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Left / Right / Mono switch

post #1 of 13
Thread Starter 

 

I'm building a switch for headphones to switch between stereo, mono, left-only mono, and right-only mono operation. Obviously the stereo position just passes the signal through. 

 

Since I normally use headphones with an impedance around 30ohms the mono position should have a 20 ohm resistor (or thereabouts) in series with the left + and right + lines before they are summed, correct?

 

With the left-only and right-only mono should I add a resistor between the + contact for the side not being hooked up to the headphones and the ground? If so, what should the value be?

 

 

Thanks!

post #2 of 13

This?


Edited by nikongod - 5/24/11 at 7:06am
post #3 of 13
Quote:
Originally Posted by nikongod View Post

This?



I am wondering if there is a better way to do this.

I imagine a rotary switch like that is expensive.

 

 

Hmmmm

EDIT: I can't even find a switch like that...


Edited by nullstring - 5/24/11 at 8:44am
post #4 of 13

Im pretty sure you can get one from radio-shack for like $3.50

post #5 of 13
Quote:
Originally Posted by nikongod View Post

Im pretty sure you can get one from radio-shack for like $3.50

 

I can't find one on their website.

Or even on mouser for less than $35
 

I did find this.. but it'll cost you $18 after shipping

http://www.stewmac.com/shopby/item/3516?utm_source=google&utm_medium=shopping&utm_campaign=2011-05-gp%22

 

It seems like you should be able to do this with less than four poles..


Edited by nullstring - 5/24/11 at 8:55am
post #6 of 13
post #7 of 13


You need four poles for this.

post #8 of 13
Thread Starter 
Quote:
Originally Posted by nikongod View Post

This?


I've seen this, and it's alright, except I had to redesign it to make it a fully balanced system (if you are clever that means you only need two more poles making it a 6P5T switch). I think you can do an unbalanced switch with 3 poles, but I'd have to think about it for a bit longer.

 

The other issue there is I'm not sure it's the healthiest thing in the world for an amp to leave one channel completely unused, and the resistors seem a bit small for some amps. This design is done better, but the images are gone and I don't think j-curve posts anymore.

 

So my question is still unanswered.

 

Quote:
Originally Posted by nullstring View Post



 

I can't find one on their website.

Or even on mouser for less than $35
 

I did find this.. but it'll cost you $18 after shipping

http://www.stewmac.com/shopby/item/3516?utm_source=google&utm_medium=shopping&utm_campaign=2011-05-gp%22

 

It seems like you should be able to do this with less than four poles..

 

Rotary switches are expensive. You can get a Grayhill that meets the specs for around $20, which is a fine price considering the high quality of those switches.

post #9 of 13

If you add those 20ohm resistors to your mono set up you have now changed the impedence to 50 ohms. As long as your amp is common ground and not isolated I believe you can just conect your right and left positives together with no adverse effects. As far as your left and right setups put a 30 ohm resister to ground to keep the loads the same but use at least a 2 watt resistor on the unused channel. To do this cheaply you should be able to do this with three spdt toggle switches...remember if the ground is common you only have to switch the positives.

post #10 of 13

 

Quote:
Originally Posted by tjohnusa View Post

As long as your amp is common ground and not isolated I believe you can just conect your right and left positives together with no adverse effects. 


I think you need to flip phase in 1 channel for this to work as a mono-system. As configured, it plays the DIFFERENCE in the 2 channels. If both left and right are perfectly in phase (imagine you played a mono song through it) no sound would come out.

 

On that note, I think this circuit will work nicely and cost very little. 

 

LRmono.jpeg

 


Edited by nikongod - 5/25/11 at 5:11pm
post #11 of 13
Thread Starter 

It's actually the opposite of what you are saying nikongod. This gets into mid-side processing. If you sum the left and right together you get what most think of as a mono signal, which is all of the in phase information (what our brain reconstructs as the center image). If you take the difference of the left and ride sides you get the side channel, which is obviously all the out of phase information (what our brain reconstructs as all the sounds at the peripheries). These are used in mixing/mastering. In the mixing process it allows you to do things like artificially make the stereo image wider or narrower for various tracks, while in mastering it allows you to pull out and work with things like vocals or snares which are almost always panned to the center.

 

As for changing the impedance, shorting the two drivers together causes them to be in parallel, which decreases the total impedance (2 / impedance of one side), and since it's an AC signal you need some resistance added or else you will get shorts between the two channels. Most headphone amplifiers now are designed to deal with very low impedances so resistors are put infront of the amp to buffer the outputs, but some are not, so crossing the channels without a buffer is considered to be a no-no. (Some reading: http://www.rane.com/note109.html )

post #12 of 13

I definitely agree that summing the 2 channels creates a mono signal (or a reasonable approximation of one), but I inferred from tjohnusa's post that he was saying that disconnecting the ground at the headphones accomplishes this goal. It does not. Floating the ground cancels all of the in phase signals between the 2 channels. 

 

It would not be too difficult to modify my circuit to include a ground-float switch at the headphones. Replace the center switch (the SPST in my schematic) with a DP3T(on, off, on) wire 1 switch as shown, and the other to switch ground. 

post #13 of 13

What I was thinking was to keep all grounds common and switch only the positives.

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