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INA149使用问题

Other Parts Discussed in Thread: INA149

如规格书第44页所示,INA149的输入输出关系为Vout=(+IN)-(-IN)+ Vref

如规格书第4页,按照5V单电源供电,Vref 2.5V的接法,1.5=< Vout =< 3.5

得出 -1=<(+IN)-(-IN)=< +1

这个明显跟手册中的输出差分范围不符。手册第四页显示差分范围在1.5~3.5V之间;是否哪里不对?

另外当我们应用Vref不等于2.5V时,比如1.25V,仍用单独的5V供电,+IN电压范围是否可以如下计算:

假设运放正负极输出端口电压为X,

(+IN – X)/380 = (X- Vref)/19

X= (+IN + 20*Vref)/21

1.5=< X =< 3.5

31.5-20*Vref=< +IN =< 73.5- 20*Vref

假设Vref= 1.25V

6.5=< +IN =< 48.5

1.5=< Vout =< 3.5

1.5=<(+IN)-(-IN)+ Vref=< 3.5

1.5 - Vref =<(+IN)-(-IN)=< 3.5 – Vref

即可定出-IN的范围。

请帮忙确认下上述计算实际输入范围的过程是否正确,谢谢!