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

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.

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 (V_o)** is related to the **input voltage (V_i)** and the **duty cycle (D)** of the chopper. The formula is given by:

  V_o = V_i / (1 - D)

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

  D = 1 - (V_i / V_o)

• 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 (T_on) is equal to the blocking period (T_off).
• Given the blocking period (T_off) is 80 µs, the pulse width (T_on) is also:

  T_on = 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 (T_on)** to the total time period (T) of the chopper.
  • It is expressed as:

    D = T_on / (T_on + T_off)

  • A higher duty cycle means a higher output voltage in a step-up chopper.
• Blocking Period (T_off)
  • 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.

V out(avg)=V source ×Duty Ratio

Where: V 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:
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

when switch S₁ ON → ν₀ = ν_s = 20V

D₂ ON → ν₀ = 0 volt

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

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

Hence, the correct option is (B).

Concept:

The circuit diagram of a boost converter is shown below.

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.

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,

Concept:

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

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

where, V_o = Output voltage 

V_in = Input voltage

D = Duty cycle

The duty cycle is given by:

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

T_ON is the pulse width of the output.

Calculation:

Given, V_o = 500 V

V_in = 220 V

T_off = 80 μs

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

D = 0.56

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

T_on = 101.8 μs

Concept:

The circuit diagram of a boost converter is shown below.

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

Duty Cycle is defined as:

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

T_ON = D × T

Putting on the respective values:

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

T_OFF = T – T_ON = 0.2 – 0.08 = 0.12 msec

CASE I : When Switch is ON

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

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

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

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

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

ΔI = 0.8 A

CASE II: Checking whether we have continuous or discontinuous conduction by finding T_{OFF new}

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

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

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

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

T_OFF = 4 × 10^-5 = 0.04 msec

As T_(OFF)new < T_OFF, conduction will be discontinuous.

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

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

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

P_avg = V_s.I_s(avg) = 12 W

Output power is calculated as:

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

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

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

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

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.

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

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

Where V₀ is the output voltage and V_s 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

V_s = 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 (I_o),

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

R = 10 Ω

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

The average source current (I_s) given by,

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

V_s = I_s × R_s

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

Important Points

Concept:

For a buck-boost converter,

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

Calculation:

Given that,

Source voltage (V_DC) = 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 V_DC = 32V,

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

⇒ 3(1 - D) = 2D

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

When V_DC = 72V,

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

⇒ 2(1 - D) = 3D

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

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.

• When the chopper is ON, supply voltage VS is applied to the load .i.e. V₀ = V_S, 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 – V_S. 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 = V_S + Ldi/dt which s greater than the input voltage.

Current changes from I_max to I_min.

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.

Concept:

The circuit diagram of a boost converter is shown below.

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

- Io ∝ 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 (V_o) = 48 V

input voltage (V_s) = 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,

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

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

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. V_L = 0
• The average current across the capacitor is zero. I_C = 0

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.

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

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

• 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,

Using KVL,

peak voltage across switch = 24 + 36 = 60 V