Question

The 3-phase modulating waveforms \( v_a(t), v_b(t), v_c(t) \), used in sinusoidal PWM in a Voltage Source Inverter (VSI) are given as: \[ v_a(t) = 0.8 \sin(\omega t) \quad v_b(t) = 0.8 \sin\left(\omega t - \frac{2\pi}{3}\right) \quad v_c(t) = 0.8 \sin\left(\omega t + \frac{2\pi}{3}\right) \] where \( \omega = 2\pi \times 40 \, {rad/s} \) is the fundamental frequency. The modulating waveforms are compared with a 10 kHz triangular carrier whose magnitude varies between +1 and -1. The VSI has a DC link voltage of 600 V and feeds a star connected motor. The per phase fundamental RMS motor voltage, in volts, is closest to:

• A. 169.71
• B. 300.00
• C. 424.26
• D. 212.13

Hint:

In sinusoidal PWM, the fundamental RMS output voltage per phase is given by: \[ V_{ph,RMS} = \frac{m_a \cdot V_{dc}}{2\sqrt{2}} \] where \( m_a \) is the modulation index and \( V_{dc} \) is the DC link voltage.

Answer 1

The Correct Option is A

Given: 
- Modulation index \( m_a = 0.8 \) 
- DC Link Voltage \( V_{dc} = 600 \, {V} \) 
- Fundamental phase voltage (peak) in sinusoidal PWM is: \[ V_{ph,peak} = \frac{m_a \cdot V_{dc}}{2} = \frac{0.8 \cdot 600}{2} = 240 \, {V} \] - Convert to RMS: \[ V_{ph,RMS} = \frac{V_{ph,peak}}{\sqrt{2}} = \frac{240}{\sqrt{2}} \approx 169.71 \, {V} \] Therefore, the per phase RMS motor voltage is: \[ \boxed{169.71 \, {V}} \]