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I want to terminate a cable in a 9-pin D plug. The design I’m working on needs a resistor and two capacitors at the terminals of the D plug.

I can’t work out how to successfully work Pin 7 – I need to join together the resistor, a capacitor and a wire to Pin 2. The thinnest wire I have is 24 gauge, 0.511 mm, and the diameter of the solder pots is 1.5mm. Diagram 1 below shows the plug wiring enter image description here

My plan is to join the wire, R1 and C2 together using a lash splice.

enter image description here

This is proving really hard to do in practice. Only one wire can fit into the solder pot; the other two have to be joined 3mm above the tip of the wire going into the solder pot.

So - is there a better way to do it?

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  • \$\begingroup\$ One note of caution when trying these solutions. The plastic surrounding the pins in a typical D-sub connector is often quite heat sensitive and if you do much soldering, can often move the pins and ruin the connector. Besides trying to use minimal heat, minimal time and minimal force on the pins; it often helps to plug an opposite-sex D-sub connector into the one you are soldering to help disperse the heat and help hold the pins in place. It also helps to clamp both of these into a small vise. \$\endgroup\$
    – Tut
    Commented Sep 24, 2015 at 19:06

3 Answers 3

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If this is to be repeated many times, consider making a small PCB that fits between the two rows of cups.

Another technique is to solder the lead of a component into the pot, and bent the lead of the next coponent in a small circle (loop) around the wire of the first component.

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  • \$\begingroup\$ I did consider putting the components on a Verboard, but the d plug case is metal and I would have had to use sillicone sealer to electrically isolate the circuit. That said, a PCB is definitely something to consider for the future. \$\endgroup\$ Commented Sep 23, 2015 at 8:12
  • \$\begingroup\$ A veroboard won't do as the strips won't match the solder cups of the pins. (And the upper and lower sides must be shifted with respect to each other). \$\endgroup\$ Commented Sep 23, 2015 at 14:31
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Wrap the free lead of the capacitor around the lead of the resistor, as close to the resistor body as possible, solder them together and trim off the rest of the capacitor lead so you are left with just the lead from the resistor. Solder it to the inside of pin 7. Then solder a short piece of wire from the inside of pin 2 over to the top of pin 7, crimping it to the resistor lead. Then reflow solder over both the wire and the pin.

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  • \$\begingroup\$ Hi, this is a good idea but I think in practice it's not so easy because I still have to put a yellow wire into pot 2. I tried putting a loop of wire around Pins 7 and 2 so that the pots are still free. There's enough clearance between the wires and the other pots (1mm), and I have some heatshrink that should cover the wire anyway. \$\endgroup\$ Commented Sep 23, 2015 at 8:16
  • \$\begingroup\$ @MusicForChameleons Actually my original post suggested putting wires around the outside of the pots (click on "edited..." to see) but then I didn't think there would be enough clearance so I changed it. \$\endgroup\$
    – tcrosley
    Commented Sep 23, 2015 at 8:21
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The easiest solution is to install a bare wire jumper between pins 2 & 7. Take a piece of bare, solid wire and form into a "U" shape. Make the legs of the "U" the right spacing to slip into pins 2 & 7. Cut the legs of the "U" to about 1/4" (6mm) long or so and slip the wire loop into both pins 2 & 7. Solder the wire loop to the pins.

Now hook and solder all of the other wires into the loop that is thus formed.

That is: the resistor, capacitor, wire leads all are connected to the wire loop that is connecting pin 2 to pin 7.

Use relatively heavy solid wire for the jumper: 18 or 20 AWG will work well.

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