Timer design based on 51 microcontrollers

The above answer is fine.

When using timer method 1, the calculated value of (65536-x) exceeds 256, that is, the 16-bit timing mode of the timer, so it is necessary to send 8 bits higher to th0 and lower bits to tl0, so to calculate the high 8 bits and the low 8 bits of (65536-x), (65536-x) 256 is to calculate the high 8 bits, and (65536-x)%256 is to calculate the low 8 bits, so it is written.

th0 = (65536-x) 256 (or th1).

tl0 = (65536-x)%256 (or tl1).

This is actually very simple, you only need to know one condition: crystal frequency.

Because the machine cycle calculation formula of 51 single-chip microcomputer is 12 crystal oscillator frequency MHz, if the crystal oscillator is 12mhz, then the machine cycle is 12 (12 10 6) = 1us, that is to say, the counter does not count once 1us.

Generally speaking, under the 12MHz crystal oscillator, the full count of the 16-bit timer is 65536 times, each time 1us, the maximum can be timed, then 65536 minus how much is the timing, such as (65536-x) 256, that is, the timing x microseconds.

So if the crystal oscillator is not 12MHz, multiply the above x by 12 crystal oscillator, and the value obtained is the timing duration. The principle is the same.

Feel free to reply to me if you don't understand something.

That's right. Although the timer timing is more accurate than the program delay, it does not mean that there are no errors. In any case, errors are always objective. Here are two additional points:

1. There is an error in the original program. The maximum count value of timer mode 0 is 2 13=8192us, and the initial installation value is e018h=57368us, that is, the initial loading value of timing is much greater than the maximum counting value.

If you want to time 1ms=1000us in mode 0, the initial installation value should be 8192-1000=7192=1c18h, and the execution of other statements can be ignored because it is 2 3us level (the three statements in the factor interrupt program take only 6us). You may ask: why not convert these times into the initial value of the timer?

The answer is that it is not necessary. No matter how it is converted, there are errors, not to mention that the time of the machine cycle is also estimated.

2. If you want to be more accurate, you can use method 2 (automatic reload of the initial value).

You're right about what you understand.

Regarding the reloading of the 51 timer, the typical solution is dynamic reloading. That is, when you renew the initial value of the count, you should subtract the count from the total number to be loaded from the time when the previous overflow was ready to perform the initial value of the reload.

For example, if the interrupt interval is 1000 counts, the initial value of the reload is -1000. In the interrupt, -1000 cannot be reloaded directly, but -(1000-th tl), that is, the reload value is -1000+th tl.

isr_t0:

push psw

clr tr0

xch a, tl0

add a, #low(-1000+7)

xch a, tl0

xch a, th0

addc a, high(1000+7) tr0 has been paused for 7 periods and needs to be added.

xch a, th0

set tr0

CPL can execute the events that the user wants to do, but it keeps taking too long.

pop psw

RetiOf course, it can't be completely said that the book is wrong, only that the examples in the book are not rigorous.