Honored ContributorHi,
I would like to use an external VCXO and I guess amplifier/filter to sync a 10MHz clk to a GPS 1 HZ signal. Can anyone suggest if this is possible and if so where would I start looking for information. Cheers.
Honored ContributorBasically, you have to design the PLL according to your requirements. A CPLD can make the digital part of it. Because of the huge frequency ratio, I guess that a digital PLL, using a Sigma-Delta DAC to control the VCXO may be meaningful, particularly if your intending low phase noise.
It's mainly a matter of good system design, the type of utilized digital logic is of secondary importance to my opinion. And it's almost clear, that the PLL components present in most FPGA are of no use for this application.Honored ContributorNot long ago I designed a phase comparator in stratix II for an external PLL. The phase comparator sends charge pump pulses(of different duty cycles) and it worked. This phase comparator is certainly implementable in CPLD
My clocks were not that far apart but you can get a counter to divide yours. If you are interested in this phase comparator I can post the designHonored ContributorHi FvM,
Thanks for the reply, My maths level is just short of Laplace and Fourier but I'm learning. --- Quote Start --- It's mainly a matter of good system design, the type of utilized digital logic is of secondary importance to my opinion. And it's almost clear, that the PLL components present in most FPGA are of no use for this application. --- Quote End --- Can you recommend a reference for good PLL system design??Honored ContributorHi Kaz,
Thanks , that would be good. I assume you used an RC filter to integrate? /average the pulses.Honored ContributorHi,
My design of the phase comparator is based on this basic idea: clk1 sets a flipflop to ‘1’ at its edge. clk2 sets another flipflop to ‘1’ at its edge both flipflops are then reset asynchronously by their ANDed output (after some delay). Thus the duration of each Q pulse will indicate the distance between edges of the two clks. In practice you don’t need to check each and every clk edge as this may be excessive leading to a swing effect, you can choose how often by using counters. These counters will also help detect the lock state.The value of counters is implementation dependant according to clk ratio and your low pass filter.. In my case the ratio was 2 and I used two counters: 0 ~ 7 on fast clk 0 ~ 3 on slow clk cp_p & cp_n are the Q outputs of flipflops and used for the charge pump. (Adjust it’s sense as required). cp_p_d & cp_n_d are the delayed version for reset(I used wiring to bidirectional IO and back) Rememebr to use fast io registers for Q outputs your clk ratio is too high and you may need to play around for best counting and filter cutoff The following code section may help ------------------------------------------------------------------------------------------------------------ process -- counter on clk1 begin wait until clk1 = ‘1’ clk1_cnt <= clk1_cnt + 1; end process; process -- counter on clk2 begin wait until clk2 = ‘1’ clk2_cnt <= clk2_cnt + 1; end process; process (clk1, clr) -- clk1 flip-flop begin if (clr = '1') then cp_n <= '0'; elsif rising_edge (clk1) then if (clk1_cnt = 7) then cp_n <= '1'; end if; end if; end process; process (clk2, clr) -- clk2 flip-flop begin if (clr = '1') then cp_p <= '0'; elsif rising_edge (clk2) then if (clk2_cnt = 3) then cp_p <= '1'; end if; end if; end process; -- reset when both outputs are high clr <= cp_p_d and cp_n_d; -- PLL lock detector process begin wait until clk1 = ‘1’; if (clk2_cnt = 3) then if (clk1_cnt = 7) or (clk1_cnt = 0) then locked <= '1'; else locked <= '0'; end if; end if; end process ; ----------------------------------- good luckHonored ContributorHi,
Not being a VHDL expert, does the following look accurate to you. use library ieee; use ieee.std_logic_1164.all; use ieee.std_logic_unsigned.all; ----------------------------------------- entity PhaseDetector is port( clk1 : in STD_LOGIC; clk2 : in STD_LOGIC; cp_n_d : in STD_LOGIC;-- delayed?? cp_p_d : in STD_LOGIC;-- delayed?? cp_n : inout STD_LOGIC; cp_p : inout STD_LOGIC; locked : out STD_LOGIC ); end PhaseDetector; architecture behav of PhaseDetector is signal clr : STD_LOGIC ; signal clk1_cnt: integer range 0 to 255; -- clk1 counter signal clk2_cnt: integer range 0 to 7; -- clk2 counter begin process -- counter on clk1 begin wait until clk1 = '1';-- clk1 is VCXO signal clk1_cnt <= clk1_cnt + 1; end process; process -- counter on clk2 begin wait until clk2 = '1'; -- clk2 is PPS signal clk2_cnt <= clk2_cnt + 1; end process; process (clk1,clr) -- clk1 flip-flop begin if (clr = '1') then cp_n <= '0'; elsif rising_edge (clk1) then if (clk1_cnt = 7) then -- adjustment value cp_n <= '1'; end if; end if; end process; process (clk2,clr) -- clk2 flip-flop begin if (clr = '1') then cp_p <= '0'; elsif rising_edge (clk2) then if (clk2_cnt = 3) then -- adjustment value cp_p <= '1'; end if; end if; end process; -- reset when both outputs are high clr <= cp_p and cp_n; -- PLL lock detector process begin if (clk2_cnt = 3) then if (clk1_cnt = 7) or (clk1_cnt = 0) then locked <= '1'; else locked <= '0'; end if; end if; end process ; end behav;Honored ContributorHi,
There are two issues here, one related to actual design and the other related to vhdl 1) your fast clk is 10MHz, slow is 1Hz. So without doubt your ratio is very high. You need to look at edge of clk1Hz and compare it with clk10MHz only when both edges are expected to lock. i.e. if slow counter is 0~3 then fast counter will be 0~(2^26-1). So only when reaching maximum counts you need to drive the relevant flipflops.(max count = 3 or 40 million-1) Clearly the design will take 4 sec per test so you expect the lock to take few seconds, unless you have means of multiplying clk1Hz(FPGA PLL can do that). 2) The code: you can reduce the ports to: port( clk1 : in STD_LOGIC; clk2 : in STD_LOGIC; cp_n : inout STD_LOGIC; cp_p : inout STD_LOGIC; locked : out STD_LOGIC ); you declare the delay nodes internally and you must drive them then use them in the AND statement. cp_n_d <= cp_n; cp_p_d <= cp_p; clr <= cp_n_d and cp_p_d; your flipflop drive is ok except for count values. So is the detection logic; Always check at the extremes of your counters(not my counters) Good luck
