Hello,
I was wondering if I can power a non-used semidigital casio watch (model AQ 230) which runs on SR920W batteries using my arduino. I would like to integrate the watch in a project, maybe not in the sense that i control the digital buttons etc. but just to power it up and have it as a design feature. I think that it should be possible but I am still a beginner in electronics tinkering.
If it is possible, how would I go about doing it? Attached you can find an image of the back of the watch.
Cheers!
That battery delivers 1.55V normally; I see no reason you can't power the thing from an external power source. E.g. by using a regulator such as the RT9013-15 which delivers 1.5V.
wvmarle:
That battery delivers 1.55V normally; I see no reason you can't power the thing from an external power source. E.g. by using a regulator such as the RT9013-15 which delivers 1.5V.
Thanks for the answer, I will order a couple of regulators and start playing around with this. The only tricky part for me would be the soldering unto the plate of the watch where the battery would have been connected. Do you have any tips regarding this?
Cheers!
You could make a simple shunt regulator from 2 series diodes like 1N914, and a limiting resistor. Like a Zener circuit. That would produce about 1.5V, you can tweak the voltage a bit with the bias resistor. The watch takes so little current that the resistor value can be quite high - maybe even 10k ohms.
I wonder if you can solder onto those connectors in the first place. Good chance it's stainless steel.
Note that the regulator I mentioned is tiny, grain of rice kind of tiny, and of course it needs input and output capacitors for stability. You may be able to find it on a breakout board. Otherwise look for a regulator with adjustable output on breakout board, makes prototyping a lot easier. Or just use an alkaline 1.5V battery until you build your own PCB with the regulator in place.
aarg:
You could make a simple shunt regulator from 2 series diodes like 1N914, and a limiting resistor.
True - that's even simpler. Should work fine, those watches take a few µA or so. Maybe add a small cap for stability? 100nF or so is probably enough.
wvmarle:
True - that's even simpler. Should work fine, those watches take a few µA or so. Maybe add a small cap for stability? 100nF or so is probably enough.
Thanks aarg and wvmarle for the suggestion. I will try to build the circuit from your schematic wvmarle! And in this case the VCC could be the arduino 5V power provider or should I use the 3.3V? The diodes in your schematic, could they be replaced with the 1N4007 diode and maybe use a different resistor? Here is a link for the specs of the 1N4007:
Could I then simply connect a wire from this circuit to the battery socket of the watch?
Cheers!
pingi_j:
Thanks aarg and wvmarle for the suggestion. I will try to build the circuit from your schematic wvmarle! And in this case the VCC could be the arduino 5V power provider or should I use the 3.3V?
Either will work just fine.
The idea is: a small current flows through R1, D1, D2. The voltage drop over D1, D2 is fixed and more or less independent of the current - about 0.6V for each diode at these very low currents. Just about any regular diode will do; 1N4007 is fine as well, a little higher voltage drop iirc. See data sheet for details. If you use Schottky diodes, you need 3-4 of them due to the lower drop.
This current is about 0.2-0.3 mA, way more than what the watch will take. As long as the current the watch takes is less than this 0.2 mA you're fine. The excess current is drained by the diodes, making this circuit actually extremely inefficient.
Could I then simply connect a wire from this circuit to the battery socket of the watch?
Yes.
Of course: do measure the voltage with your multimeter before doing so, to make sure everything is working properly.
wvmarle:
Either will work just fine.
The idea is: a small current flows through R1, D1, D2. The voltage drop over D1, D2 is fixed and more or less independent of the current - about 0.6V for each diode at these very low currents. Just about any regular diode will do; 1N4007 is fine as well, a little higher voltage drop iirc. See data sheet for details. If you use Schottky diodes, you need 3-4 of them due to the lower drop.
This current is about 0.2-0.3 mA, way more than what the watch will take. As long as the current the watch takes is less than this 0.2 mA you're fine. The excess current is drained by the diodes, making this circuit actually extremely inefficient.
Yes.
Of course: do measure the voltage with your multimeter before doing so, to make sure everything is working properly.
wvmarle:
Either will work just fine.
The idea is: a small current flows through R1, D1, D2. The voltage drop over D1, D2 is fixed and more or less independent of the current - about 0.6V for each diode at these very low currents. Just about any regular diode will do; 1N4007 is fine as well, a little higher voltage drop iirc. See data sheet for details. If you use Schottky diodes, you need 3-4 of them due to the lower drop.
This current is about 0.2-0.3 mA, way more than what the watch will take. As long as the current the watch takes is less than this 0.2 mA you're fine. The excess current is drained by the diodes, making this circuit actually extremely inefficient.
Yes.
Of course: do measure the voltage with your multimeter before doing so, to make sure everything is working properly.
I would like to start with thanking everyone involved for putting up with my many questions! Attached to this post you can find the circuit I have built from this schematic. The yellow cable on the right would be my link from POWER -> WATCH. Do I just connect this to the battery slot of the watch and how can I plug the watch to a common GND with the rest of the circuit?
Cheers!
Common ground is the negative pole of the battery.
Have you considered making a battery replacement connector from a thin piece of wood with thin copper/brass foil on either side. Solder a #30 wire to each foil and connect to your 1.5 volt source.
Paul
Paul_KD7HB:
Have you considered making a battery replacement connector from a thin piece of wood with thin copper/brass foil on either side. Solder a #30 wire to each foil and connect to your 1.5 volt source.
Paul
Paul, I have not considered this but it sounds like a really good idea! Thanks, I like the idea of having then made a power source for that type of battery.
Cheers!
