EE534 VLSI Design System Summer 2004 Lecture 06: Static CMOS inverter (CHAPTER 5)

EE534VLSI Design System

Summer 2004

Lecture 06: Static CMOS inverter

(CHAPTER 5)

Review: CMOS Inverter VTC

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5

Vin (V)

Vou

t (V

)

NMOS offPMOS res

NMOS satPMOS res

NMOS satPMOS sat

NMOS resPMOS sat NMOS res

PMOS off

VDD

Vout

CL

NMOS offPMOS in non sat

NMOS in sat

PMOS in non sat

NMOS in sat

PMOS in sat

NMOS in non sat

PMOS in sat

NMOS in nonsat

PMOS off

Vout = VDS

Dra

in c

urre

nt I D

S

Vin=2V

VCC

Vin=1V

Vin=3V

Vin=4V

Review: CMOS Inverter: VTC

Vout = VDS

Dra

in c

urre

nt I D

S

Vin=2V

VCC

Vin=1V

Vin=3V

Vin=4V

Vou

t

Vin1 2 3 40

VCC

PMOS NMOS

• Output goes completely to Vcc and Gnd• Sharp transition region

CMOS Inverter: Switch Model of Dynamic Behavior

VDD

Rn

Vout

CL

Vin = V DD

VDD

Rp

Vout

CL

Vin = 0

Gate response time is determined by the time to charge CL through Rp (discharge CL through Rn)

CMOS inverter operation

NMOS transistor: Cutoff if Vin < VTN

Linear if Vout < Vin – VTN

Saturated if Vout > Vin – VTN

PMOS transistor Cutoff if (Vin-VCC) < VTP → Vin < Vcc+VTP

Linear if (Vout-VCC)>Vin-Vcc-VTP → Vout>Vin - VTP

Sat. if (Vout-VCC)<Vin-Vcc-VTP → Vout < Vin-VTP

Vin Vout

Vcc

CMOS Static Inverter design

consideration

CMOS inverter design consideration

The CMOS inverter usually design to have, (i) VTN =|VTP|(ii) K´n(W/L)=K´p(W/L)

But K´n> K´p (because n>p)How equation (ii) can be satisfied?This can be achieved if width of the PMOS is made

two or three times than that of the NMOS device. This is very important in order to provide a symmetrical VTC, results in wide noise margin.

VCC

VCC

Vin

Vout

kp=kn

kp=5kn

kp=0.2kn

• Increase W of PMOS kp increases VTC moves to right

• Increase W of NMOS kn increases VTC moves to left

• For VTH = Vcc/2 kn = kp

Wn 2Wp

CMOS inverter design consideration (cont.)

CMOS inverter design consideration (cont.)

VscmVScm

LWLW

p

n

n

p2

2

580/230

np LW

LW

5.2

1

invertersymmetricn

pR k

kK

CMOS inverter design consideration (cont.)

Impact of Process Variation on VTC Curve

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5

Vin (V)

Vou

t (V

)

Nominal

Good PMOSBad NMOS

Bad PMOSGood NMOS

Process variations (mostly) cause a shift in the switching threshold

Effects of Vth adjustment Result from changing kp/kn ratio:

Inverter threshold VTH Vcc/2 Rise and fall delays unequal Noise margins not equal

Reasons for changing inverter threshold Want a faster delay for one type of

transition (rise/fall) Remove noise from input signal:

increase one noise margin at expense of the other

Symmetrical properties of the CMOS inverter

Example:

Definition of Noise Margins

VIH

VIL

UndefinedRegion

"1"

"0"

VOH

VOL

NMH

NML

Gate Output Gate Input

Noise Margin High

Noise Margin Low

Concept of Noise Margins

NML=VIL-VOL (noise margin for low input)NMH=VOH-VIH (noise margin for high input)

VI

ILOLILL VVVNM ILIHCCIHOHH VVVVVNM

Noise margin calculations

Vout = VDS

Dra

in c

urre

nt I D

S

Vin=2V

VCC

Vin=1V

Vin=3V

Vin=4V

CMOS inverter: VIL

KCL: IDp=IDn

Differentiate and set dVout/dVin to –1

Solve simultaneously with KCL to find VIL

2,,,0,

2,0, 2

22 pDSpDSpTpGSp

nTnGSn VVVV

kVVk

2,0

2,0 2

22 CCoutCCoutpTCCinp

nTinn VVVVVVV

kVVk

in

outCCoutCCout

in

outpTCCinpnTinn dV

dVVVVVdVdVVVVkVVk ,0,0

CCpTILoutpnTILn VVVVkVVk ,0,0 2

R

nTRCCpToutIL k

VkVVVV

1

2 ,0,0 p

nR k

kk

CMOS inverter: VIH

KCL:

Differentiate and set dVout/dVin to –1

Solve simultaneously with KCL to find VIH

2,0,2

,,,0, 22

2 pTpGSp

nDSnDSnTnGSn VV

kVVVVk

2,02

,0 22

2 pTCCinp

outoutnTinn VVV

kVVVVk

pTCCinpin

outoutout

in

outnTinn VVVk

dVdVVV

dVdVVVk ,0,0

pTCCIHppTIHoutn VVVkVVVk ,0,02

R

nToutRpTCCIH k

VVkVVV

1

2 ,0,0

p

nR k

kk

CMOS inverter: VIL and VIH for Ideal VTH

(Symmetrical, Kn=Kp) Assuming VT0,n=-VT0,p, and kR = 1,

02381

TCCIL VVV

02581

TCCIH VVV

DDIHIL VVV

ILOLILL VVVNM

ILIHCCIHOHH VVVVVNM

(symmetrical inverter)

Example 5.4

CMOS inverter: VTH

KCL:

Solve for VTH = Vin = Vout

2,0,2

,0, 22 pTpGSp

nTnGSn VV

kVVk

2,02

,0 22 pTCCinp

nTinn VVV

kVVk

R

pTCCR

nT

TH

k

VVk

VV

11

1,0,0

p

nR k

kk

CMOS inverter: Ideal VTH (Symmetrical, Kn=Kp)

Ideally, Vth = VCC/2

Assuming VT0,n = VT0,p,

R

pTCCR

nT

TH

k

VVk

VV

11

1,0,0

p

nR k

kk

2

,0

,0, 2

2

nTCC

pTCCidealR VV

VVk

1, idealRk

5.2

p

n

n

p

LWLW

nn LW

LW

5.2

For ideal symmetrical inverter required that