Headphone Impedance

Eagle156

2[H]4U
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Jan 21, 2006
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If lower impedance headphones are easier to drive, then why are there high impedance cans?:confused: Does higher impedance make it sound better? It's strange because for mics, the cheaper mics are high impedance and the more expensive stuff is low and it's the opposite way for headphones.
 
Impedance has nothing to do with sensitivity.
Where did you ever get the idea that low impedance headphones are easier to drive then high impedance headphones? ;)
 
It's not nearly so simple. Actually high impedance headphones are usually called "easy" to drive and low impedance "hard" by amp manufactures. While low impedance phones require less voltage to get loud, they require more current. This leads to problem like hiss is systems, weak bass and so on. It is fairly easy to make an output that drives something high impedance, much harder to make something drive low impedance.

As a simple example look at the difference between a soundcard and power amplifier. For normal listening levels, power amps are generally putting out somewhere in the realm of 1 volt. Well, soundcards have no problem doing that either. However if you hook your soundcard to your speakers it won't work. Why? Because soundcards are made to drive low current, high impedance systems. Line inputs are generally in the realm of 20k Ohms or more. Very little current. Speakers, however, are very low impedance, usually 8 Ohms or less. That means you need waaay more current at a given voltage level. At 1 volt a 20k Omh line input needs only 0.05 mA of current. An 8 ohm speaker at the same 1 volt needs 125 mA of current.

Opamps are good at driving voltage, not so good at current. So when you load them with a low impedance it works, to a point, but you get more noise, less bass, more distortion and so on. You need to buffer the output of the system to properly drive low impedance phones.

About the best you can get for direct soundcard usage is actually a mid impedance kind of phone. Say in the 60-120 Ohm range. Those are low enough that the voltage output form the card should be sufficient to get to high enough levels to make you happy, but high enough that the opamps are going nuts with the current requirement.

Really the best idea is just to get a headphone amp when possible. Get something that can do a higher voltage output for high impedance phones and is buffered for low impedance phones.

After all, you wouldn't try to hook speakers straight to a soundcard, why would you do it with headphones? Headphones are just small speakers. In either case, an amp is really the way to go.
 
Impedance has nothing to do with sensitivity.

It is not directly related but it has something to do, AFAIK the sensitivity in headphones is the sound pressure level that the driver can produce when is fed with a 1mW signal, the impedance can affect how easy is for a circuit to match the sensitivity. Of course, with a carefully designed circuit, impedance can be a non-issue.
 
http://tangentsoft.net/audio/ppa/amp/bench/

Tangent notes the Senns are the "easy" test, the Gardos are the "hard" test.

Not surprising from a design standpoint. If you are dealing with a real high impedance target, an opamp alone is fine. If you want to drive something lower impedance, you need something that handles the current load better.
 
If you want to drive something lower impedance, you need something that handles the current load better.

Are you really suggesting that my Ety ER6i's need more current then my HD650s? :confused:
 
Yes.

Simple math. Watts = voltage * current, voltage = current * impedance (V=IR, Ohm's law). So the lower impedance a device is, the more current it takes to drive it to a given level.

If you have really low impedance phones, they'll be mostly driven by current, not voltage. In fact you'd probably discover that you can just buffer the line outputs of your device and that'll be plenty loud as full line level is enough voltage to make them loud. If you have really high impedance phones they are mostly driven by voltage, you'll need a larger voltage swing to get them to go louder, but they won't put a heavy current load on the source.

This is also why power amps are rated in terms of the minimum ohm load they'll take. They don't give a maximum rating because for all intents and purposes there isn't a maximum. They've got no problem supplying less current, they'll just supply less power overall before maxing out. However more current is a problem, there is a max they can handle. Thus they are rated for minimum safe impedance.

If you take two 8 ohm speakers and hook them in to two channels of an amp, then take them and hook them in parallel to one channel, you'll find that their volume level is the same, if the amp can handle a 4 ohm load. The reason is that the amp is driving the same amount of voltage in either case, however when you hook both in it halves the impedance and thus doubles the current output.
 


The current draw with an Xp amp driving the Etys to ear slipping levels is only 28mA
With the Sennheisers it is 36mA.

Shot a big ass hole in your simple math eh?

Edit:
I am looking at it on the scope and I know it is not oscillating so....
 
At an equal voltage? Sorry man, but if you want to try to convince me you've found a situation where Ohm's law doesn't apply, I'm not buying. At a given voltage, the current is determined by the impedance of the system. That's just how it is.

Now if you are talking equal SPL then that depends on other factors like the efficiency of the headphone, the frequency response, and the distance from the eardrum.

However when you are talking power, V=IR is what dictates it.
 
If you have really low impedance phones, they'll be mostly driven by current, not voltage. ...

If you have really high impedance phones they are mostly driven by voltage, ....

Technically, all dynamic headphones are driven by current, only electrostatic headphones are driven by voltage.
 
Some low impedance cans can be hooked in headphone Amp and prvides greater quality than unamped like Sennhesier HD595 with 50 ohm, also there is some opamp such as LM4562 can drive high impedance cans resonably.
 
At an equal voltage? Sorry man, but if you want to try to convince me you've found a situation where Ohm's law doesn't apply, I'm not buying. At a given voltage, the current is determined by the impedance of the system. That's just how it is.

Now if you are talking equal SPL then that depends on other factors like the efficiency of the headphone, the frequency response, and the distance from the eardrum.

However when you are talking power, V=IR is what dictates it.

It does apply, however, you can't put a fixed voltage output unless you want to limit it's power.
P=IV is the correct equation to use, but the flaw is your treating V as a constant variable which isn't true which is what MisterX was trying to point out. Different circuits behave differently and so the value of V is not constant. Both I and V are variables in headphones. In EE they teach your the superposition method using the Norton and Thevenin(sp?) equivalent circuits. You have to perform both circuit analysis before you can use the superposition method to analyze a circuit.
 
god i wish i had a better understanding of electricty but it still somehow elludes me. anyways, i have the denon ah-d950 (discontinued) which have a 30 ohm rating. now true i dont have a true headphone amp to test them with but i used my sound output from my pc, my father's old reciever amp and my sandisk mp3 player powered by a single AAA battery, and they all sounded virtually the same to me.

i figured the low impedence meant they were "easier" to drive and get the same quality but apparently thats not the case. now i'm really confused as to why they sound the same coming from a reciever/pc as compared to my AAA battery powered mp3 player... unless that reciever and my pc output the same non-amp'ed output as my mp3 player?
 
V is variable, but only in so far as the signal you are feeding. Your source outputs a voltage wave, which the amp increases or decreases that goes to the headphones. The amp will then output the necessary current to drive that, if it can. If not, the voltage output will sag. Same deal as a wall socket, or anything else. The sound wave that you want to output is defined by voltage level. In an ideal situation there'd be a 1:1 correlation between voltage level and driver movement. Thus the current is dictated by the impedance. Of course if the amp can't supply the necessary current, the voltage has to drop, hence why having an amp that can supply enough current is essential, especially in terms of low impedance phones.

Thus back to the original statement. Feed a test tone, say a 1kHz sinewave at 1 volt output in to high and low impedance phones. The low impedance ones will draw more current.
 
Of course.

So you are saying that at an equal voltage, you have a system with lower impedance drawing more current? Well that'd be a violation of Ohm's Law. That would be, to put it mildly, a major shakeup in electronics. I am going to say one of three things is happening here:

1) You aren't measuring the current correctly. Remember I am talking about the current going from the amp, to the headphones. If you aren't measuring at that point, you aren't measuring the right thing.

2) Your voltage levels aren't consistent. You should again be checking that at the output. Use a sinewave as a test to ensure that the RMS level doesn't vary with time.

3) There's an impedance oddity. Impedance varies with frequency in any kind of audio drivers, headphones included. It is possible you are at a valley in the Sennheiser curve and a peak in the Ety, though I find it highly unlikely that they'd ever cross given the difference.

Most likely you are simply measuring wrong. For current you need to be in line with the amp and the headphones to be measuring it properly.
 
lol
Do you really think you need to tell me how to measure the current draw of an amplifier I designed?
 
lol
Do you really think you need to tell me how to measure the current draw of an amplifier I designed?

I didn't know you designed an amplifier, but you don't appear to be measuring what I'm talking about. I am not talking about the current draw of an amp, as that only partially relates to its headphones. For example the PPA I have draws .11amps pretty much regardless of what you do with it, because it is fairly heavily biased.

I am talking about the current output to the headphones. Lower impedance phones need more current at a given voltage, that is just how it works. It is also what is relevant for how hard an amp has to work to drive headphones and in what way. High current loads are hard for many opamps, they are better at voltage. Hence why many headphone designs have buffers (or equivilient discreet phase).

I'm going to assume there's just a miscommunication problem here, as I have a hard time believing someone who designed an amp doesn't understand Ohm's law. So to repeat: I am talking about the output current of the amp, to the headphones.
 
Thus back to the original statement. Feed a test tone, say a 1kHz sinewave at 1 volt output in to high and low impedance phones. The low impedance ones will draw more current.

Your applying all of the equations by themselves right, but missing the big picture.
P=IV
V=IR
P= I^2 * R
and yadda yadda....

however, the efficiency of the drivers effects what is being measured. There is no such thing as an "easy" or "hard" to drive headphone/speaker. Only when looked into the context of not having enough "current" does that statement even make sense.

Remember the overall big picture is that Maxwell's equations and the conservation of energy holds true. However, simple ohm's law has nothing to do with conservation of energy which is what your arguing for. An unmeasured variable doesn't mean you can ignore it all the time. In this case I think measuring the sound output level density, area of the output, heat created by the transducers(I think that is what it is called)/drivers is a bit much....

Something low impedance can draw a lower current than something of a higher impedance and still output more sound. In the case of the etymotics vs. the HD650, if we were able to measure the power output reliably, the etymotics would simply output less power. The only reason they could sound the same loudness is because the method of sound conduction is more efficient. It could also produce less waste heat from the conversion of electrical energy to mechanical energy.
 
Sigh. I think I'm just going to give up here. Perhaps I am not communicating clearly, whatever the case, it isn't working. You can perhaps go chat with Tangent over on headfi about it and maybe he can do a better job.

I was simply trying to explain that low impedance phones aren't easier to drive. They take less voltage to drive loud, but more current, at least when they are equally efficient. Remember that low impedance doesn't imply more efficiency. Efficiency also depends on driver material, magnet assembly, enclosure, and so on. For example my ATH-A700s have a quoted efficiency of about 102dB and are rated at 64 Ohms. Koss UR-18s are rated at 92dB but are only 32 Ohm.

Also I was trying to explain that opamps aren't ideal for driving current, especially the ones often selected for audio. Thus they have an easier time when presented with a high impedance load.

At any rate, I'll stop trying now, as I don't think I'm doing a good job of explaining myself.
 
No no, your doing good. It is just that your missing the fact that your not looking at the circuit from a whole.

Take this simple circuit problem
I have a 1V battery and a resistor at 1Ohm hooked up together.

Using V=IR, the current that flows through the circuit should be 1A.

What you were suggesting is that through some "magical" trick I am able to force 2Amps through the circuit is where your logic broke down. What happens then is that either the resistor or the battery is going to be competing with each other. If the resistor can pass 2Amps then it probably won't heat up and itself become a 2V voltage source causing the battery to be charged with 2 volts or explode when the battery heats up due to it's internal resistances and capacity for holding a charge. On the other hand, if the resistor can't handle 2 amps going through it, it will start to heat up and you would be looking at whether or not the resistor could handle 2 watts of power, or be able to dissipate only 1 watt of power as heat. The other 1 watt left is already taken care of because of the resistors normal operation.

I think tangent wouldn't disagree with anything said here except for the blanket statement that low impedance phones are "hard" to drive and high impedance phones are "easy" to drive. He may have used those words for lack of a better terminology. Opamps can be current or voltage limited. With opamps, if your voltage limited, you simply increase the rail voltage by adding another battery in series to each other. This is easy to do and you can increase voltage until you reach the opamps voltage specifications before releasing the magic smoke. However, if your current limited with an opamp, there isn't much you can do to get more current because opamps are made using FET's as opposed to using BJT's. Most semiconductors are better modeled as constant current sources. Tangent was simply using "hard" because he knew that the opamp could with a distinct possibility not provide enough current to fully saturate the lower impedance headphones.

And technically, high impedance loads are always the best. You get maximum power transfer between your amplifying circuit and load while minimizing the transmission losses that occur in between.
 
That post was not really intended as a correction, it was more like here, check this out. :D
 
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