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Old 12th July 2011, 04:05 PM   #31
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Jan, I think he's trying to achieve noise reduction with an output attenuator, so in that respect what he wants to do won't work. I also tried to explain to him that while you can use an attenuator that way, it is terribly impracticle, just because you can doesn't mean you should, and that's why we use preamps with volume controls.

Mike
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Old 13th July 2011, 03:49 AM   #32
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Michael Bean
If you had read what I had stated about the power dissipated being a function of the current passing through it you would have noted that your math would have been invalidated if there were a way to avoid losing the power to ground. You did not directly address this possibility and I daresay that you sidestepped it with the math you provided. Lesser minds will typically fall back to elementary explanations when their mind is forced to bend into an unfamiliar contortion but let's stop sniping and asking each other to reread things we have already stated we read and continue this exploratory discussion. If the attenuator is adding noise of 30uV which is amplified 20X, then the whole system minus the attenuator at the input must be comparatively less noisy at the output because of the much larger S/N ratio, no? Okay, let's recount the positions; the most annoying problem of attenuating the output signal is the power dissipation according to both you and janneman. You both reject any possibility of the attenuator being a garden variety pot and both agree that this type of attenuator at the output cannot be used without burning power into ground. Have I understood both of your positions clearly? Now, does this same problem exist for all other types of attenuation?

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Old 13th July 2011, 05:03 AM   #33
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OK, i'm confused. I thought you were talking about attenuating the signal between the amplifier output and the speaker, but in your example where you say "the attenuator is adding noise of 30uV which is amplified 20X" I guess you're talking about an attenuator placed before the amplifier? I need clarification on this point.

Mike
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Old 13th July 2011, 07:33 AM   #34
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yldouright, you can ignore me, physics, electronics, tradition, etc. etc. But we won't go away...
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Old 13th July 2011, 03:46 PM   #35
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sofaspud
I hope I'm not giving the impression that I'm averse to learning. I just don't want to learn something wrong or incompletely. I assure that I've no intention of ignoring anything useful. I hope others reading this agree.

Michael Bean
Don't be confused. I was showing you why an attenuator after amplification might have a sonic benefit by comparison.

janneman
What prevents us from having an attenuating pot in series without a ground at the output?
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Old 13th July 2011, 04:54 PM   #36
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OK, so the attenuator is after the amplifier. So please explain to me how "If the attenuator is adding noise of 30uV which is amplified 20X" is true if it's after the amplifier, it would have to be before the amplifier for that to happen. Your statements are contradictory.
Also, a potentiometer that is not connected to ground as a voltage divider (diagram 3 in post #11) is called a rheostat or variable resistor (refer to diagram 2), and will drop votage depending on how much current it's conducting. As was already pointed out, adding resistance to the circut will alter the damping factor, and will change with different settings. Power will be dissapated as heat in any case.
I mean no disrespect, but it seems you don't understand Ohms Law. Google it and study, and the answers you seek will be self evident.

Mike
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Old 13th July 2011, 07:50 PM   #37
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Michael Bean
You are absolutely right, you are confused. I can only assume this is because you don't understand the comparative example I presented to you. Please reread my earlier post and tell me if you get it now. While you are trying to become more familiar with the English language, I will try and become more familiar with Ohm's Law. I will repeat that I never said there would be no power dissipated in an attenuator at all times when placed at the amp output. I proposed the maximum power dissipation would be roughly half the amp power output because no current will flow at full attenuation and there is nearly zero resistance at full throttle. I asked if this position was refuted and still haven't received a direct reply. There is nothing contradictory in anything I have stated and the aim of this thread is still straight even though it has been buffeted by comments aimed to throw it off track.
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Old 13th July 2011, 08:40 PM   #38
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I was only trying to help. There is a huge store of knowlege and experience available here free for the taking. If you want further assistance from anyone else I suggest you drop the attitude and insults, stop questioning other's credentials and intelligence. Good luck, I'm outta here.

Mike
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Old 14th July 2011, 12:19 AM   #39
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Using only an in-line resistor, as suggested by the OP:
If you reduce the loudness to 1/2 (-3dB) you need an 8 Ohm resistor, 32 Ohms to go to 1/4 and 80k Ohms to go to -40dB.
Show me the pot (no, not BC bud)! E
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Old 14th July 2011, 02:44 AM   #40
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mickeymoose
I can't think of a traditional pot with a wiper that can work like this. No high power amp driving a typical speaker can be controlled this way but the situation could change if we had a line of speakers like in a stadium setting. Okay, I'm ready to rule out wiper pots what about switches?
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