10K Pot for the BOSOZ

[raincheck]

I know this has been discussed but Id like a real world answer....
Im planning on using 10k attenuators(instead of 5k) on the output for volume control.

What is a long cable run????

If I have 15 foot balanced interconnects between the preamp and Aleph 2 mono blocks will the 10 k pots cause issues??

Can anyone who has done this please comment.....Im thinking a 15 foot cable run is pretty normal(or short) cable
length.

Thanks!!!

[thylantyr]

I'm new here, what is BOSOZ?

[2Bak]

http://www.passdiy.com/pdf/balzenpre.pdf

[Steve Eddy]

Originally posted by raincheck
I know this has been discussed but Id like a real world answer....
Im planning on using 10k attenuators(instead of 5k) on the output for volume control.

Just curious, why are you opting for 10k instead of 5k? Do you have any choice in the matter?

What is a long cable run????

If I have 15 foot balanced interconnects between the preamp and Aleph 2 mono blocks will the 10 k pots cause issues??

Can anyone who has done this please comment.....Im thinking a 15 foot cable run is pretty normal(or short) cable
length.

It's not solely a matter of length. The ultimate issue is one of high frequency rolloff which results from the combination of output impedance and the shunt capacitance of the cable.

Given the much higher output impedance that the 10k pot will result in, a 15 foot run could be problematic if the cable capacitance is high enough. What type of cable will you be using?

Also, have you considered a transformer-based attenuator? It would allow you to preserve the low output impedance of the active stage outputs and make cable issues largely moot. And of course it provides ground isolation between the preamp and power amp.

se

[Peter Daniel]

Originally posted by Steve Eddy




Also, have you considered a transformer-based attenuator? It would allow you to preserve the low output impedance of the active stage outputs and make cable issues largely moot. And of course it provides ground isolation between the preamp and power amp.

se

The ones I saw were rather expensive.
http://www.audio-consulting.ch/sr_pot.htm

[Steve Eddy]

Originally posted by Peter Daniel
The ones I saw were rather expensive.
http://www.audio-consulting.ch/sr_pot.htm

The Silver Rock is a whole freakin' preamp. I'm talking about the raw transformers. Such as the <a href="http://www.sowter.co.uk">Sowter</a> 9335 or the <a href="http://www.stevens-billington.co.uk">Stevens & Billington</a> TX-102.

They're still not exactly cheap, but DIY isn't necessarily about cheap, beyond what one can save by doing something themselves versus a pre-manufactured solution.

se

[Nelson Pass]

Keep in mind that the maximum source impedance of a
voltage dividing pot as seen by the input which follows
the wiper is 1/4 the value of the pot.

A 10 Kohm pot with the wiper at the mid point (max impedance)
will be 2.5K.

If a source impedance is before the CW pin of the pot, then
it must be added to the pot value and then you divide by 4

example: 2 Kohm source, 10 Kohm pot. Maximum source
impedance seen by the next stage looking at the wiper is
(10K + 2K) / 4 = 3 K ohms.

[Steve Eddy]

Originally posted by Nelson Pass
Keep in mind that the maximum source impedance of a
voltage dividing pot as seen by the input which follows
the wiper is 1/4 the value of the pot.

A 10 Kohm pot with the wiper at the mid point (max impedance)
will be 2.5K.

If a source impedance is before the CW pin of the pot, then
it must be added to the pot value and then you divide by 4

example: 2 Kohm source, 10 Kohm pot. Maximum source
impedance seen by the next stage looking at the wiper is
(10K + 2K) / 4 = 3 K ohms.

Yup.

And if you don't mind, I'd like to add to that and illustrate just how that comes to be as it seems understanding output imepdance when potentiometers are involved seems rather a mystery to many when it needn't be.

<center>
<img src="http://www.q-audio.com/images/outputz.jpg">
</center>

Figure A is the starting point.

Vs is the idealized voltage source of the preamp, Rs is its realworld output impedance, Rx is the resistance on the clockwise side of the wiper, Ry is the resistance on the counterclockwise side of the wiper, and Zout is the ultimate output impedance all things considered.

I'll use the same numbers as Nelson did for his example, using a 10k linear pot with the wiper at midpoint.

Rs = 2,000 ohms
Rx = 5,000 ohms
Ry = 5,000 ohms

Rs and Rx are in series so that simplifies to a single resistance, Rs' which is equal to Rs + Rx, leaving us with figure B.

To simplify further, since an ideal voltage source has an impedance of 0, it can be replaced with a short circuit taking us to figure C and we see that we end up with two parallel resistances which we can further simplify to a single resistance.

Rs' = Rs + Ry = 2,000 + 5,000 = 7,000
Ry = 5,000

Rs'' = 1/(1/Rs' + 1/Ry) = 1/(1/7,000 + 1/5,000) = 1/(0.000143 + 0.0002) = 1/0.000343 = 2,915

Rounding up, 3,000 ohms.

se

[jh6you]

Steve

How about this way:

Rs + Rx + Ry = R total.
When the R is divided into two and they are parallel,
the maximum impedance value must be: (Rs+Rx+Ry)/4

Isn't it...?

JH

[Steve Eddy]

Originally posted by jh6you
Steve

How about this way:

Rs + Rx + Ry = R total.
When the R is divided into two and they are parallel,
the maximum impedance value must be: (Rs+Rx+Ry)/4

Isn't it...?

Sure, that gets you there too.

What I was intending to do was take it a past a simple mechanical process (i.e. just plugging in numbers) to more of a visualization as to how it ultimately comes to be that way. Plugging in numbers gets you a result, but not necessarily the concept behind it.

It's like the kids in school who know how to plug numbers into a calculator and get a result, but don't necessarily understand the underlying concepts.

So I just threw up this example for those who may have more of an interest.

se