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Altera_Forum's avatar
Altera_Forum
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9 years ago

Floating point addition in vhdl

Hii

I want to add 4 floating point no.what will be the result range.I am using

library ieee_proposed;

use ieee_proposed.fixed_pkg.all;

this package.For example if I want to add two floating point no then the result range is

C= A+ B -- range of C is (max(A'right ,B'right)+1 downto min (A'left ,B'left))

and it is working fine .But when I am doing

C = A+B+D+E;

What will be the new C range.

If I am following the same as above I am getting an error

"D:/214ee1411/vhdl codes/floatarith/mul_1test/fmul_1.vhd" Line 50: Expression has 11 elements ; expected 9

Please help me with this.

Thank you

11 Replies

  • Altera_Forum's avatar
    Altera_Forum
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    9 years ago

    --- Quote Start ---

    You are using fixed point?

    use ieee_proposed.fixed_pkg.all;

    --- Quote End ---

    Thank you for your reply. Yes I am using fixed point.For example I want to add

    s1 <=((to_ufixed (50,7,0))+(to_ufixed (50,7,0))+(to_ufixed (50,7,0))+ (to_ufixed (50,7,0)));

    Here is there any formulation to calculate the range of s1?
  • Altera_Forum's avatar
    Altera_Forum
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    9 years ago

    --- Quote Start ---

    Yes,

    Please see table at

    http://electro-logic.blogspot.it/2015/11/fpga-numeri-virgola-fissa-seconda-parte.html

    where there are sizing rules

    --- Quote End ---

    Thank you for the link.I am having this table.Here it is given for A and B only.My question is for adding 4 times..as if I am taking the s1 range (8 downto 0) as per table.It is showing error which I have posted above.

    When I an going for s1 range (10 downto 0 ) it is working fine.So my question is how to decide this range if I want to add multiple times.
  • Altera_Forum's avatar
    Altera_Forum
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    9 years ago

    Just break it down in to multiple 2 input additions. There are 3 additions, so the output needs to be 3 bits bigger than the largest input.

  • Altera_Forum's avatar
    Altera_Forum
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    9 years ago

    --- Quote Start ---

    Just break it down in to multiple 2 input additions. There are 3 additions, so the output needs to be 3 bits bigger than the largest input.

    --- Quote End ---

    Thank you all
  • Altera_Forum's avatar
    Altera_Forum
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    9 years ago

    --- Quote Start ---

    Just break it down in to multiple 2 input additions. There are 3 additions, so the output needs to be 3 bits bigger than the largest input.

    --- Quote End ---

    Excuse me for being a bit of a bore, but if you add up 4 values, an extension with 2 bits suffices ...
  • Altera_Forum's avatar
    Altera_Forum
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    9 years ago

    --- Quote Start ---

    Excuse me for being a bit of a bore, but if you add up 4 values, an extension with 2 bits suffices ...

    --- Quote End ---

    Algorithmically yes - but not syntactically. Each addition in the fixed point library increases the bit size.
  • Altera_Forum's avatar
    Altera_Forum
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    9 years ago

    --- Quote Start ---

    Algorithmically yes - but not syntactically. Each addition in the fixed point library increases the bit size.

    --- Quote End ---

    I haven't used fixed-point types yet.

    But if we do it as you said: break it down into multiple 2 input additions:
     result := (a + b) + (c + d)

    2 bits extra will suffice.
  • Altera_Forum's avatar
    Altera_Forum
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    9 years ago

    Yes, now you've added brackets, it will only need 2 extra bits.

    But a+b+c+d will break down into: 

    
result := (((a+b) + c) + d);

    which does require 3 extra bits.