Volume, Power, Voltage, Current, and Amplifier Output Impedance in Planar Headphones

[manbear]

I'd like to check with the folks of Sound Science to make sure I'm understanding these concepts correctly. Let's imagine we have two of the same planar magnetic headphones (how about an HE-400 with a flat impedance of 35 ohms), hooked up to two different amps. Amp #1 is a SS speaker amp with 300 ohm series resistors on the outputs. It has an output impedance of ~300 ohms and a max power of ~120 mw into a 35 ohm load. Amp #2 is a SS headphone amp with <1 ohm output impedance and a max power of 1.2 W into the same load.

These numbers are just for example though -- the whole point is that I don't think they make any difference if neither amp is clipping. I am aware that some measurements have put the impedance of the HE-400 at 50 ohms, not 35, but that shouldn't affect any of the following. 

Suppose we play white noise through the amps and adjust the volume knobs until volume is equal through both headphones and neither amp is clipping (say 80db).  Let's take some measurements across the drivers of the headphones. 

By my logic:

1) Since volume is equal, the voltage through both headphones must be equal. 
2) Since both headphones have the same impedance at all frequencies, and that impedance is constant across all frequencies (being planar headphones), Ohm's law tells us that the current through the headphones is the same. 
3) Take the same voltage, the same current, and the same impedance, and use Ohm's law again to conclude that power through both headphones is the same. 
4) Since the impedance of the headphones is constant with respect to frequency, damping factor does not affect the sound and amp #1 is not "boomier" or whatever than amp #2. 

Corollary:

If an iPod doesn't clip at a given volume, then it is putting the exact same power, voltage, and current through the HE-400 as, say, a Cavalli Liquid Gold or Headamp GS-X at the same volume. 


Questions:

Have I made a mistake anywhere? 

Can the output impedance of some amplifiers (e.g. tube amps) vary with respect to frequency and how would this affect 4)? Is there any reason to the think that an SS speaker amp with non-inductive resistors on the outputs would have a changing output impedance? 

 

 

[xnor]

Yes, yes, yes and yes.
 
 
 
If we're nitpicking then output impedance always varies a little bit due to real-world imperfections, like a rise from 20 mOhm at 20 Hz to 40 mOhm at 20 kHz, but I guess we can agree that something like that is negligible (0.005 dB "roll-off"...).
 
At the low end it can rise drastically if you have small DC blocking capacitors on the output. Fortunately such amps seem to be getting rarer and rarer.
 
 
Also, I'd hope that higher end headphone amps have a lower output impedance, so they'd actually deliver less power into dynamic headphones (~100 Hz impedance peak) if normalized by volume at 1 kHz or even 500 Hz.

 

[Chris J]

I'd like to check with the folks of Sound Science to make sure I'm understanding these concepts correctly. Let's imagine we have two of the same planar magnetic headphones (how about an HE-400 with a flat impedance of 35 ohms), hooked up to two different amps. Amp #1 is a SS speaker amp with 300 ohm series resistors on the outputs. It has an output impedance of ~300 ohms and a max power of ~120 mw into a 35 ohm load. Amp #2 is a SS headphone amp with <1 ohm output impedance and a max power of 1.2 W into the same load.



These numbers are just for example though -- the whole point is that I don't think they make any difference if neither amp is clipping. I am aware that some measurements have put the impedance of the HE-400 at 50 ohms, not 35, but that shouldn't affect any of the following. 



Suppose we play white noise through the amps and adjust the volume knobs until volume is equal through both headphones and neither amp is clipping (say 80db).  Let's take some measurements across the drivers of the headphones. 



By my logic:



1) Since volume is equal, the voltage through both headphones must be equal. 

2) Since both headphones have the same impedance at all frequencies, and that impedance is constant across all frequencies (being planar headphones), Ohm's law tells us that the current through the headphones is the same. 

3) Take the same voltage, the same current, and the same impedance, and use Ohm's law again to conclude that power through both headphones is the same. 

4) Since the impedance of the headphones is constant with respect to frequency, damping factor does not affect the sound and amp #1 is not "boomier" or whatever than amp #2. 



Corollary:

If an iPod doesn't clip at a given volume, then it is putting the exact same power, voltage, and current through the HE-400 as, say, a Cavalli Liquid Gold or Headamp GS-X at the same volume. 



Questions:



Have I made a mistake anywhere? 



Can the output impedance of some amplifiers (e.g. tube amps) vary with respect to frequency and how would this affect 4)? Is there any reason to the think that an SS speaker amp with non-inductive resistors on the outputs would have a changing output impedance? 




 

 


I agree.
If the iPod is not even clipping on peaks for a given SPL, then the headphones will receive the same amount of power whether we are talking about an iPod or a Liquid Gold.
Obviously, the liquid Gold can give us higher SPL when we turn the volume up because it has a higher power rating, (more output current, more output voltage).
Obviously an iPod is not the best amp for driving inefficient headphones.

Obviously a big SS headphone amp like the Bryston HPA can output more voltage, more current and hence more power than the lowly iPod but if we are using (for example) a high efficiency 'phone like the Grado SR-225i then the only thing the Bryston should do for us is "sound better".

 

[xnor]

Well I just skimmed through that HE-400 thread and it seems that people like mlxx are confusing quite a few things.
 
 
Ohm's law et al. are perfectly fine, if you actually understand them.
 
V = Z / (Zout + Z) * Vin
 
(where Vin is the voltage the amp outputs into an infinitely high resistance)
 
I = V / Z
P = V*I = V^2 / Z
 
 
So given a certain voltage (volume) at a given frequency the resulting power is fixed.

 

[manbear]

That first formula is the voltage divider formula, correct? I know that increasing the output impedance of an amp, all other things being equal, will reduce the power into the headphones. Which is why I added resistors to my speaker amp to get some range on the volume knob. It's funny though, with 250 ohm resistors I'm only getting ((20*35)/(250+35))^2/35 = 172 mw, or if we use the 50 ohm spec, I'm getting ((20*50)/(250+50))^2/50 = 222 mw. Which is on the low end for "recommended" power into the HE-400. So the guy with massively overpowered amp is actually using less power than most. Kind of ironic... 

I think the primary sticking point in the discussion was the concept of assuming equal volume and measuring across the headphones drivers, or "load" in the diagram below. This already takes into account the fact that increasing output impedance reduces power into the load. I think mlxx was imagining that measurements are being taken across both the "output impedance" and "load" resistors. From that reference, it would be true that you'd have different amounts of power to get the same volume.

 

 

[xnor]

To reach the 110 dB SPL target (as described in headphones-sensitivity-impedance-required-v-i-p-amplifier-gain) at 500 Hz according to IF measurement you need about 2.2 V. That's less than 100 mW into 50 ohms.
Add 3.5 dB excess gain and we arrive at roughly your 222 mW (or 3.3 V).
 
 
Output power specs into a certain load can be accurate, but sometimes the manufacturer is playing tricks with the units or does some rounding magic to arrive at numbers that are not possible in the real world.
 
 
Btw, have you shorted the negative terminals or did you recable the headphone?

 

[manbear]

 
Output power specs into a certain load can be accurate, but sometimes the manufacturer is playing tricks with the units or does some rounding magic to arrive at numbers that are not possible in the real world.
 
Btw, have you shorted the negative terminals or did you recable the headphone?

Hmm, yeah, perhaps that explains why my Little Dot MKIII that supposedly had 100 mw into 32 ohms clipped at high volumes in the HE-400. 

I made a TRS adapter with the sleeve connected to just one of the negative terminals and alligator clips connected to resistors on the positive terminals. Warning: do not touch. LOL.

I need to learn how to solder...

 

[xnor]

It's quite easy to see that you will get clipping with the LD:
100 mW into 32 ohm => 1.8 Vrms
 
Smallest gain is 3x.
1.8 / 3 = 0.6 Vrms on the amp's input. If you go higher than that you should get clipping when you turn up the volume, maybe already a bit before that.
 
The gain options don't seem to be well thought-out...
 
 
 
Anyway, yeah that does look a bit adventurous.

Make sure you don't short anything when moving the cable..

 

[Chris J]

  That first formula is the voltage divider formula, correct? I know that increasing the output impedance of an amp, all other things being equal, will reduce the power into the headphones. Which is why I added resistors to my speaker amp to get some range on the volume knob. It's funny though, with 250 ohm resistors I'm only getting ((20*35)/(250+35))^2/35 = 172 mw, or if we use the 50 ohm spec, I'm getting ((20*50)/(250+50))^2/50 = 222 mw. Which is on the low end for "recommended" power into the HE-400. So the guy with massively overpowered amp is actually using less power than most. Kind of ironic... 

I think the primary sticking point in the discussion was the concept of assuming equal volume and measuring across the headphones drivers, or "load" in the diagram below. This already takes into account the fact that increasing output impedance reduces power into the load. I think mlxx was imagining that measurements are being taken across both the "output impedance" and "load" resistors. From that reference, it would be true that you'd have different amounts of power to get the same volume.

 

 
Hey man,
So you are starting off with a 50 Watt @ 8 Ohms per channel amplifier.
This outputs a maximum of 20 Volts RMS.
Now you put a 250 Ohm resistor in series with the output of the amp (increasing output impedance to 250 Ohms).
So your maximum output voltage is 2.45 Volts.
 
I have a lowly, little Matrix M Stage headphone amp, maximum output voltage is approx. 8 Volts RMS.
 
You don't have to be a rocket scientist to see that you will get more power from the Matrix than you will from the 50 Watt power amp with a 250 Ohm resistor in series, LOL!
There's more to it than that, of course, the two amps will both current limit at some point, but the Matrix will still output more power into the 32 Ohm 'phones.
It's rated at 400 mW into 60 Ohms.

 

[xnor]

Well he can lower the resistors or just skip them altogether and push 8 watts into his headphones if he likes... the headphones won't though.

 
 
I can see you touching the volume control, *clack*: "oh, not again" (headphones blown).

 

[manbear]

@chris j

Yes, I am aware that my setup is a bit silly. It's even more silly to compare it to the M-stage, though, for a number of reasons. I can easily and cheaply change the resistors to get as much power as I want, up to 8 watts. The resistors are there to attenuate noise, but I'm in the process of experimenting with lower impedances to see how much attenuation I really need -- and how much power I really need. I think I already have enough. I don't know if I even want your 8v output... But I could have it, and more, if I wanted. LOL

Last time I checked, my amp cost less than an M-stage. Who knows, it might even sound better too.

I'm just kidding you. I mostly got this amp so I could play around with something. And because I might get an HE-6. Or speakers. Who knows....

If anything, this amp has taught me that I need less power than I thought I did. Even though I got it because I wanted more. I'm still trying resistor values that will give me 500 mw, 1000 mw and 2000 mw to see if I can hear any difference, or if the peace of mind from having enough more than enough power for transients or music with high dynamic range is worth the tradeoff in volume knob travel. I've used the amp with no resistors, but the noise floor bothers me too much that way. The HE-400 didn't mind, but my apple earbuds sure did...

[VIDEO]http://vimeo.com/80125357[/VIDEO]

 

[bigshot]

The "more is better" theory is the worst theory ever.

 

[Chris J]

  The "more is better" theory is the worst theory ever.

 
Over 11,000 posts.........

 

[xnor]

He didn't say he's better because of his post count.

 

[castleofargh]

but that is just looking at how many volt we need/how loud it gets, nothing about current.  he400 might need more power than some because it combine low efficiency and low impedance. where we often only need to look strong for either V or A so real power is not mandatory.