# Phase Controlled Converters Solved Example

A separately excited DC motor has the following parameters: 220V, 100A and 1450 rpm. Its armature has a resistance of 0.1 Ω. In addition, it is supplied from a 3 phase fullycontrolled converter connected to a 3-phase AC source with a frequency of 50 Hz and inductive reactance of 0.5 Ω and 50Hz. At α = 0, the motor operation is at rated torque and speed. Assume the motor brakes re-generatively using the reverse direction at its rated speed. Calculate the maximum current under which commutation is not affected.

Solution

We know that,

$$V_{db}=3\sqrt{\frac{2}{\pi }}\times V_{L}-\frac{3}{\pi }\times R_{b}\times I_{db}$$

Substituting the values, we get,

$220=3\sqrt{\frac{2}{\pi }}\times V_{L}-\frac{3}{\pi }\times 0.5\times 100$

Therefore,

$V_{L}=198V$

Voltage at rated speed = $220-\left ( 100\times 0.1 \right )=210V$

At the rated speed, the regenerative braking in the reverse direction,

$=3\sqrt{\frac{2}{\pi }}\times 198\cos \alpha -\left ( \frac{3}{\pi }\times 0.5+0.1\right )\times I_{db}=-210V$

But $\cos \alpha -\cos \left ( \mu +\alpha \right )=\frac{\sqrt{2}}{198}\times 0.5I_{db}$

For commutation to fail, the following limiting condition should be satisfied.

$\mu +\alpha \approx 180^{\circ}$

Therefore, $\quad \cos \alpha =\frac{I_{db}}{198\sqrt{2}}-1$

Also,

$\frac{3}{\pi }I_{db}-\frac{3\sqrt{2}}{\pi }\times 198-\left ( \frac{3}{\pi }\times 0.5+0.1 \right )I_{db}=-210$

This gives, $\quad 0.3771I_{db}=57.4$

Therefore, $\quad I_{db}=152.2A$