Single Phase Voltage Source Inverters MCQ Quiz in বাংলা - Objective Question with Answer for Single Phase Voltage Source Inverters - বিনামূল্যে ডাউনলোড করুন [PDF]

The correct answer is option 2): 207 V

Concept:

A single-phase full bridge inverter

RMS value of the fundamental component of voltage is given by: (V_o)RMS = \({4V_s\over \sqrt{2}\pi}\)

Calculation:

V_s = 230 V

V₀ = 230 × 4 × \({1\over \sqrt{2}\pi}\)

= 207 V

1ϕ full bridge voltage source inverter:

Working of 1ϕ VSI:

Case 1: When  T1T₂ is ON

Vo > 0 and Io > 0

Case 2: When  T₃T₄ is ON

Vo < 0 and Io < 0

Case 3: When  D1D₂ is ON

Vo > 0 and Io < 0

Case 4: When  D₃D₄ is ON

Vo < 0 and Io > 0

The waveform of output voltage and current for a 1ϕ full-wave inverter is given as:

If the load is purely inductive in nature then:

\(I_o = {1\over L}\int V_odt\)

If V_o is a square waveform, then the integration of the square wave will be triangular.

So, the waveform of the current will be triangular in shape.

For the given inductive load, the voltage and current waveforms are shown below.

\(L\frac{{dio}}{{dt}} = {V_s}\)

\(0 \le t \le \frac{\beta }{\omega },{i_o}\left( t \right) = - {I_p} + \frac{{Vs}}{L}t\)

\(t = \frac{\beta }{\omega },{i_o} = {I_p} \Rightarrow {I_p} = - {I_p} + \frac{{{V_s}}}{L}\frac{\beta }{\omega }\)

\({I_{pp}} = 2{I_p} = \frac{{\beta \;{V_s}}}{{\omega L}}\)

\( = \frac{{2A}}{3} \times \frac{{200}}{{100\pi \times 0.05}} = 26.66\;A\)

Concept:

In a half-bridge inverter connected to pure inductive load, the Fourier representation of load current is given by

\({i_0}\left( t \right) = \mathop \sum \limits_{n = 1,3,5} \frac{{2{V_{dc}}}}{{{n^2}\pi {\omega _0}L}}\sin n\left( {{\omega _0}t - \frac{\pi }{2}} \right)\)

The RMS value of the nth harmonic component is given by,

\({i_{on}} = \frac{{2{V_{dc}}}}{{{n^2}\pi {\omega _0}L}} \times \frac{1}{{\sqrt 2 }}\)

Calculation:

The 5^th harmonic component of load current is,

\({i_{o5}} = \frac{{2{V_{dc}}}}{{{5^2}\pi {\omega _0}L}} \times \frac{1}{{\sqrt 2 }}\)

The 7^th harmonic component of load current is,

\({i_{o7}} = \frac{{2{V_{dc}}}}{{{7^2}\pi {\omega _0}L}} \times \frac{1}{{\sqrt 2 }}\)

\( \Rightarrow \frac{{{i_{o5}}}}{{{i_{o7}}}} = \frac{{{7^2}}}{{{5^2}}} = 1.96\)

⇒ i_o5 ≈ 2 i_o7

The output voltage for a signal phase full bridge inverter is

\({{V}_{o}}=\underset{n=1,3,5}{\overset{\infty }{\mathop \sum }}\,\frac{4{{V}_{s}}}{n\pi }\sin n\omega t\)

\({{V}_{01}}=\frac{4{{V}_{s}}}{\pi \sqrt{2}}=\frac{4\times 12}{\pi \times \sqrt{2}}=10.8~V\)

Primary voltage of transformer (E_p) = 10.8 V

Secondary voltage of transformer (E_s) = 230 V

Secondary current \(\left( {{I}_{s}} \right)=\frac{230}{100}=2.3~A\)

\(\frac{{{E}_{P}}}{{{E}_{s}}}=\frac{{{I}_{s}}}{{{I}_{P}}}\)

⇒ Primary current \(\left( {{I}_{p}} \right)={{I}_{s}}\times \frac{{{E}_{s}}}{{{E}_{p}}}\)

\(=2.3\times \frac{230}{10.8}=48.98~A\)

Each thyristor conducts for half cycle. RMS value of thyristor current.

\({{I}_{rms}}=\frac{I}{\sqrt{2}}=\frac{48.98}{\sqrt{2}}=34.63~A\)

The Correct answer is option (1)

Concept:

Single Phase half-bridge inverter:

Fig: Single phase half bridge inverter

Case 1: During \(0

Q₁ ON, and Q₂ OFF

\(V_∘ = \frac {V_s}{2}\)

Case 2: During \(\frac{T_∘ }{2}

Q₁ OFF, and Q₂ ON

\(V_∘ =- \frac{V_s}{2}\ \)

Single phase full bridge inverter:

Case 1: During \(0

T₁ and T₂ are ON

V_∘ = V_s

Case 2: During \(\frac{T}{2} 

T₃  and T₄ are ON

V_∘ = - V_s

Conclusion:

From the above waveform and expression, we can conclude that the output voltage of the single phase half bridge inverter is half of the single phase full bridge inverter.

Concept:

The output voltage is controlled by varying the pulse-width 2d. This shape of the output voltage is called the Quasi-square wave.

Fourier series of this waveform is

\({V_o} = \mathop \sum \limits_{n = 1,3,5}^\infty \frac{{4{V_s}}}{{nπ }}\sin \frac{{nπ }}{2}\sin nd\sin n\omega t\)

The output of a 1-phase inverter bridge for 120 conduction mode is given. Therefore 3^rd and triple harmonics in the output voltage are absent.

For 120° conduction mode 2d = 120° 

d = 60° = π/3

In the above voltage expression, there is a sine expression

 \(sin\;nd =sin\; n{\pi \over 3}\)

for n = 3, 9, 15... (triplen harmonics)

\(sin\; n{\pi \over 3} = 0\)

So V_o = 0 for triplen harmonics.

Power delivered to the load. P₀ = V₀₁ I_or cos ϕ

V₀₁ is rms value of fundamental output voltage

I_0r is rms value of load current

\({V_{01}} = \frac{{4\;{V_{dc}}}}{{\sqrt 2 \;\pi }}\)

Given that, V_dc = 400 V

\({V_{01}} = \frac{{4 \times 400}}{{\sqrt 2 \times \pi }} = 360.12\;V\)

Power delivered, P₀ = V₀₁ I_or cosϕ

\(= 360.12 \times \frac{{200}}{{\sqrt 2 }} \times \cos 30^\circ \)

= 44.1 kW

Concept:

For single-phase full bridge inverter

\(\begin{array}{l}{V_0} = \mathop \sum \limits_{n = 1,3,5}^ \propto \;\frac{{4{V_s}}}{{n\pi }}\sin n\omega t\;volts\\ {i_0} = \mathop \sum \limits_{n = 1,3,5}^ \propto \;\;\frac{{4{V_s}}}{{n\pi .{z_n}}}\sin \left( {n\omega t - \theta n} \right)\;Amps\\ where\;Zn = {\left[ {{R^2} + {{\left( {n\omega L - \frac{1}{{n\omega c}}} \right)}^2}} \right]^{\frac{1}{2}}}\\ {\theta _n} = Ta{n^{ - 1}}\left[ {\frac{{n\omega L - \frac{1}{{n\omega c}}}}{R}} \right]rad \end{array}\)

Calculation:

The RMS value of the fundamental voltage 

\(\begin{array}{l} {V_{o1}} = \frac{{2\sqrt 2 {V_{dc}}}}{\pi }\\ = \frac{{2\sqrt 2 \times 48}}{\pi } = 43.22\ V \end{array}\)

Concept:

In a half-bridge inverter with R-L load, the RMS value of steady state current is given by

\({I_0} = \frac{{{V_s}}}{{2Z}} = \frac{{{V_s}}}{{2\sqrt {{R^2}\; +\; {{\left( {\omega L} \right)}^2}} }}\)

Calculation:

Given that, source voltage (V_s) = 100 V

The load resistance (R) = 20 Ω

Inductance = 100 mH

Frequency (f) = 50 Hz

\({I_0} = \frac{{100}}{{2\sqrt {{{\left( {20} \right)}^2}\; + \;{{\left( {2\pi\; \times \;50\; \times \;100\; \times \;{{10}^{ - 3}}} \right)}^2}} }} = 1.34\)

The peak value of load current = 1.34 × √2 = 1.90