Not necessary correct, but to really simplify things up: 8 ohm = extremely low, 16 ohm = low, 32 ohm = common, 64 = semi-high,  >100 ohm = high.

Thanks goodvibes, but sorry for the newbie question again.
 
So the lower impedance is harder to drive? or just harder to control? (which if i may assume that's why all the high end iem or headphone are high impedance???)
 
And another one, a DAP that can drive low impedance fairly well, does that mean it would do bad on high impedance? and vice versa.
 
Thanks, again sorry for the beginner q.
 

Most people get this wrong. Amplifiers (including small ones in portable devices) have no problem providing the required voltage. Their limitation is current. If your amplifier isn't capable of providing enough current, the voltage sags, compromising the sound. Because this happens preferentially at certain frequencies, this affects your frequency response. Typically, but not always, it craps out first in the bass.
 
So the best answer is, low-impedance headphones are harder to drive.
 
For high-impedance headphones (those that are 300-600 ohms), you need to look at sensitivity; impedance is irrelevant. If the sensitivity is high enough, it will play loud enough, and that's all you need to worry about. High-sensitivity, high-impedance headphones are very easy to drive.
 
Jim Austin

Ok, thanks for the reply guys, but what is sensitivity? how do you check that one out?

Quote:

jimaustin said:

Most people get this wrong. Amplifiers (including small ones in portable devices) have no problem providing the required voltage. Their limitation is current. If your amplifier isn't capable of providing enough current, the voltage sags, compromising the sound. Because this happens preferentially at certain frequencies, this affects your frequency response. Typically, but not always, it craps out first in the bass.
 
So the best answer is, low-impedance headphones are harder to drive.
 
For high-impedance headphones (those that are 300-600 ohms), you need to look at sensitivity; impedance is irrelevant. If the sensitivity is high enough, it will play loud enough, and that's all you need to worry about. High-sensitivity, high-impedance headphones are very easy to drive.
 
Jim Austin

Click to expand...

 
Thanks Jim for stating this clearly.  Your post should be a sticky around here.  I've been befuddled by all the posts claiming that higher impedance = harder-to-drive because in home speaker land I've always known the exact opposite.  Seen that claim so much in fact that I thought that maybe the measurement meant something different here in Head-Fi!
 

Quote:

newakgguy said:

 
Thanks Jim for stating this clearly.  Your post should be a sticky around here.  I've been befuddled by all the posts claiming that higher impedance = harder-to-drive because in home speaker land I've always known the exact opposite.  Seen that claim so much in fact that I thought that maybe the measurement meant something different here in Head-Fi!
 

Click to expand...

but it's not correct. You run out of voltage driving high impedance cans. A high output impedance amp loses bandwidth driving low impedance phones even if there's plenty of current as the characteristic is there at any volume level. When you run out of current or voltage, you don't lose extremes, you clip with lots of distortion.
 
 

Quote:

goodvibes said:

but it's not correct. You run out of voltage driving high impedance cans. A high output impedance amp loses bandwidth driving low impedance phones even if there's plenty of current as the characteristic is there at any volume level. When you run out of current or voltage, you don't lose extremes, you clip with lots of distortion.
 
 

Click to expand...

Just curious then goodvibes, why the difference in Head-Fi from Home-Fi?  In "Home-Fi" its definitely the low impedance speakers that are more difficult to drive.
 

Its combination of impedance and sensitivity.  Impedance impedes current, and how much current is being drawn at a voltage(current depends on voltage drop at impedance) is how much power if being applied to the load to give of the dB/mW.  W is unit of power.  P=IV.
 
Therefore to simply for an easy to understand example, at fixed voltage(its alternating I know), more impedance, means less current, means less power to the headphones.  Make sense?

Then the more sensitivity, the less current required to reach a certain volume level, then with lower sensitivity, the more current required to reach a certain volume level.
 
The distance of the driver from your ear matters a little bit, too.

Incorrect.  sensitivity is dB/mW  W is Power.  P = IV  I is current.  You have to take voltage into consideration.  It's all about Power yo.  You dig?!
 
Quote:

mad max said:

Then the more sensitivity, the less current required to reach a certain volume level, then with lower sensitivity, the more current required to reach a certain volume level.
 
The distance of the driver from your ear matters a little bit, too.

Click to expand...

 

Quote:

high_q said:

 It's all about Power yo.  You dig?!
 

Click to expand...

What is wrong with you?
 

Quote:

newakgguy said:

Just curious then goodvibes, why the difference in Head-Fi from Home-Fi?  In "Home-Fi" its definitely the low impedance speakers that are more difficult to drive.
 

Click to expand...

They generate more heat in the amp because they pull more current but they are also generally more efficient just like in headphones. It's exactly the same The output in HIFi amps is generally about 0.1 ohm and no solid state amp will show roll off driving any 4 ohm speaker. Once you get below 4 ohms in the bass it just gets, as I said before, closer yet to a short where less current capable amps become less happy. I never said that high impedance headphones are harder to drive. That's effeciency related but you may run out of voltage. I clearly stated that the optimum load would be high effeciency with high impedance. You seem to only hear bits and pieces and are missing the concept. A good solid state amp will double it's watts as you half the load impedance untill it becomes current straved. I fail to see what you think is different.
 
 

 
Quote:

I never said that high impedance headphones are harder to drive.

Click to expand...

 
Well, maybe. But when I wrote something that you now appear to be agreeing with, you said I was wrong. Here's what I wrote:
 
Quote:

 
So the best answer is, low-impedance headphones are harder to drive.
 
For high-impedance headphones (those that are 300-600 ohms), you need to look at sensitivity; impedance is irrelevant. If the sensitivity is high enough, it will play loud enough, and that's all you need to worry about. High-sensitivity, high-impedance headphones are very easy to drive.

Click to expand...

 
You responded:
 
Quote:

but it's not correct.

Click to expand...

 
and then you wrote, in your latest post:
 
Quote:

I never said that high impedance headphones are harder to drive. That's efficiency related but you may run out of voltage. I clearly stated that the optimum load would be high efficiency with high impedance.

Click to expand...

 
(spelling corrected) Which is remarkably similar to what I wrote, quoted above. Specifically, that for high sensitivity / efficiency headphones, high impedance is easier to drive. (If you wish to take issue with my use of "sensitivity" instead of "efficiency", fine, but in this context the difference is technical and not practical.) The point is that if, as I wrote, you look first at sensitivity (or efficiency; whichever the manufacturer chooses to report), and you find it's high enough, then you only need to worry about LOW impedance, not HIGH.
 
But you did write one thing that I believe is incorrect. You wrote,
 
Quote:

 
A high output impedance amp loses bandwidth driving low impedance phones even if there's plenty of current as the characteristic is there at any volume level. When you run out of current or voltage, you don't lose extremes, you clip with lots of distortion.

Click to expand...

 
But this doesn't make sense in terms of basic physics. First, let's talk about current. Because impedance is frequency-dependent, the impedance will be lower at certain frequencies than at others. Consequently, the amp will crap out first at those frequencies. What happens is the voltage sags.
 
OK, so what about voltage? Up to the point where the current demand exceeds the amplifier's capabilities, the amplifier is essentially a pure voltage source. The higher the impedance, the lower the current demand. SO INTO HIGH IMPEDANCE 'PHONES, THE AMP IS EVEN CLOSER TO BEING A PURE VOLTAGE SOURCE.  The maximum voltage output is set by the amplifier's design; up to that maximum, the output tracks the input very closely, which is exactly what you want. Because the maximum voltage capability of the amplifier -- a set maximum -- is the only limitation in output, this is indeed volume (or volume-setting) dependent, and it does not depend on frequency. When the amplifier tries to generate an output signal that's larger than the amplifier can provide, it clips. This could be a pure 50 Hz sine wave, a pure 10 kHz sine wave, or anything in between (or beyond); it makes no difference.
 
This is my last post in this thread. People will believe what they want to believe. But, so that people can make up their own minds here, I'll report credentials. I have a PhD in physics, and I write for Stereophile.
 
Thanks for your attention.
 
Jim Austin
 
 
 
 
 
 

To clear it up, let me explain it better.  Maybe I should add it to wiki when I get time. 
 
First, definitions from Chu Moy:
 
 
[size=small] Sensitivity:  A measure of headphone efficiency in dBs SPL per milliwatt of input. A low number means that the headphones need more power to sound as loud as those which have a higher sensitivity. Headphones for portables need to be fairly sensitive because of the lower power output of portable stereos. Modern dynamic headphones have sensitivity ratings of 90 dB or more. When shopping for portable headphones, look for a sensitivity rating of 100 dB or greater.[/size]
[size=small] Impedance:  A measure of headphone load on an amplifier and stated in ohms. This factor is less important with solid state amplifiers, which can drive most headphone impedances, but can be significant with tube amplifiers, which are more sensitive to load impedances. Both consumer and professional headphones generally have impedances of less than 100 ohms. There are professional models rated at 200 ohms or more to minimize loading effects on distribution amplifiers which are often drive a whole bank of headphones at one time. Be aware that very high impedance phones may require more power - on the order of Watts instead of milliWatts.[/size]
[size=small]  [/size]
[size=small] First, the electrical property relationships.[/size]
[size=small]  [/size]
[size=small] Ohm's law[/size]
 
[size=small]

An AC supply applying a voltage 

, across a load 

, driving a current 

.​

​

[/size]

[size=small] Main article: Ohm's law[/size]

[size=small] The meaning of electrical impedance can be understood by substituting it into Ohm's law.^[5][6][/size]

[size=small]

[/size] [size=small]  [/size] [size=small]  [/size] [size=small]  [/size] [size=small]  [/size] [size=small]  [/size] [size=small]  [/size] [size=small]  [/size] [size=small]  [/size] [size=small]  [/size] [size=small]  [/size] [size=small]  [/size]

Complex impedance

[size=small] Impedance is represented as a complex quantity 

 and the term complex impedance may be used interchangeably; the polar form conveniently captures both magnitude and phase characteristics,[/size]

[size=small]

[/size] [size=small] where the magnitude 

 represents the ratio of the voltage difference amplitude to the current amplitude, while the argument 

 gives the phase difference between voltage and current and 

 is the imaginary unit. In Cartesian form,[/size]

[size=small]

[/size] [size=small] where the real part of impedance is the resistance 

 and the imaginary part is the reactance 

.[/size]

[size=small] Where it is required to add or subtract impedances the cartesian form is more convenient, but when quantities are multiplied or divided the calculation becomes simpler if the polar form is used. A circuit calculation, such as finding the total impedance of two impedances in parallel, may require conversion between forms several times during the calculation. Conversion between the forms follows the normal conversion rules of complex n[/size]

[size=small]  [/size]

[size=small] Power[/size]

 

Basic calculations using real numbers

[size=small] A perfect resistor stores no energy, so current and voltage are in phase. Therefore there is no reactive power and P = S. Therefore for a perfect resistor[/size]

[size=small]

[/size]

[size=small] For a perfect capacitor or inductor there is no net power transfer, so all power is reactive. Therefore for a perfect capacitor or inductor:[/size]

[size=small]

[/size]

[size=small]

[/size]

[size=small] Where X is the reactance of the capacitor or inductor.[/size]

[size=small] If X is defined as being positive for an inductor and negative for a capacitor then we can remove the modulus signs from Q and X and get[/size]

[size=small]

[/size]

[size=small]  [/size]

[size=small]  [/size]

[size=small] The relationships:[/size]

[size=small]  [/size]

[size=small]      In the end, how loud the headphones become, comes down to Power applied to the load(headphones).  If you look at all headphones specification ithere would be a dB/mW value.  dB is the logorithmic unit of sound power coming out of the headphones or simply loudness.  So, for example if a particular headphones were 124dB/mW, a mili Watt of power applied to the headphones, would output 124dB of loudness.[/size]

[size=small]  [/size]

[size=small]     Now, lets look at the power equation.  General equation for power is P = IV.  So power is a factor of current and voltage.  Therefore, there is dependency on both, just because current is high doesn't mean the headphones will get loud because voltage may be very low.  [/size]

[size=small]  [/size]

[size=small]       Now lets get into impedance.  You can think of headphones as a impeding device that adds impedance to the load at the output of the amp.  So the relationship is, given an applied voltage, the current flow through the headphones(load) will depend on how much impedance there is.  Lets say for example, you want to drive it a certain dBm(dB/mW) level.  And impedance is Z.  P= IV = V^2/Z.  Since its inversely proportional, at fixed V, higher Z will lower the power, and thus lower the loudness. Also, P = IV = I^2Z because V=IZ.  In this case, Power is proportional to impedance, and depends on square of current.  Two ways of looking at power.  Since, when we raise the volume, we are expanding the voltage swing, its more applicable to use P = V^2/Z.  [/size]

[size=small]  [/size]

[size=small]    Now, lets think about headphone specs.  You will notice some headphones are high impedance, but yet get loud, and some low impedance, but hard to get it loud.  Even if the impedance is high, but sensitivity high, it will be louder than lets say low impedance, low sensitivity because the loudness is in dB scale.  dB/mW.  Sensitivity will have a lot more weight to the loudness because it is in logorismic scale dB than impedance, which is a decimal value.  So, if you want something that gets loud without much power, look at the sensitivity more than impedance because P=V^2/Z.  So therefore, V could scales much greater(since it is squared) if V is high enough, but in headphone signals V is typically less than decade, so it will not factor too high compared to say headphone impedance of 300 or 600 ohm.  Still, since sensitivity depends on Power in dB scale, it would have have much greater significance than impedance values. Its mili-Watt or thousandth versus impedance of maximum possible in the hundreds(since divided, hundredths).  mili-Watt will give you typical 95-125 dB of loudness, so I would pay special attention to sensitivity.[/size]

[size=small]  [/size]

[size=small]    You guys are right, the impedance is at particular frequency because the frequency graph is not linear.  Usually highes impedance value is picked out for specs.  You have to look at the graph to determine how the headphones impede current at all frequencies to know its characteristics.[/size]

[size=small]  [/size]

[size=small] I hope this clears everything up.[/size]

[size=small]  [/size]

[size=small]  [/size]

[size=small]  [/size]

 
Quote:

jimaustin said:

 
 
Well, maybe. But when I wrote something that you now appear to be agreeing with, you said I was wrong. Here's what I wrote:
 
 
You responded:
 
 
and then you wrote, in your latest post:
 
 
(spelling corrected) Which is remarkably similar to what I wrote, quoted above. Specifically, that for high sensitivity / efficiency headphones, high impedance is easier to drive. (If you wish to take issue with my use of "sensitivity" instead of "efficiency", fine, but in this context the difference is technical and not practical.) The point is that if, as I wrote, you look first at sensitivity (or efficiency; whichever the manufacturer chooses to report), and you find it's high enough, then you only need to worry about LOW impedance, not HIGH.
 
But you did write one thing that I believe is incorrect. You wrote,
 
 
But this doesn't make sense in terms of basic physics. First, let's talk about current. Because impedance is frequency-dependent, the impedance will be lower at certain frequencies than at others. Consequently, the amp will crap out first at those frequencies. What happens is the voltage sags.
 
OK, so what about voltage? Up to the point where the current demand exceeds the amplifier's capabilities, the amplifier is essentially a pure voltage source. The higher the impedance, the lower the current demand. SO INTO HIGH IMPEDANCE 'PHONES, THE AMP IS EVEN CLOSER TO BEING A PURE VOLTAGE SOURCE.  The maximum voltage output is set by the amplifier's design; up to that maximum, the output tracks the input very closely, which is exactly what you want. Because the maximum voltage capability of the amplifier -- a set maximum -- is the only limitation in output, this is indeed volume (or volume-setting) dependent, and it does not depend on frequency. When the amplifier tries to generate an output signal that's larger than the amplifier can provide, it clips. This could be a pure 50 Hz sine wave, a pure 10 kHz sine wave, or anything in between (or beyond); it makes no difference.
 
This is my last post in this thread. People will believe what they want to believe. But, so that people can make up their own minds here, I'll report credentials. I have a PhD in physics, and I write for Stereophile.
 
Thanks for your attention.
 
Jim Austin
 
 
 
 
 
 

Click to expand...

That's not the part I reponded to. It was this "Amplifiers (including small ones in portable devices) have no problem providing the required voltage. Their limitation is current. If your amplifier isn't capable of providing enough current, the voltage sags, compromising the sound. Because this happens preferentially at certain frequencies, this affects your frequency response."  Either you don't understand what you said here or specifically chose to ignore it and it's completely wrong and contradicts the running out of voltage for high impedances. Also, frequency response isn't effected by power. Power response is and because music is dominated by low frequency energy, in practical terms, only the impedance at bass frequencies matter for load when discussing power responce. You can et rolled highs with impedance mismatches but it has nothing to do with supply lag. Power response is also very different than frequrency responce as everthing will remain as linear as at low volume until a wattage limit is reached. At that point you no longer need to worry about frequency response as you're just clipping. That's the physics of it. You clearly just don't get it and even reitterated the misunderstanding. Look at amplifier measurements from the mag and they will follow my description.
 
 Quoting the parts you got right and bringing out the Grammar police doesn't make your entire post correct or mine in any way wrong if you didn't change the context. LOL Why do these always turn personal instead of just trying to get good info out there in layman's terms. I'll take good info over a PHD any day so you can just getback up on that high horse...... I have some credentials too but what does that have to do with cost of bread?
 
 
Good stuff High_Q.