Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11EE201 Lecture 40 P. 1MPT for ac circuits+-Rs+vL(t)-jXLRLIjXsVsPractical sourceAdjustable LoadIf a practical source (a voltage source in series with elements having an impedance Zs=Rs + jXs) is connected to a load (ZL = RL + jXL), the max. power transfer is delivered when ZL=Zs*Maximum Power TransferEE201 Lecture 40 P. 2I = Vs(Rs + RL) + j (Xs + XL)P = Pav = |I|2RL = Vs2RL(Rs + RL)2 + (Xs + XL)2PRL=Vs2[(Rs + RL)2 + (Xs + XL)2 – 2RL(Rs + RL)] [(Rs + RL)2 + (Xs + XL)2]2PXL Vs2[-2RL(Xs + XL)][(Rs + RL)2 + (Xs + XL)2]2== 0 = 0When ZL = Zs*RL = RsXL = -XsThe maximum average power isPmax = Vs2,eff4RsSolution:IS= 1 0oω = 1 rad / seccos(t)A1ΩIR- +ZL4H2IRExample: Find the load impedance for maximum power transfer.EE201 Lecture 40 P. 3EE201 Lecture 40 P. 4Using KCL,IR + ISC = 1 0o cos(t)A1ΩIR- +4H2IR+-IscCos(t)A1ΩIR- +4H2IR+-vocvoc = 1IR +2IR = 3IR = 3 0oFind VocFind ISCEE201 Lecture 40 P. 51.5351.536.003scocthIVZZth = 3 + j4For maximum power transferZL =Zth*ZL = 3 – j4KVL: 1IR +2 IR = ISC * jωL 3IR = ISC * j4 IR = 1/3 (Isc* j4)IR + ISC = (1 + j4/3)Isc(1+j4/3)Isc = 1 0oIsc = 0.6 -53.1oEE201 Lecture 40 P. 6Solution: The Thevenin equivalent impedance is Zth= R + jωLZth = 10Ω + j(10)(2)Zth = 10 + j20The load impedance that maximizes power transfer isZL = 10 – j20 = R – j/(ωC)10 0oA10Ω2H10ΩCExample: Find value for C to deliver maximum average power to the load at ω = 10 rad / sec.ωC = 1/20C = 1/(20*10) = 1/200 = 0.005C = 5mFExample: The radio receiver shown in figure (a) is connected to an antenna. The antenna intercepts the electromagnetic waves from a broadcast station operating at 1 MHz. For circuit analysis purposes, the antenna is represented by a Thevenin equivalent circuit shown in figure (b).(a) Find the input impedance Rin + jXin of the receiver if maximum power is to be transferred from the antenna to the receiver.(b) Under the condition of part (a), find the magnitude of the voltage across the receiver terminals and the average power delivered to the receiver.EE201 Lecture 40 P. 7Radio receiverAntenna(a)EE201 Lecture 40 P. 8Antenna equivalent circuitReceiver input equivalent circuit+_14.6 mV rms at 1 MHz21 -j1,070 RinjXin(b)Part (a) From the maximum power transfer theorem, the answers are Rin = 21 and Xin = j1070 .Part (b) Since the reactances in the circuit have been “tuned out,” the input current to the receiver is simply 14.6/(21 + 21) = 0.348 mA. The input impedance has the magnitude|Zin| = √212 + 10702 = 1070.2 Therefore, the magnitude of the voltage across the receiver terminals is 0.348 × 1070.2 = 372.4 mV (rms). The power transferred from the antenna to the receiver is 0.3482 × 21 = 2.54 W.Example 11.13A fixed load resistance RL = 100 k, representing the input resistance of an amplifier, is connected to the source of figure 11.22b in example 11.12 through a passive coupling network, i.e., a network that does not generate average power, as shown in figure 11.23.(a) Show that the maximum voltage that can be developed across RL is 0.504 V.(b) Show that the coupling network shown in figure 11.23 achieves this maximum voltage across RL.EE201 Lecture 40 P. 9+--j107021 14.6 mV rms at1 MHzZin+VL___400.9 H109.8 pFLCRL100kFixed source Coupling network Fixed loadFigure 11.23 Maximum power transfer through a coupling network.EE201 Lecture 40 P. 10SOLUTIONPart (a) The available power from the source is 2.54 W. If all of the power is delivered to RL, then the voltage VL must beVL = √Pmax RL = √2.54 × 10-6 × 100,000 = 0.504 VPart (b) The input impedance Zin of the coupling network with load must be the conjugate of the source impedance. Specifically,Zin = jL + 1 jC + 1 RLEE201 Lecture 40 P. 11Substituting = 106, L = 400.9 × 10-6, C = 109.8 × 10-12, and R = 100 × 103 into this expression yieldsZin = 21 + j1070 WWhich is indeed the conjugate of the source impedance. Since Zin is the conjugate-matched to the source impedance, the maximum power of 2.54 W is transferred to the coupling network. Since the coupling network consists of L and C, neither of which consumes average power, the 2.54 W of average power must be transferred out of the coupling network and into the load resistance. The voltage across the load resistor VL is given by VL = √P RL = √2.54 × 10-6 × 100,000 = 0.504 VThis verifies that the coupling network of figure 11.23 enables the largest voltage to appear across the load
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