您好,欢迎访问三七文档
当前位置:首页 > 商业/管理/HR > 质量控制/管理 > PSIM练习01-双闭环的直流电动机速度控制器的设计-译文及仿真结果
PSIM01 M.Brugel20071 1. 2. PID 3. 1. 2. 3. PID 4. 1. 2. 3. 4. 5. 6. LAPLACE 1. mod_DC_Chopper.pdf 2. mod_DC_Motor.pdf 3. PIPI_order1.pdf 4. PSIMAC SweepACSweep_Tutorial.pdf 50 0100ms 1. 2. 3. 4. 5. 6. 7. 1. 1.1 PSIMMAXON 2260 215maxon2260.pdf Cr=K1ΩK1=0.0005Nm/(rad/s) ‐ 1. V1500rpm V=14.5V 95%TrIa Tr83ms Ia=1.08A 1.2 / PSIMTF1TF2TF1TF2PSIMmaxon2260.pdfmod_DC_Motor.pdf TF2kf=5.15μNm/(rad/s) 2. 14.5V Tr=80ms Ia=1.08A 5A3ms 1.3 ‐ 5%2.5% 124VαV=14.5V mod_DC‐Chopper.pdf α=14.45/24=0.60420~10VVcom Vcom=0.604*10=6.04V 3. Um Um14.4V SIMVIEWIamax=1.148AIamin=1.002AIa_avg=1.07A ΔCe=(1.148-1.002)/1.07=13.6%5%mod_DC_Chopper.pdfα=0.5 L=2.8mH Δimax=E*Th/(4*L)E=24;Th=1/40kHzΔimax=5%*iav=0.05*1.08L=2.8mH 4. 12.8mH Δimax=35mA Δimax/ia_avg=0.035/1.08=3.2%5%20.92m2.8‐0.92=1.88mH Δimax=1.1063‐1.0588=0.0475A=47.5mA Δimax/ia_avg=0.0475/1.08=4.4%5% 2. 2.1 2.1.1 mod_DC_Motor.pdf 1T()(1)(1)mIfeemssksτττ+=′′++ 22222()T()()()()11(1)(1)mmeemsksUsJsfLsRkkJLRJLfkRfsskRfkRfksττΩΩ==+++=++++++=′′++ 22222()T()()()()1()11()11(1)(1)ImmeemememfeememRfemfeemIsJsfsUsJsfLsRkfsRfkRfsskRfskskssksτττττττττττττ+==++++=++++++=++++=′′++ 2mkkkRf=+ 2ffkkRf=+ 2RfRfkkRf=+ eτ′emτ′2()10eememRfesksττττ+++= 21,2()()412emRfeemRfeeemkksττττττ−+±+−= 2()()42emRfeemRfeeemeeememRfeekkkττττττττττττ+−+−′=≈+≈ 2()()42()(1)emRfeemRfeeememeememRfeemRfeemRfekkkkkτττττττττττττττ+++−′=≈+−+≈+− 22()[()4]4emRfeemRfeeemeemeemkkττττττττττ+−+−′′== τm=J/fτ'e≈τe=Ltotal/Rτ'em≈τem=τm*R*f/(k2+R*f)τm τemτe m11/mωτω== 1/eeωτ= 1/ememωτ= 21/1/eemeemωττττ′′== G0=2.4= 24V / 10VK0=AG0 VswpVswp fm=13mHzfe=122.4Hz1τm = J/f = 63.5e‐6 / 5.17e‐6 = 12.3s 1113(Hz)22212.3mmmfmωππτπ====× 2 τe = Ltotal/R = (0.92e‐3+2.8e‐3) / 2.85 = 1.3 ms 11122.4(Hz)2221.3eeefmωππτπ====× 3 τem = τm*R*f / (k2 + R*f) = 12.3 * 2.85 * 5.17e‐6 / (73e‐3*73e‐3 +2.85*1.57e‐6) = 0.03398 (s) 114.6837(Hz)220.03398ememfπτπ===× 4 211150.458(rad/s)0.00130.03398eemωττ===× 22150.45824(Hz)22fωππ=== 1Hz1kHz 5. ‐20dB/ 2.1.2 PIPITi = 1 / ω2 95%tr=1msTr 03rtω= ω0 ω2Ti f2=100Hzω2= 628rad/s Ti = 1/ω2 = 1.6ms f2=24Hzω2= 150rad/s Ti = 1/ω2 = 6.7ms PI 6. PITi=1/ω2Kp=1ωωωπKp = 10(13.4/20) = 4.677 Ti = 6.7msPI4.7 TikpPI 7. PI 1msPITikP PI 2.2 2.2.1 8. PIVcons5V1500rpmKtac Ktac = 5V / (1500/60*2*π) = 0.031831 (V/rad/s) = 31.831mV/rad/s VconVretv PI_order1.pdfVconVretvH(s)=1Ksτ+τK τ≈128ms(63)K = 23 / 5 = 4.6 2.2.2 PI_order1.pdf1C(s)=A(1)iTs+ATiTr100ms H(s)C(s) 22200()()1F(s)=11()()11211iiiHsCsTsKATHsCsTssKAKAasssτλωω+=+++++=++ 012iiKATKAKATλτωτ+== 03rTλω≈ 036611rBTKAKττλω≈==++ Ti 66*0.128116.680.1006.681.454.6BrBKTKAKτ≈−=−==== 1Ti λ=1 22244(1)(1)40.1286.68(16.68)0.05858msBiBKAKTKAKsττ==++××=+== 2Ti Ti λ 1Ti0.5τ=64msλ= 1Ti0.75τ=96ms λ 1Tiτ=128ms 9. PI 5VTi58ms64ms96ms128ms 3. 3.1 Iamax=6.7A 3A 10. PI 3.2 11. ‐ Vcons = 10V3000rpm24V 36V PSIMs‐Tutorial_ACSweel.pdf
本文标题:PSIM练习01-双闭环的直流电动机速度控制器的设计-译文及仿真结果
链接地址:https://www.777doc.com/doc-5111733 .html