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1010 Chapter 15 Differential Amplifiers and Operational Amplifier Design The amplification factor of the MOSFET is given by 1 2K, (1+aVps) and solving for K, yields Ip 1+Vps 0.01 (556) 0.01 1+(11.5 V) 100 μΑ = 2.49 V2 This value of K, is achievable using either discrete components or integrated circuits. In Fig. 15.52, the required gate voltage Vg is 21p VGG = IDRS + VGs = 3.60 + VTN + Kn 2(0.2 mA) 2.49 mA = 3.60 V+1 V + = 5.00 V V2 If the current in the bias resistors is limited to 10 percent of the drain current, then 15 V 5.00 V R3 + R4 = 20 μΑ = 750 k2 and R3 = -750 k2 = 250 kN 15 V The nearest 1 percent values from Appendix A are R3 = 249 k2 and R4 = 499 k2 with Rs = 18.2 k2. The final design appears in the figure below. Check of Results: A recheck of the math indicates that our calculations are corect. SPICE can now be used to verify our design and the results appear below. Discussion: For the MOS source, we can use a larger set of gate bias resistors, since the output resistance of the current source does not depend on RGG. Computer-Aided Analysis: Now, we can check our hand design using SPICE with VTO = 1 V, KP = 2.49 mA/V², and LAMBDA = 0.01 V-'. In the circuit shown here, zero-value source Vo is added to directly measure the output current I, and to provide a source that can be used to find Rout with a SPICE Transfer Function analysis. The results using the 1 percent resistor values are Rout = 11.3 M2 with Io = 198 µA and Iss = 219 µA, which meet all the design specifications. This could be a good point to do a Monte Carlo analysis to explore the influence of tolerances on the design. Also, more complex SPICE models can be used to double check the design. lo R42 499 kn Vo=0 R3 249 kn Rs 18.2 kn -Vss =-15 V

1010 Chapter 15 Differential Amplifiers and Operational Amplifier Design The amplification factor of the MOSFET is given by 1 2K, (1+aVps) and solving for K, yields Ip 1+Vps 0.01 (556) 0.01 1+(11.5 V) 100 μΑ = 2.49 V2 This value of K, is achievable using either discrete components or integrated circuits. In Fig. 15.52, the required gate voltage Vg is 21p VGG = IDRS + VGs = 3.60 + VTN + Kn 2(0.2 mA) 2.49 mA = 3.60 V+1 V + = 5.00 V V2 If the current in the bias resistors is limited to 10 percent of the drain current, then 15 V 5.00 V R3 + R4 = 20 μΑ = 750 k2 and R3 = -750 k2 = 250 kN 15 V The nearest 1 percent values from Appendix A are R3 = 249 k2 and R4 = 499 k2 with Rs = 18.2 k2. The final design appears in the figure below. Check of Results: A recheck of the math indicates that our calculations are corect. SPICE can now be used to verify our design and the results appear below. Discussion: For the MOS source, we can use a larger set of gate bias resistors, since the output resistance of the current source does not depend on RGG. Computer-Aided Analysis: Now, we can check our hand design using SPICE with VTO = 1 V, KP = 2.49 mA/V², and LAMBDA = 0.01 V-'. In the circuit shown here, zero-value source Vo is added to directly measure the output current I, and to provide a source that can be used to find Rout with a SPICE Transfer Function analysis. The results using the 1 percent resistor values are Rout = 11.3 M2 with Io = 198 µA and Iss = 219 µA, which meet all the design specifications. This could be a good point to do a Monte Carlo analysis to explore the influence of tolerances on the design. Also, more complex SPICE models can be used to double check the design. lo R42 499 kn Vo=0 R3 249 kn Rs 18.2 kn -Vss =-15 V

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
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What is the minimum drain voltage for which MOSFET Min the circuit on page 1010 remains saturated?

1010 Chapter 15 Differential Amplifiers and Operational Amplifier Design The amplification factor of the MOSFET is given by 1 2K, (1+aVps) and solving for K, yields Ip 1+Vps 0.01 (556) 0.01 1+(11.5 V) 100 μΑ = 2.49 V2 This value of K, is achievable using either discrete components or integrated circuits. In Fig. 15.52, the required gate voltage Vg is 21p VGG = IDRS + VGs = 3.60 + VTN + Kn 2(0.2 mA) 2.49 mA = 3.60 V+1 V + = 5.00 V V2 If the current in the bias resistors is limited to 10 percent of the drain current, then 15 V 5.00 V R3 + R4 = 20 μΑ = 750 k2 and R3 = -750 k2 = 250 kN 15 V The nearest 1 percent values from Appendix A are R3 = 249 k2 and R4 = 499 k2 with Rs = 18.2 k2. The final design appears in the figure below. Check of Results: A recheck of the math indicates that our calculations are corect. SPICE can now be used to verify our design and the results appear below. Discussion: For the MOS source, we can use a larger set of gate bias resistors, since the output resistance of the current source does not depend on RGG. Computer-Aided Analysis: Now, we can check our hand design using SPICE with VTO = 1 V, KP = 2.49 mA/V², and LAMBDA = 0.01 V-'. In the circuit shown here, zero-value source Vo is added to directly measure the output current I, and to provide a source that can be used to find Rout with a SPICE Transfer Function analysis. The results using the 1 percent resistor values are Rout = 11.3 M2 with Io = 198 µA and Iss = 219 µA, which meet all the design specifications. This could be a good point to do a Monte Carlo analysis to explore the influence of tolerances on the design. Also, more complex SPICE models can be used to double check the design. lo R42 499 kn Vo=0 R3 249 kn Rs 18.2 kn -Vss =-15 V
Transcribed Image Text:1010 Chapter 15 Differential Amplifiers and Operational Amplifier Design The amplification factor of the MOSFET is given by 1 2K, (1+aVps) and solving for K, yields Ip 1+Vps 0.01 (556) 0.01 1+(11.5 V) 100 μΑ = 2.49 V2 This value of K, is achievable using either discrete components or integrated circuits. In Fig. 15.52, the required gate voltage Vg is 21p VGG = IDRS + VGs = 3.60 + VTN + Kn 2(0.2 mA) 2.49 mA = 3.60 V+1 V + = 5.00 V V2 If the current in the bias resistors is limited to 10 percent of the drain current, then 15 V 5.00 V R3 + R4 = 20 μΑ = 750 k2 and R3 = -750 k2 = 250 kN 15 V The nearest 1 percent values from Appendix A are R3 = 249 k2 and R4 = 499 k2 with Rs = 18.2 k2. The final design appears in the figure below. Check of Results: A recheck of the math indicates that our calculations are corect. SPICE can now be used to verify our design and the results appear below. Discussion: For the MOS source, we can use a larger set of gate bias resistors, since the output resistance of the current source does not depend on RGG. Computer-Aided Analysis: Now, we can check our hand design using SPICE with VTO = 1 V, KP = 2.49 mA/V², and LAMBDA = 0.01 V-'. In the circuit shown here, zero-value source Vo is added to directly measure the output current I, and to provide a source that can be used to find Rout with a SPICE Transfer Function analysis. The results using the 1 percent resistor values are Rout = 11.3 M2 with Io = 198 µA and Iss = 219 µA, which meet all the design specifications. This could be a good point to do a Monte Carlo analysis to explore the influence of tolerances on the design. Also, more complex SPICE models can be used to double check the design. lo R42 499 kn Vo=0 R3 249 kn Rs 18.2 kn -Vss =-15 V
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