寻找图片的1分钟延迟 [英] looking for 1 minute delay for pic

问题描述

会员成员
我正在尽力将图片延迟1分钟，我正在配置设备
在启用内部振荡器和看门狗时，OPTION_REG = 10001000.
我想知道我是否还应该对行的总数进行计数并将其乘以1/64以获取所有指令所需的时间
需要您的精美评论

Hellow members
i am trying my best to delay the pic for 1 minute,i am configuring the device
at internal oscilator and watchdog enabled,the OPTION_REG=10001000.
i wonder if i should ALSO count total numbers of lines and multiply it by 1/64 to get time required for all instruction
need your fine comments

	movlw	b'10001000'
	movwf	OPTION_REG
.
.
.
		movlw	.1
		movwf	N_Minute
                call    DelayiM
.
.
.
DelayiM	sleep
		sleep	     
loopiiMM	sleep
		movlw .60
		movwf Seconds
loopi2M			sleep
			sleep
			movlw .52 
                        movwf j		

loopj2M					sleep	
					clrwdt
					decfsz	j,1	
					goto loopj2M
			decfsz	Seconds,1
			goto loopi2M
decfsz N_Minute,1
goto loopiiMM			
return

推荐答案

我已经忘记了如何创建延迟.

但是我记得我曾经做过一些计算.
首先了解您的图片时钟速度是多少.
如果您使用的是4Mhz crysta，您将获得1Mhz时钟速度.我完全忘记了为什么.
因此，在1Mh​​z中，您将获得1,000,000个周期.
因此编写一个函数以获得255 * 2 + 2周期

I have forgotten how I created delay.

But I can remember that I used to do some calculation.
first learn what is your pic clock speed.
if you are using 4Mhz crysta you will get 1Mhz clock speed. I totally forgot why.
so in 1Mhz you are getting 1,000,000 cycle.
so write a function to get 255*2+2 cycle

//i forgot how to declare a function 
;if you call this function with 255 data then you will get 255*2+2=512 cycle
func1 macro TIME //calling this function will give 255 cycle.
   DECFSZ TIME, 1 ; this will give 255 cycle
   NOP ; this will give 255 cyle
END

因此，显然您已创建了0.000512秒的延迟.您只有几步之遥，只有一秒钟的时间:)
现在，请剩下的工作……

so apparently you have created a delay of 0.000512 seconds. you are few steps away to make it one seconds :)
now please do the rest...

我认为，如果不认真看一下微控制器的文档，谁都不会知道.这是一个非常不寻常的问题，因为这里的大多数人都专门研究高级languanges.所以我想Google是您的朋友:)

I dont think anybody could know that without a serious look at the documentation form the microcontroller. THis is highly unusual question, as most people here have specialized in higher languanges. So I guess google is your friend at this one :)

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