• [模拟][模电][面试][运放]仪表放大器


    前言

    昨天访问量还是29万1千多,今天就变成了28万3千,CSDN又在倒退了!!!


    目录

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    框图

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    • 虚短:放大器的正负输入假设短路,两个端口电位相同
    • 虚断:放大器的正负输入假设断路,两个端口没有电流输入,输出也是断路

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    所以上面放大器两端电压为 u A u_A uA,下面放大器两端电压为 u B u_B uB

    假设两个放大器输出分别是 V 1 , V 2 V_1,V_2 V1,V2

    右边放大器两端电压假设为 V ′ V' V

    V ′ = R 5 R 4 + R 5 V 2 V'=\frac{R_5}{R_4+R_5}V_2 V=R4+R5R5V2

    R 3 R_3 R3电流为 I = V ′ − V 1 R 3 = 1 R 3 ( R 5 R 4 + R 5 V 2 − V 1 ) I = \dfrac{V' - V_1}{R_3} = \dfrac{1}{R_3}( \dfrac{R_5}{R_4+R_5}V_2 - V_1) I=R3VV1=R31(R4+R5R5V2V1) ,其中假设电流往左流(因为右边的放大器是反向比例放大器)

    输出电压 u 0 = ( R 3 + R f ) 1 R 3 ( R 5 R 4 + R 5 V 2 − V 1 ) = ( 1 + R f R 3 ) ( R 5 R 4 + R 5 V 2 − V 1 ) 1 ◯ u_0 = (R_3+R_f)\dfrac{1}{R_3}( \dfrac{R_5}{R_4+R_5} V_2 - V_1) = (1 + \dfrac{R_f}{R_3})( \dfrac{R_5}{R_4 + R_5}V_2 - V_1 ) \qquad \textcircled{1} u0=(R3+Rf)R31(R4+R5R5V2V1)=(1+R3Rf)(R4+R5R5V2V1)1

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    该电流从 u 0 u_0 u0直接逆时针流到 R 5 R_5 R5,再流到地

    I = V 1 − u A R 1 = u B − V 2 R 2 I = \frac{ V_1 - u_A }{R_1} = \frac{ u_B - V_2 }{R_2} I=R1V1uA=R2uBV2

    推导得到两个式子

    { R 1 R 5 R 4 + R 5 V 2 + u A R 3 = V 1 ( R 1 + R 3 ) ( R 2 R 5 R 4 + R 5 + R 3 ) V 2 = u B R 3 + R 2 V 1

    {R1R5R4+R5V2+uAR3=V1(R1+R3)(R2R5R4+R5+R3)V2=uBR3+R2V1" role="presentation" style="position: relative;">{R1R5R4+R5V2+uAR3=V1(R1+R3)(R2R5R4+R5+R3)V2=uBR3+R2V1
    R4+R5R1R5V2+uAR3=V1(R1+R3)(R4+R5R2R5+R3)V2=uBR3+R2V1

    推导得到 V 1 , V 2 V_1,V_2 V1,V2

    { V 1 = u A ( R 2 R 5 + R 3 ( R 4 + R 5 ) ) R 3 + u B ⋅ R 1 R 3 R 5 ( R 2 R 5 + R 3 ( R 4 + R 5 ) ) ( R 1 + R 3 ) − R 1 R 2 R 5 V 2 = − u A ⋅ R 2 R 3 ( R 4 + R 5 ) + u B ⋅ R 3 ( R 1 + R 3 ) ( R 4 + R 5 ) ( R 2 R 5 + R 3 ( R 4 + R 5 ) ) ( R 1 + R 3 ) + R 1 R 2 R 5

    {V1=uA(R2R5+R3(R4+R5))R3+uBR1R3R5(R2R5+R3(R4+R5))(R1+R3)R1R2R5V2=uAR2R3(R4+R5)+uBR3(R1+R3)(R4+R5)(R2R5+R3(R4+R5))(R1+R3)+R1R2R5" role="presentation" style="position: relative;">{V1=uA(R2R5+R3(R4+R5))R3+uBR1R3R5(R2R5+R3(R4+R5))(R1+R3)R1R2R5V2=uAR2R3(R4+R5)+uBR3(R1+R3)(R4+R5)(R2R5+R3(R4+R5))(R1+R3)+R1R2R5
    V1=(R2R5+R3(R4+R5))(R1+R3)R1R2R5uA(R2R5+R3(R4+R5))R3+uBR1R3R5V2=(R2R5+R3(R4+R5))(R1+R3)+R1R2R5uAR2R3(R4+R5)+uBR3(R1+R3)(R4+R5)

    V 1 , V 2 V_1,V_2 V1,V2代入 1 ◯ \textcircled{1} 1

    u 0 = − ( 1 + R f R 3 ) ⋅ u A [ − ( R 1 + R 3 ) T 2 + R 2 R 5 ( R 1 + R 3 ( R 1 + R 3 ) ) T + R 1 R 3 R 2 2 R 5 2 ] + u B [ T ⋅ R 3 2 R 5 ( R 1 + R 3 ) + R 3 R 5 2 R 1 R 2 ( 2 R 1 + R 3 ) ] T 2 ( R 1 + R 3 ) 2 − ( R 1 R 2 R 5 ) 2 u_0 = - (1 + \frac{R_f}{R_3}) \cdot \\ \frac{ u_A[ - (R_1 + R_3)T^2 + R_2R_5( R_1 + R_3(R_1 +R_3))T + R_1 R_3 R_2^2 R_5^2 ] + u_B[ T \cdot R_3^2 R_5 (R_1 + R_3) + R_3R_5^2 R_1R_2(2R_1 + R_3) ] }{ T^2 (R_1+R_3)^2 - (R_1R_2R_5)^2} u0=(1+R3Rf)T2(R1+R3)2(R1R2R5)2uA[(R1+R3)T2+R2R5(R1+R3(R1+R3))T+R1R3R22R52]+uB[TR32R5(R1+R3)+R3R52R1R2(2R1+R3)]
    T = R 2 R 5 + R 3 ( R 4 + R 5 ) T = R_2R_5 + R_3(R_4 + R_5) T=R2R5+R3(R4+R5)

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  • 原文地址:https://blog.csdn.net/weixin_41374099/article/details/126599309