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I am trying to understand relay circuits better. I have two circuits in question below.

enter image description here enter image description here

Left Image I understand Pins 2 and 9 are the coil contacts. I would wire a microcontroller outpin pin to a transistor and Power Supply (Pin 2) to control the coil activation (Ground Pin 9). I don't understand what would connect to Pin 5 and 6 since I am used to seeing just one Pin at the source of relay circuits like Pin 1 in the Right Image.

Right Image Pins 2 and 5 are the coil contacts. I would wire my microcontroller and Transistor as a switch circuit similarly to above explanation. Power supply + end to Pin 1 and Load to Pin 3 and - end of Load to - end of Power supply, is that correct?

I believe Right Image is more simple to understand but not sure at all about Left Image. If both my explanations are wrong please correct me. Also, if possible can you explain how once the coil is activated how the lever/switch moves? Based on the orientation in the circuit images I don't see how the lever moves to either pin. Unless the image is not the correct orientation of the mechanical components inside the relay. Finally, is it clear if these circuits are latching or non-latching or should the datasheet tell me that?

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    \$\begingroup\$ If it was latching, the coil would definitely have polarity. It looks like you have either an AC relay or a reed relay here, which don't need a specific direction of current to actuate. The contacts move due to the magnetic field produced by the coil. Pins 5 and 6 are just shorted together, connect either or both to your load. \$\endgroup\$
    – Hearth
    Dec 14, 2018 at 0:43
  • \$\begingroup\$ Note also that pins 1(L) and 4(R) are the normally-closed (NC) contacts, and pins 10(L) and 3(R) are normally-open (NO). \$\endgroup\$
    – Nate S.
    Dec 14, 2018 at 1:00
  • \$\begingroup\$ In your left drawing, pins 5 and 6 both connect to the moving contact of the relay - you may use either pin for your external connection. The two pins are probably provided just because that relay package has that many pin positions available. Electrically, the two relays are identical - pins 5 and 6 of the left relay are equivalent to pin 1 of the right relay. \$\endgroup\$
    – Peter Bennett
    Dec 14, 2018 at 7:34
  • \$\begingroup\$ Thank you @Hearth for that explanation of Pin 5 and 6 setup. \$\endgroup\$
    – SChand
    Dec 14, 2018 at 20:05
  • \$\begingroup\$ Thank you @NateStrickland. I was unsure on definitions for NC and NO but that is starting to clear things up for me. \$\endgroup\$
    – SChand
    Dec 14, 2018 at 20:06

1 Answer 1

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Pins 5 and 6 of the left relay are connected together internally. You can connect them together externally or not.

The internal schematic does not directly represent the internal configuration (which is 3-dimensional). For example, in the right hand relay, the normally closed contact is above the moving contact which is above the normally open (Z-axis, out of the screen). Photo from this website:

enter image description here

Monostable relays are always shown in the de-energized state. So the relay with no power on it, has pin 1 connected to 5&6 (left ) and pin 1 connected to 4 (right). With the coil energized, pin 10 is connected to 5&6 (left) and pin 1 is connected to 3 (right).

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