Hello,

I have a 2v device that just detects if the circuit is open or closed and acts upon it. I use an illuminated pushbutton to open/close the circuit but the 2v is too low to light it up. I added an extra 3v using a CR2032 coin battery but the device receives the extra 3v and is permanently closed. Can't find a button with double-pole contacts so I was thinking of adding a relay.

I have very little knowledge of relays and have used only big mechanical ones, for this application I would like to use a SSR due to its small size since it needs to fit inside this tiny button case, but I don't want to complicate it further by using a PCB or any other solution that would increase cost and labor, I would just like to solder a tiny cheap relay and be done with it.

Can you help me by pointing out a specific relay I can use? A link would be fantastic.

Thanks!

 

Hello,

I have a 2v device that just detects if the circuit is open or closed and acts upon it. I use an illuminated pushbutton to open/close the circuit but the 2v is too low to light it up. I added an extra 3v using a CR2032 coin battery but the device receives the extra 3v and is permanently closed. Can't find a button with double-pole contacts so I was thinking of adding a relay.

I have very little knowledge of relays and have used only big mechanical ones, for this application I would like to use a SSR due to its small size since it needs to fit inside this tiny button case, but I don't want to complicate it further by using a PCB or any other solution that would increase cost and labor, I would just like to solder a tiny cheap relay and be done with it.

Can you help me by pointing out a specific relay I can use? A link would be fantastic.

Thanks!

View attachment 317370
View attachment 317372

I would ditch the relay idea and get a double pole push button illuminated switch and switch the light on independently.

 

Welcome to AAC.

The really idea is a non-starter. If you can’t get a DPxT switch that fits, then switch the negative wires to the 2V “device” and the switch’s LED instead of the positive.

You can use your coin cell to power the LED, and let the mystery device’s 2V stay 2V because they will each be connected directly to their respective targets and will not interact on the negative (0V) side.

 

Some more info re your "device" would help.
For instance, what is it powered from? What current does it require?
2V is an odd value so I would think that is not correct.
A relay is not the way to go. We can help if you supply full details.

 

Thank you for your replies.

The device is a breakout box, it is connected to a micro computer (possibly a raspberry pi) with a 8 pole mini din. The breakout box is used for aquariums and has multiple inputs where you can connect all sorts of gadgets like buttons, water sensors, etc. Theres an app where you can monitor these inputs and program them to control other equipment connected to the computer. The voltage is low because it just needs to detect if the circuit is closed or open, thats enough to change the status of the input.

Breakout Box image:



App interface with the breakout box inputs (you can rename them):


I can't find an illuminated pushbutton with double poles anywhere, among many others I've looked at Digikey and Mouser. Besides being a latching button it has to be very small, be lit with very little voltage and it has to be orange. I think just finding this button was lucky enough. If I could find one with double poles it would be the best solution though, I would love to simplify my life. If you know of somewhere I could find something similar please let me know and I will gladly incorporate it.

Any solution is welcome.

Thanks.

 

Welcome to AAC.

The really idea is a non-starter. If you can’t get a DPxT switch that fits, then switch the negative wires to the 2V “device” and the switch’s LED instead of the positive.

You can use your coin cell to power the LED, and let the mystery device’s 2V stay 2V because they will each be connected directly to their respective targets and will not interact on the negative (0V) side.

I wired it like the diagram below because the light has to come on when the button is pressed. If I connect positive and negative to their respective contacts the button will light up when its on the depressed state. Just another layer of complexity. So I'm not sure if your suggestion will work as desired, I will give it a try though and let you know. Thanks.

 

Do you have any technical documentation on it that we can look at? And the make and model so we may find an instruction book.
Just out of interest, the 2V , have you measured that? And is it the voltage across the input terminals?

The voltage is low because it just needs to detect if the circuit is closed or open

The voltage being low is not necessary value for detecting inputs. To detect an input, the voltage can be just about anything. 5V, 24V 240VAC........ It depends on the device.
How do you interface this with a Raspberry Pi? The Mini DIN 8 must plug into some sort of interface.

 

I've never seen any technical documentation or a book with specifications about the breakout box. It also doesnt come with any printed info in the package when you purchase it.

There is some information about it on this pdf, on page 108, although its an older version of the breakout box:
https://www.neptunesystems.com/downloads/docs/Comprehensive_Reference_Manual.pdf

There is however an official forum by the brand where you can search for more information:
https://forum.neptunesystems.com/forum.php

There's also a video on youtube made by the manufacturer:

I have measured the voltage and it reads 2V. Each terminal reads 2V. You can find that information on the forum as well.

Heres the back of the computer where you connect the mini DIN:


Its called Apex and you can get more information about it on the manufacturers website:
https://www.neptunesystems.com/a3-apex-series/#

Let me know if I can give you more information.

Thanks.

 

Without seeing a data sheet on this switch and given the contacts are rated for 3A @ 125VAC and 1.5A @ 250VAC, my guess - and that's exactly what it is - a guess - is that the two large contacts are a switch that is independent of the LED. It looks like you can use the LED from any sufficient voltage to indicate when the switch is either ON or OFF, depending on how you make the connections. As others are saying, there are better ways than a relay. Part of the problem with a relay is the amount of current it will consume. Even momentary activation can be draining and hard on other things. And like others are asking, a clearer picture of what you're trying to accomplish will better aid us to give you the best approach at solving your problem(s) or reaching your goal(s).

Also, it looks like your diagram is bypassing the switch via the 3V battery. Let me make an attempt at a schematic of what I THINK you're after. I'll be back with that shortly.

 

OK, I don't know if your lighted toggle switch is a switch or a push button. You DID say:

I use an illuminated pushbutton to open/close the circuit

So I'm assuming there might be a built in resistor inside the switch. Can you tell us if the switch was designed to operate on 12V DC? If so then it probably has a resistor inside. If not then you need to add a resistor to the outside, but one not so big as to cause the load (the 2V device) to not operate properly.

Here's my sketch: The two vertical rectangles represent the switch legs on the bottom. I have no idea how they are wired internally but I'm assuming it's one or the other type of switch and probably the LED with built in resistance. But again, this is purely a guess.


Also, I don't know if the LED is supposed to remain lit or remain off as an indicator of some sort, or if the LED is just supposed to light up when you press the button. This is why we ask for more clarity. Too much guesswork results in wasted time.

 

Here is something for you to try...
First, test to see if the breakout box will sense an input via a diode connected as below.



If it does, than you can wire up the inputs with an external 5V power supply, for example, and old phone charger, as shown here.

I should have redrawn it without the second diode but I'm lazy

The diode stops the 5V feeding into the breakout box. Make sure the diode stripe (cathode) is pointing away from the input terminal.

If this works ok, no special switch is needed. Just a switch contact.

 

Here's a link to the pushbutton page:

https://www.adafruit.com/product/1444

Initially, wiring up this button to the breakout box conventionally was only lighting up the button when it was in the unpressed (up) position. But the website I got the button at (adafruit.com) has a forum which I then used to try to switch the wiring to light up the button when pressed (down).

There, one of the support agents taught me how to wire the button like the diagram below so it could be achieved:


This worked but the button would barely light up due to the low 2V voltage provided by the breakout box. Thats when I came up with the idea of adding another bit of voltage to light up the button brightly using a CR2032 3V battery. It was enough to light the button brightly but led me to another problem, the way I wired it the 3V was constantly being fed to the breakout box and it would think the button was permanently pressed and the status on the app wouldn't switch from closed.

He then gave me a couple of ideas to work it out:

"1) Use a double-pole switch. I don't know of any illuminated pushbuttons with double-pole contacts. But you might be able to find one at Mouser or DigiKey.

2) Use your current switch plus a relay.

3) Use your current switch plus a transistor circuit.

4) Use your current switch plus a small microcontroller."

Tried the first suggestion but couldn't find a double pole button like I wanted. So went on to the next suggestion, tried finding a relay but got lost in the relay world with so many different kinds/specs. My knowledge of electronics is limited, thats when I posted my question here.

Meanwhile the person in the adafruit.com forum posted another solution. He recommended using another one of his products: https://www.adafruit.com/product/4409

Heres the diagram he posted:



I believe it might work but at first glance it would immediately ramp up the cost of my product at least by another $6.95, which would be cost prohibitive considering the current cost and the fact that its just a simple illuminated pushbutton. Not to mention the labor, additional parts, size, etc. Its just too much for too little.

Here's a link to my post on the adafruit.com forum:

https://forums.adafruit.com/viewtopic.php?p=1007137#p1007137

 

OK, I don't know if your lighted toggle switch is a switch or a push button. You DID say:

So I'm assuming there might be a built in resistor inside the switch. Can you tell us if the switch was designed to operate on 12V DC? If so then it probably has a resistor inside. If not then you need to add a resistor to the outside, but one not so big as to cause the load (the 2V device) to not operate properly.

Here's my sketch: The two vertical rectangles represent the switch legs on the bottom. I have no idea how they are wired internally but I'm assuming it's one or the other type of switch and probably the LED with built in resistance. But again, this is purely a guess.
View attachment 317417

Also, I don't know if the LED is supposed to remain lit or remain off as an indicator of some sort, or if the LED is just supposed to light up when you press the button. This is why we ask for more clarity. Too much guesswork results in wasted time.

• Its a pushbutton
• Yes I believe there's a resistor inside the button as the support person from adafruit.com mentions: "With just 2v to the LED, you can probably skip the current-limiting resistor." https://forums.adafruit.com/viewtopic.php?p=1007137
• More information on the pushbutton can be found here: https://www.adafruit.com/product/1444
• The LED in the latching pushbutton should remain on while the button is in the pressed position and turn off only when pushed again (released), to give a visual indication to the operator that it is on so her remembers to turn it off. As this button may be used to turn off vital equipment in the aquarium temporarily (for water changes for instance), forgetting it on might have catastrophic results.
• I wish nothing but to not waste precious time, in fact I tried to give just enough info I thought would be needed to help as I figured that giving too much information right away would be unnecessary and therefore exactly a waste of time. I just wasn't sure know how much you needed. But please let me know what you need and I'll try my best.

I appreciate your time, knowledge and willingness to help. Really.

Thanks.

 

Here is something for you to try...
First, test to see if the breakout box will sense an input via a diode connected as below.

View attachment 317418

If it does, than you can wire up the inputs with an external 5V power supply, for example, and old phone charger, as shown here.
View attachment 317419
I should have redrawn it without the second diode but I'm lazy

The diode stops the 5V feeding into the breakout box. Make sure the diode stripe (cathode) is pointing away from the input terminal.

If this works ok, no special switch is needed. Just a switch contact.

Nice idea, thank you very much. Your sketch is great, if you were lazy you wouldn't have done anything.

But I need the solution to be contained in the button case, I cant rig the breakout box with a power supply. Power outlets are scarce in this scenario and cable management is a nightmare inside a small cabinet with over 20 different pieces of equipment, huge cable lengths and water hazard.

 

It will be easy to fit it all in the switch box. You just need to find a source of +5V and that is most probably available from your control box. It does not need to be at the switch.

(Cover the wire joint to the diode with heat shrink tubing. )
Then run a 3 wire lead to the switch box instead of a 2 wire lead.

 

OK, I don't know if your lighted toggle switch is a switch or a push button. You DID say:

So I'm assuming there might be a built in resistor inside the switch. Can you tell us if the switch was designed to operate on 12V DC? If so then it probably has a resistor inside. If not then you need to add a resistor to the outside, but one not so big as to cause the load (the 2V device) to not operate properly.

Here's my sketch: The two vertical rectangles represent the switch legs on the bottom. I have no idea how they are wired internally but I'm assuming it's one or the other type of switch and probably the LED with built in resistance. But again, this is purely a guess.
View attachment 317417

Also, I don't know if the LED is supposed to remain lit or remain off as an indicator of some sort, or if the LED is just supposed to light up when you press the button. This is why we ask for more clarity. Too much guesswork results in wasted time.

I get nothing, please check if I screwed up setting it up. Thanks.

 

 

It will be easy to fit it all in the switch box. You just need to find a source of +5V and that is most probably available from your control box. It does not need to be at the switch.
View attachment 317429
(Cover the wire joint to the diode with heat shrink tubing. )
Then run a 3 wire lead to the switch box instead of a 2 wire lead.

This seemed to work though I barely get any voltage back to the breakout box, my multimeter read from 0-2v, but maybe its not very precise. I believe I need a steady 2v back to make sure the computer receives a reliable signal.



Apart from that I still need that extra voltage to come from inside the button case (batteries), I cant take voltage from the other inputs as they need to be free to hookup other sensors. Anyway, is it possible to do this with 3v instead of 5v? Or maybe I can use two CR2032 batteries and ramp the voltage up to 6V, I can manage to fit both inside the button housing. Like this:


What do you think?

 

View attachment 317434

Just tried it, but the button is always on, pressed down or not. I need it to light up and stay lit only when on the pressed position (down).

I still get only 1V on my multimeter, its probably not good to measure such small voltage, it doesnt even have a decimal point.








Anyway, I need a source of voltage that can be contained in the button housing (batteries) I think a CR2032 is a good option since it has a higher voltage and capacity. I could double it though to 6V.

 

Can you read the breakout box input status on your application software to see if it reads the switch correctly?
And if you have a spare USB socket, that could be used to supply the +5V. A couple of CR2032 batteries will not last long.
For a test, connect your multi meter, (on mA) between the GND and a breakout box input to measure the current an operated switch draws. If the current is low, a Germanium diode could be used as i has a lower forward voltage drop so the input will be closer to GND when on.

Another test could be to wire a variable resistor between the input and gnd, then measure the max voltage across it when the input is read as on. That wau it tells you how much headroom you have.