Power Electronics Test 7

Concept:

In a single-phase full-bridge inverter, the maximum RMS value of output voltage is given by

\({V_{0n}} = \frac{1}{{\sqrt 2 }}\frac{{4{V_{dc}}}}{{n\pi }}\)

Calculation:

Given that, input voltage (V_dc) = 600 V

The fundamental component of the output voltage is,

\({V_{01}} = \frac{1}{{\sqrt 2 }}\frac{{4 \times 600}}{\pi } = 540\;V\)

The 3^rd harmonic component of the output voltage is,

\({V_{01}} = \frac{1}{{\sqrt 2 }}\frac{{4 \times 600}}{{3\pi }} = 180\;V\)

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

Concept:

In a three-phase square wave inverter, the fundamental magnitude (rms) of PWM inverter’s output pole voltage will be less than 0.45 E_dc

Calculation:

Given that the supply voltage (E_dc­) = 600 V

The fundamental pole voltage = 0.45 E_dc = 0.45 × 600 = 270 V

Power drawn, \({P_d} = {V_{01}}{I_{01}}\cos \phi\)

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

\({I_{01}} = \frac{{200}}{{\sqrt 2 }}mA\)

\({P_d} = \frac{{4 \times 220}}{{\pi \times \sqrt 2 }} \times \frac{{200 \times {{10}^{ - 3}}}}{{\sqrt 2 }}\cos 45^\circ = 19.8\;W\)

RMS value of the fundamental output voltage

\({V_{01}} = \frac{{2{V_s}}}{{\sqrt 2 \pi }}\)

Fundamental power in the output \(= \frac{{V_{01}^2}}{R}\)

Given that, \(\frac{{{V_{dc}}}}{2} = 110\)

\(\Rightarrow {V_{dc}} = 220\;V\)

\({V_{01}} = \frac{{2 \times 220}}{{\sqrt 2 \pi }} = 99.03\;V\)

\(P = \frac{{{{\left( {99.03} \right)}^2}}}{{10}} = 980.78\;W\)

Given that, R_L = 20 Ω

V_dc = 600 V

rms load voltage V_L = 500

Average power absorbed by the load \(=\frac{V_{rms}^{2}}{R}\)

\(=\frac{{{\left( 500 \right)}^{2}}}{20}=12.5~kWac\)

V_dcI_s = Average power absorbed by load

⇒ (600) (I_s) = 12.5 × 10³

⇒ I_s = 20.83 A