Question

Fig. 3 shows an IPT circuit. Derive an equation giving the impedance of the secondary (Zs), separated into its real and reactive parts. Show all steps in the derivation. 5. Ip Lp Lout Compensating network Vs VPR Primary Secondary side side Fig. 3 Find an equation for Zs (from question 5), reflected across to the primary part of the circuit (i.e. find Zs'). Separate Zs into its real and reactive parts. Show all steps in the derivation 6. The equation for Zs' (from question 6), should have a reactive part. State the effect that will have on compensation that is applied to the primary side. 7. Based on the equation for Zs, (from question 6), and with Ls of 10 μΗ, Cs of 253 nH and R of 50 2, state the real and imaginary parts of the secondary impedance reflected to the primary side. Operation is at 100 kHz. 8.

Answer 1

Consider the IPT circuit Compensating network RL C Primary Secondary side side (5) Determine the secondary impedance 1 R,|| jaC Z, = jaL 1 jос, = jaL 1 jоC, T = jL 1 joR,C -a R,LC j1-joR,C 1joR,C X 1-jR,C R - о' R L,C, -joR,C, (R, - 0' R_L,C.) + jol, + o'R_L,C, 1+ RiC 2 S 1+R / +RCL -RiC 1+@RC Z2

(6) The transformation ratio of the transformer is VPR К- The transformer secondary impedance referred to the primary side is. PR Z %3D К'Z, %3 Z. Vs L, +RCL-RiC, jo PR Z2 = } 1+ @R$C_ 1+RC2 VS (7) L + @R>C?L - R;C, 1+RC The reactive part of Z' is PR V The compensation network converts the reactive network with impedance Z, +Z, into resistive The impedance seen by the source V, is resistive because of the compensating network (8) L +RC^L-R?C, - jo 1+RC R, Z2 = 1+ aRC 27TX 100 x 103 = 6283 18.5 =27f Substitute the values. 50 1+628318.5x 502 x 253x 10 10x 10628318.52x 502 x VРR Z= у, (253x10)(10x10)- -502 x253x10 j6283 18.5 1+628318.52x 502 x 253 x109 VР. 1+628318.5 50j5.807) x 502 x 253x 10 V pz2.51x103(50+ j5.807) V PR V (0.125 j0.0146) Z2 =