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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
则 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=R3V′−V1=R31(R4+R5R5V2−V1) ,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+R5R5V2−V1)=(1+R3Rf)(R4+R5R5V2−V1)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=R1V1−uA=R2uB−V2
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
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
把 V 1 , V 2 V_1,V_2 V1,V2代入 1 ◯ \textcircled{1} 1◯式
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[T⋅R32R5(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)