Originally Posted by J Bones
Why is it again that a tube amp is better designed for a set of cans with much higher resistance than another pair of cans? I really don't think the Q701's (which by the way, if I didn't mention it earlier, are the set I own) are all that resistant...but either way, I've been looking into something other than my Digizoid, and the Schiit Lyr has thus caught my attention. Now, I actually decided on it well before I started hearing that a tube amp sometimes works best with a pair of cans with a higher impedance but never why. Would this for some reason not work with the Q701's? The way I originally looked at it is that the output impedence of the Lyr is much lower than some others and would still work with the Q701 no matter what (not concerning other factors like THD or power output etc).
I'm assuming this is also why you mention that lower impedance headphones are typically driven better by low impedance amps (often those that are solid state too)...
There are a couple of reasons tube amps are better for high-impedance phones. One is Ohm's Law, or rather, a variation of it. Higher-impedance loads require more voltage for the same amount of power than low-impedance ones. Without digressing into a lot of Ohm's Law, algebra, and an explanation of the relationships, let's consider some real world examples.
Let's say for a given efficiency headphone, you want to be able to produce music peaks or transients that take 100mW of power. BTW, everything is reduced to "power." It matters little to try and figure out how much current an amp can supply or how much voltage and then relate that to the headphone separately. The fact is that the headphone needs power, period. Its efficiency determines how much and its impedance what combination of voltage and current it will pull.
At 100mW and with 32 ohm headphones, you need 0.056 Amps. Using a variation of Ohm's Law, we can find out that at the same time and with 32 ohms impedance, the voltage for a music peak to produce that 100mW excursion is 1.79 V. That's too much for a single AA battery, but two AA batteries can handle it without an issue (2 x 1.5V = 3V). So, a little portable amp with a couple of AA batteries could handle things just fine. (There are penalties in many circuits - minimum voltage for opamps, etc. that will burn some of the available voltage, but this is close enough for an example.) This is why you see so many headphones touted for portable devices by stating that they're low impedance. Some are down to 25 and 18 ohms these days, not just 32.
On the other hand - assuming we have a 300 ohm headphone that is as efficient as a 32 ohm headphone, you only need 0.018 Amps. However, the voltage required for that power goes up to 5.4V. That's more voltage than can be produced by a couple of batteries - it takes 4, at least. Or worse, it's more voltage than is available from a PC's USB power supply. It's one reason that the CMoy was always built with a 9V battery, not a AA, AAA, or multiples of those batteries - it wasn't just for low weight.
None of this is absolute, averaged, or RMS power - it's musical peak power (in our example). This means that if the voltage supply in the amp is unable to produce the actual voltage difference, it will clip, period - and added distortion most likely occurs well before that. Things get much worse as you go up in power - for 300 ohms impedance, 7.8V are needed for 200mW, 9.5V for 300mW, etc. Meanwhile, only 5.7V are needed for 32 ohms at a full watt (17V are needed for 300 ohms). Remember, for every 3dB of sound level increase, a doubling of power is required. That may sound like a lot of dB's when looking at typical headphone efficiencies, but these are transient peaks that we want to retain, not overall listening volume.
Anyway, the high voltage required for most tubes means that all that extra voltage swing is readily available for high-impedance phones. It's also a reason you see so many tubes for electrostatics - it's much easier to produce all that high voltage that's needed (many solid-state parts are well-beyond their safety ratings).
The second reason tubes may be better for high-impedance (and by corollary, worse for low-impedance phones) is the issue of blocking caps. In tube amps, transformers are often quite expensive (and very heavy). Therefore, you often see tube amps at the low end (and some high-end) with some variation of OTL (Output-Transformer-Less) or hybrid design. These amps have what's called a DC offset at the output. For some OTL tube amps, this voltage may be as much as what's seen at the tubes' plates - a couple of hundred volts or more. Your headphones would be fried instantly if you plugged them in without protection. So, capacitors are used on the amplifer output to block that DC voltage. BTW, if insufficient time is allowed for these caps to charge on power-up: THUMP! It's why relay-delays are often used. The same offset issues and blocking cap requirements can be true of many solid-state amps that don't use feedback or have some other discrete buffer output that develops offset.
Well, the use of blocking capacitors at the output of an amp sets up an RC circuit along with the impedance of the headphone. An RC circuit forms what is otherwise known as a high-bandpass circuit. The frequency output is dependent on the circuit characteristics - low frequencies will be blocked.
Figuring out which low frequencies will be blocked usually involves determining the cutoff frequency of the RC circuit that results. This tells you where the frequency drops by 3dB given a certain output capacitor value and headphone impedance. Common large capacitor sizes might be 220uf, 470uf, 680uf, or 1000uf. For 220uf and a 300ohm headphone, the cutoff frequency is 2.4Hz - no problem. However, for a 32 ohm headphone, the cutoff frequency with 220uf is 22Hz - now we're getting into audibility, because phase distortions are introduced at frequencies much higher than that. Plus, this is the -3dB point - the bass actually begins to fall off between 50 and 100 Hz - maybe not so bad for a pair of headphones, but horrible for an amplifier.
So what does this mean, in practical terms, for a tube amp with low impedance vs. high impedance phones? Well, at the higher voltages for many tube amps, only electrolytic caps are available in these sizes and even then - totally unavailable or practical at the high capacitor uf ratings. This gets momumentally worse for low impedance cans than for high ones. So, expect to lose bass with OTL tube amps and low-impedance phones. It depends on the actual circuits with tube hybrids and whether or not feedback is introduced, but many may have the same issue. As for feedback, many people think feedback sounds bad, so it could have diminishing returns if used for reducing offset (typically in solid-state circuits).
So, bottom line: tube amps produce much more voltage (much easier to produce power at high impedances) and OTL tube amps will have DC blocking caps that may cut the bass at lower headphone impedances. Both facts combine to make tube amps often more suitable for high-impedance phones. There are exceptions, to be sure. Many hybrids perform outstandingly well with low impedance phones and of course, a transformer-coupled tube amp is going to be optimized for a range of impedances (why many of them have high and low-Z switches).
EDIT: P.S. I see the posting of Beyer's marketing explanation for low-impedance vs. high-impedance headphones. They make great headphones, but there's no consistency between quality and impedance in headphones and amplifiers. One does not imply a higher quality than another - although with Beyer, it seems to work out that way.
BTW, Beyer is of course correct when they state real power with AC (the music signal) must be measured in RMS. However, I wasn't trying to digress that far. Rather, I was trying to convey that the DC voltage in an amp (short of buck-boost circuits which add distortion) can't be created out of nothing. If the peak of the amplified signal wave wants to hit + and - 5V, the power supply at those points in the circuit have to have +and- 5V, period. If the power supply doesn't supply that voltage, it can't hit that peak, no matter what.
Edited by tomb - 1/18/13 at 6:57am