Barry Porter "Net EQ"

[john12ax7]

I would need to check the schematic, but usually if an EQ is designed with center tapped pots it will not work the same without them. Some issues could be different BW and interactions among bands, the center tap often ensures that a band is fully off at the 0dB position. Without the center tap sweeping the frequency of a band can alter the response of adjacent bands even when it's set flat.

 

[weiss]

I think I said I was surprised it worked without the center tap pots. I am no circuit designer and i never tried it without center tapped pots. To get the sections going i used two 5K resistors with a lead to ground to simulate the center tapped pot on the schematic. Then with another lead I could either get maximum or minimum gain. That works in the absence of a center tapped pot and without committing to a rotary switch. That was the context in which I made the commment.

Yes absoultely. I only wanted to inform others, that CT pots are not needed to make this EQ run properly. I spent a lot of time looking for the right potentiometers. It is absolutely not my intention to accuse anyone.

 

[john12ax7]

I only wanted to inform others, that CT pots are not needed to make this EQ run properly.

Have you checked this against the center tapped version? I would be very surprised if performance was the same. More likely is that the EQ works to a satisfactory level, but not as the designer intended.

 

[weiss]

Have you checked this against the center tapped version? I would be very surprised if performance was the same. More likely is that the EQ works to a satisfactory level, but not as the designer intended.

That would be interesting know. Didn't have the chance yet to try it out but i'm already satisfied how it sounds.

 

[karloff70]

Seems Barry's own notes said this: "If untapped pots are used, the sections will interact quite severely, and the constant Q characteristic will be lost."

 

[weiss]

Seems Barry's own notes said this: "If untapped pots are used, the sections will interact quite severely, and the constant Q characteristic will be lost."

Thanks for pointing it out! Can't believe i failed to read that.

Indeed. I realize the problem now. The bands interact and influence each other.. I take back my statement...
Has nobody been able to source the CT potentiometers yet?

 

[karloff70]

I am looking to build this thing in completely joined stereo, one set of pots for both sides, and all the freq values like Barry had it, no mods to start. All pots, no switches. What type of pots would I use? Any help appreciated.

 

[weiss]

I am looking to build this thing in completely joined stereo, one set of pots for both sides, and all the freq values like Barry had it, no mods to start. All pots, no switches. What type of pots would I use? Any help appreciated.

Not sure if that is even possible. The EQ already uses 3 gang potentiometers, that means you would have to find 6 gang pots ?

 

[karloff70]

It's definitely possible, these guys did it:

 

[JMan]

Surely those four (3)-deck pots would become two (6)-deck pots in this case, not eight, right? Or am I misunderstanding what is being done?

 

[Gold]

I am looking to build this thing in completely joined stereo, one set of pots for both sides, and all the freq values like Barry had it, no mods to start. All pots, no switches. What type of pots would I use? Any help appreciated.

The bell frequencies need a two gang pot. If you want to use the shelf sections you need to use separate pot elements. If you want ganged stereo you need four gang pots for frequency select. If you want the shelving sections included you have to use six gang pots on the Hi and Lo sections.

 

[Gustav]

Thanks for pointing it out! Can't believe i failed to read that.

Indeed. I realize the problem now. The bands interact and influence each other.. I take back my statement...
Has nobody been able to source the CT potentiometers yet?

There is a way to use a standard pot, if you use it in conjunction with a switch.

Set up the pot, so you have the ground on pin 1, and the wiper on pin 2. Wire pin 3 to + or - via a switch. You can use the switch to set if you are boosting or cutting.


Gustav

 

[Gustav]

I am looking to build this thing in completely joined stereo, one set of pots for both sides, and all the freq values like Barry had it, no mods to start. All pots, no switches. What type of pots would I use? Any help appreciated.

For the gain

4 x dual gang center tapped 10K

Shelving filters are in series with the parallel bands section, and while its set up to be either/or, it doesnt have to be.
The addition of a layer on the high/low band is an interface related/space saving decision.

If you separate the shelving filters, you will need.

4 x 4 gang 10K for the frequencies for all bands.
2 x 2 gang 10K for the shelving

If you dont, you will need

2 x 4 gang 10K for the low mid/high mid bands
2 x 6 gang 10k for the low/high bands

I cant remember the value of Q pots, but probably 100K?

4 x 2 gang 100K

Gustav

 

[abbey road d enfer]

Seems Barry's own notes said this: "If untapped pots are used, the sections will interact quite severely,

Actually, I don't see it at all. The only difference is the Boost/Cut taper is slightly different (more cramped at extremities) and probable noise penalty with the non-tapped pots. But there is no interaction, in the sense that the response resulting from two adjacent bands is the sum of the response of each band. At least, that's what the simulation says. Maybe someone can check on the actual product.

and the constant Q characteristic will be lost."

FWIW...
What is the benefit of constant Q? In practice, it means that the higher the boost/cut, the larger the BW. Is it desirable? My personal taste if for constant BW.

 

[Gold]

FWIW...
What is the benefit of constant Q? In practice, it means that the higher the boost/cut, the larger the BW. Is it desirable? My personal taste if for constant BW.

I haven’t plotted anything out but it behaves like a constant bandwidth EQ. The bandwidth sounds independent of gain. A constant Q EQ is useful when there isn't a separate control for BW/Q.

 

[weiss]

Actually, I don't see it at all. The only difference is the Boost/Cut taper is slightly different (more cramped at extremities) and probable noise penalty with the non-tapped pots. But there is no interaction, in the sense that the response resulting from two adjacent bands is the sum of the response of each band. At least, that's what the simulation says. Maybe someone can check on the actual product.

I tried it with untapped pots. The bands are interacting with each other: e.g. a boost in the high section results in a cut in the hi mid section and so on. The frequency curve gets messed up.

Gustav said:
There is a way to use a standard pot, if you use it in conjunction with a switch.

Set up the pot, so you have the ground on pin 1, and the wiper on pin 2. Wire pin 3 to + or - via a switch. You can use the switch to set if you are boosting or cutting.


Gustav

Thanks for the info but that's not really suitable in my case. I would have to make a new front panel.

I haven’t plotted anything out but it behaves like a constant bandwidth EQ. The bandwidth sounds independent of gain. A constant Q EQ is useful when there isn't a separate control for BW/Q.

With untapped pots, sweeping the frequency of the Net eq at fixed gain changes the Q throughout the sweep. I think that is what Barry meant.

 

[john12ax7]

For a given frequency, where all you are changing is boost/cut what is distinguishing Q vs BW? They should be inversely proportional, so fixed one implies the other is fixed as well.

 

[weiss]

For a given frequency, where all you are changing is boost/cut what is distinguishing Q vs BW? They should be inversely proportional, so fixed one implies the other is fixed as well.

Yes they are proportional. Q = f0 / BW

 

[abbey road d enfer]

Yes they are proportional. Q = f0 / BW
View attachment 82707

I could demonstrate why this is wrong, just suffice to say what happens when the boost is less than 3 dB?
Don't tell me it's in every electricity book, so it must be right...
This formula applies strictly to responses that tail asymptotically to minus infinity.
The response shown is that of a bi-quad, that tails to unity.

 

[abbey road d enfer]

I haven’t plotted anything out but it behaves like a constant bandwidth EQ. The bandwidth sounds independent of gain.

So, which one is it, constant BW or constant Q? Constant Q sees BW increasing with the amount of B/C, not so much with constant BW.
Anyway, an EQ is not a 2nd-order filter, it's a bi-quad, i.e. the combination of two antagonistic terms, so actually there are two Q"s: simplifying to one single Q* is a shortcut that shouldn't be sanctified.
When I say I favour constant BW, you're invited to take it with a pinch of whatever, because no proper definition of BW for a bi-quad has ever been agreed upon. In most practical cases, the actual BW varies with the amount of B/C, even when using a constant BW design.
That's why the wrong notion of Q for describing an EQ's behaviour has become the de facto standard, with the highest number indicating the narrower filter.

*Actually, the only case where the Q is the same for both terms is when the EQ is flat.