Explain simply what they did below? Figure I shows a complete op-amp circuit with an output load RL. The design has thefollowing properties.VDD = 5 V, VSS = 0 V and VR = 2.5 V.L for all transistors (NMOS and PMOS) = 1 µm.For the NMOS B= 0.002 A/V² when W = 10 um.For the PMOS B= 0.002 A/V² when W = 30 µm.The NMOS threshold voltage is 0.5 V and the PMOS threshold voltage is -0.5 V.The bias current in RB = 100 µA.You may assume that all the transistors are in saturation and that the Early voltage is 20 V.VDDM14M15에 M3예 M7M11M12M16M17M10CMvi1 d M1M18M2 v2M19M9RLMBM13RBMAM5MBVRVSSFigure 1. A three-stage op-amp with bias reference and Miller compensation.Question 1. DC bias calculations.300 um. What will their respective VGs be at quiescence andj. W12 = 100 µm and W13the resulting Ips in each case?k. What is the quiescent power dissipation of the amplifier? 2*10021 DS9V GS9+V,+ 0.5 = 0.72 V.%3D0.00421 DS102*100u+V,-0.5 = -0.72 V.VGS10 =0.004()We observe the following:W13 = 300 µm and W12 = 100 µm hence their B is the same and = 0.02 A/V².The voltage difference between the gates of M12 and M13 isVGS12- VGS13 = VG89 – VGS10 = 0.72 + 0.72 V = 1.44 V.The circuit is well matched so we can expect that the 1.44 V will be evenly distributedbetween M12 and M13. Also, since we expect vo = VR at quiescence we can say that there isno current in Rị and VGs12 = 0.72V and VGS13 = -0.72 V.Because there is no current in RL we can say that IDs12 = -IDS13-The quiescent current in the output stage at quiescence is therefore:Ips = 0.5B(VGs - V² = 0.5*0.02*(0.72-0.5) = 484 µA.Because the output stage is biased so as to be always on the amplifier is operating as a classАB type.%3D(k)The total quiescent current is:Ips14 + Ips15 + Ips3 + Ips7 + IpsSii + Ips12100μ+100μ+200μ+ 1 00μ+ 100μ+484μ = 1084 μΑ.%3DThe quiescent power dissipation is therefore 5*1.084m = 5.4 mW.

Here it is asked to explain the given solution. The given solution is: We have corrected the…

2 *100u 21 DS9 +V V Gs9 + 0.5 = 0.72 V. 0.004 2*100u 0.5 = -0.72 V. DS10 V GS10 +V, %3D 0.004 (j) We observe the following: W13 = 300 µm and W12 = 100 µm hence their B is the same and = 0.02 A/V². The voltage difference between the gates of M12 and M13 is VGS12 - VGS13 = VGs9 – VGS10 = 0.72 + 0.72 V = 1.44 V. %3D The circuit is well matched so we can expect that the 1.44 V will be evenly distributed between M12 and M13. Also, since we expect vo = VR at quiescence we can say that there is no current in RL and VGS12 = 0.72V and VGS13 = -0.72 V. %3D %3D Because there is no current in RL we can say that Ips12 = -IDS13. The quiescent current in the output stage at quiescence is therefore: Ips = 0.5B(VGs - V² = 0.5*0.02*(0.72-0.5) = 484 µA. %3D Because the output stage is biased so as to be always on the amplifier is operating as a class AB type. %3D (k) The total quiescent current is: Ips14 + Ips15 + Ips3 + Ips7 + Ips11 + Ips12 %3D = 100µ+100µ+200µ+100µ+100µ+484µ = 1084 µA. The quiescent power dissipation is therefore 5*1.084m = 5.4 mW.:

2 100u 21 DS9 +V V Gs9 + 0.5 = 0.72 V. 0.004 2100u 0.5 = -0.72 V. DS10 V GS10 +V, %3D 0.004 (j) We observe the following: W13 = 300 µm and W12 = 100 µm hence their B is the same and = 0.02 A/V². The voltage difference between the gates of M12 and M13 is VGS12 - VGS13 = VGs9 – VGS10 = 0.72 + 0.72 V = 1.44 V. %3D The circuit is well matched so we can expect that the 1.44 V will be evenly distributed between M12 and M13. Also, since we expect vo = VR at quiescence we can say that there is no current in RL and VGS12 = 0.72V and VGS13 = -0.72 V. %3D %3D Because there is no current in RL we can say that Ips12 = -IDS13. The quiescent current in the output stage at quiescence is therefore: Ips = 0.5B(VGs - V² = 0.50.02(0.72-0.5) = 484 µA. %3D Because the output stage is biased so as to be always on the amplifier is operating as a class AB type. %3D (k) The total quiescent current is: Ips14 + Ips15 + Ips3 + Ips7 + Ips11 + Ips12 %3D = 100µ+100µ+200µ+100µ+100µ+484µ = 1084 µA. The quiescent power dissipation is therefore 5*1.084m = 5.4 mW.:

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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