How to find out a relay's "trigger" voltage

Hey there!
I currently have several "old" (read: unused) relays laying around that I'd like to trigger using my Arduino Uno.
The issue with all of them is that they've got no datasheets available or if they have, I can't tell from the sheets at what voltage the relay triggers. Since they're all quite small and perfectly fit on a standard breadboard, I don't assume they would consume more than the 5V my arduino could provide.
I already checked the connections using my multimeter and there's definitely current flowing through the line it's supposed to run through when unenergized. The actual relay I'm talking about is an AZ8-1CH-24DE (http://www.digchip.com/datasheets/parts/datasheet/540/AZ81CH24DE-pdf.php).

Attached is the wiring scheme that's printed onto the relay.

And here's how I've wired the whole thing:

1 - +5V from Arduino (always on)
2 - LED1, which connects to arduino's GND
3 - LED2, which connects to arduino's GND
4 - arduino GND
5 - +5V from Arduino (control pulse)

I've also tried switching 4 and 5 around, thinking I inverted the two pins, but whatever I do, I can't get the relay to switch from LED1 being on to LED2 being on.

Would be glad if someone could help me out with this. I'm kinda new to the whole thing, but I though I understood how these things *should* work..
Also, if anyone comes to the conclusion that the relay will definitely need a higher (12V?) voltage to "trip", could you also tell me how I can control that relay from my arduino while only having my +5V pin from the arduino available? I thought about transistors, but maybe there's another way.

Best regards,

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Christoph680 (author) 6 years ago
Okay, so I just got myself the 4013 flip flops.
I've hooked it up to the +5V of my arduino, but somehow nothing's happening..
Attached is my current wiring. Here's what I'd expect should happen:

1) D is attached to +5V and Gnd, D = HIGH
2) Clk is attached to +5V and Gnd, Clk = HIGH
3) S is attached to Gnd, S = LOW
4) R is attached to Gnd, R = LOW

Since both - D and Clk - are HIGH, I'd expect the LED to turn on (also to measure an output voltage of around 4-5V, which I don't). The LED is connected to Q.
I'd also expect the LED to stay on when I remove the Clk wire.
Additionally, if I connect R to +5V and Gnd, I'd expect the LED to turn off and stay off.

None of that is actually happening. Upon powering the circuit, the LED stays off, no matter what wires I connect/disconnect. If I connect the LED to NQ, it turns on, as expected.

Is there anything I'm missing here?

Christoph680 (author) 6 years ago
Okay, thanks again guys ;)
Now I'm stuck on a little issue here and would like to get your opinion on that matter..

I am trying to create some sort of selection mechanism, where a motor is driving a shaft that is continuously searching for an active IR emitter in a vertical array of emitters.

Now, depending on what button the user presses (there's 12 in total) I need one of the 12 corresponding emitters to become active and stay active until the detector mounted on the shaft sends the correct signal to my arduino.

Now it's not about getting the correct signal to my arduino.. it's rather that it seems somewhat expensive (thinking about maybe expanding the mechanism to support 24 or even more selections) to solve it only using relays.

I thought about an array of 12 relays that get triggered by the pushbuttons and are running self-sufficient (think about bistable relais) until they get sent a signal to shut off.
While this works in theory and in practice, like I said, it seems quite expensive to me to pay around €2 for each additional selection I want to add..

Could anyone think of another way of how to achieve that selection mechanism (possibly thinking about expanding it easily in the future too)? It doesn't have to use IR-sensors.. that's just the cheapest parts I gathered so far.. solenoids are riddiculously expensive over here in Germany and I would still need to use some additional detection mechanism. I also thought about electric magnets, but as you can see, the main issue with this is the wiring of a 12-relay-array. I'd love to know if there's an alternative to using that much relays.. maybe I'm just not seeing it..

So this thing scans up and down until it sees one emitter ?

Easy. only needs two relays, and a couple of safety limit switches. Here's a diagram. Where I've got switches marked up and down, connect them to the arduino via transistors. IN SERIES with them, add your two limit switches.

You can get fancy, and take the signal on the BOTTOM of the relay to the arduino (suitably conditioned), so you can check the state of the relay and therefore whether you have hit the stop.

If the "up relay" is on, according to your code, AND the voltage on the bottom of the relay is 0V (roughly), then the limit switch is closed, and your motor is moving up.

If the "up relay" is on and the volts on the bottom of the relay is NOT 0V, your up limit switch is active, and you need to go DOWN again.
Christoph680 (author)  steveastrouk6 years ago
Thanks a lot ;) But like I said, the issue isn't finding the one emitter, the issue is getting that one emitter to emit using as less components as possible.

Say I have 12 possible selections. The "scanner" consists of one detector.
The user makes a selection on one of 12 pushbuttons. These are wired to the 12 emitters. If a pushbutton is hit, a bistable relay turns on and powers the emitter until the "scanner" has found its position and sends the stop signal, which switches power off the relay again.

My question was if there's a simpler way to do this basic button -> relay -> emitter connection than using 12 bistable relays, which would cost a bit.

Yes, much cheaper - use flip-flops, one per emitter. Press the push button to set the flip flop From Andy Collinson's most excellent cicuit snippet  website
I found the attached link. 

It will do exactly what you want: scaling it to 12 inputs is easy.
Christoph680 (author)  steveastrouk6 years ago
Thanks! I've actually already thought about using flip flops but I couldn't find the correct ic number.. duh

Okay, so the 4013 if I'm reading it correct can actually receive up to +18V of supply voltage as well as output -0.5 to VDD (e.g. +18), which means there wouldn't be a need for any kind of transistor aggregating voltage, is that correct?

Also, the S pin on the IC is pretty much the counterpart for the R pin, right? So while I could use D to save 0/1 I can also use R to override D in any case with 0 and S to override D in any case with 1... hope I get it at some point :D
You're getting there....
gmoon6 years ago
Even if you buy 5V relays, be mindful of a couple things:

1) AVR (arduino) I/O pins can supply/sink about 40mA of current. Exceed that draw, and you'll fry the pin.

As you suspected, adding a transistor here is the best approach for driving a typical (magnetic coil) relay.

2) Being Inductive loads, relays create spikes during operation. You'll need a diode "clamp" across the coil to prevent damaging the drive circuit.

If you can find an appropriate 5V solid-state relay, that could solve both problems...
kelseymh6 years ago
Look at page 2 of the data sheet; what you're after is the coil voltage.

Each part number is different; your "AZ8-1CH-24DE" (far right column, row 6) has a nominal coil voltage of 24V at 1.28 kOhm, a maximum of 50V, and a minimum ("must operate") coil voltage of 15.6V. It's a medium duty (6A contact) epoxy-sealed unit.
Christoph680 (author)  kelseymh6 years ago
Which means I'm gonna at least need those 15.6V to trigger the relay?
Well, if that's so, I should probably just buy some smaller ones.. they're cheap anyway.

Thanks for the quick help!

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