Headphone amp using 6SN7 + 6080?

You can build an OTL easily with a 6SN7 gain stage and then a 6080 cathode follower. Doubling the 6080 would halve your Zout indeed.

The BH Crack is pretty much a simple 12au7 gain stage with a 6080 follower running on a 170V B+.

It would be pretty easy to design a very simple OTL with the 6SN7 + 6080
 
Apologies for reviving an old thread, but I've been studying the Wheatfield HA-2 for some time and am wondering why the output capacitor is so large (47uF). Pete Millett himself even suggested boosting it to 100uF earlier in this thread.

Wouldn't the charge time of a cap that large have a negative effect on transient response and sound quality? What if one were to substitute a smaller output cap with a larger load resistor? As it is, the 47uF cap with 4.7K load resistor has a low pass filter of ~0.7Hz:

159155 / 4700 / 47 = 0.72 Hz

What if we use something like a 0.33uF cap with a 470K load resistor?

159155 / 470000 / 0.33 = 1.026 Hz

Wouldn't the quicker charge time of the 0.33uF cap be beneficial here? Just trying to understand.

Thanks.
 
Apologies for reviving an old thread, but I've been studying the Wheatfield HA-2 for some time and am wondering why the output capacitor is so large (47uF). Pete Millett himself even suggested boosting it to 100uF earlier in this thread.

Wouldn't the charge time of a cap that large have a negative effect on transient response and sound quality? What if one were to substitute a smaller output cap with a larger load resistor? As it is, the 47uF cap with 4.7K load resistor has a low pass filter of ~0.7Hz:

159155 / 4700 / 47 = 0.72 Hz

What if we use something like a 0.33uF cap with a 470K load resistor?

159155 / 470000 / 0.33 = 1.026 Hz

Wouldn't the quicker charge time of the 0.33uF cap be beneficial here? Just trying to understand.

Thanks.

Too small and it will roll off low frequency.

The size also depends on the current required by the headphones. As the capacitor charge capacity is what causes current movement across the headphone speaker.

For low impedance you'll see over 3000uF. For high impedance you don't need so much current and so a smaller cap will work.
 
Apologies for reviving an old thread, but I've been studying the Wheatfield HA-2 for some time and am wondering why the output capacitor is so large (47uF). Pete Millett himself even suggested boosting it to 100uF earlier in this thread.

Wouldn't the charge time of a cap that large have a negative effect on transient response and sound quality? What if one were to substitute a smaller output cap with a larger load resistor? As it is, the 47uF cap with 4.7K load resistor has a low pass filter of ~0.7Hz:

159155 / 4700 / 47 = 0.72 Hz

What if we use something like a 0.33uF cap with a 470K load resistor?

159155 / 470000 / 0.33 = 1.026 Hz

Wouldn't the quicker charge time of the 0.33uF cap be beneficial here? Just trying to understand.

Thanks.

Your headphones are in parralel with the load resistor hence the large capacitor requirement for good low end performance. 47uF is good for 300ohm but 100uF would suite a wider range of headphones.
If you'd use it as a preamp and drive a tube grid then 0.33uF could be enough, not for driving a few hundred ohm.
 
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PRR

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Wouldn't the quicker charge time of the 0.33uF cap be beneficial here?
The output cap charge "should not change" over an audio cycle. Or not-much.

Taking 20Hz as a goal this leads to 33uFd for 300r, 100uFd for 100r, 300uFd for 33r, 1000uFd for 8r.

However this neglects amplifier impedance. Many amps have near zero output impedance. But many 6080-type headphone amps are more like 100r Zout. (And we like it that way.) Then since the total resistance in the output circuit can't go below say 132r, there may be little point in more than 100uFd. Actually since 32r is a heavy load on a 100r tube, these amps are more for 300r phones, and we can figure on 400r, and 20uFd is OK, 47uFd is very good.
 
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The output cap charge "should not change" over an audio cycle. Or not-much.

I suspect there's a difference between output cathode voltage 0V (ie futterman) and an elevated cathode output (ie single ended) given Q=CV and Q=It.

So a DC offset and the lowest frequency (including non ideal ESR) should give the tools to work out capacitance for the lowest frequency you want.
 
Did I miss your explanation of how DC offset is factored-in?

I suspect I was thinking in terms of a constant DC offset between output and headphones relates to a constant amount of charge on the film. You then need enough charge capacity (capacitance) for both the DC offset and the largest AC wave form (ie the lowest frequency). I think that is what my thinking was at the time.