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I am trying to make a JK flip-flop in ActiveHDL environment. I want to make it with logic gates.

It should look like this:

enter image description here

This is my code:

--nand3.vhd
library ieee;
use ieee.std_logic_1164.all;

entity nand3 is 
    port(
    A,B,C: in std_logic;
    D: out std_logic
    );
end entity;

architecture nand3 of nand3 is 
begin
    D <= not(A and B and C);
end architecture;


--nand2.vhd
library ieee;
use ieee.std_logic_1164.all;

entity nand2 is 
    port(
    A,B: in std_logic;
    C: out std_logic
    );
end entity;

architecture nand2 of nand2 is 
begin
    C <= not (A and B);
end architecture;


--JK.vhd
library ieee;
use ieee.std_logic_1164.all;

entity JK is 
    port(
    J,K: in std_logic;
    Q, NQ: out std_logic;
    CLK: in std_logic
    );
end entity;

architecture JK of JK is
component nand3 
    port(
    A,B,C: in std_logic;
    D: out std_logic
    );
end component;

component nand2
    port(
    A,B: in std_logic;
    C: out std_logic
    );
end component;  

signal s1, s2: std_logic;

begin
    C11: nand3 port map(NQ,J,CLK,S1);
    C12: nand3 port map(Q,K,CLK,S2);
    C21: nand2 port map(S1,NQ,Q);
    C22: nand2 port map(S2,Q,NQ);
end architecture;

The problem is that i am getting this error when I compile JK.vhd file:

# Error: COMP96_0411: JK.vhd : (31, 22): Actual of mode "out" can not be assigned to formal "A" of mode "in"
# Error: COMP96_0411: JK.vhd : (32, 22): Actual of mode "out" can not be assigned to formal "A" of mode "in"
# Error: COMP96_0411: JK.vhd : (33, 25): Actual of mode "out" can not be assigned to formal "B" of mode "in"
# Error: COMP96_0411: JK.vhd : (34, 25): Actual of mode "out" can not be assigned to formal "B" of mode "in"

I think the problem is the backward signals.


Edit1 Improved JK.vhd file. It compiles, but it still does not work:

--JK.vhd
library ieee;
use ieee.std_logic_1164.all;

entity JK is 
    port(
    J,K: in std_logic;
    Q, NQ: out std_logic;
    CLK: in std_logic
    );
end entity;

architecture JK of JK is

    component nand3 port(
        A,B,C: in std_logic;
        D: out std_logic
    );
    end component;

    component nand2 port(
        A,B: in std_logic;
        C: out std_logic
    );
    end component;

signal s1, s2: std_logic;

-- internal copies of output signals
-- initialize
signal Q_int: std_logic := '1';
signal NQ_int: std_logic := '0';

begin
    C11: nand3 port map(NQ_int,J,CLK,S1);
    C12: nand3 port map(Q_int,K,CLK,S2);
    C21: nand2 port map(S1,NQ_int,Q_int);
    C22: nand2 port map(S2,Q_int,NQ_int);

    -- Assign values to output ports
    Q <= Q_int;
    NQ <= NQ_int;
end architecture;
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  • 1
    \$\begingroup\$ Why do you want to make it with logic gates? Is there a particular reason? \$\endgroup\$ – Francesco Conti Apr 25 '15 at 16:18
  • 1
    \$\begingroup\$ By the way, your circuit diagram would not be something I would call a flip-flop (a flip-flop is clock-edge driven). It is a latch. \$\endgroup\$ – rioraxe Apr 26 '15 at 5:33
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Let's look at one example:

  1. In your entity JK, we have an output signal named Q.
  2. In older versions of VHDL, you are not allowed to read from output signals.
  3. You tie Q to one of the inputs of C12.
  4. You can't do that, because we can't read the value of Q because it is an output signal.

Here's an example of how to fix it:

library ieee;
use ieee.std_logic_1164.all;

entity JK is port(
    J,K: in std_logic;
    Q, NQ: out std_logic;
    CLK: in std_logic
    );
end entity;

architecture JK of JK is
    component nand3 port(
        A,B,C: in std_logic;
        D: out std_logic
    );
    end component;

    component nand2 port(
        A,B: in std_logic;
        C: out std_logic
    );
    end component;  

    signal s1, s2, s3, s4: std_logic;

begin
    C11: nand3 port map(S3,J,CLK,S1);
    C12: nand3 port map(S4,K,CLK,S2);
    C21: nand2 port map(S1,S3,S4);
    C22: nand2 port map(S2,S4,S3);

    NQ <= S3;
    Q  <= S4;
end architecture;

I'm sure we could create better names for these S[1234] variables though.

| improve this answer | |
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  • \$\begingroup\$ It compiles now, but the Q and NQ outputs have always U value (unknown value). How could I solve this? \$\endgroup\$ – Thomason Apr 25 '15 at 15:12
  • \$\begingroup\$ @Thomason: Assuming all of your inputs have non U values, that wouldn't make much sense to me. Perhaps you can upload a waveform that includes all of the signals (output, input and S[1234])? \$\endgroup\$ – Bill Lynch Apr 25 '15 at 15:39
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The problem is that in VHDL, you cannot use the same signal simultaneously as an output port and an internal signal. You need to create an internal signal first, which you use for your internal feedback, and then also assign that signal to the output port of the module.

architecture JK of JK is

-- component declarations omitted

signal s1, s2: std_logic;

-- internal copies of output signals
signal Q_int, NQ_int: std_logic;

begin
    C11: nand3 port map(NQ_int,J,CLK,S1);
    C12: nand3 port map(Q_int,K,CLK,S2);
    C21: nand2 port map(S1,NQ,Q_int);
    C22: nand2 port map(S2,Q,NQ_int);

    -- Assign values to output ports
    Q <= Q_int;
    NQ <= NQ_int;

end architecture;
| improve this answer | |
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  • \$\begingroup\$ It compiles, but the outputs always have U value. \$\endgroup\$ – Thomason Apr 25 '15 at 15:14
  • \$\begingroup\$ Yes. Since the outputs start out unknown, it's impossible to get a known value at the output of either C11 or C12. You'll have to either force C21 and C22 to known states, or provide a direct reset port to get the flip-flop into a known state. \$\endgroup\$ – Dave Tweed Apr 25 '15 at 15:18
  • \$\begingroup\$ I initialized those internal signals, but the output values are still unknown. Please take a look at my Edit1. \$\endgroup\$ – Thomason Apr 25 '15 at 15:31
  • \$\begingroup\$ Yes, the initialization of the signal in the top-level module does not override the "U" coming out of the lower-level NAND module. It would probably be better to switch C21 and C22 to use the 3-input NAND gate, and use the extra inputs as direct (asynchronous) \$\overline{SET}\$ and \$\overline{RESET}\$ signals. \$\endgroup\$ – Dave Tweed Apr 25 '15 at 15:34
  • \$\begingroup\$ Why should I use asynchronous inputs like set and reset? If they are not active I do not think that the output values will get 1s and 0s. \$\endgroup\$ – Thomason Apr 25 '15 at 15:40

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