Chapter1<1>
DigitalDesignandComputerArchitecture,2nd Edition
Chapter1
DavidMoneyHarrisandSarahL.Harris
Chapter1<2>
• Background• TheGamePlan• TheArtofManagingComplexity• TheDigitalAbstraction• NumberSystems• LogicGates• LogicLevels• CMOSTransistors• PowerConsumption
Chapter1::Topics
Chapter1<3>
• Microprocessorshaverevolutionizedourworld– Cellphones,Internet,rapidadvancesinmedicine,etc.
• Thesemiconductorindustryhasgrownfrom$21billionin1985to$300billionin2011
Background
Chapter1<4>
• Purposeofcourse:– Understandwhat’sunderthehoodofacomputer– Learntheprinciplesofdigitaldesign– Learntosystematicallydebugincreasinglycomplexdesigns
– Designandbuildamicroprocessor
TheGamePlan
Chapter1<5>
• Abstraction• Discipline• TheThree–y’s
– Hierarchy– Modularity– Regularity
TheArtofManagingComplexity
Chapter1<6>
• Hiding details when they aren’t important
focu
s of
this
cou
rse
programs
device drivers
instructionsregisters
datapathscontrollers
addersmemories
AND gatesNOT gates
amplifiersfilters
transistorsdiodes
electrons
Abstraction
Chapter1<7>
• Intentionally restrict design choices • Example: Digital discipline
– Discrete voltages instead of continuous– Simpler to design than analog circuits – can build more sophisticated
systems– Digital systems replacing analog predecessors:
• i.e., digital cameras, digital television, cell phones, CDs
Discipline
Chapter1<8>
• Hierarchy
• Modularity
• Regularity
TheThree-y’s
Chapter1<9>
• Hierarchy– A system divided into modules and submodules
• Modularity– Having well-defined functions and interfaces
• Regularity– Encouraging uniformity, so modules can be easily reused
TheThree-y’s
Chapter1<10>
• Hierarchy– Three main modules:
lock, stock, and barrel– Submodules of lock:
hammer, flint, frizzen, etc.
Example:TheFlintlockRifle
Chapter1<11>
• Modularity– Function of stock: mount
barrel and lock– Interface of stock: length
and location of mounting pins
• Regularity– Interchangeable parts
Example:TheFlintlockRifle
Chapter1<12>
• Mostphysicalvariablesarecontinuous– Voltageonawire– Frequencyofanoscillation– Positionofamass
• Digitalabstractionconsidersdiscretesubset ofvalues
TheDigitalAbstraction
Chapter1<13>
• DesignedbyCharlesBabbagefrom1834–1871
• Consideredtobethefirstdigitalcomputer
• Builtfrommechanicalgears,whereeachgearrepresentedadiscretevalue(0-9)
• Babbagediedbeforeitwasfinished
TheAnalyticalEngine
Chapter1<14>
• Twodiscretevalues:– 1’sand0’s– 1,TRUE,HIGH– 0,FALSE,LOW
• 1and0:voltagelevels,rotatinggears,fluidlevels,etc.
• Digitalcircuitsusevoltage levelstorepresent1and0
• Bit: Binarydigit
DigitalDiscipline:BinaryValues
Chapter1<15>
• Born to working class parents• Taught himself mathematics and
joined the faculty of Queen’sCollege in Ireland
• Wrote An Investigation of the Lawsof Thought (1854)
• Introduced binary variables• Introduced the three fundamental
logic operations: AND, OR, andNOT
GeorgeBoole,1815-1864
Chapter1<16>
537410 =
10's column
100's column
1000's column
1's column
11012 =
2's column
4's column
8's column
1's column
• Decimal numbers
• Binary numbers
NumberSystems
Chapter1<17>
537410 = 5 × 103 + 3 × 102 + 7 × 101 + 4 × 100five
thousands
10's column
100's column
1000's column
threehundreds
seventens
fourones
1's column
11012 = 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 = 1310oneeight
2's column
4's column
8's column
onefour
notwo
oneone
1's column
• Decimal numbers
• Binary numbers
NumberSystems
Chapter1<18>
• 20 =• 21 = • 22 =• 23 =• 24 = • 25 = • 26 =• 27 =
• 28 =• 29 = • 210 =• 211 =• 212 = • 213 = • 214 =• 215 =
PowersofTwo
Chapter1<19>
• 20 = 1• 21 = 2• 22 = 4• 23 = 8• 24 = 16• 25 = 32• 26 = 64• 27 = 128• Handy to memorize up to 29
• 28 = 256• 29 = 512• 210 = 1024• 211 = 2048• 212 = 4096• 213 = 8192• 214 = 16384• 215 = 32768
PowersofTwo
Chapter1<20>
• Decimal to binary conversion:– Convert 100112 to decimal
• Decimal to binary conversion:– Convert 4710 to binary
NumberConversion
Chapter1<21>
• Decimal to binary conversion:– Convert 100112 to decimal– 16×1 + 8×0 + 4×0 + 2×1 + 1×1 = 1910
• Decimal to binary conversion:– Convert 4710 to binary– 32×1 + 16×0 + 8×1 + 4×1 + 2×1 + 1×1 = 1011112
NumberConversion
Chapter1<22>
• N-digitdecimalnumber– Howmanyvalues?– Range?– Example:3-digitdecimalnumber:
• N-bitbinarynumber– Howmanyvalues?– Range:– Example:3-digitbinarynumber:
BinaryValuesandRange
Chapter1<23>
• N-digitdecimalnumber– Howmanyvalues?10N– Range?[0,10N - 1]– Example:3-digitdecimalnumber:
• 103 =1000possiblevalues• Range:[0,999]
• N-bitbinarynumber– Howmanyvalues?2N– Range:[0,2N - 1]– Example:3-digitbinarynumber:
• 23 =8possiblevalues• Range:[0,7]=[0002 to1112]
BinaryValuesandRange
Chapter1<24>
Hex Digit Decimal Equivalent Binary Equivalent
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
A 10
B 11
C 12
D 13
E 14
F 15
HexadecimalNumbers
Chapter1<25>
Hex Digit Decimal Equivalent Binary Equivalent
0 0 0000
1 1 0001
2 2 0010
3 3 0011
4 4 0100
5 5 0101
6 6 0110
7 7 0111
8 8 1000
9 9 1001
A 10 1010
B 11 1011
C 12 1100
D 13 1101
E 14 1110
F 15 1111
HexadecimalNumbers
Chapter1<26>
• Base 16• Shorthand for binary
HexadecimalNumbers
Chapter1<27>
• Hexadecimal to binary conversion:– Convert 4AF16 (also written 0x4AF) to binary
• Hexadecimal to decimal conversion:– Convert 0x4AF to decimal
HexadecimaltoBinaryConversion
Chapter1<28>
• Hexadecimal to binary conversion:– Convert 4AF16 (also written 0x4AF) to binary– 0100 1010 11112
• Hexadecimal to decimal conversion:– Convert 4AF16 to decimal– 162×4 + 161×10 + 160×15 = 119910
HexadecimaltoBinaryConversion
Chapter1<29>
• Bits
• Bytes&Nibbles
• Bytes
10010110nibble
byte
CEBF9AD7least
significantbyte
mostsignificantbyte
10010110least
significantbit
mostsignificant
bit
Bits,Bytes,Nibbles…
Chapter1<30>
• 210 =1kilo ≈ 1000(1024)• 220 =1mega ≈ 1million(1,048,576)• 230 =1giga ≈ 1billion(1,073,741,824)
LargePowersofTwo
Chapter1<31>
• Whatisthevalueof224?
• Howmanyvaluescana32-bitvariablerepresent?
EstimatingPowersofTwo
Chapter1<32>
• Whatisthevalueof224?24 × 220 ≈16million
• Howmanyvaluescana32-bitvariablerepresent?22 × 230 ≈4billion
EstimatingPowersofTwo
Chapter1<33>
37345168+
10110011+
• Decimal
• Binary
Addition
Chapter1<34>
37345168+8902
carries 11
10110011+1110
11 carries
• Decimal
• Binary
Addition
Chapter1<35>
10010101+
10110110+
• Add the following 4-bit binary numbers
• Add the following 4-bit binary numbers
BinaryAdditionExamples
Chapter1<36>
10010101+1110
1
10110110+10001
111
• Add the following 4-bit binary numbers
• Add the following 4-bit binary numbers
Overflow!
BinaryAdditionExamples
Chapter1<37>
• Digital systems operate on a fixed number of bits
• Overflow: when result is too big to fit in the available number of bits
• See previous example of 11 + 6
Overflow
Chapter1<38>
• Sign/Magnitude Numbers• Two’s Complement Numbers
SignedBinaryNumbers
Chapter1<39>
• 1 sign bit, N-1 magnitude bits• Sign bit is the most significant (left-most) bit
– Positive number: sign bit = 0– Negative number: sign bit = 1
• Example, 4-bit sign/mag representations of ± 6:+6 =- 6 =
• Range of an N-bit sign/magnitude number:
{ }1
1 2 2 1 0
2
0
: , , , ,
( 1) 2n
N N
na i
ii
A a a a a a
A a=
=
L
Sign/MagnitudeNumbers
Chapter1<40>
• 1 sign bit, N-1 magnitude bits• Sign bit is the most significant (left-most) bit
– Positive number: sign bit = 0– Negative number: sign bit = 1
• Example, 4-bit sign/mag representations of ± 6:+6 = 0110- 6 = 1110
• Range of an N-bit sign/magnitude number:[-(2N-1-1), 2N-1-1]
{ }1
1 2 2 1 0
2
0
: , , , ,
( 1) 2n
N N
na i
ii
A a a a a a
A a=
=
L
Sign/MagnitudeNumbers
Chapter1<41>
• Problems:– Addition doesn’t work, for example -6 + 6:
1110 + 011010100 (wrong!)
– Two representations of 0 (± 0):1000 0000
Sign/MagnitudeNumbers
Chapter1<42>
• Don’t have same problems as sign/magnitude numbers:– Addition works– Single representation for 0
Two’sComplementNumbers
Chapter1<43>
( )2
11
0
2 2n
n in i
iA a a
=
= +
• Msb has value of -2N-1
• Most positive 4-bit number:• Most negative 4-bit number:• The most significant bit still indicates the sign
(1 = negative, 0 = positive)• Range of an N-bit two’s comp number:
Two’sComplementNumbers
Chapter1<44>
( )2
11
0
2 2n
n in i
iA a a
=
= +
• Msb has value of -2N-1
• Most positive 4-bit number: 0111• Most negative 4-bit number: 1000• The most significant bit still indicates the sign
(1 = negative, 0 = positive)• Range of an N-bit two’s comp number:
[-(2N-1), 2N-1-1]
Two’sComplementNumbers
Chapter1<45>
• Flip the sign of a two’s complement number• Method:
1. Invert the bits2. Add 1
• Example: Flip the sign of 310 = 00112
“TakingtheTwo’sComplement”
Chapter1<46>
• Flip the sign of a two’s complement number• Method:
1. Invert the bits2. Add 1
• Example: Flip the sign of 310 = 001121. 11002. + 1
1101 = -310
“TakingtheTwo’sComplement”
Chapter1<47>
• Take the two’s complement of 610 = 01102
• What is the decimal value of 10012?
Two’sComplementExamples
Chapter1<48>
• Take the two’s complement of 610 = 011021. 10012. + 1
10102 = -610
• What is the decimal value of the two’s complement number 10012?
1. 01102. + 1
01112 = 710, so 10012 = -710
Two’sComplementExamples
Chapter1<49>
+01101010
+11100011
• Add 6 + (-6) using two’s complement numbers
• Add -2 + 3 using two’s complement numbers
Two’sComplementAddition
Chapter1<50>
+0110101010000
111
+1110001110001
111
• Add 6 + (-6) using two’s complement numbers
• Add -2 + 3 using two’s complement numbers
Two’sComplementAddition
Chapter1<51>Copyright © 2012 Elsevier
• Extend number from N toM bits (M > N) :– Sign-extension– Zero-extension
IncreasingBitWidth
Chapter1<52>
• Sign bit copied to msb’s• Number value is same
• Example 1:– 4-bit representation of 3 = 0011– 8-bit sign-extended value: 00000011
• Example 2:– 4-bit representation of -5 = 1011– 8-bit sign-extended value: 11111011
Sign-Extension
Chapter1<53>
• Zeros copied to msb’s• Value changes for negative numbers
• Example 1:– 4-bit value = 00112 = 310
– 8-bit zero-extended value: 00000011 = 310
• Example 2:– 4-bit value = 1011 = -510
– 8-bit zero-extended value: 00001011 = 1110
Zero-Extension
Chapter1<54>
-8
1000 1001
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1010 1011 1100 1101 1110 1111 0000 0001 0010 0011 0100 0101 0110 0111 Two's Complement
10001001101010111100110111101111
00000001 0010 0011 0100 0101 0110 0111
1000 1001 1010 1011 1100 1101 1110 11110000 0001 0010 0011 0100 0101 0110 0111
Sign/Magnitude
Unsigned
Number System RangeUnsigned [0, 2N-1]Sign/Magnitude [-(2N-1-1), 2N-1-1]Two’s Complement [-2N-1, 2N-1-1]
For example, 4-bit representation:
NumberSystemComparison
Chapter1<55>
• Performlogicfunctions:– inversion(NOT),AND,OR,NAND,NOR,etc.
• Single-input:– NOTgate,buffer
• Two-input:– AND,OR,XOR,NAND,NOR,XNOR
• Multiple-input
LogicGates
Chapter1<56>
NOT
Y = A
A Y01
A Y
BUF
Y = A
A Y01
A Y
Single-InputLogicGates
Chapter1<57>
NOT
Y = A
A Y0 11 0
A Y
BUF
Y = A
A Y0 01 1
A Y
Single-InputLogicGates
Chapter1<58>
AND
Y = AB
A B Y0 00 11 01 1
AB Y
OR
Y = A + B
A B Y0 00 11 01 1
AB Y
Two-InputLogicGates
Chapter1<59>
AND
Y = AB
A B Y0 0 00 1 01 0 01 1 1
AB Y
OR
Y = A + B
A B Y0 0 00 1 11 0 11 1 1
AB Y
Two-InputLogicGates
Chapter1<60>
XNOR
Y = A + B
A B Y0 00 11 01 1
AB Y
XOR NAND NOR
Y = A + B Y = AB Y = A + B
A B Y0 00 11 01 1
A B Y0 00 11 01 1
A B Y0 00 11 01 1
AB Y A
B Y AB Y
MoreTwo-InputLogicGates
Chapter1<61>
XNOR
Y = A + B
A B Y0 00 11 01 1
AB Y
XOR NAND NOR
Y = A + B Y = AB Y = A + B
A B Y0 0 00 1 11 0 11 1 0
A B Y0 0 10 1 11 0 11 1 0
A B Y0 0 10 1 01 0 01 1 0
AB Y A
B Y AB Y
1001
MoreTwo-InputLogicGates
Chapter1<62>
NOR3
Y = A+B+C
B C Y0 00 11 01 1
AB YC
A0000
0 00 11 01 1
1111
Multiple-InputLogicGates
AND3
Y = ABC
AB YC
B C Y0 00 11 01 1
A0000
0 00 11 01 1
1111
Chapter1<63>
NOR3
Y = A+B+C
B C Y0 00 11 01 1
AB YC
A0000
0 00 11 01 1
1111
10000000
• Multi-inputXOR:Oddparity
Multiple-InputLogicGates
AND3
Y = ABC
AB YC
B C Y0 00 11 01 1
A0000
0 00 11 01 1
1111
00000001
Chapter1<64>
• Discretevoltagesrepresent1and0• Forexample:
– 0=ground (GND)or0volts– 1=VDD or5volts
• Whatabout4.99volts?Isthata0ora1?• Whatabout3.2volts?
LogicLevels
Chapter1<65>
• Range ofvoltagesfor1and0• Differentrangesforinputsandoutputstoallowfornoise
LogicLevels
Chapter1<66>
WhatisNoise?
Chapter1<67>
• Anythingthatdegradesthesignal– E.g.,resistance,powersupplynoise,couplingtoneighboringwires,etc.
• Example: agate(driver)outputs5Vbut,becauseofresistanceinalongwire,receivergets4.5V
Driver ReceiverNoise
5 V 4.5 V
WhatisNoise?
Chapter1<68>
• Withlogicallyvalidinputs,everycircuitelementmustproducelogicallyvalidoutputs
• Uselimitedrangesofvoltagestorepresentdiscretevalues
TheStaticDiscipline
Chapter1<69>
Driver Receiver
ForbiddenZone
NML
NMH
Input CharacteristicsOutput Characteristics
VO H
VDD
VO L
GND
VIH
VIL
Logic HighInput Range
Logic LowInput Range
Logic HighOutput Range
Logic LowOutput Range
LogicLevels
Chapter1<70>
Driver Receiver
ForbiddenZone
NML
NMH
Input CharacteristicsOutput Characteristics
VO H
VDD
VO L
GND
VIH
VIL
Logic HighInput Range
Logic LowInput Range
Logic HighOutput Range
Logic LowOutput Range
NMH = VOH – VIH
NML = VIL – VOL
NoiseMargins
Chapter1<71>
VDD
V(A)
V(Y)
VOH VDD
VOL
VIL, VIH
0
A Y
VDD
V(A)
V(Y)
VOH
VDD
VOL
VIL VIH
Unity GainPoints
Slope = 1
0VDD / 2
Ideal Buffer: Real Buffer:
NMH = NML = VDD/2 NMH , NML < VDD/2
DCTransferCharacteristics
Chapter1<72>
ForbiddenZone
NML
NMH
Input CharacteristicsOutput CharacteristicsVDD
VO L
GND
VIHVIL
VO H
A Y
VDD
V(A)
V(Y)
VOHVDD
VOL
VIL VIH
Unity GainPoints
Slope = 1
0
DCTransferCharacteristics
Chapter1<73>
• In1970’sand1980’s,VDD =5V• VDD hasdropped
– Avoidfryingtinytransistors– Savepower
• 3.3V,2.5V,1.8V,1.5V,1.2V,1.0V,…• Becarefulconnectingchipswithdifferentsupplyvoltages
Chipsoperatebecausetheycontainmagicsmoke
Proof:– ifthemagicsmokeisletout,thechipstopsworking
VDD Scaling
Chapter1<74>
Logic Family VDD VIL VIH VOL VOHTTL 5 (4.75 - 5.25) 0.8 2.0 0.4 2.4
CMOS 5 (4.5 - 6) 1.35 3.15 0.33 3.84
LVTTL 3.3 (3 - 3.6) 0.8 2.0 0.4 2.4
LVCMOS 3.3 (3 - 3.6) 0.9 1.8 0.36 2.7
LogicFamilyExamples
Chapter1<75>
g
s
d
g = 0
s
d
g = 1
s
d
OFF ON
• Logic gates built from transistors• 3-ported voltage-controlled switch
– 2 ports connected depending on voltage of 3rd– d and s are connected (ON) when g is 1
Transistors
Chapter1<76>
• Nicknamed “Mayor of Silicon Valley”
• Cofounded Fairchild Semiconductor in 1957
• Cofounded Intel in 1968• Co-invented the integrated
circuit
RobertNoyce,1927-1990
Chapter1<77>
Silicon Lattice
Si SiSi
Si SiSi
Si SiSi
As SiSi
Si SiSi
Si SiSi
B SiSi
Si SiSi
Si SiSi
-
+
+
-
Free electron Free hole
n-Type p-Type
• Transistors built from silicon, a semiconductor• Pure silicon is a poor conductor (no free charges)• Doped silicon is a good conductor (free charges)
– n-type (free negative charges, electrons)– p-type (free positive charges, holes)
Silicon
Chapter1<78>
n
p
gatesource drain
substrate
SiO2
nMOS
Polysilicon
n
gate
source drain
• Metal oxide silicon (MOS) transistors: – Polysilicon (used to be metal) gate– Oxide (silicon dioxide) insulator– Doped silicon
MOSTransistors
Chapter1<79>
n
p
gatesource drain
substrate
n n
p
gatesource drain
substrate
n
GND
GNDVDD
GND
+++++++- - - - - - -
channel
Gate = 0
OFF (no connection between source and drain)
Gate = 1
ON (channel between source and drain)
Transistors:nMOS
Chapter1<80>
• pMOS transistorisopposite– ONwhenGate=0– OFFwhenGate=1
SiO2
n
gatesource drainPolysilicon
p p
gate
source drain
substrate
Transistors:pMOS
Chapter1<81>
gs
d
g = 0
s
d
g = 1
s
d
gd
s
d
s
d
s
nMOS
pMOS
OFF ON
ON OFF
TransistorFunction
Chapter1<82>
• nMOS:passgood0’s,soconnectsourcetoGND
• pMOS:passgood1’s,soconnectsourcetoVDD
pMOSpull-upnetwork
outputinputs
nMOSpull-downnetwork
TransistorFunction
Chapter1<83>
VDD
A Y
GND
N1
P1
NOT
Y = A
A Y0 11 0
A Y
A P1 N1 Y
0
1
CMOSGates:NOTGate
Chapter1<84>
VDD
A Y
GND
N1
P1
NOT
Y = A
A Y0 11 0
A Y
A P1 N1 Y
0 ON OFF 1
1 OFF ON 0
CMOSGates:NOTGate
Chapter1<85>
A
B
Y
N2N1
P2 P1
NAND
Y = AB
A B Y0 0 10 1 11 0 11 1 0
AB Y
A B P1 P2 N1 N2 Y0 00 11 01 1
CMOSGates:NANDGate
Chapter1<86>
A
B
Y
N2N1
P2 P1
NAND
Y = AB
A B Y0 0 10 1 11 0 11 1 0
AB Y
A B P1 P2 N1 N2 Y0 0 ON ON OFF OFF 10 1 ON OFF OFF ON 11 0 OFF ON ON OFF 11 1 OFF OFF ON ON 0
CMOSGates:NANDGate
Chapter1<87>
pMOSpull-upnetwork
outputinputs
nMOSpull-downnetwork
CMOSGateStructure
Chapter1<88>
How do you build a three-input NOR gate?
NORGate
Chapter1<89>
B
CY
A
NOR3Gate
Chapter1<90>
How do you build a two-input AND gate?
OtherCMOSGates
Chapter1<91>
AB Y
AND2Gate
Chapter1<92>
• nMOS pass1’spoorly• pMOS pass0’spoorly• Transmissiongateisabetterswitch
– passesboth0and1well
• WhenEN =1,theswitchisON:– EN =0andA isconnectedtoB
• WhenEN =0,theswitchisOFF:– A isnotconnectedtoB
A B
EN
EN
TransmissionGates
Chapter1<93>
• Replacepull-upnetworkwithweak pMOStransistorthatisalwayson
• pMOS transistor:pullsoutputHIGHonlywhennMOS networknotpullingitLOW
Y
inputs nMOSpull-downnetwork
weak
Pseudo-nMOS Gates
Chapter1<94>
Pseudo-nMOS NOR4
A BY
weak
C D
Pseudo-nMOS Example
Chapter1<95>
• Cofounded Intel in 1968 with Robert Noyce.
• Moore’s Law:number of transistors on a computer chip doubles every year (observed in 1965)
• Since 1975, transistor counts have doubled every two years.
GordonMoore,1929-
Chapter1<96>
• “Iftheautomobilehadfollowedthesamedevelopmentcycleasthecomputer,aRolls-Roycewouldtodaycost$100,getonemillionmilestothegallon,andexplodeonceayear...”
– RobertCringley
Moore’sLaw
Chapter1<97>
• Power=Energyconsumedperunittime– Dynamicpowerconsumption– Staticpowerconsumption
PowerConsumption
Chapter1<98>
• Powertochargetransistorgatecapacitances– Energyrequiredtochargeacapacitance,C,toVDD isCVDD
2
– Circuitrunningatfrequencyf:transistorsswitch(from1to0orviceversa)atthatfrequency
– Capacitorischargedf/2timespersecond(dischargingfrom1to0isfree)
• Dynamicpowerconsumption:
Pdynamic =½CVDD2f
DynamicPowerConsumption
Chapter1<99>
• Powerconsumedwhennogatesareswitching
• Causedbythequiescentsupplycurrent,IDD(alsocalledtheleakagecurrent)
• Staticpowerconsumption:
Pstatic =IDDVDD
StaticPowerConsumption
Chapter1<100>
• Estimatethepowerconsumptionofawirelesshandheldcomputer– VDD =1.2V– C =20nF– f=1GHz– IDD =20mA
PowerConsumptionExample
Chapter1<101>
• Estimatethepowerconsumptionofawirelesshandheldcomputer– VDD =1.2V– C =20nF– f=1GHz– IDD =20mA
P =½CVDD2f +IDDVDD
=½(20nF)(1.2V)2(1GHz)+(20mA)(1.2V)
=(14.4+0.024)W≈14.4W
PowerConsumptionExample