# Calculating Pulse per minute in a FPGA

Discussion in 'VHDL' started by Cory Shol, May 28, 2013.

1. ### Cory SholGuest

Hi all,

Another problem from the two year FPGA newbie.

I am working on a new work project that does all this magnificent things and now I am a little stumped on a rather easy to understand problem.

I need to calculate how many pulses per minute of an input to display out of a Seven segment display.

I have completed all the user interface stuff and it works fine etc...

The problem is with the calculation.

I am running at 25 Mhz clock. 40 ns period.

Basically I wait for a rising edge of the first pulse and reset my counter. Every rising edge of my 25 Mhz clock I increment the counter until the next rising edge of the pulse.

Basically the Pulse can be as slow as 15 ppm to 1000 ppm. So if you have a 1000 ppm pulse inputting you would see 1500000 25 Mhz clock counts.

The hard part is converting 1500000 counts into a PPM.

I know the math is easy : 60 /[(count)*(Period of Clk)] = PPM

60/ (1500000 *(0.00000004))= 1000 ppm

Doing division in VHDL seems to be tough.

I have one algorithm idea of doing basically the basic of all basics.

which is: 60/clkPeriod = 1.5 billion
initially do this:

newCount <= newCount - count;
quotient <= quotient + 1;

Then after initial do this:

if(newCount >= Count) then
newCount <= newcount -count;
quotient <= quotient + 1;
else
end if;

Are there any other easy algorithms that are relatively easy to implement? Or a standard way the VHDL community does divisions?

Thanks

Cory
Cory Shol, May 28, 2013

On Tue, 28 May 2013 10:01:02 -0700 (PDT)
Cory Shol <> wrote:

> Hi all,
>
> Another problem from the two year FPGA newbie.
>
>
> I am working on a new work project that does all this magnificent things and now I am a little stumped on a rather easy to understand problem.
>
> I need to calculate how many pulses per minute of an input to display out of a Seven segment display.
>
> I have completed all the user interface stuff and it works fine etc...
>
> The problem is with the calculation.
>
> I am running at 25 Mhz clock. 40 ns period.
>
> Basically I wait for a rising edge of the first pulse and reset my counter. Every rising edge of my 25 Mhz clock I increment the counter until the next rising edge of the pulse.
>
> Basically the Pulse can be as slow as 15 ppm to 1000 ppm. So if you have a 1000 ppm pulse inputting you would see 1500000 25 Mhz clock counts.
>
> The hard part is converting 1500000 counts into a PPM.
>
> I know the math is easy : 60 /[(count)*(Period of Clk)] = PPM
>
> 60/ (1500000 *(0.00000004))= 1000 ppm
>
> Doing division in VHDL seems to be tough.
>
> I have one algorithm idea of doing basically the basic of all basics.
>
> which is: 60/clkPeriod = 1.5 billion
> initially do this:
>
> newCount <= newCount - count;
> quotient <= quotient + 1;
>
> Then after initial do this:
>
> if(newCount >= Count) then
> newCount <= newcount -count;
> quotient <= quotient + 1;
> else
> end if;
>
>
> Are there any other easy algorithms that are relatively easy to implement? Or a standard way the VHDL community does divisions?
>
> Thanks
>
> Cory
>
>

Stupid question: if you're trying to count how many pulses per minute
you get, why don't you just block off 1 minute chunks of time, and see
how many pulses you get in each chunk?

--
Rob Gaddi, Highland Technology -- www.highlandtechnology.com
Email address domain is currently out of order. See above to fix.

3. ### Cory SholGuest

On Tuesday, May 28, 2013 12:19:31 PM UTC-5, Rob Gaddi wrote:
> On Tue, 28 May 2013 10:01:02 -0700 (PDT)
>
> Cory Shol <> wrote:
>
>
>
> > Hi all,

>
> >

>
> > Another problem from the two year FPGA newbie.

>
> >

>
> >

>
> > I am working on a new work project that does all this magnificent things and now I am a little stumped on a rather easy to understand problem.

>
> >

>
> > I need to calculate how many pulses per minute of an input to display out of a Seven segment display.

>
> >

>
> > I have completed all the user interface stuff and it works fine etc...

>
> >

>
> > The problem is with the calculation.

>
> >

>
> > I am running at 25 Mhz clock. 40 ns period.

>
> >

>
> > Basically I wait for a rising edge of the first pulse and reset my counter. Every rising edge of my 25 Mhz clock I increment the counter until the next rising edge of the pulse.

>
> >

>
> > Basically the Pulse can be as slow as 15 ppm to 1000 ppm. So if you have a 1000 ppm pulse inputting you would see 1500000 25 Mhz clock counts.

>
> >

>
> > The hard part is converting 1500000 counts into a PPM.

>
> >

>
> > I know the math is easy : 60 /[(count)*(Period of Clk)] = PPM

>
> >

>
> > 60/ (1500000 *(0.00000004))= 1000 ppm

>
> >

>
> > Doing division in VHDL seems to be tough.

>
> >

>
> > I have one algorithm idea of doing basically the basic of all basics.

>
> >

>
> > which is: 60/clkPeriod = 1.5 billion

>
> > initially do this:

>
> >

>
> > newCount <= newCount - count;

>
> > quotient <= quotient + 1;

>
> >

>
> > Then after initial do this:

>
> >

>
> > if(newCount >= Count) then

>
> > newCount <= newcount -count;

>
> > quotient <= quotient + 1;

>
> > else

>

>
> > end if;

>
> >

>
> >

>
> > Are there any other easy algorithms that are relatively easy to implement? Or a standard way the VHDL community does divisions?

>
> >

>
> > Thanks

>
> >

>
> > Cory

>
> >

>
> >

>
>
>
> Stupid question: if you're trying to count how many pulses per minute
>
> you get, why don't you just block off 1 minute chunks of time, and see
>
> how many pulses you get in each chunk?
>
>
>
> --
>
> Rob Gaddi, Highland Technology -- www.highlandtechnology.com
>
> Email address domain is currently out of order. See above to fix.

Because The input of the pulse can change at any time. So if it changed within the minute period you wouldn't have the correct PPM.
Cory Shol, May 28, 2013
4. ### Cory SholGuest

On Tuesday, May 28, 2013 1:34:58 PM UTC-5, Cory Shol wrote:
> On Tuesday, May 28, 2013 12:19:31 PM UTC-5, Rob Gaddi wrote:
>
> > On Tue, 28 May 2013 10:01:02 -0700 (PDT)

>
> >

>
> > Cory Shol <> wrote:

>
> >

>
> >

>
> >

>
> > > Hi all,

>
> >

>
> > >

>
> >

>
> > > Another problem from the two year FPGA newbie.

>
> >

>
> > >

>
> >

>
> > >

>
> >

>
> > > I am working on a new work project that does all this magnificent things and now I am a little stumped on a rather easy to understand problem.

>
> >

>
> > >

>
> >

>
> > > I need to calculate how many pulses per minute of an input to display out of a Seven segment display.

>
> >

>
> > >

>
> >

>
> > > I have completed all the user interface stuff and it works fine etc...

>
> >

>
> > >

>
> >

>
> > > The problem is with the calculation.

>
> >

>
> > >

>
> >

>
> > > I am running at 25 Mhz clock. 40 ns period.

>
> >

>
> > >

>
> >

>
> > > Basically I wait for a rising edge of the first pulse and reset my counter. Every rising edge of my 25 Mhz clock I increment the counter until the next rising edge of the pulse.

>
> >

>
> > >

>
> >

>
> > > Basically the Pulse can be as slow as 15 ppm to 1000 ppm. So if you have a 1000 ppm pulse inputting you would see 1500000 25 Mhz clock counts.

>
> >

>
> > >

>
> >

>
> > > The hard part is converting 1500000 counts into a PPM.

>
> >

>
> > >

>
> >

>
> > > I know the math is easy : 60 /[(count)*(Period of Clk)] = PPM

>
> >

>
> > >

>
> >

>
> > > 60/ (1500000 *(0.00000004))= 1000 ppm

>
> >

>
> > >

>
> >

>
> > > Doing division in VHDL seems to be tough.

>
> >

>
> > >

>
> >

>
> > > I have one algorithm idea of doing basically the basic of all basics.

>
> >

>
> > >

>
> >

>
> > > which is: 60/clkPeriod = 1.5 billion

>
> >

>
> > > initially do this:

>
> >

>
> > >

>
> >

>
> > > newCount <= newCount - count;

>
> >

>
> > > quotient <= quotient + 1;

>
> >

>
> > >

>
> >

>
> > > Then after initial do this:

>
> >

>
> > >

>
> >

>
> > > if(newCount >= Count) then

>
> >

>
> > > newCount <= newcount -count;

>
> >

>
> > > quotient <= quotient + 1;

>
> >

>
> > > else

>
> >

>
> > > finalanswer <= quotient;

>
> >

>
> > > end if;

>
> >

>
> > >

>
> >

>
> > >

>
> >

>
> > > Are there any other easy algorithms that are relatively easy to implement? Or a standard way the VHDL community does divisions?

>
> >

>
> > >

>
> >

>
> > > Thanks

>
> >

>
> > >

>
> >

>
> > > Cory

>
> >

>
> > >

>
> >

>
> > >

>
> >

>
> >

>
> >

>
> > Stupid question: if you're trying to count how many pulses per minute

>
> >

>
> > you get, why don't you just block off 1 minute chunks of time, and see

>
> >

>
> > how many pulses you get in each chunk?

>
> >

>
> >

>
> >

>
> > --

>
> >

>
> > Rob Gaddi, Highland Technology -- www.highlandtechnology.com

>
> >

>
> > Email address domain is currently out of order. See above to fix.

>
>
>
> Because The input of the pulse can change at any time. So if it changed within the minute period you wouldn't have the correct PPM.

Also it is a requirement to update after every pulse.
Cory Shol, May 28, 2013
5. ### rickmanGuest

On 5/28/2013 1:01 PM, Cory Shol wrote:
> Hi all,
>
> Another problem from the two year FPGA newbie.
>
>
> I am working on a new work project that does all this magnificent things and now I am a little stumped on a rather easy to understand problem.
>
> I need to calculate how many pulses per minute of an input to display out of a Seven segment display.
>
> I have completed all the user interface stuff and it works fine etc...
>
> The problem is with the calculation.
>
> I am running at 25 Mhz clock. 40 ns period.
>
> Basically I wait for a rising edge of the first pulse and reset my counter. Every rising edge of my 25 Mhz clock I increment the counter until the next rising edge of the pulse.
>
> Basically the Pulse can be as slow as 15 ppm to 1000 ppm. So if you have a 1000 ppm pulse inputting you would see 1500000 25 Mhz clock counts.
>
> The hard part is converting 1500000 counts into a PPM.
>
> I know the math is easy : 60 /[(count)*(Period of Clk)] = PPM
>
> 60/ (1500000 *(0.00000004))= 1000 ppm
>
> Doing division in VHDL seems to be tough.
>
> I have one algorithm idea of doing basically the basic of all basics.
>
> which is: 60/clkPeriod = 1.5 billion
> initially do this:
>
> newCount<= newCount - count;
> quotient<= quotient + 1;
>
> Then after initial do this:
>
> if(newCount>= Count) then
> newCount<= newcount -count;
> quotient<= quotient + 1;
> else
> end if;
>
>
> Are there any other easy algorithms that are relatively easy to implement? Or a standard way the VHDL community does divisions?

Can you say, "look up table"?

--

Rick
rickman, May 29, 2013
6. ### GaborGuest

On 5/28/2013 1:01 PM, Cory Shol wrote:
> Hi all,
>
> Another problem from the two year FPGA newbie.
>
>
> I am working on a new work project that does all this magnificent things and now I am a little stumped on a rather easy to understand problem.
>
> I need to calculate how many pulses per minute of an input to display out of a Seven segment display.
>
> I have completed all the user interface stuff and it works fine etc...
>
> The problem is with the calculation.
>
> I am running at 25 Mhz clock. 40 ns period.
>
> Basically I wait for a rising edge of the first pulse and reset my counter. Every rising edge of my 25 Mhz clock I increment the counter until the next rising edge of the pulse.
>
> Basically the Pulse can be as slow as 15 ppm to 1000 ppm. So if you have a 1000 ppm pulse inputting you would see 1500000 25 Mhz clock counts.
>
> The hard part is converting 1500000 counts into a PPM.
>
> I know the math is easy : 60 /[(count)*(Period of Clk)] = PPM
>
> 60/ (1500000 *(0.00000004))= 1000 ppm
>
> Doing division in VHDL seems to be tough.
>
> I have one algorithm idea of doing basically the basic of all basics.
>
> which is: 60/clkPeriod = 1.5 billion
> initially do this:
>
> newCount <= newCount - count;
> quotient <= quotient + 1;
>
> Then after initial do this:
>
> if(newCount >= Count) then
> newCount <= newcount -count;
> quotient <= quotient + 1;
> else
> end if;
>
>
> Are there any other easy algorithms that are relatively easy to implement? Or a standard way the VHDL community does divisions?
>
> Thanks
>
> Cory
>
>

Even updating 1000 times per minute you have a long time to make the
calculation. If 1000 ppm is the highest rate, you could even do a
simple successive subtraction loop in 1000 cycles or about 40
microseconds with a 25 MHz clock.

On the other hand unless you're really short of resources, why not
just use a division IP core? I know Xilinx has it in Coregen, and
would assume other vendors offer one, too.

--
Gabor
Gabor, May 29, 2013
7. ### chrisabeleGuest

On 5/28/2013 2:36 PM, Cory Shol wrote:
> On Tuesday, May 28, 2013 1:34:58 PM UTC-5, Cory Shol wrote:

<snip>
>> On Tuesday, May 28, 2013 12:19:31 PM UTC-5, Rob Gaddi wrote:
>> Stupid question: if you're trying to count how many pulses per minute
>> you get, why don't you just block off 1 minute chunks of time, and see
>> how many pulses you get in each chunk?
>>
>>> On Tue, 28 May 2013 10:01:02 -0700 (PDT)
>>> Cory Shol <> wrote:

>>
>> Because The input of the pulse can change at any time. So if it changed within the minute period you wouldn't have the correct PPM.

>
> Also it is a requirement to update after every pulse.
>

If you're display is seven segment then I have to assume it's for people
to read. A display that changes at 1KHz (your specified maximum rate)
would be very difficult for most people to make sense of. One solution
would be to average the PPM values that you collect over some period
that's closer to human perception scale and display that. Rob's
suggestion of simply accumulating pulses for one minute is a very simple
direct way to achieve that. Or you could accumulate pulses for one
second and display that value multiplied by 60. You'd have to how to
handle the 15 to 60 PPM range, but still the problem would be much easier.

As usual starting out with a good specification of what's really needed
makes developing a solution more efficient.
chrisabele, May 29, 2013