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Old 10th July 2011, 04:23 AM   #11
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Maybe I'm not understanding what your trying to say, but an attenuator works by dissipating the power, as wasted heat, that would otherwise be dissipated by the speaker. Do you understand that an attenuator in this type of application would have to be a resistor or a resistive voltage divider capable of handling what ever wattage the amp is putting out? The only way there's no power dissipation here is if there's no power output from the amp, as in attenuating the input of the amp, in which case the output attenuator becomes redundant and unnecessary. Maybe a picture would help.

Output attenuator.JPG


The top graphic is a "standard" setup, where the pre-amp attenuates the signal, and all the power from the amp is dissipated in the speaker, and the amount of power output depends on what is put into the amp, i.e."volume control".
The middle graphic is with a resistor in the "Hot" speaker lead, if the resistor is equal to the nominal impedance of the speaker, half the power is dissipated by the resistor and the other half by the speaker by current division.
The bottom graphic has a voltage divider in the speaker circuit, it works similarly to the single resistor version except that instaed of dividing current, it divides voltage. In either of the of the output attenuator circuits, significat power is wasted as heat by the resistors, you can't break the laws of physics.

Mike
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Old 10th July 2011, 08:46 AM   #12
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Quote:
Why would you need an additonal heatsink for the output signal control? Isn't the output power 0 Watts when fully attenuated or fully on and only half of the amps rated power when at half power output?
In the proposed scenario, the output power is always whatever the amplifier is capable of. Normally it's ~all dissipated by the speaker as mechanical energy. To attenuate the output means some of that power is dissipated somewhere other than the speaker. And it will dissipate as heat. "Fully attenuated" means the attenuator is at max power.
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if the amp is unattenuated there is no power dissipated in the attenuator and if output signal is 0 or completely attenuated then no current is flowing and again, no power is dissipated in the attenuator, correct?
Incorrect. The current is flowing in the attenuator...
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Old 10th July 2011, 12:26 PM   #13
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Michael Bean
Thanks for the diagrams but they don't address the question of how you came up with the 90% figure. If you read the earlier posts you would have noted that I didn't say there wasn't any power dissipated in the attenuator but I did assert that it was manageable. When did I break the laws of physics?

sofaspud
I don't understand where I'm incorrect, please explain: Why would the metal heat up if the electrons are not moving through it? If fully attenuated means there is no output signal then there is no current flowing past the attenuator and consequently no power dissipation. There would be the voltage potential of the rails but the electrons won't move past it so no heat. I imagine an attenuator like a water valve. If it's shut then nothing flows, are you asserting that there is power dissipated when there is no current?

Last edited by yldouright; 10th July 2011 at 12:35 PM.
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Old 10th July 2011, 02:07 PM   #14
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First off: Attenuation at the output of an amp is a bad idea. You would loose the benefits of the damping of the low impedance of the amp. The speaker would do what it likes to do, not what it is told (sort of like my son).
Second: These attenuators are available and are used in commercial sound systems where you have zoned level controls. Soundquality is of secondary importance. These attenuators come as wire-wound pots, L and T pads and tapped transformers.
Third: Yes there is heat generated, even with no output to the speaker. As long as there is a signal on the output of the amp the signal is disssipated through the control, generating heat
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Old 10th July 2011, 04:02 PM   #15
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I think this would be useful to read:

What is Gain Structure?

jan didden
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Old 10th July 2011, 05:01 PM   #16
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The 90% figure I used was an example I gave you to show how much power the attenuator would need to dissipate. To keep insisting there's little or no curret flow is wrong, and the water and valve analogy you gave isn't the right way to look at it. I'll elaborate further for you. Let's assume the amp is putting out 100watts into an eight ohm load in the "standard " setup as in graphic #1, and for the sake of simplicity let's also assume the amp is a perfect voltage source and will keep the output voltage constant regardless of whatever load is connected to it. The math behind that using "Ohms Law" is: 100watts X 8ohms = (output voltage squared) or 800, the sqare root of 800 = 28.28volts. 28.28volts / 8 ohms = 3.535amps. To verify, 28.28volts X 3.535amps = 100watts, or 28.28volts squared = 800 / 8ohms = 100watts. So 28.28volts across 8ohms will give 3.535amps of current flow. Now let's assume you want to decrease the appearent loudness by 1/2. The human ear works logarithmically not linearly, so to reduce the apparent sound level to 1/2, power must decrease by 90%, or to 1/10 its original value. Now we know what we need to do to reduce sound level by 1/2. Using graphic #2, to reduce power to the speaker from 100watts to 10 watts would require a decrease in voltage to 8.94volts. To verify: 8.94volts squared = 80 / 8ohms = 10watts, and 8.94volts / 8ohms = 1.12amps. That means you need to drop 19.34volts of the original 28.28volts across the resistor in the "hot" speaker lead. To calculate resistor value: 19.34volts X 1.12 amps = 21.7ohms. 19.34volts squared = 374 / 21.7ohms =17.24watts dissipated in the resistor. Now you're probably thinking; 10watts in the speaker plus 17.24watts in the resistor equals only 27.24watts, right? Where did the rest of the watts go? Here's how it works: with the added 21.7ohm resistor, the total load on the amp is now 29.7ohms, so 28.28volts squared / 29.7ohms = 26.9watts total(figures are off a little due to rounding errors). Also, as was pointed out in another post, adding extra resistance to the speaker circuit will alter the response of the speaker. So, to conclude; using an output attenuator is the hard and wasteful way to lower sound level, and it will alter the sound of the system. Sorry to have to tell you this but if it made sense to do it your way, that is how it would be done.

Mike
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Old 10th July 2011, 08:23 PM   #17
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mickeymoose
The damping problems introduced by attenuating at the outputs are a real issue but that can't they be compensated with speaker design? I believe not all speakers like heavy damping, is this not so?

janneman
How are you my old friend? You are the pragmatist you always were. I did read and believe I understood that article but all the examples given were with small signals so the issue of power dissipation in the attenuator was not to be found within. Am I incorrect when I state that the power dissipated in the attenuator is the current flowing squared times the resistance?

Michael Bean
I believe I understand what you are saying but I am not convinced that power is dissipated in the attenuator unless there is current flowing through it that is being scrubbed (ie: resisted) so that is why I found the 90W out of 100W amp incredulous. It appears to me, the maximum power dissipated in the attenuator is around half the output power because the other part will dissipate in the speaker, am I wrong?
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Old 10th July 2011, 08:31 PM   #18
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The rheostat attenuation should be rather obvious at this point. The water valve analogy will work though for the standard potentiometer volume control, but I don't think your plumbing is correct. If the wiper is the valve, then fully open all the flow goes out of the wiper. When closed all the flow goes to ground. But there's still flow. An attenuator is not a switch ie causes no current flow.
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It appears to me, the maximum power dissipated in the attenuator is around half the output power because the other part will dissipate in the speaker, am I wrong?
That's true only when the impedance of the attenuator and speaker are equal. Reduce the amount dissipated by the speaker and that amount gets dissipated by the attenuator. Power in = power out.
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Last edited by sofaspud; 10th July 2011 at 08:42 PM.
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Old 10th July 2011, 08:36 PM   #19
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Quote:
Originally Posted by yldouright View Post
[snip]janneman
How are you my old friend? You are the pragmatist you always were. I did read and believe I understood that article but all the examples given were with small signals so the issue of power dissipation in the attenuator was not to be found within. Am I incorrect when I state that the power dissipated in the attenuator is the current flowing squared times the resistance?
[snip]
Thanks for the kind words, I'm fine!
The reason I pointed to that article is because it should be clear that using a volume-control type of attenuator at the speaker side is a very bad idea. It only has disadvantages, except maybe the theoretical advantage that it attenuates noise together with the signal thus not deteriorating the S/N ratio.
But in a reasonbly well designed power amp, S/N ratio is never a problem.

Indeed power dissipated is I*V and since V=I*R, power is also I^2*R.

jan didden
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Old 10th July 2011, 08:40 PM   #20
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Quote:
Originally Posted by yldouright View Post
[snip] If fully attenuated means there is no output signal then there is no current flowing past the attenuator and consequently no power dissipation. There would be the voltage potential of the rails but the electrons won't move past it so no heat. I imagine an attenuator like a water valve. If it's shut then nothing flows, are you asserting that there is power dissipated when there is no current?
No but there is a signal across the top to the bottom (ground) of the attenuator itself which is where the current flows through, even if the wiper is at ground so no current goes to the speaker.
This is where the analogy to the water valve breaks down, unfortunately.

jan didden
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