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#1 |
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diyAudio Member
Join Date: Apr 2005
Location: Leuven
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What's the story on tube amps, power and speaker impedance? When looking at the formula for calculating output power [ Po = (Ve-Va)*(Ve-Va)/(8*loadimpedance) ] one would conclude that the amp has twice the power when coupled to a 4ohm speaker than compared to a 8ohm (using the same opt winding ratio). However I've also read that in the case of a tube amp, contrary to SS, the power increases but doesn't really double.
Or let's put it another way: I have a 2 watts single-ended amp and a diy line array, now wired as a 16 ohm speaker and connected to 8ohm output taps. If I would rewire the speaker to a 8 ohms; would it play twice as loud (thanks to the 3db gain in sensitivity) at the same volume pot setting? Parker audio speakers apparently use this trick (2 ohms speakers) to get more spl out of the 1.5 watt decware zens (probably at the cost of additional amp distortion and reduced damping factor) BTW I'm well aware of the fact that a speaker doesn't have a constant impedance Cheers, Simon |
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#2 | |||
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diyAudio Member
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#3 |
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diyAudio Member
Join Date: Apr 2004
Location: Plainsboro, NJ
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2 Ohm Parker speakers should be used ONLY with DECWARE SE84 amps. The SE84 is atypical for a tube amp. It has single floating taps on the O/P trafos and power production will increase as the load decreases, down to 2 Ohms. I would hope Parker speakers have a "flat' impedance curve.
FWIW, I run a pair of monoblock strapped SE84Bs into 8 Ohm nominal speakers made from 94 dB. efficient drivers. That setup works well. Approx. 8 WPC is a good match for 94 dB. speakers.
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Eli D. |
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#4 | |
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diyAudio Member
Join Date: Apr 2005
Location: Leuven
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Allright, I get the picture after little thinking. So the way is to half the (relative) primary impedance of the opt when drawing the loadline for a 4ohms load and then calculate the power for 8 ohms load?
Quote:
Thanks!! Simon |
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