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Old 29th May 2002, 09:46 PM   #1
Bill F. is offline Bill F.  United States
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Default Speaker impedence and amp THD--the connection is...?

This is doubtlessly a very basic question, so I ask it humbly: Why do amplifiers tend to create less distortion into higher-impedence speaker loads?

I have never heard a concise answer to this, and I am terribly curious.

Am I right in assuming this rule applies, in varying degrees, to all amps?

Does it have to do with the fact that, because amps are voltage-driven devices, a wider voltage swing equals a lower noise floor, all else being the same?

Are there other advantages to an amp whose power output is predominantly voltage?

For helping me grasp this basic concept, I thank you.

Bill
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Old 29th May 2002, 10:35 PM   #2
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Most of the distortion in an output stage is
current related, and distortion usually progresses
geometrically with output current.

The result is that half the impedance will probably draw
twice the current, and have maybe 4 times the distortion.
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Old 29th May 2002, 10:40 PM   #3
Bill F. is offline Bill F.  United States
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Thanks, Nelson.

Is there a quick-and-easy answer to *why* distortion rises geometrically with output current?

Assuming a proper design, does current supply not perfectly track the output voltage (ignoring, for a moment, the complex load of a loudspeaker)? If not, why not?

Thanks again,
Bill
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Old 29th May 2002, 11:41 PM   #4
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The problem's primarily in the output stage; transistors
increase distortion, often non-linearly, with current.

This can be minimized by paralleling additional pairs
of transistors to divide the current among the output
pairs and by selecting transistors that are more linear
over a wider current range. Precise tracking of
bias current with Class B and AB amplifiers is
helpful because linearity is much worse at cutoff.

Those aren't all the answers, nor very precise ones;
Douglas Self goes into a fair bit of detail on his
web page:

http://www.dself.demon.co.uk/dipa.htm

Happy reading!
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Old 29th May 2002, 11:50 PM   #5
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Default Re: Speaker impedence and amp THD--the connection is...?

Quote:
Originally posted by Bill F.
Why do amplifiers tend to create less distortion into higher-impedence speaker loads?
In a typical tube amp this would be due to less output transformer between the tube & speaker.

dave
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Old 30th May 2002, 02:00 AM   #6
Bill F. is offline Bill F.  United States
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Thanks for the replies.

Of course, to be perfectly honest, the reason I'm wondering goes back to my transducer design exercises. Fiddling with 4" voice coils (and another altogether unique and tip-top secret driver topology ) it seems to me that 32ohms, 64ohms, etc. would be a cinch to wind. Using Nelson's rule of thumb, the results could be a 90% + reduction in current-related amplifier distortion, across the board.

Sounds like a no-brainer to me. Is tradition (and, of course, marketing) the only things keeping manufacturers from giving us drivers of this impedance?

Bill
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Old 30th May 2002, 02:09 AM   #7
JoeBob is offline JoeBob  Canada
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To get enough power into speakers of that impedence would require alot more voltage than usually used for amplifiers, therefore you'd have to design/build an amplifier just for your driver...
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Old 30th May 2002, 06:10 AM   #8
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Quote:
Originally posted by JoeBob
you'd have to design/build an amplifier just for your driver...
Sounds like a good speaker for a tube amp or an OTL -- you can just buy those. Some of the low impedance OPTs (1.5-3.5k) would serve well for tubes that like higher impedance (6-12k).

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Old 30th May 2002, 03:36 PM   #9
Bill F. is offline Bill F.  United States
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I've been considering building a pair of SOZs. How would I adapt the SOZ for, say, a 64ohm speaker load? Along with higher supply voltage, should bias resistor values double or quadruple? The results should include significantly less distortion and higher thermal efficiency, right?

Bill
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Old 31st May 2002, 08:27 PM   #10
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To have the same power as 8 ohms with 64 ohms, you
would multiply the the voltage by the square root of the
ratio (sqr(8)=2.8) and divide the current by the same (2.8).

The resistor values would be multiplied by the ratio itself (8).
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