Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8EE201 Lecture 6 P. 1Dependent SourcesDependent sources (from Lect. 3)_+_+++__v (t) = vxVCVS CCVSVCCS CCCSv (t) = rm ixi (t) = gm vxi (t) = ixEE201 Lecture 6 P. 2transconductance gmmho VCCSvoltage gain -- VCVScurrent gain -- CCCStransfer resistance rmohm CCVSExample: Find Req when gm equals (a) 0.5 m-mho, (b) 1.0 m-mho, (c) 2 m-mho+__+ V1IS1kVSgmV1Solution: From KCL, IS + gmVS = I1k = VS / 1kIS = [ (1/1k) - gm] VSReq = VS / IS = [ (1/1k) - gm]-1Answers(a) 2 k(b) open(c) -1 kEE201 Lecture 6 P. 3+_Example:_+_+ Vin168 Vout Iin Iout34779.8 VinFind: Req of ind. source, Iin, Iout, Vout, voltage gain.Solution:(a) Req = 3 + 47 = 50 (b) Iin = 0.04 V / 50 = 0.0008 A = 0.8 mA(c) Iout = (Gout / Geq) x 79.8 Vin Vin = (47 / 50) 0.04 V = 0.0376 V; Iout = 2A(d) Vout = 2A x 8 = 16 V(e) Voltage gain = Vout/Vin = 425.540 mV+_Example:2 3 I1 I1 Iout3 2 10 V EE201 Lecture 6 P. 4+_ I2 5 I2_+ VoutSolution:(a) I1 = 10 V/ 2 = 5 AI2 = (3 x 5A)/ 3 = 5 AIout = 5 I2 = 25 AVout = Iout x 2 = 50 VFind: (a) the output voltage and output current(b) the voltage gain(c) the power delivered by each sourceEE201 Lecture 6 P. 5Solution (cont):(b) Voltage gain = Vout/Vin = 50 V/10 V = 5(c) Power delivered by sourcesIndependent voltage sourceP = I12 x R1 = (5A)2 x 2 = 50 WDependent voltage sourceP = I22 x R2 = (5A)2 x 3 = 75 WDependent current sourceP = Iout2 x Rout = (25A)2 x 2 = 1250 WEE201 Lecture 6 P. 6+_Example:3 6A I1 I318AV2 I2 0.2I1_+V1 2A24 0.3I1Find: (a) the voltage drop V1.(b) the voltage drop V2. I4_+V3 ABCDEEE201 Lecture 6 P. 7Solution:Solve for currents first.I1 = 6 A (by inspection)KCL @ B to find I22A + 0.2I1 = 2A + 1.2A = I2 I2 = 3.2AKCL @ E to find I36A – 0.2I1 – 0.3I1 – I3 = 0I3 = 3AKCL @ C to find I4I2 + I3 = I4I4 = 6.2 AEE201 Lecture 6 P. 8Solution (cont):Use KVL around closed path B-C-E-B to find V1VBC + VCE + VEB = 02I2 – 3I3 – V1 = 0 = 6.4 – 9 – V1V1 = -2.6 VUse KVL around closed path E-C-D-E to find V3VEC + VCD + VDE = 0 9 + 4I4 + V3 = 0 = 9 + 24.8 + V3 V3 = -33.8 VUse KVL around closed path A-E-D-A to find V2VAE + VED + VDA = 0 V2 – V3 + 8 = 0 = V2 + 33.8 + 8 V2 = -41.8
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