Stepped Pan Controls

GroupDIY Audio Forum

Help Support GroupDIY Audio Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

earthsled

Well-known member
Joined
Dec 1, 2009
Messages
405
Location
Asheville, NC USA
Hi all,

I'd like to get some ideas about building a balanced stereo pan using a rotary switch. So, something like a stepped attenuator, but a stereo pan control instead. I'm considering a 12-position switch but I'm not sure how many decks I would need. I'm also not sure on the circuit or how to calculate the resistor values.

My idea is to build a number of these pans for a passive summing device. For instance, a unit with 24 balanced inputs, pan controls (no volumes), and left/right balanced outputs.

Thanks for your help!
 
Ancient Germans used that approach. http://audio.kubarth.com/rundfunk/index.cgi
 
earthsled said:
Hi all,

I'd like to get some ideas about building a balanced stereo pan using a rotary switch. So, something like a stepped attenuator, but a stereo pan control instead. I'm considering a 12-position switch but I'm not sure how many decks I would need. I'm also not sure on the circuit or how to calculate the resistor values.

My idea is to build a number of these pans for a passive summing device. For instance, a unit with 24 balanced inputs, pan controls (no volumes), and left/right balanced outputs.

Thanks for your help!
basically, you need log-antilog attenuators with -3dB at 12 o'clock. Two wafers for unbalanced. Four for balanced.
You could also use a single-wafer (unbalanced) S-law but the noise performance is always compromised.
 
If this is a summing device, you might just want to switch  A, B B  for each channel which requires a 4p3T switch which is 1 deck and easy to get a hold of and will only cost you about 2 bucks.
If you are summing outs from a DAW you can do the fine panning in the box, you assign a summing channel to A and the next summing channel to B  for a stereo Bus.  this is SOP at least.
AFAIK there is no sonic penalty for doing fine panning and level control  inside the box, it's the summing where you get benefit from a box like this.
 
abbey road d enfer said:
basically, you need log-antilog attenuators with -3dB at 12 o'clock. Two wafers for unbalanced. Four for balanced.
You could also use a single-wafer (unbalanced) S-law but the noise performance is always compromised.

Thanks for your comments!

Okay, with a four-wafer switch, I could build two balanced ladder network attenuators - one for right and one for left.

I have a basic schematic for a balanced ladder network attenuator, but I think it's linear. How should I calculate the resistor values for the log-antilog tapers?
 

Attachments

  • Ladder Network.jpg
    Ladder Network.jpg
    72.6 KB · Views: 152
You have to calculate the law.
First you want to decide for the so-called pan-law: do you want -3dB (constant power) at 12 o'clock, or -6dB (for perfect mono compatibility) or -4.5dB (as most music production mixers do)?
Then you need to calculate the attenuation of every step. There are many possibilities there, see http://www.harmony-central.com/articles/tips/panning_laws/
The pan-law in DAW's is very sophisticated, compared to analog mixers. Most of analog mixers use dual lin pots with law-steering resistors; the result is what it is...But I think it's close enough to the desired curve; I never heard any one really complain about the lack of accuracy of the pan-law on an SSL.
 
abbey road d enfer said:
You have to calculate the law.
First you want to decide for the so-called pan-law: do you want -3dB (constant power) at 12 o'clock, or -6dB (for perfect mono compatibility) or -4.5dB (as most music production mixers do)?
Then you need to calculate the attenuation of every step. There are many possibilities there, see http://www.harmony-central.com/articles/tips/panning_laws/

Okay, well my first attempt at calculating the pan law is based on the formulas for constant power I found here:
http://www.rs-met.com/documents/tutorials/PanRules.pdf

The results give me decimals for each step (left and right) ranging from 0 to 1 that conform to the sine and cosine curves. (So far, so good). Then, I converted the resulting decimals (g) into decibels using the formula 20*LOG(g).

This gives me -3.01 dB in the center (which seems to be right). But, on the extreme end of attenuation, I'm only getting about -23 dB before the channel is switched off. Does this seem correct? (see the attached jpg)



 

Attachments

  • Decibel Calculations.jpg
    Decibel Calculations.jpg
    31.3 KB · Views: 43
Just to add some notes, I found this discussion which offers some alternate formulas for calculating pan-rule values (nice graphs too): http://forum.cockos.com/showthread.php?p=454087#post434029

I double-checked my values and got the exact same results, so I'm pretty confident in that department.

Now the hard part--I can't seem to find any information on how to calculate resistor values for a fully-balanced, stepped attenuator with a taper.

Can anyone out there point me in the right direction?


 
That's the problem with the sine/cosine law, the attenuation lacks resolution at the extremes, that's why some use a square law or a squared cosine law. Even in DAW's, the resolution at extremes is not very good.
 
abbey road d enfer said:
That's the problem with the sine/cosine law, the attenuation lacks resolution at the extremes, that's why some use a square law or a squared cosine law. Even in DAW's, the resolution at extremes is not very good.

Will the stereo image seem evenly-spaced using the sine/cosine law? Or, do you think there will there be a noticeable jump at the extreme left/right step?
 
Well, I think I may have figured out some usable resistance values for this project...

I've converted to using balanced L-pad (or U-pad) design with fixed series and variable shunt resistors.
Something like this:

image81.jpg


With this configuration, I can use a 2-deck switch instead of a 4-decks to make a balanced pan control. (2-deck switches are much more readily available and cost effective).

I found the necessary equations for calculating U-pads at this website:http://www.uneeda-audio.com/pads/

I decided to go with fixed 1k series resistors based on the diagram below. The input and output impedance ratings seem like they will work well for a line-level passive summing device.

attballpad.gif


So now, more questions: I'm not sure how to split the input to the right and left sides of the pan. Should I parallel the mono input signal to two sets of series resistors?

Also, on the output side of the pans, do I need any additional resistors for summing or isolation?

Thanks for your help everyone!

 
I've been waiting for this for a while, my dream summing box is just 24 channels of mono summing with a pan pot on each channel, and maybe 2 inserts on the stereo buss for comp and eq. I'm going to follow this one closely
 
earthsled said:
I decided to go with fixed 1k series resistors based on the diagram below. 
That may be your first problem. I guess you found out that your first attenuation step is something like 0.3-0.4dB. If you want to use 1k resistors, the load impedance must be higher than 50kohms.
So now, more questions: I'm not sure how to split the input to the right and left sides of the pan. Should I parallel the mono input signal to two sets of series resistors?
Yes, no problem, as long as the source impedance is low enough (100ohms max), because at either extreme, the whole pan pot system has an input impedance of 2k.
Also, on the output side of the pans, do I need any additional resistors for summing or isolation?
No, in fact, as I said above, you should make sure there is no additional load.
 
That may be your first problem. I guess you found out that your first attenuation step is something like 0.3-0.4dB. If you want to use 1k resistors, the load impedance must be higher than 50kohms.

Using an 11-step switch with a -3dB sinusoidal pan law, I'm getting a range of about 0.1 dB to 16 dB of attenuation at the extremes.

I'm glad you pointed out the 1k value as an issue - I really need to understand this better. So, excuse me for asking a fundamental question, but how should I calculate the min source and max load impedance of these pan-attenuators? Also, how do I calculate for 8-channels vs. 24-channels?

I would like this pan-bus unit to interface well with modern DAW equipment. A Digi 002 has an output impedance of 50 ohms and an input impedance of 10M (on the line inputs). From what I understand, these specs are pretty forgiving - allowing for a wide range of equipment to be used. So, for modern equipment, what impedance specs should I shoot for?

I noticed that this passive summing schematic is using 5k resistors on the inputs:
http://www.forsselltech.com/downloads/schematics/8chsum_2.pdf
Would 5k be a better choice for series resistors?

I really appreciate your comments!
 
earthsled said:
That may be your first problem. I guess you found out that your first attenuation step is something like 0.3-0.4dB. If you want to use 1k resistors, the load impedance must be higher than 50kohms.

Using an 11-step switch with a -3dB sinusoidal pan law, I'm getting a range of about 0.1 dB to 16 dB of attenuation at the extremes.
haven't done the calc, but don't forget you want one extreme at minus infinity.
I'm glad you pointed out the 1k value as an issue - I really need to understand this better.
The big dilemma here is that in order to produce fractional dB's of attenuation, the ratio of the series resistance (2x 1k in your example) to the mixing resistors (bus injection resistors) must be quite small. If you really want 0.1dB attenuation the mixing resistors must be at least 85k. Then the bus load would be 7k, which is too high for obtaining a good noise factor from a mic amp. But would be good for a dedicated balanced virtual-earth mixer based on 5534's. Anyway, I think the 0.1dB position should be discarded, being totally inaudible. Then the next position is 0.4dB; this would favor the choice of 20k mixing resistors. The bus load would be 1.6k, which many mic preamp will be content with.
So, excuse me for asking a fundamental question, but how should I calculate the min source and max load impedance of these pan-attenuators?
The input impedance must be at least equal to the min. recommended load impedance of the preceding equipment. 50 ohms is the actual source impedance, but min. rec. Z is probably 600-2000R. The min impedance presented by the attenuator is when the shunt resistor is zero; then the impedance is 1200R, which should be adequate for most equipment.
Also, how do I calculate for 8-channels vs. 24-channels?
No difference. But you'll have to crank up the gain of the make-up amp.
I would like this pan-bus unit to interface well with modern DAW equipment. A Digi 002 has an output impedance of 50 ohms and an input impedance of 10M (on the line inputs).
Wouldn't it be 10k ? Anyway, the line input impedance is irrelevant, since mixing involves attenuating all signals a lot , then you neeed a mic input or a dedicated virtual-earth mixing node.
I noticed that this passive summing schematic is using 5k resistors on the inputs:
Would 5k be a better choice for series resistors?
With 5k, the bus load is down to 200R, which will give marginally better noise figure for the make-up amp, but then, the series resistor would be 150R and the load to the sources would be 300R, which most equipment will choke on.
 
I think the 0.1dB position should be discarded, being totally inaudible.
I understand about discarding the 0.1dB step. I can omit the shunt resistance in this position.
Then the next position is 0.4dB; this would favor the choice of 20k mixing resistors. The bus load would be 1.6k, which many mic preamp will be content with.
I'm getting about 39k for the shunt resistance at 0.44dB.(see attached JPG)
The min impedance presented by the attenuator is when the shunt resistor is zero; then the impedance is 1200R, which should be adequate for most equipment.
Is there a formula for calculating this?
Anyway, the line input impedance is irrelevant, since mixing involves attenuating all signals a lot , then you neeed a mic input or a dedicated virtual-earth mixing node.
Okay so, I'll need to have a gain-makeup of some sort following the summing bus. Would an LA-2A offer enough gain make-up?

I am very grateful for your advice!


 

Attachments

  • dB-Ohms.jpg
    dB-Ohms.jpg
    61.8 KB · Views: 91
earthsled said:
I think the 0.1dB position should be discarded, being totally inaudible.
I understand about discarding the 0.1dB step. I can omit the shunt resistance in this position.
You'll have to anyway, pls read thereafter
Then the next position is 0.4dB; this would favor the choice of 20k mixing resistors. The bus load would be 1.6k, which many mic preamp will be content with.
I'm getting about 39k for the shunt resistance at 0.44dB.(see attached JPG)
Remember the mixing resistors are in parallels with the shunt resistor. So in fact you have to correct your results. That means at the initial position already has 0.4dB attenuation (with 2x1k series and 2x20k injection res). So you have to correct your table for +0.4dB attenuation on each position.
The min impedance presented by the attenuator is when the shunt resistor is zero; then the impedance is 1200R, which should be adequate for most equipment.
Is there a formula for calculating this?
No formula needed. The two series resistors end up in series; 600 + 600 = 1200.
Anyway, the line input impedance is irrelevant, since mixing involves attenuating all signals a lot , then you neeed a mic input or a dedicated virtual-earth mixing node.
Okay so, I'll need to have a gain-makeup of some sort following the summing bus. Would an LA-2A offer enough gain make-up?
The attenuation with 24 channel is roughly 1/24, which is ca. 28dB. Is the LA 2A capable of 28dB gain?
 
Ahh! I think I've got it now.

The "injection/mixing resistors" are a fixed value in parallel with the shunt resistors
(I was thinking they were the same thing).

Please have a look at the attached JPG. Is this correct?

 

Attachments

  • injection-res.jpg
    injection-res.jpg
    70.1 KB · Views: 81
No. The bus injection resistors are the resitors that go from the pan arrangement to the bus.
They are seen as a load by the pan attenuator.
 

Attachments

  • pan attenuation001.jpg
    pan attenuation001.jpg
    82.9 KB · Views: 155
Thanks for your drawing -- this helps a lot!

Using my shunt values with the formulas you provided, I recalculated the attenuation per step.

(This table shows the left side only)

 

Attachments

  • new-values.jpg
    new-values.jpg
    82.1 KB · Views: 58

Latest posts

Back
Top