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Old 8th May 2007, 02:18 AM   #101
dimitri is offline dimitri  United States
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Quote:
However I am now stuck with a .3 ohm resistor in each output of my power amps.
I presume, noninductive ones?
 
Old 8th May 2007, 02:50 AM   #102
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Yes, they parallel my Bybee devices.
 
Old 8th May 2007, 02:55 AM   #103
dimitri is offline dimitri  United States
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Quote:
Yes, they parallel my Bybee devices.

where you got .6Ohm bybee? Mine are 25mOhm
or you mean .3Ohm in parallel with bybee?
 
Old 8th May 2007, 02:56 AM   #104
Hartono is offline Hartono  Indonesia
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Old 8th May 2007, 03:18 AM   #105
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I have OLD Bybee devices. They use .3 ohm.
 
Old 8th May 2007, 03:47 AM   #106
dimitri is offline dimitri  United States
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Again the same trick. You had already accumulated with Erno all valuable old Toshiba JFETs. The leftovers become 10 times more expensive than before. And now OLD bybees.
 
Old 8th May 2007, 03:51 AM   #107
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Quote:
Originally posted by john curl
.5uH only for Halcro. Why don't you folks stop talking and start calculating?
John,

Are you referring to the large coil that looks like it is made of
copper tubing and has a large brass fitting at one end?

I believe it is also to block external RF from entering the amp.

Interesting amp for sure, however it looks like the much cheaper
balanced Bruno Putzeys amp is pretty close in measured performance.

cheers

Terry
 
Old 8th May 2007, 03:53 AM   #108
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Calculate the inductance from the MEASURED 20KHz impedance as measured by 'Stereophile'. What is wrong with everybody? Are there no engineers here?
 
Old 8th May 2007, 04:05 AM   #109
andy_c is offline andy_c  United States
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Quote:
Originally posted by john curl
Calculate the inductance from the MEASURED 20KHz impedance as measured by 'Stereophile'. What is wrong with everybody? Are there no engineers here?
John, I get 0.78uH. I got this from:

L = sqrt(0.14^2 - 0.1^2) / (2 * pi * 20e3)

R = 0.1 and

sqrt(R^2 + X^2) = 0.14 @ 20 kHz and

X = omega * L (neglect the parallel R)

This is for the DM88
 
Old 8th May 2007, 04:12 AM   #110
GK is offline GK  Australia
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Quote:
Originally posted by john curl
.5uH only for Halcro. Why don't you folks stop talking and start calculating?

So what John? It's still a large air wound inductor, isn't it?
You have made the claim than any output inductor compromises an amplifiers sonic ability. Are 0.5uH inductors now all of a sudden OK in your opinion?

Cheers,
Glen
 

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