Download Building Blocks of Digital Circuits Storage Elements

So, Why Binary?
Electronic computers represent information as
voltage levels
 To make computer hardware simple and reliable,
computers represent information in binary form

Building Blocks of Digital Circuits
» example: voltages greater than 3V are interpreted as
representing one value (called “1”), voltages less than 2V
are interpreted as representing another value (called “0”)
 Digital
circuits, binary logic and noise
flip flops and latches
 Decoders and multiplexors
 Memory and Lookup tables (LUTs)
 Field Programmable Gate Arrays
 Tristate buffers
 In
 Gates,

Computers (like all electronics) are affected by noise
Noise margin is amount
of noise that a system can
tolerate and still correctly
identify a logic high or low
» smaller power supply voltages
lead to smaller noise margins
» but also improve performance
and reduce power requirements
AND Gate X
Y
X
OR Gate
Y
XOR Gate
Inverter
5v
High
 Logic
3v
1v Undefined
Low
0v
noise margin noise margin
1V
3V
Storage Elements
is basic storage element
» when C=1, Q=D (and Q’=D’)
» when C=0, Q and Q’ don’t change
D
flip flop stores value at D input
when clock rises
» most widely used storage element for
digital circuits
» can be implemented with two latches
 If
D input of latch changes as C goes
from 1 to 0, output value is uncertain
X
Timing Diagram
X
Y
X+Y
X’Y+XY’
X’
XY
X+Y
X Y
X’
gates “compute” elementary binary functions
» output of an AND gate is “1” when both of its inputs are
“1”, otherwise the output is zero
» similarly for OR gate, XOR gate and inverter
High
 Timing
diagram shows how output values change
over time as input values change
‹#›
 D-latch
X
Y
XY
4v
Undefined 2v
Low
‹#›
Basic Logic Gates
» various sources (nearby signal changes, thermal
vibrations of molecules in semiconductor materials, . . . )
» in computers, noise can cause binary signals to be
misinterpreted

In practice, this is rarely done
» requires more complex circuits
» circuits are more susceptible to noise, hence less reliable
Noise in Computer Systems

principle, could use more voltage levels
» example: 0 to .75V represents “0”, 1 to 1.75V represents
“1”, 2 to 2.75V represents “2”, and so forth
‹#›
Review Questions
D
Q
C
Q
D
Q


>C Q
D
C
Q
» similarly for flip flop
» output may oscillate or remain at
intermediate value (metastability)
» to avoid metastability, keep D stable when C is changing

Explain why a circuit that uses three voltage levels is more
susceptible to noise than a circuit that uses two voltage levels
(assuming the same maximum voltage in both cases).
Draw a circuit that implements the exclusive-or function using just
AND, OR gates and inverters. Can you do it with just AND gates
and inverters (hint: consider DeMorgan’s Rule)?
What does it mean for a flip flop to be metastable? What’s the
maximum amount of time that a flip flop can be metastable?
D Q
C Q
‹#›
‹#›
1
Decoders and Muxes
Memory
2→4 decoder
1→2
decoder
A
k→2k decoder converts a
k bit binary input to a 1 on
one of 2k outputs
X0
X0
X1
X1
» can construct larger decoders
using by combining two smaller ones
» useful for selecting one of a number
of cases based on a numerical value
A0
X2
X3
A0
A
multiplexor selects connects one
of its data inputs to its output
» can also construct larger muxes from
smaller ones, or using decoders
» useful for selecting one of several stored
values
A1
4-mux
D0
D1
0
D2
D3
0
1
0
X
1
enable
 Implements
array of n numbered
storage locations
read/write
data_in
» index for a location is called its address
» each location stores a word with w bits
data_out
address
00
01
02
.
.
.
FE
FF
 Operation
» when enable and read/write are high, value stored in
location specified by address appears at data_out
» when enable is high and read/write is low, value on
data_in replaces value in location specified by address
» may operate synchronously, or asynchronously
 Can
be implemented using latches or flip flops, but
usually, uses specialized, higher density circuits
1
A0
A1
‹#›
Lookup Tables
‹#›
How Many LUTs Does it Take?
 Digital
logic is often implemented using devices called
Field Programmable Gate Arrays (FPGA)
» configurable device that can implement many different circuits
» implemented using “programmable” logic elements and wires
A
Lookup Table (LUT) can be configured for any logic
function with specified number of inputs (typically 4)
U
V
of circuit implemented with LUTs:
F
X
G
Y
Z
U
V
X
Y
A
B
C
D
Z
A
B
C
D
using configurable logic circuits, it’s useful to
have way to estimate required number of LUTs
 For circuit with n outputs, need at least n LUTs
» if using LUT4s & all functions have ≤4 inputs, n is enough
» newest FPGAs use LUT6s, so for functions with up to 6
inputs, one LUT6 is enough
» can view LUT as hardware implementation of truth table
 Example
 When
 In
LUT4
a circuit with k LUT4s, m inputs and n outputs
with k>n, k–n LUT outputs connect to LUT inputs
F
» so, m+(k–n)≤4k which means k≥(m–n)/3
» so, a 4-mux requires at least (6–1)/3=2 LUT4s and
an 8-mux requires at least (11–1)/3=4 LUT4s
A’BCD’
LUT4
ABC’+B’+CD
G
• in practice, need three LUT4s for 4-mux, seven for 8-mux
‹#›
Xilinx FPGA Organization
‹#›
Xilinx Configurable Logic Block
switch matrix
wire segments
configurable logic blocks (CLB)
IO blocks (IOB)
main
function
generators
G1
G2
G3
G4
configuration
bits
CLK EC
clock edge
select
S/R
Y
S/R C
set/reset
control
LUT4
D
PRE
>C
DIN
YQ
EC CLR
H1
 CLBs
can be connected to “passing” wires
segments connected by switch matrix
 Long wire segments used to connect distant CLBs
 Configuration information stored in SRAM bits that
are loaded when power turns on
‹#›
 Wire
F1
F2
F3
F4
LUT3
clock enable
control
1
D
LUT4
PRE
>C
XQ
EC CLR
main
function
generators
1
S/R C
flip flop
X
‹#›
2
Tri-State Buffers
Exercises
1. Consider the circuit shown below,
consisting of a D-latch and a positiveedge-triggered D flip flop.
control
A
D
Q
Q
Register 2
LD
different components to take turns using shared
output wire(s) (often called a bus)
 Also allows wire(s) to be used for both input and output
‹#›
20
40
3. Such a circuit could be implemented
with 4 LUTs. This follows from the
equation on page 11.
60
80
100
clk
A
X
Y
2. An 8:1 mux can be implemented with 7
2:1 muxes. A 16:1 mux can be
implemented with 15.
4.
A circuit that implements a 12 input AND
function can be implemented with 4
LUTs. Configure each LUT as a 4 input
AND and connect the outputs of three to
the inputs of the fourth LUT. This leaves
one spare input.
The function can be implemented using
the 4 LUT circuit shown below.
F
G
H
I
J
K
L
A
B
C
D
Y
3. Consider a circuit with 12 inputs and a
single output. What is the minimum
number of 4 input LUTs that would be
required to implement this circuit?
Give an example of a 12 input circuit
that could be implemented with this
number of LUTs.
Give a circuit using LUTs that
implements the expression shown
below.
Suppose we drive the circuit with a 50
(A+ B+C+D+E)(F+G+H+I+J)(K+L)
MHz clock, with positive transitions at
20 ns, 40 ns, 60 ns and so forth. Now,
Use as few LUTs as you can. How does
the number of LUTs compare to the
suppose input A starts low, then goes
minimum number required for a 12 input
high at 25 ns, then low at 55 ns and
high at 85 ns.
function.
Draw a timing diagram covering the
4. Show how you can implement a 4 input
time period 0 to 100 ns, showing the
mux using a decoder and four tri-state
clock and A inputs, plus outputs X and
buffers.
Y.
‹#›
 Allows
0
>C
clk
» when control input is high, output equals input
» when control input is low, output is disconnected from
input – called high impedance state
1.
D Q
C
tri-state buffer has a data input and a control input
Solutions
X
D Q
Register 3
LD
LD
Register 1
D
Q
D
data_out
data_in
A
2. How many 2:1 multiplexors are needed
to implement an 8:1 multiplexor. How
many are needed to implement a 16:1
multiplexor?
select 2
24 decoder
0
1
2
3
D0
D1
D2
D3
out
(F+G+H+I+J)
(K+L)
E
‹#›
‹#›
3