analog bicmos design practices and pitfalls phần 5 potx

... CC-CE, CC-CC and Darlington configurations can be better implemented as physically larger single transistor designs. 5. 5 CC-CE and CC-CC Amplifiers CC-CEandCC-CCconfigurationsareshowninFigure5.11.Notethat some ... that A V = V o V S = −g m r b R b + r b R o R l R o + R L A V = −  50 µA 26 mV  52 KΩ 62 KΩ  5 KΩ=−8.0 65 bandgap voltage. Consider N 1 andN 2 redrawninFigure4.3A.Recall...

Ngày tải lên: 14/08/2014, 04:21

... silicon. 2.4SimpleSmallSignalModelsforHand Calculations 2.4.1BipolarSmallSignalModel 2.4.2OutputImpedance 2.4.3SimpleMOSSmallSignalModel 2.5ChapterExercises 3CurrentSources 3.1CurrentMirrorsinBipolarTechnology 3.2CurrentMirrorsinMOSTechnology ChapterExercises 4VoltageReferences 4.1SimpleVoltageReferences 4.2VbeMultiplier 4.3ZenerVoltageReference 4.4TemperatureCharacteristicsofI c andV be 4.5B...

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... 100µA, or I C (N1) = 49 .50 5µA Then I C (N2) = I C (P 2)=49 .50 5µA and I C (P 1) =50 .4 95 A. The difference between V 1 and V 2 is V dif = (26mV )ln  50 .4 95 A 49 .50 5µA  =0.5mV saturated while the ... 1997. [3] Gray, Paul R., and Mayer, Robert G., Analysis and Design of Analog Integrated Circuits, 2nd edition, John Wiley and Sons, Inc., New York, c. 1984. [4] Millman, J...

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... 26 analog voice interfaces and routers, 50 51 benefits of, 26 Cisco Catalyst switches and, 53 54 digital voice interfaces and routers, 51 53 infrastructure for, 26 inline power options and, 54 56 queing ... cluster design, 119–124 Digital Signaling Processors (DSPs) and, 183–1 85 gatekeeper placement and configuration, 158 –1 65 Catalyst 350 0 Series switches, 53 , 5...

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... resistor. breaks down at 5 V and is modeled as a 5 V battery in series with a 65 OhmresistorasshowninFigure9.22. I 2 = 37.3 − 5 2 K = 15. 6 mA 15. 6 mA flows through the 2 K resistor and is absorbed by ... zener, represented by a 5 V battery and a 65 Ohm series resistance is V 3 = V MOSGAT E = 20 − 5 1 65 65+ 5=11V 11 V is less than the maximum allowed gate voltage of 12 .5...

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... B, and C in Equations 1 .56 , 1 .57 , and 1 .58 : I nc = A  e V be V T − e V bc V T  I pe = B  e V be V T − 1  (1 .59 ) I pc = C  e V bc V T − 1  Plugging Equations 1 .59 into Equations 1 .53 and ... of NMOS and PMOS transistors is useful in the design of analog and digital circuits. Figure 1. 15 CMOS structure. 1.6.1 Simple MOS Model A simple model for the MOS transistor...

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... York, 1988. [4] P.R. Gray and R.G. Meyer, Analysis and Design of Analog Inte- grated Circuits, 2nd edition, Wiley, New York, c. 1984, pp. 1 -5. [5] R.S. Muller and T.I. Kamins, Device Electronics for Integrated ... in- creases. T F References [1] P.R. Gray and R.G. Meyer, Analysis and Design of Analog Inte- grated Circuits, 2nd edition, Wiley, New York, c. 1984, pp. 1 -5....

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... the circuitshowninFigure3.12.AssumesuppliesareV DD =5V and ground. Reference current is 20µA. Output currents are to be 20µA,40µA ,55 µA, and 70µA. KP =50 µA/V 2 , L =5 m, V th =0.8V and Λ = 0.04V −1 . Determine the ... 98 .5 A − I c (Q1) β or 1.01I c =98 .5 A This gives I c (Q1)=97 .52 5µA. Using Kirchoff’s Voltage Law at the bases of Q1 and Q2, we find V be (Q2) = V be (Q1)+97 .52 5µA...

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... pair. Figure 5. 26 The transfer characteristic for the common-source amplifier showninFigure5. 25. Figure 5. 27 Small signal equivalent circuit for the common-source amplifier showninFigure5. 25. Figure 5. 35 ... Find voltage gain and output resistance. 14.ForthesourcefollowershowninFigure5.31A,R L =50 KΩ, K = 600 µA/V 2 and ∆V =0 .5 V . Find voltage gain and output resistance. As...

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... sinusoidal input of 5V (zero to peak). 9.UsethecircuitinFigure7.8withR L =20KΩ, R 1 =5KΩ and VCC =5V . Let β NPN = β PNP = 100 and I S(NPN) = 200E − 18A. What are V o(max) and V o(min) ? What ... −4.5V<V I < −4.2V . References [1] Baker, R. Jacob, et al., CMOS Circuit Design, Layout and Simu- lation, IEEE Press, New York, c. 1998. [2] Gray, Paul R., and Mayer, Robert G., Anal...

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