Upload
doxuyen
View
228
Download
0
Embed Size (px)
Citation preview
Basic VLSI Design Flow
SYSTEM SPECIFICATION
ARCHITECTURAL DESIGN
LOGIC DESIGN
CIRCUIT DESIGN
DEVICE DESIGNLAYOUT
(Algorithmic Level)
(Register Transfer Level)
(Gate Level)
(Transistor Level)- - - - - - - - - - - - - - - - - -
Physical Design
FABRICATIONON-WAFER TESTING
PACKAGING
CHIP TESTING
Simulations at various levels
SYSTEM SPECIFICATION
ARCHITECTURAL DESIGN
LOGIC DESIGN
CIRCUIT DESIGN
DEVICE DESIGNLAYOUT
FABRICATIONON-WAFER TESTING
PACKAGING
CHIP TESTING
(Functional Simulation)
(Block level Simulation)
(Logic Simulation)
(Circuit Simulation)
(Device Simulation)
(Process Simulation)
(Fault Simulation)
OPTIMIZATION PROBLEM
Often they are considered to be constraint-satisfying problemrather than optimization problem.
SPEED
AREA
POWER
PRIMARY OBJECTIVE FUNCTIONS
The actual cost function involving these factorsdepends on specific application.
Design challenge is to have a better trade-off.
Concept of Hierarchy
• Continent• Country• State• District• City/Town/Village• House• Room• Wall• Brick• … …
• System• Subsystem• Functional unit• Functional subunit• Flip flop/MUX/Adder• Gate• Transistor• Device structure• … …
Constraint flow
Backannotation
FLOW OF CONSTRAINTS AND BACKANNOTATIONS
System Specification(Speed, Area, Power etc.)
Leaf cell parameters( Characterization data )
OPTIMIZATION AT ALGORITHMIC LEVEL
EXAMPLE 1 : DISCRETE FOURIER TRANSFORM (DFT)O(N ) [ Unoptimized ]
FAST FOURIER TRANSFORM (FFT)O(N log N) [ Optimized ]
2
2
EXAMPLE 2 : Sum of Natural Numbers
S = 1 + 2 + 3 + . . . . . . . . + NS = N(N + 1) / 2
⇒ (N − 1) Additions⇒ 1 Increment and
1 Multiplication( Division by 2 is a mere SHIFT
in binary arithmetic )
OPTIMIZATION AT ARCHITECTURAL LEVEL
input NS = 0for i = 1 to N
S = S + inext ioutput S
N Si
input NS = 0for i = N downto 1
S = S + inext ioutput S
i S
OPTIMIZATION AT LOGIC LEVEL
20 Transistors
2:1 MULTIPLEXER
4
4
42
S
B
A Y
14 Transistors
6
6
62
S
A
B
Y
OPTIMIZATION AT CIRCUIT LEVEL
2:1 MULTIPLEXER USING TRANSMISSION GATE LOGIC
S
S
S
A
B
Y
6 Transistors(including 2 for inverting S)
S
A
A
B
B
C
C
D
D
VDD
Y
Y = (AB + C)D
16 Transistors 8 Transistors
Optimized transistor level realization of Boolean function
ABCD
Y
ANALOG VLSI
Where it differs from theDesign with discrete componentsin a PCB or a breadboard?
• In the early days of Integrated Circuits:⇒ absence of Capacitors
• At present: ⇒ various constraints:Technology [CMOS]Area [e.g. in SoC]Noise [e.g. in mixed signal design]
ANALOG VLSI
At present: various constraints:Technology [CMOS]Area [e.g. in SoC]Noise [e.g. in mixed signal design]
☺ overall improvements:Higher Packing density Low Power consumptionHigher BandwidthHigh degree of Matching
ANALOG VLSI
Where do we stand today ?Thermionic Valve
Discrete Transistor
Integrated Circuits
MSI VLSI & beyondLSI
• Millions of Transistors• Deep Submicron Technology• Quantum Devices (?)
ANALOG VLSI
Challenges for Everyone:
• System designer• Circuit designer• Device designer• Layout engineer• Fabrication team• Packaging people
♥ To push the performance limits
Capacitors
Poly-to-diffusion capacitor
Poly-to-poly capacitor
Metal-to-poly capacitor
Metal-to-metal capacitor
Concept of Process corners
nMOS
Typ (VT = 500 mV)
Fast (VT = 400 mV)
Slow (VT = 600 mV)
pMOS
Typ (⏐VT ⏐= 550 mV)
Fast (⏐VT ⏐= 440 mV)
Slow (⏐VT ⏐ = 660 mV)
vin
io
+VDD
vout
Voltage gain: Av= vout/ vin= gmN/(gdsN+ gdsP)
Vbias
P-MOS
N-MOS
CMOS Amplifier with Current Source (Active) Load
vin
io
+VDD
vout
Voltage gain: Av= vout/ vin= (gmN+ gmP)/(gdsN+ gdsP)
N-MOS
P-MOS
CMOS Push Pull Amplifier
vin
io
+VDD
vout
N-MOS
P-MOS
It is interesting to note that the same circuit can work asan analog amplifier as well as a digital inverter
vin
vout
vin
+VDD
vout
N-MOS
P-MOS
Propagation delay of the digital inverter fully dependsupon the device parameters
CL
tpLH = Rp (Cout + CL )
tpHL = Rn (Cout + CL )
voutvin gmvin roRL
vout = ( ro⏐⏐RL ) gmvin
Low frequency model of an active device
io
vin
io
Three TerminalActive Device ≡
voutvin gmvin roRL
vout = ( ro⏐⏐RL ) gmvin
Typical High frequency model of an active device
Cin
CMiller io
voutvin gmvin
roRL
High frequency model of an active deviceand its corresponding frequency response
Cin
Co
Introduces two poles: one corresponds to ( ro⏐⏐RL )Coand another rs Cin (rs: source resistance of the driver)
f
|Av|
First Pole
Second Pole