Output power vs. impedance
Mar 17, 2013 at 6:17 PM Thread Starter Post #1 of 11

J Bones

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My question here mainly comes from the real-world example I have found here in my own set-up in my room:
 
The other day I happened to stop and think about the difference in specs between the A/V receiver I have and desktop amp I have. One, my A/v receiver, puts out about 100 W/channel into 8 ohm speakers and the other, desktop amp, puts up to 6W into 32ohm load. My first thought was, what the heck, why did I ever buy an amp for my headphones that outputs a 15th of the power. Well upon looking into, I know that may factors such as output resistance (being the biggest factor) and nominal load of the speakers/headphones will come into play. I know that my desktop amp has an output resistance of 1 ohm and that my receiver is somewhere near 40 or 50 ohms in output impedance. But how does this correlate to lower power at the output and/or driving the headphones more efficiently? I've read that a voltage source ideally has an output impedance of 1 ohm or less and that the opposite, a current source, would be ideally the opposite, but I don't understand how having a lower output impedance can cause reason to believe that less power at the output (6 W) is needed to drive a pair of headphones just the same or even better than my A/V receiver. I tried to look this over using real world examples with Ohm's Law and it's variations but came to no solutions or conclusions on this. Can anyone here guide me in the right path on this? Thanks for any help you may have!
 
Mar 17, 2013 at 7:06 PM Post #2 of 11
Almost every audio amp is pretty much a voltage source driving a load in series with its own output impedance.
 
V_s = source (amp) output voltage
Z_s = source output impedance
V_L = voltage seend by load (headphones)
Z_L = load (headphones) impedance
P_L = power dissipated in load
 
V_L = V_s * Z_L / (Z_L + Z_s)
P_L = V_L^2 / Z_L, kind of
 
note that for constant output level (V_L), there's more power delivered if the load impedance is smaller
 
except not really when clipping, or it's not quite that simple, there is some small distortion, etc... but as a model for what happens on a basic level under normal operating conditions, when we're focused on just power outputs, it's pretty good.  Also, technically for a complex load the power delivered is just the real component of the complex power, but let's ignore that.
 
The output voltage that the receiver is capable of is larger than that for the headphone amp.  And it can supply that kind of voltage even with a low impedance load without clipping, whereas say the headphone amp may not do so well with 8 ohms or so, depending.  As for why, you'd need to consider the layout and design of the electronics.  Many older receivers with a headphone jack are tapping the speaker amp with some resistors, effectively increasing the output impedance for usage with headphones, reducing the level of the voltage that headphones would get when connected.  That large output impedance would not be seen when running speakers.  Some just have a dedicated headphone amp circuit.
 
 
The larger the power delivered, the louder the headphones or speakers sound.  For most headphones, most people are probably listening with average power delivered of about 1 mW (0.001 W) or less.  Depends.  For some headphones it could be a lot more (though 6W is pretty crazy for anything unless maybe it's K1000), and for some headphones and definitely some IEMs it could be a whole lot less.
 
As for the sound, depends on the performance of the amps at the listening levels, how they deal with the load.  Usually the differences are overstated though.
 
Mar 17, 2013 at 8:11 PM Post #3 of 11
Ok, in looking at this I believe I can reason that:
 
With V_L and Z_L remaining constant, a decrease in my source's output impedance would make the denominator in the above equation smaller, hence for V_L to remain the same, one of the numerator's variables (V_S or Z_L) must decrease and since Z_L is being held constant then it must hold that V_S must decrease accordingly (or in this case, to achieve same effect in driving the headphones it can decrease). Since power is obviously a combination of the voltage and current supplied, holding the sources current constant and decreasing voltage supplied by the source would result in an amp that provides less power to the load but with similar or better results? Does this seem correct?

I essentially had to write this out to make sure I understood it myself but it would seem that the above would hold true for my real world example in which the only really changing variable is the source's output impedance. Being so I believe I have it right, or at least hope so! Thanks for your help!
 
Mar 18, 2013 at 12:02 PM Post #4 of 11
Wait a minute. Your A/V receiver has an output impedance of 40-50 ohms? What receiver is this? It would be a complete mess trying to drive typical loudspeakers with such a grossly high output impedance. Sure it's not more like 40-50 milliohms?
 
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Mar 18, 2013 at 12:21 PM Post #5 of 11
Quote:
Wait a minute. Your A/V receiver has an output impedance of 40-50 ohms? What receiver is this? It would be a complete mess trying to drive typical loudspeakers with such a grossly high output impedance. Sure it's not more like 40-50 milliohms?
 
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Honestly, I could be completely wrong but I thought it was somewhere in that range. It's a Sony STR DE597 that I bought off a friend a year or so ago...maybe I should look at the specs again. 
 
EDIT: So I was way off-- I believe the output impedance is only 8 ohms, but regardless the same principle should hold true as to why my desktop amp can output less power (wattage) and achieve the same amplification for my headphones (actually it's many times better than my receiver).
 
Mar 18, 2013 at 1:16 PM Post #6 of 11
The receiver output is very very very unlikely to be 8 ohms either.  From the docs I see, it's just listing max power levels into 8 ohms.  Output impedance is a lot less than that and unlisted.
 
The receiver's headphone output could possibly be different and have higher output impedance.
 
 
For any of the headphones in your sig or most headphones out there, you're not going to be using close to the max possible output power out of something like a Lyr.  Q701 aren't even rated to handle more than 0.2W, and at those levels, they would be unbearably loud.  At equivalent listening volumes out of the receiver and then out of the headphone amp, you're getting the same V_L yes.  But none of these power calculations cover any of the other performance metrics like distortion levels, frequency response, noise levels, etc.  If you're looking to max power calculations (again, you're not even using max anyway) to find out details about sound quality, you're not really going to find anything.  The actual max output levels are not really predicted by these equations anyway.
 
The only thing here of interest with respect to sound quality is if you could find or test for (takes a couple of measurements with a multimeter) the receiver's headphone out's output impedance.
 
Mar 18, 2013 at 1:52 PM Post #7 of 11
The output impedance also indicates how much "control" your amplifier has over your speakers.

Z_Load / Z_Source = Damping Factor.

where Z_Load is your headphone impedance, and Z_Source is your amplifier output impedance.
 
Mar 18, 2013 at 1:57 PM Post #8 of 11
Ok-- but I'm not looking for a comparison in SQ between two output sources based solely on their max output power. Of course I'll never get anywhere near using the full 6 W but why is it then that nearly 1W out of the Lyr could make me go deaf/destroy my cans but close to 100W out of my receiver wouldn't make my mini speakers even close to that loud? I know using the phono output on my receiver prob doesn't allow me to garner that full 100W that the receiver is capable of, but why would one of my mini speakers or tower speaker for that matter need closer to 100W out of the receiver to be powered versus the < 0.2W needed out of the Lyr for the Q's?

Sent from my XT907 using Tapatalk 2
 
Mar 18, 2013 at 2:17 PM Post #9 of 11
Quote:
Ok-- but I'm not looking for a comparison in SQ between two output sources based solely on their max output power. Of course I'll never get anywhere near using the full 6 W but why is it then that nearly 1W out of the Lyr could make me go deaf/destroy my cans but close to 100W out of my receiver wouldn't make my mini speakers even close to that loud? I know using the phono output on my receiver prob doesn't allow me to garner that full 100W that the receiver is capable of, but why would one of my mini speakers or tower speaker for that matter need closer to 100W out of the receiver to be powered versus the < 0.2W needed out of the Lyr for the Q's?

 
Oh, you were talking about using the receiver with speakers?
 
Do you listen with speakers strapped to your head?
 
It depends on a lot of things, but there's roughly a three (two, four, five) orders of magnitude difference between power levels required for equivalent SPL at your ear, from
1.  typical speakers at say nominal 1m distance
2.  typical headphones worn on head
 
e.g. 96 dB SPL / 1 mW for headphones;  96 dB SPL / 1W @ 1 meter for speakers
 
Think of all the energy that's not going to your ear, when listening with speakers.  As for how much power they can handle, think of how much larger the speaker drivers are and need to be.
 
Mar 18, 2013 at 4:25 PM Post #10 of 11
And that may be the answer I was exactly looking for! Thanks everyone for the input, I had a feeling it was a much simpler answer than I was looking for but couldn't hone in on what it was.
 

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