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,

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

Substituting the values, we get,

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

Therefore,

$V_{L}=198V$

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

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

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

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

Also,

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

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

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