Compuertas, microcontrolador pic16f84a y plc's

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Instituto Politécnico Nacional
Escuela Superior de Ingeniería Mecánica y Eléctrica
Unidad Profesional Azcapotzalco

Electrónica Digital Aplicada

“Ejercicios”
Compuertas
Microcontrolador PIC 16F84A
PLC

Alumno:
Hernández Granados José Adrián
Boleta:
2008360689

Fecha de entrega:
Miércoles, 12 de Enero de 2011
Compuertas

1. Diseñar un Sumador Completo.
Diagrama:

Tablade verdad:
C in | B 0 | A 0 | C out | S |
| | | | |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 1 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 1 | 1 |

S=Ci + B+A
C0=BA+CiA+CiB
Circuito:

2. Diseñar un Restador Completo.
Diagrama:

Tabla de verdad:
Bi | B | A | B 0 | R |
| | | | |
0 | 0 | 0| 0 | 0 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 1 | 1 |

R=Bi + (B+A)
B0=BA+BiB+ABi
Circuito:

3. Comparador de 2 Bits, Salida 1 cuando A ≥ B.
Diagrama:

Tabla de verdad:
B 1 | B 0 | A 1 | A 0 | S |
| | | | |
0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 0 |1 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 1 |

S=B1 B0+B1 A0+B1 A1+A1 A0+B0 A1
Circuito:

4. Diseñar un Circuito Multiplicador de 2 Números, de 2Bits.
Diagrama:

Tabla de verdad:
B 1 | B 0 | A 1 | A 0 | | M 3 | M 2 | M 1 | M 0 |
| | | | | | | | |
0 | 0 | 0 | 0 | | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | | 0 | 0 | 1 | 0 |
0 | 1 | 1 | 1 | | 0| 0 | 1 | 1 |
1 | 0 | 0 | 0 | | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | | 0 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | | 0 | 1 | 1 | 0 |
1 | 1 | 1 | 1 | | 1 | 0 | 0 | 1 |

M0=A0 B0
M1=A1 B1 B0+A1 A0 B0+A1 A0 B1+A0 B1 B0
M2=B1 A1 A0+A1 B1 B0
M3=B1 B0 A1 A0Circuito:

5. Diseñar un Convertidor de Código, de Binario a Gray, de 3 Bits.
Diagrama:

Tabla de verdad:
B 2 | B 1 | B 0 | | G 2 | G 1 | G 0 |
| | | | | | |
0 | 0 | 0 | | 0 | 0 | 0 |
0 | 0 | 1 | | 0 | 0 | 1 |
0 | 1 | 0 | | 0 | 1 | 1 |
0 | 1 | 1 | | 0 | 1 | 0 |
1 | 0 | 0 | | 1 | 1 | 0 |
1 | 0 | 1 | | 1 | 1 | 1 |
1 | 1 | 0 | | 1 | 0 | 1 |
1 | 1 | 1 | |1 | 0 | 0 |

G0=B1 + B0
G1=B2 + B1
G2=B2

Circuito:

6. Diseñar un Convertidor Binario a BCD de 4 Bits.
Diagrama:

Tabla de verdad:
B 3 | B 2 | B 1 | B 0 | | D 0 | U 3 | U 2 | U 1 | U 0 |
| | | | | | | | | |
0 | 0 | 0 | 0 | | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | | 0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | | 0 | 0 | 0 | 1 | 1|
0 | 1 | 0 | 0 | | 0 | 0 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | | 0 | 0 | 1 | 0 | 1 |
0 | 1 | 1 | 0 | | 0 | 0 | 1 | 1 | 0 |
0 | 1 | 1 | 1 | | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 1 | 0 | | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 1 | | 1 | 0 | 0 | 0 | 1 |
1 | 1 | 0 | 0 | | 1 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 1 | | 1 | 0 |0 | 1 | 1 |
1 | 1 | 1 | 0 | | 1 | 0 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | | 1 | 0 | 1 | 0 | 1 |

Circuito:

7. Diseñar un Contador Gray Ascendente de 3 Bits síncrono con JK.
Diagrama:

Tabla de verdad:
Q 2 | Q 1 | Q 0 | | Q 2 * | Q 1 * | Q 0 * |
| | | | | | |
0 | 0 | 0 | | 0 | 0 | 1 |
0 | 0 | 1 | | 0 | 1 | 1 |
0 | 1 | 0 | | 1 | 1 | 0 |
0 | 1 | 1 | | 0 | 1 | 0 |...
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