The output voltage $V_{01}$ and $V_{02}$ of two $1 - Φ$ full bridge inverter are added using output transformers, as shown in figure. In order to eliminate the fifth harmonic from the output voltage, the phase angle (in degree) between $V_{01}$ and $V_{02}$ is_______.

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Solution

For $1 - ϕ$ full bridge inverter,
$V_{o n} = \sum n = 1, 3, 5 \frac{4 V_{s}}{n π} sin ω t$
$V_{01} = \sum n = 1, 3, 5 \frac{4 V_{s}}{n π} sin ω t$
$V_{02} = \sum n = 1, 3, 5 \frac{4 V_{s}}{n π} sin (ω t - ϕ)$
Where $ϕ$ is the [phase displacement of $2^{n d}$ inverter w.r.t $1^{s t}$

$V_{01} = \frac{4 V_{s}}{π} [sin ω t + \frac{1}{3} sin 3 ω t + \frac{1}{5} sin 5 ω t + . . .]$

$V_{02} = \frac{4 V_{s}}{π} [sin (ω t - ϕ) + \frac{1}{3} sin 3 (ω t - ϕ) + \frac{1}{5} sin 5 (ω t - ϕ) + . . .]$
$ϕ = \frac{π}{5}$
$5^{t h}$ harmonic $V_{05} = \frac{4 V_{0}}{5 π} sin (5 ω t - \frac{5 π}{5}) = - \frac{4 V_{s}}{5 π} sin 5 ω t$
$5^{t h}$ harmonic in resultant gets canceled when
$ϕ = \frac{π}{5} = 36^{\circ}$

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