Power Electronics Test 2

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Power Electronics Test 2

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Weekly Quiz Competition

Question 1
2 / -0.33

A DC-DC converter operates in conduction mode. It has a 36 V input voltage and it feeds a resistive load of 18 Ω. The switching frequency of the converter is 200 Hz. If the switch-on duration is 1.25 ms, which of the following is/are true?

A
For buck converter, the output voltage is 9 V

B
For boost converter, the output voltage is 48 V

C
For buck converter, load power is 4.5 W

D
For boost converter, load power is 128 W

Solution

Input voltage (V_s) = 36 V
Resistive load (R) = 18 Ω
Frequency (f) = 200 Hz
Time period (T) \( = \frac{1}{f} = \frac{1}{{200}} = 5\;ms\)
Duty ratio \( = \frac{{{T_{ON}}}}{T} = \frac{{1.25}}{5} = 0.25\)
For buck converter:
V_o = δV_s = 0.25 × 36 = 9 V
\({I_o} = \frac{9}{{18}} = 0.5A\)
Load power = 9 × 0.5 = 4.5 W
For boost converter:
\({V_o} = \frac{{{V_s}}}{{\left( {1 - \delta } \right)}} = \frac{{36}}{{0.75}} = 48V\)
\({I_o} = \frac{{48}}{{18}} = 2.67A\)
Load Power = 48 × 2.67 = 128 W
Question 2
2 / -0.33

The output voltage of a single-phase full-bridge voltage source inverter is controlled by unipolar PWM with one pulse per half cycle. The input voltage of the inverter is 100 V. For the fundamental RMS component of the output voltage to be 70% of DC voltage, which of the following is/are true?

A
The required pulse width is 51°

B
The required pulse width is 102°

C
Fundamental RMS component of the output voltage is 70 V

D
Fundamental RMS component of the output voltage is 35 V

Solution

\({V_o} = \mathop \sum \limits_{n = 1,3, \ldots .}^\infty \left\{ {\frac{{4{V_s}}}{{n\pi }}\sin \frac{{n\pi }}{2}\sin nd} \right\}\sin n\omega t\)

RMS value of the fundamental component

\({V_{01}} = \left\{ {\frac{{4{V_s}}}{\pi }\sin \frac{\pi }{2}\sin d} \right\}\frac{1}{{\sqrt 2 }}\)

\( = \frac{{2\sqrt 2 {V_s}}}{\pi }\sin d\)

Fundamental rms component of output voltage

= 60% of input DC

= 0.6 × 100 = 60 V

\( \Rightarrow 70 = \frac{{2\sqrt 2 }}{\pi } \times 100 \times \sin d\)

⇒ d = 51.03°

Pulse width = 2d = 102°
Question 3
2 / -0.33

A three-phase full-wave half-controlled converter supplies a highly inductive load with R = 10 Ω the supply is a three-phase star connected with 400 V RMS. The value of firing angle in degrees, to obtain the dc output current as 15 A is _______

A
116-117

B
1164-1171

C
1167-1176

D
1169-1172

Solution

For a three-phase semi converter, the average dc output voltage is
\({V_{dc}} = \frac{{3\sqrt 3 {V_m}}}{{2\pi }}\left( {1 + \cos \alpha } \right)\)
\({I_{dc}} = \frac{{{V_{dc}}}}{R} = 15\)
⇒ V_dc = 15 × 10 = 150 V
\( \Rightarrow 150 = \frac{{3\sqrt 3 \times \frac{{400 \times \sqrt 2 }}{{\sqrt 3 }}}}{{2\pi }}\left( {1 + \cos \alpha } \right)\)
⇒ α = 116.4°
Question 4
2 / -0.33

A single-phase half wave controlled rectifier supplied from 230 V a.c. supply is operating at α = 60°. If the load resistor is 10 Ω, the power factor at the ac source is _______

A
0.6-0.65

B
0.64-0.651

C
0.67-0.656

D
0.69-0.652

Solution

The power factor at the ac source is
\(pf = \sqrt {\frac{1}{{2\pi }}\left( {\pi \alpha + \frac{1}{2}\sin 2\alpha } \right)} \)
\( = \sqrt {\frac{1}{{2\pi }}\left( {\pi - \frac{\pi }{3} + \frac{1}{2}\sin 120^\circ } \right)} \)
= 0.633 lagging
Question 5
2 / -0.33

A step-down chopper has input voltage 200 V and o/p voltage 140 V. If the conduction time of thyristor chopper is 70 μ sec. Now if the frequency operation is increased by 25% keeping conduction time of thyristor constant than find the average value of new output voltage.

A
170-180

B
1704-1801

C
1707-1806

D
1709-1802

Solution

Concept:
For step-down chopper
Average output V₀ = α V_S
Where α = duty cycle
\(\alpha = \frac{{{T_{On}}}}{{{T_{on}} + {T_{off}}}}\)
Time period \(T = {T_{on}} + {T_{off}} = \frac{1}{f}\)
Calculation:
Given
V₀ = 140 V
V_S = 200 V
T_on = 70 μsec
\(\alpha = \frac{{{V_0}}}{{{V_S}}} = \frac{{140}}{{200}} = 0.7\)
\(\frac{{{T_{on}}}}{{{T_{on}} + {T_{off}}}} = 0.7\)
0.3 T_on = 0.7 T_off
⇒ 0.3 × 70 = 0.7 T_off
T_off = 30 μ sec
Time period T = T_on + T_off = 70 + 30
= 100 μ sec
Frequency \(f = \frac{1}{T} = \frac{1}{{100}} \times {10^6}\)
= 10 kHz
Now frequency is increased by 25%
Hence f¹ = 1.25 × 10
= 12.5 kHz
New time period \({T^1} = \frac{1}{{{f^1}}} = \frac{1}{{12.5}} \times {10^3}\)
= 80 μ sec
Conduction Time is constant
T_on = 70 μ sec
New duty cycle
\(\alpha ' = \frac{{{T_{on}}}}{{{T^1}}} = \frac{{70}}{{80}}\)
New Average output voltage
V₀ = α‘V_S
\( = \frac{{70}}{{80}} \times 200\)
= 175 V
Question 6
2 / -0.33

A 3-ϕ 180° mode bridge inverter has star connected load of R = 4 Ω and L = 25 mH. The inverter is feed from 220 V dc and its output frequency is 50 Hz. Find the fundamental component of line voltage.

A
170-173

B
1704-1731

C
1707-1736

D
1709-1732

Solution

Concept:
The Fourier analysis of phase voltage
\({{\rm{V}}_{a0}} = \mathop \sum \limits_{n = 6K \pm 1}^\infty \frac{{2{V_s}}}{{n\pi }}\sin n\omega t\)
Where K = 0, 1, 2…
∴ Fundamental component of phase voltage
\({{\rm{V}}_{a1}} = \frac{{\sqrt 2 {V_s}}}{\pi }\)
Fundamentals component of line voltage
\({{\rm{V}}_{L1}} = \sqrt 3 {V_{a1}}\)
Calculation:
\(\begin{array}{l}{{\rm{V}}_{a1}} = \frac{{\sqrt {2\;} \times 220}}{\pi } = 99.035\;V\\{{\rm{V}}_{L1}} = \sqrt 3 \times 99.035 = 171.534\;V\end{array}\)
Question 7
2 / -0.33

AN SCR requires 50 mA gate current to switch it ON. The driver circuit supply voltage is 10 V. The gate-cathode drop is about 1 V and the voltage drop across diode is also 1 V. The resistive load supplied from a 100 V supply is

A
66 Ω

B
73 Ω

C
80 Ω

D
100 Ω

Solution

Resistance firing circuit:
Diagram
The voltage drop across R is V_gk and
I₁R = V_gk = I_gmaxR
⇒ I_gmax = I₁
I = I₁ + I_gmax = 2 I_gmax
By applying KVL,
V = IR₁ + V_D + V_gk
\( \Rightarrow {R_1} = \frac{{V - {V_D} - {V_{gk}}}}{I}\)
Calculation:
Given that, V = 100, V_D = 1 V, V_gk = 1 V
I = 2 I_gmax = 2 × 50 = 100 mA
Resistive load = (supply voltage – drive transistor drop – gate cathode drop)/100mA
R₁ = (10 – 1 – 1)/100mA
⇒ = 80 Ω
Question 8
2 / -0.33

A class – A chopper circuit is supplied DC source voltage 100 V. The chopper supplies power to a series R-L load with R = 0.5 Ω and L = 1 mH. The chopper switch is ON for 1 ms in an overall period of 3 ms. The minimum value of load current is ______ (in A). Assume continuous current operation of the chopper.

A
33-34

B
334-341

C
337-346

D
339-342

Solution

Concept:
In a class – A chopper,
Average load voltage V_o = D V_dc
Minimum value of load current \(={{I}_{o}}-\frac{\text{ }\!\!\Delta\!\!\text{ }{{I}_{L}}}{2}\)
Maximum value of load current \(={{I}_{o}}+\frac{\text{ }\!\!\Delta\!\!\text{ }{{I}_{L}}}{2}\)
Average load current \({{I}_{o}}=\frac{{{V}_{o}}}{R}\)
\(\text{ }\!\!\Delta\!\!\text{ }{{I}_{L}}=\frac{{{V}_{dc}}}{L}D~\left( 1-D \right)T\)
Calculation:
Given that, Source voltage (V_dc) = 100 V
R = 0.5 Ω
L = 1 mH
T_ON = 1 ms
T = 3 ms
\(D=\frac{{{T}_{On}}}{T}=\frac{1}{3}\)
\({{V}_{o}}=100\times \frac{1}{3}=33.33~V\)
\({{I}_{o}}=\frac{{{V}_{o}}}{R}=\frac{33.33}{0.5}=66.66~A\)
\(\text{ }\!\!\Delta\!\!\text{ }{{I}_{L}}=\frac{100}{1\times {{10}^{-3}}}\times \frac{1}{3}\times \left( 1-\frac{1}{3} \right)\times 3\times {{10}^{-3}}\)
= 66.66 A
Minimum value of load current \(=66.66-\frac{66.66}{2}=33.33~A\)
Question 9
2 / -0.33

A single-phase midpoint converter supplied from 240/120 V, 50 Hz transformer is connected to a load of 15 Ω resistance and infinite inductance. If the secondary winding of the transformer has a line inductance of 20 mH, which of the following is/are true for a firing angle of 60°?

A
Average load voltage is 47.6 V

B
Overlap angle is 15°

C
Average load voltage is 62.5 V

D
Overlap angle is 7.5°

Solution

For mid-point rectifier,
\({V_{dc}} = \frac{{2{V_m}}}{\pi }\cos \alpha - \frac{{\omega L{I_{dc}}}}{\pi }\)
And V_dc = I_dc R
\( \Rightarrow R{I_{dc}} = \frac{{2{V_m}}}{\pi }\cos \alpha - \frac{{\omega L}}{\pi }{I_{dc}}\)
\({I_{dc}}\left[ {R + \frac{{\omega L}}{\pi }} \right] = \frac{{2{V_m}}}{\pi }\cos \alpha \)
\( \Rightarrow {I_{dc}} = \frac{{\frac{{2{V_m}}}{\pi }\cos \alpha }}{{\left[ {R + \frac{{\omega L}}{\pi }} \right]}}\)
\( = \frac{{\frac{{2 \times 120 \times \sqrt 2 }}{\pi }}}{{15 + \frac{{100\pi \times 20}}{{1000\pi }}}}\)
= 3.17 A
V_dc = 15 × 3.17 = 47.6 V
\(\cos \alpha - \cos \left( {\alpha + \mu } \right) = \frac{{\omega L}}{{{V_m}}}{I_{dc}}\)
\( \Rightarrow \cos \left( {\alpha + \mu } \right) = \cos 60^\circ - \frac{{2\pi \times 50 \times 20 \times {{10}^{ - 3}}}}{{169.7}} \times 3.17\)
⇒ α + μ = 67.5°
⇒ μ = 7.5°

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Power Electronics Test 2

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Power Electronics Test 2

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SHARING IS CARING

Question 1

For buck converter, the output voltage is 9 V

For boost converter, the output voltage is 48 V

For buck converter, load power is 4.5 W

For boost converter, load power is 128 W

Solution

Question 2

The required pulse width is 51°

The required pulse width is 102°

Fundamental RMS component of the output voltage is 70 V

Fundamental RMS component of the output voltage is 35 V

Solution

Question 3

116-117

1164-1171

1167-1176

1169-1172

Solution

Question 4

0.6-0.65

0.64-0.651

0.67-0.656

0.69-0.652

Solution

Question 5

170-180

1704-1801

1707-1806

1709-1802

Solution

Question 6

170-173

1704-1731

1707-1736

1709-1732

Solution

Question 7

66 Ω

73 Ω

80 Ω

100 Ω

Solution

Question 8

33-34

334-341

337-346

339-342

Solution

Question 9

Average load voltage is 47.6 V

Overlap angle is 15°

Average load voltage is 62.5 V

Overlap angle is 7.5°

Solution

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