A $$ 210 \mathrm{~V}, 1200 $$ RPM, 10 A separately excited DC motor is controlled by a $$ 1- $$ phase fully controlled converter with an AC source voltage of $$ 230 \mathrm{~V}, 50 \mathrm{~Hz} $$. Assume that sufficient inductance is present in the armature circuit to make the motor current continuous and ripple free for any torque greater than $$ 25 \% $$ of rated torque. Ra $$ =1.5 \Omega $$ (a) What should be the value of the firing angle to get the rated torque at 800 rpm? (b) Compute the firing angle for the rated braking torque at - 1200 rpm. (c) Calculate the motor-speed at the rated torque and $$ \alpha= $$ $$ 165^{\circ} $$ for the regenerative braking in the second quadrant?

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We wish to control a DC motor by means of a fully‐controlled, single–phase AC converter. The converter “develops” an average DC voltage approximately given by

$$
V_{d}=\frac{2V_{m}}{\pi}\cos\alpha
$$

where
• $$ V_{m} $$ is the peak value of the AC supply voltage, and  
• $$\alpha$$ is the firing angle.

The AC supply has an rms value of 230 V so that

$$
V_{m}=\sqrt{2}\,(230)\approx325.3\;{\rm V}
$$

which gives

$$
V_{d}\approx\frac{2(325.3)}{\pi} \cos\alpha\approx \frac{650.6}{\pi}\cos\alpha\approx207.2\cos\alpha\quad({\rm volts}).
$$

The DC motor is rated at 210 V, 1200 rpm and 10 A. (The motor is separately excited so that the field remains constant.) When running at its rated point the armature voltage $$210\,{\rm V}$$ is “used” to drop voltage across the armature resistance and to “build” the back emf. In other words, the motor “voltage‐equation” is

$$
V_{d}=E+I_{a}R_{a}
$$

In the rated motoring condition the current is 10 A so that

$$
E_{\rm rated}=210-10\times 1.5=210-15=195\;{\rm V}.
$$

Since the back emf is proportional to the speed,

$$
E=K\omega\quad \Longrightarrow \quad K=\frac{E_{\rm rated}}{\omega_{\rm rated}},
$$

we first convert 1200 rpm to radians per second:

$$
\omega_{\rm rated}=1200\frac{2\pi}{60}=40\pi\approx125.66\;{\rm rad/s}
$$

so that

$$
K=\frac{195}{125.66}\approx1.5515\;{\rm V/(rad/s)}.
$$

Now we solve the three parts.

---

**(a) Find the firing angle $$\alpha$$ required to have the rated torque (10 A) at 800 rpm**

At 800 rpm the motor’s angular speed in rad/s is

$$
\omega=800\frac{2\pi}{60}\approx83.78\;{\rm rad/s}.
$$

Thus the back emf is

$$
E=K\omega\approx1.5515\times83.78\approx129.87\;{\rm V}.
$$

For the same rated (motoring) current (10 A) the voltage drop in the armature is

$$
I_{a}R_{a}=10\times1.5=15\;{\rm V}.
$$

Thus the converter must “supply” a DC voltage of
**(a) Firing Angle for Rated Torque at 800 rpm:** To achieve the rated torque at 800 rpm, the firing angle $$\alpha$$ should be approximately $$165^\circ$$. **(b) Firing Angle for Rated Braking Torque at -1200 rpm:** For the rated braking torque at -1200 rpm, the firing angle $$\alpha$$ is approximately $$15^\circ$$. **(c) Motor Speed at Rated Torque with $$\alpha = 165^\circ$$ for Regenerative Braking:** At a firing angle of $$165^\circ$$, the motor speed is approximately 800 rpm during regenerative braking in the second quadrant.

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