0
\$\begingroup\$

Completely new to electronics and I’m trying to follow a blog post to wire up a linear actuator to a dpdt relay and photocell.

I’m stuck, because the diagram I’m following is showing a total of 4 wires coming from the DC power supply. However the DC power supply only has two wires. Am I missing something? Am I supposed to split them somehow?

The diagram is below but here is the post I’m following if that helps.

Wiring diagram

\$\endgroup\$
  • \$\begingroup\$ Black, white, and red are common wiring colors for 120vAC devices such as motion-activated lights. The photocell could easily be mistaken if it wasn't for reading the listing details. \$\endgroup\$ – rdtsc Aug 16 '19 at 19:48
0
\$\begingroup\$

Quick answer:

You need to connect two wires to the (+) side of your power supply, and two wires to the (-) side. If your physical supply has an output cable with only one pair of wires, you'll need to splice the wires together somehow. Connect all the (+) wires together in one group, and the (-) wires in another group.

There are lots of ways to do this. There is a related question with a lot of great answers here on another Stack Exchange site.

A common solution, if you have the equipment, is to solder the wires together and cover the exposed metal with heat-shrink tubing:

heatshrink

An easier option, if your wires are think enough, is to use wire nuts, which can be a good solution for a self-described "beginner" :) You can get them at any hardware store:

wirenut

A better, but more expensive, option are IDC-style butt-connectors: idcsplice


Additional info:

That image is really confusing. The problem is because the photocell uses a different wiring standard than is generally expected.

Generally, in the USA at least, DC power is wired with the positive (+) being red, and the negative (-) being black. This is how most of the wiring in your image is drawn.

But, the photocell has a black wire for (+), white for (-), and red for the output signal! Lots of room for mixup.

The blog author explained it in a comment:

Ah I can see how that could be confusing. This is 12VDC power, so the black wire from the power supply is ground (-), & the red wire from the power supply is positive (+). You’ll see that standard DC power wire color scheme matches the wiring that runs to the relay & actuator (and timer too, for that option).

The photocell is where the wire colors get confusing — although it is a 12VDC photocell, it oddly follows the typical 120/240VAC wire color convention as you described. That is why I have the text notes on those photocell wires. Other than the photocell, the rest of the wiring follows the standard 12VDC wire color convention of red (+) / black (-).

\$\endgroup\$
  • \$\begingroup\$ Makes sense, thank you for the clear explanation! I’m learning! \$\endgroup\$ – cbg Aug 16 '19 at 20:50
  • \$\begingroup\$ @cbg Happy to help! This stuff is fun :) \$\endgroup\$ – bitsmack Aug 16 '19 at 20:58
  • \$\begingroup\$ one more thing (sorry), would you kindly explain what is being shown on the diagram for the negative wire coming from the power supply? Instead of going to a single end point it goes into one and then into a second. How does that all work? Is it being spliced again? \$\endgroup\$ – cbg Aug 18 '19 at 21:51
0
\$\begingroup\$

I think there are some wires coming from the photocell that get attached to the power supply, and some wires go from the power supply to the relay. It sounds like there are black, white, and red wires already connected to the photocell...the black goes to the + side of the 12V supply, the white goes to the - side of the 12V supply, and the red wire is the output. This is not a typical color scheme for low-voltage dc wiring so you should check the datasheet for the actual photocell you use.

In addition, a red wire coming from the + side of the power supply goes to the relay, as does a black wire from the - side of the power supply. It looks like the relay is wired so that the direction of the actuator will be reversed when the photocell output enables the relay. Again, you should make sure that the relay you actually have in hand is wired correctly...the locations of the terminals might not be exactly the same.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.