Post on 02-Oct-2021
Introduction to CMOS VLSI
Design
Lecture 14: Circuit Families
David Harris, Harvey Mudd College Kartik Mohanram and Steven Levitan
University of Pittsburgh
CMOS VLSI Design 9: Circuit Families Slide 2
Outline q Pseudo-nMOS Logic q Dynamic Logic q Pass Transistor Logic
CMOS VLSI Design 9: Circuit Families Slide 3
Introduction q What makes a circuit fast?
– I = C dV/dt -> tpd ∝ (C/I) ΔV – low capacitance – high current – small swing
q Logical effort is proportional to C/I q pMOS are the enemy!
– High capacitance for a given current q Can we take the pMOS capacitance off the input? q Various circuit families try to do this…
BA
11
4
4
Y
CMOS VLSI Design 9: Circuit Families Slide 4
Pseudo-nMOS q In the old days, nMOS processes had no pMOS
– Instead, use pull-up transistor that is always ON q In CMOS, use a pMOS that is always ON
– Ratio issue – Make pMOS about ¼ effective strength of
pulldown network
Vout
Vin16/2
P/2
Ids
load
0 0.3 0.6 0.9 1.2 1.5 1.80
0.3
0.6
0.9
1.2
1.5
1.8
P = 24
P = 4
P = 14
Vin
Vout
CMOS VLSI Design 9: Circuit Families Slide 5
Pseudo-nMOS Gates q Design for unit current on output
to compare with unit inverter. q pMOS fights nMOS
Inverter NAND2 NOR2
AY
B
AY
A B
gu =gd =gavg =pu =pd =pavg =
Y
gu =gd =gavg =pu =pd =pavg =
gu =gd =gavg =pu =pd =pavg =
finputs
Y
CMOS VLSI Design 9: Circuit Families Slide 6
Pseudo-nMOS Gates q Design for unit current on output
to compare with unit inverter. q pMOS fights nMOS
Inverter NAND2 NOR2
4/3
2/3
AY
8/3
8/3
2/3
B
AY
A B 4/34/3
2/3
gu =gd =gavg =pu =pd =pavg =
Y
gu =gd =gavg =pu =pd =pavg =
gu =gd =gavg =pu =pd =pavg =
finputs
Y
CMOS VLSI Design 9: Circuit Families Slide 7
Pseudo-nMOS Gates q Design for unit current on output
to compare with unit inverter. q pMOS fights nMOS
Inverter NAND2 NOR2
4/3
2/3
AY
8/3
8/3
2/3
B
AY
A B 4/34/3
2/3
gu = 4/3gd = 4/9gavg = 8/9pu =pd =pavg =
Y
gu = 8/3gd = 8/9gavg = 16/9pu =pd =pavg =
gu = 4/3gd = 4/9gavg = 8/9pu =pd =pavg =
finputs
Y
CMOS VLSI Design 9: Circuit Families Slide 8
Pseudo-nMOS Gates q Design for unit current on output
to compare with unit inverter. q pMOS fights nMOS
Inverter NAND2 NOR2
4/3
2/3
AY
8/3
8/3
2/3
B
AY
A B 4/34/3
2/3
gu = 4/3gd = 4/9gavg = 8/9pu = 6/3pd = 6/9pavg = 12/9
Y
gu = 8/3gd = 8/9gavg = 16/9pu = 10/3pd = 10/9pavg = 20/9
gu = 4/3gd = 4/9gavg = 8/9pu = 10/3pd = 10/9pavg = 20/9
finputs
Y
CMOS VLSI Design 9: Circuit Families Slide 9
Pseudo-nMOS Design q Ex: Design a k-input AND gate using pseudo-nMOS.
Estimate the delay driving a fanout of H
q G = q F = q P = q N = q D =
In1
Ink
Y
Pseudo-nMOS1
1 H
CMOS VLSI Design 9: Circuit Families Slide 10
Pseudo-nMOS Design q Ex: Design a k-input AND gate using pseudo-nMOS.
Estimate the delay driving a fanout of H
q G = 1 * 8/9 = 8/9 q F = GBH = 8H/9 q P = 1 + (4+8k)/9 = (8k+13)/9 q N = 2 q D = NF1/N + P =
In1
Ink
Y
Pseudo-nMOS1
1 H
4 2 8 133 9H k ++
CMOS VLSI Design 9: Circuit Families Slide 11
Pseudo-nMOS Power q Pseudo-nMOS draws power whenever Y = 0
– Called static power P = I•VDD
– A few mA / gate * 1M gates would be a problem – This is why nMOS went extinct!
q Use pseudo-nMOS sparingly for wide NORs q Turn off pMOS when not in use
A BY
C
en
CMOS VLSI Design 9: Circuit Families Slide 12
Dynamic Logic q Dynamic gates uses a clocked pMOS pullup q Two modes: precharge and evaluate
1
2A Y
4/3
2/3
AY
1
1
AY
φ
Static Pseudo-nMOS Dynamic
φ Precharge Evaluate
Y
Precharge
CMOS VLSI Design 9: Circuit Families Slide 13
The Foot q What if pulldown network is ON during precharge? q Use series evaluation transistor to prevent fight.
AY
φ
foot
precharge transistor φY
inputs
φY
inputs
footed unfooted
f f
CMOS VLSI Design 9: Circuit Families Slide 14
Logical Effort Inverter NAND2 NOR2
1
1
AY
2
2
1
B
AY
A B 11
1
gd =pd =
gd =pd =
gd =pd =
Yφ
φ
φ
2
1
AY
3
3
1
B
AY
A B 22
1
gd =pd =
gd =pd =
gd =pd =
Yφ
φ
φ
footed
unfooted
32 2
CMOS VLSI Design 9: Circuit Families Slide 15
Logical Effort Inverter NAND2 NOR2
1
1
AY
2
2
1
B
AY
A B 11
1
gd = 1/3pd = 2/3
gd = 2/3pd = 3/3
gd = 1/3pd = 3/3
Yφ
φ
φ
2
1
AY
3
3
1
B
AY
A B 22
1
gd = 2/3pd = 3/3
gd = 3/3pd = 4/3
gd = 2/3pd = 5/3
Yφ
φ
φ
footed
unfooted
32 2
CMOS VLSI Design 9: Circuit Families Slide 16
Monotonicity q Dynamic gates require monotonically rising inputs
during evaluation – 0 -> 0 – 0 -> 1 – 1 -> 1 – But not 1 -> 0
φ Precharge Evaluate
Y
Precharge
A
Output should rise but does not
violates monotonicity during evaluation
A
φ
CMOS VLSI Design 9: Circuit Families Slide 17
Monotonicity Woes q But dynamic gates produce monotonically falling
outputs during evaluation q Illegal for one dynamic gate to drive another!
A X
φ Yφ Precharge Evaluate
X
Precharge
A = 1
Y
CMOS VLSI Design 9: Circuit Families Slide 18
Monotonicity Woes q But dynamic gates produce monotonically falling
outputs during evaluation q Illegal for one dynamic gate to drive another!
A X
φ Yφ Precharge Evaluate
X
Precharge
A = 1
Y should rise but cannot
Y
X monotonically falls during evaluation
CMOS VLSI Design 9: Circuit Families Slide 19
Domino Gates q Follow dynamic stage with inverting static gate
– Dynamic / static pair is called domino gate – Produces monotonic outputs
φ Precharge Evaluate
W
Precharge
X
Y
Z
A
φ
B C
φ φ φ
C
AB
W X Y Z =X
ZH H
A
W
φ
B C
X Y Z
domino AND
dynamicNAND
staticinverter
CMOS VLSI Design 9: Circuit Families Slide 20
Domino Optimizations q Each domino gate triggers next one, like a string of
dominos toppling over q Gates evaluate sequentially but precharge in parallel q Thus evaluation is more critical than precharge q HI-skewed static stages can perform logic
S0
D0
S1
D1
S2
D2
S3
D3
φ
S4
D4
S5
D5
S6
D6
S7
D7
φ
YH
CMOS VLSI Design 9: Circuit Families Slide 21
Dual-Rail Domino q Domino only performs noninverting functions:
– AND, OR but not NAND, NOR, or XOR q Dual-rail domino solves this problem
– Takes true and complementary inputs – Produces true and complementary outputs
sig_h sig_l Meaning 0 0 Precharged 0 1 ‘0’ 1 0 ‘1’ 1 1 invalid
Y_h
f
φ
φ
inputs
Y_l
f
CMOS VLSI Design 9: Circuit Families Slide 22
Example: AND/NAND q Given A_h, A_l, B_h, B_l q Compute Y_h = A * B, Y_l = ~(A * B)
CMOS VLSI Design 9: Circuit Families Slide 23
Example: AND/NAND q Given A_h, A_l, B_h, B_l q Compute Y_h = A * B, Y_l = ~(A * B) q Pulldown networks are conduction complements
Y_hφ
φ
Y_lA_h
B_hB_lA_l
= A*B= A*B
CMOS VLSI Design 9: Circuit Families Slide 24
Example: XOR/XNOR q Sometimes possible to share transistors
Y_hφ
φ
Y_lA_l
B_h
= A xor B
B_l
A_hA_lA_h= A xnor B
CMOS VLSI Design 9: Circuit Families Slide 25
Leakage q Dynamic node floats high during evaluation
– Transistors are leaky (IOFF ≠ 0) – Dynamic value will leak away over time – Formerly miliseconds, now nanoseconds!
q Use keeper to hold dynamic node – Must be weak enough not to fight evaluation
A
φH
2
2
1 kX Y
weak keeper
CMOS VLSI Design 9: Circuit Families Slide 26
Charge Sharing q Dynamic gates suffer from charge sharing
B = 0
AY
φ
x
Cx
CY
A
φ
x
Y
CMOS VLSI Design 9: Circuit Families Slide 27
Charge Sharing q Dynamic gates suffer from charge sharing
B = 0
AY
φ
x
Cx
CY
A
φ
x
Y
Charge sharing noise
x YV V= =
CMOS VLSI Design 9: Circuit Families Slide 28
Charge Sharing q Dynamic gates suffer from charge sharing
B = 0
AY
φ
x
Cx
CY
A
φ
x
Y
Charge sharing noise
Yx Y DD
x Y
CV V VC C
= =+
CMOS VLSI Design 9: Circuit Families Slide 29
Secondary Precharge q Solution: add secondary precharge transistors
– Typically need to precharge every other node q Big load capacitance CY helps as well
B
AY
φ
x
secondaryprechargetransistor
CMOS VLSI Design 9: Circuit Families Slide 30
Noise Sensitivity q Dynamic gates are very sensitive to noise
– Inputs: VIH ≈ Vtn
– Outputs: floating output susceptible noise q Noise sources
– Capacitive crosstalk – Charge sharing – Power supply noise – Feedthrough noise – And more!
CMOS VLSI Design 9: Circuit Families Slide 31
Domino Summary q Domino logic is attractive for high-speed circuits
– 1.5 – 2x faster than static CMOS – But many challenges:
• Monotonicity • Leakage • Charge sharing • Noise
q Widely used in high-performance microprocessors
CMOS VLSI Design 9: Circuit Families Slide 32
Pass Transistor Circuits q Use pass transistors like switches to do logic q Inputs drive diffusion terminals as well as gates
q CMOS + Transmission Gates: – 2-input multiplexer – Gates should be restoring
A
B
S
S
S
Y
A
B
S
S
S
Y
CMOS VLSI Design 9: Circuit Families Slide 33
LEAP q LEAn integration with Pass transistors q Get rid of pMOS transistors
– Use weak pMOS feedback to pull fully high – Ratio constraint
B
S
SA
YL
CMOS VLSI Design 9: Circuit Families Slide 34
CPL q Complementary Pass-transistor Logic
– Dual-rail form of pass transistor logic – Avoids need for ratioed feedback – Optional cross-coupling for rail-to-rail swing
B
S
S
S
S
A
B
AY
YL
L