Head-Fi.org › Forums › Equipment Forums › Headphone Amps (full-size) › The IMPEDER = Less than 10 ohms output from any piece of gear
New Posts  All Forums:Forum Nav:

The IMPEDER = Less than 10 ohms output from any piece of gear

post #1 of 63
12/20/02 at 5:01am
Thread Starter 
I'm sure I'm not the first to think of this, but here's a way to combat the effects of any evil output resistors lurking behind the headphone sockets of your integrated amp, CD player etc.

If you haven't yet burnt your fingers with a soldering iron then you couldn't hope for an easier first project than this. All you need is a 3.5mm stereo plug and two 10 ohm resistors. [Oh, and a stereo miniplug double adaptor]. If you have a multimeter you might want to check that the resistors are a good match:-


Simply solder the resistors from the left and right connections to the common (earth) tab. A pair of pliers with a rubber band around the handles will reduce the swearing. I used 1/4 watt resistors with a metal shell plug. If your headphones are less efficient than HD580's then you might consider 1/2 watt resistors. Here's how a finished impeder should look with the back shell unscrewed:-


Plug your 'phones into the double adaptor as shown on the left:-

then you can plug in The IMPEDER! On the right is the impeder impeding! (Note and beware of the crankage...)

Here's the trick: If the volume doesn't drop significantly when you plug in the impeder, then you don't need the impeder for that particular piece of gear because the output impedance is already quite low. In most cases though, output resistors range in the tens to hundreds of ohms, so plugging in the impeder will require you to crank the volume.

Here's the catch: The volume drop with the impeder may be 20dB or more. It would be most uncool to unplug the impeder without turning the volume down first!! [and I take no responsibility for any consequences... disclaimer... exclusion... limited liability... poop...].

The impeder doesn't necessarily improve the sound - you be the judge. In particular, some gear will lack either the extra headroom (voltage) or the extra current capability that the impeder demands. In those cases, clipping distortion should be obvious.

How does it work? Basically you are wasting power in exchange for lower output impedance and the tighter driver control that goes along with that. With the HD580 the bass hump is reduced and lower midrange resolution is improved.
LL
LL
LL
LL
Reply
post #2 of 63
12/20/02 at 5:03am
Thread Starter 
<the 2nd image>
Reply
post #3 of 63
12/20/02 at 5:04am
Thread Starter 
<the 3rd image>
Reply
post #4 of 63
12/20/02 at 5:06am
Thread Starter 
<the 4th image>
Reply
post #5 of 63
12/20/02 at 5:22am
What a neat , prefessional, and sporty looking job J-Curve!! Let us know how it sounds with and without the Impeder in place!!! It would be interesting to see how this would affect impedance matching as many of us think that this is the culprit lurking in the shadows and explains why low impedance cans down't work with some amps, while high impedance cans do!! And vice verse.

Wouldn't you always get a volume drop using the Senn's 300 ohm impedance with the impeder?? It would seem that approximately 30 times the amount of current would travel thru it vice the cans since you connected it in parallel with the Senns. You would have equal amounts of voltage, but way less current in the Senns. This would tax the current output several times what it ws before adding it. Interesting......
Reply
post #6 of 63
12/20/02 at 11:26am
Listening impressions! Listening impressions!

Funny, I was thinking of the exact same thing a few hours ago You beat me to it though

That's a nice adapter to use quickly on any system, but if you need the impeder, the headphone jack is probably part of a speaker amp, and it would be better in the long run to connect more directly to the loudspeaker output:
(You've seen this right?)



This basically allows you to dictate the output impedance and voltage share (hence volume) freely. No more [low output impedance = low volume]!
Reply
post #7 of 63
12/20/02 at 1:56pm
Perhaps I'm missing something, but I don't think this lowers output impedance. It makes the headphone's input impedance appear lower to the amp.

--Andre
Reply
post #8 of 63
12/21/02 at 6:52am
Yeah, I feel strange about this too, but this has been touted before as a way to reduce output impedance, and even Kevin Gilmore says that the effective output impedance of the system with the resistor in parallel with the headphones is now the resistance of the output resistor in parallel with the resistor in parallel with the phones.
Reply
post #9 of 63
12/21/02 at 9:25pm
Quote:
Originally posted by AndreYew
Perhaps I'm missing something, but I don't think this lowers output impedance. It makes the headphone's input impedance appear lower to the amp.

--Andre
Basically it goes like this:
Suppose you have 120 ohm output impedance from the amp, plus 300 for the Senns. That adds up to 420 ohms since they're in series.

Now, add 10 ohms in parallel with the 300 ohm Senns and you've reduced the output impedance from 420 ohms down to 129.6 ohm!!! You've now reduced the output impedance down almost 290 ohms!!

It doesn't reduce the impedance output of the amp, just what the amp sees in total!!
Reply
post #10 of 63
12/21/02 at 9:58pm
No...
Reply
post #11 of 63
12/21/02 at 10:30pm
Quote:
Originally posted by Joe Bloggs
No...
Yeah!! Actually the circuit would be the output resistance connected in series with the parallel network of the two other resistors, one being the headphones, the other being the impeder. It's a simple series/parallel circuit.


Not:
Quote:
...the effective output impedance of the system with the resistor in parallel with the headphones is now the resistance of the output resistor in parallel with the resistor in parallel with the phones.
Reply
post #12 of 63
12/21/02 at 10:39pm
You certainly describe the circuit correctly; but that still doesn't tell us what the output impedance is. Do you have a figure in mind? I wasn't saying that the 10ohm resistor is really in parallel with the output resistor or anything, but see http://www4.head-fi.org/forums/showt...threadid=22436 where I posted a thread specifically asking about the calculation of output impedance of such a network. The replies support the view that I am stating here. I don't understand it either
Reply
post #13 of 63
12/21/02 at 11:09pm
Good question?? Output resistance is very difficult to calculate in a circuit unless there is a physical resistor in teh circuit, or you happen to know the output resistance of a certain tube or transistor or circuit design. So, I guess what I'm saying is: you got me!!!

You can't measure resistance of a circuit with power applied, so that's out.

I guess you could measure teh output voltage from the power supply on the output stage of the circuit and then see how much actually reaches and is dropped across the headphones. That would give you total power including the amount of power across the output resistance out the output transistor/tube!!

Man, even after all those years of electronics, some things are still stumpers!!!
Reply
post #14 of 63
12/21/02 at 11:18pm
This makes clear the other thread - Oh well...

An output stage of a power amplifier is basically a trans-admittance stage. What this means is as follows:
The output impedance is high (NOT low)
Therefore, it controls the current to the device.
The amplifier converts a low voltage signal at the input to high current at the output.

(This makes sense: Remember that loudspeakers are current devices (piston position is related to current in the coil suspended in the magnatic field). They have relatively low impedance.)

While similar in construction, headphones are (usually) of a higher impedance than speakers. That makes the required driver stages more towards a voltage driver than a current driver.

The circuit in the diagram does two things:
1. reduce the impedance to a value manageable by most current stages (roughly 22Ohms).
2. reduce the voltage to the headphone (using the voltage divider - 22/2)

(Note: If you were to connect a typical headphone directly to a power amplifier, the output stage of the amplifier would "try its best" to get the current output as required by the input signal multiplied by its gain. It would do this by "giving" all the voltage necessary to get the required current. It would fail, but the headphone would see full rail voltage and expire).
Reply
post #15 of 63
12/24/02 at 1:08pm
Thread Starter 

One thing I forgot to mention...

...is that if your amplifier or soundcard produces hiss in your headphones when there's no music playing, the IMPEDER is also handy because the cut in volume will likely make that background noise inaudible.

PS: Joe, that Grado circuit is an alternative solution (or let's say, "a competing technology" ) to the IMPEDER for integrated amplifiers, but the two solutions are really for different purposes.

I want to do battle with evil output resistors, which are not used or are sufficiently low in value as to be insignificant when you're driving from speaker terminals.

Also, the IMPEDER is aimed at improving the sound from component CD players etc. which have headphone sockets but no speaker terminals.
Reply
New Posts  All Forums:Forum Nav:
  Return Home
  Back to Forum: Headphone Amps (full-size)
Head-Fi.org › Forums › Equipment Forums › Headphone Amps (full-size) › The IMPEDER = Less than 10 ohms output from any piece of gear