Concept:

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

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

Calculation:

Given that, input voltage (V_dc) = 600 V

The fundamental component of the output voltage is,

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

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

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

Concept:

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

$i_0\left(t\right)=\sum _{n=1,3,5}\frac{2V_{dc}}{n^2\pi {\omega }_{0}L}\sin n\left({\omega }_{0}t-\frac{\pi }{2}\right)$

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

$i_{on}=\frac{2V_{dc}}{n^2\pi {\omega }_{0}L}\times \frac{1}{\sqrt{2}}$

Calculation:

The 5^th harmonic component of load current is,

$i_{o5}=\frac{2V_{dc}}{5^2\pi {\omega }_{0}L}\times \frac{1}{\sqrt{2}}$

The 7^th harmonic component of load current is,

$i_{o7}=\frac{2V_{dc}}{7^2\pi {\omega }_{0}L}\times \frac{1}{\sqrt{2}}$

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

⇒ 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, $P_d=V_{01}{I}_{01}\cos \varphi$

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

$I_{01}=\frac{200}{\sqrt{2}}mA$

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

Distortion factor = RMS value of fundamental current (I₁) / RMS value of input current (I)

In a single-phase half-bridge inverter,

The fundamental component of current with purely resistive load is

$i_1\left(t\right)=\frac{2E}{\pi R}\sin \omega t$

The RMS value of the fundamental component of load current is

$I_1=\frac{\sqrt{2}E}{\pi R}$

The RMS value of the current through the load is,

$I=\frac{E}{2R}$

Given that, E = 200 V and R = 10 Ω

$I_1=\frac{\sqrt{2}\times 200}{\pi \times 10}=9A$

$I=\frac{200}{2\times 10}=10A$

Distortion factor $=\frac{I_1}{I}=\frac{9}{10}=0.9$

RMS value of the fundamental output voltage

$V_{01}=\frac{2V_s}{\sqrt{2}\pi }$

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

Given that, $\frac{V_{dc}}{2}=110$

$\Rightarrow V_{dc}=220V$

$V_{01}=\frac{2\times 220}{\sqrt{2}\pi }=99.03V$

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

The output voltage for a signal phase full bridge inverter is

$V_o=\underset{n=1,3,5}{\sum ^{\mathrm{\infty }}}\frac{4V_s}{n\pi }\sin n\omega t$

$V_{01}=\frac{4V_s}{\pi \sqrt{2}}=\frac{4\times 12}{\pi \times \sqrt{2}}=10.8V$

Primary voltage of transformer (E_p) = 10.8 V

Secondary voltage of transformer (E_s) = 230 V

Primary turns (N_p) = 10

$\frac{E_p}{E_s}=\frac{N_p}{N_s}$

$\Rightarrow N_s=N_p\times \frac{E_s}{E_p}$

$=10\times \frac{230}{10.8}=212.96\approx 213$

Given that, R_L = 20 Ω

V_dc = 600 V

rms load voltage V_L = 500

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

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

V_dcI_s = Average power absorbed by load

⇒ (600) (I_s) = 12.5 × 10³

⇒ I_s = 20.83 A

For 180° conduction mode:

• Line voltage remains V_dc for 120°
• Phase voltage remains $\frac{2V_{dc}}{3}$ for each 60°

For 120° conduction mode:

• Phase voltage remains $\frac{V_{dc}}{2}$ for 120°
• Line voltage remains $\frac{V_{dc}}{2}$ for 60° 

Hence, the given waveform represents the load phase voltage when the inverter is operating in 120° conduction mode.

For a star-connected load, the phase voltage is $V_{RN}=\frac{V_{RY}}{\sqrt{3}}$

From the Fourier series expansion for V_RY, the Fourier series for i_R(t) for an R-L load is given by

$i_R\left(t\right)=\frac{2\sqrt{3}E}{\sqrt{3}\pi }\left[\frac{\sin \left(\omega t-{\theta }_1\right)}{\sqrt{R^2+{\left(\omega L\right)}^2}}-\frac{\sin \left(\omega t-{\theta }_5\right)}{5\sqrt{R^2+{\left(5\omega L\right)}^2}}-\frac{\sin \left(\omega t-{\theta }_7\right)}{7\sqrt{R^2+{\left(7\omega L\right)}^2}}+\frac{\sin \left(\omega t-{\theta }_{11}\right)}{11\sqrt{R^2+{\left(11\omega L\right)}^2}}-\dots \right]$

$\left|Z_{Ln}\right|=\sqrt{R^2+{\left(n\omega L\right)}^2}$

$\mathrm{\angle }{Z}_{Ln}={\tan }^{-1}\left(\frac{n\omega L}{R}\right)$

ω = 2πf = 314 and ωL = 314 × 20 × 10^-3 = 6.28 Ω

$Z_{Ln}=\sqrt{8^2+{\left(6.28n\right)}^2}$ and $\theta ={\tan }^{-1}\left(\frac{6.28n}{8}\right)$

$i_R\left(t\right)=12.52\sin \left(\omega t-{38}^{\circ }\right)-0.785\sin \left(5\omega t-{75.7}^{\circ }\right)-0.41\sin \left(7\omega t-{79.7}^{\circ }\right)+0.17\sin \left(11\omega t-{83.4}^{\circ }\right)+\dots$

RMS line voltage, $V_{RY\left(RMS\right)}=200\sqrt{\frac{2}{3}}=163.3V$

RMS line current, $I_{line\left(RMS\right)}=\sqrt{{\left(\frac{12.52}{\sqrt{2}}\right)}^2+{\left(\frac{0.785}{\sqrt{2}}\right)}^2+{\left(\frac{0.41}{\sqrt{2}}\right)}^2+{\left(\frac{0.17}{\sqrt{2}}\right)}^2}=8.88A$

Load power factor $\cos \varphi =\frac{3I_L^{2}R}{\sqrt{3}{V}_{RY\left(RMS\right)}{I}_L}$

$=\frac{3\times {\left(8.88\right)}^2\times 8}{\sqrt{3}\times 163.3\times 8.88}=0.75$