[MrGreen]

Quote:

Originally Posted by plonter /img/forum/go_quote.gif thank you for the detailed explenation. I did saw that formula before regarding the same issue but I probably got a head time to understand it fully because I don't exactly know what all the terms mean. so can you please explain to me the terms: voltage and current? what are they and what is the difference?

Current refers to electrical flow or the rate of electrical flow.
The ampere (SI unit for current) is the current required to produce a certain force between two parallel, infinitely long wires separated by one metre. It's a measure of flow.

Voltage refers to electrical potential difference.
The volt (SI unit for electrical potential difference) is the electrical driving force that drives conventional electric current (the flow of positive charge - i.e. the direction opposite to electron flow) in a certain direction. (I'm not 100% sure about this one, maybe 80% sure. Someone correct me please if I am wrong). It's essentially the electrical version of "potential energy".

I suggest you read wikipedia.

 

[plonter]

Quote:

Originally Posted by MrGreen /img/forum/go_quote.gif Current refers to electrical flow or the rate of electrical flow. The ampere (SI unit for current) is the current required to produce a certain force between two parallel, infinitely long wires separated by one metre. Voltage refers to electrical potential difference. The volt (SI unit for potential difference) is the electrical driving force that drives conventional electric current (the flow of positive charge - i.e. the direction opposite to electron flow) in a certain direction. (I'm not 100% sure about this one, maybe 80% sure. Someone correct me please if I am wrong). I suggest you read wikipedia.

thanks...I did read wikipedia ( a really great source for information) but it is too complicated for me to understand what they are saying...too much complicated and unknown terms.(can't blame them..it is important to all the experts out there)
but i am lucky to have you guys that can explain it to me in a more simple language.

 

[MrGreen]

It's fine. It feels good to know that taking physics in highschool wasnt a complete waste of time.

 

[Menisk]

Maybe this should be stickied somewhere. I'm sure some of the explanations in this thread would help a lot of people understand a bit more about what is required to drive headphones as opposed to the common misconception of impedance being a measure of how difficult it is to drive.

 

[Menisk]

Quote:

Originally Posted by Duggeh /img/forum/go_quote.gif There are 2000ohm headphones which sound rather pleasing right out of an ipod.

Do you have a pair of 414s Duggeh?

 

[Prog Rock Man]

Quote:

Originally Posted by plonter /img/forum/go_quote.gif ....just tell me if i am correct in the next assumption: given two headphones with the same SPL level, the more the impedance, the harder it will be to drive?

This is why I dont see how ohms is not part of the overall measurement of sensitivity. Surely in the above situation, to get the best of a headphone you would need an amp with more power ie watts.

 

[MrGreen]

Quote:

Originally Posted by Prog Rock Man /img/forum/go_quote.gif This is why I dont see how ohms is not part of the overall measurement of sensitivity. Surely in the above situation, to get the best of a headphone you would need an amp with more power ie watts.

It's related to watts (V^2/R, which is how I believe the AKG K701 sensitivity is expressed), but thats all.

 

[Prog Rock Man]

Quote:

Originally Posted by MrGreen /img/forum/go_quote.gif It's related to watts (V^2/R, which is how I believe the AKG K701 sensitivity is expressed), but thats all.

I am sorry but I dont understand that response. I am sorry if I am coming over as deliberately thick, but the answers are hopefully helping me and many others.

 

[MrGreen]

V = Volts, R = Resistance.

Dont worry about it. Just remember that ohms doesnt tell you how difficult it is to drive.

 

[Prog Rock Man]

Quote:

Originally Posted by MrGreen /img/forum/go_quote.gif V = Volts, R = Resistance. Dont worry about it. Just remember that ohms doesnt tell you how difficult it is to drive.

I appreciate your patience Mr Green. So, to clarify, ohms has nothing to do with sensitivity, the figure to look at is SPL and the higher it is the more sensitive the headphone is. So I should rearrange my list of cans as least sensitive first -

Goldring NS1000 100ohms SPL 93 (with NR off)
Grado SR80 32ohms SPL 98
Goldring NS1000 300ohms SPL 101 (with NR on)
AKG K702 62ohms SPL 105
Senn PX200 32ohms SPL 115
Senn mx500 32ohms SPL 125

and two cans with the same ohms and different SPL have different sensitivity, but two cans with the same SPL and different ohms have the same sensitivity.

 

[iancraig10]

Quote:

Originally Posted by Prog Rock Man /img/forum/go_quote.gif I appreciate your patience Mr Green. So, to clarify, ohms has nothing to do with sensitivity, the figure to look at is SPL and the higher it is the more sensitive the headphone is. So I should rearrange my list of cans as least sensitive first - Goldring NS1000 100ohms SPL 93 (with NR off) Grado SR80 32ohms SPL 98 Goldring NS1000 300ohms SPL 101 (with NR on) AKG K702 62ohms SPL 105 Senn PX200 32ohms SPL 115 Senn mx500 32ohms SPL 125 and two cans with the same ohms and different SPL have different sensitivity, but two cans with the same SPL and different ohms have the same sensitivity.

The AkG's are beggars to drive at 62 ohms.

 

[MrGreen]

Quote:

Originally Posted by Prog Rock Man /img/forum/go_quote.gif So, to clarify, ohms has nothing to do with sensitivity

At the risk of confusing you more, I'd prefer to leave it at "Sensitivity tells you how easy it is to make a headphone loud". Please be aware that loud does not equal quality (or "how easy it is to drive well").

Ohms is related to sensitivity in that watts is related to ohms (and sensitivity is given in watts). However in its truest sense sensitivity is related to efficiency.

Also, please be aware that the K701 is given in SPL/V not dB/mW. I'd convert it but I'm getting something that seems way off probably given the fact it is 4 am local time.


Like I said, I personally cannot calculate how well headphone amp x will drive headphone y, simply how loud it can make it.

 

[Duggeh]

Quote:

Originally Posted by Menisk /img/forum/go_quote.gif Do you have a pair of 414s Duggeh?

I used to.

 

[Prog Rock Man]

I suppose as well you cannot rule out the effect of the amp. My X-CAN puts out 1.3 watts and the FiiO E5 100mW. The FiiO is enough to make even the Goldrings listenable on an ipod, but the X-CAN has masses of power in reserve. I dont need to adjust the volume when I switch between K702s and the Goldrings.

 

[Prog Rock Man]

I have found this which I do understand (huurah!). It apparently was originally posted by cool_torpedo and is taken from boomanas thread on full sized headphones and a guide to newbies (hate that word!). Hopefully I have been able to higlight the main points, so the highlighting is mine;


OK, here we go, since I can't find the thread nor the reply, I'll try to explain a bit -again- how the impedance/sensitivity thing goes.

Headphones' drivers are transducers that convert AC electricity into sound. Dynamic drivers, the most common ones, use a coil -wounded wire around a hollow cylinder- into a magnetical field provided by a magnet, to convert the AC voltage into movement which is transferred to a membrane. The membrane's movement is transferred to the air particles in front of your ear. For the frequency characteristics of that vibration, and its pressure level, your brain interprets it as sound.

The voice coil of the driver has an impedance, which is the opposition it presents to the AC source (the amp or any headphone out) to the free flow of electrons thru it. The lower that impedance, the more freely the electrons travel and the closer is the scenario to a short-circuit. This means that your source of electricity needs to pump more current intensity to correctly drive the transducer. So you can take two conclussions from this:
- What makes the AC to drive any coil is its voltage. The minute variations of voltage follow the signal originally recorded.
- The current intensity is important to keep the coil excited, and you need more current intensity the lower is the impedance. There's a relation between the current voltage and the intensity which is the power measured in watts. Power is the product of the voltage and the intensity: P=V*I. This is why amps are rated for their power output and not only for their voltage capabilites.

Up to this moment there's no relation between the impedance and how loud the transducer will sound. However there's a parameter named sensitivity which tells you how loud will a transducer "sound" for a given amount of power you're feeding it. The sensitivity is rated in dB/mW for headphones, so a pair of phones delivering a SPL of 100dB/mW are more sensitive (can sound louder) than a pair rated at 90dB/mW.

So the easy or hard to drive a pair of phones is, depends on both parameters, the sensitivity and the impedance. The worst case would be a pair of phones of very low sensitivity and also a very low impedance. Why? because they'll be asking to the source more watts to sound equally loud as a more sensitive pair, and an important part of that power will be asked in the form of current intensity, which is something that most portable players, headphone outs in receivers and players, etc. aren't designed to deliver. This is the case of cans like AKG 701 or Denon D5000.
If your cans are low impedance but are very sensitive (the case of Grados and most IEMs) then despite their asking more current from the source, they still manage to sound very loud because they need very little power to do so.

Most people tend to think that low impedance equals to louder sound, but this is plainly wrong. It all depends on the sensitivity and how much power the cans need to give a high SPL. Also take into account that not all manufacturers offer their sensitivity values and not all them do in dB/mW but do in dB/mV. It's not much of a problem, you just need to convert the mV in mW knowing the phones impedance.

If anyone has any questions, please don't ask me! Thanks to torpedo and hopefully you do not mind this bastardised reproduction of the original.