关于max4080电流采样芯片内部电路问题

本帖最后由 gk320830 于 2015-3-5 13:45 编辑


文档中说,"电流从源流经采样电阻到负载,产生Vsense,因为反相输入端阻抗很高,流过Rg2的电流可以忽略(可忽略输入偏置电流),因此反相输入电压就得出了.又由于运放的开环增益迫使其同相输入端电压与反相输入相同(没懂,为什么是开环增益造成的?)因此RG1上的压降等于Vsense,内部电流镜使Irg1电流放大到IA2."描述就直接跳到这步了,我一直没明白运放输出端接一个三极管的基极,集电极有一个正反馈到运放的同相输入端是什么电路,因为学习的模电以负反馈为主,可能是我没有融会贯通吧,一直没有明白这是什么意思?难道是把电压变化转换成电流变化?作了第一级电流放大然后输出到电流镜,进行温度补偿?
具体请大家帮忙分析下吧,谢谢!
附上pdf中的描述,供大家参考!
current from the source flows through RSENSE to the load, creating a sense voltage, VSENSE. Since the internal-sense amplifier’s inverting input has high impedance, negligible current flows through RG2 (neglecting the input bias current). Therefore, the sense amplifier’s inverting input voltage equals VSOURCE - (ILOAD)(RSENSE). The amplifier’s open-loop gain forces its noninverting input to the same voltage as the inverting input. Therefore, the drop across RG1 equals VSENSE. The internal current mirror multiplies IRG1 by a current gain factor, β, to give IA2 = β  IRG1. Amplifier A2 is used to convert the output current to a voltage and then sent through amplifier A3.