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Old 12th June 2007, 05:39 AM   #1
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Default How to connect 62 ohm headphone to 8 ohm output

I just bought an AKG 701 headphone and want to use it with my tube amp (6BQ5 PP) that only has 8 ohm output terminals. Do I need another transformer?

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Old 12th June 2007, 05:48 AM   #2
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Your other option, and there might be a good reason why this is a bad idea, would be to use a power resistor network to limit the maximum voltage to the headphones. Chances are you won't want more than a fraction of a watt to go to them anyway, so the wasted energy shouldn't cause any harm.
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Old 12th June 2007, 07:58 AM   #3
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Yes, this option is also in my mind but what would be the value of the resistor? I don't think this is as simple as 62 (impedance of AKG 701) minus 8 ohms because it is impedance, not resistance.
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Old 12th June 2007, 11:14 AM   #4
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You may find this useful. Same trouble in the neighborhood.

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Old 12th June 2007, 11:44 AM   #5
Hatti is offline Hatti  Germany
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Resistor parallel to headphone:

1/8 = 1/62 + 1/x

1/x = 1/8 - 1/62

x= 9,185... Ohm 2 Watt will be enough, if your Amp only get 1.
If you have much more Power, you have to use a bigger (Watt)resistor.

The most power will go through the 9,185 (9,2) Ohm Resistor.
If this will be enough, to protect your headphone, you have to look!

cheers, Hatti
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Old 12th June 2007, 01:05 PM   #6
AndrewT is offline AndrewT  Scotland
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Hi,
a 10r or 9r1 resistor in parallel with the 62ohm headphones will load the Tube stage correctly.
But it still lets full output voltage to be fed to the headphones.

I would prefer to see a series & parallel combination to both load the output stage to match the 8ohm tapping AND reduce the voltage fed to the phones.
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Old 12th June 2007, 11:41 PM   #7
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Default resistor network (L pad)

What I do is put 8 ohms in series and 1 ohm in parallel. The net result is not exaclty 8 ohms but so what. Your speakers are not exactly 8 ohms either. You could use 7 ohms in series but I dont have that value of resistor. You can also calculate the exact values to use to maintain a specified overall resistance for a specified attenuation ( ie an L pad ) but for this application great accuracy is not required. Depending on the sensitivity of your headphones, and of your ears, you may want to tweak the value of the parallel resistor by listening tests.
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Old 13th June 2007, 12:04 AM   #8
rdf is offline rdf  Canada
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I agree with Robert's recommendation because it also lowers the static noise floor to your headphones. One expansion, if it's a triode-connected amp I would experiment with higher value resistors. The lighter load on the amp will result in lower distortion. Pentodes need the load though.
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Old 13th June 2007, 12:33 AM   #9
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Thank you for all the responses. It looks like resistor is the solution and I'll go ahead to try. Does any resistor work better than others?

Is there any good transformer available to solve this? What about a transformer that has a primary of 8 ohms and a secondary of 62 ohms? What about a transformer that matches the output from the PP 6BQ5 with a secondary of 62 ohmes? Are these transformer available in the market (e.g. Lundahl or Jensen?) or it has to be specially made?


To Robert,

A question about the connection. 8 ohms in series and 1 ohm in parallel will means the headphone is now having (62+8) ohms in parallel with 1 ohm. It doesn't sound right? Isn't it?
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Old 13th June 2007, 08:26 AM   #10
AndrewT is offline AndrewT  Scotland
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Hi,
1r0 //62ohm ~=0r98
amp sees 8r0+0r98~=8r98
The headphones look back and see 1r0//(8r0+Zout),
since Zout is fairly low the source resistance seen by the headphones is ~=0r89
The voltage felt by the headphones is approximately 0r98/(8r98)=0.109 (-19.2db). This is potentially very loud. A 10W valve amp will still send 100mW to the headphones.
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