[运算放大器系列]]二、电压转4 - 20MA电流电路分析
1.电路原理图
偶然在网上看到一个4 - 20MA转换电路原理图如下:
2. 原理分析
R
L
R_L
RL为负载,分析电流流向如上图箭头所示可以得到
假设Rloop上的压降为
V
l
V_l
Vl则:
①
V
i
−
V
+
R
1
=
V
+
−
(
V
o
−
V
l
)
R
2
\frac {V_i - V_+} {R_1} = \frac{V_+- (V_o - V_l)}{R2}
R1Vi−V+=R2V+−(Vo−Vl)
(
V
i
−
V
+
)
⋅
R
2
=
(
V
+
−
V
o
+
V
l
)
⋅
R
1
(V_i - V_+)\cdot R_2 = (V_+ - V_o + V_l)\cdot R_1
(Vi−V+)⋅R2=(V+−Vo+Vl)⋅R1
V
i
⋅
R
2
−
V
+
⋅
R
2
=
V
+
⋅
R
1
−
V
o
⋅
R
1
+
V
l
⋅
R
1
V_i \cdot R_2 - V_+ \cdot R_2 = V_+\cdot R_1 - V_o\cdot R_1 + V_l\cdot R_1
Vi⋅R2−V+⋅R2=V+⋅R1−Vo⋅R1+Vl⋅R1
V
i
⋅
R
2
+
V
o
⋅
R
1
−
V
l
⋅
R
1
=
V
+
⋅
R
1
+
V
+
⋅
R
2
V_i \cdot R_2 + V_o\cdot R_1 - V_l\cdot R_1= V_+\cdot R_1 + V_+ \cdot R_2
Vi⋅R2+Vo⋅R1−Vl⋅R1=V+⋅R1+V+⋅R2
V
i
⋅
R
2
+
V
o
⋅
R
1
−
V
l
⋅
R
1
=
V
+
⋅
(
R
1
+
R
2
)
V_i \cdot R_2 + V_o\cdot R_1 - V_l\cdot R_1= V_+\cdot (R_1 + R_2)
Vi⋅R2+Vo⋅R1−Vl⋅R1=V+⋅(R1+R2)
V
o
V_o
Vo到GND的电流关系为:
②
V
o
−
V
−
R
3
=
V
−
R
4
\frac {V_o - V_-}{R_3} = \frac {V_-}{R_4}
R3Vo−V−=R4V−
(
V
o
−
V
−
)
⋅
R
4
=
V
−
⋅
R
3
(V_o - V_-)\cdot R_4= V_-\cdot R_3
(Vo−V−)⋅R4=V−⋅R3
V
o
⋅
R
4
=
V
−
⋅
R
3
+
V
−
⋅
R
4
V_o \cdot R_4= V_-\cdot R_3 + V_-\cdot R_4
Vo⋅R4=V−⋅R3+V−⋅R4
V
o
⋅
R
4
=
V
−
⋅
(
R
3
+
R
4
)
V_o \cdot R_4= V_-\cdot (R_3 + R_4)
Vo⋅R4=V−⋅(R3+R4)
由虚短可知
V
−
=
V
+
V_-= V_+
V−=V+,令
R
3
+
R
4
=
R
1
+
R
2
R_3 + R_4=R_1 + R_2
R3+R4=R1+R2得到:
V
o
⋅
R
4
=
V
i
⋅
R
2
+
V
o
⋅
R
1
−
V
l
⋅
R
1
V_o \cdot R_4=V_i \cdot R_2 + V_o\cdot R_1 - V_l\cdot R_1
Vo⋅R4=Vi⋅R2+Vo⋅R1−Vl⋅R1
令
R
4
=
R
1
R_4=R_1
R4=R1得到:
V
l
⋅
R
1
=
V
i
⋅
R
2
V_l\cdot R_1=V_i \cdot R_2
Vl⋅R1=Vi⋅R2
V
l
=
V
i
⋅
R
2
R
1
V_l =\frac{V_i \cdot R_2}{R_1}
Vl=R1Vi⋅R2
V
l
=
I
l
⋅
R
l
o
o
p
V_l=I_l\cdot Rloop
Vl=Il⋅Rloop
得到:
③
I
l
=
V
i
⋅
R
2
R
1
⋅
R
l
o
o
p
−
−
−
约
束
条
件
(
R
3
+
R
4
=
R
1
+
R
2
,
R
4
=
R
1
)
③I_l=\frac{V_i \cdot R_2}{R_1\cdot Rloop} ---约束条件\Bigg(R_3 + R_4=R_1 + R_2,R_4=R_1\Bigg)
③Il=R1⋅RloopVi⋅R2−−−约束条件(R3+R4=R1+R2,R4=R1)
流过负载
R
L
R_L
RL的电流为流过Rloop和
R
2
R_2
R2电流之和则:
I
L
=
V
l
R
l
o
o
p
+
V
i
−
(
V
o
−
V
l
)
R
1
+
R
2
I_L=\frac{V_l}{Rloop} + \frac{V_i-(V_o-V_l)}{R_1+R2}
IL=RloopVl+R1+R2Vi−(Vo−Vl)
I
L
=
V
l
R
l
o
o
p
+
V
i
−
V
o
+
V
l
R
1
+
R
2
=
4
−
20
M
A
I_L=\frac{V_l}{Rloop} + \frac{V_i-V_o+V_l}{R_1+R2}=4-20MA
IL=RloopVl+R1+R2Vi−Vo+Vl=4−20MA
Rloop远小于
R
1
+
R
2
R_1+R2
R1+R2之和则
R
1
+
R
2
R_1+R2
R1+R2上电流可忽略不计,最后得到:
④
I
L
=
V
i
⋅
R
2
R
1
⋅
R
l
o
o
p
−
−
−
约
束
条
件
(
R
3
+
R
4
=
R
1
+
R
2
,
R
4
=
R
1
,
R
1
+
R
2
>
>
R
l
o
o
p
)
④I_L=\frac{V_i \cdot R_2}{R_1\cdot Rloop} ---约束条件\Bigg(R_3 + R_4=R_1 + R_2,R_4=R_1,R_1+R2 >> Rloop\Bigg )
④IL=R1⋅RloopVi⋅R2−−−约束条件(R3+R4=R1+R2,R4=R1,R1+R2>>Rloop)