How did they get their answers 1. Figure 1 shows the circuit diagram for a common-source amplifier. You may assumethat B = 0.01 A/N², V, = 0.6 V, Cs is negligibly small and VA = 20 V for thetransistor. All other values are shown beside the schematic.%3!a. Calculate the DC Ips for the circuit.b. Calculate gm and ro for the MOSFET.c. Draw the small-signal equivalent circuit for the amplifier.d. Derive formulas for each of the following:i. Voltage gain.ii. Input resistance.iii. Output resistance.e. Calculate the numerical values for each of the attributes in Part (d).VDD = 3 VRI = 4kR2 = 5kRD = 5kVDDR1 32 RDC2%3!C1M1outin HERS = 6kRext = 20kRextcsR2 3RSEovFigure 1VDD = 3 VR1 = 4kR2 = 6kRS = 1kVDDR1C1M1in HEC2outRexdR2 3RS 2ovFigure 2RD2 = 4kRS2 = 6kVDD = 2 VRD2Vss = 0 V2 RD1R13M2RI = 4kM1R2 = 4kRD1 = 4kV1VOUTRSI = 4kR2RS13 Cs1=RS23CS2VINFigure 3VDD = 2 VRS2 = 6kVDDVss =0 VRD1R1 = 4kS R1HT M2R2 = 4kRDI = 4kVinM1outRS1 = 4k%3DVoutR2{RS1 Tcs1RS2Vss = 0 VFigure 4 1. B= 0.01 A/V. v, = 0.6 V.C. =0 (so far as we care) and VA = 20 V.a Calculate the DC Iuso for the circuit.Ipsg Is the QUIESCENT los je, the DC cument in the circuit when there is no applica sigiaiTuhe circuit is resting, if you like). It is also known as the bias point or the operating point.There is no lo. therefore V, =VR.-=3*(5/(5+4)) = 1.67 Vo R +R.We also have Vs losRs and assuming the MOSFET is in saturation las =where Ves = Va - VVa-V.)This can be rearranged to give I =V.-„R, -V,)%3DSubstituting in the values and solving the quadratic equation for les givesLas - 149 LA or les = 212 HAOne of these solutions is wrong.The source voltage Vs = Iusks henceVs =0.89 V ar Vy = 1.27 VTherefore Vas = 0.77 V or Vos = 0.39 VThe second solution gives Vas - V, <0 therefore it cannot be trueCheck:for los = 149 HA we get Ves - V,= 0.17 V.and Vos = Vp - Vs = Veo - losRo - Vs =3 - 149u*5k -0.89 = 1.36 VSo Vps > Vas - V, hence device is in saturation.b. Calculate g, and r, for the MOSFET.5. =2 =2 0.01*149u =1.72 ms. = VAlles = 20/149u = 134 k2.c. Draw the small-signal equivalent circuit for the amplitier.outInOmgsSRDRext VoutROS CgstoVind.We can write down the following hy looking a the circuit and using Ohm's and Kirchult'slaws:L=-Soi. Voltage gain.- -G -.r, N, --(r, IR, IR)* v.ii. Input resistance,If the frequency is not very high then we can assume that there is very little current in C.hence i, = 0, thereforeV.R.R, = R, MR, R+ R,R, + R;iii. Output resistance.The output circuit for this amplifier is a pretty simple parallel combination of resistors. Ris ignored - it is outside the amplifier, henceR=, IR,- R,, (134k) is much higger than either Rp (Sk) or Reu (20). We can then approximate toR,R.i Voltage gainR +RG, =-,r, IR, IR)--R.(R, NR)=2-1.72m"(5k*20ISK+20k) = -6.88i R.R. 4kU5k = 2 2ki. R-R. since Ra sCL. therefore R 5k+R.

Answered: Image /qna-images/answer/6282055b-110c-402e-85d7-0bac33a3800b.jpg

1. Figure 1 shows the circuit diagram for a common-source amplifier. You may assume that B= 0.01 A/Vv², v, = 0.6 V, C is negligibly small and VA = 20 V for the transistor. All other values are shown beside the schematic. %3D a. Calculate the DC Ips for the circuit. b. Calculate gm and ro for the MOSFET. c. Draw the small-signal equivalent circuit for the amplifier. d. Derive formulas for each of the following: i. Voltage gain. ii. Input resistance. iii. Output resistance.

1. Figure 1 shows the circuit diagram for a common-source amplifier. You may assume that B= 0.01 A/Vv², v, = 0.6 V, C is negligibly small and VA = 20 V for the transistor. All other values are shown beside the schematic. %3D a. Calculate the DC Ips for the circuit. b. Calculate gm and ro for the MOSFET. c. Draw the small-signal equivalent circuit for the amplifier. d. Derive formulas for each of the following: i. Voltage gain. ii. Input resistance. iii. Output resistance.

EBK ELECTRICAL WIRING RESIDENTIAL

19th Edition

ISBN:9781337516549

Author:Simmons

Publisher:Simmons

Chapter25: Television, Telephone, And Low-voltage Signal Systems

Section25.1: Television Circuit

Problem 5R: From a cost standpoint, which system is more economical to install: a master amplifier distribution...

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