440_hmk2_soln

440_hmk2_soln - ECE 440 Electronics II Solutions for...

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Unformatted text preview: ECE 440 Electronics II Solutions for Homework #2 1. a. CMRR = 100/0.1 = 1000 = 60dB b. v d = v i1 - v i2 = 10sin t v i1 + v i2 = 5sin t 2 v o = Ad v d + Ac v cm = 100(10sin t) + 0.1(5sin t) = 1000.5sin t v cm = c. v d = v i1 - v i2 = 10sin t v +v v cm = i1 i2 = 5sin t + sin1000t 2 v o = Ad v d + Ac v cm = 100(10sin t) + 0.1(5sin t + sin1000t) = 1000.5sin t + 0.1sin1000t 2. a. with the two inputs grounded, the voltage at the emitters is zero, thus: -0.7 - (-15) I20K = = 0.715mA 20K I Ic1 = Ic 2 = 20K = 0.358mA 2 Vce1 = Vce 2 = 30 -10K(0.358m) - 20K(0.715m) = 12.1V b. Using the collector current found above: gm = .358 150 = 0.0138 r = = 10893 . For 26 0.0138 the difference mode there is no voltage across the 20K, thus we assume the emitters are v 2 R grounded so: v o = -Rc iB1; iB1 = d Ad = - c = -69 r 2r c. For the common mode we can split the circuit with 40K in the emitter, and: v cm Rc v o = Rc iB1; iB1 = - Ac = - = -0.248 r + ( + 1) 40K r + ( + 1) 40K d. CMRR = 69 = 276 = 48.8dB 0.25 3. a. with the two inputs grounded, the voltage at the emitters is zero, thus: 15 - 0.7 Ic1 = Ic 2 = = 0.711mA 100 + 2(10K) Vce1 = Vce 2 = 30 - 8K(0.711m) - 2(10K)(0.711m) -100(0.711m) = 10V b. Using the collector current found above: gm = .711 150 = 0.0274 r = = 5481. For 26 0.0274 the difference mode there is no voltage across the 10K, thus we assume the emitters are vd 2 Rc grounded so: v o = -Rc iB1; iB1 = Ad = - = -29.2 r + ( + 1) re 2[ r + ( + 1) re ] c. For the common mode we can split the circuit with 20K in the emitter, and: v cm Rc v o = Rc iB1; iB1 = - Ac = - = -0.395 r + ( + 1)[100 + 2(10K)] r + ( + 1)[100 + 2(10K)] d. CMRR = 29.2 = 73.9 = 37.4dB 0.0395 4. a. IQ = ID1 + ID2 = 1mA, vo = 10 RD (0.5mA) = 7 RD = 6K b. 0.5m = 0.4m(VGS - 2) VGS = 3.118 Vs = 0 - VGS = -3.118 VDS = 7 - (-3.118) = 10.118V 2 c. the device transitions out of saturation when VDS = VDS,sat = VGS VTN = 3.118 2 = 1.118. If VDS = 1.118 = 7 Vs Vs = 5.882 v1 = vcm = VGS + VS = 3.118 + 5.882 = 9V. If it is larger than 9V the devices will be in the triode region d. Using the drain current specified: gm = .5 = 0.0192 .this is wrong use MOSFET 26 formula For the difference mode there is no voltage across the 40K, thus we assume the v g R sources are grounded so: v o = -gm RD vGS ; vGS = d Ad = - m D = -57.7 For the 2 2 common mode we can split the circuit with 80K in the source, and: 6Kgm v o = -6Kgm vGS ; vGS = v cm - 80Kgm vGS vGS (1+ 80Kgm ) = v cm Ac = - = -0.075 1+ 80Kgm 57.7 CMRR = = 770 = 57.7dB 0.075 e. The difference mode gain is unchanged. The common mode gain is: 6Kgm 57.7 Ac = - = -0.00375 CMRR = = 15391 = 83.7dB 1+ 160Kgm 0.00375 f. For the MOSFETs, ro = 1/(.5m)(.1) = 20K. For the difference mode gain, since the source is grounded, if we use the ac model of the MOSFET we find that the resistor ro is in parallel with the 6K resistor. Thus, Ad = - gm (6K || ro ) = -44.3 For the common 2 mode gain, using the ac model of the MOSFET we have: 6KiD + 20K (iD - gm vGS ) + 160KiD = 0; vGS = v cm -160KiD vo 6K(20K)gm Ac = =- = .0992 v cm 6K + 20K + 20Kgm160K + 160K 44.3 CMRR = = 446 = 53dB .0992 v o = -6KiD ; 5. a. ****BIPOLARJUNCTIONTRANSISTORS NAMEQ_Q1Q_Q2 MODELQbreaknQbreakn IB2.37E062.37E06 IC3.55E043.55E04 VBE6.88E016.88E01 VBC1.14E+011.14E+01 VCE1.21E+011.21E+01 BETADC1.50E+021.50E+02 GM1.37E021.37E02 RPI1.09E+041.09E+04 b. Usingthemarkerwedeterminethattheoutputvoltageis12.12110.766=1.355Vpp or0.68Vpeak.Inthiscasevd=.01andvc=.005,thusvoshouldbe(69).01+(.248).005 =0.69. c. Usingthemarkerwedeterminethattheoutputvoltageis11.44811.443=0.005Vpp or0.0025Vpeak.Inthiscasevd=0andvc=.01,thusvoshouldbe(.248).01=0.00248. d. Notethatonlythe500Hz(differencesignal)hasanysignificantoutput.Thisiswhatwe expectfromadifferentialamplifier. ...
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