The math behind this EQ circuit using inductors.

[ruairioflaherty]

I wanna buy a roll of 1mm² copper wire, they sell it per KG, seam to be about 145 meters, but i looked at ferret inductor cores in a "hobby" shop, there are many sizes, thicknesses, diameters, etc and you're clueless again.

Anyone knows the formulas how to make them?

Lots written here over the years about DIY inductor winding, formulas and practical info - try searching

 

[Matador]

Hypothetical: taking the first band, it's series R=680, C=6.8uF, and L=1H. Let's say I measure the internal-resistance of the inductor at 320 ohms. So we effectively have R=1K. That gives a resonant frequency of 61Hz, and a Q of 0.38.

If I reduce L down to 333mH, to keep the same resonant frequency, I increase C up to 22uF. This gives a frequency of 58Hz (close enough). If the L resistance drops down to 100 ohms (2/3 less winding), then if I decrease R down to 220 ohms, so the total series R is 330 ohms, then I end up with the same Q of 0.38.

Would this new band sound the same, all other things being equal?

 

[ruffrecords]

Hypothetical: taking the first band, it's series R=680, C=6.8uF, and L=1H. Let's say I measure the internal-resistance of the inductor at 320 ohms. So we effectively have R=1K. That gives a resonant frequency of 61Hz, and a Q of 0.38.

If I reduce L down to 333mH, to keep the same resonant frequency, I increase C up to 22uF. This gives a frequency of 58Hz (close enough). If the L resistance drops down to 100 ohms (2/3 less winding), then if I decrease R down to 220 ohms, so the total series R is 330 ohms, then I end up with the same Q of 0.38.

Would this new band sound the same, all other things being equal?

The maximum boost/cut would increase due to the lower series R but the shape of the curve should be identical.

Cheers

Ian

 

[richiyobs]

The reason I asked for the link to the TubeCad page is because his single resistor ladder technique is an almost exact copy of my own passive mastering EQ. The TubeCad post is dated 14th August 202 but I published my design back in April 2017. The main difference between his EQ and mine is that mine is 100% passive and the TubeCad one is active and will only work with a balanced input signal.

With the passive EQ I think there is no option but to use an inductor but I will look into it because inductors are very expensive. One possibility is to use a bridged T notch filter instead of the inductor. Unfortunately the notch filter has a high impedance at the resonance but we need a low impedance.

Cheers

ian

Thanks for Infos.... Yes i built that eq with balanced input for the ladder as advised by the designer... Mine It s an opamp version... I m still not able with tubes ...What i found Is that without 300ohm at one side of the switch you get over -30dB in cut mode notch type... So using standard pots without limit its range you ll have notch response...
Soon i ll build your passive version and try to combine both eq... Let s see...
Best

 

[Analog_Fan]

After reading: sourcing audio inductors in 2022 (for equalizers)

One could simulate inductors.

An audio circuit collection, Part 3
https://www.ti.com/jp/lit/an/slyt134/slyt134.pdf
By Texas Instruments.



Just need to translate capacitor value in to usable.
2E-05
6.4E-06
1.5E-07
2.0E-08
8.2E-09
3.9E-10

 

[abbey road d enfer]

One could simulate inductors.

An audio circuit collection, Part 3
https://www.ti.com/jp/lit/an/slyt134/slyt134.pdf

Note that these simulated inductors have one side grounded, so you would need to alter the disposition of R-L-C.

 

[Matador]

This topology offers boost and cut:

So if I'm thinking about this correctly:

1) The pot(s) are across the input terminals, which means they are in parallel with the differential input resistance of the op-amp. Assuming the input signal magnitude allows the op-amp to do it's thing, it will keep the differential voltage across the pot at zero, meaning no current through the pot. Hence it shouldn't factor into the gain equation at either extreme.

2) With the pot dialed fully 'up', the circuit looks like a divider between R1 and the tone network, and the op-amp looks like a unity gain buffer. At (maximum) cut, and at the resonant frequency, the C and L cancel, meaning the gain (reduction) at that frequency will be (looking at the first band): R2/(R2+R1), or with the values listed, approximately 17dB reduction. So at each resonant frequency, the gain reduction is set by R1, and the series resistance(s) R2 through R6.

3) With the pot dialed fully 'down', the circuit looks like a regular non-inverting amplifier, with gain set by R7 and R2, or with the values listed, a gain of 16.5 (or again, roughly 17dB of boost).

Not sure how to think about it in the middle: since the feedback cancels the voltage across the pot, there can be no current into the pot, hence no voltage developed across the RLC network. One way to think about it as two identical networks: one providing the attenuation function of #2 above, with an identical network providing gain. As the pot is dialed lower, attenuation decreases as gain is increasing, and in the middle, they cancel, meaning unity gain through the pass band.

 

[Analog_Fan]

Note that these simulated inductors have one side grounded, so you would need to alter the disposition of R-L-C.

What do you mean with that?
One side grounded?

 

[Analog_Fan]

So if I'm thinking about this correctly:

1) The pot(s) are across the input terminals, which means they are in parallel with the differential input resistance of the op-amp. Assuming the input signal magnitude allows the op-amp to do it's thing, it will keep the differential voltage across the pot at zero, meaning no current through the pot. Hence it shouldn't factor into the gain equation at either extreme.

2) With the pot dialed fully 'up', the circuit looks like a divider between R1 and the tone network, and the op-amp looks like a unity gain buffer. At (maximum) cut, and at the resonant frequency, the C and L cancel, meaning the gain (reduction) at that frequency will be (looking at the first band): R2/(R2+R1), or with the values listed, approximately 17dB reduction. So at each resonant frequency, the gain reduction is set by R1, and the series resistance(s) R2 through R6.

3) With the pot dialed fully 'down', the circuit looks like a regular non-inverting amplifier, with gain set by R7 and R2, or with the values listed, a gain of 16.5 (or again, roughly 17dB of boost).

Not sure how to think about it in the middle: since the feedback cancels the voltage across the pot, there can be no current into the pot, hence no voltage developed across the RLC network. One way to think about it as two identical networks: one providing the attenuation function of #2 above, with an identical network providing gain. As the pot is dialed lower, attenuation decreases as gain is increasing, and in the middle, they cancel, meaning unity gain through the pass band.

Here is a DJ mixer




https://www.rane.com/downloads-legacy#ttm56
https://cdn.inmusicbrands.com/rane/pdf/ttm56ssch.pdfhttps://cdn.inmusicbrands.com/rane/pdf/ttm56s_data.pdfhttps://cdn.inmusicbrands.com/rane/pdf/ttm56s_data.pdf
Accelerated Slope Tone Control Equalizers
https://cdn.inmusicbrands.com/rane/pdf/acceler.pdf
Not clear with frequency are set.

 

[abbey road d enfer]

Not sure how to think about it in the middle: since the feedback cancels the voltage across the pot, there can be no current into the pot, hence no voltage developed across the RLC network. One way to think about it as two identical networks: one providing the attenuation function of #2 above, with an identical network providing gain. As the pot is dialed lower, attenuation decreases as gain is increasing, and in the middle, they cancel, meaning unity gain through the pass band.

Considering a single section, in the middle position, attenuation of the input signal is (R2+1/2 VR1)/(R2 +1/2VR1 +R1) and gain of the stage is (R7+1/2VR1+R2)/(R2+1/2VR1). Since R2=R7, it resolves to unity.
Actually, due to the interaction of the other bands, the calculation is quite difficult, but OTOH, the superposition theorem leads to the same conclusion.
This can become a mathematical nightmare. I would suggest you use a simulator, such a LTspice or TINA.
Something that must be considered is the noise gain.
It is the effective gain that is applied to the opamp's internal noise sources.
Noise gain is (R1+R7)/(VR1/5), so about 20dB.
In practice, the simulated inductors (often improperly referred to as "gyrators") introduce their dose of noise.

 

[abbey road d enfer]

View attachment 114195

What do you mean with that?
One side grounded?

Correct.

 

[abbey road d enfer]

Here is a DJ mixer

View attachment 114196

That's a very different topology. Signal is split in 3 bands, each one with an individual level control, and summed back together. It allows infinite cut of each band.
The problem is that response with all controls at mid-position is hardly neutral.

 

[Analog_Fan]

How about these topolgy's;

Would these be improved?

 

[Analog_Fan]

Hammond Manufacturing 5 H
(Choke designed for Marshall guitar amp, inductance 5 H @ 120 ma., 194 Series )

https://nl.mouser.com/ProductDetail...=sGAEpiMZZMv126LJFLh8y9WoEBpwx/PvAjK0Gd0oAW0=
€ 33,33

Hammond Manufacturing 2.6 H

https://nl.mouser.com/ProductDetail/Hammond-Manufacturing/193K?qs=5ZsRU9L2PDnLuBTfNmuAEQ==
€ 68,08

indeed, it does get very expensive and these are the only ones mouser has of these values using in "Matador" circuit.

 

[Matador]

indeed, it does get very expensive and these are the only ones mouser has of these values using in "Matador" circuit.

That is what prompted my question about switching to different inductor values.

For this circuit, I've been purchasing the Mesa inductors for $3.95:

https://store.mesaboogie.com/categories/no-category/electronic-components.html
They stock 1H, 390mH, 220mH, 68mH, and 33mH.

 

[abbey road d enfer]

How about these topolgy's;

Would these be improved?

The only advantage is that the response is perfectly flat in mid-position.
Also boost/cut and Q are slightly different.

 

[abbey road d enfer]

Hammond Manufacturing 5 H
(Choke designed for Marshall guitar amp, inductance 5 H @ 120 ma., 194 Series )

https://nl.mouser.com/ProductDetail...=sGAEpiMZZMv126LJFLh8y9WoEBpwx/PvAjK0Gd0oAW0=
€ 33,33

Hammond Manufacturing 2.6 H

https://nl.mouser.com/ProductDetail/Hammond-Manufacturing/193K?qs=5ZsRU9L2PDnLuBTfNmuAEQ==
€ 68,08

indeed, it does get very expensive and these are the only ones mouser has of these values using in "Matador" circuit.

These are "choke" inductors used in tub e mps for "cleaning " high voltage. It's part overkill, since it can hanle levels much above those in an EQ, and part inadequate because the magnetic core is not suitable for audio signals.

 

[Analog_Fan]

These are "choke" inductors used in tub e mps for "cleaning " high voltage. It's part overkill, since it can hanle levels much above those in an EQ, and part inadequate because the magnetic core is not suitable for audio signals.

mouser doesn't sell any other in this area, 5H.


Than i remembered this video.


started looking at German company's that sell metal the "stamped" metal sheet to make transformers.
But until some time next year i don't have a advanced millimeter that can measure Henry.

Maybe these small yellow "Wurth" transformers, would be suitable.

 

[Analog_Fan]

The only advantage is that the response is perfectly flat in mid-position.
Also boost/cut and Q are slightly different.

witch one?
the serial or the parallel?

 

[Analog_Fan]

That is what prompted my question about switching to different inductor values.

For this circuit, I've been purchasing the Mesa inductors for $3.95:

https://store.mesaboogie.com/categories/no-category/electronic-components.html
They stock 1H, 390mH, 220mH, 68mH, and 33mH.

Indeed, but the currently local post here in Europe sends you a letter asking for the VAT, that you need to pay up front before getting the package.
So you pay TWICE VAT, because the seller also wants VAT.
It's a mess right now.


Was looking at German companies that sell the "stamped metal sheets" for transformers.
https://www.sekels.de/fileadmin/PDF/Deutsch/24_Info_Kernbleche.pdf
apparently the make as small as 12.6 mm x 8.6 mm.