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
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
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.
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.
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.
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.
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?
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?
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.
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.
Sunsun22:
AndrewT
What I meant was 1 ohm in parallel with the 62 ohm speakers, and then 8 ohms in series with that. Andrew T has done the math.
AndrewT
Absoluely ! As well as sensitivity of headphones and of ears being factors in selecting the 1 ohm resistor, I should have added method of use. What I do is have my volume control set to what for me is a pleasant value with speakers ( for me about 80 dB), and my wish is to be able to plug in my headphones without changing the volume control. Since my amp is putting out probably less than 1 watt, the 8 and 1 ohm values work for me.
A pentode amp might object to the light load, possibly resulting in higher distortion or even instability in marginal cases. The other immediate downside of eliminating the resistive divider is noise. Headphones are much less tolerant of residual noise due to their very high voltage sensitivity compared to speakers from a typical listening distance.
8 vs 7 ohm series resistor
Sunsun22:
Yes, 7 ohms would be more accurate, but I just happen to have a bunch of 10watt 8 ohm resistors in my stash of parts. The loudspeakers connected to the amp wont have an impedance of exactly 8.00 ohms across all frequencies, so there is no reason for the headphone ciruit too either.
Perhaps you may want to vary the resistance to see if there is an optimum load for the amp. Normally the operating point is a bit of a compromise between maximum power and minimum distortion ( pentodes ), triodes may want a higher resistance as others have mentioned. But for a headphone you dont need much power, so you would not need to settle for a compromise value. Just a thought, not something I have actually tried myself.
Sunsun22:
Yes, 7 ohms would be more accurate, but I just happen to have a bunch of 10watt 8 ohm resistors in my stash of parts. The loudspeakers connected to the amp wont have an impedance of exactly 8.00 ohms across all frequencies, so there is no reason for the headphone ciruit too either.
Perhaps you may want to vary the resistance to see if there is an optimum load for the amp. Normally the operating point is a bit of a compromise between maximum power and minimum distortion ( pentodes ), triodes may want a higher resistance as others have mentioned. But for a headphone you dont need much power, so you would not need to settle for a compromise value. Just a thought, not something I have actually tried myself.
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