Stereo headphone jack: how can I make it mono?

#12

Joined Nov 30, 2010
18,224
it's sad because it's due to corporate mismanagement
There are conversations on this site about the, "why" of it. You can search for them if you want to. My impression is that as things became less repairable, electronics parts stores were dropping like flies, so RS tried to be a computer store and/or a phone store, but they didn't do that as well as their competitors, so they died.:(
 

#12

Joined Nov 30, 2010
18,224
According to Sony, your headphones have an impedance of 63 ohms. Using 475 ohms will leave you about 12% of the original voltage sent from the MP3 player....down 9.3 db for an apparent result of 1/2 the loudness you would normally expect. Seems to me that resistors in the neighborhood of 62 ohms to 100 ohms would work, too.

It's not a critical circuit. "Whatever you have in stock" sort of a solution will work.
 

AnalogKid

Joined Aug 1, 2013
11,055
You can build a passive summing adapter like this:

The series resistors (475 ohm) seem awfully high for audio output stages driving an ear bud. It will mix the channels, but also probably attenuate them a lot. Ear buds usually land in the 60 to 200 ohm range. What about something in the 47 to 75 ohm range for the resistors?

ak
 

#12

Joined Nov 30, 2010
18,224
The series resistors (475 ohm) seem awfully high for audio output stages driving an ear bud. It will mix the channels, but also probably attenuate them a lot. Ear buds usually land in the 60 to 200 ohm range. What about something in the 47 to 75 ohm range for the resistors?

ak
See also post #22. ;)
Thanks for the confirmation, AK.
 

KJ6EAD

Joined Apr 30, 2011
1,581
The designer claims ≈1k input Z and ≈250Ω output Z. Here's the source. It was designed to feed line level inputs but the topology was what I wanted:

http://www.rane.com/note109.html

I have a small Fiio headphone amplifier that I use to buffer the output if needed. It rarely is.
 
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Line input passive summers are easy. "Headphones" can mean lots of things, but you said "headphone+mp3" player which narrows things a bit. Attenuation is surely possible.

The mpg player generally means a much lower voltage out. It's all about power. P=V^2/R. Lower R increases power. Higher V increases power.

Output and input Z can be measured easily with a scope, signal generator (mp3 file) even) and a variable resistor. Output Z, Plate a resistor in series and adjust for 1/2 output. Measure the potentiometer. That will be your output Z.

Knowing the output Z and an added resistor if needed to increase the output Z) measure the voltage at the headset. Then you can figure it out.
 
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