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Compact ight sensor switch with low voltage Answered

Hi there,

I've got a led light that's operated via a magnet, if it doesn't detect a magnetic field it turns on if it does it turns off. It's a led light for car door(check picture for reference) and uses 3x AAA batteries in my case 1.2v x3 (3.6v), if you use regular batteries it's 4.5v

By mistake I've order a tiny light sensor module 3.3-5v(check picture) which obviously can't be used as a switch, not in it's present form. While I'm good with solder, I've build an ebike, I'm a computer guy but when it comes to things like this like this make me feel powerless :( ..I 've used 12v light sensor switches but that's a different topic.

So, how can I have a low powered light sensor switch that's compact and can save me from replacing the batteries frequent? ...the led light turns on even at daytime which has no point in doing that. I'd appriciate any help :D

Discussions

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Gentleheart

1 year ago

Just adding a photoresistor (LDR) between the batteries and the led would probably do the trick.

A nicer solution is an LDR with a NE555 chip. That way you could also add a timer (auto shut-off after a certain time) I could provide you with a simple schematic for this.

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Jack A Lopez

1 year ago

If I follow what you are saying, you want a light that turns on only when:

(1) the door is open AND (2) it is dark outside


Moreover the reason why you want this is simply to conserve battery energy, and thus, the money+time it costs to replace the batteries.

I have to wonder:

Do you leave your car door open for hours at a time? During daylight hours?

I am also wondering why you want to litter the pavement, and other people's minds, with the worship of these awful advertising logos?

I dunno.

Have you had any success adjusting your light sensor module, so it can tell the difference between day and night?

The way you tell is by way of that little LED labeled "DO", which stands for "digital output". You know, if that LED is on, it is indicating it senses the presence of something, presumably light level. Conversely, if that LED is off, it is indicating the absence of that same thing.

It would not hurt to actually put your multimeter probes on the pin, "DO", relative to ground pin, to see what it is doing. The voltage on that pin probably goes low, or high, at the same time the LED is on or off. It would be helpful to know which.

Also, I have to wonder how much power this module itself uses. It would be silly, and counterproductive, if it used as much power, sensing and keeping the light turned off, as the light alone used when turned on.

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Takisk1Jack A Lopez

Reply 1 year ago

"Moreover the reason why you want this is simply to conserve battery
energy, and thus, the money+time it costs to replace the batteries."

yes

"Do you leave your car door open for hours at a time? During daylight hours?"

No but I do open the door frequently so the batteries discharge for no reason

"I am also wondering why you want to litter the pavement, and other
people's minds, with the worship of these awful advertising logos?"

Litter? well more visibility, so better safety especially in a dark alley plus the fact that it's cool to have a seat logo(my car is a seat) - No different than a luxury car which doesn't project a logo but lights up the pavement

"Have you had any success adjusting your light sensor module, so it can tell the difference between day and night?"

Light sensor is not a light switch and I have no clue in how to turn it into a switch, if I know how-to I wouldn't ask :P

"It would not hurt to actually put your multimeter probes on the pin,
"DO", relative to ground pin, to see what it is doing. The voltage on
that pin probably goes low, or high, at the same time the LED is on or
off. It would be helpful to know which."

Did that already, the AO for 4.78v is 3.78v and for the DO it's the same 4.78v input 4.78v output - I need a switch not a light detection heh - Whatever the draw is, surely it's lower than when the led is active(the device has a powerful led)

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Jack A LopezTakisk1

Reply 1 year ago

Regarding DO, is it high (i.e. near Vcc, which is 4.5 or 4.78, or similar) when it senses darkness? Or is it high when it senses lightness, erm brightness, bright surrounding light? I am still not picturing that part. I mean, what surrounding light conditions cause DO go high, or low?

I am guessing what you want to do is use the existing magnetic switch (guessing it is just a reed switch) to turn on the light sensor, and then have the signal from the light sensor turn on a transistor, or something, for to turn on the spotlight-on-the-ground-LED.

By the way, would it be cheating to power this thing from the car's electrical supply? Or does it have to be powered by the 3x AA cells?

Do you happen to know how much current (like in A, or mA) the spotlight-on-the-ground-LED wants? The reason I ask, is because that will influence choice of transistor to switch that current.

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Jack A LopezTakisk1

Reply 1 year ago

That circuit might work. Although, it might be too simple.

It might work better by using two transistors instead of just one; i.e. using one transistor to turn on the other, or by using op-amps, or comparators, instead of transistors.

By the way, the word, "integrate", means to add or include. You know, like integrals in Calculus, where you add a bunch of little pieces together to get one big sum.

I'm pretty sure, "integrade", is not a word.

Also BTW, to measure current, it is necessary to put the multimeter in current measuring mode, and put it in series, in the place where you want to measure the current.

This page,

https://www.sciencebuddies.org/science-fair-projects/references/how-to-use-a-multimeter#qmultimetermeasurecurrent

shows how it is done.

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Takisk1

1 year ago

Hi there, sorry for the late reply

This item as in picture, it needs 3xAAA power, so it's not connected to the car's power

I have no idea about the current but I can check it out, if I put a multimeter and try to power that led(I still to remove it from the car thought... it's "glued" with a double-sided tape, so I won't have to be near the door trying to hold the batteries/multimeter and check the display :P) - As far as I know in the past these were "rated" at 1W ..not sure if it's true, Chinese sellers claim anything

With AO(analog output) when no light 4.25v(a 18650 source - 4.25v currently), when strong light hits the module it's about 3.4-3.7v on DO it's 4.25v no change ..weird but that's the results

The magnetic switch works as it should but the led-door shouldn't power when it's daylight(no point), I want to conserve energy during the day that's all :D

P.S. Thank you for your replies