BL(i) product vs impedance- sanity check, please...

[GordonW]

Having a hard time remembering the formulas this morning. Just wanted to make sure I was thinking correctly.

For a given motor strength (BL product), IIRC, HALVING the impedance of the voice coil should DOUBLE the effective BLi product due to doubling of the current for the same voltage input, right?

That would mean... if I converted a woofer from 8 ohms to 4 ohms (reconing), and maintained the same coil geometry, I could get twice the effective magnetic motor strength (and a proportional reduction in Qes), right?

Yes, I know that due to thicker cross-sectional area on the 4 ohm wire, I will have a slight reduction of the L part of the product (there's enough gap guage width to accommodate the thicker wire, fortunately)... but I should still be able to get a SIGNIFICANTLY higher motor strength (lower Qes) by using a 4 ohm coil... shouldn't I?

Yeah, I know, this is basic "duh" theory... but please bear with me here... trying to get a woofer to work in a smaller enclosure...

Thanks in advance...

Regards,
Gordon.

[Svante]

If you go from 8 to 4 ohms, and the amount of copper in the gap is to be the same, the cross-sectional area of the wire would be sqrt(2) times the 8 ohm version, and the length L of wire in the gap would be 1/sqrt(2) times the length of the 8 ohm version. Since the length in the gap is reduced by 1/sqrt(2), so is BL.

This is why the ratio (BL)²/Re or BL/sqrt(Re) sometimes occur in datasheets. This ratio remains nearly constant between versions with different impedances and can be used as a measure of "motor strength".

[GordonW]

Actually, if you maintained the same voice coil PACK LENGTH (winding depth), you would need to use LESS turns of wire, in a 4 ohm driver. However, to maintain 4 ohms, you'd also have to use a SMALLER gauge wire, for that shorter length.

So, the result is that the wire would be somewhat smaller than double the cross sectional area... which means the "l" factor in the Bl would go down LESS than it would, assuming twice the cross sectional area in each strand.

So, after thinking about it in this way, it is most likely that I'd get a SMALL reduction in Qes, going from an 8 ohm to a 4 ohm coil of the same dimensions... but NOT a very big one...

Regards,
Gordon.

[Svante]

Well...

A 4 ohm driver where the coil occupies the same physical space as an 8 ohm driver, would have 0.7 times the length, L, (1/sqrt(2)) and 1.4 times (sqrt(2)) the cross sectional area, S, of the wire. Thicker wire, that is. Since resistance is proportional to L/S, the resistance will be 0.7/1.4=0.5 times 8 ohm.

Qes is largely determined by the ratio electro-mechanical resistance, which is Rme = (Bl)²/Re . Så if Bl changes by a factor of 0.7 and Re is halved, the Qes value will remain the same.

There are little differences in that the 8 ohm version will need slightly more insulation between the windings, since the wire is thinner, but that difference is small.