[SoundFreaq]

From what I understand, there is no way to increase earphone resistance by use of resistors in serial. Instead you increase the amps output impedance. 
 
Suppose you have two different Headphones connected to a single amp by a Y-adaptor. 
 
HP1 = Load of 400 ohms
HP2 - Load of 25 ohms
 
Suppose the Output impedance of the amp (Z) = 2 ohms
 
What is Z?
What is Load 1?
What is Load 2?
 
Do loads 1 + 2 sum together to equal the resistance seen by the amp? Does each individual Load remain the same?
 
Is it possible to "add resistance" to one headphone by adding a second headphone load via a Y-adaptor? Or are you just increasing the amps output impedance?
 
If you don't know what I am talking about, there are two articles will really help you understand things all good audiophiles should know : )
 
Google: "Headphone Impedance Explained" and see the "output impedance" link within that article. (Unallowed to link to that site directly.)
 
Thanks for any insight

 

[stv014]

Quote:

Originally Posted by SoundFreaq /img/forum/go_quote.gif
 
Do loads 1 + 2 sum together to equal the resistance seen by the amp?

 
No, the amp will see a lower impedance, which is 1 / ((1 / 400) + (1 / 25)) = (400 * 25) / (400 + 25) = ~23.5 Ω. Since the voltage on both headphones is the same, the lower impedance one will likely be noticeably louder in this configuration. You can balance this by adding serial resistors to the lower impedance headphone, but it will then see a higher output impedance, which may affect its sound quality.
 
By the way, note that posting links to that site is not allowed on this forum, and your post may get deleted (or not, depending on your luck)

 

[xnor]

For the Y splitter which connects loads in parallel, see: http://en.wikipedia.org/wiki/Series_and_parallel_circuits#Resistors_2
This decreases the load resistance as seen by the amp to below the lowest resistance. The output impedance doesn't change except for that little resistance added by the Y splitter itself.
 
Adding resistors in series with each channel (L, R) does increase the load resistance as seen by the amp. The headphone however sees it as an increase in output impedance.

 

[SoundFreaq]

Excellent answers gents! Exactly what I was looking for. I'm trying to work on a problem and this helps. May have more questions later.
 
stv014, Thanks for the heads up about linking to that site. I hope my work-around is OK. 

 

[SoundFreaq]

Gents,
 
What if I add resistance, 75 ohms, in series between the line-out of the DAC, and the line-in of the amp? Would that change the amp's output impedance in an way?

 

[Steve Eddy]

Nope.
 
se

 

[mikeaj]

Quote:

Gents,
 
What if I add resistance, 75 ohms, in series between the line-out of the DAC, and the line-in of the amp? Would that change the amp's output impedance in an way?

 
No.  By design—and this is part of the point of having this in the first place—the input shouldn't have an impact like that on the output of the amp.
 
You're just putting resistance between the output of the DAC and the input of the amp.  Actually, the DAC already have output impedance commonly around something like 75 ohms, whereas most amps should have input impedance of thousands of ohms.  You'd just be very very slightly reducing the signal that the amp gets.  You can think of it as the DAC driving the input of the amp, which is easy to do.

 

[ultrabike]

@SoundFreaq: One might be able to play around with impedance using transformers. It's been a while and I'm not too current on it though, but if I wanted to add or subtract impedance, transformers might be a good key word to use while searching.
 
This POST might be of some use as well (probably not what you are looking for, but perhaps related)... Regards.

 

[SoundFreaq]

Hey thanks guys for the very helpful replies. Everything makes sense when explained, I just need to learn it. I'm getting the right answers to my questions here and my impedance issues are starting to make much more sense. Problem is, there doesn't seem like there's a way to fix them with commercial solutions. 
 
Thus begins my foray into DIY solutions. : )

 

[xnor]

Maybe explain what you're trying to achieve so we could help you more effectively.

 

[SoundFreaq]

Here's the story:
 
To keep a very long story short, I was hearing distortion in my very sensitive, very low impedance high-end IEMs. I definitely traced it back to the amp, the SR-71B.
 
I mistakenly thought I could increase the resistance of the earphone as seen by the amp with a serial adaptor. I bought the Etymotic ER4P > S 75 ohm converter cable, and the distortion disappeared, and the sound as a whole cleaned up dramatically. Talking about this, I was informed of my misdirection and I read up all on resistance, load, output impedance, and so on..
 
Two things I could not figure out:
 
1. Why did adding serial resistance eradicate the distortion in my IEMs from the amp?
 
2. Why the hell is Etymotic selling a 75 ohm serial resistor cable to make their 25 ohm "P" model into the "much better sounding 100 ohm "S" model. 
 
My answers to these questions after more research is that the 75 ohms happens to synergize with my IEMs, skewing the FR graph more to my liking. There are other examples of this happening with other earphones where serial resistance improves sound. And I suppose Etymotic's earphones work better at 100 ohms, and they have probably built more resistance into the "S" model somewhere. But how does adding a 75 ohm resistor to the "P" model make the earphone measure 100 ohms from 25?
 
Bottom line, I can't use my SR-71B with ultra low-impedance earphones, I'll use my P-51 for that. When I need more power, I'll step up to 71B. The 71B's output impedance must be too high for my little IEMs. But I love the amp, so I was just trying to find a way to make it work.

 

[xnor]

1) Some amps are just not very happy with low impedance loads.
Some op-amps are for example designed to drive 600 ohms min. but that doesn't stop some people from using them to drive headphones directly.
Another problem could be a very high output impedance.
 
Some IEMs have very weird impedance curves due to their crossovers, sometimes dipping to below 8 ohms..
 
2) So you don't have to buy the S model if you already have the P model. That's just a guess though. Whether it sounds better is a matter of preference, even if they say S is more accurate.
 
 
The 75 ohm resistors have to be in series. In series the total resistance is just the sum. This causes quite a bit of the voltage from the amp to drop across the resistor instead of the driver(s), which is also reflected in the specs, i.e. lower sensitivity.

 

[SoundFreaq]

Thanks,
 
all that makes sense, except for the part about the 75 ohm adaptor being in series and making the 25 ohm earphone 100 ohm. If that's just total resistance, the sum, you mean as seen from the amp only (OI)? You're not adding some resistance to the earphone and some to the OI. Right?
 
Shouldn't serial resistors change FR of the earphone? 

 

[xnor]

Let's assume you have an ideal op-amp with 0 ohm output impedance. The amp designer adds 50 ohm resistors inside the amp on the outputs for whatever reason, which increase the output impedance to 50 ohms. Even if you short the outputs to ground, the op-amp will see a 50 ohm load (the internal resistors). If you add a 50 ohm resistor as load, the total load for the op-amp now is 100 ohms.
 
Example two: assume you have an amp with 0 ohm output impedance and the output is set to exactly 1 V. Connect a 50 ohm headphone - the total output voltage of 1 V will drop across the headphone. Add 50 ohm resistors in series, now half the voltage will drop across the resistor, the other half across the headphone.
The problem is the impedance curve of the headphone. At 1 kHz the headphones' impedance may be 50 ohms, but at 100 Hz it may be 100 ohms.
Now the voltage will not split at 0.5V but at 0.333V, so the headphone now gets 0.666V. The amp is still outputting exactly 1V at either frequency, but the headphone gets a higher voltage at 100 Hz. This causes the change in the FR.

 

[SoundFreaq]

Got it, makes sense. Thank you for the detailed response.