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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)
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