CMOS Digital Integrated Circuits
Lec 6
CMOS Inverters:
Static Characteristics
1
CMOS Digital Integrated Circuits
CMOS Inverters – Static Design

Goals
• Understand the basic definition of basic circuit-level
parameters.
• Understand the VTC of a CMOS inverter.
• Understand in detail static analysis of the CMOS inverter
including circuit parameters, VOL, VOH, VIL, VIH, NMH, NML,
and Vth
2
CMOS Digital Integrated Circuits
Voltage Transfer Characteristic
The Ideal Transfer Characteristic
B=A’
A
Vout
Symbol
VDD
A
0
1
Logic “1”
output
Logic “0”
output
B
1
0
Truth Table
0
3
VDD/2
VDD
Vin
CMOS Digital Integrated Circuits
Generic Inverter
Voltage Transfer Characteristic (VTC)
• Static characteristics of an inverter represented by its voltage
transfer characteristic, a plot of Vout versus Vin over the range 0 to
Vdd. An example transfer characteristic of an CMOS inverter is
typical
Vout
VOH
dVout / dVin= -1
Vout = Vin
VOL
dVout / dVin= -1
Vin
VIL Vth VIH VOH
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CMOS Digital Integrated Circuits
Voltage Transfer Characteristic
Operating Points
• If Vin is at or near 0V (logic 0), then Vout is at or near 5V (logic 1). If Vin is
at or near 5V (logic 1), then Vout is at or near 0V (logic 0). Thus, an
inverter.
• Points at which the input and output voltages would lie after passing
through a long chain of like inverters.
• Equivalent to the long chain is a loop of two inverters.
• In the loop, each Vin is Vout of the other inverter, giving Vin1 =Vout2 and Vin2
=Vout1 .
• Solve graphically by positioning a second VTC over the given one such
that these relationships hold on the axes (The second curve is mirrored
about the 45 degree line).
• The intersection of these curves at a high output level gives output HIGH
voltage, VOH, and at the low level gives output LOW voltage, VOL.
• These two intersections are stable operating points. The intersection at the
mid-level is an unstable operating point. This is the point at which Vout =
Vin for the single inverter and is the inverter threshold voltage, Vth or Vinv.
5
CMOS Digital Integrated Circuits
Voltage Transfer Characteristic
Operating Points (Cont.)
Vout
VOH
V out  V in  V th
VOL
Vin
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CMOS Digital Integrated Circuits
Voltage Transfer Characteristic
Noise Margins
VIH
VOL
Maximum
allowable
voltage
VIL
Interconnect
Noise
VOH
Minimum
allowable
voltage
VIH
VOL
Interconnect
Noise
• Propagation of digital signals under the influence of noise
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CMOS Digital Integrated Circuits
Voltage Transfer Characteristic
Noise Margins (Cont.)
• Ability to tolerate “noise” on its inputs is an important aspect of static
behavior.
• To quantify this concept, we define two additional voltages levels,
VIH and VIL as the HIGH and LOW levels of Vin at which:
dVout/ dVin = -1
• What is the justification for this? For any positive voltage
superimposed on operating point value VOL, the voltage gain as
represented by the |slope| is less than 1 on the entire curve up VIL.
This implies that the total large signal gain over the same range is also
less than 1. Thus, the negative output voltage change superimposed
on VOH is smaller than the positive input change superimposed on VOL
that causes it.
• So a change due to noise propagating down a chain of inverters will
diminish in magnitude at it propagates as long as the magnitude of the
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CMOS Digital Integrated Circuits
Voltage Transfer Characteristic
Noise Margins (Cont.)
change is no more than VIL-VOL. This value is called the noise
margin (for LOW signal levels), NML.
• Similarly, for HIGH signal levels, the noise margin is NMH=VOHVIH. This can be represented in terms of voltage ranges as
OUTPUT
H
INPUT
VOH
H
NMH
Transition
Region
NML
L
9
VOL
VIH
VIL
L
CMOS Digital Integrated Circuits
Voltage Transfer Characteristic
Noise Margins (Cont.)
• Nominal output
Vout = f(Vin)
• Output under noise
Vout‫ = ׳‬f(Vin+Vnoise)
‫ = ׳‬f(V ) + (dV /dV )V
Vout
in
out
in
noise + hight order term (neglected)
Perturbed Output = Nominal Output + Gain  External Perturbation
• The nominal operating region is defined as the region where the gain
is less than unity!
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CMOS Digital Integrated Circuits
CMOS Static Parameters
The Inverter Circuit and Operating Regions
• The CMOS inverter is constructed by a pMOS and a nMOS
VDD
S
VGS,p=Vin-VDD G
VDS,p=Vout-VDD
Vin
G
D
D
Vout
VGS,n=Vin
VDS,n=Vout
S
• The pMOS source S and substrate B are both at VDD, so no body effect
for either MOS. For the pMOS, VGS=Vin-VDD and VDS=Vout -VDD.

Operating Regions
• nMOS Cutoff: Vin < VTn
Linear: Vin ≥ VTn, Vout < Vin – VTn
Saturation: Vin ≥ VTn, Vout ≥ Vin – VTn
• pMOS Cutoff: Vin > VDD + VTp
Linear: Vin ≤ VDD + VTp, Vout > Vin – VTp
Saturation: Vin ≤ VDD + VTp, Vout ≤ Vin – VTp
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CMOS Digital Integrated Circuits
CMOS Static Parameters
VTC and Parameters
Region
IDn = IDp,
VGSp = VGSn -VDD = Vin –VDD
VDSp = VDSn -VDD = Vout VDD.
Vout
VDD
Vin
Vout
nMOS
A
<VT0,n
B
VIL
high≈VOH Sat.
C
Vth
Vth Sat.
D
VIH
E
>VDD+VT0,p
Vout = Vin –VT0,p
A
VOH Cut-off
low≈ VOL Linear
VOL Linear
pMOS
Linear
Linear
Sat.
Sat.
Cut-off
B
Vout = Vin –VT0,n
nMOS in saturation
C
pMOS in saturation
D
both in saturation
E
Vin
VT0,p
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0
VT0,n VIL VIH VDD+VT0,p
VDD
CMOS Digital Integrated Circuits
CMOS Static Parameters
VTC and Parameters (Cont.)

The Parameters
• Due to cutoff of the nMOS and pMOS respectively,
VOH = VDD
VOL = 0
• VIL is at the -1 slope point. The VIL can be found by simultaneous
solutions, but without iteration necessary since there is no body
effect. The equations used:
-IDn(SAT) = IDp(LIN)
(kn/2)(VGS,n-VT0,n)2 = (kp/2 )[2(VGS,p-VT0,p)VDS,p-VDS,p2]
-Substitute VGS,n=Vin=VIL, VDS,n =Vout–VDD, and kR=kn/kp into the
above equation, we get
kR(VIL-VT0,n)2 =2(VIL-VDD-VT0,p)(Vout-VDD)-(Vout-VDD)2
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CMOS Digital Integrated Circuits
CMOS Static Parameters
VTC and Parameters (Cont.)
-Differentiate both sides of the above with respect to Vin , and
dVout/dVIL = -1, we obtain
VIL=(2Vout+VT0,p-VDD+kRVT0,n)/(1+kR)
-This equation can be solved by combining with the KCL equation
to obtain the numerical value of VIL.
• For VIH, the equations used are:
-IDn(LIN) = IDp(SAT)
(kn/2)[2(VGS,n-VT0,n) VDS,n-VDS,n2] = (kp/2 )(VGS,p-VT0,p)2
• By the similar way, we can solve VIH by the following two
equations
kR[2(VIH-VT0,n) Vout-Vout2] =(VIH-VDD-VT0,p)2
VIH=[VDD+VT0,p+kR(2Vout+VT0,n)]/(1+kR)
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CMOS Digital Integrated Circuits
CMOS Static Parameters
VTC and Parameters (Cont.)
•
Vth is solved by equating the currents in saturation for the two devices in
terms of Vin, and solving or Vth=Vin=Vout:
IDn,SAT(Vin) = IDp,SAT(Vin)
(kn/2)(VGS,n-VT0,n)2 = (kp/2 )(VGS,p-VT0,p)2
 (kn/2)(Vin-VT0,n)2 = (kp/2 )(Vin-VDD-VT0,p)2
V T 0,n 
V th 
1
kR
1
•
•
15
(V DD  V T 0, p )
1
kR
Note that, in the region C, if we neglect the channel-length modulation
effect, the VTC is vertical which implies infinite voltage. Therefore, the
output voltage can be any value between (Vth-VT0,n) and (Vth-VT0,p),
without violating the voltage conditions used in the analysis.
Actually not, If channel length modulation is included, the slope is high,
but less than infinite.
CMOS Digital Integrated Circuits
CMOS Static Parameters
VTC and Parameters (Cont.)
• Special case: Vth=VDD/2. For this case, the CMOS VTC approach
the ideal VTC. In this case, for VT0,n=|VT0p|
V th 
V DD

V T 0,n k R  V DD  V T 0, p
1
2
 0.5V V
 kR  
DD
 0.5V V
DD
• Therefore,
V IL

V IH 
1
8
1
8
kR
2
T 0, p
T 0 ,n



1
3V DD  2V T 0,n 
5V DD  2V T 0,n 
V IL  V IH  V DD (symmetric inverter)
which does give a quite good noise margin since,
NML=VIL
NMH=VDD-VIH=VIL
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CMOS Digital Integrated Circuits
CMOS Static Parameters
VTC and Parameters (Cont.)
• As we know to achieve this near ideal situation, kR=1.
To make kR=1
p
(W / L)n


(W / L) p
n
• Recall these W and L are effective values of the dimensions, not
drawn. Assuming equal L’s, with μn = 2 to 3 μp, Wp is typical 2 to 3
times Wn.
• How to choose the kR ratio to achieve a desired inversion threshold
voltage:
kR 
kn
kp

kn
kp
17

(
V DD  V T 0, p  V th
V th  V T 0,n
 n C ox (W / L)n
 p C ox (W / L) p

2
)
 n (W / L)n
 p (W / L) p
CMOS Digital Integrated Circuits
CMOS Static Parameters
Static (DC) Power Dissipation and Area

Static (DC) Power Dissipation
• Finally, the power dissipation of CMOS is quite small,
PDC = VDD (Ileaage+Isubthreshold)

Area
• In terms of area, a CMOS primitive gate with n inputs require 2n
devices whereas an NMOS gate requires only n+1 devices! Plus,
there are more complex local interconnections. The integration
density of fully-complementary MOS is governed by greater area
requirements whereas NMOS density is governed by power
dissipation and heat problems.
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CMOS Digital Integrated Circuits