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Magnitude Comparator

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?.

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Magnitude Comparator

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  1. Magnitude Comparator Module M5.2 Section 6.1

  2. 4-Bit Equality Detector A[3..0] Equality Detector A_EQ_B B[3..0]

  3. Magnitude Comparator A_LT_B A[3..0] Magnitude Detector A_EQ_B B[3..0] A_GT_B

  4. Magnitude Comparator How can we find A_GT_B? How many rows would a truth table have? 28 = 256!

  5. Magnitude Comparator Find A_GT_B Because A3 > B3 i.e. A3 & !B3 = 1 If A = 1001 and B = 0111 is A > B? Why? Therefore, one term in the logic equation for A_GT_B is A3 & !B3

  6. Magnitude Comparator A_GT_B = A3 & !B3 # ….. Because A3 = B3 and A2 > B2 i.e. C3 = 1 and A2 & !B2 = 1 If A = 1101 and B = 1011 is A > B? Why? Therefore, the next term in the logic equation for A_GT_B is C3 & A2 & !B2

  7. Magnitude Comparator A_GT_B = A3 & !B3 # C3 & A2 & !B2 # ….. Because A3 = B3 and A2 = B2 and A1 > B1 i.e. C3 = 1 and C2 = 1 and A1 & !B1 = 1 If A = 1010 and B = 1001 is A > B? Why? Therefore, the next term in the logic equation for A_GT_B is C3 & C2 & A1 & !B1

  8. Magnitude Comparator A_GT_B = A3 & !B3 # C3 & A2 & !B2 # C3 & C2 & A1 & !B1 # ….. Because A3 = B3 and A2 = B2 and A1 = B1 and A0 > B0 i.e. C3 = 1 and C2 = 1 and C1 = 1 and A0 & !B0 = 1 If A = 1011 and B = 1010 is A > B? Why? Therefore, the last term in the logic equation for A_GT_B is C3 & C2 & C1 & A0 & !B0

  9. Magnitude Comparator A_GT_B = A3 & !B3 # C3 & A2 & !B2 # C3 & C2 & A1 & !B1 # C3 & C2 & C1 & A0 & !B0

  10. Magnitude Comparator Find A_LT_B A_LT_B = !A3 & B3 # C3 & !A2 & B2 # C3 & C2 & !A1 & B1 # C3 & C2 & C1 & !A0 & B0

  11. ABEL Program MODULE 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];

  12. 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;

  13. 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

  14. 1 20 P>Q Vcc 1 16 2 19 B3 Vcc P0 P=Q 2 15 3 18 A<Bin A3 Q0 Q7 3 14 4 17 B2 A=Bin P1 P7 4 13 5 16 A>Bin A2 Q1 Q6 5 12 6 15 A1 A>Bout P2 P6 6 11 7 14 A=Bout B1 Q2 Q5 7 10 8 13 A<Bout A0 P3 P5 8 9 9 12 GND B0 Q3 Q4 10 11 GND P4 74LS85 74LS682 TTL Comparators

  15. Cascading two 74LS85s

  16. 1 20 P>Q Vcc 2 19 P0 P=Q 3 18 Q0 Q7 4 17 P1 P7 5 16 Q1 Q6 6 15 P2 P6 7 14 Q2 Q5 8 13 P3 P5 9 12 Q3 Q4 10 11 GND P4 74LS682 Question P_GT_Q P_EQ_Q P_LT_Q Draw a logic circuit for what is in the green box.

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