Lect2UP150_(100325) - Lecture 150 Resistor Implementations...

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Lecture 150 – Resistor Implementations and Current Sinks and Sources (3/25/10) Page 150-1 CMOS Analog Circuit Design © P.E. Allen - 201 LECTURE 150 – RESISTOR IMPLEMENTATIONS AND CURRENT SINKS AND SOURCES LECTURE ORGANIZATION Outline • Resistor implementations • Simple current sinks and sources • Improved performance current sinks and sources • Summary CMOS Analog Circuit Design, 2 nd Edition Reference Pages 126-134 Lecture 150 – Resistor Implementations and Current Sinks and Sources (3/25/10) Page 150-2 CMOS Analog Circuit Design © P.E. Allen - 201 RESISTOR IMPLEMENTATION USING MOSFETS Real Resistors versus MOSFET Resistors • Smaller in area than actual resistors • Can pass a large current through a large resistance without a large voltage drop 060526-10 i D v DS 100μA 1V MOSFET ( r ds = 100k Ω ) 100k Ω Resistor 10μA 10V AC resistance = v ds i d = 1 g ds where g ds ± ² 2 ( V GS - V T ) 2 ³ = I D
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Lecture 150 – Resistor Implementations and Current Sinks and Sources (3/25/10) Page 150-3 CMOS Analog Circuit Design © P.E. Allen - 201 MOS Diode as a Resistor AC and DC resistance: DC resistance = V DS I D = V T I D + 2 ± I D Small-Signal Load (AC resistance): 060526-11 r ds D = G S S G S g m v gs v gs + - v ds + - i d D S D = G AC resistance = v ds i d = 1 g m + g ds ² 1 g m where g m = ( V GS - V T ) = 2 I D Lecture 150 – Resistor Implementations and Current Sinks and Sources (3/25/10) Page 150-4 CMOS Analog Circuit Design © P.E. Allen - 201 Use of the MOSFET to Implement a Floating Resistor In many applications, it is useful to implement a resistance using a MOSFET. First, consider the simple, single MOSFET implementation. R AB = L K’W ( V GS - V T ) V Bias A B A B Fig. 4.2-9 R AB 100 μ A 60 μ A 20 μ A -20 μ A -60 μ A -100 μ A -1V -0.6V -0.2V 0.2V 0.6V 1V V GS =2V V GS =3V V GS =4V V GS =5V V GS =10V V GS =9V V GS =8V V GS =7V V GS =6V Fig. 4.2-95
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Lecture 150 – Resistor Implementations and Current Sinks and Sources (3/25/10) Page 150-5 CMOS Analog Circuit Design © P.E. Allen - 201 Cancellation of Second-Order Voltage Dependence – Parallel MOSFETs Circuit: Assume both devices are non-saturated i D 1 = ß 1 ± ² ² ³ ´ µ µ ( v AB + V C - V T ) v AB - v 2 2 i D 2 = ß 2 ± ² ² ³ ´ µ µ ( V C - V T ) v - v 2 2 i = i D 1 + i D 2 = ß ± ² ² ³ ´ µ µ v 2 + ( V C - V T ) v - v 2 2 + ( V C - V T ) v - v 2 2 i = 2 ß ( V C - V T ) v ± R = 1 2 ß ( V C - V T ) 060526-12 M1 M2 V C A B A B R AB + - v AB i AB i AB +- v AB V C Lecture 150 – Resistor Implementations and Current Sinks and Sources (3/25/10) Page 150-6 CMOS Analog Circuit Design © P.E. Allen - 201 Parallel MOSFET Performance Voltage-Current Characteristic: Vc=7V VDS 0 I(VSENSE) -2 -1 0 1 2 2mA 1mA -1mA -2mA W=15u L=3u VBS=-5.0V 6V 5V 4V 3V Fig. 4.1-11 SPICE Input File: NMOS parallel transistor realization M1 2 1 0 5 MNMOS W=15U L=3U M2 2 4 0 5 MNMOS W=15U L=3U .MODEL MNMOS NMOS VTO=0.75, KP=25U, +LAMBDA=0.01, GAMMA=0.8 PHI=0.6 VC 1 2 E1 4 0 1 2 1.0 VSENSE 10 2 DC 0 VDS 10 0 VSS 5 0 DC -5 .DC VDS -2.0 2.0 .2 VC 3 7 1 .PRINT DC I(VSENSE) .PROBE .END
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Lecture 150 – Resistor Implementations and Current Sinks and Sources (3/25/10) Page 150-7 CMOS Analog Circuit Design © P.E. Allen - 201 SIMPLE CURRENT SINKS AND SOURCES Ideal Current Sinks and Sources What is an ideal current sink or source?
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Lect2UP150_(100325) - Lecture 150 Resistor Implementations...

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