Headphones with a Tube Amp

[bwarden]

I'm new to this whole DIY thing, and am in the process of building my first tube amp. Because I have one speaker that needs refurbing, I wanted to be able to use headphones with the amp for a while.

I searched here and googled, and found a link at Elliot Sound Products in Australia (http://sound.westhost.com/project100.htm) for a Headphone Adapter for Power Amplifiers. The one-channel schematic is shown on the link above.

The resistance values quoted are for matching a 120 ohm impedance headphone, and a maximum 5v. My headphones are AKG K271's, with the following specs:

91 dB SPL/mW Sensitivity
200 mW Max Input Power
55 ohms Rated Impedance

The amp I am building will produce on watt or less (more on that later).

So, in trying to understand the schematic, I need to know if the following is correct:

1. To match the impedance, the R2 resistor needs to be 55 ohms? Or was R2 just coincidentally set to 120?

2. R1 and R2 together are a voltage divider.

a. If I want 200mW max, at 55 ohms, the max voltage would
be E = Sqrt(P x R) or E = Sqrt(.2 * 55)=3.3v

b. But if I produce 1W at 8 ohms, the amp voltage is E =
Sqrt(1*8) = 2.8v, so I don't need to do anything with the
voltage. Does this mean both R1 and R2 are out of the
circuit?

c. R3 seemed to be used to bring R1 and R3 back up to the 120
ohm level in the example. So if no R1 (and R2) are out, then
R3 is 55 ohms?

So, being a slightly confused newbie, I get to the 55 ohm resistor answer, I'm just a little unsure if I need to put it across the outputs, or in series, or both?

Thanks in advance for putting up with what is likely to be a stupid question.

-----------
Bill

[kevinkr]

To start I would just add some parallel resistance across the secondary of the output transformer to assure it never operates with no load attached. Something on the order of 12 - 16 ohms ought to be ok as a first approximation.

91dB/mW really means that you aren't going to need more than say 300mVrms (16mW) across the headphones to be as loud as you will ever need - this will get you in the vicinity of 102dBspl. In all likelihood you should attenuate the signal to your headphones by as much as 20dB to protect your hearing and in the event that the noise floor is not as low as you would like this will buy you the same improvement in noise levels. I'd try a ~470 resistor in series with the phones.

One other alternative would be a 10 ohm/1 ohm resistive divider driving the headphones, this will give you a low driving source impedance at the phones (1 ohm) which might or might not have any effect on damping in the headphone transducers and still provide a moderate load (~11 ohms) to the amplifier. (The 12 - 16 ohm resistor previously mentioned is not required as long as this network is hardwired across the output.)

(I have done both above and could not decide which was better, although the 11:1 attenuator was very definitely quieter on the LF noise front. My Sony HP are 64 ohms.)

Lots of experimentation may be required to get to the point where it sounds right. (Resistor type and value.)

Given the impedance and sensitivity of these phones your ears could be in jeopardy so be very cautious when testing.

This is neither a stupid nor trivial question, driving headphones well with small (se) tube amplifiers is not as simple as it seems. Hopefully I have adequately addressed the 3 major issues I've encountered doing this. (Amplifier protection, excessive spls, and noise)

[bwarden]

Kevin, thanks for the help. I tried what I guess is both at the same time - 10:1 divider, and the 470 ohm resister in series. It works, but the volume is low, and cannot reach loud even at full tilt.

So, in rereading your post, it occurred to me that maybe you didn't mean to both divide the voltage and attenuate 20 dB with the 470. Was I incorrect doing both at the same time?

Second, given the very good idea of protecting my ears, which do I fiddle to up the level a bit (your 'experimentation' comment)? The divider or the attenuation resistor, or both?

Thanks again for the help.

And also thanks to the others in this thread How to connect 62 ohm headphone to 8 ohm output working out a similar problem.

[kevinkr]

Originally posted by bwarden
Kevin, thanks for the help. I tried what I guess is both at the same time - 10:1 divider, and the 470 ohm resister in series. It works, but the volume is low, and cannot reach loud even at full tilt.

So, in rereading your post, it occurred to me that maybe you didn't mean to both divide the voltage and attenuate 20 dB with the 470. Was I incorrect doing both at the same time?

Second, given the very good idea of protecting my ears, which do I fiddle to up the level a bit (your 'experimentation' comment)? The divider or the attenuation resistor, or both?

Thanks again for the help.

And also thanks to the others in this thread How to connect 62 ohm headphone to 8 ohm output working out a similar problem.

You'd use one or the other, not both. (Using both would result in extremely low output levels.) Personally I think the divider is probably the better way to go as it drives the headphone with a source impedance of somewhat less than 1 ohm. Both approaches incidentally will provide attenuation, but the divider should result in a flatter overall frequency response depending on driver inductance and capacitance.

[bwarden]

Well, better double than half as much. I'm a real belt and suspenders kinda guy anyhow.

Removed the 470, now has plenty of volume, at roughly the volume level of the speakers when attached.

Thanks again.

Bill

[jcdrisc]

My approach for headphones is to use a MOSFET current sourced follower
as an output stage and a tube as the voltage amplifer. So plenty of drive and the harmonic profile of the tube combine ideally. Otherwise pay big bucks for an expensive transformer.

[ChrisA]

This is for a tube amp? If so I think it's best to always ensure the load impedance matches the transformer. OK you might get away with a big mis-match but just don't turn it up.

I think best to place an 8R power resistor accros the speaker terminals and then wire the headphone and a variable resistor in parallel tot he 8R load.

[ruffrecords]

[QUOTE=kevinkr;1230769]

91dB/mW really means that you aren

[ruffrecords]

Originally Posted by kevinkr

91dB/mW really means that you aren't going to need more than say 300mVrms (16mW) across the headphones to be as loud as you will ever need - this will get you in the vicinity of 102dBspl.

According to my calculator, 16mW into 55 ohms meeds 940mV:

0.94*0.94/55 = 0.016W = 16mW

Cheers

Ian

[DualTriode]

Hello All,

This post is about tube amplifiers and headphones.
It seems that headphones are most often a bolt on afterthought added to a power amplifier. Then the headphone bolt on is a one size fits all network of resistors and voltage dividers.

When thinking of a power amplifier driving a cone speaker we think of output impedance and dampening factor to control the Q of the speaker cone. It is the exception rather than the rule to discuss those thoughts in terms of headphones.

When a resistor network is placed between the power amplifier and the headphones the small load of the headphones is lost compared to the connected resistive load. There is little or no interaction between the headphone driver and the amplifier as you would see with the typical power amplifier and driven cone speaker.

It seems to maximize the performance of a tube headphone amplifier and headphone pair the amplifier and specific headphones should be designed to play together. The output transformer should match the load impedance to the plate impedance of the output tube(s) without resistors in between IMHO!

If you do not want to let out the magic smoke do not unplug the headphones when the amplifier is turned on!

Any thoughts? Are headphone connections to power amplifiers one size fits all after thoughts?

DT
All just for fun!