MOS Inverters Dr. Lynn Fuller - RIT - People · PDF file© April 3, 2017 Dr. Lynn Fuller...

© April 3, 2017 Dr. Lynn Fuller

MOS Inverters

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Rochester Institute of Technology

Microelectronic Engineering

ROCHESTER INSTITUTE OF TECHNOLOGYMICROELECTRONIC ENGINEERING

MOS Inverters

Dr. Lynn FullerWebpage: http://people.rit.edu/lffeee

Microelectronic EngineeringRochester Institute of Technology

82 Lomb Memorial DriveRochester, NY 14623-5604

Email: [email protected] webpage: http://www.rit.edu/kgcoe/microelectronic/

4-3-2017 MOS-Inverters.ppt

http://people.rit.edu/~lffeee

mailto:[email protected]

http://www.rit.edu/kgcoe/microelectronic/


MOS Inverters

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OUTLINE

IntroductionVoltage Transfer Curve (VTC)Noise MarginInverter Current vs. VinPMOS InverterNMOS InverterCMOS InverterPseudo CMOS ReferencesHomework


MOS Inverters

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INTRODUCTION

There are many ways to make an inverter. In this document we will investigate various MOS inverters, their voltage transfer curve, current, noise margin, speed etc. The inverter is the simplest logic gate to analyze and can give useful results for the comparison of different inverter designs and fabrication technologies.

SYMBOL TRUTH TABLE

VIN VOUT VOUTVIN

0 1

1 0


MOS Inverters

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INVERTER TYPES - VOUT VS VIN (VTC)

VIN

CMOS

+V

VO

SWITCH

VIN

+V

NMOS

ENHANCEMENT

LOAD

+V

VIN

VO

-V+V

+V0

0

+V0

0+V0

0

-V0

0

+V0

0

+V+V +V

VO

RESISTOR

LOAD

VIN

+V

VO

R

NMOS

V++

ENHANCEMENT

LOAD

+VV++

VIN

VO


MOS Inverters

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MORE INVERTER TYPES - VOUT VS VIN (VTC)

PMOS

ENHANCEMENT

LOAD

-V

VIN

VO

-V

-V0

0

+V0

0

+V

NMOS

ION IMPLANT

DEPLETION

LOAD

+V

VIN

VO


MOS Inverters

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VOLTAGE TRANSFER CURVE NMOS-RESISTOR LOAD

VIN VOUT

+V0

0

+VDD

ViL

Voh

VoL

Vih

VOUT

VIN

Idd

NML, noise margin low, D0 =ViL-VoLNMH, noise margin high, D1 =VoH-ViH

Slope = Gain

VinvRESISTOR LOAD

VIN

+VDD

VOUT

R

NMOS-M1


MOS Inverters

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CALCULATION OF VTC

RESISTOR LOAD

VIN

+VDD

VOUT

R

NMOS-M1

First figure out if the transistor is sub-threshold or off, Vgs < Vth and Vgd < Vth non-saturation, Vgs > Vth and Vgd > Vth saturation region, Vgs > Vth and Vgd < Vth

Note: Vin = Vgs, Vout = Vds, therefore Vgd = Vin-VoutVth might be +1volt

+V00

+V

Vth

VOUT

VIN

M1

Off

M1

Saturation

M1

Linear


MOS Inverters

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ID = µW Cox’ (Vg-Vt-Vd/2)Vd

L

CALCULATION OF VTC

Next calculate Vout = VDD – ID R using the correct equation for ID for the transistor depending on region of operation

+V00

+V

Vth

VOUT

VIN

M1

Off

M1

Saturation

M1

Linear

RESISTOR LOAD

VIN

+VDD

VOUT

R

NMOS-M1

ID

ID = µW Cox’ (Vg-Vt)2

2L

SaturationLinear (Non-Saturation)

Cox’ = Cox/Area = or/Xox


MOS Inverters

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CALCULATION OF VTC

M1 in Saturation Vout = VDD – R ID =


o = 8.85e-14 F/cm

r=3.9 for oxide

Xox = gate oxide thickness

W= width of MOSFET

L=Length of MOSFET

Vt = Threshold Voltage

V - R µW Cox’ (Vin-Vt)2

2L


MOS Inverters

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CALCULATION OF VTC

M1 in Non-Saturation Vout = VDD –ID R

Vout = VDD –ID R = VDD - R µW Cox’ (Vg-Vt-Vd/2)Vd

LKx

Vo = VDD - R Kx(Vin-Vt-Vo/2)Vo

Vo = VDD - R Kx(Vin-Vt)Vo- RKxVo2/2

0 = VDD - R Kx(Vin-Vt - 1)Vo- RKxVo2/2

a x2 + bx + c = 0

x = -b +/- b2-4ac

quadratic formula

2a


MOS Inverters

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CALCULATION OF VTC

M1 in Non-Saturation Vout = b +/- b2 -2VDD/KxR

b=(Vin – Vt + 1/KxR) 2a


MOS Inverters

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CALCULATION OF VTC

Note: Equations only valid in specific regions


MOS Inverters

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LTSPICE – RESISTOR LOAD INVERTER - VTC

Vout1 Id

Vout vs Vin – Voltage Transfer Curve (VTC)


MOS Inverters

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LTSPICE - INVERTER VTC – FOR DIFFERENT RL

R=1K

5K10k


MOS Inverters

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CADENCE PARAMETER SWEEPS

Select Parameters: from the “special” library and put on schematic. Then double click it. Select New Property. Give a Name and starting Value in Add New Property dialog box shown below.

Apply.


MOS Inverters

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Return to schematic and change the value of the resistor to {Rval} Including curly brackets

Where this is the new property name given in the attribute editor

The primary sweep is for V1, zero to 5 volts in small steps.

Parameter sweep can be a list of values as shown.


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V_V1

0V 1.0V 2.0V 3.0V 4.0V 5.0V

V(output)

0V

1.0V

2.0V

3.0V

4.0V

5.0V

R=1K

10K50K

Vin

V(output)


MOS Inverters

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LTSPICE PARAMETER SWEEP

RL= 10K

50K

100K

Add the step command to the schematic


MOS Inverters

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LTSPICE PARAMETER SWEEPS

Sweep SPICE parameter

VT0= 0.5, 1.0, 2.0

Sweep FET attribute

W= 5u, 20u, 40u

Sweep SPICE parameter

TOX= 10n, 20n, 30n


MOS Inverters

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CALCULATION OF NOISE MARGINS

Approach

Take derivative find where it is equal to -1

Typically D(V(out)) is -1 at two points

find VIL and VIH

VOH and VOL

Find point where Vin = Vout (switching voltage)

Find I

Find Voltage Gain

Find Power


MOS Inverters

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SPICE CALCULATIONS FOR NOISE MARGINS

1

2

VIL = 3.31

VIH = 6.95

VOH = 8.09

VOL = 3.37

D0 = VIL – VOL

=3.31-3.73= -0.42

D1=VOH-VIH=

=8.09-6.95=1.14

Max Gain = -1.32

RL = 1K

-1


MOS Inverters

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SPICE CALCULATIONS FOR NOISE MARGINS

1

2

VIL = 1.0

VIH = 2.91

VOH = 10.0

VOL = 0.91

D0 = VIL – VOL

=1.0-0.91= 0.08

D1=VOH-VIH=

=10-2.91=7.09

Max Gain = -7.2

RL = 10K


MOS Inverters

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LTSPICE WITH MEASURE COMMANDS

NML

NMH

10K

50K100K

View

Error Log


MOS Inverters

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PSPICE WITH MEASURE COMMANDS

1K

10K

50K


MOS Inverters

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PSPICE WITH MEASURE COMMANDS

Run SPICE

Select Trace > evaluate measurement

Size for square grid

Use snipping tool

NML, noise margin low, D0 =ViL-VoL=3.68– 1.80 = 1.88

NMH, noise margin high, D1=VoH-ViH=4.52-2.07 = 2.45

ViL = 3.68ViH = 1.80VoL = 2.07VoH= 4.52


MOS Inverters

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VTC NMOS INVERTER- NMOS ENHANCEMENT LOAD

NMOSENHANCEMENTLOAD

+V

VIN

VO

M2

M1

M2 is the switch and M1 is the load. The load limits the current when M2 is on. The load could be a resistor but an NMOS transistor with gate connected to the drain is smaller in size and also limits current. See the I-V characteristics. In the first quadrant the transistor approximates the resistor. However, Vout high is below VDD by the threshold voltage of M1

+V00

+V

Vt

VOUT

VIN

M2

Linear

M2

Off

M2 & M1

Saturation

+

V

-

V

I I

1/R

Vt

Vt

M1

ID = µW Cox’ (Vg-Vt)2

2LSaturation


W/L switch

W/L loadGain =


MOS Inverters

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DERIVATION OF GAIN EXPRESSION

+VDD

VIN

VO

M2

M1

Assume Vout = Vin and both transistors are in saturation for the steep part of the VTC. The current in M1 is equal to the current in M2 is equal. Also assume Vt is the same for both transistors.

W2/L2

W1/L1Gain =

I2 = I1µW2 Cox’/2L2 (VG-Vt)

2 = µW1 Cox’/2L1 (VG-Vt)2

W2 /L2 (VG-Vt)2 = W1/L1 (VG-Vt)

2

But, VG2 is VIN and VG1 = VO +Vt

(W2 /L2) (VIN-Vt)2 = (W1/L1) (VO +Vt -Vt)

2

Gain = d VO/d VIN

W/L switch

W/L loadGain =


MOS Inverters

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VTC NMOS INVERTER- NMOS ENHANCEMENT LOAD

Note: increasing L of the load is equivalent to increasing R

of a resistor load, Vout high is Vdd – VtM1 , Gain is shown.

G=2.2

G=5.5

G=9.5

W/L switch

W/L loadGain =


MOS Inverters

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VTC NMOS INVERTER- NMOS ENHANCEMENT LOAD AND V++ GATE BIAS

NMOS

ENHANCEMENT

LOAD V++ GATE BIAS

+V

VIN

VO

V++

M2

M1

M2 is the switch and M1 is the load. The load limits the current when M2 is on. The load could be a resistor but an NMOS transistor is smaller.

M1 is always on because the gate voltage is above the supply voltage. (Vgs is always above the threshold voltage. Vout max is the supply voltage.

The threshold voltage of M1 depends on source to substrate voltage for M1.

W/L switch

W/L loadGain =


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VTC NMOS INVERTER- NMOS ENHANCEMENT LOAD AND V++ GATE BIAS

G=2.2

G=5.5

G=9.5


of a resistor load, Vout high is Vdd, Gain is shown.

W/L switch

W/L loadGain =


MOS Inverters

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VTC NMOS INVERTER – NMOS DEPLETION LOAD

VO

+V

VIN

NMOS

DEPLETION

LOAD

+

V

-

V

I I

1/R

Vt

Vt

M2 is the switch and M1 is the load. The load limits the current when M2 is on. In the first quadrant the transistor approximates the resistor. M1 is always on because its threshold voltage is set to zero or slightly negative by ion implant. Note: transistor M1 symbol has solid line between D and S.

M2

M1W/L switch

W/L loadGain =


MOS Inverters

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* From Sub-Micron CMOS Manufacturing Classes in MicroE ~ 1um Technology

.MODEL RITSUBN7D NMOS (LEVEL=7

+VERSION=3.1 CAPMOD=2 MOBMOD=1

+TOX=1.5E-8 XJ=1.84E-7 NCH=1.45E17 NSUB=5.33E16 XT=8.66E-8

+VTH0=-1.0 U0= 600 WINT=2.0E-7 LINT=1E-7

+NGATE=5E20 RSH=1082 JS=3.23E-8 JSW=3.23E-8 CJ=6.8E-4 MJ=0.5 PB=0.95

+CJSW=1.26E-10 MJSW=0.5 PBSW=0.95 PCLM=5

+CGSO=3.4E-10 CGDO=3.4E-10 CGBO=5.75E-10)

Need a new SPICE model for the Depletion

mode NMOS. New model name and negative VTH0.

Using ion implant the VTH0 can be made negative.


MOS Inverters

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G=2.2

G=5.5

G=9.5


of a resistor load, Vout high is Vdd, Gain is shown.

DD

D

W/L switch

W/L loadGain =


MOS Inverters

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VTC PMOS INVERTER- PMOS ENHANCEMENT LOAD

PMOS

ENHANCEMENT

LOAD

-V

VIN

VO

M2 is the switch and M1 is the load. The load limits the current when M2 is on. The load could be a resistor but a PMOS transistor with gate connected to the drain is smaller in size and also limits current. See the I-V characteristics. In the first quadrant the transistor approximates the resistor. However, Vout high is below VDD by the threshold voltage of M1

M2

M1+

V

-

-V

-I I

1/R

Vt

Vt

W/L switch

W/L loadGain =


MOS Inverters

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VTC PMOS INVERTER- PMOS ENHANCEMENT LOAD

-10 Volts is Logic High

0 Volts is Logic Low

Gain = 2

Note: Supply and input V is negative

W/L switch

W/L loadGain =


MOS Inverters

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GAIN OF 2 INVERTER

Ld = 2l (20 µm)

Lu = 4 l (40 µm)Wu= 2l (20 µm)

Wd= 4l (40 µm)

(80 µm)

(50 µm)

Vout

Vdd

Gnd

Vin

W/L switch

W/L loadGain =


MOS Inverters

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INVERTER LAYOUT

L = 40µm

W = 20µm

L = 20µm

W = 50µm

W/L switch

W/L loadGain = = 2.34


MOS Inverters

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GAIN OF 3 INVERTER TEST RESULTS

Actual

Gain = -2.64

W is smaller than

drawn because of

lateral diffusion ~2u

Theoretical

Gain = - 3


MOS Inverters

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CMOS - CALCULATION OF VTC

First figure out if the transistor is

sub-threshold or off, Vgs < Vth and Vgd < Vth

non-saturation, Vgs > Vth and Vgd > Vth

saturation region, Vgs > Vth and Vgd < Vth

Note: Vin = Vgs, Vout = Vds, therefore Vgd = Vin-Vout

Vth might be +1volt

VIN

CMOS

+V

VO

NMOS

PMOS

+V0

0

+V

Vthn

VOUT

VIN

nmos

off

pmos

off

nmos & pmos

saturation

pmos sat

nmos linear

nmos sat

pmos linear

V-Vthp


MOS Inverters

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CMOS INVERTER

VIN VOUT

VIN

CMOS

+V

VO

Idd

+V0

0

+V

ViL

Voh

VoL

Vih

Imax

VOUT

VIN

Idd

Slope = Gain

Vinv

NML, noise margin low, D0 =ViL-VoLNMH, noise margin high, D1 =VoH-ViH


MOS Inverters

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LTSPICE – CMOS INVERTER

NML, noise margin low, D0 =ViL-VoL = 2.2-0.5 = 1.7

NMH, noise margin high, D1 =VoH-ViH = 4.5-2.5 = 2.0


MOS Inverters

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LTSPICE WITH MEASURE COMMANDS


MOS Inverters

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PSPICE CMOS VTC WITH MEASURE COMMANDS

Run SPICE

Select Trace >

evaluate measurement

Size for square grid

Use snipping tool

NML, noise margin low, D0 =ViL-VoL=2.66–0.40= 2.26

NMH, noise margin high, D1=VoH-ViH=4.64-2.01 = 2.63

ViL = 2.66ViH = 2.01VoL = 0.395VoH= 4.64


MOS Inverters

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COMPARISON OF 10u, 1u AND 100n CMOS INVERTERS

RITALDN3/RITALDP3L=10u W=880uL=10u W=880u

RITSUBN7/RITSUBP7Ln=1u Wn=2uLp=1u Wp=2u

EECMOSN/EECMOSPLn=180n Wp=200nLn=180n Wp=200n

VDD = 5 volts VDD = 3.3 volts VDD = 2.5 volts

Imax=100uA

Gain=-33

Imax=21uA

Gain=-6

Imax=5.4mA

Gain=-90


MOS Inverters

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PSEUDO – CMOS

There are situations where we want a large number of inputs. Rather than have CMOS where there will be many transistors in series (which will not work) we can use a single PMOS/NMOS transistor that is always on.

VA

CMOS

4 Input NOR

+V

VO

VB VC VD

VA

Pseudo CMOS

4 Input NOR

+V

VO

Idd

VB VC VD

VA

VB

VC

VD


MOS Inverters

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PSEUDO – CMOS

There are situations where we want a large number of inputs. Rather than have CMOS where there will be many transistors in series (which will not work) we can use a single PMOS/NMOS transistor that is always on.

VIN

CMOS

+V

VO

Idd

VIN

Pseudo CMOS

Inverter

+V

VO

Idd

Pseudo CMOS

4 Input NAND

+V

VO

Idd

VA VB VC VD

+V


MOS Inverters

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PSEUDO – CMOS INVERTER

No advantages over CMOS inverter

D0=1.0 , D1=3.0

W/L switch

W/L loadGain = 2/2

2/10 = 2.2=


MOS Inverters

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PSEUDO – CMOS NOR

Note: noise margin ~ D0=1.3 , D1=2.8max current drive 50uAstatic current not zero for Vout=lowgate delay ?

W/L switch

W/L loadGain = 8/2

2/8 = 4=


MOS Inverters

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CMOS NOR-4

Note: noise margin ~ D0=1.2 , D1=3.2max current drive 50uAstatic current is zero gate delay ?


MOS Inverters

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PSEUDO CMOS NAND

Sweep M7, M2 and M6 together

W/L switch

W/L loadGain = 6/2

2/6 = 3=


MOS Inverters

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PSEUDO CMOS NAND

Sweep M7 only, M2 and M6 off

W/L switch

W/L loadGain = 2/2

2/6 = 1.7=


MOS Inverters

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PSEUDO CMOS NAND

Sweep M7, M2 and M6 together

W/L switch

W/L loadGain = 6/2

2/2 = 1.7=


MOS Inverters

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PSEUDO CMOS NAND

Sweep M7 only, M2 and M6 off

Does not work L of M1 needs to be larger or W of all PMOS wider (Gain too low)

W/L switch

W/L loadGain = 2/2

2/2 = 1.0=

switches


MOS Inverters

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PSEUDO CMOS NAND

This is how to check truth table

0

0

0

0

0

1

0

1

0

0

1

1

1

0

0

1

0

1

1

1

0

1

1

1

Gain = 5.5


MOS Inverters

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REFERNCES

1. Hodges Jackson and Saleh, Analysis and Design of Digital Integrated Circuits, Chapter 4.

2. Sedra and Smith, Microelectronic Circuits, Sixth Edition, Chapter 13.

3. Dr. Fuller’s Lecture Notes, http://people.rit.edu/lffeee

http://people.rit.edu/lffeee


MOS Inverters

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HOMEWORK – MOS INVERTERS

1. Using SPICE obtain the VTC for a CMOS inverter with gate lengths of ~1um. Let the width of both transistors be 2um Determine the noise margins. Determine the maximum current and voltage gain. (Use the SPICE models given below) Make appropriate assumptions.

2. Repeat problem 1 for gate L and W of ~200nm.3. Given the layout shown below of a CMOS inverter find L, W,

AD, AS, PD, PS.4. The schematic below is for a tristate inverter. This device

should be able to make the output high or low when the enable (EN) is high. When the enable is low the inverter is effectively disconnected from the load (high impedance, Z). Use SPICE to show that this circuit operates as intended.


MOS Inverters

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INVERTER LAYOUT

Vin Vout

Vin

CMOS

+V

Vout

Idd

TRUTH TABLE

VOUTVIN

0 1

1 0

PMOS

NMOS

W = 40 µm

Lpoly = 2.0µm


MOS Inverters

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HOMEWORK – MOS INVERTERS

VIN

+V

VOEN

C

EN

Tristate Inverter

VOVIN

0 0 High Z

0 1 High Z

1 0 1

1 1 0

EN


MOS Inverters

Page 59



SPICE MODELS FOR CD4007 MOSFETS

*SPICE MODELS FOR RIT DEVICES - DR. LYNN FULLER 8-7-2015*LOCATION DR.FULLER'S WEBPAGE - http://people.rit.edu/lffeee/CMOS.htm**Used in Electronics II for CD4007 inverter chip*Note: Properties L=10u W=170u Ad=8500p As=8500p Pd=440u Ps=440u NRD=0.1 NRS=0.1.MODEL RIT4007N7 NMOS (LEVEL=7+VERSION=3.1 CAPMOD=2 MOBMOD=1+TOX=4E-8 XJ=2.9E-7 NCH=4E15 NSUB=5.33E15 XT=8.66E-8 +VTH0=1.4 U0= 1300 WINT=2.0E-7 LINT=1E-7 +NGATE=5E20 RSH=300 JS=3.23E-8 JSW=3.23E-8 CJ=6.8E-8 MJ=0.5 PB=0.95+CJSW=1.26E-10 MJSW=0.5 PBSW=0.95 PCLM=5+CGSO=3.4E-10 CGDO=3.4E-10 CGBO=5.75E-10)**Used in Electronics II for CD4007 inverter chip*Note: Properties L=10u W=360u Ad=18000p As=18000p Pd=820u Ps=820u NRS=O.54 NRD=0.54.MODEL RIT4007P7 PMOS (LEVEL=7+VERSION=3.1 CAPMOD=2 MOBMOD=1+TOX=5E-8 XJ=2.26E-7 NCH=1E15 NSUB=8E14 XT=8.66E-8+VTH0=-1.65 U0= 400 WINT=1.0E-6 LINT=1E-6+NGATE=5E20 RSH=1347 JS=3.51E-8 JSW=3.51E-8 CJ=5.28E-8 MJ=0.5 PB=0.94+CJSW=1.19E-10 MJSW=0.5 PBSW=0.94 pCLM=5+CGSO=4.5E-10 CGDO=4.5E-10 CGBO=5.75E-10)


MOS Inverters

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SPICE MODELS FOR MOSFETS

* From Sub-Micron CMOS Manufacturing Classes in MicroE ~ 1um Technology

.MODEL RITSUBN7 NMOS (LEVEL=7



+VTH0=1.0 U0= 600 WINT=2.0E-7 LINT=1E-7




*

*From Sub-Micron CMOS Manufacturing Classes in MicroE ~ 1um Technology

.MODEL RITSUBP7 PMOS (LEVEL=7



+VTH0=-1.0 U0= 376.72 WINT=2.0E-7 LINT=2.26E-7


+CJSW=1.19E-10 MJSW=0.5 PBSW=0.94



MOS Inverters

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SPICE MODELS FOR MOSFETS

*4-4-2013 LTSPICE uses Level=8

* From Electronics II EEEE482 FOR ~100nm Technology

.model EECMOSN NMOS (LEVEL=8


+TOX=5E-9 XJ=1.84E-7 NCH=1E17 NSUB=5E16 XT=5E-8

+VTH0=0.4 U0= 200 WINT=1E-8 LINT=1E-8




*

*4-4-2013 LTSPICE uses Level=8

* From Electronics II EEEE482 FOR ~100nm Technology

.model EECMOSP PMOS (LEVEL=8

+TOX=5E-9 XJ=0.05E-6 NCH=1E17 NSUB=5E16 XT=5E-8

+VTH0=-0.4 U0= 100 WINT=1E-8 LINT=1E-8




*

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