电路分析 ---- T型反馈反相比例器
1 T型反馈反相比例器
2 分析过程
- 根据虚短可知 u + = u − = 0 u_{+}=u_{-}=0 u+=u−=0
- 根据虚断可知流入运算放大器负输入端的电流为0
- 故可得 i 1 = u I − u − R 1 = u I − 0 R 1 = u I R 1 i_{1}=\cfrac{u_{I}-u_{-}}{R_{1}}=\cfrac{u_{I}-0}{R_{1}}=\cfrac{u_{I}}{R_{1}} i1=R1uI−u−=R1uI−0=R1uI
- 同时又有 i 1 = u − − u A R 2 = 0 − u A R 2 = − u A R 2 i_{1}=\cfrac{u_{-}-u_{A}}{R_{2}}=\cfrac{0-u_{A}}{R_{2}}=-\cfrac{u_{A}}{R_{2}} i1=R2u−−uA=R20−uA=−R2uA
- 根据上面两个式子可得 u A = − R 2 R 1 ⋅ u I u_{A}=-\cfrac{R_{2}}{R_{1}}\cdot u_{I} uA=−R1R2⋅uI
- 此外，还有 i 2 = 0 − u A R S E T = − u A R S E T = R 2 R 1 ⋅ R S E T u I i_{2}=\cfrac{0-u_{A}}{R_{SET}}=-\cfrac{u_{A}}{R_{SET}}=\cfrac{R_{2}}{R_{1}\cdot R_{SET}}u_{I} i2=RSET0−uA=−RSETuA=R1⋅RSETR2uI
- i 3 = u A − u o R 3 i_{3}=\cfrac{u_{A}-u_{o}}{R_{3}} i3=R3uA−uo
- 满足关系： i 1 + i 2 = i 3 i_{1}+i_{2}=i_{3} i1+i2=i3，即 ( − u A R 2 ) + ( − u A R S E T ) = u A − u o R 3 (-\cfrac{u_{A}}{R_{2}})+(-\cfrac{u_{A}}{R_{SET}})=\cfrac{u_{A}-u_{o}}{R_{3}} (−R2uA)+(−RSETuA)=R3uA−uo
- 变形得到： u o = ( R 3 R 2 + 1 + R 3 R S E T ) u A u_{o}=(\cfrac{R_{3}}{R_{2}}+1+\cfrac{R_{3}}{R_{SET}})u_{A} uo=(R2R3+1+RSETR3)uA
- 再代入 u A = − R 2 R 1 ⋅ u I u_{A}=-\cfrac{R_{2}}{R_{1}}\cdot u_{I} uA=−R1R2⋅uI，得到 u o = − 1 R 1 ( R 2 + R 3 + R 2 R 3 R S E T ) u I u_{o}=-\cfrac{1}{R_{1}}(R_{2}+R_{3}+\cfrac{R_{2}R_{3}}{R_{SET}})u_{I} uo=−R11(R2+R3+RSETR2R3)uI
该电路可以实现反相放大的高增益和高输入阻抗的兼顾。