Explain the answers 7.We use the relationshipWB= kLIf W is doubled, then so is B, thus for this circuit we have:B = 0.02 A/V?, V = 0.6 V, and VA = 20 V.Calculate the DC Ipso for the circuit.R,=3*(5/(5+4)) = 1.67 VR, + R,There is no IG, therefore Ve =Vp.Va -v.We also have Vs = IpsRs and assuming the MOSFET is in saturation Ips =where VGs = VG - Vs.This can be rearranged to give Ips =(Vc - IpgRs –V,)'2Substituting in the values and solving the quadratic equation for Ips givesIps = 156 µA or Ips = 201 µAOne of these solutions is wrong.….The source voltage Vs = IpsRs henceVs = 0.94 V or Vs = 1.21 VTherefore VGS = 0.72 V or VGs = 0.46 VThe second solution gives VGs - V, < 0 therefore it cannot be trueCheck:for Ips = 156 µA we get VGs - V = 0.12 V,and Vps = VD - Vs = VDD - IDSRD - Vs = 3 - 156µ*5k – 0.89 = 1.14 VSo Vps > VGS - V, hence device is in saturation.a. Calculate gm and r, for the MOSFET.8m = 2B1 ps = /2 *0.02 *156µ = 2.5 mSro = VA/IDs = 20/156µ = 127 kN. 7. The circuit in Figure 1 is rebuilt but the MOSFET is replaced with one that has twicethe width of the one used in Q1. Calculate the DC Ips for the new circuit and the newsmall-signal parameters.the voltagé gainVDD = 3 VR1 = 4kVDDR1 22 RDR2 = 5kC2RD = 5kRS = 6kC1M1outin HERext = 20kRextCSR2RS:ovFigure 1VDDVDD = 3 VR1 = 4kR2 = 6kRS = 1kR1C1M1in HEC2outRexdR2 3RS2ovFigure 2VDD = 2 VRD2 = 4k2 RD2VSS = 0 VRI = 4kR2 = 4kRD1RS2 = 6kR13M2M1RD1 = 4kVOUTV1RS1 = 4kR2RS13 CS1=RS23Cs2VINFigure 3VDD = 2 VRS2 = 6kVDDVsS = 0 VRI = 4kRD1R1R2 =4kRD1 = 4kM2VinHT M1butRSI = 4kVoutR2{ RS1 Tcs RS2VSS = 0 VFigure 4

Given : Beta =0.02 A / V² Vt = 0.6 V VA = 20 V IDS = Drain To Source Current…

7. We use the relationship W B= k= If W is doubled, then so is B, thus for this circuit we have: B = 0.02 A/V, V = 0.6 V, and VA = 20 V. %3D Calculate the DC Ipso for the circuit. R, There is no IG, therefore V =VpD -=3*(5/(5+4)) = 1.67 V R, + R, We also have Vs = IpsRs and assuming the MOSFET is in saturation Ips = (Ves -V,) GS where VGs = VG- Vs. This can be rearranged to give Ips =(Vc- InsRs-V,) Substituting in the values and solving the quadratic equation for Ips gives Ips = 156 µA or Ips = 201 µA %3D One of these solutions is wrong... The source voltage Vs = IpsRs hence Vs = 0.94 V or Vs = 1.21 V Therefore VGS = 0.72 V or VGS = 0.46 V %3D The second solution gives VGS - V <0 therefore it cannot be true Check: for Ips = 156 µA we get VGS - V = 0.12 V, and Vps = VD - Vs = VDD - IDSRD - Vs = 3 - 156µ*5k – 0.89 = 1.14 V %3D %3D So Vps > VGS - V, hence device is in saturation. ond r for the MOSFET.

7. We use the relationship W B= k= If W is doubled, then so is B, thus for this circuit we have: B = 0.02 A/V, V = 0.6 V, and VA = 20 V. %3D Calculate the DC Ipso for the circuit. R, There is no IG, therefore V =VpD -=3(5/(5+4)) = 1.67 V R, + R, We also have Vs = IpsRs and assuming the MOSFET is in saturation Ips = (Ves -V,) GS where VGs = VG- Vs. This can be rearranged to give Ips =(Vc- InsRs-V,) Substituting in the values and solving the quadratic equation for Ips gives Ips = 156 µA or Ips = 201 µA %3D One of these solutions is wrong... The source voltage Vs = IpsRs hence Vs = 0.94 V or Vs = 1.21 V Therefore VGS = 0.72 V or VGS = 0.46 V %3D The second solution gives VGS - V <0 therefore it cannot be true Check: for Ips = 156 µA we get VGS - V = 0.12 V, and Vps = VD - Vs = VDD - IDSRD - Vs = 3 - 156µ5k – 0.89 = 1.14 V %3D %3D So Vps > VGS - V, hence device is in saturation. ond r for the MOSFET.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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