[bigshot]

The experience and skill of the producer/engineering team? Mastering?

These two: The quality of the original recording itself, and the quality of the mastering for commercial release. All of these are the responsibility of the engineers.

 

[BobG55]

More basic questions from me...

Some songs/albums are considered to be of 'audiophile' quality and are often used as reference. What makes these records special as opposed to other supposedly high quality produced material made by reputable artists?
Gear quality? $$$?
The experience and skill of the producer/engineering team?
Mastering?

To mention a few examples: Nils Lofgren, Dire Straits, Miles Davis.

Steely Dan albums & ‘Chesky Records’ label especially.

 

[old tech]

These two: The quality of the original recording itself, and the quality of the mastering for commercial release. All of these are the responsibility of the engineers.

I would also add the combination/relationship between the artist and producer and their commitment to high quality sound. For example, as pointed out by @havarduf Dire Straits's Mark Knopfler and Neil Dorfsman's long time collaboration to producing great sounding quality production in the pop/rock genre. They had genuine interest and put in a lot of time to ensure their music was recorded, mixed and mastered with sound quality in mind. All their earlier analog and later digital recordings sound great. With Brothers in Arms, they understood that digital recordings required different thinking and experimented with microphone placings (as they found moving from analog to 16/44 digital meant that the recordings were picking up sounds which were previously masked (eg shuffling arms and feet). Their attention to detail really shows through.

 

[megabigeye]

I have a couple of basic/not-so-basic questions! Both of them power related.

What is the difference between power and gain? I recently posted that power only affects loudness, and somebody corrected (?) me by saying that gain is what affects loudness, not power. My interpretation of gain is that it's how much an amplifier amplifies, but it's still dealing with power (specifically, the voltage part of the power). I think Wikipedia agrees with me, but then they lose me on the math part. Was that person just being disagreeable, or does he know something I don't?

Second question: What is "not enough power?" If power only affects loudness, then it'd seem logical that less power just means less loudness... But I know that clipping (and presumably other problems?) can arise from a speaker/headphone being underpowered. When and why does less power go from "less loudness" to clipping/distortion/the boogeyman?

 

[sander99]

What is the difference between power and gain?

Gain usually stands for voltage gain, the ratio between output voltage and input voltage (sometimes expressed in dB, the number in dB would be 20log(Vout/Vin)).

Power = Voltage x Current

So to get power from a device it should be able to produce the required voltage and the required current (into an appropiate impedance).
For example a line level output often can output 2 Volts. However it can only do this into a suitable impedance, like an amplifier input (typical 50 kOhm I think?), so only a very smal current is needed.
If you connected a 4 Ohm speaker to that line output (better not to try I think!) not much good would happen. The line output's voltage would drop.
While if a capable amplifier drove the 4 Ohm speaker with a 2 Volt output signal - resulting in 1 Watt output - it would produce quite a lot of sound already. For example 88 dB at 1 Watt is not uncommon for a speaker.
In fact if you drove this amplifier with the 2 Volt input signal from the line output while outputting a 2 Volt signal then you here have a situation where the gain is 1 (or 0 dB), yet the amplifier is indispensable for providing the power.

 

[VNandor]

I imagine this post is why you asked about the difference between power and gain.

However, " Power does not change anything other than volume. " is a bit misleading. Technically speaking, the only thing that changes volume is gain. Whether the amplifier providing the gain in volume increase has sufficient power to respond to that gain with a particular load - that is actually the question.

Gain is a ratio of two "quantities" (I'm not sure about the correct English word), point is, the quantities have a unit of measurement. If you are looking at the ratio of voltages [measured in volts] then it's voltage gain, if you look at the ratios of power [measured in watts] then it's power gain, if you look at ratios of volume [I guess, SPL...?] then it's volume gain and so on...

something you can notice is that gain doesn't really have a unit of measurement. It is just a dimensionless number because you divide some unit of measurement with the same unit of measurement, so you would get something like [V/V] or [W/W] for the unit but anything divided by itself is just 1 although this can clarify what kind of gain you are talking about. Instead of saying this amplifier has a voltage gain of 50dB you could say this amp has a gain of 50dB V/V.

So a (more than 1) gain in volume will obviously always mean an increase in volume while a (more than 1) gain in power will only mean a gain in volume if all the other variables (like sensitivity) are staying the same.

Something else I found interesting in that post:

Moreover, the Senns have a pretty big impedance bump around 100Hz. On the HD800, the impedance goes from about 350 ohms at 10Hz, all the way to 650 ohms at 100Hz! It only averages 300 ohms from 20-20kHz. What does that mean? It means that all around the areas of mid-bass, where "warmth" generally comes from, power disappears with many solid-state amps, because they simply can't swing the voltage needed at that high impedance. A typical solid-state amp could be clipping at this point, but it depends on the gain. It might just sound bass-light or tinny.

Now, someone please correct me on this if I'm wrong but here's what I know about that topic.

Headphone amplifiers are voltage amplifiers. This means regardless of the load, frequency etc., it always tries to provide a constant ouput voltage (although the amount depends on the volume knob of course). So this means the power "disappears" because of the load not because of the amplifier (solid state or not). Since headphone amps are designed to put out a constant voltage of course if the load has a higher impedance at some frequency, it will logically draw less current at that frequency and thus less power. So the headphone amp won't have to magically output more volts into a bigger impedance (like at 100Hz with the sennheisers), it should not provide a bigger voltage swing into bigger loads. In fact, high impedance loads are less of a load for voltage amplifiers because it's easier to achieve a good voltage gain into high impedance loads since they draw less power.

 

[tomb]

I imagine this post is why you asked about the difference between power and gain.

Gain is a ratio of two "quantities" (I'm not sure about the correct English word), point is, the quantities have a unit of measurement. If you are looking at the ratio of voltages [measured in volts] then it's voltage gain, if you look at the ratios of power [measured in watts] then it's current gain, if you look at ratios of volume [I guess, SPL...?] then it's volume gain and so on...

something you can notice is that gain doesn't really have a unit of measurement. It is just a dimensionless number because you divide some unit of measurement with the same unit of measurement, so you would get something like [V/V] or [W/W] for the unit but anything divided by itself is just 1 although this can clarify what kind of gain you are talking about. Instead of saying this amplifier has a voltage gain of 50dB you could say this amp has a gain of 50dB V/V.

So a (more than 1) gain in volume will obviously always mean an increase in volume while a (more than 1) gain in power will only mean a gain in volume if all the other variables (like sensitivity) are staying the same.

Something else I found interesting in that post:

Now, someone please correct me on this if I'm wrong but here's what I know about that topic.

Headphone amplifiers are voltage amplifiers. This means regardless of the load, frequency etc., it always tries to provide a constant ouput voltage (although the amount depends on the volume knob of course). So this means the power "disappears" because of the load not because of the amplifier (solid state or not). Since headphone amps are designed to put out a constant voltage of course if the load has a higher impedance at some frequency, it will logically draw less current at that frequency and thus less power. So the headphone amp won't have to magically output more volts into a bigger impedance (like at 100Hz with the sennheisers), it should not provide a bigger voltage swing into bigger loads. In fact, high impedance loads are less of a load for voltage amplifiers because it's easier to achieve a good voltage gain into high impedance loads since they draw less power.

This is fundamentally incorrect in a number of statements. First, not all headphone amplifiers are voltage amplifiers. Yes, most are, but current amplifiers exist. Second, a "voltage amplifier" means it responds in gain changes by increasing or decreasing voltage, not current. The only time it would remain constant under load is with a test signal at a single frequency and amplitude. The volume knob has nothing to do with it, except for establishing the gain. The alternating signal source (music) then determines the resulting output voltages of the amplifier by responding to the differences in amplitude of the music source.

High impedance loads are NEVER "less of a load." Power is power and every headphone requires a certain amount for a certain dB of sound pressure. The DIFFERENCE is that impedances dictate the direction of the vector ratio (slope) of current vs. voltage.

What I tried to convey in layman's terms in the earlier post is that solid-state amplifier designs and tube amplifier designs (in a very general case) have advantages and disadvantages, one versus the other, with regard to producing voltage or current. Technically speaking, that involves the inherent impedance limitations of the circuitry.

Balanced drive headphone amplifiers were more or less invented by Tyll Hertsens, specifically to address the voltage swing limitations when driving high-impedance headphones, such as the Senn HD580 and Beyer DT880, 600 ohm version.

Similarly, Chu Moy invented the CMoy as an attempt to overcome the limitations in portable CD player amplifiers, with respect to voltage (increasing the 1.5 or 3V available in the portable CD player to 9V or 18V).

 

[tomb]

I have a couple of basic/not-so-basic questions! Both of them power related.

What is the difference between power and gain? I recently posted that power only affects loudness, and somebody corrected (?) me by saying that gain is what affects loudness, not power. My interpretation of gain is that it's how much an amplifier amplifies, but it's still dealing with power (specifically, the voltage part of the power). I think Wikipedia agrees with me, but then they lose me on the math part. Was that person just being disagreeable, or does he know something I don't?

Second question: What is "not enough power?" If power only affects loudness, then it'd seem logical that less power just means less loudness... But I know that clipping (and presumably other problems?) can arise from a speaker/headphone being underpowered. When and why does less power go from "less loudness" to clipping/distortion/the boogeyman?

Gain is simply a ratio of how much a signal is increased or decreased, in simplest terms.

Power is the energy required to respond to that gain, given the load it's trying to serve.

Gain is NOT energy, power is (or technically, work done in a time domain).

As for what is "not enough power," you have to also determine your definition of "loudness." Loudness in this area of headphone discussion, is often the response of mostly midrange peaks, where because of certain physical properties and the optimization of the human ear, less power is required than at any other frequency (generally speaking) to create the sensation of "loudness."

In high-fidelity headphone parlance, we want equal response to all frequencies, so that we can hear all details - including very dynamic and transient ones. That is where the quality of source, amplifier, and headphone lie - in those frequencies and responses away from 1KHz. The cheapest, most mundane amplifier or headphone can generate sufficient "loudness" at 1KHz. It's everything else we're interested in achieving, instead.

 

[VNandor]

Headphone amplifiers are voltage amplifiers. This means regardless of the load, frequency etc., it always tries to provide a constant ouput voltage (although the amount depends on the volume knob of course).

This is fundamentally incorrect in a number of statements. First, not all headphone amplifiers are voltage amplifiers. Yes, most are, but current amplifiers exist. Second, a "voltage amplifier" means it responds in gain changes by increasing or decreasing voltage, not current. The only time it would remain constant under load is with a test signal at a single frequency and amplitude. The volume knob has nothing to do with it, except for establishing the gain. The alternating signal source (music) then determines the resulting output voltages of the amplifier by responding to the differences in amplitude of the music source.

You are right, what I wanted to say is that a voltage amplifier tries to provide a constant gain of voltage and not an actually constant output voltage. If the amplifier is an ideal voltage amplifier (as most headphone amplifiers should be) then the voltage gain is constant across any frequencies and loads. The volume knob is what controls this gain of voltage.

Honestly, the output voltage wouldn't be constant in time even with a single tone test signal, the ratio between the momentary input and output voltage is what is supposed to be constant as you pointed out.

High impedance loads are NEVER "less of a load." Power is power and every headphone requires a certain amount for a certain dB of sound pressure. The DIFFERENCE is that impedances dictate the direction of the vector ratio (slope) of current vs. voltage.

High impedance loads are almost always less of a load for voltage amplifiers. If the amp has to output 10V into a 1kOhm load it only requires 10mA while providing 10V into a 10Ohm load requires 1A. So the amp is less likely to run out of current to achieve the requested output voltage if the impedance is high or at least high enough.

Power is voltage times current. Current is voltage divided by resistance. As a result power can be expressed as voltage squared divided by the resistance (U^2/R). Clearly, a low impedance headphone would draw more power at some arbitrarily picked voltage compared to a high impedance one. So I do think it's harder to achieve a good voltage gain for low impedance load because they require more power for that given voltage gain.

 

[megabigeye]

Thank you all for the answers so far... Though I shouldn't pretend to fully understand it all.

I often find myself looking at the introductions forum, wanting to intervene before somebody comes in shouting, "hIgH iMpEdAnCe?!? YoU nEeD mOaR pOwArS!!" (In my head they always sounds like Bobcat Goldthwait) I don't consider myself to be very knowledgeable in this stuff, so it puts me in the kinda awkward position of talking about things I don't fully understand just to hopefully thwart the loudmouths that understand even less than I do. I come to this thread to make sure I know what I'm saying isn't egregiously wrong. I appreciate both @tomb and @VNandor for also patrolling those threads. I wish Head-Fi had more of a culture of education and learning and less of a culture of opinion-as-fact.

@tomb, sorry I didn't quote your original post... I was confusing you with another poster whose input I wanted to actively avoid. (I actually once had an entire conversation with a Head-Fi'er that I was confusing with somebody else. I kept thinking, "why is this person not as smart as I thought they were?" Hehe. Oops!)

 

[tomb]

You are right, what I wanted to say is that a voltage amplifier tries to provide a constant gain of voltage and not an actually constant output voltage. If the amplifier is an ideal voltage amplifier (as most headphone amplifiers should be) then the voltage gain is constant across any frequencies and loads. The volume knob is what controls this gain of voltage.

Honestly, the output voltage wouldn't be constant in time even with a single tone test signal, the ratio between the momentary input and output voltage is what is supposed to be constant as you pointed out.

Yes, this is good and you are correct - the output voltage will always vary, since even a test signal varies from some positive to negative value, at the frequency selected of the test signal. As with most of these discussions, though, it comes down to semantics and our own interpretation of what a word means. All we can do is fall back on "convention," which is how we attempt to level discussions to some sort of understanding in science and engineering. (See below)

High impedance loads are almost always less of a load for voltage amplifiers. If the amp has to output 10V into a 1kOhm load it only requires 10mA while providing 10V into a 10Ohm load requires 1A. So the amp is less likely to run out of current to achieve the requested output voltage if the impedance is high or at least high enough.

Power is voltage times current. Current is voltage divided by resistance. As a result power can be expressed as voltage squared divided by the resistance (U^2/R). Clearly, a low impedance headphone would draw more power at some arbitrarily picked voltage compared to a high impedance one. So I do think it's harder to achieve a good voltage gain for low impedance load because they require more power for that given voltage gain.

Exactly the argument I was trying to make in another thread, but you're coming at it from the other end. You're already making the assumption that as voltage amplifiers, tube amplifiers do better with high-impedance loads than solid-state ones. You're simply assuming that most amplifiers have no trouble generating voltage. However, solid-state ones do have that trouble, because they are limited by the silicon, germanium, whatever the solid-state construct is used. Similarly, tubes are severely limited in current, because they're trying to move a charge from one point to another through empty space.

The issue is how easily the circuit is able to deliver voltage to a load with a given impedance. I will state it again just to be clear, because headphone impedances are often misconstrued as a requiring differences in power. There is no change in load with changes in impedance. "Load" is the power requirement and that does not change with impedance. The impedance dictates the direction of the vector result of Power (NOT the magnitude) in terms of voltage vs. current.

If you draw Power as a sloped line between axes of voltage and current (Y and X), the line will remain the exact same length, regardless of impedance (the slope of Y over X), but the slope - the angle of that line - will change.

P = I x R, period! You cannot create or remove the amount of energy required for the headphone. Yes, you can substitute Ohm's Law to replace I with V (or "U,", not sure what convention you got that from, maybe a typo?), but it does not change "P" for a given headphone. The only thing that changes is the ability of an amplifier to deliver current or voltage and at what ratio. That is determined by the output impedance of the amp and the impedance of the load.

 

[tomb]

Thank you all for the answers so far... Though I shouldn't pretend to fully understand it all.

I often find myself looking at the introductions forum, wanting to intervene before somebody comes in shouting, "hIgH iMpEdAnCe?!? YoU nEeD mOaR pOwArS!!" (In my head they always sounds like Bobcat Goldthwait) I don't consider myself to be very knowledgeable in this stuff, so it puts me in the kinda awkward position of talking about things I don't fully understand just to hopefully thwart the loudmouths that understand even less than I do. I come to this thread to make sure I know what I'm saying isn't egregiously wrong. I appreciate both @tomb and @VNandor for also patrolling those threads. I wish Head-Fi had more of a culture of education and learning and less of a culture of opinion-as-fact.

@tomb, sorry I didn't quote your original post... I was confusing you with another poster whose input I wanted to actively avoid. (I actually once had an entire conversation with a Head-Fi'er that I was confusing with somebody else. I kept thinking, "why is this person not as smart as I thought they were?" Hehe. Oops!)

Not to worry. It's just getting a bit difficult to explain in a world of semantics and differences of understanding with words.

I have only been trying to correct the incorrect assumptions that impedance dictates power. It only dictates the ratio needed between voltage and current to deliver that power. Every amplifier has ability to deliver power, but every amplifier has an optimum ratio of voltage vs. current that it can deliver.

What does this mean in practical terms? Solid-state tends to be able to deliver more current, but less voltage. If you think of voltage as FORCE (exactly what it is), the FORCE required for a current to flow through solid material can be quite high (how does anything move through a solid object?) So, in solid-state, made out of silicon, germanium, whatever - the FORCE through the solid material is very difficult to produce. Instead, the FORCE of applying more voltage tends to destroy the material by heating it up and burning it out.

On the other hand, the solid-nature of the substance means the molecules are very close together and it's extremely easy to develop current at ridiculously low voltages (FORCE). This is also analogous to sound travel in a solid vs. air or vacuum.

The opposite is true with tubes. Tubes work by transferring charge (electrons) through a vacuum from one pole to another. They do this through a vacuum, because establishing that charge transfer is literally blocked by any molecules in the way (why they use a vacuum in tubes). Instead, voltages can be quite high in order to develop a miniscule current, because the FORCE (voltage) has nothing blocking it but empty space.

As usual, the exact physics involved with both is technically difficult to describe in a forum post and I don't understand it any better, either. I'm certain someone can poke a lot of holes in what I said above, but I believe it's generally correct in a layman's understanding.

 

[castleofargh]

P = I x R, period! You cannot create or remove the amount of energy required for the headphone. Yes, you can substitute Ohm's Law to replace I with V (or "U,", not sure what convention you got that from, maybe a typo?), but it does not change "P" for a given headphone. The only thing that changes is the ability of an amplifier to deliver current or voltage and at what ratio. That is determined by the output impedance of the amp and the impedance of the load.

It's how we learn those equivalences in France with P, R, I and U instead of with V. But of course we use V as unit of voltage.
I have no idea why(the .fr word for voltage is "tension"...). plus I've seen on old French books that we used to use V. Somebody decided to mess with that on purpose at one point. Thanks Obama.

 

[tomb]

It's how we learn those equivalences in France with P, R, I and U instead of with V. But of course we use V as unit of voltage.
I have no idea why(the .fr word for voltage is "tension"...). plus I've seen on old French books that we used to use V. Somebody decided to mess with that on purpose at one point. Thanks Obama.

Interesting. "Tension" is a type of force, along with its reverse condition, "compression," both of which define direction of that force acting on a solid object. It's difficult to make an analogy with either for voltage. At least in America, the analogy taught in engineering for electricity is that of fluid flow - voltage is analogous to pressure, while current is analogous to flow. The existence of a pressure gradient causes flow (current). Pressure can be negative or positive, but again - hard to make that analogy with tension and compression, because we don't think of flow occurring with tension or compression.

 

[VNandor]

Exactly the argument I was trying to make in another thread, but you're coming at it from the other end. You're already making the assumption that as voltage amplifiers, tube amplifiers do better with high-impedance loads than solid-state ones. You're simply assuming that most amplifiers have no trouble generating voltage. However, solid-state ones do have that trouble, because they are limited by the silicon, germanium, whatever the solid-state construct is used. Similarly, tubes are severely limited in current, because they're trying to move a charge from one point to another through empty space.

I make that assumption because the maximum output voltage is almost always mentioned in the spec sheet of headphone amps. I have no doubt that the amps could reach the claimed voltage. Something that's not always mentioned is the impedance of the load (and the actual power drawn) when they measured the max output voltage. An amp might be able to output 10V into 1kOhms but not into 10Ohms because it lacks the power to do so.

The way I ballpark what amplifier a headphone needs is by looking at the sensitivity of the headphone which is typically in dB/mW (at 1kHz) and it's impedance (which is just an average value). I pick the desired SPL let's say 100dB SPL average is pretty loud already. Since most music is dynamic reaching an average of 100dB SPL requires to reach even higher values (let's just say 20dB headroom) depending on how dynamic the music really is. This would allow for a loud listening experience even with fairly dynamic music.

From the sensitivity, impedance and desired SPL I calculate the voltage the headphone needs to reach the required max SPL level. If the amp has a higher output voltage than I calculated there is already a chance the pairing will work (it will be loud enough for sure). So in this case one of the questions is can the amp actually output their claimed voltage into the headphone I want to use it with? The higher the impedance, the more likely it actually can. The question could be rephrased as well: I know the amp I'm looking at will have enough voltage to drive my headphones, the question is, will it have enough power to do so?

I understand that there are other requirements for the amps like low noise, low distortion but these usually aren't described in the spec sheets in a way it could be useful to me so after checking if the output voltage is enough I just bite the bullet, order the amp and hope I will like the sound.

There is no change in load with changes in impedance. "Load" is the power requirement and that does not change with impedance. The impedance dictates the direction of the vector result of Power (NOT the magnitude) in terms of voltage vs. current.

If you draw Power as a sloped line between axes of voltage and current (Y and X), the line will remain the exact same length, regardless of impedance (the slope of Y over X), but the slope - the angle of that line - will change.

P = I x R, period! You cannot create or remove the amount of energy required for the headphone. Yes, you can substitute Ohm's Law to replace I with V (or "U,", not sure what convention you got that from, maybe a typo?), but it does not change "P" for a given headphone. The only thing that changes is the ability of an amplifier to deliver current or voltage and at what ratio. That is determined by the output impedance of the amp and the impedance of the load.

I'll try to use V instead of U. Hmm, didn't you miss that equation as well? If I wanted to visualize P = V x I (power equals voltage times current) in a simple V-I graph I would think that the power is the rectangle that can be drawn after plotting the (V;I) coordinate. I think the product of two simple numbers, V and I, can be most easily visualized by the area of a rectangle (so the area corresponds to power which equals to V x I) where the sides' length are V and I because the area is calculated by multplying the sides' length anyways. If the area (power) is the given number, there are of course an infinite pair of V-I values that would draw a rectangle with the correct area just like how a certain load can correspond to an infinite pair of V-I values depending on the load's resistance.

So I think we are on the same page just looking at it differently. However there's one thing I still don't understand.

Moreover, the Senns have a pretty big impedance bump around 100Hz. On the HD800, the impedance goes from about 350 ohms at 10Hz, all the way to 650 ohms at 100Hz! It only averages 300 ohms from 20-20kHz. What does that mean? It means that all around the areas of mid-bass, where "warmth" generally comes from, power disappears with many solid-state amps, because they simply can't swing the voltage needed at that high impedance. A typical solid-state amp could be clipping at this point, but it depends on the gain. It might just sound bass-light or tinny.

I'm fairly sure headphones are meant to be driven with the same voltage gain across the frequencies, and definitely not the same power gain across the frequencies. Is this where we disagree?
If we don't disagree on that, then the idea of a headphone amp not being able to swing enough voltage at the frequencies where the impedance is high is still looking nonsense to me.
Why would it be harder for the amp to swing the same output voltage into a higher impedance compared to a lower one? Since it's actually easier to have a good output voltage into a higher impedance because that requires less power.