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AVR delay loop的计算

AVR 精确延时loop 计算

  • LDI: 1 clock cycle

  • DEC: 1 clock cycle

  • NOP: 1 clock cycle

  • LPM: 3 clock cycle

  • BRNE: 1 clock cycle if (Z == 0); 2 clock cycle if (Z != 0)

Example:

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		LDI		R16,	100
L1:
		DEC		R16
		BRNE	L1

这个小循环

  1. load R16 用 1 cycle
  2. R16 从100减到0，一共计算100次，其中，99次计算结果不为0，Z!=0，BRNE用 2 cycle，1次计算结果为0，BRNE用 1 cycle。每次自减是 1 cycle。

number of cycles=LDI+(100−1)×(BRNE(z!=0)+DEC)+1×(BRNE(z=0)+DEC)=1+(100−1)×(2+1)+1×(1+1)=100×3=300\begin{aligned} number\ of\ cycles & = LDI+(100-1)\times(BRNE_{(z!=0)}+DEC)+1\times(BRNE_{(z=0)}+DEC) \\ & = 1+(100-1)\times(2+1)+1\times(1+1) \\ & = 100\times3=300 \end{aligned}

delay 1s:

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    	ldi		r16, 	82
    	ldi		r17, 	43
    	ldi		r18, 	0
L1:		
	    dec		r18
    	brne	L1
    	dec		r17
    	brne	L1
    	dec		r16
    	brne	L1
    	lpm
    	nop

对于这个：

ncycle=(256−1)×3+2+1⏟r18从0自减到0+(43−1)×(256×3+2)⏞r18从0自减到0+2+1⏞r17⏟r17从43自减到0+(82−1)×[(256×3+2)×256+2]+2+1⏟r16从82自减到0+3+1=16000000\begin{aligned} n_{cycle} & = \underbrace{(256-1)\times3+2+1}_{r18从0自减到0}\\ &+\underbrace{(43-1)\times(\overbrace{256\times3+2)}^{r18从0自减到0}+\overbrace{2+1}^{r17}}_{r17从43自减到0}\\ &+\underbrace{(82-1)\times[(256\times3+2)\times256+2]+2+1}_{r16从82自减到0}\\ &+3+1 \\ &=16000000 \end{aligned}

用时：

116MHz∗16000000cycles=1s\frac{1}{16MHz}*16000000 cycles = 1s

怎么反推呢，知道时间，写代码？

http://darcy.rsgc.on.ca/ACES/TEI4M/AVRdelay.html

到着推的话，先算出82，再算出43，再算出256

16000000(3×256+2)×256+2+1=82.1680≈82......\frac{16000000}{(3\times256+2)\times256+2}+1 = 82.1680 \approx 82 \\ ...\\ ...

AVR