"Scott Dorsey" wrote in message
...
Newsguy wrote:
Hi there. Here's a question for the "veterans" in this newgroup. I have
an
old Shure 55s microphone (the Elvis type). This microphone uses an old
school version of a modern XLR connector known as an Amphenol MC3M. This
is
a three pin arrangement (pin 1, pin 2, pin 3).

Now, I have a brand new old-stock Amphenol MC3M connector, a brand new
modern male XLR connector, and a length of mic cable. I want to put it
all
together and make a functional mic cable. Do I simply go pin 1 XLR to pin
1
Amphenol, pin 2 to pin 2, and pin 3 to pin 3?

No. I forget the pinout offhand, but get an ohmmeter. You will find one
pin goes to the case. That one goes to pin 1 on the XLR. The other two
have no connection to the case but have fairly low resistance between
them.
Those go to pins 2 and 3.

I have tried to locate this information through Google, but I cannot find
anything that shows a proper diagram. As a sidenote, this 55s mic has a
three way impedence switch (L-M-H), and I am not sure if that factors into
the way the cable would be soldered up.

Actually, if you go to the Shure web site, and look at the discontinued
products page, you will be able to download the original data sheet for
the mike, with the schematic and everything.
--scott

--
"C'est un Nagra. C'est suisse, et tres, tres precis."

Hi Scott. Thank you very much much for your quick and informative reply. I
am on track now. However, here's one that is a little confusing: I found
an old cable in the bottom of a drawer that has the above-mentioned amphenol
on on end and a 1/4 phone jacl on the other. So, I cut off the 1/4 jack and
was going to put on a modern XLR. After I cut the cable, I noticed that
this cable only contains one inner wire and the "shield". In other words,
we have only two conductors here to connect to the modern XLR. Now, perhaps
the shield goes to pin 1 of the xlr, but then we have only one conductor
left over.

Does the one that's left go to XLR pin 2 or XLR pin 3?

By the way, I checked the Shure web site and found the document you were
referring to, thank you. But with my extremely limited electronics
knowledge, it really makes no sense to me.

Darren