this post was submitted on 20 Jul 2024
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I've got the appropriate amount of light for my microscope ring light, but now I need to put it all in an enclosure of some sort.

If I don't have a custom board to solder these to, what are my best options for connecting these she mounting them into something?

If this is too vague, please let me know if I can clarify

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[–] rushaction@programming.dev 6 points 3 months ago (1 children)

I'm no expert, but I think you might need resistors on those LEDs or they'll burn out fairly quickly. I cannot see the whole circuit, and my EE classes were a long-ass time ago. It looks like you might be driving them straight from the breadboard's power. Just wanted to give you a heads up in case it's an issue.

https://electronics.stackexchange.com/questions/28393/why-do-we-need-resistors-in-led

Good luck!

[–] sneekee_snek_17@lemmy.world 4 points 3 months ago (1 children)

I had that same thought, but I'm running 4, 3.3V diodes per parallel branch/arm/whatever, with a 12V power supply, so the calculations I've seen for determining the appropriate resistor tell me this doesn't need any.

It felt wrong, but I wanted to see if it would light up, and it did! I tried to wrap this concern into the "is it going to start on fire?" Bit lol

[–] marcos@lemmy.world 4 points 3 months ago (2 children)

Not only you always need at least a resistor, but matching the power supply voltage to the leds drop will ensure you'll have a hard time maintaining their brightness on the real world.

If you want to keep it simple, leave 1V or 2V as a margin and add a resistor to get the desired current. If you want to make it fancy, get a led driver.

[–] sneekee_snek_17@lemmy.world 1 points 3 months ago

That being said, I redid it with your input in mind. It's currently got 100U resistors and draws a good bit more than the power supply is capable of

[–] sneekee_snek_17@lemmy.world 1 points 3 months ago (2 children)

When you say "the real world" what do you mean, exactly? Because the 12V, 200mA power supply I used to power it for these pictures is likely what will ultimately power it

[–] coffeejoe@lemmy.dbzer0.com 2 points 3 months ago

You have a current limiting power supply. So the power supply transistor is acting as the resistor for you in this scenario.

[–] marcos@lemmy.world 1 points 3 months ago* (last edited 3 months ago) (1 children)

For example, as the leds on the original circuit age, you can expect a few of those lines to stop getting any current, and the others to get brighter and brighter until they fail.

That can happen very quickly too, if their temperature varies a lot (how much is a lot depends on the leds). Or if you change your power supply for one capable of more current, it's likely that everything will just fail immediately. Or, if you have a noisy one, you may severely reduce the leds lifetime.

As a rule, you can't use diodes internal resistance as a power limiter. It does not behave the same way as resistors and will surprise you on every bad way possible.

[–] sneekee_snek_17@lemmy.world 1 points 3 months ago

Huh, if I had the time I'd do some reading about it. Sounds like there's some really interesting physics going on there