Question

A single-phase full-bridge inverter fed by a 325 V DC produces a symmetric quasi-square waveform across 'ab' as shown. To achieve a modulation index of 0.8, the angle \( \theta \) expressed in degrees should be \(\underline{\hspace{2cm}}\) . 
(Round off to 2 decimal places.)
(Modulation index is defined as the ratio of the peak of the fundamental component of \( v_{ab} \) to the applied DC value.)

Hint:

For a single-phase full-bridge inverter, use the modulation index formula to relate the DC voltage and the fundamental component to calculate the angle \( \theta \).

Answer 1

Correct Answer: 50

For a full-bridge inverter with a quasi-square waveform, the modulation index \( m \) is defined as:
\[ m = \frac{V_{m}}{V_{DC}}, \] where \( V_m \) is the peak of the fundamental component of the waveform and \( V_{DC} \) is the applied DC voltage.
In this case, we are given: \[ m = 0.8 \text{and} V_{DC} = 325 \, \text{V}. \] The relationship for the modulation index in terms of the angle \( \theta \) is: \[ V_m = \frac{V_{DC}}{2} \left( \cos(\theta) + 1 \right), \] so substituting the given values: \[ 0.8 = \frac{325}{2} \left( \cos(\theta) + 1 \right). \] Solving for \( \cos(\theta) \): \[ 0.8 = 162.5 \left( \cos(\theta) + 1 \right), \] \[ \frac{0.8}{162.5} = \cos(\theta) + 1, \] \[ \cos(\theta) = \frac{0.8}{162.5} - 1, \] \[ \cos(\theta) = -0.9951. \] Now, solve for \( \theta \): \[ \theta = \cos^{-1}(-0.9951) \approx 170.50^\circ. \] Thus, the value of \( \theta \) is approximately \( 170.50^\circ \).