# 5.3: RMS Current Through Each Diode in Single-Phase Rectifier ## **Given:** - Input current waveform: **Triangular** per half-cycle - **Peak current** (\(I_{\text{peak}}\)): \(10\,A\) - **Base (width)** of triangle: \(60^\circ\) per half-cycle ## **Step 1: Convert Degrees to Radians** \[ 60^\circ = \frac{60}{180}\pi = \frac{\pi}{3} \text{ radians} \] ## **Step 2: Calculate RMS Value of Triangular Pulse** For a triangular waveform of height \(I_{\text{peak}}\) and base \(T\): \[ I_{\text{rms,triangle}} = I_{\text{peak}} \cdot \sqrt{\frac{1}{3}} \] (but this is for a triangle that occupies the **entire period**; our triangle only exists for \(60^\circ\) out of \(180^\circ\) in the half-cycle.) ### **Duty Cycle:** \[ \text{Duty cycle} = \frac{\text{Base width}}{\text{Half cycle}} = \frac{60^\circ}{180^\circ} = \frac{1}{3} \] ### **General RMS Expression:** If a current waveform is zero outside an interval of width \(\alpha\) (radians) in a half-cycle (length \(\pi\)), and is triangular with peak \(I_{\text{peak}}\): \[ I_{\text{rms,diode}} = \sqrt{\frac{1}{\pi} \int_^{\alpha} \left[I_{\text{peak}}\left(1 - \frac{2\theta}{\alpha}\right)\right]^2 d\theta} \] ## **Step 3: Compute the RMS Value** Let \(I(\theta) = I_{\text{peak}}\left(1 - \frac{2\theta}{\alpha}\right)\) for \( \leq \theta \leq \alpha\), zero otherwise. \[ I^2(\theta) = I_{\text{peak}}^2 \left(1 - \frac{2\theta}{\alpha}\right)^2 \] Integrate over \(\) to \(\alpha\): \[ I_{\text{rms}}^2 = \frac{1}{\pi} \int_^{\alpha} I_{\text{peak}}^2 \left(1 - \frac{2\theta}{\alpha}\right)^2 d\theta \] \[ = \frac{I_{\text{peak}}^2}{\pi} \int_^{\alpha} \left(1 - \frac{4\theta}{\alpha} + \frac{4\theta^2}{\alpha^2}\right) d\theta \] Integrate term-by-term: - \(\int_^{\alpha} 1\,d\theta = \alpha\) - \(\int_^{\alpha} \theta\,d\theta = \frac{1}{2} \alpha^2\) - \(\int_^{\alpha} \theta^2 d\theta = \frac{1}{3}\alpha^3\) Plug in: \[ = \frac{I_{\text{peak}}^2}{\pi} \left[ \alpha - \frac{4}{\alpha} \cdot \frac{1}{2} \alpha^2 + \frac{4}{\alpha^2} \cdot \frac{1}{3} \alpha^3 \right] \] \[ = \frac{I_{\text{peak}}^2}{\pi} \left[ \alpha - 2\alpha + \frac{4}{3}\alpha \right] = \frac{I_{\text{peak}}^2}{\pi} \left[ - \alpha + \frac{4}{3}\alpha \right] = \frac{I_{\text{peak}}^2}{\pi} \left( \frac{1}{3}\alpha \right) \] So, \[ I_{\text{rms,diode}} = I_{\text{peak}} \sqrt{\frac{\alpha}{3\pi}} \] ## **Step 4: Substitute Values** - \(I_{\text{peak}} = 10\,A\) - \(\alpha = 60^\circ = \frac{\pi}{3}\) radians \[ I_{\text{rms,diode}} = 10 \sqrt{\frac{\frac{\pi}{3}}{3\pi}} = 10 \sqrt{\frac{1}{9}} = 10 \cdot \frac{1}{3} = 3.33\,A \] --- ## **Final Answer (5.3)** **The RMS current through each diode is:** \[ \boxed{3.33\,A} \] --- # 5.4: Ripple Component in DC-Side Current (Through Capacitor) When a filter capacitor is used, the **ripple component** of the load current is the **AC component** of the input current (i.e., input current minus its average/DC value). ## **Step 1: Find Average (DC) Value of the Pulse** For a triangular pulse of height \(I_{\text{peak}}\) and base \(\alpha\) in a period \(T\): \[ I_{\text{avg}} = \frac{\text{Area of triangle in period}}{\text{period}} \] Area of triangle: \[ A = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot \alpha \cdot I_{\text{peak}} \] So, for a **full cycle** (\(2\pi\)): \[ I_{\text{avg}} = \frac{\frac{1}{2} \cdot \alpha \cdot I_{\text{peak}}}{2\pi} \] - \(\alpha = \frac{\pi}{3}\) - \(I_{\text{peak}} = 10\,A\) \[ I_{\text{avg}} = \frac{\frac{1}{2} \cdot \frac{\pi}{3} \cdot 10}{2\pi} = \frac{5\pi/3}{2\pi} = \frac{5}{6} \approx .833\,A \] --- ## **Step 2: Find RMS Value of Input Current Over Full Cycle** Earlier, we saw: \[ I_{\text{rms,total}} = I_{\text{peak}} \sqrt{\frac{\alpha}{3\pi}} = 10 \sqrt{\frac{\pi/3}{3\pi}} = 10 \cdot \frac{1}{3} = 3.33\,A \] --- ## **Step 3: Ripple RMS Value (Excluding DC Component)** The RMS value of the **ripple** (AC) component: \[ I_{\text{ripple,rms}} = \sqrt{I_{\text{rms,total}}^2 - I_{\text{avg}}^2} \] Plug in values: - \(I_{\text{rms,total}} = 3.33\,A\) - \(I_{\text{avg}} = .833\,A\) \[ I_{\text{ripple,rms}} = \sqrt{(3.33)^2 - (.833)^2} = \sqrt{11.09 - .694} = \sqrt{10.40} \approx 3.22\,A \] --- ## **Final Answer (5.4)** **The RMS value of the ripple component (through the capacitor) is:** \[ \boxed{3.22\,A} \]

# 5.3: RMS Current Through Each Diode in Single-Phase Rectifier ## **Given:** - **Input current waveform**: Triangular per half-cycle - **Peak current** (\(I_{\text{peak}}\)): \(10\,A\) - **Base (width)** of triangle: \(60^\circ\) ## **Step 1: Convert Degrees to Radians** \[ 60^\circ = \frac{60}{180}\pi = \frac{\pi}{3} \text{ radians} \] ## **Step 2: Calculate RMS Value of Triangular Pulse** For a triangular waveform of height \(I_{\text{peak}}\) and base \(T\): \[ I_{\text{rms,triangle}} = I_{\text{peak}} \cdot \sqrt{\frac{1}{3}} \] However, since our triangle only exists for \(60^\circ\) out of \(180^\circ\), calculate using duty cycle. ### **Duty Cycle:** \[ \text{Duty cycle} = \frac{60^\circ}{180^\circ} = \frac{1}{3} \] ### **General RMS Expression:** Using the triangular waveform formula: \[ I_{\text{rms,diode}} = \sqrt{\frac{1}{\pi} \int_0^{\alpha} \left[I_{\text{peak}}\left(1 - \frac{2\theta}{\alpha}\right)\right]^2 d\theta} \] ## **Step 3: Compute the RMS Value** Substituting \(I(\theta)\): \[ I^2(\theta) = I_{\text{peak}}^2 \left(1 - \frac{2\theta}{\alpha}\right)^2 \] Integrate: \[ I_{\text{rms}}^2 = \frac{I_{\text{peak}}^2}{\pi} \int_0^{\alpha} \left(1 - \frac{4\theta}{\alpha} + \frac{4\theta^2}{\alpha^2}\right) d\theta \] Calculating each term: - \(\int_0^{\alpha} 1\,d\theta = \alpha\) - \(\int_0^{\alpha} \theta\,d\theta = \frac{1}{2} \alpha^2\) - \(\int_0^{\alpha} \theta^2 d\theta = \frac{1}{3}\alpha^3\) Combine: \[ = \frac{I_{\text{peak}}^2}{\pi} \left[\alpha - 2\alpha + \frac{4}{3}\alpha\right] = \frac{I_{\text{peak}}^2}{\pi} \left[-\alpha + \frac{4}{3}\alpha\right] = \frac{I_{\text{peak}}^2}{\pi} \left(\frac{1}{3}\alpha\right) \] Final RMS for diode: \[ I_{\text{rms,diode}} = I_{\text{peak}} \sqrt{\frac{\alpha}{3\pi}} \] ## **Step 4: Substitute Values** - \(I_{\text{peak}} = 10\,A\) - \(\alpha = \frac{\pi}{3}\) Calculation: \[ I_{\text{rms,diode}} = 10 \sqrt{\frac{\frac{\pi}{3}}{3\pi}} = 10 \sqrt{\frac{1}{9}} = 10 \cdot \frac{1}{3} = 3.33\,A \] --- ## **Final Answer (5.3)** **The RMS current through each diode is:** \[ \boxed{3.33\,A} \] --- # 5.4: Ripple Component in DC-Side Current (Through Capacitor) ## **Step 1: Find Average (DC) Value of the Pulse** For a triangular pulse: \[ I_{\text{avg}} = \frac{\text{Area of triangle in period}}{\text{period}} = \frac{\frac{1}{2} \cdot \alpha \cdot I_{\text{peak}}}{2\pi} \] Substituting: - Area of triangle: \[ A = \frac{1}{2} \cdot \frac{\pi}{3} \cdot 10 = \frac{5\pi}{3} \] - Thus, \[ I_{\text{avg}} = \frac{\frac{5\pi/3}{2\pi}} = \frac{5}{6} \approx 0.833\,A \] --- ## **Step 2: Find RMS Value of Input Current Over Full Cycle** Reusing earlier derived values: \[ I_{\text{rms,total}} = 3.33\,A \] --- ## **Step 3: Ripple RMS Value (Excluding DC Component)** Calculate ripple current: \[ I_{\text{ripple,rms}} = \sqrt{I_{\text{rms,total}}^2 - I_{\text{avg}}^2} \] Substituting known values: \[ I_{\text{ripple,rms}} = \sqrt{(3.33)^2 - (0.833)^2} = \sqrt{11.09 - 0.694} = \sqrt{10.40} \approx 3.22\,A \] --- ## **Final Answer (5.4)** **The RMS value of the ripple component (through the capacitor) is:** \[ \boxed{3.22\,A} \]