Nelson Pass said:

**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