Transcript slides

Logic and Computer Design Fundamentals
Chapter 7 – Registers and Register
Transfer
Charles Kime & Thomas Kaminski
© 2008 Pearson Education, Inc.
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Propagation Delay: capacitances
introduce delay
 All physical devices have delays due to the charging
and discharging of (parasitic capacitors) and
interconnection resistance.
Crossing lines
level
tpd
t
Resistive tube
Propagation delay
flow
2
Where to these capacitors come
from?
Most are parasitic caps
Cross section of Altera Stratix EP1S25 (TSMC proces
Crossing lines
passivation
Crossing interconnections
(Source: IBM)
9 level of interconnections
M9
M8
M7
M6
M5
M4
M3
M2
M1
(Source: TSMC)
2um
3
Review: 6-3 Flip-Flop Timing
Parameters and metastability
 Edge triggered flip-flop
In
C
D
Q
C
Q
C
In (D)
tS
set-up time
th
hold time
 If one violates the set-up time and hold
time, the flip-flop can go in a metastable
state.
4
Out
Metastable behavior
 Example of metastable behavior:
metastable
Eventually, the flipflop will settle
(Oscilloscope trace)
 After a while the flip-flop will go into a stable state
(randomly).
 If this happens before the next clock edge, the actual
circuits will see a defined input.
 The longer the clock period is the less chance of
synchronization failure.
 Or use two synchronization flip-flops in series
5
Debouncing
PRACTICAL
CONSIDERATIONS
6
Practical considerations
 In many cases the input signals come from
push buttons and sensors:
• Switches exhibit bouncing
• Are not synchronous
• Give multiple input (sampled many times)
 This can give catastrophic failures of
timing problems
 Solutions:
• Debounce the switch
• Synchronizer
• One-pulse circuit (and synchronizer)
7
Bouncing of switches
 What is bouncing?
 When closing a switch, the switch contact
bounces back and forth, giving multiple
pulses
5V
Push
button
switch
chip
A
B
Causes spurious glitches on input B
Measured signal
8
Debouncing circuit
 The two resistors will keep S=R=0:
• Inputs are never floating
• Output does not bounce
5V
R
R
1
Q
S
S
2
Push switch
contact to 5V
Q
Q
Bouncing
No bouncing
SR latch
Is set
S=R=0
No change
9
Likely violation of set up times
ISSUES WITH
ASYNCHRONOUS INPUTS
10
Asynchronous inputs
 Asynchronous inputs (e.g. from sensors)
can cause timing problems if set up time is
violated: tS setup time
CLK
Signal from
sensor
This setup time violation can give problems, such as
metastability and erroneous operation:
stable
metastable
stable
 Use a synchronizer
11
Metastability and Synchronization
 A flip-flop can get into a metastable state:
• When the set-up or hold times are violated
• This can happen for asynchronous inputs (e.g. push
button; sensor input, etc)
Input
D0i
CLK
D
Sensor
asynchronous
input
Q
Q0
Q0
Q1
D1
D
Q
Both flip flops behave differently!
Q1
CLK
kick
kick
Q0
Q0
Q1
12
Metastability (continued)
 Asynchronous inputs:
• Use a synchronizing flip-flop
Synchronizer
Asynchronous
input
Di D
D0 D
Q
Q
Q0
Q
Q1
Ai
D1i D
• Now, only the synchronizer may get into the metastable
state; however, there is a good chance that by the end of
the clock period it went into one of the stable states.
13
Synchronizer
Sensor,async Di
D
Q
Ai
Synchronized output
CLK
CLK
Signal from
sensor
Synchronized output
Note: by having the sensor input enter through a single D flip-flop
the problem of timing circuits is reduced greatly (there is still a small
chance that the D FF will be unstable). By using two D flip-flops in
series one further reduces the risk that the 2 nd FF will be
metastable.
14
Chapter 7
Registers and Register Transfers
(Source: reference.findtarget.com)
15
7-1 Registers
 Register – a collection of flip-flops,
together with some combinational
logic, that performs data processing
tasks (e.g. storing, moving data, etc).
 In theory, a register is sequential
logic which can be defined by a state
table.
 More often think of a register as
storing a vector of binary values.
16
Simple Storage Register
inputs
Outputs
• A D flip-flop register loads information on every
clock cycle!
• To “store” or “load” information should be
controlled by a signal.
Active low Clear (asynch)
17
Register with Parallel Load: Clock
Gating
 Use a signal to block the clock to the register
 Load is a frequent name for the signal that
controls register storage and loading
• Load = 1: Load the values on the data inputs
• Load = 0: Store the values in the register
Clock gating:
1
Timing:
Gated clock: To C input of Flip-flops
Clock
Load
Gated Clock to FF
Extra edge!
18
Registers with Load-Controlled
Feedback
 A more reliable way to selectively load a register:




• Run the clock continuously, and
• Selectively use a load control to change the register
contents.
2-to-1 Multiplexers
Example: 2-bit register
with Load Control:
For Load = 0,
A1
loads register contents
Y1
D Q
(hold current values)
Load
In1
For Load = 1,
C
loads input values
(load new values)
A0
Y0
Hardware more complex
D Q
C
than clock gating, but
free of timing problems
In0
Clock
19
Registers with Load Control:
Example
K1=1: R1 stores the
result of
addition/subtraction
20
7-2 Register Transfers
 The data is stored in registers, which compose
the datapath.
(Source: reference.findtarget.com)
 In many cases one wants to perform a variety of
arithmetic and logical operations on a set of data
bits (e.g. a 16-bit word):
• Additions, subtraction, shifting, loading, etc.
21
Register Transfers Operations
 The type of operations on the data will be
determined by a controller, often called a control
unit.
(Source: reference.findtarget.com)
 This division between the datapath and control
unit makes the design of complex systems easier.
 Strategy: divide and conquer!
 Registers play a key role in complex digital
systems!
22
Datapath and Control Unit
Control signals
Control Unit
Status signals
 Determines which operation and
sequence of operation based on
status signals
 Control signals: activate various
operations in the datapath
 Is a large finite state machine
Datapath
 Stores data in registers
 Performs operations on
data, specified by Cntr unit
 Provide status signals
 Is defined by registers
and its operations:
Register Transfer
Operations (RTL)
 An elementary operation:
microoperation
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7-3 Register Transfer Operations
 Register Transfer Operations – The movement
and processing of data stored in registers
 Three basic components:
• set of registers
• operations
• control of operations
 Elementary Operations -- load, count, shift, add,
bitwise "OR", etc.
• Elementary operations called microoperations
Register Notation:
Ex. 8-bit register R:
16-bit registers:
MSB
7
…1 0
LSB
R
15
8 7
0
PC(H)
PC(L)
15
0
R2
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Conditional Transfer

If (K1 =1) then (R2  R1) is
shortened to
K1: (R2  R1)
where K1 is a control variable
specifying a conditional
execution
of the microoperation.
K1
R1
n
Load
R2
Clock
Clock
Clock
K1
K1
Transfer occurs
here
No Transfers Occur Here
Transfer occurs
here
No Transfers Occur Here
25
7-5 Microoperations
 Logical Groupings:
• Transfer - move data from one set of registers to
another
• Arithmetic - perform arithmetic on data in registers
• Logic - manipulate data or use bitwise logical operations
• Shift - shift data in registers
Arithmetic operations
+ Addition
– Subtraction
* Multiplication
/ Division
Logical operations (bitwise)
 Logical OR
 Logical AND
 Logical Exclusive OR
 Not
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Example Microoperations
 Add the content of R1 to the content of R2
and place the result in R1.
R1 R1 + R2
 Multiply the content of R1 by the content
of R6 and place the result in PC.
PC  R1 * R6
 Exclusive OR the content of R1 with the
content of R2 and place the result in R1.
R1  R1  R2
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Example Microoperations (Continued)
 On condition K1 OR K2, the content of R1
is Logic bitwise ORed with the content of
R3 and the result placed in R1:
• (K1 + K2): R1  R1  R3
 NOTE: "+" (as in K1 + K2) and means
“OR.” In R1  R1 + R3, + means “plus.”
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Arithmetic Microoperations
Symbolic Designation
R0  R1 + R2
R0  R1
R0  R1 + 1
R0  R2 + R1 + 1
R1  R1 + 1
R1  R1 – 1
Description
Addition
Ones Complement
Two's Complement
R2 minus R1 (2's Comp)
Increment (count up)
Decrement (count down)
 Note that any register may be specified for
source 1, source 2, or destination.
 These simple microoperations operate on
the whole word
29
Implementation of an Arithmetic
Micro operation
Conditional microoperation:
X K1 : R1  R1 + R2
X K1 : R1  R1 + R2 + 1
Condition
30
Logical Microoperations
Symbolic
Designation
R0  R1
R0  R1  R2
R0  R1  R2
R0  R1  R2
Description
Bitwise NOT
Bitwise OR (sets bits)
Bitwise AND (clears bits)
Bitwise EXOR (complements bits)
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Logical Microoperations (continued)
 Let R1 = 10101010,
and R2 = 11110000
 Then after the operation, R0 becomes:
R0
01010101
Operation
R0  R1
11111010
R0  R1  R2
(sets bits)
10100000
R0  R1  R2
(clears bits)
01011010
R0  R1  R2
(complements bits)
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Shift Microoperations
Symbolic
Designation
 Let R2 = 11001001
R1  sl R2
 Then after the
operation, R1
R1  sr R2
becomes:
R1
10010010
01100100
Description
Shift Left
Shift Right
Operation
R1  sl R2
R1  sr R2
 Note: These shifts "zero fill". Sometimes a
separate flip-flop is used to provide the data
shifted in, or to “catch” the data shifted out.
 Other shifts are possible (rotates, arithmetic).
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7-6: Microoperation on a Single
Register
 Microoperation on a Single Register
•
•
•
•
•
MUX-based transfers
Shift registers (needed for Lab)
Ripple Counter
Synchronous binary counter
Other counters
34
7-6 Microoperations on a Single
Register
 The focus is on the implementation of
microperations with a SINGLE register as
the Destination of the results
 In addition to the register there is some
combinational logic needed. This is
considered part of the register and is
called DEDICATED logic (in contrast to
SHARED logic for multiple destination
registers).
 Multiplexer-based transfer: makes use of
multiplexers to allow multiple operations
on a single destination register (See next)
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Multiplexer-Based Transfers
 Multiplexers connected to register inputs produce
flexible transfer structures
 Consider the following
Load
circuit
K
R2
K1
n
n
Load
R1
2
Load
S
0
MUX
1
n
R0
Combinational circuit
 What are the corresponding Register Transfer
operations of the following implementation?
36
Register Transfer Operations
implemented by the following circuit
• K1.K2: R0  R1
• K1.K2: R0  R2
Load
R2
K1
n
n
Load
K2
Load
S
0
MUX
1
n
R0
R1
37
Exercise
 Consider the following circuit (Note: Clocks are omitted for
clarity)
Load
R2
K2
K1
S
n
0
n
Load
MUX
1
Load
n
R0
R1
 What are the corresponding Register Transfer operations?
38
Shift Registers
 Shift Registers move data laterally within the register toward
its MSB or LSB position
 In the simplest case, the shift register is simply a set of
D flip-flops connected in a row like this:
B
A
In
DQ
DQ
C
DQ
Out
DQ
CP
 Data input, In, is called a serial input or the shift right input.
 Data output, Out, is often called the serial output.
 The vector (A, B, C, Out) is called the parallel output.
39
Shift Registers (continued)
 The behavior of the
In
serial shift register
is given in the listing
on the lower right
 T0 is the register
Clock CP
state just before
Clock
the first clock
pulse occurs
T0
 T1 is after the
T1
first pulse and
T2
before the second.
T3
 Initially unknown
T4
states are denoted by “?” T5
T6
 Complete the last three
rows of the table
B
A
DQ
In
0
1
1
0
1
1
DQ
A
?
0
1
1
C
DQ
B
?
?
0
1
Out
DQ
C
?
?
?
0
Out
?
?
?
?
1
40
Shift Registers (continued)
B
A
In
DQ
DQ
C
DQ
Out
DQ
Clock CP
CP
T0
T1
T2
T3
T4
T5
In
0
1
1
0
1
1
T6
1
A
?
0
1
1
B
?
?
0
1
C
?
?
?
0
Out
?
?
?
?
0
1
1
1
0
1
1
1
0
0
1
1
41
Parallel Load Shift Registers
MUX
 By adding a mux
DA
between each shift register
A
stage, data can be
D
shifted or loaded
IN
Q
 If SHIFT is low,
A and B are
SHIFT
replaced by the data on DA
CP
and DB lines, else data
shifts right on each clock.
 By adding more bits, we
can make n-bit parallel load
shift registers.
SHIFT: Q  sr Q
SHIFT’: Q  D
 Register Transfer
Operation:
DB
B
D
Q
42
Parallel Load Shift Registers with Load
Serial input
D0
2
1
0
D
Q
S1 S0
Shift
Load
D1
Control
Shift
Load
D2
D0
C
D
Q
S1 S0
Use a multiplexer with 3 inputs:
Serial input
2
1
0
2
1
0
S1 S0
=
2
1
0
Q1
C
D
S1 S0
Shift
Load
Q0
Q
Q2
C
Shift
Load
43
Timing
Clock
Load
(Shift = 0)
Di
Qi
Output changes here
Register Transfer Operation:
Shift:
Q  sl Q
Shift.Load: Q  D
44
Bidirectional shift register
45
Counters
 Counters are sequential circuits which "count"
through a specific state sequence. They can
count up, count down, or count through other
fixed sequences. Two distinct types are in
common usage:
 Ripple Counters
• Clock is connected to the flip-flop clock input on the LSB
bit flip-flop
• For all other bits, a flip-flop output is connected to the
clock input, thus circuit is not truly synchronous
• Output change is delayed more for each bit toward the
MSB.
• Resurgent because of low power consumption
 Synchronous Counters
• Clock is directly connected to the flip-flop clock inputs
• Logic is used to implement the desired state sequencing
46
Ripple Counter
 How does it work?
• When there is a positive
Clock
edge on the clock input
of A, A complements
• The clock input for flipflop B is the complemented
output of flip-flop A
Reset
• When flip A changes
from 1 to 0, there is a
positive edge on the CP
clock input of B
A
causing B to
complement
B
0
1
A
D
CR
D
B
CR
2
3
0
1
47
Ripple Counter (continued)
 The arrows show the
cause-effect relationship from the prior
slide =>
 The corresponding
sequence of states =>
CP
A
B
0
1
2
3
0
1
(B,A) = (0,0),(0,1), (1,0), (1,1), (0,0), (0,1), …
Each additional bit, C, D, …behaves like bit B,
changing half as frequently as the bit before it.
 For 3 bits: (C,B,A) = (0,0,0), (0,0,1), (0,1,0), (0,1,1),
(1,0,0), (1,0,1), (1,1,0), (1,1,1), (0,0,0), …
48
Delays in a Ripple Counter
 Starting with C = B = A = 1, the next clock
increments the count to (C,B,A) = 0; thus
from (111) to (000)
 In fine timing detail: The clock to output
delay tPHL causes an increasing
delay from clock edge for tPHL
each stage transition. CP
• Thus, the count “ripples”
from least to most
A 1
significant bit.
• For n bits, total worst case
B 1
delay is n tPHL.
C
1
tPHL
0
0
0
tpHL
1
0
0
1
1
0
49
Ripple Counter (continued)
 These circuits are called ripple counters
because the changes “ripple” through the
chain of flip-flops, i. e., each transition
occurs after a clock-to-output delay from
the stage before.
 Disadvantages:
• Slow operation due to the delays
• Intermediate results with can give rise to
unreliable operation in digital circuits
 Advantages:
• Simple hardware
• Low power consumption
50
Synchronous Counters
 To eliminate the "ripple" effects, use a common
clock for each flip-flop and a combinational circuit
to generate the next state.
 For an up-counter,
use an incrementer =>
Comb. Logic
Incrementer
A3
S3
D3 Q3
A2
S2
D2 Q2
A1
S1
D1 Q1
A0
S0
D0 Q0
Clock
51
Synchronous counter
Clock
Q0
1
1
1
Q1
1
1
Q2
Q0: changes at every clock
EN
D
Q
Q0
When EN=1: XOR inverts and Q0=D’
C
Q0
Q1: changes when Q0 = 1:
D
EN
Q1
C
Q1
Q2: changes when Q0 and Q1 = 1
Q
D
Q
Q2
C
52
Synchronous counter – parallel
gating
Incrementer
•Replace AND carry
chain with ANDs =>
in parallel
•Advantages
•Reduces path delays
•Called parallel gating
•Like carry lookahead
Symbol:
Carry Out CO
53
Synchronous Counter with an
Arbitrary Sequence
 Start with the state
diagram or the state
table
 Decide which flipflops to use
 Find the
combinational logic
(inputs to the flipflops)
 Draw or build the
circuit
000
110
001
101
010
100
54
Design Exercise: Arbitrary
Sequence
000
110
001
101
010
100
55
Arbitrary Count sequence: next state
equation
 3 Flip-flops required
 Find the next state equations: DC, DB, DA.
DA
Be careful with
Table sequence
DB
DA = A’B+AB’
= AB
DC = B’C’
DB = C
A
B
0 1
x 0
0 1
x 0
C
A
B
0 0
x 1
1 1
x 0
C
DC
A
B
1 0
x 0
1 0
x 0
C
56
Counter schematic
DA = A’B+AB’
= AB
DB = C
DC = B’C’
What about unused states?
Complete the state diagram
57