Upload
trozn
View
220
Download
0
Embed Size (px)
Citation preview
8/3/2019 124 Lecture 3
1/22
ECE 124AECE 124A
Prof. Kaustav BanerjeeElectrical and Computer Engineering
E-mail: kaustav ece.ucsb.edu
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
1
8/3/2019 124 Lecture 3
2/22
NOR GateNOR Gate
2 PMOS must be in series..
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
2
2 NMOS must be in parallel.
8/3/2019 124 Lecture 3
3/22
8/3/2019 124 Lecture 3
4/22
CMOS 3CMOS 3--input NOR Implementationinput NOR Implementation
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
4
8/3/2019 124 Lecture 3
5/22
Combinational LogicCombinational Logic
Output state X should always beavoided: static power
Output state Z is of relevance incertain gates such as multiplexers
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
5
8/3/2019 124 Lecture 3
6/22
Compound GatesCompound Gates Needs 20 transistors.Y= A.B + C.D
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
6
8/3/2019 124 Lecture 3
7/22
Compound GatesCompound GatesY= A.B + C.D
Pull-down (when is Y =0?)
Pull-up (when is Y=1?)
Need 8 transistors
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
7
8/3/2019 124 Lecture 3
8/22
8/3/2019 124 Lecture 3
9/22
Pass TransistorsPass Transistors
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
9
8/3/2019 124 Lecture 3
10/22
Transmission Gates:Transmission Gates:Pass Transistors in ParallelPass Transistors in Parallel
Both 0 and 1 passed strongly
s
s
Double Rail Logic: both the control input and its complement is required
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
10
8/3/2019 124 Lecture 3
11/22
Tristate BufferTristate Buffer
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
11
8/3/2019 124 Lecture 3
12/22
Tristate BufferTristate Buffer
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
12
8/3/2019 124 Lecture 3
13/22
Transmission Gate as Tristate BufferTransmission Gate as Tristate Buffer
Non-restorin :
input-signalwill slowly
anumber of
stages
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
13
8/3/2019 124 Lecture 3
14/22
Tristate Buffer as InverterTristate Buffer as Inverter
directly connectedto Vdd or GND
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
14
8/3/2019 124 Lecture 3
15/22
Multiplexer (MUX)Multiplexer (MUX)Connects one of n inputs to the output.
Used as data selectorsencoders
2:1 MUX
A
BY Y
C
D
21
inputs
1 output
inputs
ss1
s21 Select signal
2 Select signals
Y = As +Bs Y = As1s2 + +Bs1s2 + Cs1s2 + Ds1s2
12n m is a minterm of the
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
15
In general, 2n inputs will have n select signals
0k
kkImY
n control variable andIk is the corresponding
data input
8/3/2019 124 Lecture 3
16/22
Multiplexer (MUX)Multiplexer (MUX)
Y = D1.S +D0.S
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
16
8/3/2019 124 Lecture 3
17/22
NonNon--restoring MUXrestoring MUX
Y = D1.S +D0.S
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
17
8/3/2019 124 Lecture 3
18/22
Inverting and Restoring MUXInverting and Restoring MUX
S/S = 0/1
= : = = . .
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
18
8/3/2019 124 Lecture 3
19/22
A 4:1 MUXA 4:1 MUX
Using three 2:1 MUXs
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
19
8/3/2019 124 Lecture 3
20/22
Static CMOS SummarStatic CMOS Summar In static circuits at every point in time (except when switching)
the out ut is connected to either GND or V via a lowresistance path.
fan-in of n(or n-inputs) requires 2n(nN-type + nP-type) devices
-NMOS sizes (since no conflict between pull-up and pull-downnetworks)
No ath ever exists between Vdd and GND: low static ower
Fully-restored logic: (NMOS passes 0 only and PMOS passes1 only
= , = -down network is ON) for Y=0 (node is fully discharged)
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles
20
8/3/2019 124 Lecture 3
21/22
LatcheLatche
CLK=1: D to Q
CLK=0:Holdsstate of Q
As long asCLK remainshi h: D to Q
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles21
8/3/2019 124 Lecture 3
22/22
FliFli Flo FloCombines two latches:
ne +ve sens t ve s ave an one ve
sensitive latch (master)Edge Triggered FF or Master-Slave FF Master Slave
CLK=0: D to QM
QM
QM = D
Slave holds previous value
QM
of QCLK=1: master cant sampleinput and holds value of D
Kaustav BanerjeeLecture 3, ECE 124A, VLSI Principles22
Slave opens and QM=(D) =Q