Lecture 18 OUTLINE Continue small signal analysis Logic

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Lecture #18 OUTLINE • Continue small signal analysis • Logic functions • NMOS logic

Lecture #18 OUTLINE • Continue small signal analysis • Logic functions • NMOS logic gates • The CMOS inverter Reading • Rabaey et al. : Chapter 5. 2 • Hambley: Chapter 7. 1 -7. 2 EECS 40, Spring 2004 Lecture 18, Slide 1 Prof. Sanders

Digital Signals • For a digital signal, the voltage must be within one of

Digital Signals • For a digital signal, the voltage must be within one of two ranges in order to be defined: VDD “ 1” VOH VIH undefined region VIL “ 0” VOL • Positive Logic: – “low” voltage logic state 0 0 Volts increasing voltage – “high” voltage logic state 1 EECS 40, Spring 2004 Lecture 18, Slide 2 Prof. Sanders

Logic Functions, Symbols, & Notation NAME “NOT” “OR” “AND” SYMBOL A A B EECS

Logic Functions, Symbols, & Notation NAME “NOT” “OR” “AND” SYMBOL A A B EECS 40, Spring 2004 NOTATION F F F Lecture 18, Slide 3 F=A TRUTH TABLE A F 0 1 1 0 F = A+B A B 0 0 0 1 1 F 0 1 1 1 F = A • B A B 0 0 0 1 1 F 0 0 0 1 Prof. Sanders

“NOR” “NAND” “XOR” (exclusive OR) A B A B EECS 40, Spring 2004 F

“NOR” “NAND” “XOR” (exclusive OR) A B A B EECS 40, Spring 2004 F F F Lecture 18, Slide 4 F = A+B A B 0 0 0 1 1 F 1 0 0 0 F = A • B A B 0 0 0 1 1 F 1 1 1 0 F=A+B A B 0 0 0 1 1 F 0 1 1 0 Prof. Sanders

NMOS Inverter (“NOT” Gate) Voltage-Transfer Characteristic v. OUT Circuit: VDD F A i. D

NMOS Inverter (“NOT” Gate) Voltage-Transfer Characteristic v. OUT Circuit: VDD F A i. D v. IN = VDD 0 VT VDD v. IN VDD/RD increasing v. GS = v. IN > VT 0 EECS 40, Spring 2004 v. GS = vin VT Lecture 18, Slide 5 VDD v. DS A F 0 1 1 0 Prof. Sanders

Noise Margins Definition of Input Levels Definition of Noise Margins logic swing Vsw VOL

Noise Margins Definition of Input Levels Definition of Noise Margins logic swing Vsw VOL VOH Noise margin high Noise margin low EECS 40, Spring 2004 Lecture 18, Slide 6 Prof. Sanders

NMOS NAND Gate • Output is low only if both inputs are high VDD

NMOS NAND Gate • Output is low only if both inputs are high VDD RD F A Truth Table A B 0 0 0 1 1 B EECS 40, Spring 2004 Lecture 18, Slide 7 F 1 1 1 0 Prof. Sanders

NMOS NOR Gate • Output is low if either input is high VDD RD

NMOS NOR Gate • Output is low if either input is high VDD RD F A B Truth Table A B 0 0 0 1 1 EECS 40, Spring 2004 Lecture 18, Slide 8 F 1 0 0 0 Prof. Sanders

Disadvantages of NMOS Logic Gates • Large values of RD are required in order

Disadvantages of NMOS Logic Gates • Large values of RD are required in order to – achieve a low value of VOL – keep power consumption low à Large resistors are needed, but these take up a lot of space. • One solution is to replace the resistor with an NMOSFET that is always on. EECS 40, Spring 2004 Lecture 18, Slide 9 Prof. Sanders

The CMOS Inverter: Intuitive Perspective SWITCH MODELS CIRCUIT VDD G VDD S Rp D

The CMOS Inverter: Intuitive Perspective SWITCH MODELS CIRCUIT VDD G VDD S Rp D VOUT VIN D G VOL = 0 V VOH = VDD Rn S Low static power consumption, since V one MOSFET is always off in steady state. IN EECS 40, Spring 2004 VOUT = VDD Lecture 18, Slide 10 VIN = 0 V Prof. Sanders

CMOS Inverter Voltage Transfer Characteristic N: sat P: sat VOUT N: off P: lin

CMOS Inverter Voltage Transfer Characteristic N: sat P: sat VOUT N: off P: lin VDD C N: sat P: lin A B D E N: lin P: sat N: lin P: off 0 0 EECS 40, Spring 2004 VDD Lecture 18, Slide 11 VIN Prof. Sanders

CMOS Inverter Load-Line Analysis V GS VIN = VDD + VGSp IDn=-IDp increasin g

CMOS Inverter Load-Line Analysis V GS VIN = VDD + VGSp IDn=-IDp increasin g VIN = 0 V p =V IN - V DD VOUT = VDD + VDSp + – – VDSp=VOUT-VDD + IDn=-IDp VIN = VDD increasing VIN 0 VDD 0 VDSp = VDD EECS 40, Spring 2004 VOUT=VDSn VDSp = 0 Lecture 18, Slide 12 Prof. Sanders

CMOS Inverter Load-Line Analysis: Region A V VIN VTn GS IDn=-IDp p =V IN

CMOS Inverter Load-Line Analysis: Region A V VIN VTn GS IDn=-IDp p =V IN - V DD + – – VDSp=VOUT-VDD + IDn=-IDp 0 VDD 0 EECS 40, Spring 2004 Lecture 18, Slide 13 VOUT=VDSn Prof. Sanders

CMOS Inverter Load-Line Analysis: Region B V VDD/2 > VIN > VTn GS IDn=-IDp

CMOS Inverter Load-Line Analysis: Region B V VDD/2 > VIN > VTn GS IDn=-IDp p =V IN - V DD + – – VDSp=VOUT-VDD + IDn=-IDp 0 VDD 0 EECS 40, Spring 2004 Lecture 18, Slide 14 VOUT=VDSn Prof. Sanders

CMOS Inverter Load-Line Analysis: Region D VDD – |VTp| > VIN > VDD/2 IDn=-IDp

CMOS Inverter Load-Line Analysis: Region D VDD – |VTp| > VIN > VDD/2 IDn=-IDp V GS p =V IN - V DD + – – VDSp=VOUT-VDD + IDn=-IDp 0 VDD 0 EECS 40, Spring 2004 Lecture 18, Slide 15 VOUT=VDSn Prof. Sanders

CMOS Inverter Load-Line Analysis: Region E V VIN > VDD – |VTp| GS IDn=-IDp

CMOS Inverter Load-Line Analysis: Region E V VIN > VDD – |VTp| GS IDn=-IDp p =V IN - V DD + – – VDSp=VOUT-VDD + IDn=-IDp 0 VDD 0 EECS 40, Spring 2004 Lecture 18, Slide 16 VOUT=VDSn Prof. Sanders