Question

A 1-phase Half controlled rectifier supplied with voltage v=300sin 314t, and load resistance is 100Ω. Find Average output voltage Load current (For α=60° and α=100°).

Answer 1

To find the average output voltage and load current of a 1-phase half-controlled rectifier, we can use the following formulas for a given firing angle \(\alpha\).

### Given:
- Input voltage \( v(t) = 300 \sin(314 t) \)
- Load resistance \( R = 100 \, \Omega \)
- Firing angles \(\alpha = 60^\circ\) and \(\alpha = 100^\circ\)

### 1. **Average Output Voltage \(V_{dc}\)**

The average output voltage for a 1-phase half-controlled rectifier is given by:

\[ V_{dc} = \frac{V_m}{\pi} \left( \cos \alpha + \frac{\pi - \alpha}{2} \sin \alpha \right) \]

where \( V_m \) is the maximum value of the input voltage.

For the given input voltage \( v(t) = 300 \sin(314 t) \), the peak voltage \( V_m \) is 300 V.

#### For \(\alpha = 60^\circ\):

1. Convert \(\alpha\) to radians: \(\alpha = 60^\circ = \frac{\pi}{3} \text{ radians}\).
2. Substitute \( \alpha = \frac{\pi}{3} \) into the formula:

\[ V_{dc} = \frac{300}{\pi} \left( \cos \frac{\pi}{3} + \frac{\pi - \frac{\pi}{3}}{2} \sin \frac{\pi}{3} \right) \]

\[ V_{dc} = \frac{300}{\pi} \left( \frac{1}{2} + \frac{2\pi/3}{2} \cdot \frac{\sqrt{3}}{2} \right) \]

\[ V_{dc} = \frac{300}{\pi} \left( \frac{1}{2} + \frac{\pi \sqrt{3}}{6} \right) \]

\[ V_{dc} = \frac{300}{\pi} \left( \frac{1}{2} + \frac{0.912}{6} \right) \]

\[ V_{dc} \approx \frac{300}{\pi} \times 0.695 \]

\[ V_{dc} \approx 66.4 \text{ V} \]

#### For \(\alpha = 100^\circ\):

1. Convert \(\alpha\) to radians: \(\alpha = 100^\circ = \frac{100\pi}{180} = \frac{5\pi}{9} \text{ radians}\).
2. Substitute \( \alpha = \frac{5\pi}{9} \) into the formula:

\[ V_{dc} = \frac{300}{\pi} \left( \cos \frac{5\pi}{9} + \frac{\pi - \frac{5\pi}{9}}{2} \sin \frac{5\pi}{9} \right) \]

\[ V_{dc} = \frac{300}{\pi} \left( \cos \frac{5\pi}{9} + \frac{4\pi/9}{2} \sin \frac{5\pi}{9} \right) \]

\[ V_{dc} = \frac{300}{\pi} \left( -0.766 + \frac{2\pi/9 \cdot 0.766}{2} \right) \]

\[ V_{dc} \approx \frac{300}{\pi} \left( -0.766 + 0.135 \right) \]

\[ V_{dc} \approx \frac{300}{\pi} \times -0.631 \]

\[ V_{dc} \approx -60.2 \text{ V} \]

### 2. **Load Current \(I_{dc}\)**

The load current is given by:

\[ I_{dc} = \frac{V_{dc}}{R} \]

#### For \(\alpha = 60^\circ\):

\[ I_{dc} = \frac{66.4}{100} \]

\[ I_{dc} \approx 0.664 \text{ A} \]

#### For \(\alpha = 100^\circ\):

\[ I_{dc} = \frac{-60.2}{100} \]

\[ I_{dc} \approx -0.602 \text{ A} \]

### Summary:

- **For \(\alpha = 60^\circ\)**:
  - Average output voltage \(V_{dc} \approx 66.4 \text{ V}\)
  - Load current \(I_{dc} \approx 0.664 \text{ A}\)

- **For \(\alpha = 100^\circ\)**:
  - Average output voltage \(V_{dc} \approx -60.2 \text{ V}\)
  - Load current \(I_{dc} \approx -0.602 \text{ A}\)

Answer 2

To analyze a 1-phase half-controlled rectifier circuit with a resistive load, you need to determine the average output voltage and load current for different firing angles (\(\alpha\)).

### Given:
- Supply Voltage: \( V_s = 300 \sin(314t) \)
- Load Resistance: \( R = 100 \, \Omega \)
- Firing Angles: \( \alpha = 60^\circ \) and \( \alpha = 100^\circ \)

### Analysis of Half-Controlled Rectifier

A half-controlled rectifier circuit consists of a single SCR (Silicon Controlled Rectifier) and a diode. The output voltage and current depend on the firing angle \(\alpha\).

**1. Peak Voltage Calculation:**

The given supply voltage is \( V_s = 300 \sin(314t) \), where the angular frequency \( \omega \) is \( 314 \) rad/s.

The peak voltage \( V_m \) is given by:
\[ V_m = 300 \, \text{V} \]

**2. Average Output Voltage Calculation:**

The average output voltage \( V_{o_{avg}} \) for a half-controlled rectifier is given by:
\[ V_{o_{avg}} = \frac{V_m}{\pi} \left( \pi - \alpha \right) \]

where \(\alpha\) is the firing angle in radians.

Convert \(\alpha\) from degrees to radians:
\[ \alpha_{rad} = \frac{\alpha \times \pi}{180} \]

**For \(\alpha = 60^\circ\):**
\[ \alpha_{rad} = \frac{60 \times \pi}{180} = \frac{\pi}{3} \]

Substitute \(\alpha_{rad}\) into the formula:
\[ V_{o_{avg}} = \frac{300}{\pi} \left( \pi - \frac{\pi}{3} \right) \]
\[ V_{o_{avg}} = \frac{300}{\pi} \left( \frac{2\pi}{3} \right) \]
\[ V_{o_{avg}} = 300 \times \frac{2}{3} \]
\[ V_{o_{avg}} = 200 \, \text{V} \]

**For \(\alpha = 100^\circ\):**
\[ \alpha_{rad} = \frac{100 \times \pi}{180} = \frac{5\pi}{9} \]

Substitute \(\alpha_{rad}\) into the formula:
\[ V_{o_{avg}} = \frac{300}{\pi} \left( \pi - \frac{5\pi}{9} \right) \]
\[ V_{o_{avg}} = \frac{300}{\pi} \left( \frac{4\pi}{9} \right) \]
\[ V_{o_{avg}} = 300 \times \frac{4}{9} \]
\[ V_{o_{avg}} = 133.33 \, \text{V} \]

**3. Load Current Calculation:**

The load current \( I_{o} \) is given by:
\[ I_{o} = \frac{V_{o_{avg}}}{R} \]

**For \(\alpha = 60^\circ\):**
\[ I_{o} = \frac{200}{100} \]
\[ I_{o} = 2 \, \text{A} \]

**For \(\alpha = 100^\circ\):**
\[ I_{o} = \frac{133.33}{100} \]
\[ I_{o} = 1.333 \, \text{A} \]

### Summary

- **For \(\alpha = 60^\circ\):**
  - Average Output Voltage: \( 200 \, \text{V} \)
  - Load Current: \( 2 \, \text{A} \)

- **For \(\alpha = 100^\circ\):**
  - Average Output Voltage: \( 133.33 \, \text{V} \)
  - Load Current: \( 1.333 \, \text{A} \)