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Passive Headphone Attenuator
Old 9th July 2011
  #1
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Passive Headphone Attenuator

I'm trying to build a simple passive headphone attenuator. It will be used with a mono headphone output and is only going to be used for a click track. It doesn't need to sound good, it just needs to work.

Because I am having trouble finding a 500ohm pot that is higher than 1/2 watt I figured I would build a simple 500ohm series stepped attenuator. It is all 51ohm 1watt resistors and uses a 12 position rotary switch from parts express.

At full volume the series resistance will be 0Ω and the shunt resistance will be 561Ω. All the way down the series resistance will be 561Ω and the shunt resistance will be 0Ω.

Of course I also have a 51Ω resistor in series with the input to keep it from completely shorting.

I realize this will probably have screwy impedance's for the headphone amp, but I really don't care what it sounds like because it is just a click track.

I want it to give me quite a big range of reduction, do these figures look ok?

this is the amount of gain reduction in DB for each step taking into acount the safety resistor

  1. a little over 20db
  2. 17ish
  3. 13ish
  4. 11ish
  5. 8.5ish
  6. 7ish
  7. 5.5
  8. 4.5
  9. 3ish
  10. 2.5ish
  11. 1.5ish
  12. 1ish
Will it work? Please help
Attached Thumbnails
Passive Headphone Attenuator-102_0005.jpg  
Old 9th July 2011
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Or could I use this 1/2 500ohm pot with a 1 or 5 watt resistor in series on the input? Say 100ohms?
Old 9th July 2011 | Show parent
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I'm trying to work out exactly what you are doing, as there are a couple of discrepancies.

First, you can't work out the attenuation in this sort of circuit without knowing the impedance of the headphones. The values in the attenuator are too close to typical headphone impedances that the final result could be anywhere.

What you drew cannot result in the figures you calculated. In particular, position 0 couples the input resistor to ground and thus leads to infinite attenuation.

The circuit is backwards if a potential divider network is desired. There are a number possible ways you can couple up something like this, but I can't reproduce your numbers from any of them. As above, one must know the headphone's impedance.
Old 9th July 2011 | Show parent
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Quote:
Originally Posted by a zombie ➑️
Or could I use this 1/2 500ohm pot with a 1 or 5 watt resistor in series on the input? Say 100ohms?
Your link goes to a 50K pot.

Try Digikey.com, search for this part number: CT2234-ND
Old 9th July 2011 | Show parent
  #5
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500 Ohm Potentiometer

oops, here is the 500Ω.

Quote:
In particular, position 0 couples the input resistor to ground and thus leads to infinite attenuation.
Oops, so I would want the 51Ω resistor between position 1 and ground.

Quote:
The circuit is backwards if a potential divider network is desired.
I thought it was normal to do input to wiper, CCW to ground and output to CW. I guess input to CW, output to wiper, and ground to CCW would make a more sense in a way.

Quote:
First, you can't work out the attenuation in this sort of circuit without knowing the impedance of the headphones. The values in the attenuator are too close to typical headphone impedances that the final result could be anywhere.
I'll use this with different headphones depending on the situation.

in the practice space I'll use it with 7506's and on stage a variety of earbud style phones.

I just need this to work with a variety of headphones...

Also do i really need to use 1 watt resistors or can I get by with 1/2 or 1/4 watt?
Old 9th July 2011 | Show parent
  #6
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I could also do a ladder type attenuator, this would allow me to make a an L-pad on every step. But how would I calculate the Rshunt and Rseries value for the best input and output impedance. (I'm not good at impedance yet so go slow).
Old 9th July 2011 | Show parent
  #7
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maybe one of these?
L-Pad 15W Mono 3/8" Shaft 8 Ohm
Old 9th July 2011 | Show parent
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OK, we still need to start with the headphone's impedance. What you are trying to build isn't something that can be designed in isolation from both the source and sink of power.

7506 is 63 ohms, earbuds are typically less, often 32 ohms. But this gets us a useful ballpark. It doesn't need to be very precise in this case, but needs to be well enough defined that the device is actually useful. Call it 50 Ohms.

What I suspected you were trying to build is actually about correct. But the circuit and assumptions were a bit screwed up. What you are going to build is potential divider with a "log faking" resistor. However, with care, the log faking resistor turns out to be the headphone itself. A nice treatment of the attenuator, as applied to a volume control (and hence only worried about large impedances and no power delivery) is on Rod Elliott's site. ESP - A Better Volume Control

The key is that for a linear potentiometer to result in a useful audio taper, empirically the pot needs to have about 10 time the impedance as the faking resistor. If you look at the circuit you see that a simple 500ohm potential divider feeding your headphones works out about right. But your circuit was upside down. Wiper goes to headphones.

The only need for a safety resistance is in the case where the output is full up, and you pull the headphones partly out of the jack - something what can lead to the plug shorting a TRS or TS jack.

You could build this circuit with a switch and a pile of resistors. It would have the advantage of being essentially bullet proof. And no, you can use standard quarter watt resistors. Even under the most stressful circumstances you will have either melted your brain or the headphones before the attenuator gets into trouble. You could also use a rotary potentiometer. However the reliability in this application, unless a wire wound was used may be a lot poorer.

The impedance calculation assumes you are going to build two channels. If you are connecting the two together into mono, you are probably going to need to drop everything to 250 Ohms.
Old 9th July 2011 | Show parent
  #9
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Quote:
Originally Posted by Francis Vaughan ➑️
OK, we still need to start with the headphone's impedance. What you are trying to build isn't something that can be designed in isolation from both the source and sink of power.

7506 is 63 ohms, earbuds are typically less, often 32 ohms. But this gets us a useful ballpark. It doesn't need to be very precise in this case, but needs to be well enough defined that the device is actually useful. Call it 50 Ohms.

What I suspected you were trying to build is actually about correct. But the circuit and assumptions were a bit screwed up. What you are going to build is potential divider with a "log faking" resistor. However, with care, the log faking resistor turns out to be the headphone itself. A nice treatment of the attenuator, as applied to a volume control (and hence only worried about large impedances and no power delivery) is on Rod Elliott's site. ESP - A Better Volume Control

The key is that for a linear potentiometer to result in a useful audio taper, empirically the pot needs to have about 10 time the impedance as the faking resistor. If you look at the circuit you see that a simple 500ohm potential divider feeding your headphones works out about right. But your circuit was upside down. Wiper goes to headphones.

The only need for a safety resistance is in the case where the output is full up, and you pull the headphones partly out of the jack - something what can lead to the plug shorting a TRS or TS jack.

You could build this circuit with a switch and a pile of resistors. It would have the advantage of being essentially bullet proof. And no, you can use standard quarter watt resistors. Even under the most stressful circumstances you will have either melted your brain or the headphones before the attenuator gets into trouble. You could also use a rotary potentiometer. However the reliability in this application, unless a wire wound was used may be a lot poorer.

The impedance calculation assumes you are going to build two channels. If you are connecting the two together into mono, you are probably going to need to drop everything to 250 Ohms.
wow thanks so much, I'm just going to use the left to drive both headphone speakers. Does this make the load impedance of the 7506 only 31.5Ω and the earbuds 16Ω? making our average of 50Ω only 25Ω.

So I would need a 250Ω stepped attenuator. If I do a 250Ω 12 step attenuator with 12 24Ω resistors would I get a good range?

Here is a new circuit, It has a 24Ω on the input because I will be plugging and unplugging at full gain.
Attached Thumbnails
Passive Headphone Attenuator-102_0006.jpg  
Old 10th July 2011 | Show parent
  #10
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bump, will this work?

how do I figure out the attenuation in DB?
Old 10th July 2011 | Show parent
  #11
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Looks pretty right to me now.

To work out the attenuation you simply need to work out the parallel impedance due to the shunt leg of the potentiometer and the headphones, and then put that in series with the series leg of the potentiometer. Note that your diagram gets the shunt values out by one slot. Wiper to the first position means that the value is zero.

So: parallel impedance Ry = 1/(1/Rpot_shunt + 1/Rhp)
(Note that this is undefined at Rpot_shunt = zero, but effectively zero)

output ratio= Ry/(Rpot_series + Ry)
output in dB = 20log(output ratio) (undefined for ratio of zero, but essentially infinite)

Assuming a series safety resistor of 25 Ohms, I get:
no output, -26.4, -23.2, -21.3, -19.8, -18.3, -16.7, -14.9, -12.8, -10.2, -6.4

These values are derived from all resistances being the same - heaphones, attenuator resistors and shunt resistor. They will remain the same if you scale the design to other values.

Note that the safety resistor limits you to 6db when at maximum. If you really cared about that, you could replace the resistor with a polyswitch. But it probably isn't worth the bother.

This assumes a zero impedance (i.e. voltage source) amplifier which is pretty much what they all are.
However, big caveat. You will have a significant problem of you are driving this from the headphone output of a typical bit of gear. I forgot about this, and assumed the output was from from a typical power amp. The vast majority of headphone outputs include a 120 ohm series resistor. This is a standard recommended value. But not all gear has one. Most pro gear does, almost no audiophile gear does. So, it depends upon what you will hook this up to. You may need to check the schematic.

If you wanted to make this work with a driver that has the 120 ohm resistor the resistance values will need to be worked out very differently. The simple neat empirical idea of a linear pot tens times the faking resistor used here simply won't work because of that output resistor. Indeed there probably isn't an easy formula. Trial and error perhaps. The formula above however are still valid.

Last edited by Francis Vaughan; 10th July 2011 at 09:07 AM.. Reason: formatting, spelling
Old 10th July 2011 | Show parent
  #12
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I think the -6db max will be just fine. Hopefully -26db will be enough to bring it down to a reasonable level.

Am I right in saying that the only way to extend the range would be to use a many different resistor values to recreate an audio taper, or to get a switch with more positions?

So I will need essentially 12 24Ω resistors and a 12 position rotary switch.

Here's a simulation of the attenuator. The chart on the bottom shows watts across one side of the headphones. If you click through the switches one by one you can see a logarithmic curve of attenuation form on the chart. So it seems this will work assuming that amount of attenuation is good.
Old 10th July 2011 | Show parent
  #13
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Nice sim. I'd not seen that site before.

What you have here is already an audio taper. Somewhat counter-intuitive, since it started with a linear potentiometer, but the neat way that the additional parallel "log faking" resistor ends up with a profile that is quite suited to audio is very neat. Sufficiently neat that many people prefer this approach to the use of conventional log or audio taper pots for gain control. However it isn't nearly as accurate as a proper switched ladder network. Which is why you see such things ubiquitous in serious gear. But here is it enough that you have a useful control range.

Extending the range significantly would really need you to recalculate the resistance values. Starting with the current set of all identical, and then perturbing them (probably whilst still keeping the total value much the same) might be a good approach. Doing this with a spreadsheet would make life reasonably easy. (Aside. This is the sort of thing that a simulated annealing optimiser would be good for.)

Do make sure about the possibility of a series output resistor existing in what you drive this with. If it is there things won't work nearly as well.

Editing to add:

I had a little play with a spreadsheet to see how bad a 120 Ohm output resistor is. The answer - if you are making a mono attenuator with both HP channels in parallel the loss of control range is pretty bad. However a few things occurred to me. Making a stereo attenuator and tweaking the resistor values turns out to end up with a quit tolerable result - I could get the steps pretty close to 3db right across the range (one step was 2db, but in all, not bad.) This would be good for monitoring purposes. The other point is that if this really is only ever going to be used for click tracks, then there is no specially good reason why the headphones could not be wired in series. Without rewiring the headphones the only way of doing this is to drive tip and ring and leave the sleeve unconnected. This has the effect of reversing the phase of one channel. Ordinarily this would be unacceptable, but for a click track all that will happen is that you will hear the click as a wide image instead of a click in the middle of your head. You might actually prefer this.
Old 10th July 2011 | Show parent
  #14
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Quote:
Do make sure about the possibility of a series output resistor existing in what you drive this with. If it is there things won't work nearly as well.
Hmm... series output resistor. Is that just a current limiter to protect the device? So I would just need to drop said series resistance to 24Ω and then take out my safety resistor, right?

I can't seem to find any technical specifications for the device I am driving the phones with. It is a Yamaha RS7000 Sequencer.

I guess I should just order the parts and go from there... if it doesn't work for the reason you mentioned how hard would it be to fix?

Quote:
I had a little play with a spreadsheet to see how bad a 120 Ohm output resistor is. The answer - if you are making a mono attenuator with both HP channels in parallel the loss of control range is pretty bad. However a few things occurred to me. Making a stereo attenuator and tweaking the resistor values turns out to end up with a quit tolerable result - I could get the steps pretty close to 3db right across the range (one step was 2db, but in all, not bad.) This would be good for monitoring purposes. The other point is that if this really is only ever going to be used for click tracks, then there is no specially good reason why the headphones could not be wired in series. Without rewiring the headphones the only way of doing this is to drive tip and ring and leave the sleeve unconnected. This has the effect of reversing the phase of one channel. Ordinarily this would be unacceptable, but for a click track all that will happen is that you will hear the click as a wide image instead of a click in the middle of your head. You might actually prefer this.
Parts Express doesn't have a dual pole 12pos switch. I found this one at mouser http://www.taiwanalpha.com/english/p_e_147.htm it is made by alpha, and is cheap. I am on a tight budget and can't afford $40 for switches. Do you think the alpha would be ok? I would then want to build two 500ohm attenuators right?

edit just tested the 120series, it pretty much ruins everything...
Old 11th July 2011 | Show parent
  #15
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I'm sure that switch would be fine.

It might be worth rewinding a bit here. If budget is an issue, and the use case pretty limited, I might be inclined to go for a cheap and cheerful solution and see how you go. This depends a bit on the presence of an output resistor, however I would be surprised if your sequencer did not have one. You could try visual inspection of the circuit board to confirm. Otherwise it depends upon what test gear you have to hand. Measuring the output impedance isn't hard, but needs the ability to measure an AC signal that the sequencer is able to emit. (Simple really. You need to get the sequencer running ouputting as steady a signal as you an get it to do, and measure the amplitude of the output of the headphones. A CRO would be ideal, a multimeter with AC setting would probably be OK, even driving into a DAW and looking at the amplitude could work. Then, changing nothing else, couple a 120ohm resistor in parallel with the output. If the level drops 6db (i.e. to half) there is a 120 ohm resistor in the output. In principle no matter what the output impedance of the headphone jack, this method allows you to determine what that impedance is.)

Cheap and cheerful solution is simply getting a 1k ohm linear potentiometer, coupling it up as a potential divider and driving the tip and ring of the headphones. The pot may have slightly less reliability over time than switched resistors, and having the two phones out of phase means you get a stereo image that is very wide and spacey as opposed to focussed in the middle of your head, but you can build the whole thing for less than $10, cables and all. If it proves unsatisfactory you have lost a couple of dollars on the pot, and go for the switched attenuator design.
Old 11th July 2011 | Show parent
  #16
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Quote:
Originally Posted by Francis Vaughan ➑️
I'm sure that switch would be fine.

It might be worth rewinding a bit here. If budget is an issue, and the use case pretty limited, I might be inclined to go for a cheap and cheerful solution and see how you go. This depends a bit on the presence of an output resistor, however I would be surprised if your sequencer did not have one. You could try visual inspection of the circuit board to confirm. Otherwise it depends upon what test gear you have to hand. Measuring the output impedance isn't hard, but needs the ability to measure an AC signal that the sequencer is able to emit. (Simple really. You need to get the sequencer running ouputting as steady a signal as you an get it to do, and measure the amplitude of the output of the headphones. A CRO would be ideal, a multimeter with AC setting would probably be OK, even driving into a DAW and looking at the amplitude could work. Then, changing nothing else, couple a 120ohm resistor in parallel with the output. If the level drops 6db (i.e. to half) there is a 120 ohm resistor in the output. In principle no matter what the output impedance of the headphone jack, this method allows you to determine what that impedance is.)

Cheap and cheerful solution is simply getting a 1k ohm linear potentiometer, coupling it up as a potential divider and driving the tip and ring of the headphones. The pot may have slightly less reliability over time than switched resistors, and having the two phones out of phase means you get a stereo image that is very wide and spacey as opposed to focussed in the middle of your head, but you can build the whole thing for less than $10, cables and all. If it proves unsatisfactory you have lost a couple of dollars on the pot, and go for the switched attenuator design.
Ok, gotcha, maybe I'll just grab a 1kΩ and a 500Ω linear pot as well as the stuff to make a stepped 250Ω attenuator. I should have my bases covered in that case.

One other question about the series resistor thing. If the sequencer has one then I don't need a safety resistor at all right? because if I can use the headphone as it is then I don't need a safety resistor at all. Getting rid of the safety would make the series resistance that much better.

Also IF I can find the 120Ω resistor (assuming it is there) then I could place a 30Ω resistor parallel with it bringing it down to 24Ω. Then without the safety resistor I would be set.
Old 11th July 2011 | Show parent
  #17
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That is pretty much correct.
Old 11th July 2011 | Show parent
  #18
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yessss!
Old 15th July 2011 | Show parent
  #19
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I built it, and the attenuation curve is pretty bad, it doesn't bring the gain down all that much at the lowest setting.

I think I'll need to build a log step attenuator. Can anyone recommend DB cut per step for a 12 step attenuator?
Old 15th July 2011 | Show parent
  #20
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Which one did you build? Did you get an attenuation commensurate with predictions and it is simply not enough for practical use, or is the attenuation not as predicted?
Old 15th July 2011 | Show parent
  #21
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It seems to follow the prediction of how it would behave when there is a series resistor on the output of the sequencer. This resistor causes the linear attenuator to behave like a linear attenuator, which is how my pot is behaving. So that means that if I build a 250Ω log attenuator I should get a good result. I think...

I found this http://homepages.tcp.co.uk/~nroberts/atten.html

which for a 250Ω pot with 4db steps gave me this table

Quote:
Step 1, Attenuation = 44 dB, Rx = 248 ohms, Ry = 2 ohms, Resistor = 2 ohms.
Step 2, Attenuation = 40 dB, Rx = 247 ohms, Ry = 3 ohms, Resistor = 1 ohms.
Step 3, Attenuation = 36 dB, Rx = 246 ohms, Ry = 4 ohms, Resistor = 1 ohms.
Step 4, Attenuation = 32 dB, Rx = 244 ohms, Ry = 6 ohms, Resistor = 2 ohms.
Step 5, Attenuation = 28 dB, Rx = 240 ohms, Ry = 10 ohms, Resistor = 4 ohms.
Step 6, Attenuation = 24 dB, Rx = 234 ohms, Ry = 16 ohms, Resistor = 6 ohms.
Step 7, Attenuation = 20 dB, Rx = 225 ohms, Ry = 25 ohms, Resistor = 9 ohms.
Step 8, Attenuation = 16 dB, Rx = 210 ohms, Ry = 40 ohms, Resistor = 15 ohms.
Step 9, Attenuation = 12 dB, Rx = 187 ohms, Ry = 63 ohms, Resistor = 23 ohms.
Step 10, Attenuation = 8 dB, Rx = 150 ohms, Ry = 100 ohms, Resistor = 37 ohms.
Step 11, Attenuation = 4 dB, Rx = 92 ohms, Ry = 158 ohms, Resistor = 58 ohms.
Step 12, Attenuation = 0 dB, Rx = 0 ohms, Ry = 250 ohms, Resistor = 92 ohms.
Pretty cool, so I just need go get those resistors and I can give it another go. Does this seem like a good plan?
Old 15th July 2011 | Show parent
  #22
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Sadly you are still in trouble because of the source impedance - that (probably) 120 ohm resistor. Nor does that calculator take into account the headphone impedance. It is assuming a zero source impedance and an infinite sink impedance - which is fine for say an attenuator for setting a level in a bit of gear that is fed by and fed into a couple of op-amps.
Old 15th July 2011 | Show parent
  #23
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damn, so what can I do?
Old 15th July 2011 | Show parent
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Ha! I have an idea. I think a different topology might get you what you need. We shunt the headphones with a variable resistor, and use the 120 ohm resistor as a given part of the circuit. Let me whip up a spreadsheet....
Old 15th July 2011 | Show parent
  #25
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Quote:
Originally Posted by Francis Vaughan ➑️
Ha! I have an idea. I think a different topology might get you what you need. We shunt the headphones with a variable resistor, and use the 120 ohm resistor as a given part of the circuit. Let me whip up a spreadsheet....
sounds promising, I'll stay up a bit longer for that
Old 15th July 2011 | Show parent
  #26
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OK, this is pretty quick, so there might be mistakes.

Lets use the series resistor, and to some extent we are turning the output of the sequencer into a current source. To be safe we add even more resistance in the unit - this makes it safer, and actually helps in the new case.

The attenuator goes in parallel with the headphones. It is simply a variable resistance. The potential divider is formed by the parallel impedance of the heapdhones along with the variable resistor as the lower resistor, and the 120 ohm output resistor plus another resistor in the unit making up the top.

I have chosen the resistor values from the E12 series, they don't need to be precision, they are running across the decades pretty quickly. Actually getting the low values will be hard - you will need to parallel resistors up. This design gets you ten steps with 6db between steps at most settings, mostly pretty good jumps.

So - the additional safety resistor - 68 ohms.

[Edited to tweak the numbers.]

Resistors in the variable resistor go:
0.1, 0.1, 0.22, 0.39, 0.82, 1.5, 3.3, 8.2, 27, 560

I get attenuation steps of:

-6.0, -6.4, -5.6, -5.9, -5.4, -5.7, -5.9, -6.0, -6.0, - that is 0 to -53db in close to 6db steps. Tweaking with E96 series values could get you as close as you wish. Almost certainly not worth the effort. Note, these values are for a stereo headphone connection. Ie into 63 ohms. I can try a mono version, into 32 ohms. It will be harder, and the maximum loudness will be 6db lower.

Note a caveat - at full volume the attenuation is about 3db down on what you are getting out of the headphone socket. Not as much as I first feared, this is simply because the 120 ohm resistor in the output does all the damage even if there is no attenuator.

And some more updates.

A mono version is not too bad actually. Keeping a 68 ohm series resistor one can get 5 to 6db steps at most positions with E24 series values. reasonably well spread. Only 4db drop at the first step.

0.1,
0.1,
0.18,
0.33,
0.68,
1.5,
3.3,
8.2,
33,
open

with attenuation steps of:

-6.0,
-5.5,
-5.3,
-5.6,
-5.9,
-5.7,
-5.4,
-5.3,
-4.0,

Last edited by Francis Vaughan; 15th July 2011 at 01:47 PM.. Reason: Tweaked the numbers.
Old 15th July 2011 | Show parent
  #27
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Ok, I think I'll give this a shot. So the pot just shunts the output to ground. So the more resistance of the shunt the higer output.

So do I wire it like a series attenuator. or does each step flow through it's respective resistor and then to ground?

so like this http://www.falstad.com/circuit/#%24+...5E-105+0+-1%0A

why only ten steps? why not 12?
Old 16th July 2011 | Show parent
  #28
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Quote:
Originally Posted by a zombie ➑️
Ok, I think I'll give this a shot. So the pot just shunts the output to ground. So the more resistance of the shunt the higer output.
Yup. Note how the highest setting is actually open, with no resistor. (In the stereo version a 560 ohm resistor in that position turned out to neatly even out the steps, but is vastly larger than the headphone's impedance so is only slightly tweaking things.)

The more I look at it, the more obvious it is that this is the right circuit.

Quote:
So do I wire it like a series attenuator. or does each step flow through it's respective resistor and then to ground?
The numbers were worked assuming a series connection, so the total resistance of the variable resistor comes from the sum of those resistors not shorted.

Quote:
why only ten steps? why not 12?
No reason at all. I think I forgot you had a 12 step switch :-) However you will be struggling to get many more steps. You need to add them at the low resistance end, so 0.047, and 0.027 ohms. The last two resistors end up the same value, the rest halving. You may find getting a length of nichrome wire and using length of resistance wire is the only viable way of getting resistances this low. It will also probably not be worth the effort.


I hope my number are good, I'm very good at making trivial mistakes of out by one errors in any code I write.
Old 16th July 2011 | Show parent
  #29
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good news, I just built it and it works wondefully. I forgot to install the 68Ω series resistor, but it still works just fine. I'm post some photos in a bit
Old 16th July 2011 | Show parent
  #30
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Excellent! heh

The only reason for the 68 ohm resistor is safety. The device can short out the output of anything it is connected to otherwise. Which could be rather unfortunate. 68 ohm is a compromise, but should be hard to kill. Coupled to a 100 watt power amp it might not live all that long with the amp cranked up, but for most purposes it should both be safe, and cause little issue with the operation of the attenuator.

Thinking about this a bit more it strikes me that the numbers are not too far off being able to create a generally useful design. The trick is to note that the only place where the resistor values get weird is when the resistance of the attenuator is in the vicinity of the impedance of the headphones. Which is typically when the volume control is up loud. A control can be built that can cope with most headphones - say the range 32 to 600 ohms, with some small variability in the attenuation, except that for low impedance headphones the top couple of settings will be close to useless, or provide only slight attenuation. But the overall utility of the device will remain. With a 12 position switch one should still have a useful range of settings available no matter what happens.

When I get bored I might throw some numbers together.

I can't imagine this is anything new. I get the feeling I'm simply working through a previously well worn, but perhaps recently forgotten, path.
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