I have some Motorola Walkie Talkie TKLR T7 (seems only available in Hong Kong, cannot search much information), was trying to connect a pair of them to a relay. Not that I did not succeed, but I encountered some strange problem.

4 wire of brown green red blue colour

The earphone jack is 3 parts 2.5mm diameter 14mm length (Motorola M2 standard?), it is connected to the Speaker, PTT/mic, ground. When I cut open the wire, there is four separate wire of four colours (it was found out later that 2 actually connected to the same ground of the jack, so it is still 3 connections). Not knowing which wire is which, with the jack plugged in, I tried to short exhaustive combinations of 2 wires, by twisting them together or by using crocodile clamps, to init the push to talk signal, but the icon never showed up on the walkie screen.

Desperate, I actually peeled all plastic off a jack, and directly soldered some plastic coated copper wire to the 3 connection points, and then when I short 2 of them (ground and PTT), the push to talk icon is showing on the walkie.

So I can conclude that when I was using the 4 coloured wires, I did not make them connect, So I thought the coloured coating might be cloth like insulators. I tried to unwind the threads, only to find out that the coating seemed to be very thin metallic wires twisted around some transparent rip cord like threads. And again no matter how I try to short the metallic like wires or the rip cord like thread, I cannot get the push to talk signal nor the icon on the walkie screen.

So can someone tell me what kind of wires are these? and where is the conducting metal part? is it the coloured thread or the transparent thread in the middle? if it is the coloured threads, how did they prevent themselves from shortcircuiting each other in the long wire? was there some filler that was destroyed when I rip the wire apart? Or was there some simple physics I have missed? And why did I fail to short circuit them?

  • \$\begingroup\$ They might be enamelled copper wire wound over some insulating thread. \$\endgroup\$
    – Simon B
    Aug 4, 2017 at 14:19
  • \$\begingroup\$ @SimonB It does looks so, however I don't understand why they do not short circuit when wrapped in the actual cable, is my earlier filler assumption correct? could they be touching each other near the root without me noticing so my short circuit operation failed? \$\endgroup\$ Aug 4, 2017 at 14:50
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    \$\begingroup\$ The enamelling is a very thin insulation. Transformer windings use enamelled wire, because it's finer than plastic-insulated wire. You need to scrape the enamel off to make contact with the wire. \$\endgroup\$
    – Simon B
    Aug 4, 2017 at 14:57
  • \$\begingroup\$ @SimonB I see what you mean, you reckon that if I peel off some colour from the wires it will show copper? \$\endgroup\$ Aug 4, 2017 at 15:07
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    \$\begingroup\$ Just heat the pretty coloured wires in a blob of molten solder on the tip of your iron (or in a solder pot). After a couple of seconds the varnish burns off. It's self-fluxing so it leaves a tinned wire end.. \$\endgroup\$
    – user16324
    Aug 4, 2017 at 20:52

1 Answer 1


Turning the comments into an answer...

The fine wires will be enameled copper wire. The enamel is a thin coat of insulation applied to the wire. It's normally translucent, and can be made in different colors. The enamel is thinner than conventional plastic insulation.

To make contact with it, for soldering or screw terminals, just scrape off the enamel at the ends.


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