Design a Binary-Coded Hexadecimal decoder (logic circuit) for the Hexa digits A through F,

abdullah7saman wrote:

(...) ccompanied by each other or by the shop director.
• The worker can enter this room only with any one of the rest.
• More than two persons are not permitted to enter this room together.
Design a combinational logic circuit that will allow access by producing a logic 1 when the above conditions are met.

To design the combinational logic circuit, let's assign a binary variable to each person:

D (Director) - A
C (Cashier) - B
E (Employee) - C
W (Worker) - D

We can represent the permissions as a truth table, where 1 indicates access granted:

| A | B | C | D | Access |
|---|---|---|---|--------|
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 |
| 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 1 | 0 |
| 1 | 1 | 1 | 1 | 0 |

Now, we can find the minimized Boolean expression for the "Access" column using the Karnaugh map or using a software tool. The minimized expression is:

Access = A'B'C'D + AB'C'D' + A'BC'D' + A'B'CD'

You can now implement this expression using AND, OR, and NOT gates in a combinational logic circuit. If you need help with the implementation, please let me know.

abdullah7saman wrote:

Design a Binary-Coded Hexadecimal decoder (logic circuit) for the Hexa digits A through F, assuming that the digits O through 9 are of no interest anymore (don't care case). Assume that the inputs are in the range 1010 to 1111 and that the output characters appear in the seven-segment display in the following way: ABCDEF if you tell me how to create circuit by using or , and gate

To design a Binary-Coded Hexadecimal decoder for the hexadecimal digits A through F, we'll need to first create truth tables for each segment of the seven-segment display. We can then use Karnaugh Maps to simplify the expressions and implement the circuit using AND, OR, and NOT gates.

Here's a breakdown of the input codes for the hexadecimal digits A through F and the segments that should be on (1) or off (0):

| Input (DCBA) | A | B | C | D | E | F | G |
|--------------|---|---|---|---|---|---|---|
| 1010 (A) | 1 | 1 | 1 | 0 | 1 | 1 | 0 |
| 1011 (B) | 0 | 1 | 1 | 1 | 1 | 0 | 1 |
| 1100 (C) | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1101 (D) | 1 | 1 | 1 | 1 | 0 | 0 | 1 |
| 1110 (E) | 1 | 0 | 0 | 1 | 1 | 1 | 0 |
| 1111 (F) | 1 | 0 | 0 | 0 | 1 | 1 | 0 |

Now let's create Karnaugh Maps for each segment and derive the simplified expressions:

A segment:

A = D' + A'

B segment:

B = DC + AC

C segment:

C = D'C + AC

D segment:

D = AD' + D'C'

E segment:

E = A'D' + A'D

F segment:

F = A'D' + A'C

G segment:

G = AD' + DC

Now we can build the circuit using AND, OR, and NOT gates:

1. A = OR (AND (NOT D, NOT A), D, A)
2. B = OR (AND (D, C), A, C)
3. C = OR (AND (NOT D, C), A, C)
4. D = OR (AND (A, NOT D), NOT D, NOT C)
5. E = OR (AND (NOT A, NOT D), NOT A, D)
6. F = OR (AND (NOT A, NOT D), NOT A, NOT C)
7. G = OR (AND (A, NOT D), D, C)

Connect the inputs (DCBA) to the appropriate gates in each expression, and you'll have a decoder circuit for hexadecimal digits A through F using AND, OR, and NOT gates.

abdullah7saman wrote:

(...) R (AND (D, C), A, C)
3. C = OR (AND (NOT D, C), A, C)
4. D = OR (AND (A, NOT D), NOT D, NOT C)
5. E = OR (AND (NOT A, NOT D), NOT A, D)
6. F = OR (AND (NOT A, NOT D), NOT A, NOT C)
7. G = OR (AND (A, NOT D), D, C) can you give me the last circuit of function if we have a 4bit digit 7segement decoder

To create a 4-bit 7-segment decoder, you'll first need to create a truth table for all 4 input bits (A, B, C, D) and their corresponding 7-segment output values (a, b, c, d, e, f, g). Here's a simplified truth table for a common anode 7-segment display:



We already have the circuits for a, b, c, d, e, f, and g from the given expressions. Now let's simplify the expression for segment 'h' (the decimal point).

Since we're only dealing with a 4-bit 7-segment decoder, the decimal point is not used. Therefore, we can simplify the 'h' segment as always OFF:

8. H = 0

Now your 4-bit 7-segment decoder circuit will include the expressions 1-7 for segments a-g and the expression 8 for segment h.