如何校准 ti INA219 传感器？ （在 C 中）

Question

所以我根据this计算了校准寄存器值：

我的分流电阻值为 0.01 欧姆。即使我根据我的分流值校准它，我的电流也会关闭

所有电流显示错误值（例如它应该显示7.8mA，它显示1mA

以下是我的代码：

#include <unistd.h>
#include <stdlib.h>
#include <stdio.h>
#include <errno.h>
#include <endian.h>
#include <string.h>
#include <time.h>
#include <getopt.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <sys/ioctl.h>
#include <fcntl.h>
#include <linux/i2c-dev.h>

#define CONFIG_REG          0
#define SHUNT_REG           1
#define BUS_REG             2
#define POWER_REG           3
#define CURRENT_REG         4
#define CALIBRATION_REG     5


#define INA_ADDRESS1         0x40 
#define INA_ADDRESS2         0x41



int interval = 60;
int i2c_bus = 1;
int i2c_address1 = INA_ADDRESS1;
int i2c_address2 = INA_ADDRESS2;

int handle;

int i2c_read( void *buf, int len )
{
    int rc = 0;

    if ( read( handle, buf, len ) != len )
    {
        printf( "I2C read failed: %s\n", strerror( errno ) );
        rc = -1;
    }

    return 0;
}


int i2c_write( void *buf, int len )
{
    int rc = 0;

    if ( write( handle, buf, len ) != len )
    {
        printf( "I2C write failed: %s\n", strerror( errno ) );
        rc = -1;
    }

    return rc;
}

int register_read( unsigned char reg, unsigned short *data )
{
    int rc = -1;
    unsigned char bite[ 4 ];

    bite[ 0 ] = reg;
    if ( i2c_write( bite, 1 ) == 0 )
    {
        if ( i2c_read( bite, 2 ) == 0 )
        {
            *data = ( bite[ 0 ] << 8 ) | bite[ 1 ];
            rc = 0;
        }
    }

    return rc;
}

int register_write( unsigned char reg, unsigned short data )
{
    int rc = -1;
    unsigned char bite[ 4 ];

    bite[ 0 ] = reg;
    bite[ 1 ] = ( data >> 8 ) & 0xFF;
    bite[ 2 ] = ( data & 0xFF );

    if ( i2c_write( bite, 3 ) == 0 )
    {
        rc = 0;
    }

    return rc;
}


int get_voltage( float *mv )
{
    short bus;

    if ( register_read( BUS_REG, (unsigned short*)&bus ) != 0 )
    {
        return -1;
    }

    *mv = ( float )( ( bus & 0xFFF8 ) >> 1 );
    return 0;
}


int get_current( float *ma )
{
    short shunt;

    if ( register_read( SHUNT_REG, &shunt ) != 0 )
    {
        return -1;
    }

    *ma = (float)shunt / 10;
    return 0;
}


void show_current( void )
{
    float ma;

    if ( get_current( &ma ) )
    {
        fprintf( stderr, "Error reading current\n" );
        return;
    }
}


void show_voltage( void )
{
    float mv;

    if ( get_voltage( &mv ) )
    {
        fprintf( stderr, "Error reading voltage\n" );
        return;
    }
    printf( "%4.0f\n", mv );
}


void show_voltage_current( void )
{
    float mv, ma;

    if ( get_current( &ma ) || get_voltage( &mv ) )
    {
        fprintf( stderr, "Error reading voltage/current\n" );
        return;
    }


    else
    {
        printf( "Load Voltage: %g  Current: %4.1fmA\n", (mv/1000), (ma) );

    }
}


int main( int argc, char *argv[] )
{
    char filename[ 20 ];
    register_write(CALIBRATION_REG, 4096);



    snprintf( filename, 19, "/dev/i2c-%d", i2c_bus );
    handle = open( filename, O_RDWR );
    if ( handle < 0 )
    {
        fprintf( stderr, "Error opening bus %d: %s\n", i2c_bus, strerror( errno ) );
        exit( 1 );
    }

    if ( ioctl( handle,  I2C_SLAVE, i2c_address1 ) < 0 )
    {
        fprintf( stderr, "Error setting address %02X: %s\n", i2c_address1, strerror( errno ) );
        exit( 1 );
    }

        show_voltage_current();
        sleep(1);


    if ( ioctl( handle,  I2C_SLAVE, i2c_address2 ) < 0 )
    {
        fprintf( stderr, "Error setting address %02X: %s\n", i2c_address2, strerror( errno ) );
        exit( 1 );


        show_voltage_current();
        sleep(1);
    }
    return 0;
}

Answer 1

查看https://github.com/flav1972/ArduinoINA219 作为参考。

您应该在电流/功率值可靠之前设置状态和校准寄存器

尝试将预期电流至少更改为最大电流的 1/4，并记住设置有效的分流电压。

在我的测试板上，如果我将预期电流定义为 1.6A 和 max_current 16A，我的结果是错误的，但如果我将其更改为 4A，结果似乎还可以

为我工作： Rshunt=0.1R, Vbus=16V, Imax=1.6A, Iexp=1.6A, Rshunt=0.01R,Vbus=16V,Imax=16A,Iexp=4..16A