Guys can someone explain to me how to get started on this problem. Right now l am stuck l don't know where to start:


Design a circuit which shows the letter "F" (shown below) on a seven-segment display when the input 4-bit binary number has the number 3 as one of its factors and the letter "n" (shown below) when it does not.


eg. 4 has the following factors 1, 2, and 4, therefore display should show an 'n'.

 

I think I would tackle this question by drawing up a truth table. Your 7 segments are normally numbered a - g (see a manufacturers data sheet for a typical 7-segment display) and your digital inputs are A,B,C,D.

I notice from your diagram that some segments are always on and some are always off. You can disregard these (hard-wire them) so you truth table already becomes simpler.

Your truth table will have 16 rows:- 1 for every possible combination of inputs ABCD. In the 7-segment columns, fill in what you want your segments to be doing.

From the truth table produce logic equations. Simplify them. Draw logic circuit.

 

JDT has a good point. A truth table is the best was to start this problem.

Post your stab at a truth table that represents what your class assignment is asking you to do and I am sure a member will be able to guide you to the next step in the process.

Despite your confusion, this si an elementary problem as you will see once you have worked through it.

hgmjr

 

In the real life using say a micro controller i would have created a lookup table But I understand this is a academic problem. I agree with the others create a truth table. Then use Boolean algebra or a karnaugh map to simplify your expression. You will find help here http://www.allaboutcircuits.com/vol_4/index.html

 

Guys now that l got the truth table l have 7 expressions which need to be solved. Which program can l use ?

 

Guys now that l got the truth table l have 7 expressions which need to be solved. Which program can l use ?

Post your truth table so we can check it out.

hgmjr

 

I have uploaded the truth table on a word document and jpeg image.

 

I'm confused. Your original statement of the problem was that the 7-segment display should display an F when the input was evenly divisible by 3 and when the input was not dividible by 3 the 7-segment display should contain a lower-case "n".

If that is the correct interpretation then I would expect your truth table to have only to outcomes. Your truth table appears to contain a few other characters in addition to the "n" and the "F".

What have I missed?

hgmjr

 

I'm confused. Your original statement of the problem was that the 7-segment display should display an F when the input was evenly divisible by 3 and when the input was not dividible by 3 the 7-segment display should contain a lower-case "n".

If that is the correct interpretation then I would expect your truth table to have only to outcomes. Your truth table appears to contain a few other characters in addition to the "n" and the "F".

What have I missed?

hgmjr

He's giving the 7-segment outputs. (Note the A..G on the output columns.) There should be, however, only two sets of outputs: 1,0,0,0,1,1,1 ('F') and 0010101 ('n')

--Rich

 

If we inspect things closer some segments are turned on all the time. The upper part of the "F" can be controlled by one signal. And to draw the "n" we we have to turn off the upper part of the "F" and only control the last segment (the last leg) of the "n" . But your thinking is correct

 

He's giving the 7-segment outputs. (Note the A..G on the output columns.) There should be, however, only two sets of outputs: 1,0,0,0,1,1,1 ('F') and 0010101 ('n')

--Rich

Yes. That was the point that I was trying to make.

hgmjr