Half-Controlled Bridge Converter: Input Current Ratio Analysis

The problem asks for the ratio of the RMS value of the fundamental component of the input current to the RMS value of the total input current for a single-phase half-controlled bridge converter feeding an inductive load with ripple-free current. The triggering angle is given as $\alpha = 60^\circ$.

Input Current Waveform and RMS Value

For a single-phase half-controlled bridge converter supplying a highly inductive load (resulting in ripple-free load current $I_L$), the input AC source current ($i_s(t)$) waveform can be described over a cycle ($2\pi$) as follows:

• $i_s(t) = I_L$ for $\alpha \le \omega t \le \pi$
• $i_s(t) = -I_L$ for $\pi+\alpha \le \omega t \le 2\pi$
• $i_s(t) = 0$ elsewhere

The RMS value of the total input current ($I_{s,rms}$) is calculated over the full cycle:

$ I_{s,rms}^2 = \frac{1}{2\pi} \left[ \int_{\alpha}^{\pi} I_L^2 d(\omega t) + \int_{\pi+\alpha}^{2\pi} (-I_L)^2 d(\omega t) \right] $

$ I_{s,rms}^2 = \frac{I_L^2}{2\pi} \left[ (\pi - \alpha) + (2\pi - (\pi+\alpha)) \right] = \frac{I_L^2}{2\pi} [(\pi - \alpha) + (\pi - \alpha)] $

$ I_{s,rms}^2 = \frac{I_L^2}{2\pi} [2(\pi - \alpha)] = \frac{I_L^2 (\pi - \alpha)}{\pi} $

$ I_{s,rms} = I_L \sqrt{\frac{\pi - \alpha}{\pi}} $

Substituting $\alpha = 60^\circ = \frac{\pi}{3}$:

$ I_{s,rms} = I_L \sqrt{\frac{\pi - \frac{\pi}{3}}{\pi}} = I_L \sqrt{\frac{\frac{2\pi}{3}}{\pi}} = I_L \sqrt{\frac{2}{3}} $

RMS Value of the Fundamental Component ($I_{s1,rms}$)

The RMS value of the fundamental component ($I_{s1,rms}$) is derived from the Fourier series coefficients ($a_1, b_1$) of the input current waveform $i_s(t)$. The formula is $I_{s1,rms} = \frac{\sqrt{a_1^2 + b_1^2}}{\sqrt{2}}$.

The coefficients are:

$ a_1 = -\frac{2 I_L}{\pi} \sin(\alpha) $

$ b_1 = \frac{2 I_L}{\pi} (1 + \cos(\alpha)) $

Thus, the RMS value of the fundamental component is:

$ I_{s1,rms} = \frac{2 I_L}{\pi} \sqrt{1 + \cos(\alpha)} $

Substituting $\alpha = 60^\circ = \frac{\pi}{3}$:

$ I_{s1,rms} = \frac{2 I_L}{\pi} \sqrt{1 + \cos\left(\frac{\pi}{3}\right)} = \frac{2 I_L}{\pi} \sqrt{1 + 0.5} = \frac{2 I_L}{\pi} \sqrt{1.5} $

Ratio Calculation

The required ratio is $\frac{I_{s1,rms}}{I_{s,rms}}$:

$ \text{Ratio} = \frac{\frac{2 I_L}{\pi} \sqrt{1.5}}{I_L \sqrt{\frac{2}{3}}} = \frac{2 \sqrt{1.5}}{\pi \sqrt{\frac{2}{3}}} $

$ \text{Ratio} = \frac{2 \sqrt{\frac{3}{2}}}{\pi \sqrt{\frac{2}{3}}} = \frac{2 \times \frac{\sqrt{3}}{\sqrt{2}}}{\pi \times \frac{\sqrt{2}}{\sqrt{3}}} = \frac{2\sqrt{3}}{\sqrt{2}} \times \frac{\sqrt{3}}{\pi\sqrt{2}} = \frac{2 \times 3}{\pi \times 2} = \frac{3}{\pi} $

Calculating the numerical value:

$ \text{Ratio} \approx \frac{3}{3.14159} \approx 0.9549 $

Rounded to 3 decimal places, the ratio is 0.955.