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Old 10th February 2008, 05:51 AM   #11
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I'm a little bit confused. Why doesn't a "fixed l-pad" structure work at the amp input if you want to prevent excess input?
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Old 10th February 2008, 07:55 AM   #12
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Quote:
Originally posted by danielwritesbac
I'm a little bit confused. Why doesn't a "fixed l-pad" structure work at the amp input if you want to prevent excess input?
I'm sorry, I'm not quite familiar with what a "fixed l-pad" is... Is there something I'm missing here? I'm going to end up putting some kind of attenuator in the signal path anyway, I figure I might as well stick it in the preamp and get a lot of gain control flexibility, which will be nice to have with my headphones as well.
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Old 10th February 2008, 07:55 AM   #13
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Sorry for the double post...it's not letting me edit. But I just learned that a fixed l-pad is a big fancy name for "voltage divider." And you're right, I don't see any reason why that wouldn't work. That saves me a decent chunk of money as well, since I don't have to buy another ALPS, another knob, and all that jazz.

I think that I'll go ahead and lower my preamp gain while I'm at it, because it seems to be a bit too high anyway. I'll probably bring it down to about the 5-6 range. What kind of resistors and what kind of attenuation should I be looking for here? I see a lot of stuff floating around about carbon vs. metal film, and I'm not quite sure what to believe. Would I want something like a 0.5k and a 1k resistor, or would I need more attenuation than that? I know how to make a voltage divider, but I'm pretty inexperienced when it comes to the right types of components and the specific values that would be appropriate for audio. Thanks for the help.

If it makes any difference, I think I'm going to go with the audiosector stereo classic kit that uses LM3875's. I read around and people seem to prefer it a bit. It's also cheaper.
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Old 10th February 2008, 08:22 AM   #14
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Instead of a potentiometer at the amplifier input?

The sum of two resistors would be between 20k and 30k. This represents the 25k potentiometer shown as an optional component.

There's no need to modify the preamp, and every reason not alter it, if you already like the way it sounds.

Well, I think that's right. Can someone double-check my figures?

EDIT: Alpa A20K Dual Gang isn't bad for a potentiometer. You could use it, play with it, and find your most frequently preferred setting. Then you could remove it, measure it, and replace it with resistors that represent the same setting.
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Old 10th February 2008, 09:47 AM   #15
AndrewT is offline AndrewT  Scotland
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Hi,
the disadvantage of using a low attenuation fixed resistor voltage divider is that either the input impedance is low and difficult to drive or the output impedance is high and becomes a high Rs that the following equipment may not tolerate well.
High attenuation factors avoid most of this compromising of Zin and Rs

eg. 10 +1k1 gives -20db, with Zin~=11k and Rs~=1k.
but 10k + 5k1 gives -9.4db, with Zin~=15k and Rs~=3k4
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Old 10th February 2008, 07:57 PM   #16
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Quote:
Originally posted by danielwritesbac
Instead of a potentiometer at the amplifier input?

The sum of two resistors would be between 20k and 30k. This represents the 25k potentiometer shown as an optional component.

There's no need to modify the preamp, and every reason not alter it, if you already like the way it sounds.

Well, I think that's right. Can someone double-check my figures?

EDIT: Alpa A20K Dual Gang isn't bad for a potentiometer. You could use it, play with it, and find your most frequently preferred setting. Then you could remove it, measure it, and replace it with resistors that represent the same setting.
No, I'll lower the gain of the preamp AND put something at the power amp input. The preamp already sounds great, but I'm changing it because 1) I like to tweak things and 2) the volume in my headphones is comfortable at about 1/6 of the volume pot's rotation. It'll just give me a larger range of useful knob turnage.

I was thinking I might use smaller resistors because I heard that less current means more resistor noise. Maybe that's only true in the Megaohm range? Having low resistor values isn't a problem in terms of driving current, since my preamp is a class A headphone amplifier capable of driving loads in the 16ohm range.

I'll check out the pot you mentioned, thanks.

AndrewT, could you explain why a resistor divider at the input would affect the output resistance?
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Old 10th February 2008, 09:51 PM   #17
BWRX is offline BWRX  United States
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Quote:
Originally posted by johan851
I was thinking I might use smaller resistors because I heard that less current means more resistor noise. Maybe that's only true in the Megaohm range?
Noise Analysis - Resistor Example
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Old 10th February 2008, 10:03 PM   #18
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Great link. Thanks BWRX.
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Old 11th February 2008, 09:08 AM   #19
AndrewT is offline AndrewT  Scotland
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Quote:
Originally posted by johan851
could you explain why a resistor divider at the input would affect the output resistance?
Hi,
use the diagram in that link that BWRX sent you.
Add the source resistance for the preamp to V1. let Rs=1r0
Rlp1=Zin=22k

With r1=0r0 and r2=1M the power amp sees at it's input Rs'= [R1+Rs]//R2
The preamp sees Zin of the power amp //R2 and before it gets there it also passes through R1 so Zin'=R1+[R2//Zin]

Now set you resistors to attenuate by that 9.4db mentioned earlier.
R1=10k, R2=5k1
Rs'=[R1+Rs]//R2=[10k+1r0]//5k1=3378.
Zin'=R1+[R2//Zin]=10k+[5k1//22k]=14k14
Adding the 9.4db attenuator has changed Rs' from 1r0 to 3k4 and changed Zin from 22k to 14k

If you adjust the attenuating resistors downwards then Rs' and Zin' both become lower.
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Old 11th February 2008, 07:25 PM   #20
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Thanks for walking through that. Now that I figured out what your resistor notation means (I'm used to doing this on paper, were I would just say 14.14kohm instead of 14k14) I understand what you're saying.

So what you're saying is that the higher the impedance seen by the preamp is (Zin') the more noise I'm going to have. But I also need to make sure that Zin' doesn't go so low that my preamp can't drive it. Correct?

I'm saying that since my preamp is designed to drive loads as low as around 20r0, it makes sense to have lower resistor values in the attenuator so that I reduce the amount of noise produced by the attenuator. That doesn't mean I want extremely low values...maybe R1 = 5k1, R2 = 10k.

The point I'm still missing, I think, is why Rs' matters. Does the impedance at the amp input / preamp output as seen by the power amp actually matter?
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