output error clocks from PLL

Altera_Forum · ‎12-22-2010

HI! Everyone!

I met a strange question before. the source clock is 48Mhz, So I want to use a PLL to get c0(CLK_25MHz) and c1(CLK_100MHz).But when I use Modelsim and SIgnalTap II for verification,the result made me surprised.

The following is Modelsim simulation printscreen,I know It's right:

and this is the SignalTap II printscreen:

You can see the Duty Cycle is different, SO I wonfer what happened,can you tell me?

************************************************** **

PS: the following is my HDL source and testbench code:

HDL source code:

LIBRARY IEEE;

USE IEEE.STD_LOGIC_1164.ALL;

USE IEEE.STD_LOGIC_UNSIGNED.ALL;

USE IEEE.STD_LOGIC_ARITH.ALL;

ENTITY SYSTEM_CONTROL IS

PORT(--INPUT SIGNALS

CLK : IN STD_LOGIC; --FPAG输入时钟信号48MHz

RST_n : IN STD_LOGIC; --系统复位信号

--OUTPUT SIGNALS

SYS_RST_n: OUT STD_LOGIC; --系统全局复位信号，低有效

CLK_25M : OUT STD_LOGIC; --PLL输出25MHz时钟

CLK_100M : OUT STD_LOGIC --PLL输出100MHz时钟

);

END ENTITY SYSTEM_CONTROL;

ARCHITECTURE behave OF SYSTEM_CONTROL IS

SIGNAL LOCKED: STD_LOGIC; --PLL输出有效标志位，高表示PLL输出有效

SIGNAL PLL_RST: STD_LOGIC; --PLL复位信号，高有效

SIGNAL RST_r1: STD_LOGIC; --PLL复位寄存信号r1

SIGNAL RST_r2: STD_LOGIC; --PLL复位寄存信号r2

SIGNAL SYS_RST_nr0:STD_LOGIC; --全局系统复位寄存信号r0，低有效

SIGNAL SYS_RST_nr1: STD_LOGIC; --全局系统复位寄存信号r1，低有效

SIGNAL SYS_RST_nr2: STD_LOGIC; --全局系统复位寄存信号r2，低有效

SIGNAL CLK_100M_r: STD_LOGIC; --PLL输出100MHz时钟寄存信号

COMPONENT PLL IS

PORT

(

areset : IN STD_LOGIC := '0';

inclk0 : IN STD_LOGIC := '0';

c0 : OUT STD_LOGIC ;

c1 : OUT STD_LOGIC ;

locked : OUT STD_LOGIC

);

END COMPONENT;

BEGIN

--***********************************************

--PLL复位信号产生，高有效

--异步复位，同步释放

PROCESS(CLK,RST_n,RST_r1)

BEGIN

IF RST_n = '0' THEN

RST_R1 <= '1';

ELSE

IF CLK'EVENT AND CLK = '1' THEN

RST_r1 <= '0';

END IF;

END PROCESS;

PROCESS(CLK,RST_n,RST_r2,RST_r1)

BEGIN

IF RST_n = '0' THEN

RST_r2 <= '1';

ELSE

IF CLK'EVENT AND CLK = '1' THEN

RST_r2 <= RST_r1;

END IF;

END PROCESS;

PLL_RST <= RST_r2;

--***********************************************

--系统复位全局信号产生，低有效

--异步复位，同步释放

SYS_RST_nr0 <= RST_n AND LOCKED; --系统复位直到PLL有效输出

PROCESS(CLK_100M_r,SYS_RST_nr0,SYS_RST_nr1)

BEGIN

IF SYS_RST_nr0 = '0' THEN

SYS_RST_nr1 <= '0';

ELSE

IF CLK_100M_r'EVENT AND CLK_100M_r = '1' THEN

SYS_RST_nr1 <= '1';

END IF;

END PROCESS;

PROCESS(CLK_100M_r,SYS_RST_nr0,SYS_RST_nr2,SYS_RST _nr1)

BEGIN

IF SYS_RST_nr0 = '0' THEN

SYS_RST_nr2 <= '0';

ELSE

IF CLK_100M_r'EVENT AND CLK_100M_r = '1' THEN

SYS_RST_nr2 <= SYS_RST_nr1;

END IF;

END PROCESS;

SYS_RST_n <= SYS_RST_nr2;

--***********************************************

--例化PLL产生模块

PLL_inst : PLL PORT MAP

(

areset => PLL_RST, --PLL复位信号,高电平复位

inclk0 => CLK, --PLL输入时钟，25MHz

c0 => CLK_25M, --PLL输出25MHz时钟

c1 => CLK_100M_r, --PLL输出100MHz时钟

locked => LOCKED --PLL输出有效标志位，高表示PLL输出有效

);

CLK_100M <= CLK_100M_r;

END ARCHITECTURE behave;

*************************************************

Testbench code:

LIBRARY ieee;

USE ieee.std_logic_1164.all;

ENTITY SYSTEM_CONTROL_vhd_tst IS

END SYSTEM_CONTROL_vhd_tst;

ARCHITECTURE SYSTEM_CONTROL_arch OF SYSTEM_CONTROL_vhd_tst IS

-- constants

CONSTANT CLK_PRD:TIME:= 21 NS; --the source clock is 48MHz

-- signals

SIGNAL CLK : STD_LOGIC;

SIGNAL CLK_25M : STD_LOGIC;

SIGNAL CLK_100M : STD_LOGIC;

SIGNAL RST_n : STD_LOGIC;

SIGNAL SYS_RST_n : STD_LOGIC;

COMPONENT SYSTEM_CONTROL

PORT (

CLK : IN STD_LOGIC;

CLK_25M : OUT STD_LOGIC;

CLK_100M : OUT STD_LOGIC;

RST_n : IN STD_LOGIC;

SYS_RST_n : OUT STD_LOGIC

);

END COMPONENT;

BEGIN

i1 : SYSTEM_CONTROL

PORT MAP (

-- list connections between master ports and signals

CLK => CLK,

CLK_25M => CLK_25M,

CLK_100M => CLK_100M,

RST_n => RST_n,

SYS_RST_n => SYS_RST_n

);

CLK_init : PROCESS

-- variable declarations

BEGIN

CLK <= '1';

WAIT FOR CLK_PRD/2;

CLK <= '0';

WAIT FOR CLK_PRD/2;

END PROCESS CLK_init;

RST_n_INIT : PROCESS

-- optional sensitivity list

-- ( )

-- variable declarations

BEGIN

RST_n <= '0';

WAIT FOR CLK_PRD*4;

RST_n <= '1';

WAIT;

END PROCESS RST_n_INIT;

END SYSTEM_CONTROL_arch;

Altera_Forum · ‎12-22-2010

Hi,

SignalTap captures data on the sampling clk. Ideally you use same sampling clk as data clk, Or integer multiple of it since timing problems and cross domain issues apply to this tool as to logic. so I believe you are sampling your clk wrongly.

Altera_Forum · ‎12-22-2010

kaz,

Thanks for your reply, In the Design, I used the ouput clcok c1(CLK_100MHz) to sample data, But the sampling clock I used in the SignalTap II is the source clock(48MHz), I can't understand If there is any relation between the two sampling clock used in different area, So Can you make it more clear,many thanks!

Altera_Forum · ‎12-22-2010

sampling clock must be equal to twice or more integer multiple of 100 and 25MHz otherwise you get timing problems in the signaltap buffer(as any dc fifo). Moreover you are violating sampling theorem (Nyquist will get angry??)

Altera_Forum · ‎12-22-2010

In theory and as long as sampling clk is twice or more its data then it should be ok. i.e it does not need be integer strictly. In practice I always sample my data on its own clk and get more consistent results. I use several signaltap instants per file if required. I don't trust the signaltap clk domain cross over when data clk and sampling clk are not related.

Altera_Forum · ‎12-22-2010

wow, I got it,your reply is reliable and humor! thanks!