Finding the Performance Parameters of a Dc–dc Converter Drive A dc separately excited motor is powered by a dc–dc converter (as shown in Figure 14.15a), from a 600-V dc source. The armature resistance is Ra = 0.05 Ω. The back emf constant of the motor is Kv = 1.527 V/A rad/s. The average armature

Question

Finding the Performance Parameters of a Dc–dc Converter Drive

A dc separately excited motor is powered by a dc–dc converter (as shown in Figure 14.15a), from a 600-V dc source. The armature resistance is [latex]R_a = 0.05 Ω[/latex]. The back emf constant of the motor is [latex]K_v = 1.527 V/A rad/s[/latex]. The average armature current is [latex]I_a = 250 A[/latex]. The field current is [latex]I_f = 2.5 A[/latex]. The armature current is continuous and has negligible ripple. If the duty cycle of the dc–dc converter is 60%, determine (a) the input power from the source, (b) the equivalent input resistance of the dc–dc converter drive, (c) the motor speed, and (d) the developed torque.

Accepted Answer

[latex]V_s = 600 V, I_a = 250 A[/latex], and k = 0.6. The total armature circuit resistance is [latex]R_m = R_a = 0.05 Ω[/latex].

a. From Eq. (14.33),

[latex]P_o = V_a I_a = kV_s I_a[/latex] (14.33)

[latex]P_i = kV_s I_a = 0.6 × 600 × 250 = 90 kW[/latex]

b. From Eq. (14.35)

[latex]R_{eq}=\frac{V_s}{I_s}=\frac{V_s1}{I_ak}[/latex] (14.35)

[latex]R_{eq} = 600/1250 × 0.62 = 4 Ω[/latex].

c. From Eq. (14.32),

[latex]V_a = kV_s[/latex] (14.32)

[latex]V_a = 0.6 × 600 = 360 V[/latex]. The back emf is

[latex]E_g = V_a - R_m I_m = 360 - 0.05 × 250 = 347.5 V[/latex]

From Eq. (14.2),

[latex]E_g = K_vωI_f[/latex] (14.2)

the motor speed is

[latex]\omega =\frac{347.5}{1.527 imes 2.5} =91.03 rad/s[/latex] or [latex]91.03 imes \frac{30}{\pi }=869.3 rpm[/latex]

d. From Eq. (14.4),

[latex]T_d = K_t I_f I_a[/latex] (14.4)

[latex]T_d = 1.527 × 250 × 2.5 = 954.38 N · m[/latex]

Question 14.8: Finding the Performance Parameters of a Dc–dc Converter Driv...

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