1. Home
  2. Embedded Programming
  3. How to generate a signal on Xilinx Spartan II

How to generate a signal on Xilinx Spartan II

Hi,

I wish to generate a frequency of approx 400 Hz using Xilinx Spartan II(200 MHz)and send the 1 bit signal to a speaker output and hope to hear some noise. My VHDL code, tested on PeakVHDL simulator does generate the waveform and is pasted at the far bottom. The problem is that the code does not compile on Xilinx because "WAIT for 2.5 ns" is not supported on Xilinx Spartan II for a process. What would be the simplest way out to generate 400 approx Hz on a Xilinx 200MHz device? I have used

200MHz/(2 to the power of 19) = 382 Hz approx. (Use MSB of 19 bits of STD_LOGIC_VECTOR)

Another thing which has confused me is: If I wish to write an entity(below) for Spartan II, does the programmer worry about generating the signal for "clk" input? Or simply connect it to the correct pin of FPGA and I should get the signal of 200MHz?

ENTITY some_entity IS PORT (clk : IN BIT); END some_entity;

For my code tested on PeakVHDL, I have generated the 200MHz signal using a test bed(music_tester) and then modified it to 400Hz.

I apologise if the question is basic.

Thanks in advance

ENTITY music_tester IS PORT (clk : OUT STD_LOGIC; freq : IN STD_LOGIC); END music_tester;

ARCHITECTURE behavioral OF music_tester IS BEGIN process BEGIN clk

Reply to
Rakesh Sharma
Loading thread data ...

The "WAIT..." statement is not synthesisable. You simply need a counter (ripple counter will do) to divide the clock down to 400 Hz or so. Write the VHDL for a toggle flip-flop and string lots of them together - not very elegant but it should work.

Leon

Reply to
Leon Heller

The best solution is to use a phase accumulator (aka Direct Digital Synthesis). All that is is a fancy name for an accumulator that adds a fixed increment value to itself on every clock cycle. The MSB of the accumulator is the clock output, and the output frequency is related to the master clock frequency by:

fo = fc*n/(2^k)

where: fo = output frquency fc = master clock frequency n = increment value (integer) k = number of bits in accumulator.

For example, you have a 200 MHz master clock, and you need the output to be 400 Hz. Lets say we use an 32 bit accumulator, so n = 400/200M * (2^32) = 8589.9, which rounded is 8590. Plugging that back in, you have an output frequency of 400.003 Hz. The number of bits in the accumulator sets the resolution of the frequency setting you can have. The output will have a maximum jitter of +/- 1/2 clock period of the master clock.

The circuit is just an accumulator that always adds the constant n to itself. This works by taking advantage modulo arithmetic. Basically , you keep accumulating the fractional part and throwing away the integer part of the accumulated sum. Each time the sum crosses over to the next integer, the integer part is dropped.

Rakesh Sharma wrote:

19);

STD_LOGIC);

--

--Ray Andraka, P.E. President, the Andraka Consulting Group, Inc.

401/884-7930 Fax 401/884-7950 email snipped-for-privacy@andraka.com
formatting link

"They that give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety." -Benjamin Franklin, 1759

Reply to
Ray Andraka

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.