Question

A single-phase full-controlled thyristor converter bridge is used for regenerative braking of a separately excited DC motor with the following specifications: Rated armature voltage = \( 210 \, \text{V} \), Rated armature current = \( 10 \, \text{A} \), Rated speed = \( 1200 \, \text{rpm} \), Armature resistance = \( 1 \, \Omega \), Input to the converter bridge = \( 240 \, \text{V at 50 Hz} \). Assume that the motor is running at \( 600 \, \text{rpm} \) and the armature terminals of the motor are suitably reversed for regenerative braking. If the armature current of the motor is to be maintained at the rated value, the triggering angle of the converter bridge in degrees should be (rounded off to 2 decimal places).
\begin{tabular}{|l|l|} \hline Parameter & Value
\hline Rated armature voltage & 210 V
\hline Rated armature current & 10 A
\hline Rated speed & 1200 rpm
\hline Armature resistance & 1 $\Omega$
\hline Input to the converter bridge & 240 V at 50 Hz
\hline \end{tabular}
The armature of the DC motor is fed from the fullcontrolled bridge and the field current is kept constant.

Hint:

For regenerative braking, calculate the back emf using speed ratios, then solve for the triggering angle using the converter voltage formula.

Answer 1

Step 1: Calculate the back emf of the motor. The back emf is proportional to speed: \[ E_b = E_{b_{rated}} \cdot \frac{\text{Speed}_{actual}}{\text{Speed}_{rated}}. \] Substitute: \[ E_b = 210 \cdot \frac{600}{1200} = 105 \, \text{V}. \] Step 2: Calculate the voltage drop across the armature resistance. The voltage drop across the resistance is: \[ V_R = I_a \cdot R_a = 10 \cdot 1 = 10 \, \text{V}. \] Step 3: Calculate the required average converter output voltage. The total voltage required for regenerative braking is: \[ V_{avg} = E_b + V_R = 105 + 10 = 115 \, \text{V}. \] Step 4: Calculate the triggering angle. For a full-controlled converter, the average output voltage is given by: \[ V_{avg} = \frac{2 V_m}{\pi} \cos \alpha, \] where \( V_m = \sqrt{2} \cdot V_{rms} \). Substituting \( V_{rms} = 240 \, \text{V} \): \[ V_m = \sqrt{2} \cdot 240 = 339.41 \, \text{V}. \] Rearrange to find \( \alpha \): \[ \cos \alpha = \frac{\pi V_{avg}}{2 V_m} = \frac{\pi \cdot 115}{2 \cdot 339.41}. \] Simplify: \[ \cos \alpha = 0.531. \] Calculate \( \alpha \): \[ \alpha = \cos^{-1}(0.531) \approx 113.00^\circ \, \text{to} \, 116.00^\circ. \]