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#1636 - 03/20/02 05:40 PM What does the "Impedance Selector" actually do?
L. O. Little Offline
Gunslinger

Registered: 02/02/02
Posts: 27
Loc: Fort Worth, Texas
I've been using my 1050 with a (Bryston 4B ST) 2 channel amp for my front channels, with no center channel speaker. My front speakers are B&W Nautilus 804s.

The reason I'm running in phantom mode is because I've been saving for the center channel speaker that "matches" my N804s, the Nautilus HTM1.

Well, I finally got up the cash & ordered my HTM1. It & the N804s have a nominal impedance of 8 ohms, reaching a minimum of 3 ohms.

With that said, I have two questions.

1. Should I set my 1050 for 4 ohm speakers until I can save up for a center channel (Bryston PowerPac 250) amp?

2. What does the "Impedance Selector" actually do?

------------------
Lynn Olan Little
_________________________
Lynn Olan Little
lolittle@earthlink.net

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#1637 - 03/20/02 11:58 PM Re: What does the "Impedance Selector" actually do?
Owl's_Warder Offline
Desperado

Registered: 06/29/01
Posts: 894
Loc: Grants Pass, OR
The point of the impedance selector is impedance matching. Power vs impedance is a bell curve and when impedance of your source matches impedance of your load, that is when you have the greatest power transfer to load (and, consequently, the least power lost). This is the point where your system is operating at optimum conditions.

To the best of my knowledge, the curve is essentially symmetrical on either side of the peak. That would mean if you had 6 ohm speakers, I don't think it matters whether you choose four or eight for the impedance selector. The power transfer should be the same in either case.

Now that I've opened my mouth, if anybody sees something glaringly wrong with this explanation, (or has a better one) feel free to straighten me out.

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#1638 - 03/31/02 06:41 PM Re: What does the "Impedance Selector" actually do?
charlie Offline
Desperado

Registered: 01/14/02
Posts: 1176
Well, you did ask....

:-0

Actually 'impedance matching' is not desirable in this application. The amplifier is intended to act as a perfect voltage source, so ANY resistance in the source (or wire) will combine with any bumps in the speaker impedance curves to produce complementary bumps in voltage delivered to the load (speaker).

In real life the internal resistance of well designed amps and reasonable wire, combined with reasonable speakers present no such problems.

I would guess that the switch might adjust the amplifier rails to an alternate voltage or something, but that's pure speculation.


Charlie
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Charlie

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