[mgrewe]

Awesome thanks everyone, I'm going to head over to radioshack and pick up some carbon film resistors cause that's what I've been using so far. I'll let you guys know if it works.

 

[mgrewe]

I ended up putting in two 100k ohm resistors per channel in series and it got better. Still only 1/8 of a turn gets me to a good listening volume. Should there have been a larger effect with adding two 100K ohm resistors in there?

 

[tomb]

Quote:

Originally Posted by mgrewe /img/forum/go_quote.gif I ended up putting in two 100k ohm resistors per channel in series and it got better. Still only 1/8 of a turn gets me to a good listening volume. Should there have been a larger effect with adding two 100K ohm resistors in there?

It depends on the efficiency of your headphones and the strength of your source. Try taking out one resistor each from each channel and see how that works (just a single 100K in each channel).

 

[mgrewe]

Quote:

Originally Posted by tomb /img/forum/go_quote.gif It depends on the efficiency of your headphones and the strength of your source. Try taking out one resistor each from each channel and see how that works (just a single 100K in each channel).

I started with a single and then moved to two resistors per channel.

I'm going from a Marantz CD5003 -> SSMH -> Denon D2000 / AKG K240 sextett. They are great for the sextetts but still pretty sensitive with the Denons.

 

[tomb]

Quote:

Originally Posted by mgrewe /img/forum/go_quote.gif I started with a single and then moved to two resistors per channel. I'm going from a Marantz CD5003 -> SSMH -> Denon D2000 / AKG K240 sextett. They are great for the sextetts but still pretty sensitive with the Denons.

Well, realize that it's counter-intuitive. With parallel resistances, the attenuation is twice as much with one resistor than with two. So, if you started with one and then went to two per channel, you were going from softer to louder.

 

[gore.rubicon]

Quote:

Originally Posted by tomb /img/forum/go_quote.gif Well, realize that it's counter-intuitive. With parallel resistances, the attenuation is twice as much with one resistor than with two. So, if you started with one and then went to two per channel, you were going from softer to louder.

Rmb your physics

1/R = 1/R' + 1/R" and so on

 

[Juaquin]

Yeah, with two 100k in parallel you get 50k. This should be quieter than no resistor.

With a single 100k you get, well, 100k. This should be even quieter.

If that still wasn't enough, you could put two 100k in series to get 200k, which would be very quiet, although I doubt that this should be necessary and I'm not sure if anyone has used that high an input resistor in the SSMH. You may want to examine the source that is providing music to the amp - is it very loud, or amplified?

 

[mgrewe]

Quote:

Originally Posted by tomb /img/forum/go_quote.gif Well, realize that it's counter-intuitive. With parallel resistances, the attenuation is twice as much with one resistor than with two. So, if you started with one and then went to two per channel, you were going from softer to louder.

Believe it or not, but I have them wired in series!

 

[tomb]

Quote:

Originally Posted by mgrewe /img/forum/go_quote.gif Believe it or not, but I have them wired in series!

Oops! Better follow Juaquin's advice and look harder at your source - is it amplified, can it be attenuated or provided with a line out, etc., etc. Also - per chance do you have a 100K volume pot? We've been assuming 50K with our advice on resistor values, but if it's a 100K pot, you multiply all those resistor suggestions X2.

 

[mgrewe]

Quote:

Originally Posted by tomb /img/forum/go_quote.gif Oops! Better follow Juaquin's advice and look harder at your source - is it amplified, can it be attenuated or provided with a line out, etc., etc. Also - per chance do you have a 100K volume pot? We've been assuming 50K with our advice on resistor values, but if it's a 100K pot, you multiply all those resistor suggestions X2.

Nope, unless it's mislabeled then I'm cruisin with a 50k pot. My source is a Marantz CD5003 (analog rca outputs in back) so not sure if that's amplified or not.

 

[Juaquin]

Hmm, I'm looking at the manufacturer's page for the CD5003 and it doesn't seem to have any particular amplification (as long as you're using the regular outputs and not the headphone out). 200K + a 50k pot is way high, though. At that level I don't know if you could hear much of anything from a lot of sources. I don't know a whole lot about audio amplifiers - would there be any negative effect if he increased the input resistors further? I mean, I know in general a good amplifier has a high input resistance but I've never dealt with anything over ~100k.

This may sound stupid, but have you checked the resistor values with a DMM to verify that they really are 100k?

 

[mgrewe]

Quote:

Originally Posted by Juaquin /img/forum/go_quote.gif Hmm, I'm looking at the manufacturer's page for the CD5003 and it doesn't seem to have any particular amplification (as long as you're using the regular outputs and not the headphone out). 200K + a 50k pot is way high, though. At that level I don't know if you could hear much of anything from a lot of sources. I don't know a whole lot about audio amplifiers - would there be any negative effect if he increased the input resistors further? I mean, I know in general a good amplifier has a high input resistance but I've never dealt with anything over ~100k. This may sound stupid, but have you checked the resistor values with a DMM to verify that they really are 100k?

Thanks for checking.

My multimeter hasn't been working while checking resistance. That feature just crapped out on me so I've been relying on Radioshack's word.

 

[Juaquin]

If you're bored, you could hook up a 9V (or whatever voltage battery) to the resistor and measure the current (if that's working). R=V/I. I'm 97% positive that you have the right resistors but you never know. Hopefully someone knows if you can use an even higher input resistance.

 

[mgrewe]

Quote:

Originally Posted by Juaquin /img/forum/go_quote.gif If you're bored, you could hook up a 9V (or whatever voltage battery) to the resistor and measure the current (if that's working). R=V/I. I'm 97% positive that you have the right resistors but you never know. Hopefully someone knows if you can use an even higher input resistance.

I know my volts are working. I'll do this once I get home cause unfortunately I just took off for college yesterday and left my baby back home. I don't have a dac yet and there's not enough room on my desk. Also, I'm letting my brother use it with the Grado's I got him for christmas. But I still would like to figure out this situation.