2
\$\begingroup\$

I'm having some trouble wiring up a camera slider where a motor moves the camera to one end of a slider and hits an NC/NO/C 3-pin switch which triggers a 12V 8-Pin DPDT relay to reverse the motor polarity and move to the other end of the slider and trigger another switch. The camera keeps "ping-ponging" back and forth as it should. I'm basically following this example: http://cheesycam.com/cheesycam-diy-auto-reverse-polarity-motorized-video-slider-update/

Here is the wiring diagram from the cheesycam post: enter image description here

The trouble is that I need to be able to control the speed of the camera, so I added a speed controller to the motor. The speed controller reduces the voltage to slow down the motor - which causes the relay to stop working once the speed/voltage is low enough.

Here is my wiring diagram: enter image description here

Is there a way to retain the speed controller without reducing the voltage (and thus making the relay stop working)? Does it make sense to use a lower voltage relay (6V DC, for example) so it will work for a wider range of speeds?

I thought I could move the speed controller directly between the motor and the relay (basically use the relay's output as the input to the speed controller, and then use the speed controller output directly to the motor), but couldn't make it work.

Here is a related post regarding the same circuit I'm trying to build, but the answer uses a 3PDT relay (and plus I'm not able to read the wiring diagram there unfortunately): Adding Speed Control for a DC Motor

I not at all experienced with wiring circuits, so any help or ideas are greatly appreciated.

**UPDATE WITH FINAL SOLUTION: @Charlie provided the solution after some back and forth. Here is the final working solution: enter image description here

Ideally, this would be wired with a 3PDT relay, but all I had on hand was a 4PDT, so the last of the 4 relay switches was not used. This is working great, and the speed controller is now independent of the relay voltage, so the relay continues to work and reverse the polarity/direction of the motor even at low speeds.

\$\endgroup\$
  • \$\begingroup\$ Speed controllers typically work by reducing the voltage to the motor (or using PWM, which is effectively reducing the voltage the motor sees due to its inductance). You need to power the relay coil directly from the 12V power source; however, the switched power should come from the speed controller. If you provide an actual part number for your relay, someone would likely be able to give you more specific implementation details. \$\endgroup\$ – Kurt E. Clothier Oct 29 '15 at 6:32
  • \$\begingroup\$ Thanks, @kurt-e-clothier. The relay is here: amazon.com/gp/product/… - Amico DC 12V Coil 8 Pin General Purpose Relay DPDT HH52P w PYF08A Socket. Makes sense that the speed controller needs to go between the relay and the motor so that the relay keeps it's voltage constant. I'll try wiring it up that way again. \$\endgroup\$ – jhohman Oct 29 '15 at 16:59
  • \$\begingroup\$ I tried moving the speed controller wiring to directly between the relay and the motor in my diagram above, but the speed controller is inexplicably making the circuit inoperable. Not sure why, but this isn't the answer either. Any other ideas? \$\endgroup\$ – jhohman Oct 29 '15 at 21:24
  • \$\begingroup\$ I think I can figure this out if I draw a schematic diagram. I will be back soon. \$\endgroup\$ – Charles Cowie Oct 29 '15 at 21:34
1
\$\begingroup\$

I think this will do what you want. You will need an additional relay or a 3-pole relay. I showed a diode across the speed control output, but that may already be in the speed control. I also showed an on/off switch. This is similar to the circuit offered for the other question, but I showed an additional relay instead of a 3-pole relay. I hope it is easier to understand the way I drew it.

enter image description here

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thanks @CharlesCowie! Now I need to figure out how this drawing translates into me wiring it up (I should have taken that circuits class in college). \$\endgroup\$ – jhohman Oct 29 '15 at 23:29
  • \$\begingroup\$ I guess I could add the numbers for the RL1 terminals. \$\endgroup\$ – Charles Cowie Oct 30 '15 at 0:23
  • \$\begingroup\$ with your diagram and the diagram from the other answer, I think I might be able to piece it together. I have a 4PDT relay on hand, so I assume I can use it and just not use the last set of terminals. Does it look like I have this labeled right? I don't have a socket for the relay, so I want to be somewhat sure of what I'm doing before soldering it up. Here is my updated and labeled diagram: joshhohman.com/assets/diagram.jpg \$\endgroup\$ – jhohman Oct 30 '15 at 0:47
  • \$\begingroup\$ I see you just updated the diagram. Thanks. I'm going to convert the numbering to my 4PDT relay and give it a shot. I'll let you know how it turns out. \$\endgroup\$ – jhohman Oct 30 '15 at 0:58
  • \$\begingroup\$ The standard way of showing NO & NC contacts on a diagram is to show the NC contacts closed and the NO contacts open, the non-energized or non-activated condition. Your diagram is opposite that. The NC contact of S1 should be used and the NO contacts of S2 and S3 should be used. You will also need to be careful about which end of travel is "forward" and which is "reverse." \$\endgroup\$ – Charles Cowie Oct 30 '15 at 1:25
0
\$\begingroup\$

I have modified it just a bit for a 3pdt.

  1. Motor spins to move carriage.
  2. Relay will change polarity to motor to cause the carriage to "ping-pong"
  3. Speed control with adjust speed.
  4. SW1 and SW2 will cause relay to be energized or non-energized.
  5. Relay needs a full 12v to operate. Speed control reduces voltage.
  6. The relay is non-energized to begin with.
  7. The carriage needs to head toward SW2 when started. Because if it hits SW1 first nothing is going to change and the motor is going to keep turning that direction tearing stuff up. The circuit to the relay is broken to start with. Breaking the negative at SW1 isn't going to change anything. So hitting SW2 first will complete the circuit to 11 of the relay and energize the relay, causing all three switches (1->4) (2->5) and (3->6) to move reversing polarity. They will be electrically held at 4, 5 and 6. Then when the carriage hits SW1 it breaks the negative, thus breaking the circuit and causing the relay to become non-energized, causing the switches to move back to the original state they were in when it started, thus reversing polarity.
  8. If carriage doesn't head toward SW2 to start switch the + and - wires to motor.
    enter image description here
| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ This isn't an answer. You are trying to ask another question. While we absolutely encourage asking of questions on this site, it does need to be asked using the "ask" button at the top of your page. Please click on the link above to "edit" your post; copy the entire thing (Ctrl-C), then delete it as an answer, so you can ask it as its own question (click "Ask," then paste [Ctrl-V] into that edit box). and welcome to Stack Exchange :) \$\endgroup\$ – Robherc KV5ROB Feb 3 '16 at 21:12
  • \$\begingroup\$ I am sorry, I did intend for this to be an answer. It expands on the knowledge that he is providing. I am 99.99% sure it is right. I just put that clause in there to allow for the possibility that it is wrong. \$\endgroup\$ – Mooneyb123 Feb 3 '16 at 21:19
  • \$\begingroup\$ Ok, then please edit/rewrite this so it is easier to tell that it is intended as an aswer, and (if possible) explain what you can of "how you got there" so that the asker (and others who find this when searching the web) can educate themselves from it. :) \$\endgroup\$ – Robherc KV5ROB Feb 3 '16 at 21:32

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.