Headphone impedance and power

[TerryS]

Not sure if this belongs here or in DIY, but I'll start here.
I'm considering designing and building a headphone amp. A very capable headphone amp. Maybe even an over the top headphone amp. Actually, it will be a 'trial run' before I go after designing a power amplifier.
My question is what range of headphone impedances and power ratings should I consider? I know headphones range anywhere from as low as 8 Ohms to as much as maybe 600 Ohms. Is higher than that common?
I currently have AKG K-701s. They are 62 Ohms, with 200 mW maximum power rating. That implies they need about 10 Vp-p signal swing and 80 mA peak current to drive them to full power. Higher impedance phones will need a higher voltage to drive them, and lower impedance phones will need more current.
Beyerdynamic DT-880 are 600 Ohm at 100 mW phones, so they need 22 Vp-p at 20 mA.
I don't see much in the way of 8 Ohm specs, but even assuming a totally absurd power rating of 2 W, the current required would be 700 mA peak. What is the maximum power level I should budget for? What are the most power hungry (least efficient) phones? I don't need to cover every possible phone, just the range I am most likely to ever want to use.

Terry

 

[Jon_Logan]

If you need to swing 22V for your 600 ohm, 20mA scenario, and you want to drive 8 ohm headphones too, then you've established what you need the amp to do.
You might want to check your assumptions about P-P voltage. There will be losses across the output devices and their emmiter/Drain resistors if used.

Your highest voltage swing will be for the Beyer DT-880's. The voltage across those appears to be ~7.75V(rms). You divide that by 0.5 to get ~3.875V(rms). Now you multiply that by approx 1.41 (+3dB) to get ~5.5V peak. You add some margin to account for losses across the output device(s) and you get about +/- 8 volt rails.

Your highest current is going to be the 8 ohm/2W, or 500mA

So, it looks like you can do it with a +/-8v supply with 500mA, or more, of current capability.

Check that, I think +/- 15 volt supplies will get you there.

 

[TerryS]

The voltage swing I was talking about is the actual swing into the phones, so like you point out, I will have to allow for the drop in the output stage. The Beyer DT-880s say 600 Ohm with 100mW max power. That means I need the 7.75 Vrms, which is almost 22V p-p or 11 V peak. So probably 15 volts supply rails. I'm not sure where you are going by dividing the rms voltage by two.
I will have to design for the highest peak current, which at 8 Ohms/2W is 700 mA. That's where it gets a little painful. If I use 15V rails and want 700 mA peak current available in class A, I will have to bias the outputs at 350mA. That means 5.25 Watts of idle power in each of the outputs (so 10.5W total). Not undoable, but I wonder if that guess at the maximum power (2W) is reasonable.

terry

 

[Jon_Logan]

The voltage swing I was talking about is the actual swing into the phones, so like you point out, I will have to allow for the drop in the output stage. The Beyer DT-880s say 600 Ohm with 100mW max power. That means I need the 7.75 Vrms, which is almost 22V p-p or 11 V peak. So probably 15 volts supply rails. I'm not sure where you are going by dividing the rms voltage by two.
I will have to design for the highest peak current, which at 8 Ohms/2W is 700 mA. That's where it gets a little painful. If I use 15V rails and want 700 mA peak current available in class A, I will have to bias the outputs at 350mA. That means 5.25 Watts of idle power in each of the outputs (so 10.5W total). Not undoable, but I wonder if that guess at the maximum power (2W) is reasonable.

terry

Yeah, the "divide by two" was a brain fart. I missed the Class A requirement. Class A is much easier( cooler) to do into 600 ohms than into 8 ohms. You want full Class A into an 8 ohm load?

It'd be fun to have a selector that switched the rail voltage and the bias, depending on what phones you wanted to plug in.

 

[TerryS]

I thought about a selector like you suggest, although if I can keep the dissipation to a level that I can deal with in a reasonably sized case, I probably won't bother.
Right now I'm looking at about 20W of heat by the time I account for power supply losses and such.

Terry

 

[TerryS]

I found this summary of specs:
http://gilmore2.chem.northwestern.edu/tech/dbohn1_table.htm
The highest power rating I see is 450mW, and the lowest impedance is 32 Ohms.
I know vintage phones were often 8 Ohms, but maybe not anymore.
The highest power number I have seen mentioned for a headphone amp is 1W @ 4 Ohms, which is 710 mA peak.
On the absurd end of the range, I see a Beyerdynamic DT 100, which is available in 400 Ohms impedance with 1000mW power handling capability. This would require 57 Vp-p signal swing! But they also list the SPL at 94dB @ 1mW. So 1000mW would be 124 dB SPL, which would cause nearly instant hearing damage, so maybe no need to be quite that nuts.

Terry

 

[Blackbeerd]

I think you're approaching this entirely wrong.

Look at the sensitivity in mV/dB or mW/dB and then figure out the highest level you want peaks in your music to be at - add in a bit of a safety factor, calculate the voltage swing you need, make sure the amplifier can handle the current needed to reach that, and you're set. I see no legitimate reason to use the headphone's maximum power handling capability as the basis for choosing an amplifier.

 

[TerryS]

mmm....logic. That isn't going to cut it it audio design.
I want to be able to drive any (or at least most) available headphones to their maximum capability. If that causes my ears to bleed, so be it. That's what the volume knob is for. ;-)

Terry

 

[Rex Aeterna]

if it for home use only, just save yourself the trouble and make a speaker box for the headphones. add the lowest possible resistors for lowest possible output impedance you feel safe with(or don't use any resistors at all if you want) and just drive your headphones from speaker outputs. with speaker outputs the headphones impedance dips and spikes will mean nothing and you will never run out of power driving headphones.

just an idea to save you the hassle of worrying bout trying to build the perfect headamp to ''power them all''.

 

[dewdude]

I did something similar to this with my console tube amps. I wanted to drive headphones with them...so I put some 8ohm non-inductive resistors across the speaker terminals...added a 100ohm dropping resistor and volume control. (This is all in a project box...not on the amp itself)

I didn't calculate how to impedence match because it was just a quickie project and I don't know much about it.

They're 32 ohm cans. The tube hum and "grid crinkle" is quite noticable...something I could maybe avoid matching the resistors better. But it works. The cans don't seem to mind...the amps don't seem to mind.

But I'm also considering building a headphone amp...so its something I'm about to study.

Sent from my SPH-M910 using Tapatalk

 

[TerryS]

In my case the point of this (if I decide to go ahead with it) is to build a super headphone amp, but also as a practice run before tackling a power amplifier. Getting up the learning curve on a headphone amp will be cheaper than the same curve on a power amp, and many of the lessons will translate. In fact, if I can make a headphone amp with low enough hum, noise and distortion, that goes a long way to being successful with a power amp design.
From the discussion above, it looks like the main difference between a good headphone amp and a power amp is the current required. The voltages actually seem pretty similar (if you want to drive high impedance phones to high levels). So the main delta will be in the output stage (and power supply of course).

terry