Question

3 - $\phi$ fully controlled rectifier $V_{avg} \propto$:

• A. $(3V_m/\pi)\cos\alpha$
• B. $(3\sqrt{3}V_m/\pi)\cos\alpha$
• C. $(6V_m/\pi)\cos\alpha$
• D. $(V_m/\pi)\cos\alpha$

Hint:

Remember the key average voltage expressions for three-phase rectifiers under continuous conduction:
1. Semi-controlled (half-controlled) 3-phase rectifier: $V_{avg} = \frac{3\sqrt{3}V_{ml}}{2\pi} (1 + \cos\alpha)$.
2. Fully controlled 3-phase rectifier: $V_{avg} = \frac{3\sqrt{3}V_m}{\pi} \cos\alpha$ (using peak phase voltage $V_m$).
Always verify whether the formula uses peak phase voltage ($V_m$) or peak line voltage ($V_{ml}$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the mathematical expression that the average output DC voltage ($V_{avg}$) of a three-phase fully controlled bridge rectifier is proportional to, as a function of the peak phase voltage ($V_m$) and the firing angle ($\alpha$).

Step 2: Key Formula or Approach:
The average output voltage ($V_{avg}$) of a three-phase fully controlled converter operating in continuous conduction mode is found by integrating the line-to-line AC input voltage over a $60^\circ$ ($\pi/3$ radians) conduction interval.

Step 3: Detailed Explanation:

• A three-phase fully controlled bridge rectifier consists of six thyristors.

• In continuous conduction mode, a new thyristor is triggered every $60^\circ$ (or $\pi/3$ radians), and at any given time, one thyristor from the positive group and one from the negative group are conducting.

• The output voltage waveform is composed of segments of the three-phase line-to-line voltages.

• Let the phase voltages be represented as $v_{an} = V_m \sin(\omega t)$, $v_{bn} = V_m \sin(\omega t - 120^\circ)$, and $v_{cn} = V_m \sin(\omega t - 240^\circ)$, where $V_m$ is the peak phase voltage.

• The peak line-to-line voltage ($V_{ml}$) is given by:
\[ V_{ml} = \sqrt{3} V_m \]

• We integrate the line voltage over a conduction period of $\pi/3$ radians, centered around the peak of the line voltage, with a firing delay angle $\alpha$:
\[ V_{avg} = \frac{3}{\pi} \int_{\pi/6 + \alpha}^{5\pi/6 + \alpha} V_{ml} \sin(\omega t + \pi/6) d(\omega t) \]

• Evaluating this definite integral yields:
\[ V_{avg} = \frac{3 V_{ml}}{\pi} \cos\alpha \]

• Substituting $V_{ml} = \sqrt{3} V_m$ into the equation:
\[ V_{avg} = \frac{3\sqrt{3}V_m}{\pi} \cos\alpha \]

• This shows that the average output voltage is directly proportional to $(3\sqrt{3}V_m/\pi)\cos\alpha$.

Step 4: Final Answer
Thus, the average output voltage is proportional to $(3\sqrt{3}V_m/\pi)\cos\alpha$, which corresponds to option (B).