125KHz HW PWM Output

[Zebryk]

Greetings!

Having problems generating a clean 50% duty cycle hardware PWM @125KHz using a PIC16F628 with a 20MHz crystal.
The below is from a 2002 Microchip AppNote: (Which specified a 16F628 in the original schematic.)

The drive signal can be generated directly by the PWM
unit on most PIC® microcontrollers such as a
PIC16F627 microcontroller. For a device operating at
20 MHz, one can obtain a 125 kHz signal by setting the
Timer2 prescaler to 1. A period of 8 μs is then obtained
by setting the PR2 register to 39. To get a 50% duty
cycle output, set CCPR1L to 14 and CCP1CON<5:4>
to <0:0>. These settings will ensure a constant carrier.
To modulate the data, one can turn the drive signal on
and off by setting and clearing the CCP1RIL bit.

My interpertation of the above code is:
PR2 = 39 ' A period of 8 µs is then obtained by setting the PR2 register to 39.
CCPR1L = 14 ' To get a 50% duty cycle output, set CCPR1L to 14 and CCP1CON<5:4> to <0:0>.
CCP1CON = %00001100 ' PWM Mode, CCP1CON<5:4> = <0:0>
T2CON = %00000100 ' For a device operating at 20 MHz, one can obtain a 125 kHz signal by setting the Timer2 prescaler to 1.

However, this results in the below:

9243

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Note that although it is exactly 125.0KHz, it is not symetrical and has an approximate duty cycle of 2.8us/5.2us * 100 = 35%

Although running on an old dev board from Bruce Reynolds, I would prefer to update to a PIC10(LF32x) dev board.
(Or something from melabs.)

Suggestions?

Jay Zebryk, W1JRZ​

[richard]

Note that although it is exactly 125.0KHz, it is not symetrical and has an approximate duty cycle of 2.8us/5.2us * 100 = 35%

maybe more like duty cycle of 2.8us/8us * 100 = 35%



CCPR1L = 14 ' To get a 50% duty cycle output, set CCPR1L to 14 and CCP1CON<5:4> to <0:0>.

or correctly
' To get a 50% duty cycle output,
CCPR1L to $14 and CCP1CON<5:4> to <0:0>.

[Zebryk]

I found this:

https://www.micro-examples.com/public/microex-navig/doc/097-pwm-calculator.html

Which produces this:

PR2 = 0b00001001 ;
T2CON = 0b00000101 ;
CCPR1L = 0b00000100 ;
CCP1CON = 0b00111100 ;

Running that results in 3.8 / (3.8 +5.2) = 47.5% which is much closer.

FYI: This is for a RFID application wherein the center frequency is important to the resonate gain of the attached antenna.

Wondering if this is an 8, 10 or 16Bit computation?
Wondering if this makes any difference consideration the even value of 125.0KHz?

Thanks,

Jay Zebryk, W1JRZ

[richard]

totally missing the point

CCPR1L = $14 not 14 ; its hex 14 or dec 20

[Ioannis]

It is a 10bit PWM system but the resolution of the PWM signal depends on the PWM period or frequency. Higher the frequency, the lower the resolution.

So, at 20MHz and a frequency signal of 125KHz, the resolution will be 7 bits only, even though the control register is 10bit wide.

Ioannis

[Zebryk]

Ioannis,

Thanks for the feedback!

I have been using the attached calculator from Mister E.

9245

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Is the Duty Step % (0.625) X 50% = 31.25% the problem?
(I am measuring 3.8uS On / 4.2uS Off)
How can I set it for 4mS On / 4mS exactly?

(FYI, I am headed towards filtering the Square wave into a Sine wave and want it to be symetrical.)

Is this even doable with the ancient 16F628?
I am only using this because I had the Dev Boards handy.
Do you remember Bruce Reynolds?
What would you recommend for a 8PIN Dip?

Peace,

Jay Zebryk

[mpgmike]

Referencing the Mister e Calculator, it looks like the best resolution you can get is 0 to 160, where 160 will be 100% Duty Cycle.

[Zebryk]

Hi,

That implies one needs to set the Duty Cycle to 80 to get a 4uS On / 4uS Off square wave?
(This does not make sense to me.)

Thanks,

Jay Zebryk

[HenrikOlsson]

Richard already covered it but it seems to have been overlooked.

The duty cycle register is 10bit wide but, as Ioannis says, the usable number of bits differs depending on the PWM period. The calculator tells you that for 125kHz (with PR2=39) varying the duty cycle register from 0 to 160 will give you a duty cycle of 0-100%.

For 50% duty cycle you obviously want to set the register to 80.

Now, remember that the register is 10bit wide and that two low order bits are in CCP1CON so what goes in CCPR1L is the top 8 bits of the 10bit value. Or, in this case 80/4=20 which, as Richard rightly points out is $14.

[richard]

Is the Duty Step % (0.625) X 50% = 31.25% the problem?(I am measuring 3.8uS On / 4.2uS Off)
How can I set it for 4mS On / 4mS exactly?

your understanding is flawed and math is incorrect

Duty is Steps x 0.625% // steps = CCPR1L x4 + CCP1CON<5:4>

so:
80x0.625% = 50% <=> steps = 20x4 + 0
56x0.625% = 35% <=> steps = 14x4 + 0