51单片机的D/A转换 TLC5615的一个程序问题 51单片机AD转换实验,电路和程序
51\u5355\u7247\u673a\u7a0b\u5e8f\u95ee\u9898 stc12c5a60s2 \u7684a/d\u8f6c\u6362\u95ee\u9898..\u597d\u5427\uff0c\u5148\u58f0\u660e\u6211\u4e5f\u6b63\u5728\u5b66\u5355\u7247\u673a\uff0c\u56de\u7b54\u53ef\u80fd\u4e0d\u4e13\u4e1a\uff0c\u56e0\u4e3aAD\u95ee\u9898\u6211\u53ea\u63a5\u89e6\u8fc7\u4e00\u6b21\u3002\u4ee5\u4e0b\u7ed3\u8bba\u6765\u7740STC12\u7684\u624b\u518c\u3002
1.\u8f6c\u6362\u7684\u5f15\u811a\u7531ADC_CONTG\u7684\u4f4e3\u4f4dCHS2\u3001CHS1\u548cCHS0\u63a7\u5236\uff0c\u4e09\u4f4d\u4e8c\u8fdb\u5236\u80fd\u8868\u793a0~7\u516b\u4e2a\u6570\uff0c\u4e0eP1\u53e3\u4e00\u4e00\u5bf9\u5e94\u3002ADC\u7684\u7ed3\u6784\u51b3\u5b9a\u4e86\u5b83\u4e00\u6b21\u53ea\u80fd\u8f6c\u6362\u4e00\u4e2a\u53e3\uff0c\u4f46\u662fSTC\u7684\u8f6c\u6362\u901f\u5ea6\u8fbe\u523010\u4e07\u6b21/\u79d2\uff0c\u4f9d\u6b21\u8f6c\u6362P1.0\u548cP1.1\u4e0e\u540c\u65f6\u8f6c\u6362\u5dee\u522b\u4e0d\u5927\u3002\u8bbe\u7f6eP1.0\u65f6\uff0cch=0x00.\u8bbe\u7f6eP1.1\u65f6\uff0cch=0x01.
2.\u662f\u6309\u7167\u4e00\u4e2a\u5b57\u8282\u8bfb\u7684\uff0c\u76f4\u63a5\u8bfbADC_RES\u7684\u8bdd\u53ea\u80fd\u8bfb\u5230\u6570\u636e\u7684\u9ad8\u516b\u4f4d\u3002\u53ea\u75288\u4f4dADC\u8f6c\u6362\u7684\u8bdd\uff0c\u8bfbADC_RES\u5c31\u53ef\u4ee5\u4e86\u3002\u8981\u60f3\u5f97\u523010\u4f4d\u6570\u636e\u5fc5\u987b\u52a0\u4e0aSendData(ADC_LOW2);
3.\u65e2\u7136send\u51fd\u6570\u63d0\u5230\u4e86SBUF\uff0c\u90a3\u4e48\u8bf4\u660e\u8f6c\u6362\u7ed3\u679c\u662f\u901a\u8fc7\u4e32\u53e3\u4f20\u5230\u7535\u8111\u4e0a\uff0c\u7528STC\u4e0b\u8f7d\u5de5\u5177\u7684\u4e32\u53e3\u8c03\u8bd5\u5e94\u8be5\u53ef\u4ee5\u770b\u5230\u8f6c\u6362\u7ed3\u679c\u3002
4.\u8fde\u7eed\u8bfb\u53d6\u53ef\u4ee5\u7528\u51fd\u6570\u5faa\u73af\u6216\u5b9a\u65f6\u5668\u5faa\u73af\u6765\u8c03\u7528\u8f6c\u6362\u51fd\u6570\u3002\u4f8b\u5b50\u561b\uff0c\u5b9a\u65f6\u5668\u6211\u8fd8\u4e0d\u4f1a\u7528\u800c\u4f60\u7ed9\u7684\u503c\u5b9a\u65f6\u5668\u4e2d\u65ad\u6cd5\u3002\u6240\u4ee5\u6211\u8bf4\u4e0b\u51fd\u6570\u7684\u65b9\u6cd5\uff0c\u5927\u6982\u662f\u8fd9\u6837\uff1a
void main{
InitADC();//\u521d\u59cb\u5316ADC
while(1){//\u6b7b\u5faa\u73af
adc_isr();//\u8f6c\u6362\u51fd\u6570
delay(xx);//\u5ef6\u65f6\u4e00\u6bb5\u65f6\u95f4\uff0c\u5426\u5219\u8f6c\u6362\u592a\u5feb\uff0c\u6570\u636e\u91cf\u592a\u5927\u3002
}
}
5.UART\u662f\u4e00\u79cd\u901a\u7528\u4e32\u884c\u6570\u636e\u603b\u7ebf\uff0c\u5728\u8fd9\u91cc\u662f\u7528\u4e8e\u4e0e\u7535\u8111\u8fdb\u884c\u4e32\u53e3\u901a\u4fe1\u7684\u3002
#include
#include
#include
// define P1.0 to check STATUS.
sbit STATUS = P1^0;
unsigned char xdata CTRL _at_ 0x2FFF;
unsigned char xdata ADSEL _at_ 0x4FFF;
unsigned char hByte;
unsigned char lByte;
void adc_Convert (void)
{ // Start a conversion with A0 and A/$C$ low.
// The convesion takes place on rising CE edge.
CTRL = 0x00;
ADSEL = 0x00;
// Wait until we have completed a conversion .
while(STATUS==1);
// Set R/$C$ with A0 low and read the low byte.
CTRL = 0x02;
hByte = ADSEL;
// Set R/$C$ with A0 high and read the high.
CTRL = 0x03;
lByte = ADSEL;
}
void main(void)
{ unsigned int delay, MSB , LSB, adc_Res;
// Initialize serial interface
SCON = 0xDA; // SCON: mode 1, 8-bit UART, enable rcvr */
TMOD |= 0x20; // TMOD: timer 1, mode 2, 8-bit reload */
TH1 = 0xFD; // TH1: reload value for 1200 baud @ 12MHz */
TR1 = 1; // TR1: timer 1 run */
TI = 1; // TI: set TI to send first char of UART */
while(1)
{ adc_Convert();
MSB=(unsigned int)(hByte << 4);
LSB=(unsigned int)(lByte >> 4);
// adc_Res now has the converted data with 12-bit resolution.
adc_Res = MSB + LSB;
// Send adc results to the serial interface
printf("ADC READINGS: %03Xh\n", adc_Res);
// simple delay - it is mcu clock dependent !
for (delay=0; delay<10000; delay++)
;
}
}
for(j=0;j<12;j++)
{
if(dat&0x8000)
din=1;
else
{ din=0;
clk=1;
dat<<=1; //左移一位，并且使其有时间读入数据
clk=0;}
你的ELSE 后面的第一条语句算其条件语句，后面2条每次循环都执行了，要在ELSE后面条件满足的几条语句加个大括号。
void input(uint dat)
{
uchar j;
clk=0;
cs=0;
dat<<=6;
for(j=0;j<12;j++)
{
if(dat&0x0800)//是不是这里啊。
din=1;
else
din=0;
clk=1;
dat<<=1; //左移一位，并且使其有时间读入数据
clk=0;
}
cs=1;
}
应该这样吧，你的for循环转换的是12位的啊，如果12位，那你输入256的结果应该是5*256/4096啊，你仔细检查一下你的位数，包括for循环、还有if(dat&0x0800)里面的常数。
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