Help with AC input for a relay

Thread Starter

EEDude

Joined Nov 18, 2008
40
I am trying to use an Omron relay G2R-1-AC120. I have four possible siganl wires that can be usedas the input to the coil of the relay. There appears to be two pairs of wires. Two of the wires when I measure with a scope show a sine wave with a peak voltage of around 240. The other two wires show a sine wave two that is really noisy and very small. I am using a series resistor in front of the input side of the coil (to create a voltage divider) to reduce the voltage to around 120rms to work with this relay. I tried using the 240VAC signal to the input side of the coil, and the small signal on the output side of the coil, but the switch never closed. When I measure the voltage after the resistor which should be what the coil sees I see about 175VAC on the scope, and about 116rms with a DVM. Seems like this should work im not sure why its not. Am I using the wrong combination of wires?? Also I tried using a variable 50K resistor, and I swept from 0 to 50K and still couldn't get the swith to close. Not sure if the frequency matters.....there is no spec for frequency for the relay. Anyone have any help/ideas that would be great thanks!!
 

beenthere

Joined Apr 20, 2004
15,819
You can always go to the Omron site and get the data sheet. It is rated for 50/60 Hz.

It is a plug-in relay, configured SPDT. The two contacts immediately under the coil are the ones that operate it.

You speak of signal wires - nothing like that appears on the datasheet. Are you sure of the part identification?
 

Thread Starter

EEDude

Joined Nov 18, 2008
40
The signal wires that will be used to control the relay were before used to control whether or not a flueorescent light bulb would be on or not. Now instead of using fluorescent light I am using an led light. So I need to take the signal wires that controlled the old light to control the input of the relay (so now instead of the AC signal turning on the flueorescent light it will energize the coil and close the relay switch). Then the output of the relay is connected to the DC power to the LED light. So when the AC input signal is energized the switch will close and the DC current will flow through the LED turning it on. When the AC signal goes away the switch opens and the LED will turn off. The AC signal is controlled on/off by the machines software which is why I need to have this new light automated the way the old one was.
 

Thread Starter

EEDude

Joined Nov 18, 2008
40
I understand the pin layout on the relay, what I am not sure about is what combination of wires do I put to the pins of the coil. Like I said there are four wires that do not have anything to do with the relay other than the fact that this is what I will be using to control when the switch opens and closes (these wires are attached to a machine and the internal software tells the machine when to send the signal). A 240VAC coil wont work because the 240Vpeak that I measure with the scope is only 170Vrms. The coils are rated at rms, so I have to get something 170 or lower. So I am asking do I use the combination of the big and small signal for the coil. The two big signals, or the two smaller signals. I already know that this relay wont work because the frequency of the signal I want to use is to large, but I would still like to know what combination of two of the four wires would I use to energize the coil if the signal frequency were 50/60Hz
 

Thread Starter

EEDude

Joined Nov 18, 2008
40
yes, and that is why the relay wont work. Frequency is to big

________ Wire 1: 240Vpeak 55Khz

________ Wire 2: ~10Vpeak 55Khz (signal is noisy)

________ Wire 3: 240VPeak 55Khz

________ Wire 4: ~10Vpeak 55Khz (signal is noisy)


There are the four wires I can choose from. Assuming the frequency wasn't so high and this relay would work what pair of these four wires would I put to the pins of the relay coil?
 
Last edited:

Bernard

Joined Aug 7, 2008
5,784
What is your referance point used to measure voltages, ground or line to line ? Might rectify " signal " and use a DC relay.
 
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