Wednesday, August 9, 2017

IPM Low-Speed Optimization

I had mentioned in a previous post that IPM's require both d and q-axis currents for optimal performance. Thanks to the motor equations, it is easy to quantify this split.

Recall that a sinusoidally-varying motor is modeled by:$$
\tau=\frac{3}{2}n_p(\lambda I_q+(L_d-L_q)I_d I_q)\\
V_d=R_s I_d-\omega L_q I_q\\
V_q=R_s I_q+\omega L_d I_d+\omega\lambda\\
$$ Suppose we have unlimited back EMF, and we wish to optimize torque per amp. There are two ways to look at this. Firstly, we could $$
\mbox{minimize }
I_d^2+I_q^2\mbox{ subject to}\\
\lambda I_q+(L_d-L_q)I_d I_q=\tau_0
$$ Or, we could $$
\mbox{maximize }
\lambda I_q+(L_d-L_q)I_d I_q\mbox{ subject to}\\
$$ As it turns out, the second current-first approach results in much easier math (we only need to solve a quadratic, not a quartic) at the expense of being somewhat less intuitive (it is unclear what current corresponds to what torque).

There are several ways to solve the second problem; we use Lagrange multipliers here. The Lagrangian is $$L(I_d,I_q,u)=\lambda I_q+(L_d-L_q)I_d I_q-u(I_d^2+I_q^2-I_0^2)$$ where \(u\), not \(\lambda\), is the multiplier.
The system of partial derivatives is $$
\frac{\partial L}{\partial I_d}=(L_d-L_q)I_q-2I_d u=0\\
\frac{\partial L}{\partial I_q}=(L_d-L_q)I_d-2I_q u+\lambda=0\\
\frac{\partial L}{\partial u}=I_0^2-I_d^2-I_q^2=0
$$ This system is easily solved by a computer algebra system or by multiplying the first equation by \(I_q\) and the second by \(I_d\), giving $$
$$ where we have picked the signs knowing that \(I_d\) is negative and \(I_q\) is positive.

Armed with this information we can make some plots. Plugging in the HSG data \(L_d=0.0006\), \(L_q=0.0015\). and \(\lambda=0.053\) (units: Henries, Volt-seconds), we have the following plot:

As expected, \(I_d\) is about the same magnitude as \(I_q\) at high currents.

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