[roll-man]

I often see posts where everybody's throwing around technical data regarding amp or source output, usually in terms of mv or mw or whatever. As you can tell, I'm no engineer nor diy-er. Hell, I play the drums...that should explain it. Well anyhow, I was studying the specs of my Denon AVR 2500 receiver, which powers my main, non-portable rig and I came across the specs for the headphone jack. It read:

284 mW (8/ohms)

So, what does it all mean? Perhaps those in the know can educate a poor fool like me and let me know how this particular spec stacks up in the realm of the headphone world. I'm really interested in learning a little more about the whole output scenario (wife rolling eyes right about now) and really appreciate your input (or is that output?)....well you know what I mean.

 

[Neruda]

it means that if a headphone has an impedance of 8 ohms, that headphone amp will be able to power it with up to 284mW. However, if the impedance is higher than that (and I've never seen a headphone with an impadance of less than 16), that power will drop the higher the impedance goes. So it won't be putting out nearly that amount of power into a pair of Sennheisers (impedance: 300 ohms) or Beyerdynamics 9impedance: 250-600 ohms) than it would a pair of grados. Does that help?

 

[roll-man]

Thanks for the response Neruda. I was starting to think that I'd posted the thread in the wrong forum... So how does that particular jack compare to, let's say a good pcd player or maybe a TA, ya know for comparison sake? Is it a piece of crap, or is fantastic, or something in between?

 

[Neruda]

power can't really tell you the quality of the jack, so I can't say how it would compare to other amps (it's most likely better than a basic pcdp though...). I sort of doubt that it's a very good jack, but I'm unfarmiliar with the amp so i can't say for sure. (The only reason I say that at all is because most headphone jacks are of poor quality.) I do believe that the power output will be very low with higher-ohm headphones, which isn't a really good thing. I actually think that the spec they gave you in that manual (or wherever you read it) is sort of misleading, since (as I said before) there are most likely no headphones out there with a 8 ohm impedance. if there are, they're certainly not among the top-tier headphones.

EDIT: removed faulty mW/mV information

 

[roll-man]

and even I can do the math with my hd 580's @ 300 ohms...not much left over power is there?

 

[Neruda]

I wish i knew how much the power reduced every time the number of ohms doubled...I think eagle_driver knows it, as many others do I'm sure. But yeah, there probably wouldn't be much left over. Still, it could be enough. Really, more power doesn't help anything unless you have the amplifier maxed or near-maxed all the time. All more power can do is make your headphones go louder (although you do need extra power; you don't want to have your amp maxed or else it will clip and distort). The amp will sound better when better electronics are used.

 

[Xevion]

Neruda - a watt isn't 2.5 volts, a watt actually is semi- independant of a volt.

This is my putput as you have given me input. Since GIGO:
HJKFHJKLFHDJKLHFEJKLHJKLF&^R*()F YUIOY&*HIDOP#HUCILHE&#OCOIO#HCIO


Just kidding

. As a former Computer Engineering major, I think I can manage ohm's law here, but it is late so my math might be off.



A watt is the actual amount of power there is

Amperage is the current - think of it as a hose - a wider hose (more amps) means more current

Volts - think of volts as speed - if you have a lot of amps then you don't have a lot of volts - the water passes through a thin pipe much faster then through a thick pipe but the amount coming out can be the same.

Resistance is how much resistance there is - think of a blocked artery. Now - if there is an area with no resistance and an area with 30 ohms of resistance, all of the current will go through where there isn't resistance. If there is a 20 ohm resistor and a 2 ohm resistor in parallel (Road fork), the 2 ohm resistor won't have as many amps passing through it.


So we need to solve some equations. Here is the information we have:

A = ?

W = .284 (1000mW =1W, 1MW = 1,000,000W, do NOT confuse them

)

O = 8

V = ?

So we need to solve for A and V knowing W and O.

Voltage = amps * ohms

Watts = volts^2 / ohms


Just plugging the numbers into the equations we see the top one can't be done yet, but the bottom one can.

.284W = V^2 / 8O

.284W * 8O = V^2

2.272 = V^2

V = 1.507

So - with an 8 ohm load the reciever outputs 284 mW, 1.507V, and ...

1.507v = A *8O

A = .188

So:
V = 1.507
W = .284
A = .188
R = 8 ohms

Now, lets assume for a second that the maximum voltage and amp levels of the reciever are achieved with the 8 ohm load (I have no idea how that works)

So now all we need to do is lose the wattage and change the
ohms and stick into equation.

V = 1.507
W = ?
A = .188
R = 300 ohms


Watts = 1.507V^2 / 300ohms

Watts = .00757 eg 7.57mW at 300 ohms

Now this can be simplified quite a bit - (Watts @ 8 ohms(.284)) / (Ohm rating of headphone / 8)


Assuming the reciever won't reach a higher voltage or amperage level with a higher impedance headphone, these are the numbers you will see.

For comparison, most PCDPs have 5mW outputs, at 32 or 16 ohms. Now - if the reciever said it was rated for .1 watt at 100 ohms, it could pump out .0334 watts on some 300 ohm HD600s for example.


While these power numbers sound like nothing, note that headphone sensitivity is extremly high as the drivers are right next to your ears. I think the HD600 has a sensitivity of 97dB at 1mW or something like that, which means they will play a 1Khz tone at 97dB with 1mW of power - which means your reciever amp won't be underpowered when driving HD600s, but I would imagine a seperate amp would be a good idea as the electronics are probably better in it, and it will probably offer somewhat more power at such a high impedance.

 

[dngl]

Thanks Xevion- your post should definitely be archived.

 

[Budgie]

The easiest way I have found to think of volts and amp is -

Volts is how hard you push.

Amps is how much flow the push creates.

Resistance is the , uh, resistance!

 

[Gluegun]

isn't resistance what pushes back?

 

[Xevion]

Uhh, I don't know how to decribe resistance. It dosen't push back, it is just resistance.

If you guys want me to delve a bit deeper into this and describe parallel and serial circuts and how they effect resistance/voltage/amperage I would be glad to - parallel circuts could explain output of an amp given 2 headphones to drive for example.

 

[Xevion]

Double post

 

[Gluegun]

Viva la Resistance!!

 

[coolvij]

Resistance is, in layman's terms, how HARD it is for the power to move, right?

 

[Xevion]

coolvij - Thats the thing, it dosen't make it "harder" for the power to move. In fact in a parallel circut with a high resistance resistor and a low resistance resistor, the high resistance resistor will be the one dissapating more heat (wattage) which is kind of the inverse of what you are thinking but isn't really. It is rather hard to explain adequately. Ah, well, later tonight I will post the equations and the numbers and you can draw your own conclusions on what to call it.