Analog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedAndrew R. NeureutherEECS 105 Microelectronics Devices and Circuits, Spring 2001Topics: Value Added by CircuitsLecture 2, February 19, 2001Revised 1/21/01Reading: (review of EE 40), HS 1, 8.2.2, 9.1Analog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedHow Does a Digital Camera Work?l Physics (semiconductor junction)» Photons => charge => voltagel Analog Circuits» Amplify, gray level conversionl Digital Circuits» Encode, store, move, playl Analog Circuits» Display driversAnalog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedModel for Photo Detectorl Film sensitivity ~ 3x104photons ∆∆QS= 3x104electronsl Junction capacitance CJ~ 30 fFl ∆∆VS= ∆∆Q/CJ= 3x104x1.6x10-19/3x10-14∆∆VS= 160 mVl Series resistance RS= 200 Ohms∆Q∆QCJVINRSVSOURCE= VBIAS+ ∆∆VS= VBIAS+ ∆∆Q/CJ+-Analog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedRFiRFVOUT∆Q∆QCJVIN-VRRSCurrent to Voltage Conversionl Op-Amp CircuitVOUT= VR-RF{(VBIAS+∆VS-VR)/RS}CurrentRemove BiasΣIi= 0KCL =>Vs+-Analog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedOp-Amps are Ideal but EE 105 is Notl Ideal Op-Amp properties» No input current (Infinite RIN!)» V-= V+(Infinite voltage gain! With feedback)l Circuit configurations give the leverage to build nearly ideal circuits from devices with less than ideal properties.Don’t forget about Op-Amps from EE 40 as in EE 105 we will use Op-Amps to study circuit concepts like frequency-response.How do we get these?Topic for EE 140Analog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedBack to the Futurel 3 MegaPixels with 3 colors requires nearly 10M Op-Amps.l If each draws 100 µµA, the battery must supply 1000A. l Solution: Analog switch array of 10 levels and 210–1024 factor of sharing.A car battery would last only 3 minutes!1024 PhotodiodesResistance and Capacitance of 10 analog switches in seriesRSA = 10*10 kΩΩ = 100 kΩΩCSA= 10*30fF = 300 fFAmplifierAnalog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedRFiRFVOUT∆Q∆QCJVIN-VRRSRSAModel For Switching and AmplifierVOUT= VR- RF{(VBIAS+∆V’S-VR)/(RS+RSA)}From 200 Ohms to 100,200 Ohms => 500X smaller signal!∆V’Sis 10 times smaller due to CSA and is now about 15 mV∆V’SCSAApproximate effect of CSAby combining it with CJ.Analog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedSimple EE 105 Amplifier∆Q∆QCJVINRSRSARINiINβiINROUTRLOADRIN= 30 kΩ β = 100 ROUT=Infinite RLOAD= 10 kΩVOUT= [∆V’S/(RS+RSA+RIN)](-β)RLOAD =115mVΣVi= 0 => iIN∆V’S = 15 mVΣIi= 0 with ROUT=Infinite =>NodeLoop∆V’SVOUTAnalog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedCircuit to Hold the Charge Longerl Problem CJDischarges Quickly T = CJ*RSA= 30fF * 100kΩ = 3*10-10Secl Solution: Add Value Through Circuit Design of High Input Resistance AmplifierAnalog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value Added∆Q∆QCJVINRSRSARINiINβiINROUTRLOADREHigh Input Impedance CircuitRE= 30 kΩ RIN EQ= RIN+(β+1)RE = 3.06 MΩ23.5 times less currentVOUT= [∆V’S/(RS+RSA+RIN EQ)](-β)RLOAD =5 mV23.5 times smaller gainNote ground connectionsΣVi= 0 => iINLoopNode∆V’SVOUTAnalog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedAdding a Second StageRSARINiINβiINROUTRE∆Q∆QCJVINRSRINiINβiINROUTRLOADSwitch First Stage Second StageVOUT= [∆V’S/(RS+RSA+RIN EQ1)](-β1)Rin2(1/Rin2 )(- β2)ROUT= 490 mVNodeNodeLoop∆V’SVOUTAnalog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedVisualizing as a Multistage AmplifierRINiINβiINROUTRINiINβiINROUTSource Switch First Stage Second StageRIN= RSAAI= 1, AO= 1ROUT= RSARIN= 30kΩβ = 100ROUT= infiniteRIN= 3.06MΩβ = 100ROUT= infinite∆∆QCJRS∆VSRLOADVOUT-1I1V2I2+1V2RSA+_V1I1Error fixed 1/21/00Analog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedVisualizing as an Equivalent Two-PortRIN= RSA+ RIN EQ1GM= [1/(RS+RSA+RIN EQ1)](-β1)Rin2 (1/Rin2 )(- β2)ROUT= Infinite (Assumed Initially)∆∆QCJRS∆VSRLOADVOUTVINRINGMVINROUTAnalog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedMultistage AmplifiersThis example from the reading in Chapter 8 this week.Analog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedClassification of Two-PortAmplifiersVoltageCurrentTransconductanceTransresistanceAnalog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedWhat Goes in the Amplifier BoxMaterial from Chapter 8 from week 11.MOSBJTAnalog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedSmall Signal Models for TransistorsMOSBJTWeek 5 Week 8Analog Integrated CircuitsOverview and Circuit Value AddedOverview and Circuit Value AddedLayout of TransistorsWeek 2 Week
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