The thing about low impedance headphones vs high impedance headphones

But if you have less volume, you also draw less current - so if you have the amp at 5 instead of 10 to get the same volume, wouldn't that balance out in terms of current drawn?

anyway, from my understanding, amp is not just about volume - the dynamics - response to rapid transients, the frequency response, etc. all matter. So I think it's more a matter of the weakest link in the chain - if you have phones that can resolve a high level of sonic detail, then a weak/poor amp will limit what the phones can reveal - so you have to try and balance it that way...

My general understanding is that high impedance phones work better with amps that can do strong voltage swings, usually tube-based for best results. Low impedance phones correspondingly are supposed to work best with amps that deliver high current, such as solid-state designs.

I'm going to take a wild guess at this. After reading forums for a few days now I think I might be on to something. Please post corrections.

Lower impedance will draw more current at given voltage than higher impedance phones. However, loudness has NOTHING to do with power alone...but power in the combination of efficiency. If you were to take two headphones/speakers where A is half the impedance of B. At any given voltage, A will draw twice the amount of current as B, but A will not necessarily sound twice as loud. Therefore if B is more efficient, it will be "easier" for the amp to drive when both are at the same loudness.

I have experienced this in home audio comparing my 8 ohm Klipsch Reference and my 4 ohm Cerwins. On the same amp, the Klipsch sound louder (and cleaner) than the cerwins at the same gain level AND are easier on my amp.

While reading this, I found this article that might help: read me. From the article: by doubling power (all things equal) you will get a 3db boost. The human ear usually associates twice the volume as a 10db difference. Therefore you need to double the power a little over three times (or 2^3 = 8 to make things simpler). All things equal, more than 8 times the power is required to make double the perceptive volume. Based on that, if there headphones/speakers were of equal power efficiency, a 50 ohm headphone would sound twice as loud as a 400 ohm headphone at the same gain level. However, the 400ohm headphones would pull about 8 times less current at that same gain level. Since I would argue that it would be unlikely that two headphones would have 8 times difference in efficiency, the higher impedance will generally be easier to drive for similar volume output.

Unfortunately, I cannot tell you anything about the HD595s because I have no experience with them.

ok....is there some unit quantifying efficiency? I have heard/read similar things, but i must say at the moment efficiency is very esoteric to me - eg i've read that even though hd595's are 50 ohm and akg k701' are 60ohm - because of the akg's are such "inefficient phones" they need proper amplification - to the point where some would only recommend desktop amps...this would be contrary to the notion that higher imp phones would be more efficient - which made me think that somehow efficiency is not related to impedance at all - maybe there are mechanical factors?
i'm in the dark with these things - headphone manufacture is very complex indeed...

Krisjan, efficiency is measured in Decibels, or dB. You'll see this expressed as how many dB of output you get with a certain amount of power input.

Impedance is a dance between the amp and transducer.

You'll see that amps have an output impedance. How closely the amp's output impedance matches the impedance of the transducer tells you how well power transfers from the amp to the transducer.

If there is an impedance mismatch, you lose power. There's a formula where you can calculate the power loss.

When you know how much power transfers from the amp to the transducer, you can use the efficiency rating to show how loud the transducer will get.

There's more to this, but you have to remember that the transducer's impedance is only half of what you need to know.

'Efficiency' isn't really the correct term to use here - 'Sensitivity' is more appropriate. As a poster above alluded to, the loudness that a driver produces is a function of both power (which is determined by voltage and impedance) and sensitivity, which is typically measured in dB @ 1mW (confusingly written as dB/mW).

Clearing up my previous post...

True it is confusing and I'm only beginning to understand it. The basic point I was trying to make is the following:

Assume:

• Voltage supplied by the amp is constant and at a consistent frequency
• There are two speakers/headphones: A and B
• Impedance(A) = 1/2 impedance(B) at the output frequency
• Both are equal in sensitivity

Results:

• A will draw two times more power than B.
• A will be somewhat louder than B but no where near twice as loud.

Unfortunately, the real world introduces many variables which make it very difficult to conclude anything about a headphone or speaker from it's specifications alone. This is especially true since there are lots of factors in the ratings.

The way I wrote my last post may have implied that a higher impedance headphone would be more sensitive, but as far as I know impedance and sensitivity are unrelated. That's exactly what krisjan pointed out. I believe sensitivity has more to do with mechanical factors such as the driver build, voice coil, magnet structure, enclosure type, etc.

The other thing, and to better answer the original post, is that "easy" has a few different meanings. One can infer that on headphone is easier to drive than another because it is louder. IMO, easy is a term reserved for the amplifier. The higher the current draw, the harder it is on the amplifier. With that said, lower impedance headphones are always harder to drive than higher impedance at X volts.

Anyone else want to chime in on this to confirm or deny any of this? (Hopefully I didn't overlook something)

EDIT:
I just noticed MoodySteve suggested the use of sensitivity after I posted this, so I changed my wording to reflect that.

The simplest way of calculating efficiency is by dividing your output by your input, then multiply by 100 to get the efficiency in percent. No amplifier is 100% efficient (especially a Class A). So, you're always going to be down by at least 20% for an 80% efficient amplifier. I think even 80% is being too generous.

If you don't multiply by by 100, you can take the log of the quotient, multiply by 20, and it will give you the overall gain of the amplifier in dB.

It typically takes an increase of ten times in power to sounds twice as loud. This is the most widely accepted factor. This argument shows up in the first few chapters of the Douglas Self amplifier book.

Usually you can bridge single ended amplifiers and get four times the output of one of the single ended amps. If you have no balanced headphones, this proves to be a pretty useless practice.

Humans hear loudness through decibels (dB), not Watts.

Does anyone know the equation for impedance mismatch power loss?

Also I agree as above stated that sens is the other part of the impedance to loudness equation. The problem with the aforementioned k701 is that they are a low sensitivity low impedance phone meaning they require a lot of power but also that they need a low output impedance amp or the amp will have to apply even more power.
Now out of curiosity when the transducer and amp impedance are mismatched what is lost, voltage, watts, or a combination of both?
Great thread!
Dave

Keep in mind that some manufacturers like to report sensitivity in dB/V instead of dB/mW to get bigger numbers and make the phones seem more sensitive.

True and that's a good point with the entire audio industry. For the most part, ratings are NOT standardized. Sure there are some, but to make specifications really useful there needs to be more regulation imo.

i always thought the output impedance of a headphone amp must be as low as possible - typically <1ohm - i see all the meier amps are like this? so then what is all this talk about matching the impedances? what would be a good match for 50ohm phones (on the amp side)

Power is voltage x current. While a high impedance headphone may require less current the headphone will demand more voltage swing. SS devices struggle delivering high voltage .... tubes struggle delivering high current. If you are using a low impedance headphone SS amps will have a much easier time driving them because they have more than adequate current delivery and are not taxed trying to deliver high voltage demands. Consequently, most small portables and soundcards will do a much better job driving a low impedance headphone. Since the majority of todays amps are SS designs a low impedance headphone WILL generally be easier to drive with such a design.

Conversely, a high impedance headphone will tax a tube amp far less as tubes are voltage devices and have more than enough voltage swing for a typical high impedance headphone like the Senn 650. However, a tube amp with a low impedance headphone like the AKG 701 can cause the amp to strain to deliver enough current.