PWM half-bridge inverter switching instants

With a frequency ratio of 1, there is one carrier triangle in each half cycle of the modulating wave, so each half cycle has a single pulse (single-pulse modulation). Reading used here: in each half cycle the triangular carrier falls from its peak $A_c$ to 0 at the middle (90°, 270°) and rises back to $A_c$, and it is compared with the magnitude of the square reference of amplitude $A_r = 0.5A_c$ ($M = A_r/A_c = 0.5$). A switch is ON while the reference exceeds the carrier.

i) Waveforms

 Ac      \        /\        /
          \      /  \      /
 0.5Ac  ---\----/----\----/---  ref |vr|
            \  /      \  /
 0           \/        \/
        0    90  180  270  360  deg
 vo
 +Vs/2      ____
    0   ___|    |____      ___
 -Vs/2               |____|
          45   135  225  315

• Positive half (reference positive): S1 is ON while the carrier is below the reference, giving $v_o = +V_s/2$.
• Negative half: S2 is ON for the same interval, giving $v_o = -V_s/2$.
• With a resistive load, when both switches are OFF the current is zero and $v_o = 0$.

ii) Switching instants

The carrier on $0$–$180°$ is $v_c = A_c\left(1 - \dfrac{\theta}{90°}\right)$ for $0 \le \theta \le 90°$, and rises symmetrically after $90°$. Setting $v_c = 0.5A_c$:

$$\theta_1 = 0.5 \times 90° = 45°, \qquad \theta_2 = 180° - 45° = 135°$$

Pulse width $2d = M \times 180° = 0.5 \times 180° = 90°$.

For example, at 120 Hz ($T = 8.333$ ms) these are: S1 ON at 1.042 ms, OFF at 3.125 ms; S2 ON at 5.208 ms, OFF at 7.292 ms.

iii) RMS output voltage

$$\begin{aligned} V_{o,rms} &= \sqrt{\frac{1}{\pi}\int_{45°}^{135°}\left(\frac{V_s}{2}\right)^2 d\theta} = \frac{V_s}{2}\sqrt{\frac{2d}{\pi}} \\ &= \frac{V_s}{2}\sqrt{\frac{\pi/2}{\pi}} = \frac{V_s}{2}\times 0.7071 = 0.3536\,V_s \end{aligned}$$

Answer: switching instants S1: 45° to 135°, S2: 225° to 315°; $V_{o,rms} = 0.3536\,V_s$.