With 51 single chip microcomputer to show the frequency of the multivibrator composed of 555 timer.-CodePudding

随着电容的不同,多谐振荡器在示波器显示的方波是不同的,周期不同频率也不同,f=1/T=0.7(R1+2R2)C,这里的R1和R2都是5k,C为c1到c5的变化,如图所示,程序编译也没有问题,为什么数码显示的总有问题,特别是当c=10uf时,按照上方的公式应该显示9.52左右,可它却显示80,不知问题出在哪,麻烦各位看一下,对于这些又不是很懂,搞了很久不知错在哪里,下边还有当c为0.1uf,10uf屏幕显示的频率,麻烦各位看一下,急啊,

CodePudding user response:

39 lines should be interrupt 0?

CodePudding user response:

reference 1st floor forever74 response:

39 lines should be interrupt 0?

To interrupt 0 are shown F000

CodePudding user response:

Computing machine cycle timer crystals is like this: 11.0592 MHz, then the machine cycle is 12 * (1/11059200)=1.09 mu m, if T=50 ms the N=50000/1.09=45872. The code interrupt initial TH0 and TL0 values are accurate? 43 TR1=0;//the interrupt enable, the line if we need to move to the front line 41,

CodePudding user response:

TMOD=0 x51; T0 timer work in timing 50 ms mode, 50 ms interrupt time, execute the statement in the interrupt function, T1 work in counting mode, TH1 high deposit produced by 555 square wave number eight, deposit TL1 low square wave number eight, and then to interrupt the value to the bb and cc, TR1=1, start the counter T1, TR1=0, counter T1 to stop working,

CodePudding user response:

reference QZJHJXJ reply: 3/f

computing machine cycle timer crystals is like this: 11.0592 MHz, then the machine cycle is 12 * (1/11059200)=1.09 mu m, if T=50 ms the N=50000/1.09=45872. The code interrupt and initial value TH0 TL0 values are accurate? 43 TR1=0;//the interrupt enable, the line if we need to move to the front line 41,

Moved to 41 before or not,