Question 3.AE.14: Current ripples due to PWM current-source inverters cause vo......
Current ripples due to PWM current-source inverters cause voltage stress in windings of machines and transformers. The configuration of Fig. E3.14.1 represents a simple winding with two turns. Each winding can be represented by four segments having an inductance L_i , a resistance R_i , a capacitance to ground (frame) C_i , and some interturn capacitances C_{ij} and inductances L_{ij}. Figure E.3.14.2 represents a detailed equivalent circuit for the configuration of Fig. E3.14.1.
a) To simplify the analysis neglect the capacitances C_{ij} and the inductances L_{ij}. This leads to the circuit of Fig. E3.14.3, where the winding is fed by a PWM current source. One obtains the 15 differential equations of Eqs. E3.14-1 to E3.14-8a,b as listed below. In this case the current through L_1 is given (e.g., is(t)) and, therefore, there are only 15 differential equations, as compared with Application Example 3.13.
\frac{d v_1}{d t}=\frac{i_s}{C_1}-\frac{G_1}{C_1} v_1-\frac{i_2}{C_1} (E3.14-1)
\frac{d_2}{d t}=-\frac{R_2}{L_2} i_2-\frac{\nu_2}{L_2}+\frac{v_1}{L_2}, (E3.14-2a)
\frac{d v_2}{d t}=\frac{i_2}{C_2}-\frac{G_2}{C_2} v_2-\frac{i_3}{C_2}, (E3.14-2b)
\frac{d i_3}{d t}=-\frac{R_3}{L_3} i_3-\frac{v_3}{L_3}+\frac{v_2}{L_3} (E3.14-3a)
\frac{d v_3}{d t}=\frac{i_3}{C_3}-\frac{G_3}{C_3} v_3-\frac{i_4}{C_3}, (E3.14-3b)
\frac{d i_4}{d t}=-\frac{R_4}{L_4} i_4-\frac{v_4}{L_4}+\frac{v_3}{L_4}, (E3.14-4a)
\frac{d v_4}{d t}=\frac{i_4}{C_4}-\frac{G_4}{C_4} v_4-\frac{i_5}{C_4} (E3.14-4b)
\frac{d i_5}{d t}=-\frac{R_5}{L_5} i_5-\frac{v_5}{L_5}+\frac{v_4}{L_5}, (E3.14-5a)
\frac{d v_5}{d t}=\frac{i_5}{C_5}-\frac{G_5}{C_5} v_5-\frac{i_6}{C_5} (E3.13-5b)
\frac{d i_6}{d t}=-\frac{R_6}{L_6} i_6-\frac{v_6}{L_6}+\frac{v_5}{L_6}, (E3.14-6a)
\frac{d v_6}{d t}=\frac{i_6}{C_6}-\frac{G_{6}}{C_6} v_6-\frac{i_7}{C_6} (E3.14-6b)
\frac{d i_7}{d t}=-\frac{R_7}{L_7} i_7-\frac{v_7}{L_7}+\frac{v_6}{L_7}, (E3.14-7a)
\frac{d v_7}{d t}=\frac{i 7}{C_7}-\frac{G_7}{C_7} v_7-\frac{i_8}{C_7}, (E3.14-7b)
\frac{d i_8}{d t}=-\frac{R_8}{L_8} i_8-\frac{v_8}{L_8}+\frac{v_7}{L_8}, (E3.14-8a)
\frac{d v_8}{d t}=\frac{i_{\mathrm{B}}}{C_8}-v_8\left(\frac{1}{Z C_8}+\frac{C_8}{C_8}\right) . (E3.14-8b)
The parameters are
L_{15}=L_{37}=0.005 ~\mathrm{mH}, L_{26}=L_{48}=0.01~\mathrm{mH}
Z is either 1 μΩ (short-circuit) or 1 MΩ (open-circuit). Assume a PWM current-step function i_s(t) as indicated in Fig. E3.14.4.
b) For the simplified equivalent circuit using Mathematica compute all state variables during the time period from t=t_{calculate} =0 to 2.1 ms at rise times of t_{rise}=5 μs and 75 μs. Plot the current is (t) for 1 ms and voltages (v_1 – v_5),\ (v_2 – v_6),\ (v_3 – v_7),\ and\ (v_4 – v_8) for t_{plot}=2 ms.
To demonstrate the sensitivity of the induced voltage stress as a function of the rise time of the current ripple, a rise time of 5 μs has been assumed first and the results based on Table E3.14.1 are plotted in Figs. E3.14.5 to E3.14.9 for Z=10^{-6}Ω, that is the considered winding is short-circuited. Note, the voltage across capacitance C_{15} (that is, the voltage between the first turn and the second turn at the end region (ER)) is very large. For t_{rise} = 75 μs at Z=10-6Ω the voltage stress is smaller (see Figs. E3.14.10 to E3.14.14).
| Table E3.14.1 Mathematica program list for Application Example 3.13 (part a) for short- circuit with Z=10^{-6}~Ω | |
| amp=5; period=200*^-6; duty=0.5; trise=5*Λ-6; Is[t_]:=If[Mod[t,period]> duty*period, 0,If[Mod[t,period]<trise,amp/ (trise) *Mod[t,period],amp-amp/ (period*duty-trise) *(Mod[t,period]-trise)]]; |
ic6=V3[0]==0; eq7=I4’[t]==–R4/L4*I4[t]–V4[t]/L4+V3[t]/L4; ic7=I4[0]==0; eq8=V4’[t]==I4[t]/C4–G4/C4*V4[t]–I5[t]/C4; ic8=V4[0]==0; eq9=I5’[t]==–R5/L5*I5[t]–V5[t]/L5+V4[t]/L5; ic9=I5[0]==0; eq10=V5’[t]==I5[t]/C5–G5/C5*V5[t]-I6[t]/C5; ic10=V5[0]==0; eq11=I6’[t]==–R6/L6*I6[t]-V6[t]/L6+ V5[t]/L6; |
| Table E3.14.1 Mathematica program list for Application Example 3.13 (part a) for short- circuit with Z=10^{-6}~Ω | |
| Plot[Is[t],{t,0,.001}, PlotRange!All,AxesLabel !{“t”,”Is(t)”}] R1=25*^–6; R2=25*^–6; R3=25*^–6; R4=25*^–6; R5=25*^–6; R6=25*^–6; R7= 25*^–6; R8=25*^–6; L1=1*^–3; L3=1*^–3; L5=1*^–3; L7=1*^–3; L2=10*^–3; L4=10*^–3; L6=10*^–3; L8=10*^–3; L15=5*^–6; L37=5*^–6; L26=10*^–6; L4 8=10*^–6; C1=.7*^–12; C3=.7*^–12; C5=.7*^–12; C7 =.7*^–12; C2=7*^–12; C4=7*^–12; C6=7*^–12; C8=7*^–12; C15=.35*^–12; C37=.35*^–12; C26=3.5*^–12; C48=3.5*^– 12; G1=1/5000; G3=1/5000; G5=1/5000; G7=1/5000; G2=1/500; G4=1/500; G6=1/500; G8=1/500; Z=1*^–6; eq2=V1’[t]==Is[t]/C1–G1/ C1*V1[t]–I2[t]/C1; ic2=V1[0]==0; eq3=I2’[t]==–R2/L2*I2[t]– V2[t]/L2+V1[t]/L2; ic3=I2[0]==0; eq4=V2’[t]==I2[t]/C2–G2/ C2*V2[t]–I3[t]/C2; ic4=V2[0]==0; eq5=I3’[t]==–R3/L3*I3[t]– V3[t]/L3+V2[t]/L3; ic5=I3[0]==0; eq6=V3’[t]==I3[t]/C3–G3/ C3*V3[t]–I4[t]/C3; |
ic11=I6[0]==0; eq12=V6’[t]==I6[t]/C6–G6/C6*V6[t]–I7[t]/C6; ic12=V6[0]==0; eq13=I7’[t]==–R7/L7*I7[t]–V7[t]/L7+ V6[t]/L7; ic13=I7[0]==0; eq14=V7’[t]==I7[t]/C7–G7/C7*V7[t]–I8[t]/C7; ic14=V7[0]==0; eq15=I8’[t]==–R8/L8*I8[t]–V8[t]/L8+V7[t]/L8; ic15=I8[0]==0; eq16=V8’[t]==I8[t]/C8–V8[t]*(1/(Z*C8)+G8/C8); ic16=V8[0]==0; sol1=NDSolve[{eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9, eq10,eq11,eq12,eq13,eq14,eq15,eq16,ic2,ic3,ic4,ic5, ic6,ic7,ic8,ic9,ic10,ic11,ic12,ic13,ic14,ic15,ic16}, {V1[t], I2[t],V2[t],I3[t],V3[t],I4[t],V4[t],I5[t],V5[t],I6[t], V6[t],I7[t],V7[t],I8[t],V8[t]},{t,0,.002}, MaxSteps!100000] Plot[Evaluate[V1[t]/.sol1] Evaluate[V5[t]/.sol1], {t,0,.002}, AxesLabel!{“Time”, “V1 – V5”}] Plot[Evaluate[V2[t]/.sol1] –Evaluate[V6[t]/.sol1], {t,0,.002}, AxesLabel!{“Time”, “V2 – V6”}] Plot[Evaluate[V3[t]/.sol1] –Evaluate[V7[t]/.sol1], {t,0,.002}, AxesLabel!{“Time”, “V3 – V7”}] Plot[Evaluate[V4[t]/.sol1] –Evaluate[V8[t]/.sol1], {t,0,.002}, AxesLabel!{“Time”, “V4 – V8”}] |
