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Verilog code for frequency divider (50 Mhz to 1 kHz)

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ph333sh

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what would the verilog code be to change a 50MHz clock to 1kHz with a reset input.
 

verilog code frequency divider

module fredivider(clk,rst,clk_out);
input clk,rst;
output clk_out;
reg counter[15];

always @(posedege clk or negedge rst)
begin
if(!rst)
counter<=16'd0;
else
if(counter==16'd5000)
counter<=16'd0;
else
counter<=counter+1;
end

assign out_clk<=counter[15;

end module







I think this code should work...
it might have syntax errors though...
 
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Re: frequency divider

a small modification to lordsatish code :

use assign out_clk = (counter == 16'd5000);

instead of assign out_clk<=counter[15;

and declare counter as reg[15:0] counter;
 

Re: frequency divider

The duty cycle in the previous code would be 50%
and this modification will give you 0.02%...
correct me if i am wrong...!
 

frequency divider

Thats true lordsatish !!!

You would get only 0.02% duty cycle with my assignment statement

But u can get 50% duty cycle with assign out_clk = (counter <= 16'd25000);
 

frequency divider

would

if(counter >= 5000)

work? would that be 1kHz?
 

Re: frequency divider

This is a Divide by 5000 counter...
So there is no point in the count value being greater than 5000...
However there will be clk_out when count value is equal to 5000...
So it will work...
 

Re: frequency divider

I think that correct value 4999!
From 0 to 5000 you have 5001 pulses
 

Re: frequency divider

Hope this modified code will give u a 1khz output clock for an input of 50Mhz clk with 50% duty cycle, correct am if am wrong


module fredivider(clk,rst,clk_out);
input clk,rst;
output reg clk_out;
reg [15:0] counter;

always @(posedege clk or negedge rst)
begin
if(!rst)
begin
counter<=16'd0;
clk_out <= 1'b0;
end
else
if(counter==16'd25000)
begin
counter<=16'd0;
clk_out <= ~clk_out;
end
else
begin
counter<=counter+1;
end
end

endmodule
 
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frequency divider

Hello, i'm trying to divide 100 MHz clock by 11 and 22 to get 9.091 MHz and 4.545 MHz clock outputs.
I'm using XC9536XL. I thought of using FDD (Double Edge Triggered Flip Flop), but it's not available to XC9536. How can i do it ? Thank's
 

Re: frequency divider

Hi,

I tried to change this code to get a larger duty cycle.


module coinsorter(clk,rst,top_motor);
input clk,rst;
output reg top_motor;
reg [15:0] counter;
always @(posedge clk or posedge rst)
begin
if(rst)
begin
counter<=16'd0;
top_motor <= 1'b0;
end
else
if(counter==16'd12500)
begin
top_motor <= 1'b1;
counter<=counter+1;
end
else
if(counter==16'd50000)
begin
counter <= 16'd0;
top_motor <= 1'b0;
end
else
begin
counter<=counter+1;
end
end
endmodule


The simulation shows that it's working but when I actually apply it to the RC servo motor it doesn't work.

Also I am trying to control this with an external switch.
I don't want the motor to run unless the switch is on, but it keeps giving me errors whenever I try to compile (The code I posted doesn't include the switch bit, obviously)
 

Maybe because your pwm frequency and/or duty cycle values are wrong?

**broken link removed**
 

mrflibble said:
Maybe because your pwm frequency and/or duty cycle values are wrong?

l
Click to expand...

Using resistors the motor now turns 360 degrees (not just a maximum of 180). I need a frequency of 1KHz to rotate it clockwise and 2KHz for anti-clockwise.
So I think the frequency is right because the motor does rotate with a duty cycle of 50% at 1KHz and 2KHz
Only when I try to change the on time does it not work.
 

Any other handy bits of information? The information in your last post is a bit hard to guess based on your first post. :p That, and we don't even know what clock frequency you use for that coinsorter module...

Also ...

The simulation shows that it's working but when I actually apply it to the RC servo motor it doesn't work.
Click to expand...

Kindly post testbench code as well. That always helps.
 

Using resistors the motor now turns 360 degrees (not just a maximum of 180). I need a frequency of 1KHz to rotate it clockwise and 2KHz for anti-clockwise.
Click to expand...
Doesn't seem to refer to a standard RC servo.
 

Wasn't going to be the first to bring that up explicitely. Thought I'd just be the first to bring that up implicitly by including **broken link removed**.

Case in point being: 50 Hz PWM frequency, pulse width on the order of 1-2 ms and all that. So that 1 kHz / 2 kHz scheme looked a bit curious....

So a bit more information on the design probably wouldn't hurt.
 

Re: verilog code frequency divider

Hey i cant get 50 per duty cycle.. how would i in this code??
 

Maybe in another 5 years this thread will have accumulated enough 1-poster fragments to start making some sense. :)
 

Dear All,

I am using clock divider where you will see verilog code in followinx text:
But I need to control the frequency using external register.

Could you please let me know how I should do it.
Thank you very much for your help in advance.

module clock_divider(my_clk,clk_div,timer_divider);
input my_clk;
input reg [31:0] timer_divider;

reg [31:0] count; // clock divider counter

output reg clk_div; // clock divider output

always @ (posedge my_clk)
begin

// if(count == 32'b(timer_divider)) // timer_divider should assign to register where I can control it
if(count ==32'b1111101000)
count <= 32'b0000; // reset to 0
else
count <= count+1; // increment counter
clk_div <= (count == 32'b0000);
end
 

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