Binary to 7-segment screen encoder

Thread Starter

alexa-lol

Joined Nov 22, 2009
5
I get this... I have made 7-segment display in 2 different ways...
1st was that I took each segment as a boolean function and than I made minimal normal form for each segment. The "screen" showed the indexes of entering minterm vector; the values form 10-15 were reserved for letter E.

It worked but it took me 65 NOT, AND and OR gates + It wasnt very efficient

2nd way was that I first made 4/11 decoder wich decoded input vector to 11 different exits. Each exiting vector had a minterm property (one of the exitst is always 1 and the others are 0)

Than I made a encoder wich encoded 11 values to 7 values...
1st I did it with a use of standard values of encoder's input vector. I combined them with OR gates.
2nd I did it with the negated input vector and with AND gates...
I found out that 2nd method is in this case better.

I am having troubles because at the lecture from Basics of digital circuits the professor said that the input of the encoder must be a vector with minterm properties, meaning only I input can be 1 and the others must be 0.

If we have this the boolean equation is in form y=m v& C
where
y is output vector
m in entering minterm vector
C coding matrix

So from where do I get an equation for first segment (D_0) . What do I have to look for? Do I have to look where the value in coding matrix is 1?
Can somebody explain me how did we get the equation for the first one and I will try to extract the method from it.

Thanks for your time
 

Thread Starter

alexa-lol

Joined Nov 22, 2009
5
I got it...
For segment 1(D_0) the value is 0 in case if the value of input vector is 0 0 0 1 or 0 1 0 0 . From here I see that the value is of first and third digit are always 0.
But here is a problem..if I would only take this the values would also be 0 when input vector would be 0 0 0 0 and 0 1 0 1 so I must exclude them. I do this with the second half of the boolean equation.

now i got it :)
 
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