[tubaman]

Is it possible to get some kind of device, between headphone jack and headphone, to change the impedance of headphones? I got a Dynalo and when using with UE 4 Pro there's a lot of background noise. The preamp section is perfectly fine, though. It's just when I use the UE 4 Pro I have this noise problem. Unfortunately I have no other headphones at hand to help isolate the problem.

 

[jc9394]

Try fleabay, it may not work on balanced armature drivers.

 

[Steve Eddy]

When you say noise, do you mean like hiss?

se

 

[nickchen]

Smart solution.

I kind of dislike loose connection DIY solutions in this sector, hence I soldered the 200 ohm resistors diectly onto the systems (DT880).

 

[tubaman]

Quote:

Originally Posted by Koyaan I. Sqatsi /img/forum/go_quote.gif When you say noise, do you mean like hiss? se

Yes, hiss is the more accurate description.

 

[tubaman]

Quote:

Originally Posted by jc9394 /img/forum/go_quote.gif Try fleabay, it may not work on balanced armature drivers.

Wow, this is an amazing response, thanks!

I guess the question is, how much impedance do I need...?

 

[Steve Eddy]

Quote:

Originally Posted by tubaman /img/forum/go_quote.gif Yes, hiss is the more accurate description.

Then your problem's not one of impedance, but of gain. The gain of your Dynalo is sufficiently high that its own self noise is being amplified to the point that it's annoying.

The solution short of reducing the gain of the Dynalo is to attenuate the Dynalo's output. While adding a simple series resistor can do this by way of forming a voltage divider with the headphone's impedance, it does so at the expense of significantly increasing the Dynalo's output impedance which isn't necessarily a good thing.

The better approach would be to use a two resistor attenuator that's comprised of a series resistance followed by a shunt resistance of a significantly lower value.

This attenuates the output while being able to still keep the output impedance low. Not as low as it would be without it, but lower than just using a series resistor.

How DIY capable are you?

se

 

[tubaman]

Quote:

Originally Posted by Koyaan I. Sqatsi /img/forum/go_quote.gif Then your problem's not one of impedance, but of gain. The gain of your Dynalo is sufficiently high that its own self noise is being amplified to the point that it's annoying. The solution short of reducing the gain of the Dynalo is to attenuate the Dynalo's output. While adding a simple series resistor can do this by way of forming a voltage divider with the headphone's impedance, it does so at the expense of significantly increasing the Dynalo's output impedance which isn't necessarily a good thing. The better approach would be to use a two resistor attenuator that's comprised of a series resistance followed by a shunt resistance of a significantly lower value. This attenuates the output while being able to still keep the output impedance low. Not as low as it would be without it, but lower than just using a series resistor. How DIY capable are you? se

Thanks for the explanation. I have zero DIY capability, unfortunately. Maybe someone on Head-fi wouldn't mind helping out a zero-DIY capacity headfier while making a few bucks?

 

[Steve Eddy]

Quote:

Originally Posted by tubaman /img/forum/go_quote.gif BTW the person who made the Dynalo just told me he has no problem using his Audio Technica which is at 64 ohm.

Regular headphones don't seem to be quite as sensitive as many IEM's.

Quote:

I guess I'm also missing the reason why the amp would work for a 64 ohm AT or a 300 ohm Sennheiser HD-650 but it would be a problem if I used a resistor to add the impedance.

Because by simply adding a series resistor, you're also increasing the output impedance of the amplifier.

Depending on the impedance characteristics of the headphone in question, driving it from a higher source impedance can do more than just result in attenuation, but can also alter the frequency response.

If the impedance of your IEM's is rather constant across the frequency band, you may be ok using just a series resistor.

I was just suggesting a means of achieving the same result but without increasing the source impedance as seen by the headphone as much as you would with just a series resistor.

se