DC chopper: input power, speed, torque, speed range

Data: $V_s = 400$ V, $R_a = 0.1\ \Omega$, $K_v = 4$ V-s/rad, $I_a = 150$ A (continuous, ripple-free), duty cycle $k = 0.6$. For a step-down chopper, $V_a = kV_s$ and the average source current is $kI_a$.

a) Input power

$$\begin{aligned} V_a &= kV_s = 0.6\times400 = 240\ \text{V} \\ P_i &= V_s(kI_a) = 400\times0.6\times150 = 36{,}000\ \text{W} = 36\ \text{kW} \end{aligned}$$

b) Motor speed

$$\begin{aligned} E_g &= V_a - I_aR_a = 240 - 150\times0.1 = 225\ \text{V} \\ \omega &= \frac{E_g}{K_v} = \frac{225}{4} = 56.25\ \text{rad/s} \\ N &= \frac{56.25\times60}{2\pi} = 537.1\ \text{rpm} \end{aligned}$$

c) Developed torque

$$T = K_vI_a = 4\times150 = 600\ \text{N-m}$$

Speed range for $k$ = 0.1 to 0.9 (same $I_a = 150$ A)

Answer: (a) 36 kW; (b) 56.25 rad/s = 537.1 rpm; (c) 600 N-m. Speed range: minimum 59.7 rpm (6.25 rad/s) at 10 %, maximum 823.6 rpm (86.25 rad/s) at 90 % duty cycle.