Replacing a device's button with an Arduino

Question

Hi,

Let me start by saying that I'm fairly new to electronics, and I may get some terminology wrong in this post. I'll do my best to ensure it makes sense, though, and please feel free to correct me where necessary as I am very keen to learn.

As the title (vaguely) suggests, I am working on a project in which I need to - for want of a better description - replace a device's 'push to make' switch with an Arduino Uno.

The project is actually the glove of a costume I am building. It is going to release 'smoke' when the wearer's hand is in specified positions. The setup isn't hugely complicated, and I believe I have everything straight in my head except for one thing. I'll start by explaining the setup...

In the glove I have an accelerometer, and with my Arduino sketch I can define gestures and recognise those gestures.

I've built a basic shield for the Arduino with a transistor so that I can control a blower fan. The Arduino and fan are powered by a 9v battery which connects to this shield. I've essentially enclosed the blower fan so that its inlet is connected to a 'vape' (you know, one of those e-cigarette type things).

So far I have got everything working, but the vape is not connected to the Arduino in any way. In fact, it hasn't even arrived yet, so while awaiting its delivery I've been considering how I might connect it all up.

The intention is to have the fan and some LEDs turn on and the vape button pressed when a specific gesture is made. The vape will heat up, the fan will suck vapor through the tubes, and the vapor will emerge from the glove. When the gesture is no longer being made, the fan turns off and the vape button is released.

I've gone through a couple of options in my head, including things like having the Arduino control a servo which mechanically presses the button on the vape, but what I'd really like is it pretty much desolder the vape's switch, remove it from the vape, and solder wire at its connections which then connect to the Arduino... or thereabouts, if you see what I mean.

So I'm here to get some advice on what would be the best way to approach this. Is it possible? Trivial? Will I need to use a transistor, a DAC, a relay, or some other method?

Any help would be greatly appreciated and I am more than happy to elaborate further if needs be.

jlangbridge · Accepted Answer

Aha... So there is a bit of logic there, this isn't a simple on/off switch (I'm not a vaper either, so I had no idea). Well, maybe the "easiest" solution would be to use a relay. There are several types, but here, a reed relay is the simplest; you don't have a lot of current or voltage to switch on and off, so let's keep things simple. To test, there are multiple shields and breakout boards withrelays. And yes, your code will look something very much like that (although these are going to be on/off pins, so stick with digitalWrite, we don't need analog outputs). Looking forward to the photos!

jlangbridge · Answer

Hi there! Welcome to the wonderful world of Arduino. With the ideas that seem to be in your head, this is probably just the beginning of your adventure It is extremely difficult to answer your question without having seen the hardware. If the "switch" is just a signal to tell the vape to work, you might not actually need any electronics at all. If, on the other hand, it is a power button, and pressing the button sends the energy to the vape, then things become slightly more complicated, but only slightly, you will probably be using a relay, or something close. Relays on the Arduino range from easy to very, very easy. Some might require a little bit of power, by adding a transistor, but you might find a model that will work with the Arduino's output current. Sorry if I can't answer that question right now, but the only answer I can give you is that I don't see any difficulty in completing your project, depending on what the button does. I don't see this as a problem, and you should be up and running with minimal fuss Just let us know when you get the hardware!

jlangbridge · Answer

Not quite! The diode will be between the relay's coil pins. Have a look at the Wikipedia page, the first schematic.

https://en.wikipedia.org/wiki/Flyback_diode

When the coil is powered, the diode is useless, but when the coil is powered down, any inverse energy caused by the magnetic field breakdown gets shorted out.

talongrayson · Answer

Hi again! The vape arrived today and oh boy are these things more complex and sophisticated than I'd assumed. The unit itself has 3 buttons, as stated before. 2 of them are used to set the amount of power used to heat the element, the other is the 'fire' button, which is the one I wish to control. I've done some tests to make sure the rig works as a whole (i.e. vapour is drawn from the vape, through the tubing, etc) so I know it all works, and I think I am now ready to hook the thing up to the Arduino via a reed relay. Well, I'm ready to (with some help) figure out how to hook it up to the Arduino, to be more precise! I've attached a photo of the switch itself (nothing terribly exciting - a 4-pin push to make) but I'm not really well versed in electronics to know what we need to figure out. I'm guessing current across the switch while the unit is on, and current across the switch while the element is being heated? In terms of wiring it to the Arduino, given how small the solder points are, I'm thinking - to avoid getting up close and personal with the soldering iron - I could tin some wire and heat the wire from further up in the hope that the tinned wire and solder on the board fuse together. And I'm thinking I could leave the switch in since running wires to the Arduino will allow the circuit to ignore the switch anyway. Does that sound ridiculous?

jc2048 · Answer

I've just looked at the protection on an Arduino pin and it turns out I am wrong. Here's pin '8' set to be an input and driven with a sine wave that goes outside the supply range (with a series resistor to limit the current). Ground is where the '2' marker is, so we are looking at negative voltages. The yellow trace is the sine wave, the blue trace is the voltage that the protection network in the chip is clamping it to. It's a much larger voltage than I would have thought (about -780mV at the centre, at a current of around 8.5mA), so I'm quite wrong about how they are doing it. Here it is with an external 1N4004 added, and the external diode is having an effect - its forward voltage is coming in at less than the protection network. Ok, that's for an input. But what if you don't turn it back to an input when you don't want the coil to be powered but just set the output to be low? In that case we get this because the MOSFET channel is bidirectional and will clamp the input to ground even when it goes negative. I've now raised the current at the centre to about 10mA and you can see that the voltage is now about -180mV. That gives a dynamic resistance for the MOSFET of about 18 ohms, which is about right for a processor IO pin.

talongrayson · Answer

Hey, thanks for the welcome and the fast reply! From what I've seen of the hardware so far, it functions roughly as follows: The vape is powered by a pair of 3.7V, 35A batteries. The vape is essentially a single button device, so you use that one button to turn the vape on or off with 5 presses in a row, and once on, you hold the same button down to heat the element which vaporises the glycerine. Typical usage would be to turn the vape on with 5 presses, then whenever the user wants to take a puff, they hold the button down and suck the vapor out (again, maybe not the correct terminology, I'm not a smoker/vaper), then release the button between puffs. I do know that the tech inside the vape is fairly sophisticated. It has timer cutoffs, temperature cutoffs, and so on. But in essence I am just looking to replace the physical switch with a virtual one - and I guess running all the vape's current through the Arduino as a means to that end. From a software point of view, when the Arduino recognises the prescribed gesture, the code will 'press' the switch 5 times (with maybe 500ms between presses, or whatever), then 'hold the button down', and when the gesture is no longer recognised the button would be 'released', and then 'pressed' 5 times again. Each press or hold would, I presume, be an analogWrite(); so the code for an on/off toggle would be: // Toggle on/off: for(int i=0; i<5; i++) { analogWrite(VAPE_PIN, HIGH); delay(20); analogWrite(VAPE_PIN, LOW); delay(500); } Anyway, I'm kinda brain dumping here... I'll come back with some photos of the hardware once I receive the thing in the post, or if I find any tech specs, I'll post those. Cheers

talongrayson · Answer

Progress so far:
I have soldered wires to 2 pins of the vape device (essentially the blue and yellow wires in the below diagram). I have run those wires to a breadboard with a switch, and that switch can now act as the FIRE switch on the device itself.

I have ordered a reed relay and intend to wire it as below. I will probably stick a 2-pin JST between the relay and the vape for convenience.

If this goes to plan, all that remains is the actual costume itself. And of course, if it works, jlangbridge answer will be marked as correct.

jlangbridge · Answer

Hi there! Sorry, the weekend was a little more hectic than expected. Looks good, but I am a little worried about one thing. I'm not too sure the Arduino can supply enough current to be able to activate the reed switch. I don't have any, so I can't check. If it doesn't work, there is something else you can try; instead of connecting one of the pins to GND, connect it to the 5V supply. You will need to change the code slightly; if the "output" is high, then no current will flow. If the "output" is low, then current will flow from the 5V rail to the pin (this is known as active low). The problem will all of this is that the MCU can only provide (and sink) a certain amount of current before damage can occur. Another thing to take into account is the use of a snubber diode; when electromagnetic fields collapse, they can sometimes generate a spike of current. Some reed relays actually have one built in.

That being said, everything looks good! You have effectively put a switch in parallel, so you should be good to go as soon as you confirm the timing is good, but that is only a software tweak.

jc2048 · Answer

"When the coil is powered, the diode is useless, but when the coil is powered down, any inverse energy caused by the magnetic field breakdown gets shorted out." I suspect that you'll find, in this case, the diode is always useless. The internal anti-static protection diode for the IO pin (it's in parallel with your external one) will have a lower Vf and will dissipate the coil's energy.

jlangbridge · Answer

The other way around, the "illogical" way. The way you put it, if you power up pin 8, it will go through the diode, and to ground. If you put the diode the other way around, then the diode will block the flow, and basically do nothing. The problem happens when you shut down the electromagnetic field. EM fields don't like change, and will fight to keep things the way they are. If you power down the relay coil, it sometimes creates an inverse, essentially putting a + on the ground pin, and a - on pin 8, where there used to be a +. As you can imagine, even if it is extremely brief, it can cause damage to the board. To counter this, we use a diode. If the current is in the "normal" position, ie you are powering the coil, then it doesn't do anything. If you shut down the coil and the coil fights back, then it will short out the difference, protecting your Arduino. I'd add a few sketches, but I can't where I am right now...

jc2048 · Answer

No. I said I suspect. I can't know for sure how Atmel have implemented the anti-static protection, but the most common way is to utilise metal-barrier diodes that arise naturally as part of the process and don't cost the manufacturer anything (they're Schottky diodes and the forward voltage would be something like 0.2V or 0.3V at the relay current). It's not good design practice to use the internal diodes like this, which I think is why the manufacturers are a bit coy about giving specs on them. In practice, for a personal project, the microcontroller would manage. With a 500ohm coil, the max current is only 10mA and with such a low repetition rate the dissipation is next to nothing. The biggest disadvantage is a side effect: the slow release of the relay. The 0.3V means that the initial dissipation is low (10mA x 0.3 = 3mW) and the current through the coil will fall slowly. So the release of the contact, when it comes, will be slow and if it's switching a high current there may be problems with the contacts after you've done it a few times.

jc2048 · Answer

The bottom output MOSFET is doing the clamping rather than the diode. [So I was half right.] It might be an idea to put the diode in. If you publish the circuit it encourages people to put the diode in (for when it actually is needed) and it'll save you from having to try and explain to all the people who will inevitably come along and tell you that there should be a diode there. If this were my project, and it were going on a shield, I'd use an external MOSFET or BJT to switch the relay (where you would most definitely would need to protect the transistor). That's because the dissipation of the coil's energy is faster and more in line with what the reed relay manufacturer was expecting [since they include a protection diode in some of their variants we can assume that the contact switching is fine when the the coil is working into the 0.7V that the diode limits to].

Replacing a device's button with an Arduino

Top Replies