Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9EE201 Lecture 14 P. 1Maximum Power TransferWhat is the value of RL that maximizes power transferred to load?pL (t) = RL [ iL(t)]2where vOC(t)_RTHvL+_ vOCpL(t) =iL(t) =+iLRLRTH + RLvOC2(t) RL (RTH + RL)2EE201 Lecture 14 P. 2To find RL for max load power, satisfy: d pL(t) d RL= 0To find the maximum power transfer, put RTH = RL into power equation, RTH + RL = 2 RL RTH = RL d pL(t) d RL= = 0vOC2(t) (RTH + RL)22 vOC2(t) RL (RTH + RL)3pL(t)max =vOC2(t) 4 RTHEE201 Lecture 14 P. 3For circuit on P. 1, what is the effect of varying RTH (RTH 0) for fixed RL (RL > 0)? Take derivative. d pL(t) d RTH= -2 RL (RTH + RL)3 vOC2(t) < 0 Maximum power transfer occurs when RTH = 0.Example: Find RL for maximum power transfer and power absorbed by load._+18 78 1A12 10 I1RLEE201 Lecture 14 P. 4Step 1: Find Thevenin equivalent circuit.RTH = 10 + 90 18 = 25 vOC = 18 I1(1/30) (1/30) + (1/78)I1 =vOC = 13 V (1 A)_++_25 13 V Thevenin equivalent circuitEE201 Lecture 14 P. 5Step 2: Equate RL = RTH and solve for power.vL = (25/50) 13 = 6.5 VpL = vL2 / RL = vL2 / RTH = 1.69 W_25 vL+_ 13 V+iL25 pL(t)max = =vOC2(t) 4 RTH 169 100pL(t)max = 1.69 WUsing equation on P.2 for maximum power transfer,EE201 Lecture 14 P. 6Example: Find RL and the maximum power transfer to the load._+18 mA 2 k 32 V5 k10 ixRL+_AStep 1: Determine ix.KCL @ A: ix + 0.018 = 10 ixix = 0.002 AStep 2: Calculate vOC and iSC.vOC = (5 k)(10 ix) = 100 ViSC = 10 ix = 0.02 AixEE201 Lecture 14 P. 7Step 3: Calculate RTH and maximum power transfer (MPT).RTH = (vOC/iSC) = (100 / 0.02) = 5 kFor MPT, RL = RTHpmax =vOC2(t) 4 RTH (100)2 4 (5000) pmax = 0.5 W=EE201 Lecture 14 P. 8Example: Find the maximum power that can be absorbed by placing a load resistor across terminals a-b.Linear circuita+_bi (A)vab (V)153045-0.10.10.2_RTHvL+_ vOC+iRLbaDraw Thevenin equivalent circuit.EE201 Lecture 14 P. 9From KVL: vab = RTH i + vOCCompare with plot:vab = 150 i + 15 RTH = 150 ; vOC = 15 VMaximum power,pmax =vOC2 4 RTH (15)2 4 (150) pmax = 0.38
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