[Steve Eddy]

Quote:

Originally Posted by linuxworks /img/forum/go_quote.gif yes, its still 'in' the circuit but it has minimal effect when its in pass-thru mode.

I guess it depends on what you mean by "effect."

Quote:

SOME effect, yes, of course. it has capactance and all the other non-ideal (for a pure R) things that real world elements have. my point is that you can *effectively* ignore the pot when its at full pass-thru (no atten) mode.

Not sure what you mean by "ignore" it. Ignore it in what respect?

Quote:

the input stage also has a Z of its own, and the pot is now in parallel with that complex (x+r) component. if the input z of the circuit is 1M or higher and the pot is 50k, say, its effect won't really load down the circuit much. there will be X issues but you'll only see that on test equipment (imho).

It would be a pretty horrible design that had the source component seeing much less than the end-to-end resistance of the pot.

se

 

[Leny]

1.
audioqueso, if you set the volume control of the amp to max and control volume by way of input level variation (from your processor) then you will achieve your objective.

Your issue will be that most amps have too much amplification factor, and hence your input must be really very low indeed or it will overload the amp. If you are using a digital volume control (on your processor) turned right down then you may loose some signal resolution because that is the way digital volume normally works. You might or might not actually notice it.

One compromise would be to set the amp volume control to mid-way, hence avoiding the normally poorly matched (left/right) low-end region of the potentiometer, and then use your processor to control volume.

(Note, you are used to amps distorting when you turn them up because the amp actually has too much input signal combined with too much internal amplification factor).

2
Steve Eddy, when a conventional volume control (/potentiometer) is set to maximum it is not electrically out of the circuit, however it simply represents an impedance in parallel with the input of the active circuit. Hence it could be replaced with a fixed resistor of the same value, hence AS A VOLUME CONTROL it IS effectively out of the circuit.

 

[Steve Eddy]

Quote:

Originally Posted by Leny /img/forum/go_quote.gif Steve Eddy, when a conventional volume control (/potentiometer) is set to maximum it is not electrically out of the circuit, however it simply represents an impedance in parallel with the input of the active circuit. Hence it could be replaced with a fixed resistor of the same value, hence AS A VOLUME CONTROL it IS effectively out of the circuit.

Y'ever consider going into politics?

se

 

[linuxworks]

Quote:

Originally Posted by Leny /img/forum/go_quote.gif One compromise would be to set the amp volume control to mid-way, hence avoiding the normally poorly matched (left/right) low-end region of the potentiometer, and then use your processor to control volume.

I didn't believe it until I saw it, but on rmaa and other analyzer tests, when the pot is at full rotation (0 atten) the crosstalk isn't quite as bad; but as you rotate and put the pot 'more into play' you see the crosstalk go up as freq goes up. a trapezoid kind of pattern.

if you remove the pot entirely you can see quite a bit more separation. if this is audble, I'm no so sure. its detectable on test equip, to be sure but I'm not convinced that its all that bad in the real world with real recorded content.

but still, if you do a test, you move the vol control/pot at max and that, at least, does what it can to remove it from the circuit. its not a total removal but its pretty close. having it midway puts it back into the equation and I would not do that unless there was no other way to atten the source level.

 

[Baldeagle58]

You know Krell built such an amp (KSA 5) at one time. You might be able to pick one up second hand somewhere.

 

[nikongod]

Quote:

Originally Posted by Baldeagle58 /img/forum/go_quote.gif You know Krell built such an amp (KSA 5) at one time. You might be able to pick one up second hand somewhere.

Total production was in the teens at best. You will also need the power supply although that's easier than finding the amp.

The Krell KSA-5 is an awesome amp though. A NYC local has a highly modified one.

 

[audioqueso]

Quote:

Originally Posted by Leny /img/forum/go_quote.gif Your issue will be that most amps have too much amplification factor...

You mean most HEADPHONE amps have too much amplification even at min. volume?

Quote:

Originally Posted by Leny /img/forum/go_quote.gif ...and hence your input must be really very low indeed or it will overload the amp.

Meaning the output of my processor, right?

 

[Leny]

Quote:

Originally Posted by audioqueso /img/forum/go_quote.gif You mean most HEADPHONE amps have too much amplification even at min. volume?

99% of headphone amps have fixed (pre-set) amplification, called 'gain'. The volume control is normally placed at the input and feeds a portion of the input signal to the amplifier stage. The amplifier stage then increases the signal level by it's pre-set gain and then it is passed on to the output socket.

Only at maximum rotation of the volume control do you have all of the input signal passed to the amplifier. At all other settings the vol control attenuates the input before it is passed to the amplifier stage, hence the vol control is often known as an 'attenuator'.

In my experience most headphones do not need a great deal of signal level to achieve high volume and virtually every amp has too much amplification (i.e. 'gain'). This means that the volume control is normally turned right down in order to attenuate the incoming signal before it is passed onto the amplification stage, otherwise the output (after the amplification stage) would simply be too great. In turn this means that only a small amount of the volume control is actually useable.


Quote:

Originally Posted by audioqueso /img/forum/go_quote.gif Meaning the output of my processor, right?

Yes.

 

[audioqueso]

Leny,

Thanks. lol You didn't really need to explain all that. I know how it works, I just didn't know if I was misunderstanding you and you meant it the other way around (meaning the output from the processor being too high instead of the amp). Thank you though. You've been very helpful.