http://www.amobbs.com/forum.php?mod=viewthread&tid=3661385&page=3#pid3803226
对R13 R16 R17做星-三角变换：RAB = R13 + R17 + R13 * R17 / R16 = 4.63KRBC = R16 + R17 + R16 * R17 / R13 = 14.03KRCA = R13 + R16 + R13 * R16 / R17 = 46.3K变换后RAB固定接于5V和地之间，可忽略。Vpwm变化范围0-5V，根据虚短原理，C点电压变化范围也是0-5V。现在问题变成：A = 5V，B = 0V，C = 0-5V，求Vout范围。先求各臂电流关系：Ir15 = (Vc - Vout) / R15Irca = (5 - Vc) / RCAIrbc = Vc / RBC根据基尔霍夫电流定律：Ir15 + Irca + Irbc = 0(Vc - Vout) / R15 + (5 - Vc) / RCA + Vc / RBC = 0最后得出：Vout = Vpwm + [(5 - Vpwm) / RCA + Vpwm / RBC] * R15代入Vpwm = 0-5V计算得出Vout范围约-1.08-8.56V。
这样考虑更简单：1. 5V,R13,R17可等效为电动势和内阻：
电动势 : 5V*(1/(1+3.3))=1.162790V
内阻 : 1*3.3/(1+3.3)=0.76744K
2. 设：U5A输入阻抗无穷，则流过R15和上述内阻的电流相同。(y-x)/10 = (x-1.162790)/(0.76744+10) = 0.0928726 x - 0.107991化简：y = 1.928726 x - 1.07991 验算：x = 5, y = 8.56372
Y = Vout
X = Vpwm
转载于:https://www.cnblogs.com/shangdawei/p/3286381.html
相关资源：电阻星型联接与三角形联接的等效变换