[bLue_oNioN]

Quote:

Originally Posted by hintermute Heh, never noted the image, thats pretty sweet. Question extension: The resistors, my understanding of electronics leads me to believe that they would just reduce the amount of power to the drivers. I dont understand anything beyond a reduce in volume by way of less .. hm, resistors only change voltage, or amperage? and how would that affect the drivers. -amperage = -volume? Man oh man, this headphone amp, hi-fi business is so much more confusing than my digital electronics class in HS taught me, but even then, outside of the and/or/not gate theory, my intellect fails me and i remain oh so dim.

Do you mean, [size=xx-small]P = V^2/R[/size] ?

 

[hintermute]

Quote:

Originally Posted by bLue_oNioN Do you mean, [size=xx-small]P = V^2/R[/size] ?

Wow, I am rusty. If in that P=amperage,
Then I think so. However.. Yeah, I think
I understand. That brings up another
though. Headphone amps, they provide
an increase in amperage, voltage, or
both? And how would both affect the
performance of the drivers?

 

[sumguy_]

P = power...

Voltage = loudness; Higher V, greater amplitude

Resistors reduce Current (Amperage) not Voltage.

This is just my general knowledge. Im not sure how this applies to headphones but Ill give it a try based on a few assumptions:

Amplifiers can output a greater voltage for loudness, but also supply greater current (thus driving the headphones better). Small amps in DAP's usually can supply the voltage, but not enough amperage to drive cans well. They may be able to drive headphones to loud volumes, but they will sound tinny/not to their full potential, unless they are very efficient.

Usually lower impandance headphones require more current (amperage) to sound their best whereas higher impendance headphones require more voltage to achieve acceptable volume.

 

[xluben]

I am in Physics which deals with electricity and magnetism at the University of Minnesota right now, so I'll try to explain as much as I understand on the topic.

The equations that we are concerned with should be:

V=IR
P=V^2/R

V=voltage (potential difference) (volts)
I=current (amps)
R=Resistance (ohms)
P=power (proportional to volume)

So, as you can see there is an inverse relationship between current (I) and resistance (R). So for a given potential difference (voltage, V) which is provided by the DAP (digital audio player), the lower the resistance the more current will flow in the circuit. Also, power is proportional to volume, and as you can see by the second equation (with voltage (V) constant) the less resistance the greater volume

So, in theory you want low resistance (impedance) headphones to get greather volume output. But the problem is that many DAP's do not have a very powerful amp, and they run out of current. What happens is, there will be a bass note, and with low resistance headphones the volume of the bass note will start out at the correct volume and quickly drop off (possibly to zero). This happens so quickly you can't actually hear that this is happening, but results in the thinner, weak bass and overall 'tininess'.

To get the best sound you need to use a resistor, so that the DAP's amp can provide plenty of current. Then you use a separate amp to increase the availabe current, and run that to your headphones.
You could also run the DAP's line out to the amp, and then (since the voltage is increased) you can get more volume with the resistor in place between the amp and headphones.

Hopefully this made sense (and is correct). I made an assumption that power is proportional to volume. We did a lab which proved power is proportional to brightness (in a light bulb), so I believe it would be the same for volume.

Also the power equation can be written in many different ways depending on what you substitute in:

P=V(V/R)
P=VI
P=I^2R

All of these equations are equivalent, and are equivalent to the second equation above. I include these in case it makes it easier to see the relationship between voltage-resistance, voltage-current, current-resistance, depending on what you are looking to solve for. If you know the max current, voltage output of your DAP or amp you could theoretically find the best resistance to get max volume.

Also, check out this post and the links it contains:

http://forums.ipodlounge.com/showthr...threadid=89618

 

[hintermute]

Awesome. Now it's actually making sense.
All your guys' assistance is appreciated.
Peace