Chopper ON time for separately excited DC motor

Assumption: field current is constant (rated), so $E = K N$ and $T = K I_a$.

Key relations

• Chopper output: $V_a = \alpha V_s$
• Load torque proportional to speed: $T_L \propto N$. Since $T = K I_a$, the armature current is also proportional to speed: $I_a = c\,N$.
• Armature equation:

$$\begin{aligned} V_a &= E + I_a R_a = K N + R_a c N = (K + R_a c)\,N \end{aligned}$$

So the armature voltage is directly proportional to speed for this load, whatever the values of $K$ and $c$. Hence

$$\frac{\alpha_2}{\alpha_1} = \frac{V_{a2}}{V_{a1}} = \frac{N_2}{N_1}$$

Duty ratio at 800 rpm

$$\begin{aligned} \alpha_2 &= \alpha_1 \frac{N_2}{N_1} = 0.9 \times \frac{800}{1200} = 0.6 \end{aligned}$$

ON time

Chopping period:

$$T = \frac{1}{f} = \frac{1}{250} = 4\ \text{ms}$$ $$\begin{aligned} T_{on} &= \alpha_2 T = 0.6 \times 4\ \text{ms} = 2.4\ \text{ms} \\ T_{off} &= 4 - 2.4 = 1.6\ \text{ms} \end{aligned}$$

Check: $V_{a1} = 0.9 \times 220 = 198$ V at 1200 rpm and $V_{a2} = 0.6 \times 220 = 132$ V at 800 rpm; $132/198 = 800/1200$.

Note: the rated data (2.2 kW, 1000 rpm) are not needed, because both the back emf and the $I_a R_a$ drop scale with speed for a load with $T_L \propto N$.

Answer: Duty ratio = 0.6, ON time $T_{on}$ = 2.4 ms (OFF time 1.6 ms).