Example 5 P47 Exercise 2.3.3 ()设计同相放大器 由虚短与虚断的概念 i=VN/RI=VP/RI=Vi/RI y=0.8V时，i1<100uA→0.8V/R,<100uA→R1>8K2 由：A,=1+R2/R1=10→R2=11.5R1≥92K2 (2)设计反相放大器：y:=-1V R =(-w)R=,/R1<20μAS →R1>50K2 由：A,=-R2/R1=-8→R2=8R1≥400K2 ◆Example 5 P47 Exercise 2.3.3 vi vo - + R2 R1 vN vP i2 i1 (1) 设计同相放大器 vi vo - + R2 R1 vN vP i1 iN iP i2 由虚短与虚断的概念 i1 = vN /R1 = vP /R1 = vi /R1 vi =0.8V时, i1 <100μA 0.8V /R1 <100μA  R1 >8KΩ 由：Av  R1 / R12   R2  . 1  92K5R11 10  (2) 设计反相放大器: vi = -1V i1 = (vi -vN )/R1 = vi /R1 <20μA  R1 >50 KΩ 由：A  R / R12v  R2 1  400K8R -8 