What kind of EQ is this?

[lion]

I'm about to complete this circuit for a friend.



I build the first section last year, which he use in his guitar as an onboard preamp set at unity gain. Now he wants the EQ section also.

It's clearly some kind of a 3 control/band filter, but I'm not clever enough to read/understand in details what it does from the schem. Can anyone throw some light on what it is - and what it does.

(My friend is into a certain type of instro-music. One of the big names in this particular genre is purportedly using the circuit (with EQ preset) to get his sound).

I'm also wondering about the gain set in the first inverting opamp stage – AFAICS it has a gain of 150! Seem quite high - even with a high loss in the 3 parallel filters?

Erik

[Keppy]

It appears to be a modified Baxandall circuit, but it's written wrong. In a Bax, you can think of the pots as panning filters from the input to the feedback loop. That's an oversimplification, but it works. More info here: http://sound.westhost.com/dwopa2.htm#baxandall. I've never seen a Bax with a midrange control, so this is kind of cool to me.

However, the schematic you posted shows the output of the last opamp directly converted to the inverting input. It will pass no signal that way. The inverting input (pin 6) should be connected to the junction of three 10k resistors and not connected to the other components as shown in the schem.

The Baxandall circuit is active, not passive, and does not suffer signal loss. The gain in the first opamp is just that, gain. This appears to be designed for lots of clipping.

Also note that the circuit is misbiased. The 100k resistor to ground following the FET booster isn't needed for "normal" operation, as the output will bias the inverting input to match the non-inverting one, which is biased to 1/2 supply. The 100k to ground ensures that the output will bias at the positive rail, with significant signal needed to get it to swing downward. There is much gating and asymmetry here if that part of the schematic is accurate. Is your friend interested in industrial, noise, or EDM?

[anotherjim]

Maybe that 100k to ground at the input is an anti pop resistor when the FET is standalone - it shouldn't be fitted with the eq.
I'm just as suspicious as Keppy over the eq wiring.
Is there more info where that scheme came from?

[PRR]

> AFAICS it has a gain of 150!

Only if E230||4.7K were a ZERO source impedance.

Not even close.

The simple gain ratio would be more like 150K / (10Kpot+JFET)   or 22. With some stray leaks, minimum gain is more like 14. Maximum gain is over 100, and I don't see why you could possibly need that much.

The leftmost 100K resistor can not be right. Assume V+ is 10V. Opamp DC bias calls for +12.5V at its output pin. In reality it will be stuck at +8V, and never move except for huge inputs.

Considering the TL072 can be rigged for a simple gain of 1-100 and a 10Meg input if desired, I don't know what the JFET et al bring to the party. The overall inversion is generally moot.

As the others say, there is an obvious error in the Bax connections.

Adding a MID pot to the Bax is a fairly old trick.

[teemuk]

Yes, the Baxandall circuit (with added mid-range) is drawn incorrectly in that schematic. What should really connect the inverting input is the node that mixes signals coming from potentiometer wipers.

The underlying operation of Baxandall circuit is crudely that of an inverting amplifier with negative feedback loop.

Gain of the circuit is approximately RF/R1 so if RF>R1 then circuit provides gain, if RF<R1 it attenuates. Unity gain, (gain=1), is acquired when RF=R1.

The Baxandall circuit merely substitutes RF and R1 resistances with impedances of the RC filters. If potentiometer is dialled to "middle" position the impedances should roughly be equal and the circuit has unity gain at the band the RC filter passes through. If the balance changes the circuit either amplifies or attenuates.

[lion]

Thanks gentlemen.

There's no info where the schem comes from, and - it's surely one of these cases where something is considered "the holy grail" to some because of said relations to certain user. When I was first asked to build the buffer my friend was very focused on the E230 transistor – in his circles it's considered the "secret" of the circuit. I couldn't source the discontinued E230, but my friend insisted and spent many weeks to finally locate two E230's in France and have them sent to me. When I had built the buffer I did two more boards with different subs for the E230 – which proved no differences in performance, much to my friends surprise and disbelieve.

Actually I did look at the active Baxandall tone circuit but reconned it was not because of two differences – the one that you all point out – and also that the rightmost side of the three 100k pots followed by a series cap/res goes to the output of the second opamp, whereas on the Bax's I could find they go to the inv input.

That is on the Bax:
mixed signal from the wipers > inverting input
mixed signal from the pot ends > opamp output

On this circuit, with the correction you suggest, it's the other way around:
signal from the wipers > opamp output   
signal from the pot ends > inverting input

Is this another error in the schem, or can it actually be either way?

I understand the leftmost 100k has to be removed.

Re the gain - as unity gain with the tone controls centered is what's needed I recon I can replace the 150k resistor with say a 100k trimpot to dial it in.

Erik

[PRR]

http://oi63.tinypic.com/34yz50z.jpg

[lion]

Thanks Paul - got it. Obviously simple from your pic. I don't know how I got pin 6 and 7 and what goes where mixed up like I did 

Erik