Software Time Delays

# Software Time Delays - o o o mov cx timeval2 mov dx...

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Software Time Delays 1. Short Time Delays o o o mov cx, timeval1 call delay1 o o o Clks delay1: dec cx 3 nop 3 jnz delay1 16/4 ret 16 a. Time of execution equation: Te| delay1 = 1/fCLK [22(CX – 1) + 10 + 16] = tCLK [22(CX – 1) +26] = tCLK [22CX – 22 + 26] = tCLK [22CX + 4] Example: Assume fCLK = 5 MHz and the desired delay = 10 milliseconds. Te = 200ns [22CX + 4] = 10ms

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CX = [(10 * 10 -3 /0.2 * 10 -6 ) – 4] / 22 = 49,996 / 22 = 2272 10 (= 8E0h) Therefore, timeval1 = 08E0h. b. What value of timeval1 would generate the shortest time delay in the delay1 routine? Longest time delay? Some thoughts: 1) The shortest time would be to pass through the delay 1 routine once without making any loops. 2) The longest time would occur when the value chosen for timeval1 would loop as many times as possible for a 16-bit number. 3) The value 0000h decremented back to 0000h is equivalent to 65,536 iterations. 2. Long Time Delays
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Unformatted text preview: o o o mov cx, timeval2 mov dx, timeval3 call delay2 o o o Clks delay2: dec cx 3 nop 3 jnz delay2 16/4 dec dx 3 nop 3 jnz delay2 16/4 ret 16 a) Time of execution equation: Te| delay2 = 1/fCLK {22(CX – 1) + 10 + [22(65535) + 10 + 22](DX – 1) + 10 + 16} = tCLK {22(CX – 1) + [1441802](DX – 1) + 36} Example: Assume fCLK = 10MHz and desired delay = 500 milliseconds. The longest count is from the (DX – 1) term. So let’s choose a value for DX to get close to the 500 ms value, then we will solve the Te equation for CX to trim into the 500 ms value. Choose DX = 4, then 100ns [1441802](4 – 1) = 0.432 sec. 100ns {22(CX – 1) + [1441802](3) + 36} = 500ms 100ns {22CX + 4325420} = 500ms CX = (5000000 – 4325420) / 22 = 30,663d (= 77C7h) Therefore, o timeval2 = 77C7h o timeval3 = 0004h...
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Software Time Delays - o o o mov cx timeval2 mov dx...

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