on}}{T_{on}+T_{off}}\)

Ton = Time for which the output is ON

Toff = Time for which the output is OFF

Also, the Time period is the sum of the ON time and OFF time, i.e.

T = Ton + Toff

Duty Cycle = \(\frac{T_{on}}{T}\)

∴ The duty cycle for repetitive waveform is defined as the ratio of ON time to Total time.

Top Step Up Choppers MCQ Objective Questions

A step-up chopper is fed with 200 V. The conduction time of the thyristor is 200 µs and the required output is 600 V. If the frequency of operation is kept constant and the pulse width is halved, what will be the new output voltage?

  1. 600 volts
  2. 300 volts
  3. 400 volts
  4. 200 volts

Answer (Detailed Solution Below)

Option 2 : 300 volts

Step Up Choppers Question 6 Detailed Solution

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Formula:

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

Where, Vo is the output voltage

Vin is the input voltage

TON is the pulse width

Application:

Given,

Vin = 200 volts

TON = 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 (TON') will be,

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

Hence,

Hence, the new value of output voltage (V0') will be,

\(V_o'=V_{in}(\frac{T}{T-T_{ON'}})=200\times (\frac{300}{300-100})=300\ volts\)

A step-up chopper is used to deliver a load voltage of 500 V from a 220 V d.c. source. If the blocking period of the thyristor is 80 μs, the required pulse width is -

  1. 50.8 μs
  2. 101.8 μs
  3. 92.4 μs
  4. 152.4 μs

Answer (Detailed Solution Below)

Option 2 : 101.8 μs

Step Up Choppers Question 7 Detailed Solution

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Concept:

The output voltage of a step-up chopper is given by:

\(V_o = {V_{in} \over 1-D}\)

where, Vo = Output voltage 

Vin = Input voltage

D = Duty cycle

The duty cycle is given by:

\(D = {T_{ON} \over T_{ON}+T_{OFF}}\)

TON is the pulse width of the output.

Calculation:

Given, Vo = 500 V

Vin = 220 V

Toff = 80 μs

\(500 = {220 \over 1-D}\)

D = 0.56

\(0.56 = {T_{ON} \over T_{ON}+80}\)

Ton = 101.8 μs

If the duty ratio of a boost converter is 50 percent, the output voltage corresponding to an input of 25 V is

  1. 50 V
  2. 30 V
  3. 40 V
  4. 25 V

Answer (Detailed Solution Below)

Option 1 : 50 V

Step Up Choppers Question 8 Detailed Solution

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Concept:

The circuit diagram of a boost converter is shown below.

F1 U.B Deepak 24.10.2019 D 10

Step Up or Boost converter is used to obtain the output voltage greater than the input voltage.

The relation between the output voltage and the input voltage is given by

\(Vo=\frac{{{V}_{s}}}{1-D}\)

Where D is the duty cycle of the chopper

Calculation:

Duty ratio = 50 % = 0.5

Input voltage (VS) = 25 V

Average output voltage (Vo) =

\(V_0 = \frac{V_s}{{1 - D}} \Rightarrow V_0 = \frac{25}{{1 - 0.5}}\)

V0 = 50 V

A dc to dc converter shown in the figure is charging a battery bank, B2 whose voltage is constant at 150 V. B1 is another battery bank whose voltage is constant at 50 V. The value of the inductor, L is 5 mH and the ideal switch, S is operated with a switching frequency of 5 kHz with a duty ratio of 0.4. Once the circuit has attained steady-state and assuming the diode D to be ideal, the power transferred from B1 to B2 (in Watt) is ___________ (up to 2 decimal places).

GATE EE 2018 Techinical 54Q images Q53

Answer (Detailed Solution Below) 12

Step Up Choppers Question 9 Detailed Solution

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Duty Cycle is defined as:

\(Duty\;Cycle = \frac{{{T_{ON}}}}{{{T_{ON}} + {T_{OFF}}}}\)

TON = D × T

Putting on the respective values:

\({T_{ON}} = 0.4 \times \frac{1}{f} = 0.4 \times \frac{1}{{5 \times {{10}^3}}} = 0.8\;msec\)

TOFF = T – TON = 0.2 – 0.08 = 0.12 msec

CASE I : When Switch is ON

F2 Subham Deepak 11.01.2020 D1

\({V_L} = L\frac{{di}}{{dt}} = 50\)

\(\frac{{di}}{{dt}} = \frac{{50}}{L} = \frac{{50}}{{5 \times {{10}^{ - 3}}}}\)

\(\frac{{{\rm{\Delta }}I}}{{DT}} = \frac{{50}}{{5 \times {{10}^{ - 3}}}}\)

\({\rm{\Delta }}I = \frac{{50}}{{5 \times {{10}^{ - 3}}}} \times DT\)

\(= \frac{{50}}{{5 \times {{10}^{ - 3}}}} \times 0.4 \times \frac{1}{{5 \times {{10}^3}}}\)

ΔI = 0.8 A

CASE II: Checking whether we have continuous or discontinuous conduction by finding TOFF new

During TOFF → Switch is open and the diode is ON.

F2 Subham Deepak 11.01.2020 D2

\({V_L} = L\frac{{di}}{{dt}} = - 100\)

\(\frac{{{\rm{\Delta }}I}}{{{T_{OFF}}}} = \left( { - \frac{{100}}{L}} \right)\)

\({T_{OFF}} = - \frac{{L \times {\rm{\Delta }}{I_L}}}{{100}} = \frac{{ - 5 \times {{10}^{ - 3}} \times 0.8}}{{100}}\)

TOFF = 4 × 10-5 = 0.04 msec

As T(OFF)new < TOFF, conduction will be discontinuous.

F2 Subham Deepak 11.01.2020 D3

\({I_L}\;or\;{I_{s\left( {avg} \right)}} = \frac{1}{2}\left[ {\frac{{{\rm{\Delta }}I \times {T_{ON}}\; + \;{\rm{\Delta }}I \times {T_{\left( {OFF} \right)new}}}}{T}} \right]\)

\( = \frac{1}{2} \times 0.8\;\left( {{{10}^{ - 3}} \times 0.12 \times f} \right)\)

Is(avg) = 0.4 × 0.12 × 103 × 5 × 103 = 0.24 A

Pavg = Vs.Is(avg) = 12 W

Output power is calculated as:

F2 Subham Deepak 11.01.2020 D4

\({I_{0\left( {avg} \right)}} = \frac{1}{2} \times \frac{{{\rm{\Delta }}I \times {T_{\left( {OFF} \right)new}}}}{T}\)

\({I_{0\left( {avg} \right)}} = \frac{1}{2} \times 0.8 \times 0.04 \times {10^{ - 3}} \times 5 \times {10^3}\)

\({I_{0\left( {avg} \right)}} = 0.4 \times 0.04 \times 5 = 0.08\;A\)

Hence, P0 = V0 × I0 = 150 × 0.08 = 12 W

Consider the boost converter shown. Switch Q operating at 25 kHz with a duty cycle of 0.6. Assume the diode and switch to be ideal. Under steady-state condition, the average resistance Rin as seen by source is ______ Ω. (Round off to 2 decimal places.)

F1 Shraddha Koda 20.02.2021 D16

Answer (Detailed Solution Below) 1.55 - 1.65

Step Up Choppers Question 10 Detailed Solution

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Concept:

Boost Converter:

It is a switch mode DC to DC converter in which the output voltage is greater than the input voltage. It is also called a step-up converter.

A typical Boost converter is shown below.

F1 U.B Madhu 08.06.20 D10

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

\(\frac{{{V_o}}}{{{V_s}}} = \frac{1}{{\left( {1 - D} \right)}}\)

Where V0 is the output voltage and Vs is the input voltage.

The average source current (Is) given by

\({I_s} = \frac{{{I_o}}}{{\left( {1 - D} \right)}} \)

Calculation:

Given,

D = 0.6

Vs = 15 volt

From the above concept,

\({V_0} = \frac{{{V_s}}}{{\left( {1 - D} \right)}} = \frac{{15}}{{\left( {1 - 0.6} \right)}} = 37.5\;volt\)

Current through the load (Io),

\({I_o} = \frac{{{V_o}}}{R}\)

R = 10 Ω

\({I_o} = \frac{{37.5}}{{10}} = 3.75\;A\)

The average source current (Is) given by,

\({I_s} = \frac{{{I_o}}}{{\left( {1 - D} \right)}} = \frac{{3.75}}{{\left( {1 - 0.6} \right)}} = 9.375\;A\)

Vs = Is × Rs

\({R_s} = \frac{{{V_s}}}{{{I_s}}} = \frac{{15}}{{9.375}} = 1.6\;{\rm{\Omega }}\)

Important Points

Converter

Circuit diagram

Output voltage

Buck converter

F1 U.B Madhu 08.06.20 D9

Vo = DVin

Boost converter

F1 U.B Madhu 08.06.20 D10

\({V_o} = \frac{{{V_{in}}}}{{1 - D}}\)

 

Buck-Boost converter

F1 U.B Madhu 08.06.20 D11

\({V_o} = - \frac{D}{{1 - D }}{V_{in}}\)

 

Cuk converter

F1 U.B Madhu 13.04.20 D19

\({V_o} = - \frac{D }{{1 - D}}{V_{in}}\)

 

The input voltage VDC of the buck-boost converter shown below varies from 32 V to 72 V. Assume that all components are ideal, inductor current is continuous, and output voltage is ripple free. The range of duty ratio D of the converter for which the magnitude of the steady-state output voltage remains constant at 48 V is

GATE IN Signals and digital 30Q Sunny.docx 20

  1. \(\frac{2}{5} \le D \le \frac{3}{5}\)
  2. \(\frac{2}{3} \le D \le \frac{3}{4}\)
  3. 0 ≤ D ≤ 1
  4. \(\frac{1}{3} \le D \le \frac{2}{3}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{2}{5} \le D \le \frac{3}{5}\)

Step Up Choppers Question 11 Detailed Solution

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Concept:

For a buck-boost converter,

\({V_0} = \left( {\frac{D}{{1 - D}}} \right){V_{DC}}\)

Calculation:

Given that,

Source voltage (VDC) = 32 V to 72 V

Output voltage, (V0) = 48 V

For a buck-boost converter,

\({V_0} = \left( {\frac{D}{{1 - D}}} \right){V_{DC}}\)

Where D is duty ratio

When VDC = 32V,

\(\Rightarrow 48 = \left( {\frac{D}{{1 - D}}} \right)32\)

⇒ 3(1 - D) = 2D

\(\Rightarrow D = \frac{3}{5}\)

When VDC = 72V,

\(\Rightarrow 48 = \left( {\frac{D}{{1 - D}}} \right)72\)

⇒ 2(1 - D) = 3D

\(\Rightarrow D = \frac{2}{5}\)

Range of D is: \(\frac{2}{5} \le D \le \frac{3}{5}\)

Find the type of chopper that produces an output voltage whose value is given by Vs/(1 - α). (where, Vs is source voltage and α is duty cycle)

  1. Step up chopper
  2. DC chopper
  3. AC link chopper
  4. Stepdown chopper

Answer (Detailed Solution Below)

Option 1 : Step up chopper

Step Up Choppers Question 12 Detailed Solution

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Chopper:

  • It is a basic static power electronics device that converts fixed DC voltage/power to variable DC voltage or power.
  • It is nothing but a high-speed switch that connects and disconnects the load from the source at a high rate to get variable or chopped voltage at the output.
  • There are mainly two types of the chopper: Step-up and Step-down choppers. This classification is based on the average DC output voltage of the chopper.
  • However, on the basis of quadrant operation, a chopper may be classified into five different types: Class-A, Class-B, Class-C, Class-D, and Class-E chopper.

 

The step-up chopper works as a step-up transformer on DC current. This chopper is used when the output DC voltage has to be made higher than the input voltage.


The working principle of a step-up chopper can be explained in the above diagram. In the circuit, a large inductor L is connected in series to the supply voltage. The capacitor maintains the continuous output voltage to the load. The diode prevents the flow of current from load to source.

F1 Savita Engineering 22-8-22 D5

  • When the chopper is ON, supply voltage VS is applied to the load .i.e. V0 = VS, and the inductor starts storing energy. At this condition load current raises from Imin to Imax.
  • When the chopper is switched OFF, the supply voltage takes the path from L – D – Load – VS. During this period the inductor discharges the stored e.m.f through diode D to the load.
     

Thus the total voltage at the load:

V0 = VS + Ldi/dt which s greater than the input voltage.

Current changes from Imax to Imin.

The average output voltage of the step-up chopper is given by

\(V_0 = \frac{V_s}{1-\alpha}\)
Step–up chopper is also known as Boost choppers. Applications of the step-up choppers include battery charging and voltage booster.

In a DC-DC boost converter, the duty ratio is controlled to regulate the output voltage at 48 V. The input DC voltage is 24 V. The output power is 120 W. The switching frequency is 50 kHz. Assume ideal components and a very large output filter capacitor. The converter operates at the boundary between continuous and discontinuous conduction modes. The value of the boost inductor (in μH) is _______.

Answer (Detailed Solution Below) 24

Step Up Choppers Question 13 Detailed Solution

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Concept:

The circuit diagram of a boost converter is shown below.

F1 U.B Deepak 24.10.2019 D 10

Step up or Boost converter is used to obtain the output voltage greater than the input voltage.

Volt - sec balance:

VS  ∝T + Vs (1 - D) – V0 (1 - D)T = 0

\({{\rm{V}}_0} = \frac{{{V_s}}}{{1 - \; \propto }}\)

Ampere – sec balance:

- Io ∝T + (IL – I0) (1 - ∝) T = 0

- I∝ T + IL (1 - ∝) T – I0 (1 - ∝) T = 0

\({{\rm{I}}_L} = \frac{{{I_0}}}{{1 - \propto }}\)    

Ripple current:

VL(ON) = Vs

\(\begin{array}{l} L\frac{{\Delta {\rm{I}}}}{{ \propto T}} = {V_s}\\ \Delta {\rm{I}} = \frac{{ \propto {V_s}}}{{FL}}\\ \Rightarrow {I_{Lmax}} = {I_L} + \frac{{\Delta {{\rm{I}}_L}}}{2}\\ = \frac{{{{\rm{I}}_0}}}{{1 - \propto }} + \frac{{ \propto {V_s}}}{{2FL}} \end{array}\)

Calculation:

Output voltage (Vo) = 48 V

input voltage (Vs) = 24 V

output power (P) = 120 W

For boost converter,

\({V_o} = \frac{{{V_s}}}{(1-\delta) }\)

\(\Rightarrow 48 = \frac{{24}}{1-\delta } \Rightarrow \delta = 0.5\)

Switching frequency (f) = 50 kHz

\(Time\;period\;\left( T \right) = \frac{1}{f} = \frac{1}{{50 \times {{10}^3}}} = 20\;\mu s\)

Avg value of supply current = Avg value of inductor current.

\(\Rightarrow {I_S} = {I_L} = \frac{P}{V} = \frac{{120}}{{24}} = 5\;A\)

\({I_{{L_{max}}}} = 2{I_L} = 10A\)

We know that,

\(current\;ripple = \frac{{{V_S}}}{L}\delta T\)

\(\Rightarrow 10 = \frac{{24}}{L} \times 0.5 \times 20 \times {10^{ - 6}}\)

⇒ L = 24 μH

A self-commutating switch SW, operated at duty cycle \(\delta\) is used to control the load voltage as shown in the figure.

Gate EE 2015 paper 1 Images-Q33

Under steady state operating conditions, the average voltage across the inductor and the capacitor respectively, are

  1. \({V_L} = 0\) and \({V_C} = \frac{1}{{1 - \delta }}{V_{dc}}\)
  2. \({V_L} = \frac{\delta }{2}{V_{dc}}\) and \({V_C} = \frac{1}{{1 - \delta }}{V_{dc}}\)
  3. \({V_L} = 0\) and \({V_C} = \frac{\delta }{{1 - \delta }}{V_{dc}}\)
  4. \({V_L} = \frac{\delta }{2}{V_{dc}}\) and \({V_C} = \frac{\delta }{{1 + \delta }}{V_{dc}}\)

Answer (Detailed Solution Below)

Option 1 : \({V_L} = 0\) and \({V_C} = \frac{1}{{1 - \delta }}{V_{dc}}\)

Step Up Choppers Question 14 Detailed Solution

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Identification of DC-DC Converter:

  • Buck converter: Inductor is connected in series with the combination of the capacitor and resistor.
  • Boost converter: Inductor is connected in series with the supply voltage source.
  • Buck-Boost converter: Inductor is connected in parallel to the supply voltage source.

 

Under steady-state operating conditions:

  • The average voltage across the inductor is zero. VL = 0
  • The average current across the capacitor is zero. IC = 0

 

 

Buck

Boost

Buck-Boost

Output voltage V0

V0 = DVdc

\({V_0} = {V_{dc}}\left[ {\frac{1}{{1 - D}}} \right]\)

\({V_0} = {V_{dc}}\left[ {\frac{D}{{1 - D}}} \right]\)

Inductor current IL

\({I_L} = {I_0} = \frac{{{V_0}}}{R}\)

\({I_L} = \frac{{{I_0}}}{{1 - D}} \)

\({I_L} = \frac{{{I_0}}}{{1 - D}}\)

Source current Is

Is = DI0

\({I_s} = \frac{{{I_0}}}{{1 - D}}\)

\({I_s} = {I_0}\left[ {\frac{D}{{1 - D}}} \right]\)

 

Given DC-DC converter is a Boost converter with duty cycle ratio δ.

The voltage across the load is the same as the voltage across the capacitor.

\({V_C} = \;{V_0} = {V_{dc}}\left[ {\frac{1}{{1 - δ }}} \right]\)

A buck-boost DC-DC converter, shown in the figure below, is used to convert \(24\;V\) battery voltage to \(36\;V\) DC voltage to feed a load of \(72\;W\). It is operated at \(20\;kHz\) with an inductor of \(2\;mH\) and output capacitor of \(1000\;\mu F\). All devices are considered to be ideal. The peak voltage across the solid-state switch (S), in volt, is ____________.

Gate EE 2016 paper 2 Images-Q20

Answer (Detailed Solution Below) 59.5 - 60.5

Step Up Choppers Question 15 Detailed Solution

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Concept:

Buck - Boost Converter: 

The buck–boost converter is a type of DC to DC converter that has an output voltage magnitude that is either greater than or less than the input voltage magnitude. It is equivalent to a Fly-back converter using a single inductor instead of a fly-back Transformer. The opposite voltage is the opposite polarity of that of the input.

The steady-state output voltage Vout of buck-boost converter is

\({V_{out}}\; = -\;{V_{in}}\left( {\frac{D}{{1\; - \;D}}} \right)\)

Here D = duty cycle

GATE IN Signals and digital 30Q Sunny.docx 20

  • When Chopper is ON, Current flows from Sending Voltage, Vs to Chopper, CH, Inductor, L and back to Vs
  • Energy is stored in the inductor during TON
  • When the chopper is OFF, the inductor stored energy now discharges through the path, load, diode D, and L.

 

Explanation:

When switch is OFF, diode D is ON as shown in below figure,

F1 Gaurav.EE 15-09-21 Savita D16

Using KVL,

peak voltage across switch = 24 + 36 = 60 V

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