如何使用PIC24读取多个ADC通道；只能获取AN0

Question

我使用PIC24读取数据，使用3个模拟输入，但只得到1个来显示正确的结果。我在互联网上到处寻找，仍然不能让代码工作。

我正在尝试读取3个模拟输入信号，但只能在AN0中读取。

目前，我正在使用加速度计获取数据并将其显示在LCD屏幕上。我实现了3种不同的方法来获取和显示数据，但只有an0有效，而an1不是正确的值。

void InitADC(int amask) {
    AD1PCFG = 0xFFF8; // select AN0, AN1, AN2 as analog inputs
    AD1CON1 = 0x00E0; // auto convert @ end of sampling, Integer Data out.
    // see Text pg. 179 & Sec. 17 on AD1CON1.
    //AD1CON2bits.CSCNA = 1;
    AD1CON3 = 0x1F01; // Tad = 2xTcy = 125ns. 31*Tad for conversion time.
    //AD1CSSL = 0xFFF7; // Scan 3 channels
    AD1CON1bits.ADON = 1; // Turn on the ADC
} // InitADC

main() {
   InitADC(0xFFF8); // initialize the ADC and analog inputs 
   char x_string [12];
   char y_string [12];
   char z_string [12];
   //TRISB = 1; // all PORTB pins as outputs
   TRISBbits.TRISB0 = 1;
   TRISBbits.TRISB1 = 1;
   TRISBbits.TRISB2 = 1;
   InitPMP(); // Initialize the Parallel Master Port
   InitLCD(); // Initialize the LCD
   float x_val;
   float y_val;
   float z_val;
   float x_axis, y_axis, z_axis;

   while (1) // main loop
   {  
    x_axis= SelectPort(0);   
    x_val= ((((x_axis * 3.3)/ 1024)-1.58)/0.380);
    
    sprintf(x_string, "X: %0.2f ", x_val);
    ms_delay(2.5);

    y_axis= SelectPort(1);  
    y_val= ((((y_axis * 3.3)/ 1024)-1.58)/0.380);
    
    sprintf(y_string, "Y: %0.2f ", y_val);
    ms_delay(2.5);

    z_axis= SelectPort(2);  
    z_val= ((((z_axis * 3.3)/ 1024)-1.58)/0.380);
    
    sprintf(z_string, "Z: %0.2f ", z_val);
    ms_delay(2.5);
}

下面是读取数据的代码：

int SelectPort(int ch)
{
//int *result; 
AD1CON1bits.ADON = 0; // Turn off the ADC to reconfigure
//result = &ADC1BUF0;

switch(ch) // set values based on the channel to use
{
    case 0: // select AN0 as analog input
        //AD1CHSbits.CH0SA=0;
        //result = ADC1BUF0;
        AD1PCFG = 0xFFFE;
        break;
    
    case 1:
        //AD1CHSbits.CH0SA=1;
        //result = ADC1BUF1;
        AD1PCFG = 0xFFFD; // select AN1 as analog input
        break;
    
    case 2:
        //AD1CHSbits.CH0SA=2;
        AD1PCFG = 0xFFFB; // select AN2 as analog input
        break;
    
    // there's only so many options here, so there's not really a default case
}

AD1CON1bits.ADON = 1; // Turn on the ADC

AD1CHS = ch; // 1. select analog input channel

AD1CON1bits.SAMP = 1; // 2. Start sampling.
while (!AD1CON1bits.DONE); //5. wait for conversion to complete
AD1CON1bits.DONE = 0; // 6. clear flag. We are responsible see text.
return ADC1BUF0; // 7. read the conversion results
}

我是PIC24的新手，需要帮助来弄清楚为什么我不能获得多个模数转换通道来读取数据。

Answer 1

更换输入端口后，应始终留出一段时间让ADC输入电压稳定下来。此外，请确保输入信号的阻抗小于10K。

分离ADC通道切换并读取其值将有所帮助，并使您有机会添加延迟。

我在PIC24数据表中找不到AD1PCFG。AD1PCFG是一个PIC32寄存器...PIC24使用ANSx和TRSx将引脚设置为模拟输入。在启动序列期间，应设置一次模拟输入。将模拟信号馈送到数字输入CMOS引脚将导致功耗增加，甚至可能导致硬件故障！

//  Set the pins as analog inputs once and for all in your setup code.
void InitADC() 
{
    // select RB0/AN0, RB1/AN1, RB2/AN2 as analog inputs
    TRISB   |= 0x07;
    ANSB    |= 0x07;

    // ...
}

void ADC_SelectInput(int ch)
{
    AD1CHS = ch & 0x0F
}

int ADC_Read()
{
    AD1CON1bits.SAMP = 1;
    while (!AD1CON1bits.DONE)
        ;
    AD1CON1bits.DONE = 0;
    return ADC1BUF0;
}

然后你的循环变成：

// ...


int adc_in;

while (1)
{  
    ADC_SelectInput(0);
    ms_delay(3);         // settling delay, you could also do something
                         // else in the meantime instead of waiting. 
                         // NOTE: the argument to ms_delay() is an integer
                         // so waiting at least 2.5 ms makes it 3. 

    // when computing floats, use float constants, not double.
    x_val  = (((ADC_Read() * 3.3f) / 1024) - 1.58f) / 0.380f;
    sprintf(x_string, "X: %0.2f ", x_val);

    ADC_SelectInput(1);
    ms_delay(3);
    y_val  = (((ADC_Read() * 3.3f) / 1024) - 1.58f) / 0.380f;
    sprintf(y_string, "Y: %0.2f ", y_val);

    ADC_SelectInput(2);
    ms_delay(3);
    z_val  = (((ADC_Read() * 3.3f) / 1024) - 1.58f) / 0.380f;
    sprintf(z_string, "Z: %0.2f ", z_val);
}