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Many AV Receivers/Stereo Integrated amps do NOT use opamp powered headphone jacks - Page 9

post #121 of 216
Quote:
Originally Posted by Sovkiller View Post
Here is an article of how to do it....hope this helps...
Thanks, that looks familiar. Since I don't listen to speakers, my wife is hard of hearing and doesn't like sound, it's even simpler for me to make.

Now another question. Having just read about damping factor, and having read earlier about resistor networks, is the reason for 120 ohm output impedance because if you lower it too much you won't have enough current or volts to run the headphones?
post #122 of 216
Quote:
Originally Posted by Sovkiller View Post
Here is an article of how to do it....hope this helps...
That's a great little table; Here are the equations they used to derive the table (warning: real R values are different from the calculated and always make sure your resistor is properly rated for the correct power output)

1. Vout = sqrt (power * 8ohm)

2. R1 = Vout * R2/(R2+R1); R2 is 120 in their example, use lower R if you want lower output imp.

3. R3 = Rout - R1*R2/(R1+R2)

for Z=10ohm with 20W amp, R1=15.3, R2=10, and R3=3.95 (GL finding a 3.95ohm resistor and with R2 =10 you are probably killing your current delivery (current travels the least resistive path), wherein the Norton equivalent Z is only R2, which is just 10ohm in this case, whereby connecting a Zload of 300ohm will result in 3.2% of the total output current going through the 300ohm load (in the 20W amp case, only 50mA in theory --- wow that sucks...)

Quote:
Originally Posted by scompton View Post
Now another question. Having just read about damping factor, and having read earlier about resistor networks, is the reason for 120 ohm output impedance because if you lower it too much you won't have enough current?
yes
post #123 of 216
Quote:
Originally Posted by tfarney View Post
I still don't get it. Perhaps it's because I don't know what the "V" is. But I get input and output. At either end, the signal from the source is running through a nest of resistors. If that sucks out detail what is the difference. And talk real slow and kinda loud. I'm dumb and I don't hear so good.

Tim
What he is trying to explain is basically that the resistors in a stepped attenator are placed in another position compared to where the resistors are placed when you attenuate the output of a power amp. A regular volume control is placed at the input where the variable resistance allows the user to change strength of the signal which is then amplified and sent out into the hedphones. Changing the volume does not change the output impedance of the amplifier circuit, as the volume control is not a part of the "output circuit".

The resistors used in changing the output of a power amp to a headphone are on the other hand placed after the regular output of the amplifier. This is because a power amp by its very nature is not designed to drive a load like a headphone. This means that you, apart from attenuating the signal, also change the output impedance of the amplifier, and hence also change the way the way the amplifier interacts with the load, that is, your headphones. Damping factor has already been explained earlier in this thread. Also take another look at the graphs on Meier's page.

Meier Audio

As you can see, changing the output impedance changes the frequency response, something which is not true of changing the resistors that the signal is passing through in a stepped attenuator. Hence we can conclude that using a recevier with resistors at the end results in a distortion of the intended sound, not necessarily restriced to frequency response. Headphones can't be made with a high output impedance in mind anymore. Let's just say "Ipod".

I hope this works in spite of my lack of technical schooling and imprecise use of some terms.
post #124 of 216
Good explanation. But all that really matters is whether the theory has any audible effects with real headphones and real music.
post #125 of 216
Quote:
Originally Posted by DefectiveAudioComponent View Post
Good explanation. But all that really matters is whether the theory has any audible effects with real headphones and real music.

Haven't noticed any.
post #126 of 216
Quote:
Originally Posted by scompton View Post
Thanks, that looks familiar. Since I don't listen to speakers, my wife is hard of hearing and doesn't like sound, it's even simpler for me to make.

Now another question. Having just read about damping factor, and having read earlier about resistor networks, is the reason for 120 ohm output impedance because if you lower it too much you won't have enough current or volts to run the headphones?
It depends on what you mean by lowering the impedance.

By reducing the resistance in R1, you get more current and voltage. To me the main issue of doing this, apart from having too much power for your headphone, is that you'd need a higher power rating resistor. Typical quality metal film resistor only goes up to 3W.

By reducing the resistance in R2, there will be less voltage and current flowing through the headphone. We want the maximum amount of voltage and current without blowing the headphone.

By reducing the R3, I don't see much issue, and I removed it altogether. The only down side is that it the power fluctuation between using headphone of difference impedance will be greater (i.e. higher headphone impedance headphone would receive even less power than with the R3 in the circuit). However it's effect is quite negligible unless very low impedance headphone is used. I tailor the resistor value to my K501, and I don't intend to have headphone of lower impedance in the future.

However, I have been asking about why is 120 ohms the standard since forever. It is more convenient to have a circuit at a higher impedance, but I hope someone can tell me what is the disadvantages of increasing the impedance further.
post #127 of 216
Quote:
Originally Posted by JohanK View Post
What he is trying to explain is basically that the resistors in a stepped attenator are placed in another position compared to where the resistors are placed when you attenuate the output of a power amp. A regular volume control is placed at the input where the variable resistance allows the user to change strength of the signal which is then amplified and sent out into the hedphones. Changing the volume does not change the output impedance of the amplifier circuit, as the volume control is not a part of the "output circuit".

The resistors used in changing the output of a power amp to a headphone are on the other hand placed after the regular output of the amplifier. This is because a power amp by its very nature is not designed to drive a load like a headphone. This means that you, apart from attenuating the signal, also change the output impedance of the amplifier, and hence also change the way the way the amplifier interacts with the load, that is, your headphones. Damping factor has already been explained earlier in this thread. Also take another look at the graphs on Meier's page.

Meier Audio

As you can see, changing the output impedance changes the frequency response, something which is not true of changing the resistors that the signal is passing through in a stepped attenuator. Hence we can conclude that using a recevier with resistors at the end results in a distortion of the intended sound, not necessarily restriced to frequency response. Headphones can't be made with a high output impedance in mind anymore. Let's just say "Ipod".

I hope this works in spite of my lack of technical schooling and imprecise use of some terms.
Thank you. Great explanation. And actually, we had this a couple of days ago. And the feedback I've been getting from off of this board is that the effect of high output impedance is changes in the FR of the phones, but that it has to be pretty high to be noticeable, and even higher to impact high-impedance phones. What was confusing me about cheesebert's response, beyond the technical nature of it, was the "loss of detail" discussion. He seemed to be saying (and I probably misunderstood him) that stepping output down through resistors actually removed information from the signal, when what it actually appears to be doing is, in most cases very slightly, is changing the FR curve I'm hearing. Which, of course, happens when I turn off the speakers and plug in the phones anyway. Which happens when I unplug HD580s and plug in K701s. Which happens when I switch from a Gilmore Lite to a Woo 6.

So the headphone out of a speaker amp, stepped down through resistors, isn't exactly the same as the output of the amp at the speaker terminals. That's ok if I like the sound. I was never going to hear it the same anyway. And I think it puts us back to where we began. The mere fact that it is a headphone jack does not make it inferior to dedicated amps. It doesn't even mean you'll hear a difference. Use your ears.

Tim
post #128 of 216
"Which, of course, happens when I turn off the speakers and plug in the phones anyway. Which happens when I unplug HD580s and plug in K701s. Which happens when I switch from a Gilmore Lite to a Woo 6."

Wow. I think that you may have just hit upon something important here.

I'd love to hear others respond. (And stay completely out of this myself.)
post #129 of 216
The values shown in the ESP link results in approximately 120 Ohms output inpedance, which is an internationally recommended standard for headphones. However, it does seriously degrade damping factor. That's a very simple fact.

And remember, just because it's a standard doesn't make it good or preferable.

So just ask yourself do you want an amp designed for the job, or a resistor-network derived compromise? Your choice!
post #130 of 216
Quote:
Originally Posted by Leny View Post
So just ask yourself do you want an amp designed for the job, or a resistor-network derived compromise? Your choice!
I want an amp that sounds good.

Seriously, it has already been discussed.
post #131 of 216
Correct me if I am wrong, but change in FR is a form of "loss of detail". Let's suppose that a 10khz component of the composite wave taken at a particular instant during a busy music passage is only -50dB, by taking another 3dB or more off of that you are basically pushing that frequency component closer toward the noise thereby reducing your S/N, and accordingly that's perceived as a loss of detail.
post #132 of 216
Quote:
Originally Posted by chesebert View Post
Correct me if I am wrong, but change in FR is a form of "loss of detail". Let's suppose that a 10khz component of the composite wave taken at a particular instant during a busy music passage is only -50dB, by taking another 3dB or more off of that you are basically pushing that frequency component closer toward the noise thereby reducing your S/N, and accordingly that's perceived as a loss of detail.
Correct me if I am wrong too.

While the signal is pushed towards the noise floor, wouldn't the noise floor be pushed down further just like the signal? I'm not questioning you, and I am not in a position to do that.

I have a question that you might be able to answer, since earlier on you touched on attenuating line level signal. I have been thinking about it for the past few days, but decided that it is a worse idea than adding resistors at the amp output. What would happen if line level signal that is fed to the the amp is attenuated to a point that headphones can be plugged directly to the speaker output without the need of resistors? As at this moment, I am also considering of making a passive preamp with a custom stepped attenuator just for this purpose.

My guess is the signal level would be so low that it would get drowned by the noise floor of the amp. However earlier on you mentioned that attenuating line level signal does not cause the loss of detail, and this leads me to my question.
post #133 of 216
Quote:
Originally Posted by odigg View Post
Just so we have more information regarding that bass behind the noise section in the second track, how does it sound through speakers? Do you get that same feeling, that the bass is hidden? I'm trying to figure out if that is the character of the amp or if we are talking about issues only with the headphones.
It's clearly not hidden with the loudspeakers. But it was not completely hidden with the cans either, it was just more clear from the XCAN. I do not have any other power amplifier to use as a reference, to see if the sound would be even more clear or more distinct.

The XCAN was more expensive than the power amplifier so it really should be somewhat better.
post #134 of 216
Quote:
Originally Posted by Navyblue View Post
Correct me if I am wrong too.

While the signal is pushed towards the noise floor, wouldn't the noise floor be pushed down further just like the signal? I'm not questioning you, and I am not in a position to do that.

I have a question that you might be able to answer, since earlier on you touched on attenuating line level signal. I have been thinking about it for the past few days, but decided that it is a worse idea than adding resistors at the amp output. What would happen if line level signal that is fed to the the amp is attenuated to a point that headphones can be plugged directly to the speaker output without the need of resistors? As at this moment, I am also considering of making a passive preamp with a custom stepped attenuator just for this purpose.

My guess is the signal level would be so low that it would get drowned by the noise floor of the amp. However earlier on you mentioned that attenuating line level signal does not cause the loss of detail, and this leads me to my question.
Unlike noise which is composed of all the frequencies with relatively harmonious amplitude, complex music is a composite of a plurality of sine waves at different frequencies with different amplitudes. A -3dB drop at 10khz in the FR will only affect the 10khz component of both the noise and the signal; the rest of the wave components are not effected. However the attenuated 10khz music signal is still measured against the entire noise spectrum not withstanding a decreased 10khz component of that noise. Accordingly the diminution of the 10khz component of the complex wave will affect the final shaping of the complex wave and that is usually experienced as a loss of detail.
post #135 of 216
I have a question on the table. The last column is labeled R1 Power. Is that measured after R1?

chesebert, the link you supplied for the damping factor calculator didn't mention headphones in particular, but I assume that for dynamic headphones, the calculator for dynamic speakers applies. Is this correct? I thought the Wikipedia article on damping factor explained the effect pretty well.

If I understand correctly, a higher damping factor helps control the largest movements of the driver, i.e. give better control in the bass. To get a high damping factor, you need a lower output impedance. It makes me wonder if the graphs in the Meier article indicate an increase in uncontrolled bass as well as a suppression of treble.


Navyblue, when you say you tailor the resistor to the K501, what do you mean? I assume you're using a value of R2 that matches or is higher than the impedance of K501, so more of the current and voltage is supplied to the headphone instead of going though R2.


DefectiveAudioComponent and pp312, I hear a difference in my DT831 between my NAD and a portable amp plugged into the wall. The DT831 has a reputation for being bass shy which it is not out of the NAD. Some of the difference I'm sure is attributable to the portable amp not being up to driving the DT831, but it does mimic the charts in the Meier article to some extent. The DT831's impedance is 250 ohm and I'm using a 150 ohm adapter, so it's 400 ohms. That is asking a lot for a portable amp, even if it's not running off of a battery. Unfortunately, I only have a portable amp for comparison.

I've been reluctant to buy a dedicated amp since I don't listen to dynamic headphones much any more. I am thinking of building a Staving Student Millet Hybrid. I like $50 and some of my time a lot better than $300+. I'm cheap.
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