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Old 26th December 2004, 05:13 PM   #1
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Default LED biased EL84 headphone amp

Hello all you wise people out there, I hope you can help me.
I've been tempted to buildthis circuit for a while now, but I've been inspired by SY's comments on LED biased output stages. Well this is what I have so far for the output stage, and I have a few questions. (please bear in mind I'm still fairly new to designing my own circuits)
How is the value of the grid input resistor derived, or if it is needed at all?
Same for the resistor between g2 and the anode on Helmut's design.
How is the value of the resistor accross the output terminals derived?

Thanks for any help,
Steve
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Old 26th December 2004, 09:03 PM   #2
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someone out there must know? Or have I missed something stupidly blatantly obvious?
Steve
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Old 26th December 2004, 09:07 PM   #3
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Quote:
Originally posted by baggystevo82
someone out there must know? Or have I missed something stupidly blatantly obvious?
Steve
What bias voltage are you trying for on the cathode?
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Old 26th December 2004, 09:10 PM   #4
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6.8v
or 4 x 1.7, that's another thing actually, my christmas effected head can't quite get round. As the voltage dropped accross the LED's isn't a product of ohms law, they just have a constant drop anyway (relatively), does that mean that the layout shown in that schematic will have the same drop but with double the current handling of 4 LED's in series on their own?
Cheers,
Steve
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Old 27th December 2004, 10:06 AM   #5
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Quote:
Originally posted by baggystevo82
does that mean that the layout shown in that schematic will have the same drop but with double the current handling of 4 LED's in series on their own?
Yep, 'think so. But I've got a 'Christmas-affected brain' too.
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Old 27th December 2004, 10:50 AM   #6
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Ahh that's good then. I'm about to have my first cup of tea in 4 days the recovery starts here, until new years...
So any ideas or clues about the resistor values?
Cheers,
Steve
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Old 27th December 2004, 11:12 AM   #7
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You're right about the LEDs- just think of them as a battery. Two strings in parallel double the current rating and halve the impedance. Figure that you'll end up with about 10 ohms- not bad!

Make sure that the 1K resistor has a nice, high current rating and you'll be good to go. A 470K value for the input resistor is probably OK, though with the bootstrapping inherent in your CF, you could drop that to 100K without harm. The grid stopper should be somewhere between 1K and 10K, with the body of the resistor as close to the tube pin as possible.
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Old 27th December 2004, 11:35 AM   #8
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Cheers SY. I've just done a bit of searching around and lots of people seem to recommend 1k grid stoppers on EL84's so I think I'll go with that. Any ideas why Helmut's uses only 100ohms here? I'm guessing I should include the 100ohm g2-plate stopper resistor too (is this the right term?), or should I leave it out and only include it if there's oscillation problems?

That resistor accross the output is still a problem for me though. Is it just there to keep the output tied to ground if no headphones are plugged in?

Oh sorry one more, if I drop the value of the grid leak resistor (I think that's its purpose here?) what value of R should I use to calculate the frequency response of the input coupling?

Cheers again!
Steve
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Old 27th December 2004, 11:49 AM   #9
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Quote:
Originally posted by baggystevo82
Any ideas why Helmut's uses only 100ohms here? I'm guessing I should include the 100ohm g2-plate stopper resistor too (is this the right term?), or should I leave it out and only include it if there's oscillation problems?
The resistor's size (when optimized) will vary from application to application. But really, it's pretty non-critical- if you use 4.7K, for example, you'll lose a tiny bit of bandwidth, which is critical only if you're a dolphin. But you'll be nearly guaranteed that the tube won't oscillate due to grid factors. Likewise, the g2 stopper- it may work fine without it, but 100 ohms in series will just eliminate the worry.

Quote:
That resistor across the output is still a problem for me though. Is it just there to keep the output tied to ground if no headphones are plugged in?
Yes.

Quote:
Oh sorry one more, if I drop the value of the grid leak resistor (I think that's its purpose here?) what value of R should I use to calculate the frequency response of the input coupling?
The grid leak is bootstrapped by the cathode resistor. So its effective value is multiplied by the feedback factor. What's the mu of a triode strapped EL84?
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Old 27th December 2004, 12:21 PM   #10
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Cheers for all the help!!

Ive found some UV LED's that have a drop of 3.2v, not sure what 'resistance' they are though.

I'm just looking for its mu now, but the mu of a 'normal' el84 is 19 I think.
Thanks,
Steve
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