Alternative to tapped inductor in 3110a or 3115a?

[robfive]

Greetings,

I have a multitude of 2225H, 2445J (with 2380 horns), and 2404H drivers that I am trying to put to use. I would like to build something similar to a 4673 clone but the tapped inductor of the 3110a and 3115a crossovers have me stumped. Is there any way around them using multiple standard inductors?

I have seen the 3110 equivalent HP filter that gets around the tapped inductor, but I need the "a" version to go with the CD 2380 horns.

I have used the Behringer CX4300 in the past but I want to keep this a passive crossover.

My next question is how much of an issue is CD compensation if the 2404Hs are blended in at 5kHz?

Thanks for any input.
Robert

[Baron030]

Hi Robert

You cannot use multiple standard inductors simulate a tapped inductor. A tapped inductor acts like a transformer. And since, there are not separate primary and secondary winding, these devices are often referred to as Autoformers or Autotransformers. I don’t know of anyone that is manufacturing these devices, so unless you can track down some JBL crossovers. You might need to wind your own coils, which is not something that should be attempted without test equipment to measure the inductance.
And due to the necessary inductive interaction between taps, I would recommend using an iron core rather than an air core design. Rewinding an Erse Super Q coil would be a good starting point for such a project.

Here is a link to the inductance ratios that would be needed:

http://www.audioheritage.org/vbullet...hread.php?5376
http://www.erseaudio.com/Products/Su.../ESQ55-16-3300


Baron030

[robfive]

Thanks for the info, Baron030. I am comfortable with modifying an existing inductor or creating a new one from scratch. I am also familiar with the theory of transformer windings but I need to read up on the mechanics of winding multiple taps. I have now read through many of the autoformer threads here but they seem to dwell on the theory more than the execution of the tapped inductors.

I guess my biggest question is do you just start with the primary inductance winding (say 3.3mH) and then wind the secondary coils over the primary according to their calculated ratios? Are the secondary coils connected to each other? Then I notice you mention that there are not seperate primary and secondary windings.

Thankfully I am not in a hurry to finish this project as it appears that I still have quite a bit of reading/learning to do.

[Baron030]

Hi Robert

Here is how I would attempt to make a tapped 3.3mH coil for the 3110a crossover. I would start with a 3.3mH Erse 16 gauge Super-Q coil and some additional 16 gauge magnet wire and an inductance meter. The reason for the additional magnet wire is that when you are removing windings from the Erse coil. The insulation is going to come off that wire because Erst cements the wires together with enamel or varnish. Start by removing windings and while measuring the inductance until the coil’s inductance is reduced to 2.97mH. Keep track of the length of the wire removed and the number of turns removed and record those values. Then continue removing windings and measuring inductance until the inductance is reduced to 2.7786mH. Again record the length and number of turns of wire removed. Then continue removing windings and measuring inductance until the inductance is reduced to 2.4717mH. The wire lead that is closest to the iron core is “Lead # 1” and the outer wire lead is “Lead #2 = -6db”. Attach some new magnet wire to “Lead #2”, it should be a few feet longer than the total amount removed. Start rewinding the coil until the inductance is back up to 2.7786mH and then form a loop marking it as “Lead #3 = -8db”. Continue rewinding until the inductance is 2.97mH and then form another loop, marking it as “Lead #4 = -10db”. And then continue rewinding until the inductance is back up to 3.3mH and marking it as “Lead #5”. While winding you will want to brush some varnish to cement the winding together. Needless to say this is all very labor intensive.

Baron030

[robfive]

Wow, thanks for the explanation. That sounds very straight-forward. I'm sure it will be a very tedious process but the concept sounds simple enough.Thanks for the info.Robert

[GrooveControl]

Has anyone reworked the 3110a to have a fixed HF gain of -10db and not use L3 and S2?

[4343]

Hi Robert

You cannot use multiple standard inductors simulate a tapped inductor. A tapped inductor acts like a transformer. And since, there are not separate primary and secondary winding, these devices are often referred to as Autoformers or Autotransformers. I don’t know of anyone that is manufacturing these devices, so unless you can track down some JBL crossovers. You might need to wind your own coils, which is not something that should be attempted without test equipment to measure the inductance.
And due to the necessary inductive interaction between taps, I would recommend using an iron core rather than an air core design. Rewinding an Erse Super Q coil would be a good starting point for such a project.

Here is a link to the inductance ratios that would be needed:

http://www.audioheritage.org/vbullet...hread.php?5376

...

Baron030

I'm not seeing the values you got.

From the chart, -6dB is supposed to be 25.1% of the original inductance or 0.8283mH. That would certainly be closer to half the wire than what I made, there's many layers on that 2.5mH I started with, and only one layer on top to get to 3.3mH. It looks like you subtracted 25.1% to get 74.9% of 3.3mH, or 2.4717mH.

I read "% of original inductance" to mean the end goal, not the amount to remove from the original inductance.

By my reading:

-8dB is 15.8% or 0.5214mH.
-10dB is 10% or 0.33mH.

>EDIT

If you look at the notes under the model, the -10dB value is in fact listed. (-10dB is 330 micro Henries = 0.33 milli Henries)

END EDIT<

Using that chart, I think it would be fun to build a 3110A with a few different steps. -6, -8, -10 could be -1, -2, -3, -6, -9, or maybe for testing, all of the them!

-1dB=79.4%=2.6202mH
-2dB=63.1%=2.0823mH
-3dB=50.1%=1.6533mH
-4dB=39.8%=1.3134mH
-5dB=31.6%=1.0428mH
-6dB=25.1%=0.8283mH
-7dB=20.0%=0.66mH
-8dB=15.8%=0.5214mH
-9dB=12.6%=0.4158mH
-10dB=10.0%=0.33mH

I'll be looking for some more wire now.

[Baron030]

Quote by 4343:
From the chart, -6dB is supposed to be 25.1% of the original inductance or 0.8283mH. That would certainly be closer to half the wire than what I made, there's many layers on that 2.5mH I started with, and only one layer on top to get to 3.3mH. It looks like you subtracted 25.1% to get 74.9% of 3.3mH, or 2.4717mH.

Hi 4343
Yes, I did simply subtract the 25.1% to get the 2.4717mH value for the -6 tap. I thought the idea would be worth trying. And in a way I am actually surprised how quickly you have been running with it. The next big question is are you getting 0.8283mH of inductance when you measure it between pins #2 and #5? If not then my logic is really flawed. And the chart’s turn ratio should be followed instead. This would mean that a 3.3mH I-Bar coil would need to be completely unwound and the total number of turns counted. And then use the bare core and a lot of fresh of wire to rewind it completely with taps. And with there is the issue of can Pin #1 be the lead closest to the iron core Or does Pin #5 needs to be closest to the iron core? The point being that does the secondary winding need to be closest to the core? Or can secondary winding be wound on top of the primary winding? If the secondary can be on the outside then a 3.3mH I-Bar when it would mean that a lot less of the coils’ original wire would need to be removed.
Baron030

[Baron030]

For those of you following this thread, you can tell I have more questions than I have answers right now.

Hi 4343, Are you getting 0.8283mH of inductance when you measure it between pins #2 and #5?

I have been digging through some boxes tonight and I have turned up almost all of the parts necessary to construct the HF section of the 3110a crossover network. I have a 16 ohm load resistor to stand in for the HF driver. A 0.02mH coil and a 3uf cap for the “Max Boost” section. I have 2-8 ohm resistors that will have to stand in for a 3.9 ohm resistor. And 2-4.3uf caps that will have to stand in for the 8uf cap. As well as 2-40 ohm resistors that will stand in for the 2-39 ohms resistors. And I have one more interesting find, a pair of Jantzen 15 gauge air core inductors. I know I am not hitting all of the part values spot on. But, it might be close enough for a good test.

I am beginning to suspect that turn-ratios might be more actuate and important then relative inductances. Now, I really don’t want to damage a 3.3mH coil for this test, because I may have some uses for them later. But, if I nick the insolation in just a few outside edges, I should be able to create some taps at those points. And then I should be to test the voltage drives to see they the match mathematical formulas or not. Unfortunately, I have to make some bathroom repairs this weekend and other projects as well. So, it might take me quite a while before I can post some CLIO results and then have some real answers.

Robfive, please post the 3110 equivalent network schematic that gets around the tapped inductor. Maybe the focus can shift to just adding the 2380 CD compensation to it.

Baron030

[robfive]

Baron030, thanks for your help in this. I know what it's like to have several projects going on at once. Please fix your bathroom before messing with this inquiry.

Here is a link to the 3110 equivalent HP x-over posted by 4313B back in 2005:

http://www.audioheritage.org/vbullet...ll=1#post52084

It looks like a straight-forward second-order filter and L-pad.

Thanks,
Robert