passive blend knob +sum of L/R circuit help needed

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Jonkan

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Joined
Jun 7, 2004
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Sweden
Hi! So I havent been doing any audio work at all the last few years because of severe tinnitus. Have forgotten even basic electronics it feels like. But now I felt the need for a simple solution to a problem I am having. I need help making a circuit that takes the output from a consumer mp3 player, allows me to blend ch1/ch2 (that is mp3 player left/right) so the user can set how much they want of each signal (basically a panpot really) and then sum that signal into mono so that it gets in both left/right headphones. It doesnt need to be overly precise, its for an “art” project with fairly lo-fi audio sources. Ideally passive circuit.

Hopefully someone here will be able to help out.

I made a sketch explaining it further.

Anyway, thank you and I hope you are having a great day.
/Jonas

 

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Do you mean something like a balance knob?

I would think you would start with a resistor across the input of each side, which would be determined by the input impedance. Then you might have a dual pot and a mix resistor for each side.
 
So you want a variable transition from being fully stereo L -> L and R -> R to fully mono L+R -> L and R+L -> R?

Pretty easy. You just need a dual potentiometer and 4 mix resistors. For passive, you want the resistance to be lowish. 1K would be good but if the source is a headphone output, 500 ohms might be even better. The 4 resistors should be about the same (say 1K).

Code:
Lin------1K------Lout
             |
Rin---       |
     |       |
     P---1K---
     |
     G

Then do the same circuit with the other channel using the other half of the dual pot.

But the taper might not be exactly what you expect (most of the transition happens suddenly at the end of the rotation). It might be that a log potentiometer gives better results. Or you might be able to "slug" the pot to give you a reverse-log taper but I'll skip that for now in the interest of simplicity.

Capiche?
 
I think the OPs source is two mono signals so with the control left he wants source one, with it in the centre he wants source 1 + source 2 and with it right he wants source 2. This is more like the balance control you see on stereo inputs to low end mixers.

Cheers

Ian
 
Jonkan said:
Ideally passive circuit.
Just make sure the potentiometer value is significantly higher than the output impedance of the source. Typically domestic players have an impedance of a few hundred ohms, so a 10-50k pot should do.
 

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ruffrecords said:
I think the OPs source is two mono signals so with the control left he wants source one, with it in the centre he wants source 1 + source 2 and with it right he wants source 2. This is more like the balance control you see on stereo inputs to low end mixers.

Cheers

Ian

Yes! This is what I need. I have a 50k log pot, would that work? What is mixing resistors? Any chance one of you pros could sketch a simple circuit so its easier to grasp?

Like you said i want to balance/blend two mono tracks coming from mp3 player, then combine into mono. 

Do i use a dual gang pot as voltage divider but wire left/right opposite?

Thanks alot
/J
 
Oh, wait i missed the schematic posted above. That would mean I only need a simple regular ole pot? 50-100k, log or lin?

 
abbey road d enfer said:
Just make sure the potentiometer value is significantly higher than the output impedance of the source. Typically domestic players have an impedance of a few hundred ohms, so a 10-50k pot should do.

This circuit works fine with a 50k lin pot, however 80% of the level is around the very edges of the potentiometer movement. Is there any way to make it less so, or do I have to build an active circuit?

Thank you,
Jonas
 
Jonkan said:
This circuit works fine with a 50k lin pot, however 80% of the level is around the very edges of the potentiometer movement. Is there any way to make it less so, or do I have to build an active circuit?

Thank you,
Jonas
Use a lower value for the pot. There is too much loading.
 
Are you coming out the headphones socket of the MP3 player? If you are then you can probably use a 1K LIN pot. Should wor4k much better than a 50K.

Cheers

Ian
 

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