Purpose of this pot in API 536 circuit?

[FIX]

Depends what linear panner you are talking about. A simple low cost single LIN pot panner will have a 6dB loss at the extremities but one using a dual LIN pot with pull up resistors has no loss at the extremities. It also has the advantage that LIN pots are much more consistent than log and revlog pots.

Cheers

Ian

If the wiper is to ground, you will always have a loss of some kind. Most log and rev log pots only have 2 or 3 overlays so they aren’t smooth. I use all linear panners with make up right under the pot.

 

[fazer]

The api used a fixed gain diff input 2520 instead of the transformer. That meant there was no line in trim. When mixing through the channel in line-in, you used the stacked concentric 1k pot for level offset to keep the faders at a unity level. This allows return to unity position when lifting or lowering gain while mixing.

 

[JW]

I use all linear panners with make up right under the pot.

Thanks for your input Paul. Can I ask what scheme you would use with a dual linear pot? 536 circuit-10Kdual pot-(8X33K+6auxes with 10Kpots that are also bused through 33K each)

So the total load from all the buses in parallel is what determines the pull up resistors correct?

So, 8 at 33K =4.1K

How do I calculate the following 6 buses that are also hanging off the DOA? (in the schematic below)

They have a 10K pot in front of a 33K resistor each. So, is it just (10K+33K)/6=7.2K Then we parallel the 4.1K and 7.2K for total bus load of about 2.6K?

 

[ruffrecords]

If the wiper is to ground, you will always have a loss of some kind. Most log and rev log pots only have 2 or 3 overlays so they aren’t smooth. I use all linear panners with make up right under the pot.

I agree, but that is not what I was saying. I was talking about a DUAL LIN pot. This would normally be 6dB down at the centre so you use pull UPS to reduce this to 3dB or so. See attached Neve standard pan pot. No wipers to ground here.

Cheers

Ian

 

[JW]

Thanks Ian. I actually had found a Neve bulletin in the mean time of the exact same circuit you just posted. Okay, how very simple, and I've arrived at 1.8K for -3dB at center pan. Same as Yamaha, whaddya know, but they have 1.8K's in wiper to ground position.

 

[ruffrecords]

Thanks Ian. I actually had found a Neve bulletin in the mean time of the exact same circuit you just posted. Okay, how very simple, and I've arrived at 1.8K for -3dB at center pan. Same as Yamaha, whaddya know, but they have 1.8K's in wiper to ground position.

That is odd. It will not work like that with LIN pots. However a rev log will be about 10% down at the mid point i.e. 1K. So the the attenuator is approximately 1.8/2.8 = -3.8dB.

Cheers

Ian

 

[JW]

Which is odd? What Yamaha did, or the way I'm using the Neve schematic you just posted, w/ 1.8K from wiper to top of pot? And, where is the 2.8 coming from? My total bus resistance?

 

[ruffrecords]

Which is odd? What Yamaha did, or the way I'm using the Neve schematic you just posted, w/ 1.8K from wiper to top of pot? And, where is the 2.8 coming from? My total bus resistance?

A typical 10K REV log pot at half way has 1K from wiper to top; that's where the 1K comes from.

What's odd would be Yamaha using LIN pots with pull downs. Sorry, I should have been clearer about that.

If I were you I would stick to the Neve circuit.

Cheers

Ian

 

[FIX]

I agree, but that is not what I was saying. I was talking about a DUAL LIN pot. This would normally be 6dB down at the centre so you use pull UPS to reduce this to 3dB or so. See attached Neve standard pan pot. No wipers to ground here.

Cheers

Ian

Right. That works. Linears are always better, as an audio taper pot is actually layers screened on top of layers, starting at different points, so they vary a lot. Cheap pots have 2, better ones have 3.

If you open up a 100mm fader, the audio track has another track along side it, that is attached to the audio track like a ladder. So, about every .5 inch, it slugs the audio track to simulate an audio taper. The more expensive the fader, the more slugs, and the older P&Gs had little grinder marks where they tuned them.

 

[ruffrecords]

Right. That works. Linears are always better, as an audio taper pot is actually layers screened on top of layers, starting at different points, so they vary a lot. Cheap pots have 2, better ones have 3.

If you open up a 100mm fader, the audio track has another track along side it, that is attached to the audio track like a ladder. So, about every .5 inch, it slugs the audio track to simulate an audio taper. The more expensive the fader, the more slugs, and the older P&Gs had little grinder marks where they tuned them.

You are absolutely right. Quality pots and faders are a breed apart and a sight to be hold (internally) . Cheap LOG pots are essentially a pair of LIN elements in series - you can see this quite clearly on the published curves for them. A typical cheap 10K log pot is 9K LIN squeezed into the first half followed by 1K LIN over the remaining half.

Cheers

Ian

 

[FIX]

and they overlap them.. some are 20k linear, then at 10am another layer starts and then at 2pm one starts.

the old Clarostat pots used in the API oscillator had the bump zone right at +4 out and you could never get it dead on…

 

[JW]

Just a bump here to ask another question. Thanks so much for all the help with this.

When I added the Neve panning arrangement to the API's output, I am in need of a 'switched in' 6.8K series resistor at the output of the opamp, in order to pad down the output in "pan off" position so it matches gain in the middle of "pan on."

I wonder if this is okay? Is it too much resistance? Isn't it the same as with pan-on anyway, which is why it matches? (5K+1.8K) I only ask because I am getting some level drop (1-2dB) when switching in more buses.

Hmmmm. Maybe I should use that original 2K pan pot that Yamaha already provided. . . as it would be less loading.

Don't know quite why it would matter though. If the minimum load that the opamp sees is 2.4K, if my earlier calculations are correct, then adding 6.8K to that wouldn't be harder for the opamp to drive, would it, just require a little more power, which it has by adjusting the feedback loop. On the receiving end, the ACA input would be seeing roughly 24 ch's of (33K + 6.8K) in parallel right? Why would this matter if the original bus resistors API used anyway, were 47K?