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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 - 156µ*5k – 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)
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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7. We use the relationship W B= k L 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. Calculate the DC Ipso for the circuit. R, =3*(5/(5+4)) = 1.67 V R, + 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,)' 2 Substituting in the values and solving the quadratic equation for Ips gives Ips = 156 µA or Ips = 201 µA 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 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 So 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 mS ro = VA/IDs = 20/156µ = 127 kN.
Transcribed Image Text:7. We use the relationship W B= k L 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. Calculate the DC Ipso for the circuit. R, =3*(5/(5+4)) = 1.67 V R, + 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,)' 2 Substituting in the values and solving the quadratic equation for Ips gives Ips = 156 µA or Ips = 201 µA 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 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 So 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 mS ro = VA/IDs = 20/156µ = 127 kN.
7. The circuit in Figure 1 is rebuilt but the MOSFET is replaced with one that has twice the width of the one used in Q1. Calculate the DC Ips for the new circuit and the new small-signal parameters. the voltagé gain VDD = 3 V R1 = 4k VDD R1 2 2 RD R2 = 5k C2 RD = 5k RS = 6k C1 M1 out in HE Rext = 20k Rext CS R2 RS: ov Figure 1 VDD VDD = 3 V R1 = 4k R2 = 6k RS = 1k R1 C1 M1 in HE C2 out Rexd R2 3 RS2 ov Figure 2 VDD = 2 V RD2 = 4k 2 RD2 VSS = 0 V RI = 4k R2 = 4k RD1 RS2 = 6k R13 M2 M1 RD1 = 4k VOUT V1 RS1 = 4k R2 RS13 CS1= RS23 Cs2 VIN Figure 3 VDD = 2 V RS2 = 6k VDD VsS = 0 V RI = 4k RD1 R1 R2 =4k RD1 = 4k M2 Vin HT M1 but RSI = 4k Vout R2 { RS1 Tcs RS2 VSS = 0 V Figure 4
Transcribed Image Text:7. The circuit in Figure 1 is rebuilt but the MOSFET is replaced with one that has twice the width of the one used in Q1. Calculate the DC Ips for the new circuit and the new small-signal parameters. the voltagé gain VDD = 3 V R1 = 4k VDD R1 2 2 RD R2 = 5k C2 RD = 5k RS = 6k C1 M1 out in HE Rext = 20k Rext CS R2 RS: ov Figure 1 VDD VDD = 3 V R1 = 4k R2 = 6k RS = 1k R1 C1 M1 in HE C2 out Rexd R2 3 RS2 ov Figure 2 VDD = 2 V RD2 = 4k 2 RD2 VSS = 0 V RI = 4k R2 = 4k RD1 RS2 = 6k R13 M2 M1 RD1 = 4k VOUT V1 RS1 = 4k R2 RS13 CS1= RS23 Cs2 VIN Figure 3 VDD = 2 V RS2 = 6k VDD VsS = 0 V RI = 4k RD1 R1 R2 =4k RD1 = 4k M2 Vin HT M1 but RSI = 4k Vout R2 { RS1 Tcs RS2 VSS = 0 V Figure 4
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