Introduction to CMOS VLSI
Design
Lecture 4:
DC & Transient Response
Greco/Cin-UFPE
(Material taken/adapted from Harris’ lecture notes)
Outline
DC Response
Logic Levels and Noise Margins
Transient Response
Delay Estimation
Activity
1) If the width of a transistor increases, the current will increase decrease not change
2) If the length of a transistor increases, the current will increase decrease not change
3) If the supply voltage of a chip increases, the maximum transistor current will
increase decrease not change
4) If the width of a transistor increases, its gate capacitance will increase decrease not change
5) If the length of a transistor increases, its gate capacitance will increase decrease not change
6) If the supply voltage of a chip increases, the gate capacitance of each transistor will
increase decrease not change
Activity
1) If the width of a transistor increases, the current will increase decrease not change
2) If the length of a transistor increases, the current will increase decrease not change
3) If the supply voltage of a chip increases, the maximum transistor current will
increase decrease not change
4) If the width of a transistor increases, its gate capacitance will increase decrease not change
5) If the length of a transistor increases, its gate capacitance will increase decrease not change
6) If the supply voltage of a chip increases, the gate capacitance
DC Response
DC Response: Vout vs. Vin for a gate
Ex: Inverter
– When Vin = 0 -> Vout = VDD – When Vin = VDD -> Vout = 0 – In between, Vout depends on
transistor size and current
– By KCL, must settle such that Idsn = |Idsp|
– We could solve equations
– But graphical solution gives more insight
Idsn Idsp
Vout VDD
Vin
Transistor Operation
Current depends on region of transistor behavior
For what Vin and Vout are nMOS and pMOS in – Cutoff?
– Linear?
– Saturation?
nMOS Operation
Cutoff Linear Saturated
Vgsn < Vgsn >
Vdsn <
Vgsn >
Vdsn >
Idsn Idsp
Vout VDD
Vin
nMOS Operation
Cutoff Linear Saturated
Vgsn < Vtn Vgsn > Vtn
Vdsn < Vgsn – Vtn
Vgsn > Vtn
Vdsn > Vgsn – Vtn
Idsp
Vout VDD
Vin
nMOS Operation
Cutoff Linear Saturated
Vgsn < Vtn Vgsn > Vtn
Vdsn < Vgsn – Vtn
Vgsn > Vtn
Vdsn > Vgsn – Vtn
Idsn Idsp
Vout VDD
Vin
Vgsn = Vin Vdsn = Vout
nMOS Operation
Cutoff Linear Saturated
Vgsn < Vtn Vin < Vtn
Vgsn > Vtn Vin > Vtn
Vdsn < Vgsn – Vtn Vout < Vin - Vtn
Vgsn > Vtn Vin > Vtn
Vdsn > Vgsn – Vtn Vout > Vin - Vtn
Idsp
Vout VDD
Vin
Vgsn = Vin V = V
pMOS Operation
Cutoff Linear Saturated
Vgsp > Vgsp <
Vdsp >
Vgsp <
Vdsp <
Idsn Idsp
Vout VDD
Vin
pMOS Operation
Cutoff Linear Saturated
Vgsp > Vtp Vgsp < Vtp
Vdsp > Vgsp – Vtp
Vgsp < Vtp
Vdsp < Vgsp – Vtp
Idsp
Vout VDD
Vin
pMOS Operation
Cutoff Linear Saturated
Vgsp > Vtp Vgsp < Vtp
Vdsp > Vgsp – Vtp
Vgsp < Vtp
Vdsp < Vgsp – Vtp
Idsn Idsp
Vout VDD
Vin
Vgsp = Vin - VDD Vdsp = Vout - VDD
Vtp < 0
pMOS Operation
Cutoff Linear Saturated
Vgsp > Vtp
Vin > VDD + Vtp
Vgsp < Vtp
Vin < VDD + Vtp Vdsp > Vgsp – Vtp Vout > Vin - Vtp
Vgsp < Vtp
Vin < VDD + Vtp Vdsp < Vgsp – Vtp Vout < Vin - Vtp
Idsp
Vout VDD
Vin
Vgsp = Vin - VDD V = V
Vtp < 0
I-V Characteristics
Make pMOS is wider than nMOS such that n = p
Vgsn5
Vgsn4 Vgsn3 Vgsn2 Vgsn1
Vgsp5 Vgsp4 Vgsp3 Vgsp2 Vgsp1
VDD -VDD
Vdsn -Vdsp
-Idsp Idsn
0
Current vs. V
out, V
inVin5
Vin4
Vin3 Vin2 Vin1 Vin0
Vin1
Vin2 Vin3 Vin4 Idsn, |Idsp|
Load Line Analysis
Vin5
Vin4
Vin3 Vin2 Vin1 Vin0
Vin1
Vin2 Vin3 Vin4 Idsn, |Idsp|
Vout VDD
For a given Vin:
– Plot Idsn, Idsp vs. Vout
– Vout must be where |currents| are equal in
Idsn Idsp
Vout VDD
Vin
Load Line Analysis
Vin0 Vin0
Idsn, |Idsp|
Vin = 0
Load Line Analysis
Vin1 Vin1
Idsn, |Idsp|
Vout VDD
Vin = 0.2VDD
Load Line Analysis
Vin2 Vin2
Idsn, |Idsp|
Vin = 0.4VDD
Load Line Analysis
Vin3 Vin3
Idsn, |Idsp|
Vout VDD
Vin = 0.6VDD
Load Line Analysis
Vin4
Vin4 Idsn, |Idsp|
Vin = 0.8VDD
Load Line Analysis
Vin5 Vin0
Vin1
Vin2 Vin3 Vin4 Idsn, |Idsp|
Vout VDD
Vin = VDD
Load Line Summary
Vin5
Vin4
Vin3 Vin2 Vin1 Vin0
Vin1
Vin2 Vin3 Vin4 Idsn, |Idsp|
DC Transfer Curve
Transcribe points onto Vin vs. Vout plot
Vin5
Vin4
Vin3 Vin2 Vin1 Vin0
Vin1
Vin2 Vin3 Vin4
Vout VDD
C Vout
0
Vin
VDD VDD
A B
D E
Vtn VDD/2 VDD+Vtp
Operating Regions
Revisit transistor operating regions
C Vout
0
V
VDD VDD
A B
D E
Vtn VDD/2 VDD+Vtp
Region nMOS pMOS A
B C D E
Operating Regions
Revisit transistor operating regions
C Vout
0
Vin
VDD VDD
A B
D E
Vtn VDD/2 VDD+Vtp
Region nMOS pMOS
A Cutoff Linear
B Saturation Linear
C Saturation Saturation
D Linear Saturation
E Linear Cutoff
Beta Ratio
If p / n 1, switching point will move from VDD/2
Called skewed gate
Other gates: collapse into equivalent inverter
Vout
0 VDD
0.5 21
p 10
n
p 0.1
n
Noise Margins
How much noise can a gate input see before it does not recognize the input?
Indeterminate Region NML
NMH
Input Characteristics Output Characteristics
VOH VDD
VOL
GND VIH
VIL
Logical High Input Range
Logical Low Input Range Logical High
Output Range
Logical Low Output Range
By definition, VIH and VIL are the operational points of the inverter where
Vout
VOH
VDD Unity Gain Points
Slope = -1
p/n > 1 Vin Vout
A high gain (g) is desirable
Logic Levels
To maximize noise margins, select logic levels at
VDD Vin Vout
VDD
p/n > 1 Vin Vout
0
Logic Levels
To maximize noise margins, select logic levels at – unity gain point of DC transfer characteristic
Vout
VOH
VDD Unity Gain Points
Slope = -1
p/n > 1 Vin Vout
Transient Response
DC analysis tells us Vout if Vin is constant
Transient analysis tells us Vout(t) if Vin(t) changes – Requires solving differential equations
Input is usually considered to be a step or ramp – From 0 to VDD or vice versa
Inverter Step Response
Ex: find step response of inverter driving load cap
(
0)
(
) )
(
o i
ut n
out
V t t t
V t V
d d
t
Vin(t)
Vout(t) Cload Idsn(t)
Inverter Step Response
Ex: find step response of inverter driving load cap
0 0
( )
( )
( )
( )
ou
DD in
t out
u t t V
d d
t
t t V t V
V
t
Vin(t)
Vout(t) Cload Idsn(t)
Inverter Step Response
Ex: find step response of inverter driving load cap
0
(
0( )
) ( (
) )
DD D
o i
D o t
n
ut u
V t
u t t V V
d d
t
t V
V
t t
Vin(t)
Vout(t) Cload Idsn(t)
Inverter Step Response
Ex: find step response of inverter driving load cap
0
0
( )
( )
( ) (
(
) )
DD DD
loa d ou
i
d t
o n
ut sn
V V
u t t V t t
V t
V d
dt C
t
I t
0
( ) out DD t
ou ds
t DD t
I n t V V
V
V V
V
t t
Vin(t)
Vout(t) Cload Idsn(t)
Inverter Step Response
Ex: find step response of inverter driving load cap
0
0
( )
( )
( ) (
(
) )
DD DD
loa d ou
i
d t
o n
ut sn
V V
u t t V t t
V t
V d
dt C
t
I t
0 2
2
0 )
( DD out DD t
n
Ids V
t t
V V V V
t
Vin(t)
Vout(t) Cload Idsn(t)
Inverter Step Response
Ex: find step response of inverter driving load cap
0
0
( )
( )
( ) (
(
) )
DD DD
loa d ou
i
d t
o n
ut sn
V V
u t t V t t
V t
V d
dt C
t
I t
0 2
2
0
)2 )
( ( )
( DD DD t
DD
out
out out out D t
n
t ds
D
I V
t t
V V V V
V V V V V
t
V t V t
Vout(t) Vin(t)
t0 t
Vin(t)
Vout(t) Cload Idsn(t)
nMOS – “1” transfer is bad
Delay Definitions
tpdr: rising propagation delay
– From input to rising output crossing VDD/2 (upper bound)
tpdf: falling propagation delay
– From input to falling output crossing VDD/2 (upper bound)
tpd: average propagation delay – tpd = (tpdr + tpdf)/2
tr: rise time
– time required for a node to charge from the 10% point to 90% ( 0.1 VDD to 0.9 VDD )
tf: fall time
Tpdr = TpHL Tcdf = TpHL
tr tf
Delay Definitions
CMOS Inverter Switching Characteristics
Define:
– Falling delay tdf = delay time with output falling – Rising delay tdr = delay time with output rising
Trajectory of a n-transistor operating point
a. Transistor is Cut off
a. Applying a Vgs=VDD b. Vout -> 0,9 to (VDD-Tth) a. VDD-Tth -> to 0.1V
Fall-time equivalent circuit
Rise-time equivalent circuit
The fall-time is faster than the rise time,
tf=tr/2
primarily due to different carrier mobilities associated with the p and n- devices:
µn = 2 µp
If the designer wants to have both n an p transistor with the same rise and fall time:
βn / βp = 1
This implies that channel width for the p-device must be increased to approximately two to three times of the n-device:
wp = 2 – 3 wn
n+ n+
p-type body
W
L tox
SiO2 gate oxide (good insulator, ox = 3.90) polysilicon
gate
Trajectory of a n-transistor operating
point
Delay Estimation
We would like to be able to easily estimate delay – Not as accurate as simulation
– But easier to ask “What if?”
The step response usually looks like a 1st order RC response with a decaying exponential.
Use RC delay models to estimate delay – C = total capacitance on output node – Use effective resistance R
– So that tpd = RC
Characterize transistors by finding their effective R
RC Delay Models
Use equivalent circuits for MOS transistors
– Ideal switch + capacitance and ON resistance – Unit nMOS has resistance R, capacitance C – Unit pMOS has resistance 2R, capacitance C
Capacitance proportional to width
Resistance inversely proportional to width
k g
s d
g
s d
kC kC
R/k kC
k g
s d
g
s
d kC
kC kC 2R/k
Capacitance
proportional to width
Resistance inversely
proportional to width
nMOS pMOS
Resistance of a uniform slab:
– R = (l/A) = (/t) (L/W) where is the resistivity in ohm-cm, t is the thickness in cm, L is the length, W is the width, and A is the cross-sectional area
– Using the concept of sheet resistance, R = Rs (L/W) where Rs is called the sheet resistance and given in ohms per square
• Rs = / t
Gate Capacitance -
Cg = oxWL/tox = oxA/tox
n+ n+
p-type body
W
L tox
SiO2 gate oxide (good insulator, ox = 3.90) polysilicon
gate
RC Delay Models
MOS device capacitances
MOS Device capacitances
– Cgs, Cgd = gate-to-channel capacitances, which are lumped at the source and the drain regions.
– Csb, Cdb = source and drain-duiffusion capacitances to bulk (or substrate) – Cgb = gate –to-bulk capacitance
1. Off region, where Vgs < Vt. There is no channel Cgs, Cgd = 0.
2. Non-saturated region, where Vgs - Vt. > Vds. Cgb = 0
3. Saturated region, where Vgs <-Vt. < Vds. Channel is heavely inverted.
The drain is pinched off causing Cgd = 0.
MOS device capacitances
RC Delay Models
pMOS capacitance
= 2x nMOS capacitance
Example: 3-input NAND
Sketch a 3-input NAND with transistor widths chosen to achieve effective rise and fall resistances equal to a unit inverter (R).
Example: 3-input NAND
Sketch a 3-input NAND with transistor widths chosen to achieve effective rise and fall resistances equal to a unit inverter (R).
Example: 3-input NAND
Sketch a 3-input NAND with transistor widths chosen to achieve effective rise and fall resistances equal to a unit inverter (R).
3 3 2 2
2
3
3-input NAND Caps
Annotate the 3-input NAND gate with gate and diffusion capacitance.
2 2 2
3 3 3
3-input NAND Caps
Annotate the 3-input NAND gate with gate and diffusion capacitance.
2 2 2
3 3
3 3C
3C 3C 2C
2C
2C 2C
2C 2C
3C 3C
2C 2C 2C
Resultant capacitance per gate = 2C+3C = 5C
Parallel capacitances 3x2C+3C = 9C
3-input NAND Caps
Annotate the 3-input NAND gate with gate and diffusion capacitance.
9C 3C 3 3C
3 3 2 2
2 5C
5C
5C
Elmore Delay
ON transistors look like resistors
Pullup or pulldown network modeled as RC ladder
Elmore delay of RC ladder
R1 R2 R3 RN
C C C C
nodes
1 1 1 2 2
...
1 2...
pd i to source i
i
N N
t R C
R C R R C R R R C
Example: 2-input NAND
Estimate worst-case rising and falling delay of 2- input NAND driving h identical gates.
h copies
2 2 2 2
B A
x
Y
Example: 2-input NAND
Estimate worst-case rising and falling delay of 2- input NAND driving h identical gates.
h copies
Example: 2-input NAND
Estimate worst-case rising and falling delay of 2- input NAND driving h identical gates.
h copies
Parallel capacitances 2C+2C+2C= 6C
Example: 2-input NAND
Estimate rising and falling propagation delays of a 2- input NAND driving h identical gates.
h copies
6C 2 2C
2 2 2
4hC B
A
x
Y
Example: 2-input NAND
Estimate rising and falling propagation delays of a 2- input NAND driving h identical gates.
h copies
6C 2 2C
2 2 2
4hC B
A
x
Y
R
(6+4h)CY
t
pdr
Example: 2-input NAND
Estimate rising and falling propagation delays of a 2- input NAND driving h identical gates.
h copies
6C 2 2C
2 2 2
4hC B
A
x
Y
R Y
t 6 4 h RC
Rising time: nMOS - “off”
pMOS - “on”
Example: 2-input NAND
Estimate rising and falling propagation delays of a 2- input NAND driving h identical gates.
h copies
6C 2 2C
2 2 2
4hC B
A
x
Y
Example: 2-input NAND
Estimate rising and falling propagation delays of a 2- input NAND driving h identical gates.
h copies
6C 2 2C
2 2 2
4hC B
A
x
Y
t
x R/2 Y
Consider that the nMOS resistance = (pMOS resistance)/2 = R/2
Example: 2-input NAND
Estimate rising and falling propagation delays of a 2- input NAND driving h identical gates.
h copies
6C 2 2C
2 2 2
4hC B
A
x
Y
2 2 2
2 6 4
7 4
R R R
t
pdfC h C
h RC
(6+4h)C R/2 2C
x R/2 Y
Delay Components
Delay has two parts – Parasitic delay
• 6 or 7 RC
• Independent of load – Effort delay
• 4h RC
• Proportional to load capacitance
Contamination Delay
Best-case (contamination) delay can be substantially less than propagation delay.
Ex: If both inputs fall simultaneously
6C 2 2C
2 2 2
4hC B
A
x
Y
R
(6+4h)CRY
3 2
t
cdr h RC
3 7C 2 2
2 2C
2C
Isolated Contacted Diffusion Merged
Uncontacted Shared Contacted Diffusion
Diffusion Capacitance
we assumed contacted diffusion on every source/drain
Good layout minimizes diffusion area
Ex: NAND3 layout shares one diffusion contact – Reduces output capacitance by 2C
– Merged uncontacted diffusion might help too
Capacitance = 2C+2C+3C = 7C
Layout Comparison
Which layout is better?
VDD A
GND
B
Y
VDD A
GND
B
Y