[Strangelove424]

I’m asking because when I switch between 32 and 600 ohm headphones I don’t have to adjust the volume at all, and they both get too loud at the same volume… kinda weird. If I switch phones on any other source/amp I have I need to do some fairly drastic gain/volume adjustments. I'm not sure how to account for this. 

 

[stv014]

A simple explanation is that the receiver has high output impedance. That is in fact fairly common, the headphone output of many receivers is driven by the speaker amplifier through a resistor or simple voltage divider. For exactly the same power output into both 32 and 600 Ω, the output impedance needs to be the geometric mean of these values (= ~138.6 Ω), but anything in the range of 80 to 240 Ω will still limit the power difference to +/- 3 dB. With the (outdated) IEC standard 120 Ω output impedance, there is also less than 1 dB difference; minimizing power variation depending on the load was in fact one of the reasons why this impedance was chosen, in addition to making headphone outputs easier and cheaper to implement.

 

[Strangelove424]

Quote:

A simple explanation is that the receiver has high output impedance. That is in fact fairly common, the headphone output of many receivers is simply driven by the speaker amplifier through a resistor to reduce the voltage.

Wouldn't that translate to a huge amount of loudness on my 32ohm cans? If I plug those (grados) into an STX at high gain or OTL tube amp they are way too loud at the same volume that the 600ohm cans are just fine. On the receiver, no adjustment required at all. Zero, zilch, none. When I plug the headphones in, there is a brief pause and an audible click, I'm not sure what that indicates as far as circuitry, however. 

 

[stv014]

Quote:

Wouldn't that translate to a huge amount of loudness on my 32ohm cans?

 
No, because the output impedance attenuates the voltage on a lower impedance load more.
 
Here is how the power output from V voltage into Rload through Rout output resistance can be calculated, assuming that no current limiting occurs:
 
P = (V * Rload / (Rload + Rout)) ^ 2 / Rload
 
With V = 7, Rout = 10 (approximate values for the Xonar STX), and Rload = 32 or 600, the power is:
 
32 Ω: 889 mW
600 Ω: 79 mW
 
So, the lower impedance headphone is 10.5 dB louder if it is equally as efficient. Now, with the output impedance increased to 120 Ω, and the voltage to 10 Vrms, the power changes as follows:
 
32 Ω: 139 mW
600 Ω: 116 mW
 
The difference has been reduced to less than 1 dB.
With really high output impedance, the 600 Ω headphone will even be louder than the 32 Ω one.

 

[Strangelove424]

I tend to be a visual learner, do you have a graph or chart I could look at?
 
Edit: Ok, I just saw your formulas. I usually hate math, but I guess I need to learn this stuff finally... let me play with the numbers for a while, maybe I can make a spreadsheet with a chart or something to visualize it better.

 

[stv014]

Here is a simple graph:

Note that the power peaks at Rload = Rout. However, with very low impedance loads, the amplifier might not actually be able to output enough current. Also, for maximum power calculation, all output impedance should be taken into account, even if it is in a negative feedback loop.

 

[mikeaj]

Took me a while (okay, a few seconds) to count those log-scale divisions and realize the load impedances shown were between 16 (?) and 600 ohms.  Can't remember the last time I saw a graph with only one value labeled on the x-axis.  

 

[Strangelove424]

Ok, thanks for the graph. That helps a bit. I'm still trying to get a grip on the formula though. I made a spreadsheet so I could plug numbers in easily, but I'm not sure what to make of P. Is this in watts? And does watts equate to dB level directly?  
 
 

P = (V * Rload / (Rload + Rout)) ^ 2 / Rload
Rload=	600			Rload+Rout=	610
Rout=	10			Voltage x Rload=	4200
Voltage=	7			VxRload/Rload+Rout^2=	47.40661
P=	0.079011

 

[xnor]

It is basically a simple voltage divider.
 
Some receivers have 470 ohm output impedance, but you can plug any number into the formula: Vout = R / (470 + R)
 
Vout = 0.5 if the headphone also has an impedance of  R = 470 ohms (-6 dB re 1 V)
 
Vout = 0.064 if R = 32 ohms (-24 dB re 1V)
 
Vout = 0.56 if R = 600 ohms (-5 dB re 1V)
 
 
So if you have a 32 ohm headphone with a sensitivity of 114 dB/V and a 600 ohm headphone with 95 dB/V they will be exactly the same volume from the receiver.

 

[xnor]

Yeah, Vout squared divided by the load resistance will give you the power in watts.
 
0.001 W = 1 mW (milliwatt)

 

[Strangelove424]

okay, sensitivity was the missing piece of the puzzle. The only problem I have now is that there are no specs for the receiver's headphone out. I don't have a stated output impedance or voltage@impedance number. Should I just deduce that based on STX specs and similiar sound levels (once sensitivity has been accounted for)?
 
edit: i'm not sure deduction would work since there are 2 variables. my head hurts. i think i'm starting to kinda get it though. 

 

[Strangelove424]

BTW, I checked sensitivity ratings and the Grados are 98db, the Beyers 96db. It's probably an audible difference, but they're very close.

 

[xnor]

That's dB/mW. In dB/V: 113 (grado) and 102 (250 ohm beyer) or 98 (600 ohm beyer).

 

[Strangelove424]

Quote:

That's dB/mW. In dB/V: 113 (grado) and 102 (250 ohm beyer) or 98 (600 ohm beyer).

I'm using the 600ohm version, but Beyer's website cites the 96dB figure in dB/mW (as Grado does) and the sensitivity level is the same for all impedances  

 

[xnor]

Yeah I converted it to dB/V for you if you want to use the simpler Vout formula. dB/V does change with different impedances, because you need higher voltage for higher impedance to reach a milliwatt.
 
Btw, most receivers have a much higher output impedance than 10 ohms. Could be 100, 120, ... 470 ohms. Maybe even higher?!