Changing my door handle puddle LED

lovemyleds

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20210515_195302.jpg
I have a 2012 Audi C7 with puddle lamps in my doorhandles. One of them was not working, so I decided to open up and investigate. I found that it was water damaged.
I want to upgrade them anyway, due to them not being very bright. After the resistors, there is 3.23V being supplied. How close does the forward voltage of the LED I am
buying have to be to that number? Also, any tips as to where I can find the brightest 2835 chip? Due to the space, it is difficult to fit anything larger. Width-wise, there
are too many things in the way. Height-wise, there is a projector in the way. I guess I could give it up if the brightness of the chip made up for the lack of it. Though It
seems width is more necessary when going up in lumens, than height...
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lightfooted

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Pretty close to higher voltage. If you use one that has a lower forward voltage you will likely have it burning up or just going out soon after installing it. Significantly lower and it will flash out as soon as it is powered up.
 

lovemyleds

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Pretty close to higher voltage. If you use one that has a lower forward voltage you will likely have it burning up or just going out soon after installing it. Significantly lower and it will flash out as soon as it is powered up.
Thanks, I was looking up LEDs for ages, the datasheet had some pretty conclusive information.
My next thought was, the 2835 LEDs are a bit limiting, I found a Cree XML 2 that will just fit in the space, though there is no room for adding a heatsink. The lights only run for 30 seconds or so, a couple of times a day. Would this be ok? Is there any point of using heatsink paste under the chip. I realise a PCB is not the best heatsink, but better than nothing?
 

lightfooted

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Thanks, I was looking up LEDs for ages, the datasheet had some pretty conclusive information.
My next thought was, the 2835 LEDs are a bit limiting, I found a Cree XML 2 that will just fit in the space, though there is no room for adding a heatsink. The lights only run for 30 seconds or so, a couple of times a day. Would this be ok? Is there any point of using heatsink paste under the chip. I realise a PCB is not the best heatsink, but better than nothing?

Unless it has some metal plating, PCB material is not a heat sink at all. It's actually designed to be an insulator, not just electrically but also thermally.
 

lovemyleds

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Unless it has some metal plating, PCB material is not a heat sink at all. It's actually designed to be an insulator, not just electrically but also thermally.

Oh wow, interesting! Any suggestions? Or is just a matter of see how I go? I've already bought them, I don't mind the time spent installing them, just wondering how to best mount them. They have two surface mounting pads, I was going to install two wires, insulate them with some hot glue and then put the led on top. Being a door handle led, it's never on for extended periods.
 

Dave_H

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Unless it has some metal plating, PCB material is not a heat sink at all. It's actually designed to be an insulator, not just electrically but also thermally.


Epoxy fiberglass thermal conductivity is about 1400 times lower than copper so it doesn't help much. It's a good electrical insulator but not specifically designed to be a poor thermal conductor, just happens to be. PCB can be and is used for heatsinking mainly due to copper.

Small white LEDs usually have forward voltage in the range 2.6v to 3.2v, divided into "bins".

If the LED has a series resistor, lower forward voltage is not likely to cause it to burn up. Current goes up somewhat, that's about it. Example, if supply is 5v, 100 ohm resistor, a 3.2v LED draws 18mA, 2.6v LED draws 24mA, these are at the extremes. That's not enough to damage a typical LED. In fact the difference will be less, as higher current drives up vf which reduces current in a resistor circuit, depends on the specific LED.

Dave
 

Dave_H

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Oh wow, interesting! Any suggestions? Or is just a matter of see how I go? I've already bought them, I don't mind the time spent installing them, just wondering how to best mount them. They have two surface mounting pads, I was going to install two wires, insulate them with some hot glue and then put the led on top. Being a door handle led, it's never on for extended periods.

If LED current is low say 60mA or less you can probably get away with wiring them how you describe.


Dave
 

lovemyleds

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Epoxy fiberglass thermal conductivity is about 1400 times lower than copper so it doesn't help much. It's a good electrical insulator but not specifically designed to be a poor thermal conductor, just happens to be. PCB can be and is used for heatsinking mainly due to copper.

Small white LEDs usually have forward voltage in the range 2.6v to 3.2v, divided into "bins".

If the LED has a series resistor, lower forward voltage is not likely to cause it to burn up. Current goes up somewhat, that's about it. Example, if supply is 5v, 100 ohm resistor, a 3.2v LED draws 18mA, 2.6v LED draws 24mA, these are at the extremes. That's not enough to damage a typical LED. In fact the difference will be less, as higher current drives up vf which reduces current in a resistor circuit, depends on the specific LED.

Dave

Thanks for the replies Dave. It is a tiny PCB, if it is even rated as one? It has a 3300 resistor on one side... And the other, I believe it was around 30, sitting under the resistor. I can try and fit a bit of aluminium underneath it. Something better than nothing? Sounds like I should be ok anyway...
 

Dave_H

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Thanks for the replies Dave. It is a tiny PCB, if it is even rated as one? It has a 3300 resistor on one side... And the other, I believe it was around 30, sitting under the resistor. I can try and fit a bit of aluminium underneath it. Something better than nothing? Sounds like I should be ok anyway...

Not sure what 3300 ohms or 30 ohms means, does not quite add up. Can you measure LED current with DMM? Perhaps post an image?

Dave
 

maxx44

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Few months ago, I've had all the same problems with my door handle, but which was installed in the front door of the house. It's happened when I was in our second home town, in Canada. I've simply looked for couple of emergency services online, and could discover this locksmith service ( link: https://emergencylocksmithvancouver.ca/ ). Locksmith specialist has arrived pretty fast and fixed everything, so my door lock started working even smoother, than it was before.
 
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DIWdiver

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If a resistor is marked with the digits '3300' it almost certainly means that the resistance is 330 x 10^0. Notice that the first three digits represent three digits of value, and the last digit represents the number of zeroes that follow (the exponent on 10, or the last digit in 'x 10^0'). In this case '330' followed by '0' zeroes is 330 ohms.

330 ohms would be a pretty reasonable resistor to use with a 12V supply and 3V LED, giving about 27 mA.

This means that the Vf of your LED is almost insignificant. Any Vf close to 3V will give you a current close to 27 mA. I can give you the math to support this if you want, but it's a lot more detail, so I won't unless requested. Don't worry if your LED is 2.7V or 3.3V, it won't matter much. The current will be around 27 mA no matter. This is right in the range of many small LEDs that operate without any heatsinking.
 

Dave_H

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If a resistor is marked with the digits '3300' it almost certainly means that the resistance is 330 x 10^0. Notice that the first three digits represent three digits of value, and the last digit represents the number of zeroes that follow (the exponent on 10, or the last digit in 'x 10^0'). In this case '330' followed by '0' zeroes is 330 ohms.

330 ohms would be a pretty reasonable resistor to use with a 12V supply and 3V LED, giving about 27 mA.

Possible, but 330 ohms is more likely marked "331". Higher precision taken to 3rd digit (which this application is not likely to require) would be 332 ohms so marked "3320".

What I am wondering is if there are any other components on the PCB such as small transistors, which might be a current regulator circuit. Common form uses the transistor base-emitter voltage (which varies over temperature, so does the current) around 0.5v to 0.7v.

Using 0.6v and 30 ohms, LED current would be 20mA which is reasonable for a small LED. 3300 ohms (if it is that) sounds like a bias resistor in the circuit.

OP's input needed.

Dave
 

DIWdiver

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Possible, but 330 ohms is more likely marked "331". Higher precision taken to 3rd digit (which this application is not likely to require) would be 332 ohms so marked "3320".

OP specifically wrote "3300 resistor" which meant either it was measured at 3300 ohms, or it is marked 3300, which means it would be 330 ohms.

Since 1% resistors are the same price as 5% resistors, at my company we no longer stock or design with 5%. Also, the 5% (E24) values are all available as 1% resistors.

I can't imagine why there would be 2 resistors unless there's other stuff too. Agreed that the OP's input is needed.
 
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