Cnet UK posts Denon D7000 hands-on report
Sep 5, 2008 at 5:16 PM Post #16 of 50
Quote:

Originally Posted by user123456 /img/forum/go_quote.gif
Before screaming troll, and before saying it needs an amp, read this:

Electrical impedance - Wikipedia, the free encyclopedia
Ohm - Wikipedia, the free encyclopedia

Just because it's a denon doesn't mean it needs an amp, science means more than brand, just because you have an high-end headphone does not mean its new model need an amp because the others needed.

How it works is, the lower the ohm/impedance the less power you need to drive an headphone, it's basic science.


I obviously know more than you about audio/video/science.



That sentence shows that you not only don't know a damn about basic science, and that I know just a bit more than you of it, but also that you don't have the least interest to learn.
Power is not the same as voltage, and there's a parameter called sensitivity which is of utmost importance to know how easy to drive a transducer is. However you can keep trolling around, I'll happily ignore any further ignorant and disrespectful post of yours.
 
Sep 5, 2008 at 5:20 PM Post #17 of 50
Quote:

Originally Posted by Cool_Torpedo /img/forum/go_quote.gif
Why? Don't you know already, despite your belonging to Head-fi for several months, that the lower the impedance the more current the driver asks from the amp?
You're very wrong if still think that low impedance is equal to high sensitivity. The Denon cans are a bad case of lowish sensitivity plus low impedance, which is very demanding on the amp. Of course they sound from a DAP or crappy computer output, and pretty loud for that matter, but SQ is so inferior than when properly fed by a decent amp...

Rgrds



Its not always true but many a times a lower impedance means Lower Loudness at a given volume level.

And i did mention that with the 25 ohm Headphones i tried today they were deteorited...so u should have read that.

Guess they had High sensitivity like many other low impedance stuff out there.
 
Sep 5, 2008 at 5:25 PM Post #18 of 50
Cool_torpedo..

most of the people tend to have a prejudice that Lower Ohm headphones sound fuller without an amp....its quite true in most cases...u r trying to say there are other factors too..but its not so common so find hard to drive Low impedance headphones.


I guess i ll have to try Denons with an Amp.

So you are basically trying to say that at same impedance two headphones can behave totally different with an Amp?
 
Sep 5, 2008 at 5:39 PM Post #19 of 50
Quote:

Originally Posted by Nocturnal310 /img/forum/go_quote.gif
So you are basically trying to say that at same impedance two headphones can behave totally different with an Amp?


He's not 'trying' to say that but it is true. He's trying to explain how you have to look at more than just impedance to determine how to best drive a headphone. Sensitivity among other things.

There was a really good topic that dispelled impedance myths. The problem is manufacturers don't always give you the desired info.
 
Sep 5, 2008 at 5:39 PM Post #20 of 50
Quote:

Originally Posted by Cool_Torpedo /img/forum/go_quote.gif
That sentence shows that you not only don't know a damn about basic science, and that I know just a bit more than you of it, but also that you don't have the least interest to learn.
Power is not the same as voltage, and there's a parameter called sensitivity which is of utmost importance to know how easy to drive a transducer is. However you can keep trolling around, I'll happily ignore any further ignorant and disrespectful post of yours.



Well said... "Dr. House!"
 
Sep 5, 2008 at 5:48 PM Post #21 of 50
Quote:

Originally Posted by Gradofan2 /img/forum/go_quote.gif
Well said... "Dr. House!"


Err... more of Wannabe Dr. House? ..u cannot compare House to anyone mortal.


You try changing your DP to Greg House and u ll notice a sudden change in your personality too..it ll be amplified.
 
Sep 5, 2008 at 5:54 PM Post #22 of 50
Quote:

Originally Posted by Nocturnal310 /img/forum/go_quote.gif
Err... more of Wannabe Dr. House? ..u cannot compare House to anyone mortal.


You try changing your DP to Greg House and u ll notice a sudden change in your personality too..it ll be amplified.



It just that House doesn't "give quarter to idle chatter, or suffer fools gladly."

"Not that there's anything wrong with that."
 
Sep 5, 2008 at 6:11 PM Post #23 of 50
Nocturnal310, just don't tell me that you've tried to plug low-impedance phones into some OTL tube amp or other amp not designed to drive low-impedance phones.


user123456, there's absolutely no need to show how stupid you are in public.
Also, tell me, do Akg k701(62ohm) need an amp or not? Koss KSC-75(64ohm) need an amp or not?
And finally, do most speakers with impedance around 4-8ohms need amp or not?
 
Sep 5, 2008 at 6:17 PM Post #24 of 50
Quote:

Originally Posted by user123456 /img/forum/go_quote.gif
Sure, but lower impedance means an headphone needs less power, didn't you read the wiki source?
.



Did you even read it? The second one clearly stated power dissapated:

P=V^2/R

So if you lower the resistance for a given voltage you actually dissapate more energy with a lower resistance. That said the impedance still is not the whole story because if it delivers a lower level of dB/mW to work it may even require more voltage.

For example if your phone is has half the impedance of another and has half the sensitivity then you end up with exactly the same volume as before and actually the amp (soundcard, whatever) is putting out 2x the power to sustain this.

The math:
Power=V^2/(R/2)=2*V^2/R = 2*the original phones with resistance R)
Sound Output= (sensitivity/2)*power=.5*sensitivity*2*V^2/R
= sensitivity*V^2/R = Same sound output as original phones

I am afraid that your blanket statement is plain wrong.
 
Sep 5, 2008 at 6:18 PM Post #25 of 50
Quote:

Originally Posted by Nocturnal310 /img/forum/go_quote.gif
Cool_torpedo..

most of the people tend to have a prejudice that Lower Ohm headphones sound fuller without an amp....its quite true in most cases...u r trying to say there are other factors too..but its not so common so find hard to drive Low impedance headphones.


I guess i ll have to try Denons with an Amp.

So you are basically trying to say that at same impedance two headphones can behave totally different with an Amp?



Many other members and I have covered this topic in other threads. Impedance is just one of the factors that we have to consider to foresee how demanding on an amp or HP output a pair of cans is.

It's true that impedance on AC is not the very same as resistance on DC, but considering that many HPs have quite flat impedance curves across the frequency response, with flat phase angles, it's quite safe equaling them when speaking of HPs.
Looking at things this way, it's also true that the higher the impedance the phones have, the more opposition they present against the current flow, so they seem to be needing more power. But this is not the whole picture. You need to see the heapdhone as a mechanical transducer which asks for electrical power to produce sound pressure level. And here is when sensitivity comes into play.

Sensitivity tells you how loud the phones will sound when applying a mW of power on them. If you have very sensitive headphones, something like 115dB/mW, it doesn't matter what's their impedance, with very little power they'll sound loud and quite good. The higher is their impedance, the less current intensity they'll ask from the output, so they can be even easier to drive than an equally sensitive pair of phones having lower impedance. Remember that power is a function of Voltage and current Intensity. More impedance asks for more voltage being all things equal. The reason why most IEMs don't need an amp to sound good is not their lowish impedance, is their high sensitivity which in most cases is well above 110dB/mW. But this high sensitivity has its tradeoff, which is that very low voltage signals, like hums and noises in the source, will be clearly heard.

As long as the sensitivity of the phones goes decreasing, you need more power to make them sound equally loud. Exactly the double of power for every 3dB you decrease the sensitivity. Hence phones rated at 97dB/mW will need 64 mW to sound at 115 dB, while the cans having that 115dB/mW sensitivity need just one miliwatt! To make matters worse, if those cans at 97dB/mW are low impedance, they'd be asking for that power more in the form of current Intensity than in voltage form.

If that's clear, then it's quite easy to understand that impedance means nothing alone, since also sensitivity should be considered, and that having a combination of low impedance and low sensitivity, is quite demanding on the power source.

Rgrds
 
Sep 5, 2008 at 6:46 PM Post #26 of 50
Yup,

it sums it up..

you can actually paste this in another post and make it a sticky.

We seriously need a Sticky on impedance & frequency response.

Since these 2 are the only terms found on Headphone packaging...many people buy headphones based on these.
 
Sep 5, 2008 at 6:54 PM Post #27 of 50
No offense, Nocturnal310, i'm thinking of you as a nice person, but i have an advice for you and it's up to you accept it or not.
Being not so knowledgeable, try not to suggest, post you opinions on matters you're not completely understand in many threads around head-fi, but rather ask more questions, search for info, try to expand your knowledge base. It's obvoius you're seriously interested in headphones, i wish you luck.
 
Sep 5, 2008 at 7:06 PM Post #28 of 50
Quote:

Originally Posted by Nocturnal310 /img/forum/go_quote.gif
Yup,

it sums it up..

you can actually paste this in another post and make it a sticky.

We seriously need a Sticky on impedance & frequency response.

Since these 2 are the only terms found on Headphone packaging...many people buy headphones based on these.



I posted a similar message on a thread titled "impedance for once and for ever" or something like that, but it died buried. Very few people is really interested in knowing why things happen and how they work, so they don't do searches and not even try to understand when a good explanation is offered.

Like QQQ I encourage you to keep reading and getting a deeper knowledge about this. It's not that difficult and would place you in a better position to understand comments from knowledgeable people.

Rgrds
 
Sep 5, 2008 at 7:09 PM Post #29 of 50
@Torpedo

Exactly! Many on these forums have been relying on a hp's impedance rating as the only factor to consider when judging it's ability to be driven when, in fact, many other factors (epecially sensitivity) account for this. It's like saying a pair of 4 ohm satellite speakers would be easier to power than 16 ohm in-ears.
 
Sep 5, 2008 at 9:32 PM Post #30 of 50
Quote:

Originally Posted by user123456 /img/forum/go_quote.gif
...

Usually the lower the ohms are the less power you need, since more ohm/impedance needs more power to drive.

But there is also sensitivity and the db/watt also got something to do with if or not an amp is needed, but for the most part the higher the ohms are the more need there is for power.

There is a confusion saying more ohm makes the can drive better, but then again 1 ohm is the same as 1 volt releasing 1 coulomb per second which equals 1 ampere. The more ampere the more power there is need for.



No no no no and no. Power is just power and once you feed a miliwatt or a watt, what tells you how loud will it sound is the driver's sensitivity. Once you have a power source capable to deliver that needed power to get a certain SPL, nothing else matters. The problem is that amp manufacturers just tell you how much power their unit can deliver, but they give that figure rated for a certain fixed impedance.
A power source might be able to deliver 1mW at 32 ohms, but if it cannot swing volts enough, it won't be able to deliver such mW onto a 300 ohms load. The same goes if the manufacturer tells you that his amp can deliver that 1 mW at 100ohms, if the amp isn't capable to deliver enough current intensity, it won't reach that mW under lower impedances.

Just look at this easy formulas. Power is equal to voltage by intensity (P=V*I) ok?
On the other hand by the well known Ohm's law, I=V/z (where z is impedance) right?
Then on the first formula you can replace I for V/z so power P=V^2/Z Hence the higher Z, the lower P, so the higher the impedance of the load, supposing you're capable to maintain the voltage constant, the lower the power you're using. That's because a higher impedance demands less current intensity (I)

Seriously, if you don't really know how it works, please stop spreading nonsense and misinformation to the rest of members that have a true interest to learn something about their systems.
 

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