Also, and I think that I have read this above, just want to make sure I understand, with a 10k pot the output impedance rises as the knob is rotated clockwise until it reaches 10k. Did I get this right?
No, Herman's relevant statement was:
In my example they were both 10K so it is 1/2 of the 10K. With the volume close to zero it rises to about 10K. and falls as the volume goes up.
In your case, with a 10K pot and a 100K amp Zin, at minimum volume control setting the input impedance (not output impedance) looking into the input side of the pot would be 10K. With the control turned all the way up, the impedance would be 10K in parallel with 100K, which is 9.1K.
The source component would see a load impedance equal to that value (dependent on the volume control setting) factored by square of the ratio of transformer primary turns to secondary turns (assuming the transformer is "ahead" of the volume control).
As I indicated a couple of posts ago, the OUTPUT impedance of the passive, which must be kept low in relation to the amp input impedance, will be at its worst case maximum when the volume control is set to the mid-point of its resistance range (assuming the source component's output impedance is small). Apart from the presence of the transformer, with a 10K pot and a 50 ohm source impedance, that output impedance would be 2.5K. The step-up transformer will raise that value a little, by stepping up the (very small) contribution of the 50 ohms. But for any reasonable turns ratio the resulting output impedance is likely to still be well under 5K, and therefore fine in relation to the amp's 100K Zin.
BTW, calculating the resistance of a parallel combination of two resistances is easy. It's just the product (multiplication) of the two numbers, divided by their sum.
If more than two resistances are in parallel, it's a little more difficult, the result being equal to the reciprocal of the sum of the reciprocals of the individual resistance numbers. (Reciprocal = the number divided into 1).
Best regards,
-- Al
