magnitude comparator
DESCRIPTION
Magnitude Comparator. Module M5.2 Section 6.1. 4-Bit Equality Detector. A[3..0]. Equality Detector. A_EQ_B. B[3..0]. Magnitude Comparator. A_LT_B. A[3..0]. Magnitude Detector. A_EQ_B. B[3..0]. A_GT_B. Magnitude Comparator. How can we find A_GT_B?. - PowerPoint PPT PresentationTRANSCRIPT
Magnitude Comparator
Module M5.2Section 6.1
4-Bit Equality Detector
EqualityDetector
A[3..0]
B[3..0]A_EQ_B
Magnitude Comparator
MagnitudeDetector
A[3..0]
B[3..0]A_EQ_B
A_LT_B
A_GT_B
Magnitude Comparator
How can we find A_GT_B?
How many rows would a truth table have?
28 = 256!
A0
A1
A2
A3
B0
B1
B2
B3
A_EQ_B
C0
C1
C3
C2
Magnitude Comparator
If A = 1001 and B = 0111is A > B?Why?
Because A3 > B3i.e. A3 & !B3 = 1
Therefore, one term in thelogic equation for A_GT_B isA3 & !B3
Find A_GT_BA0
A1
A2
A3
B0
B1
B2
B3
A_EQ_B
C0
C1
C3
C2
Magnitude Comparator
If A = 1101 and B = 1011is A > B?Why?
Because A3 = B3 and A2 > B2i.e. C3 = 1 and A2 & !B2 = 1
Therefore, the next term in thelogic equation for A_GT_B isC3 & A2 & !B2
A_GT_B = A3 & !B3 # …..
A0
A1
A2
A3
B0
B1
B2
B3
A_EQ_B
C0
C1
C3
C2
Magnitude Comparator
If A = 1010 and B = 1001is A > B?Why?
Because A3 = B3 and A2 = B2 and A1 > B1i.e. C3 = 1 and C2 = 1 and A1 & !B1 = 1
Therefore, the next term in thelogic equation for A_GT_B isC3 & C2 & A1 & !B1
A_GT_B = A3 & !B3 # C3 & A2 & !B2 # …..
A0
A1
A2
A3
B0
B1
B2
B3
A_EQ_B
C0
C1
C3
C2
Magnitude Comparator
If A = 1011 and B = 1010is A > B?Why?
Because A3 = B3 and A2 = B2 and A1 = B1 and A0 > B0i.e. C3 = 1 and C2 = 1 and C1 = 1 and A0 & !B0 = 1
Therefore, the last term in thelogic equation for A_GT_B isC3 & C2 & C1 & A0 & !B0
A_GT_B = A3 & !B3 # C3 & A2 & !B2 # C3 & C2 & A1 & !B1 # …..
A0
A1
A2
A3
B0
B1
B2
B3
A_EQ_B
C0
C1
C3
C2
Magnitude Comparator
A_GT_B = A3 & !B3 # C3 & A2 & !B2 # C3 & C2 & A1 & !B1 # C3 & C2 & C1 & A0 & !B0
A0
A1
A2
A3
B0
B1
B2
B3
A_EQ_B
C0
C1
C3
C2
Magnitude Comparator
A_LT_B = !A3 & B3 # C3 & !A2 & B2 # C3 & C2 & !A1 & B1 # C3 & C2 & C1 & !A0 & B0
Find A_LT_BA0
A1
A2
A3
B0
B1
B2
B3
A_EQ_B
C0
C1
C3
C2
ABEL ProgramMODULE magcomp4
TITLE '4-BIT COMPARATOR, R. Haskell, 9/21/02‘
DECLARATIONS
" INPUT PINS "
A3..A0 PIN 6,7, 11, 5;
A = [A3..A0];
B3..B0 PIN 72, 71, 66, 70;
B = [B3..B0];
" OUTPUT PINS "
A_EQ_B PIN 36;
A_LT_B PIN 37;
A_GT_B PIN 35;
C3..C0 NODE;
C = [C3..C0];
A0
A1
A2
A3
B0
B1
B2
B3
A_EQ_B
C0
C1
C3
C2
ABEL Program (cont.)EQUATIONS
C = !(A $ B);
A_EQ_B = C0 & C1 & C2 & C3;
A_GT_B = A3 & !B3
# C3 & A2 & !B2
# C3 & C2 & A1 & !B1
# C3 & C2 & C1 & A0 & !B0;
A_LT_B = !A3 & B3
# C3 & !A2 & B2
# C3 & C2 & !A1 & B1
# C3 & C2 & C1 & !A0 & B0;
ABEL Program (cont.)test_vectors ([A, B] -> [A_EQ_B, A_LT_B, A_GT_B])
[0, 0] -> [1, 0, 0];
[2, 5] -> [0, 1, 0];
[10, 12] -> [0, 1, 0];
[7, 8] -> [0, 1, 0];
[4, 2] -> [0, 0, 1];
[6, 6] -> [1, 0, 0];
[1, 7] -> [0, 1, 0];
[5, 13] -> [0, 1, 0];
[12, 0] -> [0, 0, 1];
[6, 3] -> [0, 0, 1];
[9, 9] -> [1, 0, 0];
[12, 13] -> [0, 1, 0];
[7, 0] -> [0, 0, 1];
[4, 1] -> [0, 0, 1];
[3, 2] -> [0, 0, 1];
[15, 15] -> [1, 0, 0];
END
TTL Comparators1 2 3 4 5 6 7
9 10 11
128
1920
1718
1516
1314
GND
Vcc P>Q P0Q0P1Q1P2Q2P3Q3
P=Q Q7P7Q6P6Q5P5Q4P4
74LS682
1 2 3 4 5 6 7 8 9
10111213141516
GND
Vcc B3A<Bin A=Bin A>Bin A>BoutA=BoutA<Bout
A3B2A2A1B1A0B0
74LS85
Cascading two 74LS85s
13
1
3 6
9
14
2
10
11
7
4
15
12
5
P0P1P2P3
Q0Q1Q2Q3
< = >
P<Q
P=Q
P>Q
13
1
3 6
9
14
2
10
11
7
4
15
12
5
P0P1P2P3
Q0Q1Q2Q3
< = >
P<Q
P=Q
P>Q
A3 A2 A1 A0 A7 A6 A5 A4
B3 B2 B1 B0 B7 B6 B5 B4
+5V A<B A=B A>B
Question
1 2 3 4 5 6 7
9 10 11
128
1920
1718
1516
1314
GND
Vcc P>Q P0Q0P1Q1P2Q2P3Q3
P=Q Q7P7Q6P6Q5P5Q4P4
74LS682
P_GT_QP_EQ_QP_LT_Q
Draw a logic circuit forwhat is in the green box.