[Brent Hutto]

I'm actually not a DIYer but I hope that this is the place to have a couple of concepts explained. I have an electrical engineering degree (early 80's vintage) but don't recall very much practical design stuff and what I knew is perhaps obsolete anyway. I'm trying to study up enough to make informed judgements about all the headphone amps out there commercially (Headroom, Meier, et. al.) and semi-commerically (Portaphile, Shellbrook, many others).

First off, for portable headphone amps the three main choices are 3V supply, 9V supply (usually made bipolar with a "rail splitter" op-amp, right) and dual 9V supply. I understand that a high impedence headphone makes a large voltage demand on the amp while a low impedence one can get by with less available voltage as long as the current is in ample supply. Just where is the break between "high" and "low" impedence. I'd suppose than a 32-ohm headphone is what the 3V amps have in mind and that you want at least dual 9V for some old 600-ohm 'phones. How about headphones in the 50-120 ohm range? It would seem unlikely to me that a 3V amp could get much output from them and even a plus or minus 4.5 volt design might not have a lot of voltage capability in reserve for a >100-ohm load. Or it is not that serious?

Second, about buffers. The cheapest and simplest portable amp designs have some sort of op-amp or custom low-voltage headphone driver IC (with a couple op-amp plus support circuitry on chip) directly driving the headphone. While acknowledging with all due respect that these amps sound great to a lot of folks, that strikes me as a really, really severe shortcut given how complex a load any dynamic transducer presents to an output amplifier. My vague memories from back in the day were that not having a buffer of some kind on the output of an op-amp was just insufficient design. Has the op-amp technology changed that much in 25 years?

How about "monolithic buffers" versus a plain old pair of bipolar discrete transisitors? Does "monolithic" mean an IC chip with a pair of transistor buffers on it? Or does it mean the buffer is on the same chip with the op-amp? Either way, what are its advantage and disadvantages relative to discrete transistors?

One other thing. Both the semi-commercial and larger vendors of headphone amps go on at some length about the high-quality electrolytic capacitors they use. Are they talking about output capacitors to keep DC offsets from reaching the headphone jack? My memory is that an electrolytic capacitor in the output path of an amplifier is a much more definite no-no than having an op-amp and no buffer. It's my understanding that you aren't going to get "hi-fi" sound with a leaky old box of electrolyte right there in the high-level signal path (maybe that's just a prehistoric prejudice that doesn't apply nowadays). Am I wrong or are they talking about some sort of power-filtering capacitors or something like that?

Thanks in advance for any kernels of wisdom you can share with a curious layman...

 

[jcx]

http://headwize.com/articles/index.htm

you will find lots of info at Headwize

but basically you may need ~100mW rms for many headphones to avoid clipping on dynamic music while only a few mW continous power into most headphones will cause hearing damage over time

iPod and other hi end portables often use a 3.6 V Li-ion battery and are supplied with 16 Ohm headphones to match the low V source, they use purpose designed headphone op amps that can deliver >100 mA only a few 100mV from the supply rails - not yesterdays typical audio op amp

but other than matching V, I capabilities to the headphone impedance, headphones are usually pretty tame loads- mostly resistive with of course the cable C load to watch out for - the "official" DIN specs include headphone amplifier output series resistance that pretty much isolates the amp from headphone Z variations

as an engineer you will find many "curious" component/circuit concerns among audiophiles - I discount most by >90% since fewer than 10% posts even give a clue that the author is applying minimal controls to their listening evaulations - A/B/X with frequency response and level matching to < 0.1 dB is the gold standard for psycoacoustic comparisons - anecdotal "evidence" from people listening to their only amp/prototype with a smoking soldering iron in hand are too subject to well established confounding factors

given the difficulty of resolving issues with psycoacoustic testing people follow fads and take otherwize good engneering principles to illogical extremes - lowering power supply impedance with execess C is one, buffering outputs with 10X greater current capacity than required while ignoring V headroom is another you will see here

 

[Brent Hutto]

Thanks for the pointer to Headwize. The "Projects" pages are a wealth of information from the perspective of several different authors/designers. I've just scratched the surface by reading a couple of them but I should be able to understand this stuff pretty well once I've made a couple passes through the whole batch (which will take a while).

I concur with your statements about hearing what you believe is supposed to be there versus testing for differences (A/B/X or at least some attempt at a controlled comparison). For my part, I figure the opposite tack makes sense. Buy an amp that seems well built with a design "as simple as possible...but no simpler" and not get carried away with the fan-boy hype. Hence my wanting to know about buffers or not and voltage requirements.

So just as a sanity check, the peak-to-peak output voltage of an audio amplifier can't be any higher than the (under load) supply voltage minus resistive losses (minimal) and perhaps voltage drops across the junctions of the output devices, right?

1) If you split a 9V battery into plus or minus 4.5V (call it plus or minus 4.0V under load) then the RMS output voltage will be something less than root-two times 8V or about five and half volts at the very most.

2) To get 100mW out with 5.5V to work with implies a maximum load impedence of about 300 ohms, no matter how low the output impedence of the amp.

3) Therefore, for higher impedence headphones no amount of brute-force current capability is going to allow for 100mW peaks if you're limited to that 9V battery as the source. That's what you mean by voltage headroom.

 

[amb]

Quote:

Originally Posted by Brent Hutto 1) If you split a 9V battery into plus or minus 4.5V (call it plus or minus 4.0V under load) then the RMS output voltage will be something less than root-two times 8V or about five and half volts at the very most. 2) To get 100mW out with 5.5V to work with implies a maximum load impedence of about 300 ohms, no matter how low the output impedence of the amp. 3) Therefore, for higher impedence headphones no amount of brute-force current capability is going to allow for 100mW peaks if you're limited to that 9V battery as the source. That's what you mean by voltage headroom.

Indeed. Not only that, very few opamps or discrete circuits could swing rail-to-rail, so the actual maximum voltage swing would actually be much less than that.

 

[robzy]

Quick follow on question, assuming 32 ohm Grado's, how much voltage and current would they need for moderate volume general music listening? (Approxiamatly of course).

Rob.

 

[steinchen]

Grados got an efficiency of 98dB/1mW, for 100dB SPL (very loud) they need 1.6mW or 225mV

 

[kevin gilmore]

Quote:

Originally Posted by Brent Hutto 1) If you split a 9V battery into plus or minus 4.5V (call it plus or minus 4.0V under load) then the RMS output voltage will be something less than root-two times 8V or about five and half volts at the very most.

Actually its peak to peak divided by 2.8 to give maximum rms volts.
In this case about 2.85.

Very few opamps actually do rail to rail, so it is more like 2.5 Vrms.

 

[tangent]

Quote:

Originally Posted by Brent Hutto what I knew is perhaps obsolete anyway

I don't believe we have invented any relevant new physics since then.

Quote:

I understand that a high impedence headphone makes a large voltage demand on the amp while a low impedence one can get by with less available voltage as long as the current is in ample supply.

That's a gross overgeneralization, and I wish people would quit propounding it. I can give several counterexamples, but just one suffices: AKG K-501 (120 ohm) and Senn HD-580 (300 ohm) both require about the same voltage for the same loudness. The AKGs are indeed going to require more current than the HD-580s, but it's wrong to get into a mode where you think "oh, lower impedance, that means more current and less voltage.

The voltage/current question is a question of driver design and such. There is no 1:1 correlation based solely on impedance.

Quote:

My vague memories from back in the day were that not having a buffer of some kind on the output of an op-amp was just insufficient design. Has the op-amp technology changed that much in 25 years?

In some ways, yes. One of the biggest new developments are "line driver" op-amps, which tend to have rail-to-rail I/O (or nearer-to than traditional designs) and high output current. The new AD8397 (chip du jour around here...) is such a design.

The main problem with going with line drivers is that you often can't find one chip that has exactly the right balance of features. There's a lot to be said for using separate buffers, so that you can swap the op-amp independently to tune the sound more precisely.

Quote:

How about "monolithic buffers" versus a plain old pair of bipolar discrete transisitors?

Fairer comparison: monolithic buffers vs. a good discrete buffer. I insist on that because at minimum an IC buffer is going to have far better biasing than a simple push-pull pair.

On those grounds, then, the answer is that the discrete design can be tuned finely to your tastes. The monolithic is what is is. Ever since the Elantec buffers went away, I've yet to find an IC buffer I'm really happy about...but of course I still use them, because discretes have other problems. Real estate, for instance.

Quote:

Does "monolithic" mean an IC chip with a pair of transistor buffers on it?

Absolutely not. Even the ancient LH00xx buffers were more complex than that. Look at some of their datasheets. The BUF634, HA5002 and OPA633 all include internal chip schematics, IIRC.

Quote:

Or does it mean the buffer is on the same chip with the op-amp?

I'm not sure exactly what you're talking about here. There are IC buffers that are closed-loop, so you could argue that there's an op-amp in there. And many high output current op-amps are used as buffers.

Quote:

Are they talking about output capacitors to keep DC offsets from reaching the headphone jack?

Depends on the design. It's increasingly rare for headphone amps to have output capacitors...unfashionable, it is.

Most often, electrolytics are used as rail capacitance these days.

Quote:

the RMS output voltage will be something less than root-two times 8V or about five and half volts at the very most.

Kevin's right...sine waves are 2.828 times higher peak-to-peak than their RMS voltages. You are thinking of peak voltage instead, which is a factor of 1.414. If you're going to talk in terms of rail-to-rail voltage, you need to think in terms of peak-to-peak output. You'd use the other method if thinking in terms of just one half of a dual supply.

 

[Brent Hutto]

You guys...thanks so much for the feedback. Sounds like my dim memory is at least correct in the broader strokes. I had a feeling I had an extra factor of two somewhere in my voltage/current ciphering.

I particularly appreciate the point about not just assuming that the nominal impedence gives a one-number summary of the voltage and current required for a headphone. Point taken.

On the buffer issue, it sounds like for most applications you can choose the correct one (assuming something close to "correct" is available) and get very good performance. To do significantly better than that you have to put in some elbow grease matching transistor type and/or individual parts for an optimized discrete buffer. If you're not going to sweat the details, go ahead and use a monolithic buffer chip.

Now I think a lot of what I read on the project pages at Headwize will actually educate me instead of getting hung up on the simple basics.

 

[SnoopyRocks]

It appears I'm a bit late and all the technical questions have been answered, leaving the opportunity to comment on some other stuff.

Quote:

Originally Posted by Brent Hutto I'm trying to study up enough to make informed judgements about all the headphone amps out there commercially (Headroom, Meier, et. al.) and semi-commerically (Portaphile, Shellbrook, many others).

Good luck with that. Unfortunately, it's probably not going to get you vary far unless the the differences are LARGE. Tech specs rarely correlate to a specific sound. Generally speaking, but with a few expections, any engineering specs stated are for marketing purposes only and are in large part intentially misleading. This can include anything from listing exotic parts to great, usually exagerated, performance figures without precisely stating the test conditions. (Yes, it does make that much difference.) Engineers refer to this as specmanship, their version of creative writing if you will. Audio, where true believers abound, makes this especially difficult because it's more akin to religion than science. It's really just a matter of discovering which half-truth you prefer. If the general public cared, let alone could actually hear a difference, things might be a bit different.

Quote:

Originally Posted by tangent I don't believe we have invented any relevant new physics since then.

Physics isn't invented as much as it is observed (curve fit). There has been a lot of curve fitting in the last 20 years, especially with regard to semiconductors.

 

[robzy]

Quote:

Originally Posted by steinchen Grados got an efficiency of 98dB/1mW, for 100dB SPL (very loud) they need 1.6mW or 225mV

Hrm, that doesnt make sense to me
P = IV
I = P/V
I = 1.6/225 (ma)

That would mean that its in the microamp range... that couldnt be right.

I would assume though that its something im doing wrong, so please help me out...

Rob.

 

[nysulli]

correct me if i'm wrong, but if you need 1.6 watt for 100 db, grado's are 32 ohm headphones

P=IV V=IR
P=I^2 R
I^2 = 1.6/32
I^2 = 0.05
I=0.2236 A

so 224 mA

V=IR
V=0.224 * 32
V=7.168 V

 

[bg4533]

Quote:

Originally Posted by nysulli correct me if i'm wrong, but if you need 1.6 watt for 100 db, grado's are 32 ohm headphones P=IV V=IR P=I^2 R I^2 = 1.6/32 I^2 = 0.05 I=0.2236 A so 224 mA V=IR V=0.224 * 32 V=7.168 V

People earlier said 1.6mW, not 1.6W. I am too tired to figure out who or what is right.

 

[steinchen]

it's not that simple since we're talking about AC here, not DC
impedance is in fact Z (complex number) and not resistance R

Quote:

Originally Posted by tangent That's a gross overgeneralization, and I wish people would quit propounding it. I can give several counterexamples, but just one suffices: AKG K-501 (120 ohm) and Senn HD-580 (300 ohm) both require about the same voltage for the same loudness. The AKGs are indeed going to require more current than the HD-580s, but it's wrong to get into a mode where you think "oh, lower impedance, that means more current and less voltage. The voltage/current question is a question of driver design and such. There is no 1:1 correlation based solely on impedance.

nevertheless, 1.6mW and 225mV -> 0.00711A = 7.11mA and not microamps

in fact you can expect Grados to draw about 20mA peak at a very loud volume levels

 

[robzy]

Quote:

Originally Posted by steinchen nevertheless, 1.6mW and 225mV -> 0.00711A = 7.11mA and not microamps

Maybe im missing something, but:

mw = mv * ma

1.5 = 225 * ma
1.5 / 225 = ma
ma = some amount of microamps.

And to follow on from that, the OPA134 has an output current capability of ~35ma. If that were true, and Grados (which as a general rule of thumb draw more current than other cans) draw 20ma max, why on earth would anyone need a buffer?

(I am sorry for the thread hijack btw, but this is kinda carrying on from the origional question...)

Rob.