ELEC 5270-001/6270-001 Spring 2009
Low-Power Design of Electronic Circuits
Power Dissipation of CMOS Circuits
Vishwani D. Agrawal
James J. Danaher Professor
Dept. of Electrical and Computer Engineering
Auburn University, Auburn, AL 36849
vagrawal@eng.auburn.edu
http://www.eng.auburn.edu/~vagrawal/COURSE/E6270_Spr09/course.html
Copyright Agrawal, 2007
ELEC5270/6270 Spring 09, Lecture 4
1
nMOS Logic (Inverters)
For logic 1 input,
continuous static
power is dissipated.
Pseudo-nMOS
Saturated-load nMOS
R. C. Jaeger and T. N. Blalock, Microelctronic Circuit Design,
Third Edition, McGraw-Hill, 2006, Chapter 6.
Copyright Agrawal, 2007
ELEC5270/6270 Spring 09, Lecture 4
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CMOS Logic (Inverter)
No static leakage
path exists for either
1 or 0 input.
F. M. Wanlass and C. T. Sah, “Mamowatt Logic using FieldEffect Metal-Oxide-Semiconductor Triodes,” IEEE International
Solid-State Circuits Conference Digest, vol. IV, February 1963,
pp. 32-33.
Copyright Agrawal, 2007
ELEC5270/6270 Spring 09, Lecture 4
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Components of Power

Dynamic

Signal transitions
 Logic
activity
 Glitches


Short-circuit
Static

Leakage
Copyright Agrawal, 2007
Ptotal =
=
Pdyn + Pstat
Ptran + Psc + Pstat
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Power of a Transition: Ptran
Ron
VDD
ic(t)
vi (t)
vo(t)
CL
R = large
Ground
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ELEC5270/6270 Spring 09, Lecture 4
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Charging of a Capacitor
R
t=0
v(t)
i(t)
C
V
Charge on capacitor, q(t)
=
C v(t)
Current, i(t)
=
C dv(t)/dt
Copyright Agrawal, 2007
=
dq(t)/dt
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C dv(t)/dt =
[V – v(t)] /R
dv(t)
V – v(t)
───
=
─────
dt
RC
dv(t)
dt
∫ ───── = ∫ ────
V – v(t)
RC
-t
ln [V – v(t)]
=
── + A
RC
i(t)
=
Initial condition, t = 0, v(t) = 0 → A = ln V
-t
v(t) =
V [1 – exp(───)]
RC
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ELEC5270/6270 Spring 09, Lecture 4
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v(t) =
i(t)
Copyright Agrawal, 2007
=
-t
V [1 – exp( ── )]
RC
dv(t)
C ───
dt
=
ELEC5270/6270 Spring 09, Lecture 4
V
-t
── exp( ── )
R
RC
8
Total Energy Per Charging Transition
from Power Supply
Etrans =
=
Copyright Agrawal, 2007
∞
∫ V i(t) dt =
0
CV
∞V
2
-t
∫ ── exp( ── ) dt
0 R
RC
2
ELEC5270/6270 Spring 09, Lecture 4
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Energy Dissipated per Transition in
Resistance
∞2
R ∫ i (t) dt
0
Copyright Agrawal, 2007
=
V ∞
-2t
R ──
∫
exp(
──
)
dt
R2 0
RC
=
1
2
─ CV
2
2
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Energy Stored in Charged
Capacitor
∞
∞
-t V
-t
∫ v(t) i(t) dt = ∫ V [1-exp( ── )] ─ exp( ── ) dt
0
0
RC R
RC
1
2
= ─ CV
2
Copyright Agrawal, 2007
ELEC5270/6270 Spring 09, Lecture 4
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Transition Power

Gate output rising transition
2
Energy dissipated in pMOS transistor = CV /2
2
 Energy stored in capacitor = CV /2


Gate output falling transition

2
Energy dissipated in nMOS transistor = CV /2
2
Energy dissipated per transition = CV /2
 Power dissipation:

Ptrans =
α
Copyright Agrawal, 2007
2
Etrans α fck =
α fck CV /2
=
activity factor
ELEC5270/6270 Spring 09, Lecture 4
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Components of Power

Dynamic

Signal transitions
 Logic
activity
 Glitches


Short-circuit
Static

Leakage
Copyright Agrawal, 2007
Ptotal =
=
Pdyn + Pstat
Ptran + Psc + Pstat
ELEC5270/6270 Spring 09, Lecture 4
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Short Circuit Power of a Transition: Psc
VDD
vi (t)
isc(t)
vo(t)
CL
Ground
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ELEC5270/6270 Spring 09, Lecture 4
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Short Circuit Current, isc (t)
VDD
VDD - VTp
Vi (t)
Volt
p-transistor
starts
conducting
n-transistor
cuts-off
Vo(t)
VTn
0
Iscmaxf
isc(t)
Isc
0
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tB
tE
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Time (ns)
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Peak Short Circuit Current
Increases with the size (or gain, β) of
transistors
 Decreases with load capacitance, CL
 Largest when CL = 0
 Reference: M. A. Ortega and J. Figueras,
“Short Circuit Power Modeling in
Submicron CMOS,” PATMOS ’96, Aug.
1996, pp. 147-166.

Copyright Agrawal, 2007
ELEC5270/6270 Spring 09, Lecture 4
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Short-Circuit Energy per Transition

Escf =
=
tE
∫tB
VDD isc(t)dt
(tE – tB) Iscmaxf VDD / 2

Escf =
tf (VDD - |VTp| - VTn) Iscmaxf / 2

Escr =
tr (VDD - |VTp| - VTn) Iscmaxr / 2

Escf = Escr = 0, when VDD = |VTp| + VTn
Copyright Agrawal, 2007
ELEC5270/6270 Spring 09, Lecture 4
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Short-Circuit Energy
Increases with rise and fall times of input
 Decreases for larger output load
capacitance
 Decreases and eventually becomes zero
when VDD is scaled down but the
threshold voltages are not scaled down

Copyright Agrawal, 2007
ELEC5270/6270 Spring 09, Lecture 4
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Short-Circuit Power Calculation
Assume equal rise and fall times
 Model input-output capacitive coupling
(Miller capacitance)
 Use a spice model for transistors


T. Sakurai and A. Newton, “Alpha-power Law
MOSFET model and Its Application to a
CMOS Inverter,” IEEE J. Solid State Circuits,
vol. 25, April 1990, pp. 584-594.
Copyright Agrawal, 2007
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Short Circuit Power
Psc =
Copyright Agrawal, 2007
α fck Esc
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Psc, Rise Time and Capacitance
VDD
Ron
vi (t)
tf
VDD
ic(t)+isc(t)
vo(t)
CL
R = large
vo(t)
tr
Ground
Copyright Agrawal, 2007
ELEC5270/6270 Spring 09, Lecture 4
vo(t)
───
R↑
21
isc, Rise Time and Capacitance
Isc(t) =
Copyright Agrawal, 2007
-t
VDD[1- exp(─────)]
vo(t)
R↓(t) C
──── = ──────────────
R↑(t)
R↑(t)
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iscmax, Rise Time and Capacitance
i
Small C
vo(t)
Large C
vo(t)
1
────
R↑(t)
iscmax
t
tf
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ELEC5270/6270 Spring 09, Lecture 4
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Psc, Rise Times, Capacitance
For given input rise and fall times short
circuit power decreases as output
capacitance increases.
 Short circuit power increases with increase
of input rise and fall times.
 Short circuit power is reduced if output rise
and fall times are smaller than the input
rise and fall times.

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ELEC5270/6270 Spring 09, Lecture 4
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Summary: Short-Circuit Power




Short-circuit power is consumed by each
transition (increases with input transition time).
Reduction requires that gate output transition
should not be faster than the input transition
(faster gates can consume more short-circuit
power).
Increasing the output load capacitance reduces
short-circuit power.
Scaling down of supply voltage with respect to
threshold voltages reduces short-circuit power;
completely eliminated when VDD ≤ |Vtp| + Vtn .
Copyright Agrawal, 2007
ELEC5270/6270 Spring 09, Lecture 4
25
Components of Power

Dynamic

Signal transitions
 Logic
activity
 Glitches


Static

Copyright Agrawal, 2007
Short-circuit
Leakage
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Leakage Power
IG
Ground
Gate
VDD
R
Source
Drain
n+
Bulk Si (p)
Isub
IPT
IGIDL
n+
ID
nMOS Transistor
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Leakage Current Components
Subthreshold conduction, Isub
 Reverse bias pn junction conduction, ID
 Gate induced drain leakage, IGIDL due to
tunneling at the gate-drain overlap
 Drain source punchthrough, IPT due to
short channel and high drain-source
voltage
 Gate tunneling, IG through thin oxide; may
become significant with scaling

Copyright Agrawal, 2007
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Subthreshold Current
Isub = μ0 Cox (W/L) Vt2 exp{(VGS –VTH ) / nVt }
μ0: carrier surface mobility
Cox: gate oxide capacitance per unit area
L: channel length
W: gate width
Vt = kT/q: thermal voltage
n: a technology parameter
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IDS for Short Channel Device
Isub= μ0 Cox(W/L)Vt2 exp{(VGS –VTH + ηVDS)/nVt}
VDS = drain to source voltage
η: a proportionality factor
W. Nebel and J. Mermet (Editors), Low Power Design in Deep Submicron
Electronics, Springer, 1997, Section 4.1 by J. Figueras, pp. 81-104
Copyright Agrawal, 2007
ELEC5270/6270 Spring 09, Lecture 4
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Increased Subthreshold Leakage
Log (Drain current)
Scaled device
Ic
Isub
0 VTH’ VTH
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Gate voltage
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Summary: Leakage Power




Leakage power as a fraction of the total power
increases as clock frequency drops. Turning
supply off in unused parts can save power.
For a gate it is a small fraction of the total power;
it can be significant for very large circuits.
Scaling down features requires lowering the
threshold voltage, which increases leakage
power; roughly doubles with each shrinking.
Multiple-threshold devices are used to reduce
leakage power.
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Technology Scaling
Scaling down 0.7 micron by factors 2 and
4 leads to 0.35 and 0.17 micron
technologies
 Constant electric field assumed

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Constant Electric Field Scaling
B. Davari, R. H. Dennard and G. G.
Shahidi, “CMOS Scaling for High
Performance and Low Power—The Next
Ten Years,” Proc. IEEE, April 1995, pp.
595-606.
 Other forms of scaling are referred to as
constant-voltage and quasi-constantvoltage.

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Bulk nMOSFET
Polysilicon
Gate
Drain
W
Source
n+
n+
L
p-type body (bulk)
SiO2
Thickness = tox
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Technology Scaling



A scaling factor (S ) reduces device dimensions as
1/S.
Successive generations of technology have used a
scaling S = √2, doubling the number of transistors
per unit area. This produced 0.25μ, 0.18μ, 0.13μ,
90nm and 65nm technologies, continuing on to
45nm and 30nm.
A 5% gate shrink (S = 1.05) is commonly applied to
boost speed as the process matures.
N. H. E. Weste and D. Harris, CMOS VLSI Design, Third Edition, Boston:
Pearson Addison-Wesley, 2005, Section 4.9.1.
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Constant Electric Field Scaling
Device Parameter
Scaling
Length, L
1/S
Width, W
1/S
Gate oxide thickness, tox
1/S
Supply voltage, VDD
1/S
Threshold voltages, Vtn, Vtp
1/S
Substrate doping, NA
Copyright Agrawal, 2007
ELEC5270/6270 Spring 09, Lecture 4
S
37
Constant Electric Field Scaling(Cont.)
Device Characteristic
β
Scaling
W / (L tox)
S
β (VDD – Vt ) 2
1/S
Resistance, R
VDD / Ids
1
Gate capacitance, C
W L / tox
1/S
Gate delay, τ
RC
1/S
Clock frequency, f
1/ τ
S
CV 2 f
1/S 2
Current, Ids
Dynamic power per gate, P
1/S 2
Chip area, A
Power density
P/A
1
Current density
Ids /A
S
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Problem: A Design Example


A battery-operated 65nm digital CMOS device is found
to consume equal amounts (P ) of dynamic power and
leakage power. The short-circuit power is negligible.
The energy consumed by a computing task, that takes
T seconds, is 2PT.
Compare two power reduction strategies for extending
the battery life:
A.
B.
Clock frequency is reduced to half, keeping all other
parameters constant.
Supply voltage is reduced to half. This slows the gates down
and forces the clock frequency to be lowered to half of its
original (full voltage) value. Assume that leakage current is
held unchanged by modifying the design of transistors.
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Solution: Strategy A. Clock
Frequency Reduction


Reducing the clock frequency will reduce
dynamic power to P / 2, keep the static
power the same as P, and double the
execution time of the task.
Energy consumption for the task will be,
Energy = (P / 2 + P ) 2T = 3PT
which is greater than the original 2PT.
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Solution: Part B. Supply Voltage
Reduction


When the supply voltage and clock frequency
are reduced to half their values, dynamic
power is reduced to P / 8 and static power to
P / 2. The time of task is doubled and the total
energy consumption is,
Energy = (P / 8 + P / 2) 2T = 5PT / 4 =1.25PT
The voltage reduction strategy reduces
energy consumption while a simple frequency
reduction consumes more energy.
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Comparing Strategies A and B
Original
A
B
Voltage
V
V
V/2
Clock Frequency
F
F/2
F/2
Task Duration
T
2T
2T
Power
2P
1.5P
0.625P
Energy
2PT
3PT
1.25P
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