Relays and flashing LEDs

Hi there

I'm teaching myself electronics and I've hit a bit of a brick wall with getting my ahead around relays.

What I'm trying to do is build is a simple relay circuit on a breadboard to witness its function, possibly by seeing two LEDs flicker or turn on / off.

The relay I'm working with is an OMRON G5V-1. It has six pins. I'm not sure as to how to orientate it on the breadboard, and where I should connect to jumper wires. I do know some theory though: a relay has two circuits, one is the electromagnet that when electrified engages an armature which pulls two contacts together to close the secondary circuit which in-turn causes the attached load (i.e. LED) to light up.

If anyone could offer any pointers they would be greatly appreciated, thanks.

At my disposal...

Voltage supply: 6V Breadboard: 880 pin LEDs Capacitors Resistors (fixed and variable) Transistors 555 IC

by dannyarcher
February 04, 2021

Depends on the exact relay that you use. Some have strict polarity, some not. Take a look at a G5PZ Omron, page 3, it has no polarity, the magnet is on pins 1 and 4 and the circuit on pins 2 and 3. I have serious doubt that each and every relay has the same configuration, as a proof of that, yours have 6 pins! Can be a SPDT switch, can be something else, totally. If you can't find the markings on yours, or you can't find its datasheet on internet, try to experiment with a ohmeter: the magnet should be close to 0 ohm between its two pins, while the circuit, if normally open, should be at a very high resistance.

by vanderghast
February 04, 2021

I just saw that you supplied the relay. G5V_1. Parge 41, you have the pins number. Pins 1, 5 and 6 are internally connected. You have to supply 80% of the rated voltage between pins 2 and 9 to close the path between pins 1 and 10.

So, place the relay across the middle separation on the breadboard ( 3 pins on each side of the separation). Pins 6, 9 and 10 are "away from you". You can check it with a ohm meter: there should be a high resistance between pins 6 and 10, while there is a little resistance between pins 1 and 5, which would be "close to you"..

Assuming a RED LED and a power supply of 6 Volt, add a resistor of 200 ohm between pins 10 and ground; add the LED with the proper orientation between 6 V rail and pins 1, 5 or 6.

Connect pin 9 to ground. If you connect pin 2 to 6V, the coil of the relay is active, internally closing the internal switch between pins 10 and pins { 1, 5, 6}, so the LED should turn on and if you then connect pin 2 to ground, the internal switch opens, so the LED should turn off. (You could exchange the role of pins 2 and 9, that relay is not polarised.) Sure, you can put pins 2 to 6 V or to ground with a flying wire that you manually plug at the right place, or through the output of your NE555.

You should be able to place a GREEN LED between the 6V rail and the existing 200 ohm resistor. With that arrangement, the green LED is ON when the red one if OFF, and vice-versa. Indeed, the two LED are now in parallel, but when the red one is on, it offers an easier path to electricity and starves the green one. When the red LED is off, the only path left if the one through the green LED, and so, it turns on.

by vanderghast
February 10, 2021
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2 Answers

Answer by PolycabIndia

To understand the circuit you need to understand how a relay works. When the relay coil has power, the switch will disconnect the power from the electromagnet and connect the power to the light bulb instead so that it will light up. ... This makes the LED turn ON. Polycab Luminaires a wide range of functional and elegant lighting solutions for everything from roads to bridges to gardens to walking and jogging tracks and many other outdoor lighting requirements that are critical for urban living. https://polycab.com/products/lighting-luminaires/

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by PolycabIndia
February 07, 2021
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Answer by febb

For example if you orient your breadboard vertically so you see divider between left and right groups of sockets goes vertically. These group of sockets connected internally horizontally. But left and right groups are electrically disconnected from each other. Orient you relay also vertically long side. You need to insert left 3 pins of you relay in left side of sockets and 3 right pins on right side. So breadboard divider separates them. Then you can use jumper wires to insert on both sides of the relay to connect to pins..

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by febb
February 09, 2021
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