Finding the Performance of a Dc–dc Converter-Fed Drive in Regenerative Braking A dc–dc converter is used in regenerative braking of a dc series motor similar to the arrangement shown in Figure 14.16a. The dc supply voltage is 600 V. The armature resistance is Ra = 0.02 Ω and the field resistance

Question

Finding the Performance of a Dc–dc Converter-Fed Drive in Regenerative Braking

A dc–dc converter is used in regenerative braking of a dc series motor similar to the arrangement shown in Figure 14.16a. The dc supply voltage is 600 V. The armature resistance is [latex]R_a = 0.02 Ω[/latex] and the field resistance is [latex]R_f = 0.03 Ω[/latex]. The back emf constant is [latex]K_v = 15.27 mV/A rad/s[/latex]. The average armature current is maintained constant at [latex]I_a = 250 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 average voltage across the dc–dc converter [latex]V_{ch}[/latex]; (b) the power regenerated to the dc supply [latex]P_g[/latex]; (c) the equivalent load resistance of the motor acting as a generator, [latex]R_{eq}[/latex]; (d) the minimum permissible braking speed [latex]\omega _{\min}[/latex]; (e) the maximum permissible braking speed [latex]ω_{\max}[/latex]; and (f) the motor speed.

Accepted Answer

[latex]V_s = 600 V, I_a = 250 A, K_v = 0.01527 V/A rad/s, k = 0.6[/latex]. For a series motor [latex]R_m = R_a + R_f = 0.02 + 0.03 = 0.05 Ω[/latex].

a. From Eq. (14.37),

[latex]V_{ch} = (1 - k)V_s[/latex] (14.37)

[latex]V_{ch} = (1 - 0.6) × 600 = 240 V[/latex].

b. From Eq. (14.38),

[latex]P_g = I_a V_s(1 - k)[/latex] (14.38)

[latex]P_g = 250 × 600 × (1 - 0.6) = 60 kW[/latex].

c. From Eq. (14.40),

[latex]R_{eq}=\frac{E_g}{I_a}=\frac{V_s}{I_a}(1-k)+R_m[/latex] (14.40)

[latex]R_{eq} = (600/250)(1 - 0.6) + 0.05 = 1.01 Ω[/latex].

d. From Eq. (14.42),

[latex]\omega _{min}=\frac{R_mI_a}{K_\upsilon I_f}[/latex] (14.42)

the minimum permissible braking speed,

[latex]\omega _{min}=\frac{0.05}{0.01527}=3.274 rad/s[/latex] or [latex] 3.274 imes \frac{30}{\pi } = 31.26 rpm[/latex]

e. From Eq. (14.43),

[latex]\omega _{max}=\frac{V_s}{K_\upsilon I_f}+\frac{R_mI_a}{K_\upsilon I_f}[/latex] (14.43)

the maximum permissible braking speed,

[latex]\omega _{max}=\frac{600}{0.01527 imes 250}+\frac{0.05}{0.01527}=160.445 rad/s[/latex] or 1532.14 rpm

f. From Eq. (14.39),

[latex]E_g= V_{ch} + R_m I_a = (1 - k)V_s + R_m I_a[/latex] (14.39)

[latex]E_g = 240 + 0.05 × 250 = 252.5 V[/latex] and the motor speed,

[latex]\omega =\frac{252.5}{0.01527 imes 250}=66.14 rad/s[/latex] or 631.6 rpm

Note: The motor speed would decrease with time. To maintain the armature current at the same level, the effective load resistance of the series generator should be adjusted by varying the duty cycle of the dc–dc converter.

Question 14.9: Finding the Performance of a Dc–dc Converter-Fed Drive in Re...

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