MC LOADING & STEP UP DEVICE

In other words, typically 47,000 ohms divided by 933 in this case, or 50 ohms.

I should have added, of course, that the 50 ohms would be less if load resistors are added.

Regards,
-- Al

Here is the excerpt from Art Dudley's write-up:

Aschenbrenner's Hommage T1 measures somewhat differently, with a primary coil impedance of about 3 ohms and a secondary coil impedance of 2.8k ohms

I assumed that the secondary coil impedance was equivalent to a secondary trannie loading that is reflected back to primary as discussed in Axel's post above. The math seemed to confirm this:

47k*2.8k / 47k+2.8k = 2,643 / 933 = 2.83 ohms

Is it coincidence that the mathematical result agrees with the "primary coil impedance of about 3 ohms" that is cited in the write-up?

Art Dudley was using "impedance ratio," in this case 933 to 1, as a round-about way of indicating the turns ratio (which is the square root of 933, since impedance is reflected between the two sides of the transformer in proportion to the square of the turns ratio).

Hence the primary and secondary coil impedances he refers to (3 ohms and 2800 ohms) differ by a factor of 933. But I'm not sure how either of those numbers are defined -- as dc resistance, or as impedance at some frequency with the other side of the transformer connected to some load, or what?

The reason for the 2.83 ohm vs. 3 ohm "coincidence" is that the 47K load impedance is an insignificant load in comparison to 2.8K, which in turn reflects back to the primary as 3 ohms, while 47K in parallel with 2.8K reflects back to the primary as 2.83 ohms.

However, it is incorrect to consider the 2.8K as being in parallel with the 47K. Within the range of its intended operating conditions (frequency and voltage), a transformer has (to a very close approximation) no impedance of its own. In Axel's example, the 18K that he assumed was in parallel with the 47K preamp impedance represented an external 18K resistor, not the transformer's secondary coil impedance.

Regards,
-- Al

Al, thank you for clarifying this:
>> In Axel's example, the 18K that he assumed was in parallel with the 47K preamp impedance represented an external 18K resistor, not the transformer's secondary coil impedance.<<<

The example aught to be clear, by looking at the basic calculation of parallel resistance: 47k input impedance parallel with an 18k load resistor on secondary.

It true, that the secondary coil DCR is ALSO reflected back into primary, again by dividing it by the square of the winding ratio.
B U T, it would be negligible in the case of my example and would be about 0.06ohm added to the calculated 13ohm of the example.
In the case of an FR XF-1 type M, (30dB) primary DCR = ~1.6 ohm, and secondary = ~ 60ohm (if I recall it correctly).

The secondary coil DCR is ALSO reflected back into primary, again by dividing it by the square of the winding ratio. B U T, it would be negligible in the case of my example and would be about 0.06ohm added to the calculated 13ohm of the example.

Yes, with the slight further clarification that the dc resistance of the secondary coil is reflected back into the primary by dividing it by the square of the winding ratio, in terms of its effects on the range of signal frequencies and voltages over which the transformer acts like a transformer. Which is why the 1.6 ohm and 60 ohm dc resistance numbers (dc being zero frequency) do not differ by a factor of 900 or so, in case anyone wonders.

Best regards,
-- Al