MTPA数理推导
MTPA数理推导
转矩方程与电流方程
Te=f(Isd,Isq)Is=L(Isd,Isq)\begin{array}{l} {T_e} = f({I_{sd}},{I_{sq}})\\ {I_s} = L({I_{sd}},{I_{sq}}) \end{array} Te=f(Isd,Isq)Is=L(Isd,Isq)
求解方案有两种, 直接代入, 然后求一元方程导数, 求得驻点, 然后取得最值; 第二种采用拉格郎日乘数法.
本文推导第二种:拉格朗日乘数法。
其实就是求多元函数, 在D封闭区间的最值问题, 而MTPA要求的, 就是二元函数.
可以将Is=L(Isd,Isq){I_s} = L({I_{sd}},{I_{sq}})Is=L(Isd,Isq)看作封闭区间, 亦可以将Te=f(Isd,Isq){T_e} = f({I_{sd}},{I_{sq}})Te=f(Isd,Isq)看作封闭区间.
亦即, 给定Te,能够求出最小的Is; 或者给定Is,能够求出最大的Te
再根本, 我们要求的是Id与Iq的对应关系.
Te=32p(φfIsq+(Ld−Lq)IsdIsq)Is2=Isd2+Isq2\begin{array}{l} {T_e} = \frac{3}{2}p({\varphi _f}{I_{sq}} + ({L_d} - {L_q}){I_{sd}}{I_{sq}})\\ {I_s}^2 = {I_{sd}}^2 + {I_{sq}}^2 \end{array} Te=23p(φfIsq+(Ld−Lq)IsdIsq)Is2=Isd2+Isq2
对上式, 分别对Isd、Isq求偏导
∂TeIsd=32p(Ld−Lq)Isq∂TeIsq=32p(φf+(Ld−Lq)Isd)∂IsIsd=2IsdIsd2+Isq2∂IsIsq=2IsqIsd2+Isq2\begin{array}{l} \frac{{\partial {T_e}}}{{{I_{sd}}}} = \frac{3}{2}p({L_d} - {L_q}){I_{sq}}\\ \frac{{\partial {T_e}}}{{{I_{s{\rm{q}}}}}} = \frac{3}{2}p({\varphi _f} + ({L_d} - {L_q}){I_{sd}})\\ \frac{{\partial {I_s}}}{{{I_{sd}}}} = \frac{{2{I_{sd}}}}{{\sqrt {{I_{sd}}^2 + {I_{sq}}^2} }}\\ \frac{{\partial {I_s}}}{{{I_{s{\rm{q}}}}}} = \frac{{2{I_{sq}}}}{{\sqrt {{I_{sd}}^2 + {I_{sq}}^2} }} \end{array} Isd∂Te=23p(Ld−Lq)IsqIsq∂Te=23p(φf+(Ld−Lq)Isd)Isd∂Is=Isd2+Isq22IsdIsq∂Is=Isd2+Isq22Isq
列出拉格朗日乘数法:
{32p(Ld−Lq)Isq+λ2IsdIsd2+Isq2=032p(φf+(Ld−Lq)Isd)+λ2IsqIsd2+Isq2=0Isd2+Isq2−Is2=0\left\{ \begin{array}{l} \frac{3}{2}p({L_d} - {L_q}){I_{sq}} + \lambda \frac{{2{I_{sd}}}}{{\sqrt {{I_{sd}}^2 + {I_{sq}}^2} }} = 0\\ \frac{3}{2}p({\varphi _f} + ({L_d} - {L_q}){I_{sd}}) + \lambda \frac{{2{I_{sq}}}}{{\sqrt {{I_{sd}}^2 + {I_{sq}}^2} }} = 0\\ {I_{sd}}^2 + {I_{sq}}^2 - {I_s}^2 = 0 \end{array} \right. ⎩⎪⎪⎨⎪⎪⎧23p(Ld−Lq)Isq+λIsd2+Isq22Isd=023p(φf+(Ld−Lq)Isd)+λIsd2+Isq22Isq=0Isd2+Isq2−Is2=0
可以看出, 对子式1、子式2分别两端除以λ\lambdaλ, 则从以Is为D封闭区间转变为以Te为D的封闭区间。
对上式求解
{32p(Ld−Lq)Isq+λ2IsdIsd2+Isq2=032p(φf+(Ld−Lq)Isd)+λ2IsqIsd2+Isq2=0Isd2+Isq2−Is2=0{32p(Ld−Lq)IsqIsq+λ2IsdIsqIsd2+Isq2=032p(φf+(Ld−Lq)Isd)Isd+λ2IsdIsqIsd2+Isq2=0Isd2+Isq2−Is2=0((Ld−Lq)Isq2−(φfIsd+(Ld−Lq)Isd2)=0(Ld−Lq)(Isq2−Isd2)=φfIsd(Isq2−Isd2)Isd=Isq=φf1(Ld−Lq)Isd−Isd20=φf1(Ld−Lq)Isd−Isd2−Isq2\begin{array}{l} \left\{ \begin{array}{l} \frac{3}{2}p({L_d} - {L_q}){I_{sq}} + \lambda \frac{{2{I_{sd}}}}{{\sqrt {{I_{sd}}^2 + {I_{sq}}^2} }} = 0\\ \frac{3}{2}p({\varphi _f} + ({L_d} - {L_q}){I_{sd}}) + \lambda \frac{{2{I_{sq}}}}{{\sqrt {{I_{sd}}^2 + {I_{sq}}^2} }} = 0\\ {I_{sd}}^2 + {I_{sq}}^2 - {I_s}^2 = 0 \end{array} \right.\\ \left\{ \begin{array}{l} \frac{3}{2}p({L_d} - {L_q}){I_{sq}}{I_{sq}} + \lambda \frac{{2{I_{sd}}{I_{sq}}}}{{\sqrt {{I_{sd}}^2 + {I_{sq}}^2} }} = 0\\ \frac{3}{2}p({\varphi _f} + ({L_d} - {L_q}){I_{sd}}){I_{sd}} + \lambda \frac{{2{I_{sd}}{I_{sq}}}}{{\sqrt {{I_{sd}}^2 + {I_{sq}}^2} }} = 0\\ {I_{sd}}^2 + {I_{sq}}^2 - {I_s}^2 = 0 \end{array} \right.\\ (({L_d} - {L_q}){I_{sq}}^2 - ({\varphi _f}{I_{sd}} + ({L_d} - {L_q}){I_{sd}}^2) = 0\\ ({L_d} - {L_q})({I_{sq}}^2 - {I_{sd}}^2) = {\varphi _f}{I_{sd}}\\ \frac{{({I_{sq}}^2 - {I_{sd}}^2)}}{{{I_{sd}}}} = {I_{sq}} = \sqrt {{\varphi _f}\frac{1}{{({L_d} - {L_q})}}{I_{sd}} - {I_{sd}}^2} \\ 0 = {\varphi _f}\frac{1}{{({L_d} - {L_q})}}{I_{sd}} - {I_{sd}}^2 - {I_{sq}}^2 \end{array} ⎩⎪⎪⎨⎪⎪⎧23p(Ld−Lq)Isq+λIsd2+Isq22Isd=023p(φf+(Ld−Lq)Isd)+λIsd2+Isq22Isq=0Isd2+Isq2−Is2=0⎩⎪⎪⎨⎪⎪⎧23p(Ld−Lq)IsqIsq+λIsd2+Isq22IsdIsq=023p(φf+(Ld−Lq)Isd)Isd+λIsd2+Isq22IsdIsq=0Isd2+Isq2−Is2=0((Ld−Lq)Isq2−(φfIsd+(Ld−Lq)Isd2)=0(Ld−Lq)(Isq2−Isd2)=φfIsdIsd(Isq2−Isd2)=Isq=φf(Ld−Lq)1Isd−Isd20=φf(Ld−Lq)1Isd−Isd2−Isq2
根据上式, 则可以求出:
Isd=−φf2(Ld−Lq)+(φf2(Ld−Lq))2−Isq2{I_{sd}} = - \frac{{{\varphi _f}}}{{2({L_d} - {L_q})}} + \sqrt {{{(\frac{{{\varphi _f}}}{{2({L_d} - {L_q})}})}^2} - {I_{sq}}^2} Isd=−2(Ld−Lq)φf+(2(Ld−Lq)φf)2−Isq2
可以从式清晰地看出, 上式成立,需要满足以下条件
{Ld≠Lq(φf2(Ld−Lq))2>Isq2\left\{ \begin{array}{l} {L_d} \ne {L_q}\\ {(\frac{{{\varphi _f}}}{{2({L_d} - {L_q})}})^2}{\rm{ > }}{I_{sq}}^2 \end{array} \right. {Ld=Lq(2(Ld−Lq)φf)2>Isq2
备注:
- 需要明确指出的是, Ld和Lq 都极可能随着电机运行工况而变化, 所以, 工程中的直接使用的可用性不高。
- 更常见的是使用查表法