[lini]

Sovkiller: I don't think so. Look at the drawing again, please - in case a, the load impedance for the amp must be 330 Ohm plus anything below 100 Ohm. In case b, the load impedance for the amp is definitely something below 100 Ohm. But the effective output impedance for the headphone is 330 Ohm in both cases, because in both cases has the signal has to pass through the 330 Ohm resistor in order to reach the phone. Case c is the only case in which the effective output impedance "seen by the headphone" is below 100 Ohm, whereas the load impedance for the amp would be some 77 Ohm plus the impedance of the headphone.

Greetings from Hannover!

Manfred / lini

 

[pbirkett]

So from what you say, lini, this speaker adapter probably wouldnt make any appreciable difference to using the inbuilt jack on my amp?

 

[Zoide]

I agree w/ the other comments about Pioneer amps. My Pioneer A-35R makes the HD580s sound very very good, according to some short testing I did.

 

[antonik]

Quote:

Originally Posted by lini antonik: Well, either my understanding of electrophysics is fundamentally flawed (which is possible ) - or you are wrong. In this particular case, I'd give the latter a 75 % chance. But I've made a little drawing for further discussion: Greetings from Hannover! Manfred / lini

Manfred - I am correct, you can work out the rest

. On your drawing "b" is not a divider and in this case the output impedance would be 330 Ohm, "a" and "c" will provide same output impedance (77 Ohm) but with different voltage levels available (and "c" is not a divider either). If you will turn the divider in "a" upside down (100 Ohm on top) it would be still 77 Ohm output impedance, thought the division ratio would change.

Alex

 

[lini]

Alex, the only thing I can work out is that you obviously fail or refuse to see the difference between output impedance and load impedance.

And just btw, b is also a divider - but it's a current divider and not a voltage/potential divider. However, I've only included it, because it fits the Arcam guy's description better than case a. Case c is not a potential divider in its strict sense, but nevertheless it works as such - and it's the only case in which the headphone would see an output impedance of ~ 77 Ohm. And if one swapped the 100 and 330 Ohm resistors in case a, the division ratio would indeed change - however, the headphone would get to see an effective output impedance of 100 instead of 330 Ohm. And for the amp, the load impedance would change from 330 Ohm plus something below 100 Ohm to 100 Ohm plus something below 330 Ohm.

Paul: Are you refering to the Grado model? Well, they use fairly low values for the resistors, so that might actually make a difference.

Greetings from Hannover!

Manfred / lini

 

[antonik]

Quote:

Originally Posted by lini Alex, the only thing I can work out is that you obviously fail or refuse to see the difference between output impedance and load impedance. And just btw, b is also a divider - but it's a current divider and not a voltage/potential divider. However, I've only included it, because it fits the Arcam guy's description better than case a. Case c is not a potential divider in its strict sense, but nevertheless it works as such - and it's the only case in which the headphone would see an output impedance of ~ 77 Ohm. And if one swapped the 100 and 330 Ohm resistors in case a, the division ratio would indeed change - however, the headphone would get to see an effective output impedance of 100 instead of 330 Ohm. And for the amp, the load impedance would change from 330 Ohm plus something below 100 Ohm to 100 Ohm plus something below 330 Ohm.

You are, unfortunately, completely wrong in your explanation for "a" . Look at it this way - voltage across the load would be half of off-load voltage if the load impedance is equal to the output impedance. Do the calculations and you''ll see that the OUTPUT impedance of the circuit "a" is ~77 Ohm in either case - with 100 or 330 Ohm at the top, thought the off-load voltage would be different.

Alex

 

[Coalescent]

 

[Jorg]

I've heard some very old Denon CDP. The sound was amazing. I can't find it anywhere and I start to forget how it looked. They all look the same... BTW I've been told that it cost a lot.

 

[lini]

Quote:

Originally Posted by antonik Look at it this way - voltage across the load would be half of off-load voltage if the load impedance is equal to the output impedance.

Uhm... Why should that be the case?

Puzzled greetings from Hannover!

Manfred / lini

 

[antonik]

Quote:

Originally Posted by lini Uhm... Why should that be the case?

Because it is one of the ways the output impedance defined?!

Imagine, that you don't know what's inside the amp. How would you measure its output impedance? There are two common ways - either as I've described - by loading the output with a known value load, putting some signal into the amp and looking at the output voltage reduction comparing to off-load condition (as the output impedance and load create a potential divider), or by putting the signal INTO the output of the measured amplifier from another amplifier through a known value resistor and look at the voltage on the output - in this case the divider consists from the external resistor and the output impedance (in the second case there should be no signal on the input of the amplifier under test).

If you have an electronics reference book, look for Thevenin's theorem - that will make things clear. In "The Art of Elecronics" by Horowitz & Hill it is on page 11.

Alex

 

[lini]

Alex, sorry man, I still don't get it. I understand your measuring approach, I guess, but I think we're missing each others' points. So let me try another example - with better values for easier "brain-only" calculation: Case a with R1 being 300 Ohm and R2 being 100 Ohm - ok? So that would give us a 3:1 divider, doesn't it? Now if we add another 100 Ohm resistor parallel to R2, R2total of R2.1 and R2.2 should be 50 Ohm, changing the divider ratio to 6:1, right? And if we added a 300 Ohm resistor instead, R2total should be 75 Ohm, giving us a 4:1 divider, wouldn't it?

Ahhhhh, now I get it - we misunderstood each other. You're correct in the way you understand it - but I am, too. Because for the headphone itself, the signal flow through R2 isn't really interesting. What it sees, is that it is getting its signal through R1 - which is significant for what I mean when I talk about "effective output impedance", you see?

Greetings from Hannover!

Manfred / lini

 

[antonik]

Quote:

Originally Posted by lini Ahhhhh, now I get it - we misunderstood each other. You're correct in the way you understand it - but I am, too. Because for the headphone itself, the signal flow through R2 isn't really interesting. What it sees, is that it is getting its signal through R1 - which is significant for what I mean when I talk about "effective output impedance", you see?

Manfred, sorry, there is no such thing as "effective output impedance", only just "output impedance", and it's value is important, as it affects the frequency response, distortion and sound quality of the headphones. For the load (headphones in our case) does not matter HOW this output impedance created - by one resistor, by two or by five. The important thing is only the equivalent value of the output impedance. Please read Thevenin's theorem.

Cheers

Alex

 

[Steve999]

76.74418604651163 ohms.

 

[bifcake]

My foot bone's connected to my leg bone
My leg bone's connected to my knee bone
My knee bone's connected to my hip bone
And that's how Bifcake was born

 

[lini]

Alex, I know what you mean, now. Anyway - an output impedance value of 77 Ohm is too high for many, if not most of the headphones out there (with just a handful exceptions). So in order to get better sound, Arcam should use lower resistor values - for example they could use 33 and 10 Ohm resistors instead, maintaining their divider ratio, but providing a much lower output impedance. Of course, the resistors would have to burn a higher wattage and be more expensive - which I'd suspect to be Arcam's real reason for picking higher resistor values.

Greetings from Hannover!

Manfred / lini