COMP 103
Lecture 10
Inverter Dynamics:
The Quest for Performance
Section 5.4.2, 5.4.3
[All lecture notes are adapted from Mary Jane Irwin, Penn State, which were adapted from Rabaey’s Digital Integrated
Circuits, ©2002, J. Rabaey et al.]
COMP103-L10.1
What is this lecture+ about? PERFORMANCE
1.
Factors affecting performance: CL , W/L, VDD,
2.
The ratio of the PMOS to NMOS could be optimized for
symmetrical tpHL & tpLH and symmetric VTC… but here we learn how
to set the ratio to optimize tp
3.
While sizing up an inverter reduces its delay, it also increase its
input capacitance – impacting the delay of the driving gate! (selfOut
loading). What’s the best sizing? In
1
Cg,1
4.
Now we can size a chain of inverters..
If CL is given
-
How should the inverters be sized?
How many stages are needed to minimize the delay?
5.
What about input slope impact (instead of a step)?
6.
What about inverters with long wire delays inbetween?
COMP103-L10.2
CL = 8 Cg,1
Inverter Propagation Delay, revisited
‰
Propagation delay is proportional to the time-constant of
the network formed by the pull-down resistor and the load
capacitance
VDD
tpHL = f(Rn, CL)
Vout = 0
Rn
CL
Vin = V DD
‰
tpHL = ln(2) Reqn CL = 0.69 Reqn CL
tpLH = ln(2) Reqp CL = 0.69 Reqp CL
tp = (tpHL + tpLH)/2 = 0.69 CL(Reqn + Reqp)/2
To equalize rise and fall times make the on-resistance of
the NMOS and PMOS approximately equal.
COMP103-L10.3
Inverter Propagation Delay, Revisited
‰
To see how a designer can optimize the delay of a gate
have to expand the Req in the delay equation
5.5
For VGS = VDD,
VDS = VDD-> VDD/2
5
tp(normalized)
4.5
tpHL = 0.69 Reqn CL
4
3.5
3
2.5
2
1.5
= 0.69 (3/4 (VDD)/IDSATn ) CL
≈ 0.52 CL / (W/Ln k’n VDSATn )
COMP103-L10.4
1
0.8
1
1.2
1.4
1.6
1.8
VDD (V)
2
2.2
2.4
Derivation of Reqn
COMP103-L10.5
Yes…. That should have looked familiar..
Same results with simulated
equivalent resistance of a
minimum size NMOS
transistor
Req (Ohm)
‰
7
x105
(for VGS = VDD,
VDS = VDD→VDD/2)
6
5
4
3
2
1
0
0.5
1
1.5
2
2.5
VDD (V)
VDD(V)
NMOS(kΩ)
PMOS (kΩ)
COMP103-L10.6
1
35
115
1.5
19
55
2
15
38
2.5
13
31
Design for Performance
Reduce CL
‰
z
internal diffusion capacitance of the gate itself
- keep the drain diffusion as small as possible
z
z
interconnect capacitance
fanout
Increase W/L ratio of the transistor
‰
z
z
the most powerful and effective performance optimization
tool in the hands of the designer
watch out for self-loading! – when the intrinsic capacitance
dominates the extrinsic load
Increase VDD
‰
z
z
z
can trade-off energy for performance
increasing VDD above a certain level yields only very minimal
improvements
reliability concerns enforce a firm upper bound on VDD
COMP103-L10.7
NMOS/PMOS Ratio
‰
Concerns:
z
z
z
symmetrical VTC
equal high-to-low and low-to-high propagation delays
speed, tp
‰
So far have sized the PMOS and NMOS so that the Req’s match (ratio
of 2-3) ⇒ symmetric VTC, equal tpHL & tpLH
‰
If speed is the only concern, reduce the width of the PMOS device!
z
‰
What happens?
There must be a ratio β = (W/Lp)/(W/Ln) that optimizes tp!!
r = Reqp/Reqn (resistance ratio of identically-sized PMOS and NMOS)
βopt = √r when wiring capacitance is negligible
COMP103-L10.8
Derivation of βopt
COMP103-L10.9
PMOS/NMOS Ratio Effects -- Simulation
-11
5 x 10
tpLH
tp(sec)
4.5
tpHL
β of 2.4 (= 31 kΩ/13 kΩ)
gives symmetrical
response
tp
4
β of 1.6 to 1.9 gives
optimal performance
3.5
3
1
2
3
β = (W/Lp)/(W/Ln)
COMP103-L10.10
4
5
Device Sizing for Performance
‰ Divide capacitive load, CL, into
z
Cint : intrinsic - diffusion and Miller effect
Cext : extrinsic - wiring and fanout
z
where tp0 = 0.69 Req Cint is the intrinsic (unloaded) delay of the gate
z
tp = tp0 (1 + Cext/Cint)
‰
Widening both PMOS and NMOS by a factor S reduces
Req by an identical factor (Req = Rref/S), but raises the
intrinsic capacitance by the same factor (Cint = SCiref)
tp = 0.69 Rref Ciref (1 + Cext/(SCiref)) = tp0(1 + Cext/(SCiref))
z
z
z
tp0 is independent of the sizing of the gate
with no load, there is no gain. The drive of the gate is totally
offset by the increased capacitance
any S sufficiently larger than (Cext/Cint) yields the best
performance gains with least area impact
COMP103-L10.11
Example of finding S
‰
Given a time budget of 4 ps, tp0 = 2ps, Cext=9 ff,
Cint = 3ff, determine the smallest S that would allow tp to
meet the timing budget.
COMP103-L10.12
Sizing Impacts on Delay
The majority of the
improvement is already
obtained for S = 5. Sizing
factors larger than 10
barely yield any extra gain
(and cost significantly
more area).
x 10-11
3.8
for a fixed load
3.6
3.4
3.2
tp(sec)
3
2.8
2.6
2.4
2.2
2
1
3
5
7
9
11
13
15
S
self-loading effect
(intrinsic capacitance
dominates)
COMP103-L10.13
Can’t study delay in isolation
‰
Simplest case studied in an inverter chain… but basics
apply (creatively) to other cases…
In
Out
Cg,1
COMP103-L10.14
1
2
N
CL
Impact of Fanout on Delay
‰
Extrinsic capacitance, Cext, is a function of the fanout of
the gate - the larger the fanout, the larger the external
load.
‰
Two stages:
First, determine the relationship between input loading Cg and output
loading Cint , both are proportional to the gate sizing. Define:
γ = Cint /Cg
Second, determine the relationship between the Cext and Cg
f = Cext/Cg (f is the effective fan-out)
‰ tp
‰
= tp0 (1 + Cext/ Cint) =
Because γ is close to 1 in most processes, the delay of an inverter is a
function of the ratio between its external load capacitance and its input
gate capacitance: f
COMP103-L10.15
Inverter Chain
‰
Our goal is to minimize the delay through an inverter
chain
In
Out
Cg,1
1
2
N
CL
The delay through the stages:of the j-th inverter stage is
tp, total = ∑tp,j = tp0 ∑ (1 + Cg,j+1/(γCg,j))
COMP103-L10.16
Sizing Inverter Chains: The Questions
‰
If Cg,1 and CL is given
z
Given a fixed number of inverter stages, how should the
inverters be sized?
z
How many stages are needed to minimize the delay? And what
sizes should they be?
COMP103-L10.17
Sizing the Inverters in the Chain
‰
How many unknowns are there? (check with the tp, total equation)
‰
Take N-1 partial derivatives, and equate to 0.
‰
Result: constraints:
z
‰
Cg, j+1 / Cg, j = Cg, j / Cg, j -1,
with j = 2, .. N
The optimum size of each inverter is the geometric mean of its neighbors –
Each gate will have the same effective fan-out and the same delay. If each
inverter is sized up by the same factor f wrt the preceding gate, then,
N
N
f = √CL/Cg,1 = √F
where F represents the overall effective fan-out of the circuit (F = CL/Cg,1)
and the minimum delay through the inverter chain is
tp = N tp0 (1 + ( √F ) / γ)
COMP103-L10.18
Example of Inverter Chain Sizing
In
Out
Cg,1
‰
1
CL = 8 Cg,1
CL/Cg,1 has to be evenly distributed over N = 3 inverters
CL/Cg,1 = 8/1
f =
COMP103-L10.19