Another Headphone Resistor Question

[megabigeye]

Hi there,
I've been using my Quickie/Quicksand primarily as a headphone amp and it's actually a pretty enjoyable listen with the Beyerdynamic DT1990, especially since I upgraded the tubes to RCA.  The two caveats are that 1) there is a fair amount of noise from the Quickie, and 2) there is too much power / not enough usable range with the volume control.

I'm wondering if a possible solution to both problems might be to swap out the Quicksand's headphone resistors for a higher value?  And if so, what value might be appropriate?  I'm already using the 1/4W 113Ω resistors.
Will using a higher value alter the sound or cause other deleterious effects?
Does changing resistors affect the speaker output?

Thanks in advance for your help!

 

[Paul Birkeland]

There are some additional mods you can do in your situation, as you have too much gain in the system and need to reduce the sensitivity a bit.

The parts you'll need:

2 x 8 ohm resistors (2-3W should be fine)
2 x 1 ohm resistors (2-3W should be fine)

Replace the existing 113R resistors with the 8R resistors.  You may have to bend the leads at the body and let the resistors poke out away from the jacks.

Connect the 1R resistors across each pair of speaker binding posts.

This will give you significant padding and better headphone damping. 

 

[megabigeye]

Okay.  Thanks.

Just to clarify, you're suggesting that I use lower impedance resistors?  (Lower resistance resistors?  That just sounds redundant, but maybe it's more right?)

Also, by "across each pair of binding posts" you mean between the red and black binding posts of each set?

I wish I understood this stuff more!

Thanks for holding my hand as I bumble along.  Much appreciated!

 

[Paul Birkeland]

Yes, that is correct.

 

[megabigeye]

Awesome!  Thank you so much, PB!  I'll give these mods a try.

Will this alter the speaker output? Either way, I'll give this a go since I'm not using the setup for speakers at the time; sounds like it'd be pretty easy to switch back.

So, I don't want to get too into the science that I'm not going to understand*, but how come the original resistor value is either 0Ω for no headphones (or low sensitivity?) and 113Ω for higher sensitivity 'phones, but then this mod calls for a low value?  The way it was working in my brain is that 0Ω is causing no resistance = more power to headphones/speakers; 113Ω causes more resistance = less power to headphones (and speakers?); consequently, 8Ω would cause only slightly lowered power to headphones.

*I actually do want the science, I just don't want to not understand it.  :  There's a sub-Reddit that's something like "explain things to me like I'm a five year-old" where things are explained so, uh, even a kid could understand.  The couple of times I've stumbled upon it, it's actually been pretty good.  (...And, no, I don't really expect you to hold my hand quite this much)

 

[Paul Birkeland]

If you remove the resistors from the speaker posts, normal amplifier operation will be restored.

The other configuration allows for the full power of the amplifier for demanding headphones, then some series resistors that look a lot like what you would find between speaker posts and headphone jack on a receiver.

For what you're doing, excessive gain is the issue and output power isn't a concern, so the best course of action is to pad down the output of the amplifier.  If you had this same complaint with speakers, I would recommend padding at the input of the power amp.

 

[megabigeye]

Okay.  I think that makes some small amount of sense to me.

Thanks again for your help!

 

[Paul Joppa]

Here's another perspective: it's all about the 1-ohm resistor in conjunction with the 8 ohms. The one ohm is in parallel with the headphone, and absorbs almost all of the current. Thus it pulls the signal voltage at the headphone down to 1/9 of the amp's output voltage. This is the voltage divider effect, also referred to as a "pad."

You could get the same attenuation by keeping the 113-ohm resistor and adding a 14 ohm resistor across the headphone. But for many headphones, especially low impedance ones, a low amplifier output impedance sounds better. By using the 8 ohm/1 ohm combination that output impedance is kept below 1 ohm, which is very low indeed.

 

[megabigeye]

Okay.  I think it's starting to make sense.  I mean, don't ask me to describe it to somebody else, or anything, but right here, right now with your words in front of me...  those words aren't just a jumble.
I swear I'm not such a dullard most of the time (well, that's what my mom tells me, anyway. "Ken, you're not such a dullard most of the time.").

Okay.  Doing a little bit of sleuthing, it looks like the 113Ω currently mounted are Vishay/Dale 1% metal film.  I assume that I want to go with something similar for 1Ω and 8Ω?  And do I absolutely need exactly 8 ohms, or is 8.06 within a tolerably tolerable tolerance?
Also, what the heck is the difference between all those 1 ohm resistors, other than price?  Will any of them do?

Oof.  How anybody keeps all of this information in their head boggles the mind.

Thanks again for your help.

 

[Paul Birkeland]

The values aren't super critical.

For the page of 1 ohm resistors, you have four 2W parts and three 3W parts.

One of the parts is on a reel (cut tape), and other differences are in the metallurgy of the leads.