DIY inductors

[Yazoo]

Do any of you wind your own inductors? This doesn't really relate to wah inductors. It's more about inductors used in power breaks like the Marshall SE100 which uses a combination of different inductors to give 0, 6db, 12db or 18db reduction while being able to function as a dummy load for up 100 Watts for an 8 Ohm output. If my maths is correct, this means the inductors need to be rated for up to around 3.5 Amps.

Would it be feasible to wind your own inductors for this while not costing an arm and a leg? I've done a lot of Googling for information and if you know of a step-by-step guide on how to do this, I would be very grateful. 

[PRR]

Bust-open a SE100 and see what they did, if it looks like good DIY.

Without knowing the values you want, but knowing speaker impedance, I think you need BIG air-core (lot of copper but easy figuring); or iron-core, which is a total pain to calculate. Even having a sample in hand is not enough, because these will probably be gapped iron, and you can't be sure of the gap spacing or pressure. (They may be adjusted in the factory with a rawhide hammer.)

BTW: should be "brake", which may aid searching.

[Rob Strand]

Do any of you wind your own inductors? This doesn't really relate to wah inductors. It's more about inductors used in power breaks like the Marshall SE100 which uses a combination of different inductors to give 0, 6db, 12db or 18db reduction while being able to function as a dummy load for up 100 Watts for an 8 Ohm output. If my maths is correct, this means the inductors need to be rated for up to around 3.5 Amps.

It's 10 to 15 years since I've looked at the details of the Marshall stuff.  It pretty much works like you said.  I think it tries to form a complex divider with the speaker (using a scaled speaker model from RLC's for the dividers and loads) and also present the correct impedance to the amp.   As a result it uses a heap of RLC networks.    The Marshall power brake use a different method it has a single load and an auto transformer.   Drops the cost and space of all the inductors at the expense of an auto transformer.

The small inductors (330u, 500u, 750u) can easily be wound with air cores.  .

The 1mH inductor could be air-cored,  a ferrite rod, or a gapped iron core.

The 12.5mH inductor would probably need to be a gapped iron core.

Be warned, there are two tapped inductors used on that unit.  The circuit marks these as 12.5mH but it is not clear what terminals is 12.5mH and what the inductance is across the other terminals; they might not be 50% tapped.   An additional factor is the coupling between the coils.   Even if only one side of those LCR network is active the L and C on the other arm is still active and impact in the final L and C values.   I would try to calculate the required LC values based on scaling the impedance in each air.   That would give you the LC value required if the coupling was not present.  You could even use non-tapped inductors.   I suspect Marshall used tapped inductors to reduce the number of inductors, since they are probably iron cored.

The question is what DC resistance do you want.  If the DC resistance is too high the unit won't perform correctly and the inductors could get hot.  Too low and you are making the inductors large an expensive.

For air coils you can look up many sites on calculating Multilayer coils (like coil32, Steve Moshier's NBS formulas, or the Wheeler multilayer formulas).   The air cores only care about the DC resistance.  You don't have to worry about saturation.

For iron cores you will need to read a lot of books.  You need to consider DC resistance and also core saturation.  Both those determine the size of the core and size of the gap.

Ferrite rods are tricky.  Very little theoretical info.  There were a few sites giving crude formulas and I posted some ways to estimate saturation in a thread on this forum a few years back.

So before you start you need to determine:
- DC resistances
- maximum currents
- the couple/split of the coupled "12.5mH" inductors.
-------------------
EDIT:
It just occurred to me there's no added resistors to set the Q of the parallel LC network.  That means the DC resistance of the inductors in the parallel LC networks have been tweaked to give the correct Q (the Q which matches the impedance of a speaker in sealed enclosure.)

[Rob Strand]

LN, AC etc refer to the points mark L, A, C etc. on the SE100 schematic.

1) By scaling the impedances in the LN arm I derived the matching L and C values for the arms with the coupled inductors.  ***** These values are when there is no coupling between the inductors. ***** ie. like if you uses all separate caps and inductors.

                     Master               
Component    LN        AC     CN    FH        HN
R1s                8.2      4.7      10     6.8        3
R2p               33       15       33     27        12
Ls [mH]          1         0.5      1       0.75       0.33
Cp [uF]          160      279.1  131.2  192.9    437.3
Lp [mH]          12.5     7.14    15.20  10.34   4.56
fr[Hz]             112.5   112.7   112.7  112.7   112.7

2) From this we can work out that the inductors have turns ratio.  It is *not* centre tapped.  It is tapped such that the inductance from the tap to one outside terminal is twice that from the tap to the other outside terminal.
That means the turns ratio is: 
large L tap to outside is 1 :   Small L tap to outside   = 1 : 1/  sqrt(2)   = 1 : 0.414

3) Now, for the tricky part.   If you think of the tapped inductor as a transformer, the total capacitance across the inductor terminals with 330uF terminal is 330uF *plus* the effective capacitance of the 160uF due to the transformer load.   That ends up looking like about 650uF.    In other words, if you see a value of 330uF in the coupled inductor case it ends-up looking twice that value.

4) More details to confuse the story.   When I calculate the effective capacitances in each case using the doubling rule from (3) we find the total capacitance does not agree with the table calculated in (1).

What we find is in order to get the same resonant frequency with the actual cap values in the circuit we need smaller inductances than that calculated in step (1).

We need:
Inductor 1:  CN = 12.5mH and AC = 6.25mH
Inductor 2:  FH = 6.25mH and HN = 3.13mH

So one of those marked "12.5mH" is actually 12.5mH across the tap and the outside.
However the other is half that.

Note the smaller inductance wire in parallel with the larger capacitance - so the resonant frequency stays the same.

Comments:
- using the coupled inductors lets you use smaller capacitors and reduces the number of cores.

These calculations are fiddly so it's easy to make an error.   I would check over what I have done before progressing any further.

[Rob Strand]

3) Now, for the tricky part.   If you think of the tapped inductor as a transformer, the total capacitance across the inductor terminals with 330uF terminal is 330uF *plus* the effective capacitance of the 160uF due to the transformer load.   That ends up looking like about 650uF.    In other words, if you see a value of 330uF in the coupled inductor case it ends-up looking twice that value.

Just a warning:

In the circuit the coupled inductors are part of the *same* attenuator.    I'm pretty sure that does more than just make the caps larger since there is transformer action going between the input of the attenuator and the output of the attenuator.    That complicates the whole thing.   In fact I'm not even sure you can get the attenuator to work with the coupling present so I wonder how those inductors are actually constructed!

The separated components for the Lp Cp network I calculated in step 1 are OK.   (However, keep in mind the existing part values for R1s, R2p and Ls don't quite scale.   That's due to the discrete component values chosen by Marshall.)
------------------------------------------------------
EDIT:
OK, it *is* possible to get the attenuators working with coupled inductors.    You have to treat the attenuator as an auto transformer.    The amount of attenuation determines the turns ratio.  This is different to the uncoupled inductors (ie. case (1)) where the amount of attenuation determines the inductance ratio.    I know how to go about it but I haven't got time right now to write out a nice solution.

[Yazoo]

It has become obvious to me that trying to wind my own inductors for this is just beyond my abilities unfortunately. Thanks for the advice - it has saved me spending money on something I would not be able to successfully do.

I have gone ahead and started on the cab sim section of the SE100 which I will be able to use with line level signals, so fingers crossed that this sounds good.

[Rob Strand]

It has become obvious to me that trying to wind my own inductors for this is just beyond my abilities unfortunately. Thanks for the advice - it has saved me spending money on something I would not be able to successfully do.

I suppose there are two problems:

The first is making the inductors.  You can get around this by buying off the shelf parts.  To keep costs down it's best to get inductors with a DC resistance suited to your requirements.

The second problem is what inductors to buy.  The schematic is deceptive as there's not enough info to actually build a unit from it.  The DC resistance details for the inductors are missing from the circuit - admittedly some positions in the circuit are less critical than others.    The details of the possibly coupled inductors are missing. 

I did play with the coupled inductors.    I couldn't get the coupled inductor case to exactly match the non-coupled inductor case.  However, it was certainly possible to get a more than good-enough solution (which was not easy as far as calculations go).   The weird thing is, in all my calculations the cap values for the parallel C+L networks did not match the values on the schematic.   That says to me someone's tinkered with the L and C values, possibly with the aim of using smaller inductors and making the design more cost effective.

In the end the question of what the circuit actually is could not be derived though "obvious" calculations.    So if I was going to build that thing I'd probably just start from scratch and work out my own values.