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pacman91
Joined: 17 Jun 2011
Posts: 28
Location: Malaysia
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| I got question about the PWM |
 Posted: Sat Jun 18, 2011 11:02 am |
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I copy few example from other forum
I'm newbie, forgive me if i ask stupid question =.=
1.
| Quote: | Here is an example of the code the spreadsheet produces for a PWM frequency of 16KHz on a midrange PIC running at 8MHz.
Code:
setup_timer_2(T2_DIV_1, 124, 1 );
setup_ccp1( CCP_PWM );
set_pwm1_duty( 250L ); // square wave output - 50% duty |
2.
| Quote: | #include <16F877.h>
#fuses XT, NOWDT, NOPROTECT, BROWNOUT, PUT, NOLVP
#use delay(clock = 4000000)
main()
{
output_low(PIN_C1); // Set CCP2 output low
output_low(PIN_C2); // Set CCP1 output low
setup_ccp1(CCP_PWM); // Configure CCP1 as a PWM
setup_ccp2(CCP_PWM); // Configure CCP2 as a PWM
setup_timer_2(T2_DIV_BY_16, 124, 1); // 500 Hz
set_pwm1_duty(31); // 25% duty cycle on pin C2
set_pwm2_duty(62); // 50% duty cycle on pin C1
while(1); // Prevent PIC from going to sleep (Important !)
} |
3.
| Quote: | setup_timer_2(t2_div_by_16,0xff,16); The timing works like this. 1/clock frequency gives you your period (4/4000000=1us) . Then, I have a pre-scale of 16 (.000001*16=16us) Then I have a post-scale of 16 (.000016*16=256us). The period is 0xff or 255. This is how many times the timer can loop before an overflow (.000256*255=65.28ms or 65,280us). 1/.06528=15.32Hz. This is the slowest we can make this timer.
set_pwm1_duty(512); The duty is the ratio of high to low. A duty of 1 is about 1% high and 99% low. This makes an LED very dim. 512 is 50% high and 50% low. It’s the equivalent of this:
output_high(LED); delay_ms(8); output_low(LED); delay_ms(8); //loop this forever
A duty cycle of 1023 is 1% high and 99% low. The LED is very bright!
set_timer_2(0); This is to start the timer, and where to start it. |
4.
| Quote: | #use delay(clock=20000000)
void main(void)
{
setup_timer_2(T2_DIV_BY_1,24,1);
setup_ccp1(CCP_PWM);
setup_ccp2(CCP_PWM);
set_pwm1_duty(20); //80%
set_pwm2_duty(20); //80%
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i confuse
2 and 4 the duty cycle
| Quote: | setup_timer_2(T2_DIV_BY_16, 124, 1);
set_pwm1_duty(31); 124*25% = 31
set_pwm2_duty(62); 124*50% = 62 |
| Quote: | setup_timer_2(T2_DIV_BY_1,24,1);
setup_ccp1(CCP_PWM);
setup_ccp2(CCP_PWM);
set_pwm1_duty(20); //80%
set_pwm2_duty(20); //80% |
compare with 1 and 3
| Quote: | setup_timer_2(T2_DIV_1, 124, 1 );
setup_ccp1( CCP_PWM );
set_pwm1_duty( 250L ); 50%?? why not 124*50%=62?? |
| Quote: |
setup_timer_2(t2_div_by_16,0xff,16);
set_pwm1_duty(512); 50%?? why not 0xff = 255, 255*50% = 128??
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I doing a project about the three axis accelerometer to control the servo motor, so i need know how to use the PWm or timer.....confusing |
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PCM programmer
Joined: 06 Sep 2003
Posts: 21708
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pacman91
Joined: 17 Jun 2011
Posts: 28
Location: Malaysia
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| Posted: Sat Jun 18, 2011 1:21 pm |
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o??
just now 1,2,3,4 have some different is because of the bits?? |
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PCM programmer
Joined: 06 Sep 2003
Posts: 21708
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pacman91
Joined: 17 Jun 2011
Posts: 28
Location: Malaysia
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| Posted: Sat Jun 18, 2011 1:29 pm |
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http://www.ccsinfo.com/forum/viewtopic.php?t=26751
| Quote: | So if you use the same parameter value in both 8-bit and 10-bit mode,
then the perceived effect will be that the duty cycle is divided by 4
when you use a 16-bit parameter. If that's controlling a motor, the
maximum speed will be only 1/4 of what it was. (Assuming a 1:1
relationship between duty cycle and speed).
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question1
mean that the case 2 and 4 is 8bits, case 1 and 3 is 10bits?
question2
| Quote: | setup_timer_2(T2_DIV_1, 124, 1 );
setup_ccp1( CCP_PWM );
set_pwm1_duty( 250L ); // square wave output - 50% duty |
124*4 = 496
496*50% = 248
why it is 250L? |
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pacman91
Joined: 17 Jun 2011
Posts: 28
Location: Malaysia
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| Posted: Sat Jun 18, 2011 2:29 pm |
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ok i learned about it thx
in case i want to control a servo motor
period = 20ms, pull up 2ms, pulldown 18ms, how to write the code?
..........2ms
...........___........18ms..............___............................
_____|......|______________|......|_____________
..........<------------20ms--------> |
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PCM programmer
Joined: 06 Sep 2003
Posts: 21708
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pacman91
Joined: 17 Jun 2011
Posts: 28
Location: Malaysia
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| Posted: Sat Jun 18, 2011 9:40 pm |
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wohooo..thx |
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pacman91
Joined: 17 Jun 2011
Posts: 28
Location: Malaysia
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| Posted: Sun Jun 19, 2011 1:45 am |
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setup_adc(ADC_CLOCK_DIV_32);
for example crystal= 20Mhz
TAD(time for Analog to Digital???) = 1/20Mhz * 32 = 0.16 microseconds
Requires min. 12 TAD for 10 bit...Why ned 12 in 10bit?? how about 8bit??
Min. conversion time approx. 2 microsecond?? Mean 12 * 0.16microseconds = 1.92 = 2microseconds? convert A/D need 2 microseconds? only in 10 bit? how about 8 bit?
why we want to put DIV_32?
can we put DIV_1,2,3,4,........100???? |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19504
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| Posted: Sun Jun 19, 2011 2:51 am |
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Data sheet, data sheet, data sheet.
The ADC has a maximum clock rate _it_ can run at. Depends on the chip, but typically 500KHz for the older chips up to a couple of MHz for the later chips. The division brings the clock rate from your CPU down to a rate the ADC can use.
Division ratios allowed _are in the data sheet_ typically limited range of binary values only, so perhaps 1,2,4,8,16,32,64 - again _read the data sheet_. For your chip, only /2, /8, and /32 are available.
The ADC performs a 'successive approximation' conversion. Effectively 'is the value greater than half range', then select this half of the range, and ask if the value is in the top/bottom half of this range, then repeat again and again. One test for each bit in the output value. Two more clock cycles, one to disconnect the input source, and set everything up, and one to reconnect at the end. Hence 12 clocks for a 10bit conversion. Generally, the converter performs a complete (10bit) conversion, even if you are only reading 8bits of the result. Read up on successive approximation ADC's.
The 'min conversion time' being referred to in the data sheet, is _per bit_. It is the minimum value for Tad - so for the example you give, 2uSec per bit = 500KHz. However it is _not_ actually 2uSec for your chip.
From the 16F877 data sheet:
"For correct A/D conversions, the A/D conversion clock (Tad) must be selected to ensure a minimum Tad time of 1.6uSec.". 625000Hz
Now step to 20MHz. /8, gives 20000000/8 = 2500000Hz. Too fast.
20000000/32 = 625000Hz. OK
Tad is then 1.6uSec, and a conversion takes 19.2uSec
A repeated conversion takes longer than this. First, the capacitor used for the conversion needs to recharge - Tacq. 19.72uSec for your chip, and also the value needs to be read out of the registers. Also on this chip there is a 2*Tad delay between completing the reading, and the input capacitor being reconnected. So the fastest 'repeated' conversion, would probably be about 19.2+19.72+3.2uSec per conversion. Just over 42uSec/conversion. About 23500 conversions/second. Trying to get better than perhaps 20000conversions per second on this chip would probably be a waste of effort - latter chips have faster ADC's...
You need to look at some of the Microchip application notes about the ADC, and read the data sheet.
Best Wishes |
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pacman91
Joined: 17 Jun 2011
Posts: 28
Location: Malaysia
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| Posted: Sun Jun 19, 2011 3:20 am |
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| ok thx bro =) |
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