Additional gates
•
•
We’ve already seen all the basic Boolean operations and the associated
primitive logic gates.
There are a few additional gates that are often used in logic design.
– They are all equivalent to some combination of primitive gates.
– But they have some interesting properties in their own right.
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Additional Gates and Decoders
1
Additional Boolean operations
Operation:
Expressions:
Truth table:
NAND
(NOT-AND)
(xy)’ = x’ + y’
NOR
(NOT-OR)
XOR
(eXclusive OR)
(x + y)’ = x’ y’
x  y = x’y + xy’
x
y (xy)’
x
y
(x+y)’
x
y
xy
0
0
1
0
0
1
0
0
0
0
1
1
0
1
0
0
1
1
1
0
1
1
0
0
1
0
1
1
1
0
1
1
0
1
1
0
Logic gates:
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Additional Gates and Decoders
2
NANDs are special!
•
The NAND gate is universal: it can replace all other gates!
– NOT
–
–
(xx)’ = x’
[ because xx = x ]
((xy)’ (xy)’)’ = xy
[ from NOT above ]
((xx)’ (yy)’)’ = (x’ y’)’
=x+y
[ xx = x, and yy = y ]
[ DeMorgan’s law ]
AND
OR
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Additional Gates and Decoders
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Making NAND circuits
•
•
The easiest way to make a NAND circuit is to start with a regular,
primitive gate-based diagram.
Two-level circuits are trivial to convert, so here is a slightly more
complex random example.
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Additional Gates and Decoders
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Converting to a NAND circuit
•
Step 1: Convert all AND gates to NAND gates using AND-NOT symbols,
and convert all OR gates to NAND gates using NOT-OR symbols.
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Additional Gates and Decoders
5
Converting to NAND, concluded
•
Step 2: Make sure you added bubbles along lines in pairs ((x’)’ = x). If
not, then either add inverters or complement the input variables.
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Additional Gates and Decoders
6
NOR gates
•
•
•
The NOR operation is the dual of the NAND.
NOR gates are also universal.
We can convert arbitrary circuits to NOR diagrams by following a
procedure similar to the one just shown:
– Step 1:
Convert all OR gates to NOR gates (OR-NOT), and all AND gates to
NOR gates (NOT-AND).
– Step 2:
Make sure that you added bubbles along lines in pairs. If not, then
either add inverters or complement input variables.
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Additional Gates and Decoders
7
XOR gates
•
•
•
A two-input XOR gate outputs true when exactly one of its inputs is
true:
x
y
xy
0
0
0
0
1
1
1
0
1
1
1
0
x  y = x’ y + x y’
XOR corresponds more closely to typical English usage of “or,” as in
“eat your vegetables or you won’t get any pudding.”
Several fascinating properties of the XOR operation:
x0=x
xx=0
x  1 = x’
x  x’ = 1
x  (y  z) = (x  y)  z
xy=yx
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Additional Gates and Decoders
[ Associative ]
[ Commutative ]
8
More XOR tidbits
•
•
The general XOR function is true when an odd number of its arguments
are true.
For example, we can use Boolean algebra to simplify a three-input XOR
to the following expression and truth table.
x  (y  z)
= x  (y’z + yz’)
= x’(y’z + yz’) + x(y’z + yz’)’
= x’y’z + x’yz’ + x(y’z + yz’)’
= x’y’z + x’yz’ + x((y’z)’ (yz’)’)
= x’y’z + x’yz’ + x((y + z’)(y’ + z))
= x’y’z + x’yz’ + x(yz + y’z’)
= x’y’z + x’yz’ + xyz + xy’z’
•
[ Definition of XOR ]
[ Definition of XOR ]
[ Distributive ]
[ DeMorgan’s ]
[ DeMorgan’s ]
[ Distributive ]
[ Distributive ]
x
y
z
xyz
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
1
0
1
0
0
1
XOR is especially useful for building adders (as we’ll see on later) and
error detection/correction circuits.
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Additional Gates and Decoders
9
XNOR gates
•
•
Finally, the complement of the XOR function is the XNOR function.
A two-input XNOR gate is true when its inputs are equal:
3/24/2016
x
y (xy)’
0
0
1
0
1
0
1
0
0
1
1
1
(x  y)’ = x’y’ + xy
Additional Gates and Decoders
10
Design considerations, and where they come from
•
•
•
•
Circuits made up of gates, that don’t have any feedback, are called
combinatorial circuits
– No feedback: outputs are not connected to inputs
– If you change the inputs, and wait for a while, the correct outputs
show up.
• Why? Capacitive loading:
– “fill up the water level” analogy.
So, when such ckts are used in a computer, the time it takes to get
stable outputs is important.
For the same reason, a single output cannot drive too many inputs
– Will be too slow to “fill them up”
– May not have enough power
So, the design criteria are:
– Propagation delay (how many gets in a sequence from in to out)
– Fan-out
– Fan-in (Number of inputs to a single gate)
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Additional Gates and Decoders
11
Prime implicants
•
•
•
•
The challenge in using K-maps is selecting the right groups. If you don’t
minimize the number of groups and maximize the size of each group:
– Your resulting expression will still be equivalent to the original one.
– But it won’t be a minimal sum of products.
What’s a good approach to finding an actual MSP?
First find all of the largest possible groupings of 1s.
– These are called the prime implicants.
– The final MSP will contain a subset of these prime implicants.
Here is an example Karnaugh map with prime implicants marked:
Y
W
1
1
0
0
1
1
1
0
0
0
1
1
0
0
0
1
X
Z
June 17, 2002
Basic circuit analysis and design
12
Essential prime implicants
Y
W
1
1
0
0
1
1
1
0
0
0
1
1
0
0
0
1
X
Z
•
•
•
If any group contains a minterm that is not also covered by another
overlapping group, then that is an essential prime implicant.
Essential prime implicants must appear in the MSP, since they contain
minterms that no other terms include.
Our example has just two essential prime implicants:
– The red group (w’y’) is essential, because of m0, m1 and m4.
– The green group (wx’y) is essential, because of m10.
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Basic circuit analysis and design
13
Covering the other minterms
Y
W
1
1
0
0
1
1
1
0
0
0
1
1
0
0
0
1
X
Z
•
•
•
Finally pick as few other prime implicants as necessary to ensure that
all the minterms are covered.
After choosing the red and green rectangles in our example, there are
just two minterms left to be covered, m13 and m15.
– These are both included in the blue prime implicant, wxz.
– The resulting MSP is w’y’ + wxz + wx’y.
The black and yellow groups are not needed, since all the minterms are
covered by the other three groups.
June 17, 2002
Basic circuit analysis and design
14
Practice K-map 2
•
Simplify for the following K-map:
Y
W
0
1
1
0
0
0
1
0
1
1
1
1
0
1
X
1
0
Z
June 17, 2002
Basic circuit analysis and design
15
Solutions for practice K-map 2
•
Simplify for the following K-map:
Y
W
0
1
1
0
0
0
1
0
1
1
1
1
0
1
X
1
0
Z
All prime implicants are circled.
Essential prime implicants are xz’, wx and yz.
The MSP is xz’ + wx + yz.
(Including the group xy would be redundant.)
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Basic circuit analysis and design
16
I don’t care!
•
•
•
You don’t always need all 2n input combinations in an n-variable function.
– If you can guarantee that certain input combinations never occur.
– If some outputs aren’t used in the rest of the circuit.
We mark don’t-care outputs in truth tables and K-maps with Xs.
x
y
z
f(x,y,z)
0
0
0
0
0
0
1
1
0
1
0
1
0
1
X
0
1
1
1
1
0
0
1
1
0
1
0
1
0
1
X
1
Within a K-map, each X can be considered as either 0 or 1. You should
pick the interpretation that allows for the most simplification.
June 17, 2002
Basic circuit analysis and design
17
Example: Seven Segment Display
Input: digit encoded as 4 bits: ABCD
a
Table for e
f
b
Assumption: Input
represents a legal
g
e
c
digit (0-9)
d
CD
AB
00
01
11
10
00
1
0
0
1
01
0
0
0
1
11
X
X
X
X
10
1
0
X
X
CD’ + B’D’
June 17, 2002
Basic circuit analysis and design
A
B
C
D
e
0
0
0
0
0
1
1
0
0
0
1
0
2
0
0
1
0
1
3
0
0
1
1
0
4
0
1
0
0
0
5
0
1
0
1
0
6
0
1
1
0
1
7
0
1
1
1
0
8
1
0
0
0
1
9
1
0
0
1
0
X
X
X
X
X
X
X
X
X
X
X
X
18
Example: Seven Segment Display
a
f
Table for a
b
g
e
d
c
CD
AB
00
00
1
01
01
11
10
1
1
1
1
1
11
X
X
X
X
10
1
1
X
X
A + C + BD + B’D’
June 17, 2002
Basic circuit analysis and design
A
B
C
D
a
0
0
0
0
0
1
1
0
0
0
1
0
2
0
0
1
0
1
3
0
0
1
1
1
4
0
1
0
0
0
5
0
1
0
1
1
6
0
1
1
0
1
7
0
1
1
1
1
8
1
0
0
0
1
9
1
0
0
1
1
X
X
X
X
X
X
X
X
X
X
X
X
19
Practice K-map 3
•
Find a MSP for
f(w,x,y,z) = m(0,2,4,5,8,14,15), d(w,x,y,z) = m(7,10,13)
This notation means that input combinations wxyz = 0111, 1010 and 1101
(corresponding to minterms m7, m10 and m13) are unused.
Y
W
1
1
0
1
0
1
x
0
0
x
1
0
1
0
X
1
x
Z
June 17, 2002
Basic circuit analysis and design
20
Solutions for practice K-map 3
•
Find a MSP for:
f(w,x,y,z) = m(0,2,4,5,8,14,15), d(w,x,y,z) = m(7,10,13)
W
1
1
0
1
0
1
x
0
Z
Y
0
x
1
0
1
0 X
1
x
All prime implicants are circled. We can treat X’s as 1s if we want, so
the red group includes two X’s, and the light blue group includes one X.
The only essential prime implicant is x’z’. The red group is not essential
because the minterms in it also appear in other groups.
The MSP is x’z’ + wxy + w’xy’. It turns out the red group is redundant;
we can cover all of the minterms in the map without it.
June 17, 2002
Basic circuit analysis and design
21
Summary
•
•
K-maps are an alternative to algebra for simplifying expressions.
– The result is a minimal sum of products, which leads to a minimal
two-level circuit.
– It’s easy to handle don’t-care conditions.
– K-maps are really only good for manual simplification of small
expressions... but that’s good enough for CS231!
Things to keep in mind:
– Remember the correct order of minterms on the K-map.
– When grouping, you can wrap around all sides of the K-map, and your
groups can overlap.
– Make as few rectangles as possible, but make each of them as large
as possible. This leads to fewer, but simpler, product terms.
– There may be more than one valid solution.
June 17, 2002
Basic circuit analysis and design
22
Basic circuit design
•
•
The goal of circuit design is to build hardware that computes some
given function.
The basic idea is to write the function as a Boolean expression, and
then convert that to a circuit.
Step 1:
Figure out how many inputs and outputs you have.
Step 2:
Make sure you have a description of the function, either as a
truth table or a Boolean expression.
Step 3:
Convert this into a simplified Boolean expression. (For this course,
we’ll expect you to find MSPs, unless otherwise stated.)
Step 4:
Build the circuit based on your simplified expression.
June 17, 2002
Basic circuit analysis and design
23
Design example: Comparing 2-bit numbers
•
•
Let’s design a circuit that compares two 2-bit numbers, A and B. The
circuit should have three outputs:
– G (“Greater”) should be 1 only when A > B.
– E (“Equal”) should be 1 only when A = B.
– L (“Lesser”) should be 1 only when A < B.
Make sure you understand the problem.
– Inputs A and B will be 00, 01, 10, or 11 (0, 1, 2 or 3 in decimal).
– For any inputs A and B, exactly one of the three outputs will be 1.
June 17, 2002
Basic circuit analysis and design
24
Step 1: How many inputs and outputs?
•
•
Two 2-bit numbers means a total of four inputs.
– We should name each of them.
– Let’s say the first number consists of digits A1 and A0 from left to
right, and the second number is B1 and B0.
The problem specifies three outputs: G, E and L.
•
•
Here is a block diagram that shows the inputs and outputs explicitly.
Now we just have to design the circuitry that goes into the box.
June 17, 2002
Basic circuit analysis and design
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Step 2: Functional specification
•
•
•
•
For this problem, it’s probably easiest
to start with a truth table. This way,
we can explicitly show the relationship
(>, =, <) between inputs.
A four-input function has a sixteenrow truth table.
It’s usually clearest to put the truth
table rows in binary numeric order; in
this case, from 0000 to 1111 for A1,
A0, B1 and B0.
Example: 01 < 10, so the sixth row of
the truth table (corresponding to
inputs A=01 and B=10) shows that
output L=1, while G and E are both 0.
June 17, 2002
A1
A0
B1
B0
G
E
L
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
0
0
0
1
0
0
0
1
1
0
0
1
1
1
0
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
0
1
1
1
0
0
1
1
0
0
0
1
0
0
0
0
Basic circuit analysis and design
26
Step 3: Simplified Boolean expressions
•
Let’s use K-maps. There are three functions (each with the same inputs
A1 A0 B1 B0), so we need three K-maps.
B1
A1
0
1
1
1
0
0
1
1
0
0
0
0
B1
0
0
A0
1
0
B0
G(A1,A0,B1,B0) =
A1 A0 B0’ +
A0 B1’ B0’ +
A1 B1’
June 17, 2002
A1
1
0
0
0
0
1
0
0
0
0
1
0
B1
0
0
A0
0
1
B0
E(A1,A0,B1,B0) =
A1’ A0’ B1’ B0’ +
A1’ A0 B1’ B0 +
A1 A0 B1 B0 +
A1 A0’ B1 B0’
Basic circuit analysis and design
A1
0
0
0
0
1
0
0
0
1
1
0
1
1
1
A0
0
0
B0
L(A1,A0,B1,B0) =
A1’ A0’ B0 +
A0’ B1 B0 +
A1’ B1
27
Step 4: Drawing the circuits
G = A1 A0 B0’ + A0 B1’ B0’ + A1 B1’
E = A1’ A0’ B1’ B0’ + A1’ A0 B1’ B0 + A1 A0 B1 B0 + A1 A0’ B1 B0’
L = A1’ A0’ B0 + A0’ B1 B0 + A1’ B1
LogicWorks has gates
with NOTs attached
(small bubbles) for
clearer diagrams.
June 17, 2002
Basic circuit analysis and design
28
Testing this in LogicWorks
•
•
Where do the inputs come from? Binary switches, in LogicWorks
How do you view outputs? Use binary probes.
probe
switches
June 17, 2002
Basic circuit analysis and design
29
Example wrap-up
•
•
•
Data representations.
– We used three outputs, one for each possible scenario of the
numbers being greater, equal or less than each other.
– This is sometimes called a “one out of three” code.
K-map advantages and limitations.
– Our circuits are two-level implementations, which are relatively easy
to draw and follow.
– But, E(A1,A0,B1,B0) couldn’t be simplified at all via K-maps. Can you
do better using Boolean algebra?
Extensibility.
– We used a brute-force approach, listing all possible inputs and
outputs. This makes it difficult to extend our circuit to compare
three-bit numbers, for instance.
– We’ll have a better solution after we talk about computer
arithmetic.
June 17, 2002
Basic circuit analysis and design
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Summary
•
•
•
•
Functions can be represented with expressions,
truth tables or circuits. These are all equivalent,
and we can arbitrarily transform between them.
Circuit analysis involves finding an expression or
truth table from a given logic diagram.
Designing a circuit requires you to first find a
(simplified) Boolean expression for the function
you want to compute. You can then convert the
expression into a circuit.
Next time we’ll talk about some building blocks
for making larger combinational circuits, and the
role of abstraction in designing large systems.
June 17, 2002
Basic circuit analysis and design
31