Power, Electronic Systems, Applications and Resources on Electrical and Electronic Project-Thesis: Single-Phase Bridge Inverter

In this circuit four switches are used and the DC supply centre-tap is not required. Switches Q1 and Q2 are switched together while switches Q3 and Q4 are switched together alternately to Q1 and Q2 in a complementary manner. The four feedback diodes D1-D4 conduct currents as indicated in the figure below.

The output load voltage alternates between +Vs when Q1 and Q2 are on and -Vs when Q3 and Q4 are on, irrespective of the direction of current flow. It is assumed that the load current does not become discontinuous at any time. In the following analysis we assume that the load current does not become discontinuous at any time, same as for the half-bridge circuit.

The output RMS voltage

${V_{o(rms)}} = \sqrt {\frac{2}{\pi }\int_0^\pi {V_S^2d\theta } } = {V_S}$

The fundamental RMS output voltage obtained from

${V_{o1(rms)}} = \frac{{4{V_S}}}{{\sqrt 2 \pi }} = 0.90{V_S}$

Fourier series of output voltage

${v_o}(t) = \sum\limits_{n = 1,3,5,....}^\infty {\frac{{4{V_S}}}{{n\pi }}} \sin n\omega t$

$= 0{\rm{ }}for{\rm{ }}n = 2,4,...$

Bridge inverters are preferred over other arrangements in higher power ratings
With the same dc input voltage, output voltage is twice that of the half-bridge inverter

Example-1
A single-phase half-bridge inverter has a dc input voltage of Vs = 440V.
Determine-
(a) The rms output voltage at the fundamental frequency, V1 [ V1,rms=198.07V]
(b) The harmonic factor of the lowest-order harmonic. [HF = 0.333]
Example-2
R=10, L=31.5 mH, C=112 µF, f₀=60 Hz, V_s=220 V, ω= 2πf= 377 rad/s,
X_L=jnωL= j11.87 nΩ, X_c= j/nωc = (-j23.68)/n Ω
$\left| {{Z_n}} \right| = \sqrt {{R^2} + {{\left( {n\omega L - \frac{1}{{n\omega C}}} \right)}^2}} = \sqrt {{{10}^2} + {{\left( {11.87n - \frac{{23.68}}{n}} \right)}^2}}$
${\theta _n} = {\tan ^{ - 1}}\left( {\frac{{11.87n}}{{10}} - \frac{{23.68}}{{10n}}} \right)$
${v_o}(t) = \sum\limits_{n - 1,3,5,...}^\infty {\frac{{4{V_S}}}{{n\pi }}\sin n\omega t}$
$= 0{\rm{ }}for{\rm{ }}n = 2,4,...$
${i_o}(t) = \frac{{{v_o}(t)}}{{\left| {{Z_n}} \right|\angle {\theta _n}}}$
$= \sum\limits_{n = 1,3,5,...}^\infty {\frac{{4{V_S}}}{{n\pi \sqrt {{R^2} + {{\left( {n\omega L - \frac{1}{{n\omega C}}} \right)}^2}} }}\sin (n\omega t - {\theta _n})}$

Power electronic converter-DC-AC CONVERTERS-Single-Phase Bridge Inverter

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2 comments:

VirajJuly 27, 2019 at 6:34 PM
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