"PDRUNEN" wrote in message
...
Hi Group,
Got a 706 which has a 500 or 600 ohm mic input impedance. I have a D104
which
best I can tell has a high output impedance.
If I use a 60Hz step down transformer, say, 120V in gives 10v out, the
turn
ratio is 120/10 or 12:1. Given that I attach the mic output to the
primary and
the rig on the secondary, this should step down the impedance by a factor
of
12^2 or 144 so that a D104 with 50K on the output is now seen as about 400
ohm.
Here are a few questions,

The other, more practical questions are answered by others, but:



2: Is the calculation of the impedance correct?

YES. Power remains the same on both sides of the transformer. If "V"
drops by a factor of X, the "I" increases by the same factor. R (or Z) = V
/ I

Z Pri = Vin / Iin
Z Sec = Vout / Iout
Vout = Vin / X
Iout = Iin * X
Zsec = (Vin/X) / (Iin*X)

= Vin/Iin / X / X
= Z Pri / X^2


3: Would the audio as seen on the secondary be reduced by a factor of 12

Yes.

such that the audio would be weak?


The low Z mics have less output voltage than Hi Z mics and since you can
simply use a mic transformer to go between the two, you shouldn't have a
level problem.

--
Steve N, K,9;d, c. i My email has no u's.