Internal signals in vhdl

von Hareesh M. (Company: Mindteck) (hareeshp)

2017-08-23 06:26

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Hi,
How do i access the internal signals in the top module to testbench 
program.The change in internal signals causes main signals to change.

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Re: Internal signals in vhdl

von Lothar M. (Company: Titel) (lkmiller) (Moderator)

2017-08-23 06:55

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Hareesh M. wrote:
> How do i access the internal signals in the top module to testbench
> program.
Wht do you mean with "access"? Just to view the signals? Or to force 
them to a desired state?

And most important: What simulator do you use?

> The change in internal signals causes main signals to change.
Indeed thats usually the desired behaviour of almost every HDL design.

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Re: Internal signals in vhdl

von Hareesh M. (Company: Mindteck) (hareeshp)

2017-08-23 06:59

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Lothar M. wrote:
> Hareesh M. wrote:
>> How do i access the internal signals in the top module to testbench
>> program.
> Wht do you mean with "access"? Just to view the signals? Or to force
> them to a desired state?
>
> And most important: What simulator do you use?
>
>> The change in internal signals causes main signals to change.
> Indeed thats usually the desired behaviour of almost every HDL design.

i want to force them to a desired state, and forcing them causes the 
changes in other signals too. Currently i'm using modelsim simulator.

2017-08-23 06:59: Edited by User

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Re: Internal signals in vhdl

von Hareesh M. (Company: Mindteck) (hareeshp)

2017-08-23 07:16

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Hareesh M. wrote:
> Lothar M. wrote:
>> Hareesh M. wrote:
>>> How do i access the internal signals in the top module to testbench
>>> program.
>> Wht do you mean with "access"? Just to view the signals? Or to force
>> them to a desired state?
>>
>> And most important: What simulator do you use?
>>
>>> The change in internal signals causes main signals to change.
>> Indeed thats usually the desired behaviour of almost every HDL design.
>
> i want to force them to a desired state, and forcing them causes the
> changes in other signals too. Currently i'm using modelsim simulator.

i'm writing the main code and testbench for the same
////////////////////////////////////////////////////////////////////

  library ieee;
  use ieee.std_logic_1164.all;
    package vhdl_pkg is
      constant HOTS_WAP:   std_logic_vector(2 downto 0):= "111";
    end vhdl_pkg;
  library ieee;
  use ieee.std_logic_1164.all;
  use ieee.numeric_std.all;
  use ieee.std_logic_unsigned.all;
  library work;
  use work.vhdl_pkg.all;
  Entity hotswap is
    port(ifc_clk: in std_logic;
        BD_SEL_PLD: in std_logic:= '1';
        HS_PG00D_PLD:in std_logic:= '1';
        HS_FLT_PLD: in std_logic:= '1';
        ifc_ad0_7: inout std_logic_vector(2 downto 0);
        ifc_cs3_n: in std_logic;
        ifc_oe_n: in std_logic;
        ifc_we_n: in std_logic;
        ifc_a27_31: in std_logic_vector(2 downto 0)
  end hotswap;
  Architecture mpu of hotswap is
    signal ifc_dout:   std_logic_vector(2 downto 0);
    signal cpld_cs:   std_logic;
    signal cpld_oe:   std_logic;
    signal cpld_we:  std_logic;
    signal flt_reg:   std_logic;
    signal hotswap_pgood : std_logic;
    signal bd_sel_pld_reg : std_logic;
    signal pon_rst_n:  std_logic;
    process(ifc_clk) 
    begin
      if(rising_edge(ifc_clk)) then
        if(pon_rst_n = '1') then
          cpld_cs <= '0';
          cpld_oe <= '0';      
        else
          cpld_cs <= ifc_cs3_n;
          cpld_oe <= ifc_cs3_n and ifc_oe_n;
          cpld_we <= ifc_cs3_n and ifc_we_n;
        end if;
      end if;
    end process;
    process(ifc_clk)
      begin      
      if((cpld_cs and cpld_oe)= '1') then
        case ifc_a27_31 is
          when HOTS_WAP =>
            ifc_dout(0) <= HS_FLT_PLD;
            ifc_dout(1) <= BD_SEL_PLD; 
            ifc_dout(2) <= HS_PG00D_PLD;
          when others =>
            ifc_dout <= "000";
        end case;
      end if;
     end process;
     ifc_ad0_7 (0) <= ifc_dout(0);
     ifc_ad0_7 (1) <= ifc_dout(1);
     ifc_ad0_7 (2) <= ifc_dout(2);
  end mpu;

/////////////////////////////////////////////////////////////////////
and the testbench is

library ieee;
use ieee.std_logic_1164.all;
entity hot_swap is
end entity ;
architecture hotswapping of hot_swap is
 component hotswap 
  port( ifc_clk:   in std_logic;
        BD_SEL_PLD: in std_logic:= '1';
        HS_PG00D_PLD: in std_logic:= '1';
              HS_FLT_PLD: in std_logic:= '1';
              ifc_ad0_7: inout std_logic_vector(2 downto 0);
              ifc_cs3_n: in std_logic;
              ifc_oe_n: in std_logic;
              ifc_we_n: in std_logic;
              ifc_a27_31: in std_logic_vector(2 downto 0)
 end component;
  signal  ifc_clk:    std_logic;
  signal  BD_SEL_PLD:  std_logic:= '1';
  signal  HS_PG00D_PLD: std_logic:= '1';
  signal  HS_FLT_PLD:  std_logic:= '1';
  signal  ifc_ad0_7:   std_logic_vector(2 downto 0);
  signal  ifc_cs3_n:   std_logic;
  signal  ifc_oe_n:    std_logic;
  signal  ifc_we_n:    std_logic;
  signal  ifc_a27_31:  std_logic_vector(2 downto 0);
  constant time_tb:   time:= 10 ms;
signal pon_rst_n : std_logic;
  uut: hotswap port map(
  ifc_clk => ifc_clk,
  BD_SEL_PLD =>  BD_SEL_PLD,
  HS_PG00D_PLD => HS_PG00D_PLD,
  HS_FLT_PLD  => HS_PG00D_PLD,
  ifc_cs3_n =>  ifc_cs3_n,
  ifc_oe_n  =>  ifc_oe_n,
  ifc_we_n  => ifc_oe_n,
  ifc_a27_31 =>  ifc_a27_31);
clk: process
    ifc_clk <= '1';
    wait for time_tb/2;
    ifc_clk <= '0';
    wait for time_tb/2;  
     end process;
stimlus: process
  wait for 5 ms;
  pon_rst_n <= '0';
  wait for 5 ms;
  pon_rst_n <= '1';
  wait for 5 ms;
  ifc_a27_31 <= "111";
       end process;

2017-08-23 07:25: Edited by Moderator

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Re: Internal signals in vhdl

von Lothar M. (Company: Titel) (lkmiller) (Moderator)

2017-08-23 07:44

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Usually you do NOT force an internal signal because you will not be able 
to do that in a real design. So apply the stimuli data (inputs) in a way 
you get the desired internal signals.

A few words to that design:
At all the simulation of this code will not match the reality due to a 
wrong and incomplete sensitivity list. Here in simulation it will look 
like the process depends on the clock, but in real life it is only a 
completely combinatorial process:

    process(ifc_clk)
      begin      
      if((cpld_cs and cpld_oe)= '1') then
        case ifc_a27_31 is
          when HOTS_WAP =>
            ifc_dout(0) <= HS_FLT_PLD;
            ifc_dout(1) <= BD_SEL_PLD; 
            ifc_dout(2) <= HS_PG00D_PLD;
          when others =>
            ifc_dout <= "000";
        end case;
      end if;
     end process;

And much more worse: you implement a latch here.
That is 1. not necessary, 2. not desired and 3. no good design practice.

I suggest to write it without any process (using the concatenate 
operator '&') this way:

      ifc_dout <= HS_PG00D_PLD & BD_SEL_PLD & HS_FLT_PLD when 
                        cpld_cs='1' and cpld_oe='1' and ifc_a27_31=HOTS_WAP
                  else "000";


And one word to that here:

if((cpld_cs and cpld_oe)= '1') then

Thats not a fine trick. Write it this way, it is "more usual" and better 
readable:

if (cpld_cs='1' and cpld_oe='1') then

If you want you can skip the parenthesis also:

if cpld_cs='1' and cpld_oe='1' then



BTW: Pls use the [vhdl] tokens around VHDL code (as described above 
every text edit box).

2017-08-23 07:50: Edited by Moderator

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Re: Internal signals in vhdl

von Hareesh M. (Company: Mindteck) (hareeshp)

2017-08-23 07:55

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thanks for your suggestion. Actually the values of HS_PG00D_PLD, 
BD_SEL_PLD and HS_FLT_PLD might change while execution.The width of 
ifc_dout is 7, for simulation i made it 3. That's why i wrote like that. 
And i'll change the line as per your suggestion if((cpld_cs and 
cpld_oe)= '1') then.
and what about the testbench?

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Re: Internal signals in vhdl

von Lothar M. (Company: Titel) (lkmiller) (Moderator)

2017-08-23 08:17

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Hareesh M. wrote:
> Actually the values of HS_PG00D_PLD, BD_SEL_PLD and HS_FLT_PLD might
> change while execution.
Then those signals must be in the sensitivity list!

If you want to march on with the process, then this is the correct 
sensitivity list:

    process(HS_PG00D_PLD, BD_SEL_PLD, HS_FLT_PLD, cpld_cs, cpld_oe, ifc_a27_31)
      begin      
      if((cpld_cs and cpld_oe)= '1') then
        case ifc_a27_31 is
          when HOTS_WAP =>
            ifc_dout(0) <= HS_FLT_PLD;
            ifc_dout(1) <= BD_SEL_PLD; 
            ifc_dout(2) <= HS_PG00D_PLD;
          when others =>
            ifc_dout <= "000";
        end case;
      end if;
     end process;


Hareesh M. wrote:
> and what about the testbench?
You already startet stimuli generation. Just go on with all the possible 
states:

stimlus: process
  wait for 5 ms;
  pon_rst_n <= '0';
  wait for 5 ms;
  pon_rst_n <= '1';
  wait for 5 ms;
  ifc_a27_31   <= "111";
  ifc_ad0_7    <= "ZZZ";
  BD_SEL_PLD   <=  '1';
  HS_PG00D_PLD <=  '0';
  HS_FLT_PLD   <=  '0';
  ifc_cs3_n    <=  '0'; -- low active
  ifc_oe_n     <=  '0'; -- low active
  ifc_we_n     <=  '0'; -- low active
  wait for 1 us;
  assert ifc_ad0_7="100" report "Read not correct!" severity failure;
  wait for 1 us;
  ifc_a27_31   <= "111";
  ifc_ad0_7    <= "ZZZ";
  BD_SEL_PLD   <=  '1';
  HS_PG00D_PLD <=  '1';
  HS_FLT_PLD   <=  '0';
  ifc_cs3_n    <=  '0'; -- low active
  ifc_oe_n     <=  '0'; -- low active
  ifc_we_n     <=  '0'; -- low active
  wait for 1 us;
  assert ifc_ad0_7="110" report "Read not correct!" severity failure;
  wait for 1 us;
  -- and so on
end process;


BTW1:
obviously all of your select signals are low active due to the _n at the 
end of their names. Therefore this is rubbish:

          cpld_cs <= ifc_cs3_n;              -- low active --> high active??
          cpld_oe <= ifc_cs3_n and ifc_oe_n; -- cpld_oe is low active!
          cpld_we <= ifc_cs3_n and ifc_we_n; -- cpld_we also!

But at all this complete "control signal process" is not necessary at 
all. Instead it is leading to a kind of clocked bus handling. But that 
you will find out in simulation further on.
A hint: you MUST set the inout bus to HiZ when you don't use it for 
output!

BTW2:
That looks very, very confusing to me:
ifc_ad0_7: inout std_logic_vector(2 downto 0);
Must it be ifc_ad2_0 instead?

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Re: Internal signals in vhdl

von Ricky T. (Company: Toshiba) (rickyterzis)

2017-08-25 18:42

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Hi...i am a new user here.As per my knowledge you do NOT force an 
internal signal because you will not be able. To do that in a real 
design. So apply the stimuli data in a way you get the desired internal 
signals.

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