Step Up Choppers MCQ Quiz in বাংলা - Objective Question with Answer for Step Up Choppers - বিনামূল্যে ডাউনলোড করুন [PDF]

In a boost converter, for continuous conduction mode

\(K > \frac{{2fL}}{R}\)

K > (1 – D)²D

For the minimum value of K,

\(\frac{d}{{dt}}\left[ {{{\left( {1 - D} \right)}^2}D} \right] = 0\)

\( \Rightarrow D = \frac{1}{3}\)

The minimum value of K is \( = {\left( {1 - \frac{1}{3}} \right)^2} \times \frac{1}{3} = 0.148\)

Given that, f = 100 kHz

Inductance, L = 10 μH

\(K > \frac{{2fL}}{R}\)

\( \Rightarrow 0.148 > \frac{{2 \times 100 \times {{10}^3} \times 10 \times {{10}^{ - 6}}}}{R}\)

⇒ R > 13.51 Ω

Minimum load current \({I_0} = \frac{{{V_o}}}{R} = \frac{{60}}{{13.51}} = 4.44\;A\)

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:

Input voltage (VS) = 5 V

Average output voltage (Vo) = 10 V

\(10 = \frac{5}{{1 - D}}\)

\( \Rightarrow D = \frac{1}{2}\)

Given that, input voltage (V_s) = 25 V

Output voltage (V_o) = 10 V

Duty cycle (D) \(= \frac{{{V_o}}}{{{V_s}}} = \frac{{10}}{{25}} = 0.4\)

Switching frequency (f) = 25 kHz

Peak to peak ripple current (ΔI) = 0.5 A

Peak to peak ripple voltage (ΔV) = 15 mV

\({\rm{\Delta }}I = \frac{{{V_o}\left( {{V_s} - {V_o}} \right)}}{{fL{V_s}}}\)

\( \Rightarrow 0.5\; = \frac{{10\;\left( {25 - 10} \right)}}{{25\; \times \;{{10}^3}\; \times \;L \;\times\; 25\;}}\)

⇒ L = 480 μH

\({\rm{\Delta }}{V_o} = \frac{{{V_o}\left( {1 - D} \right)}}{{8LC{f^2}}}\)

\( \Rightarrow 15 \times {10^{ - 3}} = \frac{{10\;\left( {1 - 0.4} \right)}}{{8\; \times \;480\; \times\; {{10}^{ - 6}}\; \times \;C \;\times\; {{\left( {25 \;\times\; {{10}^3}} \right)}^2}}}\)

⇒ C = 166.67 μF

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:

Buck-Boost Regulator:

• It provides negative polarity output with respect to common terminal of the input voltage & the output voltage can be either higher or lower than the input voltage.
• It can be obtained from the cascading of a step-down converter & step-up converter.
• The ratio of output to input voltage can be expressed as:
   

\(\frac{{{V_o}}}{{{V_{in}}}} = - \left( {\frac{\alpha }{{1 - \alpha }}} \right)\)

α = Duty cycle

Also, the output voltage ripple is given by:

\({\rm{\Delta }}{V_o} = \frac{{\alpha {I_o}}}{{Cf}}\)

α = Duty cycle

C = Capacitor

f = Switching frequency

I_o = Average ouput current

Calculations:

Given V_i = 12 V

Duty Cycle = 0.25

Switching frequency (f) = 25 kHz

L = 125 μH and C = 220 μF

The average output voltage will be:

\({V_o} = - \left( {\frac{\alpha }{{1 - \alpha }}} \right){V_i} = \frac{{ - \left( {0.25} \right)}}{{\left( {0.75} \right)}}\left( {12} \right)\;V\)    

V₀ = - 4 V

Similarly, the output voltage ripple ΔV_o is given by:

\({\rm{\Delta }}{V_o} = \frac{{\alpha {I_o}}}{{Cf}} = \frac{{0.25\; \times \;1.25}}{{220\; \times \;{{10}^{ - 6}}\; \times \;25\; \times \;{{10}^{ + 3}}}} = + \;56.8\;mV\)   

Voltage equation across inductance is given by

\({v_L} = L\frac{{dI}}{{dt}}\)

Calculation:

Circuit diagram for the boost chopper and its equivalent circuit with conducting switch open are shown below.

Applying KVL in the above circuit

V_s – V_L – V₀ = 0

⇒ V_s – V₀ = V_L

Rate of change of inductive current  is 

\({V_L} = L\frac{{d{I_L}}}{{dt}}\)

\({V_s} - {V_0} = L\frac{{d{I_L}}}{{dt}}\)

\(\frac{{d{I_L}}}{{dt}} = \frac{{{V_s} - {V_0}}}{L}\)

Calculation:

V_o = I_oR_L = 3 × 10 = 30 V

V_b = V_o (1 – D)

\(1 - D = \frac{{12}}{{30}} = 0.4\)

D = 0.6

Periodic time \(T = \frac{1}{f} = \frac{1}{{50 \times {{10}^3}}} = 20\;\mu s\)

On time = DT = 0.6 × 20 × 10^-6 = 12 μs 

over an on-time the change in battery current will be

\({\rm{\Delta }}i = \frac{{{V_L}}}{L} \times Ton\)

\({\rm{\Delta }}i = \frac{{12}}{{20 \times {{10}^{ - 6}}}} \times 20 \times {10^{ - 6}} = 7.2\;A\)

Concept:

When the transistor is switched on, current ramps up in the inductor and energy is stored.

When the transistor is switched off, the inductor voltage reverses and acts together with the battery voltage to forward bias the diode, transferring energy to the capacitor.

When the transistor is switched on again, load current is maintained by the capacitor, energy is stored in the inductor and the cycle can start again.

The value of the load voltage is increased by increasing the duty cycle or the on-time of the transistor.

During the transistor on-period, assuming an ideal inductor and transistor

V_b = V_L = Ldi/dt

V_b/s = (sL) i(s)

i(s) = V_b/s²L

i(t) = (V_b/L)t = kt

Over an on-time of T_ON, the change of battery current will be

Δi = k T_ON

During the off-period

V_o = V_b + V_L = V_b + L(di/dt) = V_b + L(Δi/Δt)

= V_b + L(V_b/L)T_ON /T_off = V_b(1 + T_ON /T_off)

Let the duty cycle

D = T_ON T, and T_off = T – T_ON = T(1 - D). Now

V_o = V_b (1 – DT/T(1 – D))

which simplifies to

V_o = V_b/(1 - D)

In an ideal circuit, V_bI_b = V₀I₀. Therefore, from eq.

I_b = (V_o/V_b)I_b = I_o/(1 – D)

I_b = I_b + Δi/2

I_bo = I_b – Δi/2

When the steady variation of battery current has been reached, the variation will be between l_o and I₁, as shown in.

Concept:

In a boost converter, the output voltage is given by,

\(\frac{V_0}{V_s}=\frac{1}{1−D}\)

Calculation:

Input dc voltage (V_s) = 360 V

Output voltage (V₀) = 400 V

⇒ \(\frac{400}{360}=\frac{1}{1-D}\)

⇒ 400 - 400 D = 360

⇒ 400 D = 40

D = 0.1

Output power = V_sI_s = 400 × I₀ = 4 kW

⇒ I0 = 10 A

From the power balance equation, input power is equal to output power.

V_sI_s = V_oI_o

or, 360 Is = 4000

Is = 11.1  A

Neglecting ripple in Is

IRMS = \(I_s \times \sqrt{D}=11.1\times \sqrt{0.1}\approx3.5\ A\)

Concept

At the edge of the continuous conduction, the value of inductance becomes minimum and such value of inductance is known as the critical inductance of the converter.

At the boundary of continuous conduction, I_L(min) = 0

\(I_L{(min)}=I_{L(avg)}-{Δ I_L\over 2}\)

\(0=I_{L(avg)}-{Δ I_L\over 2}\)

\(I_{L(avg)}={Δ I_L\over 2}\)                ..........(i)

Calculation of ΔI_L

When the switch is ON:

\(-V_s+V_L=0\)

\(V_s=V_L\)

\(L{di_L\over dt}=V_s\)

\(\Delta I_L={V_sD\over Lf}\)

Putting this value in equation (i):

\(I_{L(avg)}={V_sD\over 2Lf}\)

For the boost converter,  \(I_{L(avg)}={I_{o(avg)}\over 1-D}\)

\({I_{o(avg)}\over 1-D}={V_sD\over 2Lf}\)

\({V_{o(avg)}\over R(1-D)}={V_sD\over 2Lf}\)

For the boost converter,  \(V_{o(avg)}={V_{s}\over 1-D}\)

\({V_{s}\over R(1-D)^2}={V_sD\over 2Lf}\)

The value of the minimum inductance (Lmin) for continuous current for a boost converter is given by:

Lmin = \(\rm\frac{D(1−D)^2 R}{2 f}\)

Concept

Buck converter:

The output voltage of the buck converter is given as follows:

Vo = D Vin

Calculation:

V₀ = DV_s

\(\rm D=\frac{T_0w}{T_0w+T_{df}}\)

\(\rm D =\frac{20}{20+10}=\frac{20}{30}=0.666\)

V₀ ⇒ 0.66 × 60

V₀ ⇒ 40 V

output voltage ⇒ 40 V.A