test3sp01_sols

test3sp01_sols - 2 RTH = R || ( R + R) = R 3 RTH RTH VTH a...

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Unformatted text preview: 2 RTH = R || ( R + R) = R 3 RTH RTH VTH a + - b If a-b are short-circuited the current will be the short circuit current and V SC I ab = R a VS + - b R Vb Source transformation Define reference node + - RI S R Vab = VS − Vb Vab = VS − R RI S = VTH 3R TH RTH SC PVs = VS (− I ab ) Vs[V] 12.000 24.000 9.000 Is[mA] 5.000 6.000 4.000 R[kOhm] 12.000 15.000 9.000 Rth[kOhm] 8.000 10.000 6.000 Voc[V] -8.000 -6.000 -3.000 Isc[mA] -1.000 -0.600 -0.500 P(Vs)[mW] 12.000 14.400 4.500 6.000 2.000 15.000 10.000 -4.000 -0.400 2.400 * RL = RTH PMX 2 VTH = 4 RTH 2 é1 ù P1k = ê VTH [mW ] ë1 + RTH RTH VTH Vs + - aI R I" R Vs + - KVL KCL : I '+ aI ' = 0 I ' = 0 + V =V =V R ' OC VOC RTH = S TH VOC 2R = I SC 1 + a − KCL : I "+ aI " = I SC VS 2R V = (1 + a ) S 2R KVL : I " = aI" I SC I SC 1kΩ b To determine the Thevenin equivalent R I' + - a R[kOhm] Vs[V] a Voc[V] Isc[mA] Rth[kOhm] Pmx[mW] P(1k)[mW] 6 5 12 12 20 24 2 4 0.5 12 20 24 3 10 1.5 4 2 16 9 50 9 5.76 44.444 1.9931 3 12 2 12 6 2 18 16 KCL : I P = −bI x + IP = Ro abVS2 PI = V y (−bI x ) = R1 RTH VTH = 0 a RTH = b VS Ro R1 = I P R1 + abRo Vy Ro V y = VS I x = − Vs abVs + Ro R1 Vs[V] a b Ro[kOhm] R1[kOhm] Ip[mA] P_I[mW] Rth[kOhm] Vth 40 0.8 0.2 10 2 7.2 128 5.56 0 36 0.5 0.1 12 1 4.8 64.8 7.5 0 24 36 0.5 0.2 0.1 0.5 12 9 1 1 3.2 7.6 28.8 130 7.5 4.74 0 0 aV y R1 ...
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