LMV852说明书中的应用，关于电路的输出电压，不知道怎么推导出来的，总算不对，求教！

Pressure  Sensor  Example
The  configuration  shown  in Figure55 is simple,and is very useful for the read out of pressure sensors.With two
op amps in this application,the dual LMV852 fits verywell.
The op amp configured as a buffer and connected at the negative output of the pressure sensor prevents the
loading of the bridge by resistor R2.The buffer also prevents the resistors of the sensor from affecting the gain of
the following gain stage.Given the differential output voltage Vs of the pressure sensor,the output signal of this
op amp configuration,VOUT, equals:

To align the pressure range with the full range of an ADC,the power supply voltage and the span of the pressure
sensor are needed.For this example a power supply of 5V is used and the span of the sensor is 100mV.
When a 100? resistor is used for R2,and a 2.4k? resistor is used for R1,the maximum voltage at the output is
4.95V and the minimum voltage is 0.05V.This signal is covering almost the full input range of the ADC.Further
processing can take place in the microprocessor following the ADC.

我的计算思路，接跟随器的电压为Vdd/2,结果算出
Vout=Vdd/2-Vs(1+R1/R2)
跟示例的结果有不少出入，请老师帮忙，谢谢！

VS只是差模信号，您把电桥输出信号中的共模信号，然后用叠加定理推导。

例如

第一个通道的输出信号：(VDD/2 - VS/2)

第二个通道的同向输入端：(VDD/2 + VS/2)

Vout = (VDD/2 + VS/2)*(1+R1/R2) + (VDD/2 - VS/2)*(-R1/R2) = VDD/2 + VS(1+2R1/R2)

您这一说我就明白了，其实问题的关键是您把桥左右两边的电压分别认定为V1=(VDD/2+Vs/2)，V2=(VDD/2-Vs/2)了！

问题1：我不知道为什么要这么认定（或如此假设）？或许测压力的时候很方便？

问题2：我用这个图中的原理测温度时一般把桥右侧的分压固定为V2=VDD/2，桥的左下电阻为测温电阻，左侧分压视为V1=VDD/2+Vs，
 左右侧的压差为Vs，根据输出电压反过来求Vs，从而求得测温电阻阻值大小。我这种方式是否可行呢？实际操作中右侧电压不一定是
VDD/2，但是一个固定值。 

也可以认定左边为vdd/2，右边为VDD/2  -  vs,这样的结果是Vout=vdd/2  +  r1/r2  *  vs

我感觉按照你的问题而中所说的也是可以的，总之就是求出测温电阻或者应变片电阻上的电压变化