Pan circuit for passive mixer

[Newmarket]

"Howland" current pump.
It's kind of unfortunate that the same name applies for a supplier of plumbing supplies (Howland pump & supplies).

Howland indeed. I blame productive text

 

[Liutmod]

Thank you Ian, so you suggest to use 10k Ohm pot for inputs, 50k Ohm pots for pan (plus two 22k ohm resistors) and 50k Ohm pot for aux sends, with 47k Ohm bus feed resistors both for the channels and for the aux send. It's right?

 

[JohnRoberts]

Howland indeed. I blame productive text

Back decades ago, I just called them "synthesized current sources". Here is a thread with more discussion from a few years ago (not passive summing). summing large amounts of channels.

JR

 

[Liutmod]

Hello, the aux send 47k Ohm potentiometers must be linear or logarithmic?

 

[ruffrecords]

Thank you Ian, so you suggest to use 10k Ohm pot for inputs, 50k Ohm pots for pan (plus two 22k ohm resistors) and 50k Ohm pot for aux sends, with 47k Ohm bus feed resistors both for the channels and for the aux send. It's right?

Yes.

And in answer to your other question you can use either LOG or LIN pots for the AUX sends. If you use LOG pots your the signal will be about 20dB down with the pot at half way. The trouble with LOG pots is they vary a lot between samples so there could be several dB difference between pots at identical settings. LIN pots are much more consistent. For this reason at Neve we only used LIN pots. If you slug them with a resistor from wiper to ground that is equal to half the pot value, the response will be 10dB down with the pot set at half way and it will be much more consistent across the channels This is how all AUX sends were done on Neve mixers back in the 70s. So, since you already have a 47K bus resistor which is effectively connected from wiper to ground then all you have to do is use a 100K LIN pot and you have a Neve style AUX send.

Cheers

Ian

 

[Liutmod]

Thak you Ian, so, for aux sends intead of 50k Ohm pots I can use 100k Ohm lin pots with te same value of 47k Ohm bus resistors?
When you say that 47 k Ohm bus resistors are connected from wiper to ground, this happens when the pots are at zero, it is so, right?
In this way I can use 10k Ohm linear pot for level, putting a 5k Ohm resistor from wiper to ground?
Thank you again

 

[ruffrecords]

Thak you Ian, so, for aux sends intead of 50k Ohm pots I can use 100k Ohm lin pots with te same value of 47k Ohm bus resistors?

Correct.

When you say that 47 k Ohm bus resistors are connected from wiper to ground, this happens when the pots are at zero, it is so, right?

It happens however the pot is set. You said you have 18 inputs. So when you send a signal down one of the 47K bus resistors there are 176 other 47K resistors attached to it on the bus. The other end of each of those is a pot of some kind. Even it is is a 100K pot its maximum source resistance will be about 25K so worst case each of the other 17 inputs adds 4&K +25K to the bus. 17 of these in parallel is 4K2. So you 47K bus resistor is connected to ground via the bus by a resistance of just 4K2. So it is effecitvely grounded as far as working out the law of the pot is concerned

In this way I can use 10k Ohm linear pot for level, putting a 5k Ohm resistor from wiper to ground?
Thank you again

Yes This will give 10dB down at the mid point. However, you need to remember that, as far as the load the pot applies to its source is concerned, when the pot is turned right up its resistance is 10K in parallel with 5K. When it is turned right down it is 10K

Cheers

Ian

 

[Liutmod]

Thank you Ian
With 47k ohm resistors the aux output should have an impedance of 47.000/16=2.937,5 Ohm
If I wanto to balance this output, do you think I can use a 10k:10k tranformer?
Alternatively I could put an unbalanced to balanced opamp circuit with which I could recover the bus loss, too

 

[ruffrecords]

Thank you Ian
With 47k ohm resistors the aux output should have an impedance of 47.000/16=2.937,5 Ohm
If I wanto to balance this output, do you think I can use a 10k:10k transformer?

This depends on what you want to connect it to and how far away it is.

Alternatively I could put an unbalanced to balanced opamp circuit with which I could recover the bus loss, too

This is a better option

Cheers

Ian

 

[Liutmod]

Thanks a lot Ian

 

[Tubetec]

You might not require the abillity to pan on each and every channel as some might already be stereo pairs ,
A simple mono/stereo switch per pair of inputs might be useful ,in mono mode both are fed equally to L/R bus , stereo and the sources are fed individually to L/R outputs .

 

[Liutmod]

What you say is right, but I wanto to make a mixer with traditional controls.

Years ago I tried a passive mixer made by Sm Pro Audio, and I don't remember having any problems.
The one I want to make has an additional effect send, which is a second mixer inside the mixer, but that shouldn't be a problem.

 

[Liutmod]

Hello, instead of pan potentiometer I thought to use an LCR rotary switch connecting to ground the unused resistor in left and right positions.
I've found this schematic, but I can't figure out how to connect the unused resistors to ground.
Does anyone have a schematic that allows this?

 

[Tubetec]

I think I remember one of Ians mixers uses the LCR switching arrangement to keep things simple ,

 

[ruffrecords]

Hello, instead of pan potentiometer I thought to use an LCR rotary switch connecting to ground the unused resistor in left and right positions.
I've found this schematic, but I can't figure out how to connect the unused resistors to ground.
Does anyone have a schematic that allows this?

You need to turn the switch the other way round and switch the source for each resistor. Attached is a snippet from the routing section of a mixer I am building S2A and S2B do the LCR switching.

Cheers

Ian

 

[Liutmod]

As it is not too clear for me, is it possible you have a layout similar to the one I posted?

 

[thor.zmt]

Hi everyone. After a few years of inactivity a friend convinced me to help him to build a passive mixer with one fader and pan control per channel.
The inputs are unbalanced
I want to realize the pan circuit with a double pot and resistors and I would like to know if anyone knows of a pcb to make it
A thousand thanks

I had a quick read of the thread. Some comments.

Modern gear will easily drive 1kOhm, so why not use a 1k level control? Maybe a stepped attenuator for repeatability?

You don't need a huge range for this to balance the levels of individual tracks, so a 24 position switch would likely do fine, in 1dB steps. Or even a 12 position switch in 2dB steps.

One could add a center neutral toggle switch as 0dB/-24dB/-oo to expand range and have true mute if needed.

For pan, a "pot" solution is messy. On the other hand using an 11 position rotary switch could give (say) 4 steps between L & C and another 4 strips between C & R. I can think of many situations where I personally would like more control than L/C/R.

Just switch a suitable resistor chain into the sum node to get the right level balance.

Be aware that any passive pan without buffering will introduce crosstalk. How much depends on the impedance ratios. This is why commonly mixing amplifiers have "Zero Ohm" input impedance.

How much of an issue this is will depend and should be analysed. Excel is your friend.

Using a pair of output transformers with a large step-up ratio can strongly mitigate this, by presenting a low load impedance to the summing node and then near noiselessly adding gain back.

It might also add a tiny bit of character and "glue" to the mix. Crosstalk and interactions between controls will be minimised.

Alternatively, using a 2 channel level control with the pan ahead of this level control could minimise the crosstalk issue.

In order to keep the interaction between controls low you need a fairly high ratio between summing resistor and level/pan control. High resistance values add noise.

This is why modern mixers with large numbers of channels use virtual earth mixing. It massively reduces headaches.

I might be tempted to make it a semi-passive mixer with summing amps to give virtual earth summing. A low impedance passive frontend makes sense.

Modern Op-Amp's like OPA1656 have low noise, low distortion and high current drive that make the job fairly painless.

Still another angle would be to use two channels with the same signal input and have a L/R level control that allows both panning and level control.

It would be a fairly standard passive mixer, with pairs of controls that can be used so each control gets either channel 1 or channel 2 input and are assigned permanently as L/R.

Select input 1 for both, you have level & pan for input 1; select input 2 for both, you have level & pan for input 2; select 1 for L and 2 for R and you have a stereo pair. For even more flexibility you could also select each control L/C/R.

The totally opposite angle would be of course to discard all and any controls and use the DAW to control levels and panning and just do a passive stereo summing system, with enough channels of DA.

Modern DAC Chips exist that have sufficient dynamic range and directly output +8dBu (e.g. PCM5102).

Add off the shelf 8-channel USB Interface PCB's with I2S output (e.g. minidsp) and cascade enough of these using a USB hub, voila, 16 stereo pairs and passive mixer in a box.

Or (shock, horror, gasps, some in the in the audience call for a mob with torches and pitchforks) just mix in the box and use an interesting analogue mastering Suite to add glue and fettle the final sound just so.

Thor

 

[ruffrecords]

As it is not too clear for me, is it possible you have a layout similar to the one I posted?

Assuming we are looking at the switch from the rear, then:

1. Connect the input signal to 1b, 1c, 2a and 2b
2. Connect 1a and 2c ro 0V
3. Connect left bus feed resistor to 1p and right bus feed resistor to 2p

Cheers

Ian

 

[Liutmod]

Works perfectly, thanks Ian!

 

[thor.zmt]

Be aware that any passive pan without buffering will introduce crosstalk. How much depends on the impedance ratios. This is why commonly mixing amplifiers have "Zero Ohm" input impedance.

How much of an issue this is will depend and should be analysed. Excel is your friend.

....

In order to keep the interaction between controls low you need a fairly high ratio between summing resistor and level/pan control. High resistance values add noise.

Ok, I quickly made a spreadsheet based on the Passive Forsell Mixer, which also comes commercially as SM Pro Audio PM8.

Excel is attached as Zip. I think I got calculations right, numbers look ok to me. But it may have incorrect formulas.

For those without excel, here a screen scrape:



What this calcuates are levels, impedances and noise based on the shown schematic scraped from Fred Forsells PDF.

Assumptions are:

Signal is "0dBFS DAC Output", I assume a Pro Audio output taken between Pin 1 & Pin 2 without shorting Pin 3 to ground, so 1/2 of the common 8V / appx. +20dBu.

Level is set at -6dB (this is included in attenuation calculation)

Pan is set to center

Sum Amp assumes a Op-Amp with noise rated as nV|/Hz.

As is the post sum amp noise is dominant in SNR.

Setting gain to full and using 8V input improves SNR by appx. 12dB

Thor