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Impedance Adapters/Cables | Explained & Listed - Page 3

post #31 of 103
Quote:
Originally Posted by Steve Eddy View Post


Ok. You'd said "Since you have added a damping resistor across the headphone you have done two things..." I thought you meant it was the 220 ohm resistor across the headphones. Sorry.

 

 

Ok. So if he's got a 27 ohm resistor in parallel with the headphones, that means it's also across the amplifier's output. So the effective output impedance of the amplifier, i.e. the source impedance seen by the headphones, will be much lower still.

 

 

If we consider the headphone to be the load, then the headphone is effectively seeing an amplifier with an output impedance of about 25 ohms, i.e. 100 + 220 in parallel with 27.

 

se

 


No problem.

 

If he has the 27 ohm resistor in parallel with the headphones, then the headphones appear to be a 24 ohm load.

For this parallel combination, obviously the 27 ohm resistor will pull more output current than the 250 ohm headphone.

 

The headphone amp appears to have a 320 ohm output impedance because the 220 ohm resistor is in series with the headphone amp's original output impedance of 100 ohms.

So we have increased the output impedance.

STV014's experiment added a 27 ohm damping resistor across the headphone, he used this damping to counter the effects of the high output impedance (assuming the headphone you are using needs damping!).

 

Another way to look at this is: assume the headphone amp is an ideal voltage source with ALL the other impedances outside the headphone amp.

The IDEAL voltage source (i.e.headphone amp in this discussion) has Zero output impedance.

Then the load on the IDEAL headphone amp appears to be 100 + 220 + 24 = 344 ohms.  So we may now calculate how much TOTAL current is drawn from the amp.

With a little bit of calculation (using Voltage divider principle) and power calcs we can also see that the efficiency of this whole network is very poor, i.e very little output voltage is actually applied across the headphone.  And most of the output current is flowing thru a 27 ohm resistor!

 

I can very easily see how this confuses everyone!confused_face_2.gif

 

C

 


 

 

post #32 of 103
Thread Starter 

You have that correct, i am a bit lost, I am going to reread this, but only later when I have time, but please continue to discuss it :) Chris J your input has been very valuable, thank you.

 

It also may be worth stating the risks involved here, again, where others can read it.

post #33 of 103
Quote:

Originally Posted by Chris J View Post

 

If he has the 27 ohm resistor in parallel with the headphones, then the headphones appear to be a 24 ohm load.

 

The headphone appears to be a 24 ohm load relative to what exactly?

 

With the configuration given, the amplifier sees a load of 220 ohms plus the parallel combination of 27 and 250 ohms, or about 244 ohms.

 

Quote:
For this parallel combination, obviously the 27 ohm resistor will pull more output current than the 250 ohm headphone.

 

It won't be much. You're only talking about a 6 ohm difference. Given, say, a 2 volt output, the 250 ohm headphone will draw 5.7mA (accounting for the amplifier's 100 ohm output impedance). With the headphone and the resistor network, it will draw 5.8mA. So you're talking about a difference of 0.1mA.

 

Quote:

The headphone amp appears to have a 320 ohm output impedance because the 220 ohm resistor is in series with the headphone amp's original output impedance of 100 ohms.

 

No, the headphone amp appears to have an output impedance of 24.9 ohms. You're leaving out the 27 ohm resistor that is across the amplifier's output. And that's what, in this case anyway, largely defines the headphone amplifier's effective output impedance.

 

Quote:
So we have increased the output impedance.

 

No. The output impedance has not been increased. From the perspective of the headphones, which is the only perspective of any meaning here, the amplifier's output impedance drops from 100 ohms to 24.9 ohms.

 

Quote:
STV014's experiment added a 27 ohm damping resistor across the headphone, he used this damping to counter the effects of the high output impedance (assuming the headphone you are using needs damping!).

 

You keep overlooking the fact that the 27 ohm resistor is not just across the headphone but also across the headphone amp's output. And as such, it defines the headphone's output impedance as seen by the headphone.

 

Quote:
Another way to look at this is: assume the headphone amp is an ideal voltage source with ALL the other impedances outside the headphone amp.

 

Which is precisely how I was looking at it.

 

Quote:
The IDEAL voltage source (i.e.headphone amp in this discussion) has Zero output impedance.

 

Yes.

 

Quote:
Then the load on the IDEAL headphone amp appears to be 100 + 220 + 24 = 344 ohms

 

Yes.

 

Quote:
So we may now calculate how much TOTAL current is drawn from the amp.

 

Yes.

 

Quote:
With a little bit of calculation (using Voltage divider principle) and power calcs we can also see that the efficiency of this whole network is very poor, i.e very little output voltage is actually applied across the headphone.  And most of the output current is flowing thru a 27 ohm resistor!

 

Yes.

 

But since when was the issue efficiency?

 

The issue was your claim that a resistor network could not reduce the effective output impedance of the amplifier. And that's simply not the case as even basic circuit analysis here shows.

 

If you take your ideal voltage source and add a 100 ohm resistor in series with its output and you've got an amplifier with a 100 ohm output impedance. Add the 220 ohm series and 27 ohm shunt resistors and you now have an amplifier with a 24.9 ohm output impedance just as sure as I'm sitting here. That's the source impedance that the headphone will see and it will behave just as it would if it were being driven from an amplifier without the network and an inherent output impedance of 24.9 ohms.

 

se

 

 


Edited by Steve Eddy - 3/27/12 at 9:39pm
post #34 of 103
Thread Starter 

Just to note the efficiency claim comes from stv014 who is actually using that adapter, I am sure he also noted that he needs to increase his volume levels.

post #35 of 103
Quote:
Originally Posted by WiR3D View Post

Just to note the efficiency claim comes from stv014 who is actually using that adapter, I am sure he also noted that he needs to increase his volume levels.


Yes, he mentioned that in his post with the graphs. With an amp with 100 ohm output impedance, a resistive network with a series 220 ohm resistor and a shunt 27 ohm resistor and 250 ohm headphones, you lose 20dB compared to without the network.

 

se

 

 

 

post #36 of 103
Thread Starter 

Steve Eddy, chris and i had a nice chat in private, and (I am going to paraphrase, which means it could be incorrect) he mentioned something about certain amps being designed to have high output impedances as a safety factor, and that lowering it can be dangerous?

 

Also the TI headphone amp in the Asus STX is designed for a 10 ohm output resistance, so could lowering it be bad for my sound card?

post #37 of 103
Quote:
Originally Posted by WiR3D View Post
Also the TI headphone amp in the Asus STX is designed for a 10 ohm output resistance, so could lowering it be bad for my sound card?


Not if the sound card still sees a reasonably high impedance load. That is why the adapter includes a serial resistor as well, not just a parallel one. The output impedance of the amplifier itself is not changed, only what is seen from the headphone. Let's say we have an adapter with a 24 Ohm serial, and a 4.7 Ohm parallel resistor, the headphone impedance is 25 Ohm, and the card has an output impedance of 10.5 Ohm. This means that:

  - the amplifier sees a load of 10.5 + 24 + 1/(1/4.7+1/25) = 38.46 Ohm (i.e. the headphone+adapter combination is ~28 Ohm)

  - the headphone sees an output impedance of 1/(1/4.7+1/(10.5+24)) = 4.14 Ohm

  - the damping factor is increased from 2.38 to 6.04

  - maximum voltage on the headphone is (with 7 Vrms output from the TPA6120): 7 / (1/4.7+1/25) / 38.46 = 0.72 Vrms

  - this is enough for a maximum SPL of 90+20*log10(0.72/0.064) = 111 dB (more power?)

  - the maximum current (with sine output) is 0.72 * (1/4.7+1/25) = 0.182 Arms

  - power dissipation on the serial resistor is 24 * 0.182^2 = 0.795 W (at least 1 W resistor is needed)

  - check here for some idea on how the TPA6120 is likely to perform with a 28 Ohm load

  - the worst case power dissipation on the TPA6120, with +/- 12V supply voltage and 6 V square wave output, is 6^2 / 38.46 = 0.936 W per channel. This may be of concern if the chip is not well cooled, although in practice with realistic music signals and listening volumes the power is likely to be less

Hopefully the power supply on the STX (particularly the negative rail) is good enough to actually output enough current at the maximum voltage with this load, this cannot be verified from the available information.

By the way, this headphone+adapter combination is only about twice as efficient overall as the Hifiman HE-6.

 

post #38 of 103
Quote:
Originally Posted by WiR3D View Post

Steve Eddy, chris and i had a nice chat in private, and (I am going to paraphrase, which means it could be incorrect) he mentioned something about certain amps being designed to have high output impedances as a safety factor, and that lowering it can be dangerous?

 

 

Well, one example of that would be my TEAC A-H500 50 watt integrated amplifier. The headphone output is driven straight off the loudspeaker outputs through a pair of 390 ohm resistors. That's so that if you've got the amp cranked up and are pumping 50 watts into a pair of 8 ohm speakers, if you plug in say, a pair of 50 ohm LCD-2's, they'll only be driven by about 100mA instead of 566mA. However using the impedance adapter described here wouldn't lower the amp's output impedance in the same way as if those 390 ohm resistors were replaced with 24.9 ohm resistors. In other words, Chris J's concerns are unfounded.

 

se

 

 

 


 

 

post #39 of 103
Thread Starter 
Quote:
Originally Posted by Steve Eddy View Post

 

Well, one example of that would be my TEAC A-H500 50 watt integrated amplifier. The headphone output is driven straight off the loudspeaker outputs through a pair of 390 ohm resistors. That's so that if you've got the amp cranked up and are pumping 50 watts into a pair of 8 ohm speakers, if you plug in say, a pair of 50 ohm LCD-2's, they'll only be driven by about 100mA instead of 566mA. However using the impedance adapter described here wouldn't lower the amp's output impedance in the same way as if those 390 ohm resistors were replaced with 24.9 ohm resistors. In other words, Chris J's concerns are unfounded.

 

se

 

Thank you once again for the input,

And i wouldn't say unfounded, reading it over again he was more specifically referring to changing the resister before the output jack, if there is one.


But now that I have the equations I can update the main post. Thank you stv014

 

Thank you

 

post #40 of 103
Thread Starter 

updated, when i get time i will redo the equations into step by step fashion, but not now

post #41 of 103
Quote:
Originally Posted by WiR3D View Post

 

Thank you once again for the input,

And i wouldn't say unfounded, reading it over again he was more specifically referring to changing the resister before the output jack, if there is one.

 


Ah, ok. Yes, if you're talking about reducing the output impedance of an amplifier that's using an output resistor to limit output by reducing or removing that resistor, that's a whole other matter and agree with Chris J on that.

 

se

 

 

post #42 of 103
Thread Starter 
Quote:
Originally Posted by Steve Eddy View Post


Ah, ok. Yes, if you're talking about reducing the output impedance of an amplifier that's using an output resistor to limit output by reducing or removing that resistor, that's a whole other matter and agree with Chris J on that.

 

se


I am adding you to the thanks list :)

 

post #43 of 103
Quote:
Originally Posted by WiR3D View Post


I am adding you to the thanks list :)

 



Oh don't bother. Just another mouth to feed. biggrin.gif

 

se

 

 

post #44 of 103
Quote:
Originally Posted by Steve Eddy View Post

 

Well, one example of that would be my TEAC A-H500 50 watt integrated amplifier. The headphone output is driven straight off the loudspeaker outputs through a pair of 390 ohm resistors. That's so that if you've got the amp cranked up and are pumping 50 watts into a pair of 8 ohm speakers, if you plug in say, a pair of 50 ohm LCD-2's, they'll only be driven by about 100mA instead of 566mA. However using the impedance adapter described here wouldn't lower the amp's output impedance in the same way as if those 390 ohm resistors were replaced with 24.9 ohm resistors. In other words, Chris J's concerns are unfounded.

 

se



Looks like you read 50% of the thread and understood 85% of it.

In addition, there is a difference between output impedance and reactive load damping.

 

post #45 of 103
Quote:
Originally Posted by Chris J View Post


Looks like you read 50% of the thread and understood 85% of it.

In addition, there is a difference between output impedance and reactive load damping.

 


Not from the perspective of the headphones there isn't. Sorry.

 

se

 

 

 

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