极客工坊

 找回密码
 注册

QQ登录

只需一步,快速开始

查看: 10567|回复: 0

8F328U在使用 delayMicroseconds()時會很不準

[复制链接]
发表于 2019-4-18 14:58:27 | 显示全部楼层 |阅读模式
本帖最后由 eddiewwm 于 2020-3-2 22:26 编辑

實測 8F328U在使用 delayMicroseconds()時會很不準,會快很多,delay 1000us時,得出的約是 765us。

可參考網上的文章 https://www.arduino.cn/thread-12468-1-1.html 作出改進。

LGT 原來在 wiring.c 中的設定更改為:
  1. /* Delay for the given number of microseconds.  Assumes a 1, 8, 12, 16, 20 or 24 MHz clock. */
  2. void delayMicroseconds(unsigned int us)
  3. {
  4.         // call = 4 cycles + 2 to 4 cycles to init us(2 for constant delay, 4 for variable)

  5.         // calling avrlib's delay_us() function with low values (e.g. 1 or
  6.         // 2 microseconds) gives delays longer than desired.
  7.         //delay_us(us);

  8. #if F_CPU >= 32000000L
  9.         // for the 32 MHz clock for the aventurous ones, trying to overclock

  10.         // zero delay fix

  11.         if (us < 1) return; //  = 3 cycles, (4 when true)

  12.         // the following loop takes a 1/6 of a microsecond (4 cycles)
  13.         // per iteration, so execute it six times for each microsecond of
  14.         // delay requested.
  15.         us = (us << 3) + 2*us + us/2 + us/8; // ? cycles

  16.         // account for the time taken in the preceeding commands.
  17.         // we just burned 22 (24) cycles above, remove 5, (5*4=20)
  18.         // us is at least 6 so we can substract 5
  19. //        us -= 5; //=2 cycles

  20. #elif F_CPU >= 24000000L
  21.         // for the 24 MHz clock for the aventurous ones, trying to overclock

  22.         // zero delay fix
  23.         if (!us) return; //  = 3 cycles, (4 when true)

  24.         // the following loop takes a 1/6 of a microsecond (4 cycles)
  25.         // per iteration, so execute it six times for each microsecond of
  26.         // delay requested.
  27.         us *= 6; // x6 us, = 7 cycles

  28.         // account for the time taken in the preceeding commands.
  29.         // we just burned 22 (24) cycles above, remove 5, (5*4=20)
  30.         // us is at least 6 so we can substract 5
  31.         us -= 5; //=2 cycles

  32. #elif F_CPU >= 20000000L
  33.         // for the 20 MHz clock on rare Arduino boards

  34.         // for a one-microsecond delay, simply return.  the overhead
  35.         // of the function call takes 18 (20) cycles, which is 1us
  36.         __asm__ __volatile__ (
  37.                 "nop" "\n\t"
  38.                 "nop" "\n\t"
  39.                 "nop" "\n\t"
  40.                 "nop"); //just waiting 4 cycles
  41.         if (us <= 1) return; //  = 3 cycles, (4 when true)

  42.         // the following loop takes a 1/5 of a microsecond (4 cycles)
  43.         // per iteration, so execute it five times for each microsecond of
  44.         // delay requested.
  45.         us = (us << 2) + us; // x5 us, = 7 cycles

  46.         // account for the time taken in the preceeding commands.
  47.         // we just burned 26 (28) cycles above, remove 7, (7*4=28)
  48.         // us is at least 10 so we can substract 7
  49.         us -= 7; // 2 cycles

  50. #elif F_CPU >= 16000000L
  51.         // for the 16 MHz clock on most Arduino boards

  52.         // for a one-microsecond delay, simply return.  the overhead
  53.         // of the function call takes 14 (16) cycles, which is 1us

  54.         if (us < 1) return; //  = 3 cycles, (4 when true)

  55.         // the following loop takes 1/4 of a microsecond (4 cycles)
  56.         // per iteration, so execute it four times for each microsecond of
  57.         // delay requested.

  58.         us = (us << 2) + us + us/4 + us/17  ; // ? cycles

  59.         // account for the time taken in the preceeding commands.
  60.         // we just burned 19 (21) cycles above, remove 5, (5*4=20)
  61.         // us is at least 8 so we can substract 5
  62. //        us -= 1; // = 2 cycles,

  63. #elif F_CPU >= 12000000L
  64.         // for the 12 MHz clock if somebody is working with USB

  65.         // for a 1 microsecond delay, simply return.  the overhead
  66.         // of the function call takes 14 (16) cycles, which is 1.5us
  67.         if (us <= 1) return; //  = 3 cycles, (4 when true)

  68.         // the following loop takes 1/3 of a microsecond (4 cycles)
  69.         // per iteration, so execute it three times for each microsecond of
  70.         // delay requested.
  71.         us = (us << 1) + us; // x3 us, = 5 cycles

  72.         // account for the time taken in the preceeding commands.
  73.         // we just burned 20 (22) cycles above, remove 5, (5*4=20)
  74.         // us is at least 6 so we can substract 5
  75.         us -= 5; //2 cycles

  76. #elif F_CPU >= 8000000L
  77.         // for the 8 MHz internal clock

  78.         // for a 1 and 2 microsecond delay, simply return.  the overhead
  79.         // of the function call takes 14 (16) cycles, which is 2us
  80.         if (us <= 2) return; //  = 3 cycles, (4 when true)

  81.         // the following loop takes 1/2 of a microsecond (4 cycles)
  82.         // per iteration, so execute it twice for each microsecond of
  83.         // delay requested.
  84.         us <<= 1; //x2 us, = 2 cycles

  85.         // account for the time taken in the preceeding commands.
  86.         // we just burned 17 (19) cycles above, remove 4, (4*4=16)
  87.         // us is at least 6 so we can substract 4
  88.         //us -= 4; // = 2 cycles

  89. #else
  90.         // for the 1 MHz internal clock (default settings for common Atmega microcontrollers)

  91.         // the overhead of the function calls is 14 (16) cycles
  92.         if (us <= 16) return; //= 3 cycles, (4 when true)
  93.         if (us <= 25) return; //= 3 cycles, (4 when true), (must be at least 25 if we want to substract 22)

  94.         // compensate for the time taken by the preceeding and next commands (about 22 cycles)
  95.         us -= 22; // = 2 cycles
  96.         // the following loop takes 4 microseconds (4 cycles)
  97.         // per iteration, so execute it us/4 times
  98.         // us is at least 4, divided by 4 gives us 1 (no zero delay bug)
  99.         us >>= 2; // us div 4, = 4 cycles
  100.         

  101. #endif

  102.         // busy wait
  103.         __asm__ __volatile__ (
  104.                 "1: sbiw %0,1" "\n\t" // 2 cycles
  105.                 "brne 1b" : "=w" (us) : "0" (us) // 2 cycles
  106.         );
  107.         // return = 4 cycles
  108. }


复制代码

針對 8F328U 板用的是 16MHz主頻 作出了以下的更改
#elif F_CPU >= 16000000L
.......
us = (us << 2) ; // x5 us, = 7 cycles

delay 1000us 時得到的是 :
us = (us << 2);                 // x  ==> 765us
us = (us << 2) + us;        // x  ==> 954us
us *= 6;                         // x  ==> 1140us

在不作任何改動下,就祇能乘以系數 1.31 作補償了。
回复

使用道具 举报

您需要登录后才可以回帖 登录 | 注册

本版积分规则 需要先绑定手机号

Archiver|联系我们|极客工坊

GMT+8, 2024-4-20 08:32 , Processed in 0.048124 second(s), 17 queries .

Powered by Discuz! X3.4 Licensed

Copyright © 2001-2021, Tencent Cloud.

快速回复 返回顶部 返回列表