CODEWITHSUNDEEP
vhdl
mahmood
mahmood

Reputation: 24805

RS latch with VHDL

I have wrote a simple RS latch with VHDL and tried to synthesis it with ISE. The synthesizer added a D flip flop where the D input is grounded and my (S)et and (R)eset inputs are treated as preset and clear inputs. I expected to see NAND gates only. Why it added a flip flop while there is no need for that? Also why the D input is connected to ground? 

entity rs is
Port ( r : in  STD_LOGIC;
       s : in  STD_LOGIC;
       q : inout  STD_LOGIC);
end rs;

architecture Behavioral of rs is
begin
 process( r, s )
 begin
    if r = '1' then
        q <= '0';
    elsif s = '1' then
        q <= '1';
    else
        q <= q;
    end if;
 end process;
end Behavioral;

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Upvotes: 0

Views: 2145

Answers (2)

JHBonarius
JHBonarius

Reputation: 11291

An FPGA does not contain NAND gates. Even though the gates are shown in the schematic from ISE, their combined function is in fact implemented in LookUp Tables (LUTs). The behavior of a latch function only using LUTs is different from what you would have if you would implement a latch with NAND gates.

To realize latch functions, a Xilinx FPGA has to utilize "storage elements" (an actual available physical structure in the FPGA) to emulate their behavior. These elements have predictable behavior. The storage elements offer set and reset inputs. Together with some logic you can create latch-like behavior.

You could look at the PAR output instead of the synthesis output to see how it is realized.

Upvotes: 2

scary_jeff
scary_jeff

Reputation: 4384

Your observation that a flip flop has been used is not correct. The element is labelled ldcp, and this is a transparent latch. Looking at your code, a latch is what you have described. The D input is connected to ground, because all your process ever does is set or clear the latch; you have not described a 'load' operation, so the latch does not use its D input.

Upvotes: 2

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