[Dreamslacker]

I need to know how you guys calculate how much current & voltage a pair of headphones need from the output of an amp to sound optimal. ie. Like how you guys came out with the numbers for the K1000's in another thread.

I mean, does this come from graphs or is there a formula to calculate from the nominal impedance of the cans?

 

[Dreamslacker]

Perhaps I didn't phrase it properly. But what I'm interested in is how I can figure out how many volts (rms) a pair of cans need.

I can calculate the current from the impedance & SPL ratings. Roughly...

 

[mtlin12]

You must check another important spec.~ sensitivity.

For AKG K-1000 it is only 74dB/mW, but as my K-501 it's 94dB/mW ,
it means you nead 10X power to get the same volume,
they are the same impedance ~120£[, so you need much more current
for the headphones, or you have to drive them at higher voltage.

 

[jeffreyj]

Quote:

Originally posted by Dreamslacker Perhaps I didn't phrase it properly. But what I'm interested in is how I can figure out how many volts (rms) a pair of cans need....

You need to know the sensitivity of the cans, which is usually given as dB of output for some standard input (like 1mW or 0dBV or 2.83Vrms or about a dozen other things, it seems, which are "standard").

If it's given as 90dB @ 1mW, then you use the equation that converts the ratio between two power levels into decibels:

10*LOG(P2/P1)

If it's given as 100dB @ 2.83Vrms, then you use the equation for voltage ratios:

20*LOG(V2/V1)

Either way, you still need to know the impedance of the cans to determine how much current and voltage will be required to achieve a given SPL level. For example, say you cans put out 90dB with 1mW input and have a nominal impedance of 300 ohms. Employ Ohm's Law to get the voltage/current combination: V^2 = 300 * 0.001, or 0.548V; and I=P/V, or 1.842mA.

Not exactly burning down the house, here, but 90dB isn't exactly the sound pressure level desired for Block Rockin' Beats. You want something along the lines of 105dB, or 15dB more.

Invert the appropriate dB equation from above to convert a known change in dB back to a power or voltage ratio. First divide the dB by 10 (for power) or 20 (for voltage. This gives you the exponent you will raise 10 to to get the ratio of power or voltage:

10/15=1.5 10/20=0.75
10^1.5 = 31.6 10^0.75=5.62

So 31.6*1mW = 31.6mW while 5.62*0.548 = 3.08V and 3.08V across 300R requires 10.27mA.

 

[jeffreyj]

Quote:

Originally posted by mtlin12 You must check another important spec.~ sensitivity. For AKG K-1000 it is only 74dB/mW, but as my K-501 it's 94dB/mW , it means you nead 10X power to get the same volume, they are the same impedance ~120£[, so you need much more current for the headphones, or you have to drive them at higher voltage.

Actually, you need 100X more power; the 10X multiplier applies to the increase in voltage (or current) necessary to drive these amazingly inefficient cans.

 

[mtlin12]

Yes, it should be 100X power, you're right.

 

[Dreamslacker]

OKay.. Thankz for the great help guys..

I should have realised that those are the formula's needed.. I had all of them in my head.. It just didn't occur to me that they could be used for this application. =)

 

[morsel]

RaneNote 100 - Understanding Headphone Power Requirements

This RaneNote has a table of headphones and their power requirements. I have an older paper copy that included a cool graph, but they seem to have sacked the graph in the process of updating the table to cover more recent models. Sorry, I don't have a scanner.

 

[ServinginEcuador]

There is another way, albeit a bit harder a one that will cause you to think and grow. When you know the impedance of the cans in question, and the volume and wattage needed to acheive that ovlume, you can figure out the rest.

Lets' start with some basic electronics:

Electricity is equal to the Current times the Resistance. E=IR
Power is equal to the Voltage times Current. P=IE
Power is equal to the Current squared times the Resistance P=I^R
Power is equal to the Voltage squared divided by Resistance. P=E^2/R

To figure out what is going on in this circuit you need to rearrange some equations to figure out voltage or current. To do that we need to only use what we have, power and resistance, to figure out what we need. So, after taking some Tylenol, here's what I came up with:

Since P=I^2 R, and we have R and P, we can transpose the equation to find our that I= square root of P/R. By inserting a few numbers into the equation we find that at 74dB of sound we have: I=Square root of 1mW/120 ohms, or .0029 amps of current.

Since we now have of Current, we can figure out the Voltage by taking E=IR. So, .0029amps X 120 ohms gives us a voltage of .346V.


Let's say you're a volume freak and want to get about 110dB of sound from these boys. Now let's see how much current and voltage we would need to run them this hard:

First, in order to get to this power level we have to do the logs to get there. If we need 1mW to get 74dB, we would now need 4096mW to get that volume. So, let's start plugging these things in:

Again, I=sw root of the P/R. So, we now have I=4096mW/120, and now have exactly .185mA of current.

To figure out voltage we take E=IR, and come up with Voltage= .185*120=22.17V.


So, in order to start out with 74dB of sound we need to have .0029 amps of current and .346V of electricity.

To get to an ear piercing 110dB we need a LOT more of both, or 185mA of current and 22.17V of electricity.

Easy, huh?

 

[tangent]

Just one other thing...the behavior of headphones can't be summed up in a single number (dB/mW), so what you calculate using these methods just gets you in the right ballpark. You'll need to listen to find out the exact requirements of the headphones in your situation.

Just one monkey-wrench thrown in the works is that the impedance of headphones isn't constant over frequency.

 

[mtlin12]

Quote:

Originally posted by tangent Just one other thing...the behavior of headphones can't be summed up in a single number (dB/mW), so what you calculate using these methods just gets you in the right ballpark. You'll need to listen to find out the exact requirements of the headphones in your situation. Just one monkey-wrench thrown in the works is that the impedance of headphones isn't constant over frequency.

Yes, I think moving coil type Headphones impedance could be much lower than the spec. at some frequencies, so it is not very easy to drive them well, usually I like to tset them with wide
dynamic range and very loud classical music.

 

[MadCow]

I wonder if anyone has created a tool to ease the calculation of voltage and current based on input parameters such as SPL, sensitivity, impedence, etc, similar to what photographers have for calculating exposure? That would be very useful for the not-math-inclined....

 

[logwed]

Quote:

I wonder if anyone has created a tool to ease the calculation of voltage and current based on input parameters such as SPL, sensitivity, impedence, etc, similar to what photographers have for calculating exposure? That would be very useful for the not-math-inclined....

Not really, because as JeffreyJ said, the different manufacturers all seem to use a different metric for sensitivity. There is no industry standard for that measurement, sadly. Also, I'm not sure if one would be useful, because as Tangent said, the impedance and sensitivity vary greatly across the frequency range.