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Simulation of Logic Circuits and ALU Example Using 7408 and 7404 Components

Dive into the fascinating world of digital logic with this comprehensive simulation showcasing the fundamental operations of logic gates and an Arithmetic Logic Unit (ALU). This simulation leverages the 7408 AND gate and 7404 NOT gate to demonstrate logical operations and arithmetic addition. Explore how binary numbers are manipulated through combinational logic circuits, with a focus on truth tables and boolean algebra. Designed for students and enthusiasts, this guide includes essential insights into how digital circuits operate effectively.

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Simulation of Logic Circuits and ALU Example Using 7408 and 7404 Components

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  1. Simulatsioon Stiimulid Resultaadid 1 1 1 1 1 1 0 0 1 0 0 1 1 0 0 1 0 0 . . . . . . 1 1 0 1 1 0 Objekt (loogikaskeem) T. Evartson

  2. T. Evartson

  3. x2x1 00 01 11 10 00 1 1 0 0 x4x3 01 0 0 0 1 11 0 0 0 0 10 1 0 0 0 T. Evartson

  4. Simulatsioon 7408 +5v 7408 X2 R e s u l t a a t S t i i m u l i d 7404 X4 7432 e X1 X3 GND 10 10 10 10 10 10 10 10 10 10 X1 0 1 0 1 0 1 0 1 0 1 X2 0 0 1 1 0 0 1 1 0 0 X3 0 0 0 0 1 1 1 1 0 0 X4 0 0 0 0 0 0 0 0 1 1 e 1 0 1 0 0 0 1 0 1 0 9 5 4 6 0 1 2 3 7 8 T. Evartson

  5. Indikaatori segmendi e juhtimine T. Evartson

  6. Simulatsioon +5v 7408 R e s u l t a a d i d x2 7404 & S t i i m u l i d 7432 x1 e 1 1 7408 & 7404 x3 1 GND 10 10 10 10 10 10 10 10 10 10 X1 0 1 0 1 0 1 0 1 0 1 X2 0 0 1 1 0 0 1 1 0 0 X3 0 0 0 0 1 1 1 1 0 0 e 1 0 1 0 0 0 1 0 1 0 9 5 4 6 0 1 2 3 7 8 T. Evartson

  7. Simulatsioon +5v 7408 X2 R e s u l t a a t S t i i m u l i d 7432 7404 e X1 7408 7404 X3 GND 10 10 10 10 10 10 10 10 10 10 X1 0 1 0 1 0 1 0 1 0 1 X2 0 0 1 1 0 0 1 1 0 0 X3 0 0 0 0 1 1 1 1 0 0 e 1 0 1 0 0 0 1 0 1 0 9 5 4 6 0 1 2 3 7 8 T. Evartson

  8. ALU näide I +5v a0 SN74S153 S b0 0 M 1 EN s´0 MUX 0 n0 1 y0 k0 2 e0 3 EN s´1 a1 0 n1 0 y1 1 b1 k1 2 e1 3 0 M = 0 S = 0 Y = A + B S = 1 Y = R1 (A) M = 1 S = 0 Y = A B S = 1 Y = B GND V T. Evartson

  9. ALU näide I I AND NOT Liitmine Nihutamine 1+1=2 A&B NOT (B) 1+2=3 R1(A) 10 10 10 10 10 10 10 10 a0 1 1 1 1 0 1 0 0 a1 0 0 1 0 0 1 1 1 b0 1 0 0 0 1 1 1 0 b1 0 1 0 0 1 0 1 0 m 0 0 0 0 1 1 1 1 s 0 0 1 1 0 0 1 1 y0 0 1 0 0 1 0 1 1 1 1 0 0 0 1 1 0 y1 T. Evartson

  10. Lühike EDwin XP juhend T. Evartson

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