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EE4800 CMOS Digital IC Design & Analysis
Lecture 4
DC and Transient Responses, Circuit Delays
Zhuo Feng
4.1
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Outline
■ Pass Transistors
■ DC Response
■ Logic Levels and Noise Margins
■ Transient Response
■ RC Delay Models
■ Delay Estimation
4.2
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
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
4.3
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
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
4.4
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Pass Transistors
■ We have assumed source is grounded
■ What if source > 0?
VDD
► e.g. pass transistor passing VDD
■ Vg = VDD
VDD
► If Vs > VDD-Vt, Vgs < Vt
► Hence transistor would turn itself off
■ NMOS pass transistors pull no higher than VDD-Vtn
► Called a degraded “1”
► Approach degraded value slowly (low Ids)
■ PMOS pass transistors pull no lower than Vtp
4.5
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Pass Transistor Ckts
VDD
VDD
VDD
VDD
VDD
Vs = VDD-Vtn
Vs = |Vtp|
VDD-Vtn VDD-Vtn
VDD
VDD-Vtn
VDD
VSS
4.6
VDD
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
VDD-2Vtn
VDD-Vtn
DC Response
■ DC Response: Vout vs. Vin for a gate
■ Ex: Inverter
► When Vin = 0 ->
Vout = VDD
► When Vin = VDD
->
Vout = 0
VDD
► In between, Vout depends on
transistor size and current
Idsp
Vin
Vout
► By KCL, must settle such that
Idsn
Idsn = |Idsp|
► We could solve equations
► But graphical solution gives more insight
4.7
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Transistor Operation
■ Current depends on region of transistor behavior
■ For what Vin and Vout are NMOS and PMOS in
► Cutoff?
► Linear?
► Saturation?
4.8
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
NMOS Operation
Cutoff
Vgsn < Vtn
Linear
Vgsn > Vtn
Saturated
Vgsn > Vtn
Vdsn < Vgsn – Vtn
Vdsn > Vgsn – Vtn
VDD
Vin
Idsp
Idsn
4.9
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Vout
NMOS Operation
Cutoff
Vgsn < Vtn
Linear
Vgsn > Vtn
Saturated
Vgsn > Vtn
Vdsn < Vgsn – Vtn
Vdsn > Vgsn – Vtn
VDD
Vgsn = Vin
Vdsn = Vout
4.10
Vin
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Idsp
Idsn
Vout
NMOS Operation
Cutoff
Vgsn < Vtn
Vin < Vtn
Linear
Vgsn > Vtn
Vin > Vtn
Vdsn < Vgsn – Vtn
Vout < Vin - Vtn
Saturated
Vgsn > Vtn
Vin > Vtn
Vdsn > Vgsn – Vtn
Vout > Vin - Vtn
VDD
Vgsn = Vin
Vdsn = Vout
4.11
Vin
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Idsp
Idsn
Vout
PMOS Operation
Cutoff
Linear
Saturated
Vgsp >
Vgsp <
Vgsp <
Vdsp >
Vdsp <
VDD
Vin
Idsp
Idsn
4.12
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Vout
PMOS Operation
Cutoff
Linear
Saturated
Vgsp > Vtp
Vgsp < Vtp
Vgsp < Vtp
Vdsp > Vgsp – Vtp
Vdsp < Vgsp – Vtp
VDD
Vin
Idsp
Idsn
4.13
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Vout
PMOS Operation
Cutoff
Linear
Saturated
Vgsp > Vtp
Vgsp < Vtp
Vgsp < Vtp
Vdsp > Vgsp – Vtp
Vdsp < Vgsp – Vtp
VDD
Vgsp = Vin - VDD
Vdsp = Vout - VDD
4.14
Vtp < 0
Vin
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Idsp
Idsn
Vout
PMOS Operation
Cutoff
Vgsp > Vtp
Vin > VDD + Vtp
Linear
Vgsp < Vtp
Vin < VDD + Vtp
Vdsp > Vgsp – Vtp
Vout > Vin - Vtp
Saturated
Vgsp < Vtp
Vin < VDD + Vtp
Vdsp < Vgsp – Vtp
Vout < Vin - Vtp
VDD
Vgsp = Vin - VDD
Vdsp = Vout - VDD
4.15
Vtp < 0
Vin
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Idsp
Idsn
Vout
I-V Characteristics
■ Make PMOS is wider than NMOS such that
bn = bp
Vgsn5
Vgsn4
Idsn
Vgsn3
-Vdsp
Vgsp1
Vgsp2
-VDD
0
VDD
Vdsn
Vgsp3
Vgsp4
-Idsp
Vgsp5
4.16
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Vgsn2
Vgsn1
Current vs. Vout, Vin
Idsn, |Idsp|
Vin0
Vin5
Vin1
Vin4
Vin2
Vin3
Vin3
Vin4
Vin2
Vin1
Vout
4.17
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
VDD
Load Line Analysis
■ For a given Vin:
► Plot Idsn, Idsp vs. Vout
► Vout must be where |currents| are equal in
Idsn, |Idsp|
Vin0
Vin5
Vin1
Vin4
Vin
Idsp
Idsn
Vin2
Vin3
Vin3
Vin4
Vin2
Vin1
Vout
4.18
VDD
VDD
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Vout
Load Line Analysis
■ Vin = 0
Vin0
Idsn, |Idsp|
Vin0
Vout
4.19
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
VDD
Load Line Analysis
■ Vin = 0.2VDD
Idsn, |Idsp|
Vin1
Vin1
Vout
4.20
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
VDD
Load Line Analysis
■ Vin = 0.4VDD
Idsn, |Idsp|
Vin2
Vin2
Vout
4.21
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
VDD
Load Line Analysis
■ Vin = 0.6VDD
Idsn, |Idsp|
Vin3
Vin3
Vout
4.22
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
VDD
Load Line Analysis
■ Vin = 0.8VDD
Vin4
Idsn, |Idsp|
Vin4
Vout
4.23
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
VDD
Load Line Analysis
■ Vin = VDD
Vin0
Idsn, |Idsp|
Vin5
Vin1
Vin2
Vin3
Vin4
Vout
4.24
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
VDD
Load Line Summary
Idsn, |Idsp|
Vin0
Vin5
Vin1
Vin4
Vin2
Vin3
Vin3
Vin4
Vin2
Vin1
Vout
4.25
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
VDD
DC Transfer Curve
■ Transcribe points onto Vin vs. Vout plot
Vin0
Vin5
VDD
A
Vin1
Vin4
Vin2
Vin3
Vin3
Vin4
Vin2
Vin1
Vout
4.26
VDD
B
Vout
C
D
0
Vtn
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
VDD/2
Vin
E
VDD+Vtp
VDD
Operating Regions
■ Revisit transistor operating regions
Region
NMOS
PMOS
A
Cutoff
Linear
B
Saturation
Linear
C
Saturation
Saturation
D
Linear
Saturation
E
Linear
Cutoff
VDD
A
B
Vout
C
D
0
Vtn
VDD/2
Vin
4.27
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
E
VDD+Vtp
VDD
Beta Ratio
■ If bp / bn  1, switching point will move from VDD/2
■ Called skewed gate
■ Other gates: collapse into equivalent inverter
VDD
bp
 10
bn
Vout
2
1
0.5
bp
 0.1
bn
0
Vin
4.28
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
VDD
Noise Margins
■ How much noise can a gate input see
before it does not recognize the input?
Output Characteristics
Logical High
Output Range
VDD
Input Characteristics
Logical High
Input Range
VOH
NMH
VIH
VIL
Indeterminate
Region
NML
Logical Low
Output Range
VOL
GND
4.29
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Logical Low
Input Range
Logic Levels
■ To maximize noise margins, select logic levels at
► unity gain point of DC transfer characteristic
Vout
Unity Gain Points
Slope = -1
VDD
VOH
b p/b n > 1
Vin
VOL
Vin
0
Vtn
4.30
Vout
VIL VIH VDD- VDD
|Vtp|
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
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
4.31
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Inverter Step Response
■ Ex: find step response of inverter driving load cap
Vin (t )  u(t  t0 )VDD
Vout (t  t0 ) 
Vin(t)
dVout (t )

dt
4.32
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Vout(t)
Cload
Idsn(t)
Inverter Step Response
■ Ex: find step response of inverter driving load cap
Vin (t )  u(t  t0 )VDD
Vout (t  t0 )  VDD
Vin(t)
dVout (t )

dt
4.33
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Vout(t)
Cload
Idsn(t)
Inverter Step Response
■ Ex: find step response of inverter driving load cap
Vin (t )  u(t  t0 )VDD
Vout (t  t0 )  VDD
Vin(t)
dVout (t )
I dsn (t )

dt
Cload


I dsn (t )  


4.34
Vout
Vout
t  t0
 VDD  Vt
 VDD  Vt
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Vout(t)
Cload
Idsn(t)
Inverter Step Response
■ Ex: find step response of inverter driving load cap
Vin (t )  u(t  t0 )VDD
Vin(t)
Vout (t  t0 )  VDD
dVout (t )
I dsn (t )

dt
Cload

0


2
b
I dsn (t )  
V

V


DD
2

V (t )
 b VDD  Vt  out 2
 
4.35
Vout(t)
Cload
Idsn(t)
t  t0
Vout  VDD  Vt
 V (t ) V  V  V
 out
out
DD
t

Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Inverter Step Response
■ Ex: find step response of inverter driving load cap
Vin (t )  u(t  t0 )VDD
Vout (t  t0 )  VDD
Vin(t)
Vout(t)
Cload
dVout (t )
I dsn (t )

dt
Cload

0


2
b
I dsn (t )  
V

V


DD
2

V (t )
 b VDD  Vt  out 2
 
4.36
Idsn(t)
Vin(t)
t  t0
Vout  VDD  Vt
 V (t ) V  V  V
 out
out
DD
t

Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Vout(t)
t0
t
Delay Definitions
■ tpdr: rising propagation delay
► From input to rising output crossing VDD/2
■ tpdf: falling propagation delay
► From input to falling output crossing VDD/2
■ tpd: average propagation delay
► tpd = (tpdr + tpdf)/2
■ tr: rise time
► From output crossing 0.2 VDD to 0.8 VDD
■ tf: fall time
► From output crossing 0.8 VDD to 0.2 VDD
4.37
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Delay Definitions
■ tcdr: rising contamination delay
► From input to rising output crossing VDD/2
■ tcdf: falling contamination delay
► From input to falling output crossing VDD/2
■ tcd: average contamination delay
► tpd = (tcdr + tcdf)/2
4.38
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Simulated Inverter Delay
■ Solving differential equations by hand is too hard
■ SPICE simulator solves the equations numerically
► Uses more accurate I-V models too!
■ But simulations take time to write
2.0
1.5
1.0
(V)
Vin
tpdf = 66ps
tpdr = 83ps
Vout
0.5
0.0
0.0
200p
400p
600p
800p
t(s)
4.39
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
1n
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
► Depends on average current as gate switches
4.40
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Effective Resistance
■ Shockley models have limited value
► Not accurate enough for modern transistors
► Too complicated for much hand analysis
■ Simplification: treat transistor as resistor
► Replace Ids(Vds, Vgs) with effective resistance R
▼ Ids = Vds/R
► R averaged across switching of digital gate
■ Too inaccurate to predict current at any given time
► But good enough to predict RC delay
4.41
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
RC Delay Model
■ 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
d
g
d
k
s
s
kC
R/k
2R/k
g
g
kC
kC
d
k
s
s
4.42
kC
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
kC
g
kC
d
RC Values
■ Capacitance
► C = Cg = Cs = Cd = 2 fF/mm of gate width
► Values similar across many processes
■ Resistance
► R  6 KW*mm in 0.6um process
► Improves with shorter channel lengths
■ Unit transistors
► May refer to minimum contacted device (4/2 l)
► Or maybe 1 mm wide device
► Doesn’t matter as long as you are consistent
4.43
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Inverter Delay Estimate
■ Estimate the delay of a fanout-of-1 inverter
2C
R
A
2 Y
2
1
1
2C
2C
2C
Y
R
C
R
C
C
Delay = 6RC
4.44
2C
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
C
C
Delay Model Comparison
4.45
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
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).
2
2
2
3
3
3
4.46
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
3-input NAND Caps
■ Annotate the 3-input NAND gate with gate and
diffusion capacitance.
2
2
2
3
3
3
4.47
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
3-input NAND Caps
■ Annotate the 3-input NAND gate with gate and
diffusion capacitance.
2C
2
2C
2C
2C
2
2C
2
2C
3C
3C
3C
4.48
2C
2C
2C
3
3
3
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
3C
3C
3C
3C
3-input NAND Caps
■ Annotate the 3-input NAND gate with gate and
diffusion capacitance.
2
2
3
5C
5C
5C
4.49
2
3
3
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
9C
3C
3C
Elmore Delay
1v
t
V (t )  1  e RC

V(t)
R
C
Time Constant=RC

What is the time constant for more complex
circuits?
A
B
Inv 1
4.50
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
C
Inv 2
Elmore Delay (Cont.)
■ ON transistors look like resistors
■ Pullup or pulldown network modeled as RC ladder
■ Elmore delay of RC ladder
t pd 

Ri to sourceCi
nodes i
 R1C1   R1  R2  C2  ...   R1  R2  ...  RN  CN
R1
4.51
R2
R3
C1
C2
RN
C3
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
CN
Elmore Delay (Cont’d)

Resistance-oriented Formula:
Tdelay 

Ri Cdownstream,i
i on path
Tdelay,4=R1(C1+C2+C3+C4+C5)+R2(C2+C4+C5)+R4C4
4.52
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Example: 2-input NAND
■ Estimate worst-case rising and falling delay of 2-
input NAND driving h identical gates.
2
4.53
2
A
2
B
2x
Y
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
h copies
Example: 2-input NAND
■ Estimate rising and falling propagation delays of
a 2-input NAND driving h identical gates.
2
2
A
2
B
2x
4.54
6C
Y
4hC
2C
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
h copies
Example: 2-input NAND
■ Estimate rising and falling propagation delays of
a 2-input NAND driving h identical gates.
2
2
A
2
B
2x
Y
4hC
6C
h copies
2C
Rising
Delay
R
Y
(6+4h)C
t pdr  6  4h  RC
4.55
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Example: 2-input NAND
■ Estimate rising and falling propagation delays of
a 2-input NAND driving h identical gates.
2
2
A
2
B
2x
6C
Y
4hC
h copies
2C
Falling
Delay
x
R/2
R/2
2C
Y
(6+4h)C
t pdf   2C   R2    6  4h  C   R2  R2    7  4h  RC
4.56
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Delay Components
■ Delay has two parts
► Parasitic delay
▼ 6 or 7 RC
▼ Independent of load
► Effort delay
▼ 4h RC
▼ Proportional to load capacitance
4.57
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Contamination Delay
■ Best-case (contamination) delay can be
substantially less than propagation delay.
■ Ex: If both inputs fall simultaneously
2
2
A
2
B
2x
R R
Y
(6+4h)C
4.58
6C
Y
4hC
2C
tcdr   3  2h  RC
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
Diffusion Capacitance
■ We assumed contacted diffusion on every s / d.
■ Good layout minimizes diffusion area
■ Ex: NAND3 layout shares one diffusion contact
► Reduces output capacitance by 2C
► Merged un-contacted diffusion might help too
2C
Shared
Contacted
Diffusion
2C
Isolated
Contacted
Diffusion
Merged
Uncontacted
Diffusion
2
2
3
3
3C 3C 3C
4.59
2
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
3
7C
3C
3C
Layout Comparison
■ Which layout is better?
VDD
A
VDD
B
Y
GND
4.60
A
B
Y
GND
Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis
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