# BCD to 7 segment display logic minimisation

1. Jan 10, 2017

### fonz

1. The problem statement, all variables and given/known data

A 7-segment display is used to show a decimal digit and it is driven from 4-bit input. Each bar is assumed to light up when a logical 1 is applied to it.
1. Draw the truth table to drive segment (d) of the display.

2. Using the truth table you have obtained, draw a Karnaugh map to find the minimised logic expression in the 1-st canonical form (SOP, i.e. series of AND terms ORed together).

3. Convert the expression from 2 to a function which uses only NAND gates and draw the equivalent circuit diagram in Logic Circuit and generate its truth table to compare with 1.

2. Relevant equations

De-Morgan's laws:$$(A + B)' = A' \cdot B'$$$$A' + B' = (A \cdot B)'$$
3. The attempt at a solution

See attached for truth table and k-map.

Canonical Form:$$A + C'D + CD' + A'B'$$

Applying De-Morgan's Theorem:$$A'BCC'DD'$$

but since $C \cdot C' = 0$ and $D \cdot D' = 0$ then this all reduces to 0.

I'm not sure whether it is my K-Map minimisation or application of De-Morgan's laws that is wrong?

#### Attached Files:

• ###### segment_d.png
File size:
35.6 KB
Views:
48
2. Jan 10, 2017

### Staff: Mentor

3. Jan 10, 2017

### fonz

Yes that is correct thanks berkeman.

4. Jan 10, 2017

### Staff: Mentor

And this part of the problem statement implies you should only show 0-9 and blank on any number 10 and up? Or are you supposed to display A-F too...

5. Jan 10, 2017

### fonz

In the example that was provided for segment c, the decimal values 10 to 15 are shown as 'don't care' terms so I have kept it consistent with the example.

6. Jan 10, 2017

### Staff: Mentor

I'm confused though. You should end up with 7 truth tables, one for each segment drive, right?

7. Jan 10, 2017

### fonz

Yes in total, although this homework is only to derive the logic circuit for segment d.

8. Jan 10, 2017

### Staff: Mentor

Ah, okay. I missed that if you said it in your OP. Let me check your work now...

9. Jan 10, 2017

### Staff: Mentor

10. Jan 10, 2017

### fonz

I did think this originally but then I checked the example solution that was provided in the homework material and found this.

#### Attached Files:

• ###### segment.png
File size:
8.5 KB
Views:
72
11. Jan 10, 2017

### Staff: Mentor

Ah, fair enough. The issue I see is the blue rectangle and green rectangle that you have around 3 terms each in your K-map. I don't think you can do that...

http://www.ee.surrey.ac.uk/Projects/Labview/minimisation/graphics/g3.gif

12. Jan 10, 2017

### fonz

Great thanks for pointing that out, still got the same issue though:

Canonical:

$A + BC'D + B'C + B'D' + CD'$

Applying De-Morgan's law:

$A'BBB'CC'C'DDD'$

Now the $BB'$, $CC'$ and $DD'$ terms all equal zero.

#### Attached Files:

• ###### segment_d.png
File size:
36.2 KB
Views:
80
13. Jan 12, 2017

### Staff: Mentor

That's not what I'm getting for a corrected version of your K-map. Can you post the corrected K-map and show your groupings?

14. Jan 13, 2017

### fonz

I think I am missing the $A'$ from the third minterm.

$A + BC'D + A'B'C +B'D' + CD'$

#### Attached Files:

• ###### segment_d.png
File size:
36.6 KB
Views:
59
15. Jan 13, 2017

### Staff: Mentor

Your 3rd minterm does include A', and I get the same minterms as you now.