Guys,
How would I incorporate an extra jack for 32 ohm headphones?
This is designed for 100ohms.
Also would like an output to use this as a linestage if possible.
Pete Millet used the 6080 in a couple of his projects, the HA-1 and HA-2 headphone amps, the data is still available on his personal archive of the Wheatfield Audio website at http://www.pmillett.com/wheatfield.htm
How would I incorporate an extra jack for 32 ohm headphones?
This is designed for 100ohms.
Also would like an output to use this as a linestage if possible.
Pete Millet used the 6080 in a couple of his projects, the HA-1 and HA-2 headphone amps, the data is still available on his personal archive of the Wheatfield Audio website at http://www.pmillett.com/wheatfield.htm
Hi!
You'll want to use at least 250uF as your output dc decoupling capacitor instead of the 47uF in the schematic. 600uF would be even better for 32 ohms.
It should be fine as a line stage too. Just put in a switch so you can choose between a big 250+uF cap for headphones, and a small 0.1 to 0.47uF to feed another amplifier.
Do a search on "Cheapfield" here if you're interested in my own follies. Finishing the HA2 up will be my next project. I just moved to a well used house, so all my DIY efforts are doing there for now.
You'll want to use at least 250uF as your output dc decoupling capacitor instead of the 47uF in the schematic. 600uF would be even better for 32 ohms.
It should be fine as a line stage too. Just put in a switch so you can choose between a big 250+uF cap for headphones, and a small 0.1 to 0.47uF to feed another amplifier.
Do a search on "Cheapfield" here if you're interested in my own follies. Finishing the HA2 up will be my next project. I just moved to a well used house, so all my DIY efforts are doing there for now.
I thought my big 6.3V tranny would be up to supplying the heater on a 6080 and 6sl7gt but i was wrong.
It's voltage is dragged down to 5V when I put these 2 in parallel.
I have a 4Amp 12.6 V tranny but I understand I can only put 2 6.3V tubes in series when they draw the same current.
Is there is a method to put the 6080 and 6ls7gt in series from a 12.6V tranny?
It requires the addition of a shunt resistor across one of the heaters.
This is a new concept to me. Would you be able to point me in the right direction?
Thks
Yup, to run them in series requires similar resistances from each.
So... your 6080 takes 2.5 A, so you have to make the 6SL7 take 2.5 A also. Here goes...
Using Ohms law; 2.5 A from 6.3 V requires a resistance of 2.52 Ohms.
Your 6SL7 presently draws 0.3 A which means that it has a resistance of 21 Ohms.
To turn 21 Ohms into (approx.) 2.52 Ohms you can strap a 3 Ohm resistor across it. The result is a sum total of 2.6 ohms, which is close enough.
The 'excess' current of (2.5 - 0.3 =) 2.2 A will flow through this 3 Ohm resistor. The 3 Ohm resistor will dissipate approximately 14 Watts, which is considerable.
So you need a 3 Ohm resistor with a rating of at least 30 Watts to be safe.
So... your 6080 takes 2.5 A, so you have to make the 6SL7 take 2.5 A also. Here goes...
Using Ohms law; 2.5 A from 6.3 V requires a resistance of 2.52 Ohms.
Your 6SL7 presently draws 0.3 A which means that it has a resistance of 21 Ohms.
To turn 21 Ohms into (approx.) 2.52 Ohms you can strap a 3 Ohm resistor across it. The result is a sum total of 2.6 ohms, which is close enough.
The 'excess' current of (2.5 - 0.3 =) 2.2 A will flow through this 3 Ohm resistor. The 3 Ohm resistor will dissipate approximately 14 Watts, which is considerable.
So you need a 3 Ohm resistor with a rating of at least 30 Watts to be safe.
So you need a 3 Ohm resistor with a rating of at least 30 Watts to be safe.
Excellent explanation. cheers.
Looks like it might be easier to get a new tranny rather than a 30W resistor that will heat my whole apartment.
or the easier way would have been to say that you are looking for a resistor that passes 2.2 amps at 6.3 volts.
E=IR goes to R=E/I
R=6.3/2.2
R=2.9 ohms
P=EI=6.3x2.2
P=13.9 watts
As you can see, use of a shunt resistor is best suited to heaters that are more closely matched.
E=IR goes to R=E/I
R=6.3/2.2
R=2.9 ohms
P=EI=6.3x2.2
P=13.9 watts
As you can see, use of a shunt resistor is best suited to heaters that are more closely matched.
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