filterinductormutual-inductance
I want to design an higher order LC filter for filtering out a power signal. In the theoretical design, the inductors are supposed to be on separate cores. But in practice, it will be very costly and space consuming. Is it possible to wind the inductors on the same core (e.g.; a toroidal core)? If it is possible, what must be the proper dot-convention between these inductors?
simulate this circuit – Schematic created using CircuitLab
Your assumption that the core flux cancels out to zero is not valid because you don't have perfect coupling between L1 and L2. If you had, then the inductance for the series-additive connected inductor should measure 120 (not 90) uH, because inductance is proportional to the square of the number of turns, and you're essentially doubling the number of turns. Likewise, you would measure no inductance for the series-opposing connection. Note that even if there was no coupling you would still measure 60 uH for the series-additive pair, so it seems there is a fair amount of leakage flux.
In the opposing flux configuration, the effect on the circuit is similar to replacing the coupled inductor with two separate inductors having much lower inductance values. As a result, the current ripple is higher, and the associated core losses are higher.
You must consider the load, which the loudspeaker creates for your LC filter. It changes the response of the filter. Below are results of the simulation performed with LTspice for different values of L and C (keeping almost constant the LC product):
As you can see, from the tested values 330nF, 33µH gives the best results (I have added this set of values basing on results of simulation).
\begin{array}{|c|c|c|} L & C & \text{color} \\ \hline 220\text{μH} & 47\text{nF} & \text{violet}\\ 33\text{μH} & 330\text{nF} & \text{turqoise}\\ 22\text{μH} & 470\text{nF} & \text{red}\\ 2.2\text{μH} & 4.7\text{μF} & \text{blue}\\ 220\text{nH} & 47\text{μF} & \text{green}\\ \end{array}
From the point of view of the loudspeaker (R1) we have a parallel RLC circuit. Its Q factor is defined by the formula: $$Q = R_1\sqrt{\dfrac{C_1}{L_1}}$$
For C1=330nF, L1=33µH the Q factor is equal to 0.8
The optimal solution will be the one, for each Q is equal to 1.
Best Answer
No, this will not give you what you want because all the inductors will be coupled at nearly 98%. Any voltage across L1 will be mirrored across L2, L3 etc..
The filter topology shown is a low pass type so I'm assuming the power signal you wish to remove is higher than what you wish to pass. The higher frequency power signal will, due to the first filter capacitor being a low impedance to that frequency, appear largely across L1 and either inverted or of the same phase across L2 and L3. No matter what you do, with near 100% coupling and the capacitors being low impedance to the power signal you wish to remove, the final inductor will produce virtually the same power signal at the output as seen at the input minus a few percent due to winding losses etc..