Study on STC12 delay function

This writing method is to nest assembly in C. The exact delay time = 2*t*T+5*T, T is one machine cycle. For 8051,

If t=1, then the execution of this function should be 7us. If it is STC12, then the delay of this delay function should be: if t=1, the precise delay = 2*1*1/12+5*1/12=（7/12）us.

I checked the STC12 manual yesterday. The manual describes the clock in a very general way. One sentence is: Compared with the execution time of 8051 instructions, there are 12 instructions in total, which can be executed in one clock, and the operation speed is increased by 8 to 12 times at the same frequency.

Should it be (7/8)us? Instead of (7/12)us?

This is a uint type function and the time is approximate

Then you can use delay time = 8*t*T

It should be 8*1*(1/12)=8/12

According to the program analysis, it is conjectured that the delay function should be 1us when t=1, which is the most reasonable.

Using an oscilloscope to test, it takes about 5.8ms to execute a delay (5000). 51 It takes about 40064us to execute a delay (5000). Well, using an oscilloscope to measure the frequency change of the IO port is 85.47HZ, so a delay (5000) is (1/85.47)/2*1000=5850us

Here, t is set to uint type, which has errors. In addition, the execution of LED=1;LED=0; will also consume time and result in errors.

Finally, it is concluded that this delay statement takes about 7us to execute once on 51 and 1us on 12. The oscilloscope test passed! It is consistent with my calculation.