Determining the Optimized Gains and Time Constants of the Current and Speed Loop Controllers The motor parameters of a converter-fed dc motor are 220 V, 6.4 A, 1570 rpm, Rm = 6.5 Ω, J = 0.06 kg–m2, Lm = 67 mH, B = 0.087 N–rn /rad/s, Kb = 1.24 V/rad/s. The converter is supplied from a Y-connected

Question

Determining the Optimized Gains and Time Constants of the Current and Speed Loop Controllers

The motor parameters of a converter-fed dc motor are [latex]220 V, 6.4 A, 6.4 A, 1570 rpm,R_m=6.5\Omega ,J=0.06 kg−m^2,L_m= 67 mH,B= 0.08N−rn/rad/s,K_b=1.24 V/rad/s.[/latex]. The converter is supplied from a Y-connected 230 V, three-phase ac source at 60 Hz. The converter can be assumed linear and its maximum control input voltage is [latex]V_{cm} = ±10 V[/latex]. The tachogenerator has the transfer function [latex]G_\omega (s)=0.074/(1+0.002s)[/latex]. The speed reference voltage has a maximum of 10 V. The maximum permissible motor current is 20 A. Determine (a) the gain [latex]K_r[/latex] and time constant [latex]τ_r[/latex] of the converter, (b) the current feedback gain [latex]H_c[/latex] , (c) the motor time constant [latex]τ_1, τ_2[/latex], and [latex]τ_m[/latex], (d) the gain [latex]K_c[/latex] and the time constant [latex] au _c[/latex] of the current controller, (e) the gain [latex]K_i[/latex] and the time constant [latex] au _i[/latex] of the simplified current loop, and (f) the optimized gain [latex]K_s[/latex] and the time constant [latex]τ_s[/latex] of the speed controller.

Accepted Answer

[latex]V_{dc} = 220 V, I_{dc} = 6.4 A, N = 1570 rpm, R_m = 6.5 \Omega [/latex], [latex]J = 0.06 kg−m^2, L_m = 67 mH, B = 0.087 N-m /rad/s, Kb = 1.24 V/rad/s, V_L = 220 V, f_s = 60 Hz, V_{cm }=± 10 V 20 A, I_{a(max)} = 20 A, K_ω = 0.074, t_ω= 0.002 s[/latex].

a. The phase voltage is [latex]V_s=\frac{V_L}{\sqrt{3} }=\frac{220}{\sqrt{3} }=127.02 V[/latex]

The maximum dc voltage is [latex]V_{dc(max)} = K_rV_{cm} = 29.71 × 10 = 297.09 V[/latex]

The converter control voltage is [latex]V_c=\frac{V_{dc}}{V_{dc(max)}}V_{cm}=\frac{220 imes 110}{297.09}=7.41 V[/latex]

Using Eq. (14.88),

[latex]K_r=\frac{3\sqrt{3} V_m}{\pi V_{cm}} =\frac{3\sqrt{3} \sqrt{2} }{\pi V_{cm}}V_s =2.339\frac{V_s}{V_{cm}}[/latex] (for three−phase converter) (14.88)

[latex]K_s=\frac{2339V_s}{V_{cm}}=\frac{2.339 imes 127.02}{10}=29.71 V/V[/latex]

Using Eq. (14.91a)

[latex] au _r=\frac{360/(2 imes 6)}{360}T_s= \frac{1}{12} imes \frac{1}{f_s}[/latex] (for three-phase converter) (14.91a)

[latex] au _r=\frac{1}{12f_s}=\frac{1}{12 imes 60}=1.39 ms[/latex]

b. The current feedback gain is [latex]H_c=\frac{V_c}{I_{a(max)}}=\frac{7.41}{20}=0.37 V/A[/latex]

c. Using [latex]B_t = B[/latex] and Eq. (14.99),[latex]K_m=\frac{B}{K_b^2+R_mB}=\frac{0.087}{1.24^2+6.5 imes 0.087}=0.04[/latex]

Using Eq. (14.100),

[latex]-\frac{1}{ au _1};-\frac{1}{ au _2}=\frac{-\left(\frac{R_m}{L_m}+\frac{B}{J} ight)\pm \sqrt{\left(\frac{R_m}{L_m}+\frac{B}{J} ight)^2 -4\left(\frac{R_mB+K_b^2}{JL_m} ight) } }{2}[/latex] (14.100)

[latex]r_1 = -5.64[/latex] ; [latex] au _1=\frac{-1}{r_1} =0.18 s[/latex]

[latex]r_2 = -92.83[/latex] ; [latex] au _2=\frac{-1}{r_2} =0.01 s[/latex]

[latex] au _m=\frac{J}{B} =\frac{0.06}{0.087}=0.69 s[/latex]

d. The time constant of the current controller is [latex] au _c= au _2=0.01 s[/latex]

Using Eq. (14.110), [latex]K_c=\frac{ au _1 au _2}{2 au _r}\left(\frac{1}{K_mK_rH_c au _m} ight)=\frac{0.18 imes 0.01 }{0.04 imes 29.71 imes 0.37 imes 0.69}=2.19[/latex]

[latex]K_c=\frac{ au _1 au _2}{2 au _r}\left(\frac{1}{K_mK_rH_c au _m} ight)[/latex] (14.110)

Using Eqs (14.112) to (14.115),

[latex] au _i=\frac{ au _3}{1+K_1}[/latex] (14.112)

[latex] au _3= au _1+ au _r[/latex] (14.113)

[latex]K_i=\frac{1}{H_c}\left(\frac{K_1}{1+K_1} ight)[/latex] (14.114)

[latex]K_1=\frac{K_mK_cK_rH_c au _m}{ au _c}[/latex] (14.115)

[latex]K_1=\frac{K_mK_cK_rH_c au _m}{ au _c}=\frac{0.4 imes 2.19 imes 29.71 imes 0.37 imes 0.69}{0.01}=63.88[/latex]

[latex]K_i=\frac{1}{H_c}\left(\frac{K_1}{1+K_1} ight)=\frac{1}{0.37}\left(\frac{63.88}{1+63.88} ight)=2.66[/latex]

[latex] au _3= au _1+ au _r=0.18 + 0.00139 = 0.18 s[/latex]

[latex] au _i=\frac{ au _3}{1+K_1}=\frac{0.18}{1+63.88}=2.76 ms[/latex]

e. Using Eqs (14.118) and (14.119),

[latex] au _e= au _i+ au _\omega [/latex] (14.118)

[latex]K_\omega =\frac{KiK_bH_\omega }{B au _m} [/latex] (14.119)

[latex] au _e= au _i+ au _\omega =2.76 imes 10^{-3} + 2 imes 10^{-3} = 4.76 ms[/latex]

[latex]K_\omega =\frac{KiK_bH_\omega }{B au _m}=\frac{2.66 imes 1.24 imes 0.074}{0.087 imes 0.69}=4.07[/latex]

Using Eqs (14.121) and (14.122)

[latex]K_s=\frac{1}{2K_\omega au _e}[/latex] (14.121)

[latex] au _s=4 au _e[/latex] (14.122)

[latex]K_s=\frac{1}{2K_\omega au _e}=\frac{1}{2 imes 4.07 imes 4.76 imes 10^{-3}}=25.85[/latex]

[latex] au _s=4 au _e=4 imes 4.76 imes 10^{-3} = 19.02 ms[/latex]

Question 14.14: Determining the Optimized Gains and Time Constants of the Cu...

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