Pan Pot Circuit

[80hinhiding]

.

 

[ruffrecords]

See attached file.

Cheers

Ian

 

[abbey road d enfer]

I apologize if this has been covered a bunch of times.  My understanding of this is you can use a linear pot, either single or dual gang for a pan pot. 

I want to understand what's going on.  If I use a 10K linear pot, by having it in the center the signal is already attenuated by 5K of resistance?  Turning one way or the other off center adds resistance to one side and therefore lowers the db level for one side?  How do I achieve this middle is truly middle ground? Typically a pot would be from max resistance to 0 ohm.  Hmm..

The situation you describe is that of a basic DJ mixer fader. A single pot pan-pot operates quite differently, as can be seen in Ian's paper. A noticeable issue is the way the signal fades on the opposite side; let's say you pan right, the left channel doesn't change much until reachin 50% right, then decreases quite abruptly in teh last percents of rotation. That may or may not be an issue. Also, wiper resistance limits the amount of cut-off; not a real problem when the pot is new, but gets worse over time. Add the problem of needing buffering (well, some mixers do without, but they're crappy). There are too many good reasons to use a dual pot.
Now I wouldn't want to discourage you in experimenting; it's good education.

 

[Studio_318]

https://drive.google.com/file/d/0B_n67A1hN3qtaFdhNW8zcmpkZDg/view?usp=sharing

Cheers

Ian

Thanks for this info!

 

[ruffrecords]

I responded to this before the groupdiy.com site went down.  Thanks gentlemen for the help.

Ian, great document!  I think your -3db comes from the 10K + 25K being 1.4 times 25K?  Not quite sure why there's no level loss when hard panned and would like to get a grip on that concept.

At the mid point there is 25K either side of the pot wiper. The top half has 10K in parallel with it so the effective resistance is 7K14. The bottom half has the 47K bus resistor in parallel with it so the effective resistance is 16K3.

Hence the attenuation is 16K3 / (16K3 +7K14) = 0.695 = -3.15dB

When hard panned the wiper is right at the top of the pot. The 47K bus resistor is in parallel with the 50K pot so the load is about 24K but there is no attenuation.

Also wondering why you fed the bus with 47K?  Are 50K pots the best for this or is this something you can do at lower resistance, it just depends on your circuit needs?

Adam

As discussed in the document, the tube line amp can drive a load of not less than 2400 ohms. In a typical mixer this line output might be driving a pan pot and several AUX buses plus a direct output. All these loads in parallel must not be less than 2400 ohms. To ensure a reasonable number of buses and a pan pot and a direct out can be driven it is necessary to use 47K bus resistors. If the line amp could drive a lower resistance load or there was more than one line amp available, then the bus resistors and pots could also be lower in value.

Cheers

Ian

 

[rjuly]

https://drive.google.com/file/d/0B_n67A1hN3qtaFdhNW8zcmpkZDg/view?usp=sharing

Cheers

Ian

Thanks Ian - Just what I was looking for...

One question about the dual gang pot - do the two resistive elements need to be arranged in opposite directions withing the pot, or does one simlpy wire the two sides inversely?

Regards,
Richard

 

[ruffrecords]

Thanks Ian - Just what I was looking for...

One question about the dual gang pot - do the two resistive elements need to be arranged in opposite directions within the pot, or does one simply wire the two sides inversely?

Regards,
Richard

They are standard linear pots. The slugging resistors create the pan law and ensure each one is 3dB to 3.5dB down in the centre position. Notice how in the dual gang pot schematic it looks like I have drawn one pot upside down. This is down so the wipers of both pots move in the same direction on the schematic as the pot is operated. So you wire the signal to the normal top of the first pot and to the bottom of the other one. Also the bottom of the first pot is connected to the top of the other one and this point goes to analogue 0V as in the attached picture (note this does not show the fixed resistors which need to be added from the input to wiper of each pot.

Cheers

Ian

 

[rjuly]

They are standard linear pots. The slugging resistors create the pan law and ensure each one is 3dB to 3.5dB down in the centre position. Notice how in the dual gang pot schematic it looks like I have drawn one pot upside down. This is down so the wipers of both pots move in the same direction on the schematic as the pot is operated. So you wire the signal to the normal top of the first pot and to the bottom of the other one. Also the bottom of the first pot is connected to the top of the other one and this point goes to analogue 0V as in the attached picture (note this does not show the fixed resistors which need to be added from the input to wiper of each pot.

Cheers

Ian

That's the business - thank you.
- Richard

 

[rjuly]

I am planning to use the smart pan circuit on my current console project, but only with pan pots on four of the eight channels, the other four only assigning L,R or both. I take it I can use a simpler circuit for the channels without pan pots, simply having the sitch connect either side or both - that is to say the center level (or centre) will be the same for the channels with and without pan pots.

- Richard

 

[ruffrecords]

I am planning to use the smart pan circuit on my current console project, but only with pan pots on four of the eight channels, the other four only assigning L,R or both. I take it I can use a simpler circuit for the channels without pan pots, simply having the sitch connect either side or both - that is to say the center level (or centre) will be the same for the channels with and without pan pots.

- Richard

That should be fine.

Cheers

Ian