quick update on my DIY LED lamps

[Amanda P]

I've found that waterproof isn't too important - rain isn't very conductive. I tend to use insulated automotive crimp terminals, and run one side of the circuit through the frame. Never had any electrical problems with this.

Switches are vulnerable to corrosion, though, so it's worth paying a bit extra for switches with a high IP protection level.

 

[gbb]

]I've found that waterproof isn't too important - rain isn't very conductive[/COLOR]. I tend to use insulated automotive crimp terminals, and run one side of the circuit through the frame. Never had any electrical problems with this.

Switches are vulnerable to corrosion, though, so it's worth paying a bit extra for switches with a high IP protection level.

While trying to learn about all this UPhil, i did read in the manufacturers specs that the LED must be protected from moisture prior to use. I assume it affects the LEDs life in some way. They even recommend baking an LED if it has been exposed to moisture prior to use.
Once fitted and soldered, it may be a different matter, but....

 

[gbb]

I've now realised that powering a single LED isnt as easy as it seems, maybe thats why a double lamp setup seems more popular.

Ideally 3.7 to 5 volts for a single LED. So you're looking at 3x 1.2 volt rechargeable batteries. I'd rather not push it to the 5v max with 4x 1.2v batts..i've already blown one LED by connecting a 6v battery in my haste...and promptly blew the LED in an instant

So, 3x 1.2 v batts, but ideally you want cells at least, to get any decent mAh, then a 3 cell holders not that easy to get. Minor problems, but....

Or am i being over cautious...is it ok to use 4 x 1.2 volt. I know its still underrated, but close to the max permissible ? Any idea anyone ?

A double LED setup obviously requires 7.4 to 10V. Much easier to get a battery with a decent rating (the kind my RC car uses is perfect, self contained, easy to charge etc etc etc)

The lamp i'm building looks quite good at the mo.....i'll post some pics when its done. Shame i didnt think to take photos of it in its stages of assembly.

 

[Amanda P]

While trying to learn about all this UPhil, i did read in the manufacturers specs that the LED must be protected from moisture prior to use. I assume it affects the LEDs life in some way. They even recommend baking an LED if it has been exposed to moisture prior to use.
Once fitted and soldered, it may be a different matter, but....

Well, I've just learned something, then.

I thought you were commenting on in-line connectors for wires. These won't normally be sitting in a puddle.

But it sounds like you were thinking of a connector that results in the light housing itself being sealed, to keep the LED dry. That's probably easier to do, and would be wise anyway, because water could accumulate inside a light casing, and would damage the tiny, short-sensitive components of a driver/buckpuck/voltage regulator/constant current gadget or whatever you're using.

A grommet is probably adequate, otherwise perhaps a blob of silicon sealant around the wire entry hole.

 

[02GF74]

I've now realised that powering a single LED isnt as easy as it seems, maybe thats why a double lamp setup seems more popular.

Ideally 3.7 to 5 volts for a single LED. So you're looking at 3x 1.2 volt rechargeable batteries. I'd rather not push it to the 5v max with 4x 1.2v batts..i've already blown one LED by connecting a 6v battery in my haste...and promptly blew the LED in an instant

No mate, that is all wrong. ^^^^ You need to see the voltage/current graph for a LED to get a grip of what is required.

The important factor to destroying (or not) a LED is the current through it and from the graph you would see the voltage that should be applied across it.

The difficulty is that it is unlikely you will have a battery or voltage source that can supply exactly that voltage. Also due to manufacturing tolerances, this can vary for different LEDs.

But there are two solutions;

1. connect LED to a higher voltage battery in series with a drop resistor - this limits the current to the LED. This is not efficient. Also the current through the LED changes as battery voltage changes, i.e. LED dims as the juice in battery starts to run out.

2. use an electronic driver, like the ones I have listed. These supply the correct current without the need of a drop resistor plus will tolerate a wider range of battery voltage.

So basically you cannot connect a battery directly to a LED and expect it to work well, if at all, as you have found out.

 

[gbb]

Thanks 02GF...this maybe touches on a point i'd been struggling with.
So...can you supply one LED (3.7 to 5 volt Vf) with a battery of say 6 volts, so long as it's through (in my case) a current control driver.

I think (maybe) i've been reading Vf and thinking it was the maximum voltage you can apply.
Is Vf the actual voltage it will allow to pass forward....not the voltage range it can work within

Sorry if my explanations a bit awry...

 

[02GF74]

here is the graph

I am not sure for what colour LED it is for as it will be different for blue/red/green/white ....

Basically the manufacturer states the maximum current that you should put through the LED and also state the forward voltage in that case. Knowing these 2 values, you can work out what resistor you need if connecting to a voltage higher than that.

For a driver, all you need to know is the maximum current for it and ensure that the LED can handle it.

In my case, the 3 W and 5 W drivers will supply approx 0.75 A and 0.9 mA; my LEDs can handle 1 A so no problems there. The driver is specified for a voltage range it will work, lets say this is 6 to 12 V; if I connect to a 7.2 V NiMH battery, then all will work with no other intervention.

Vf by convention is the voltage dropped across the diode, as in the horizontal axis in the chart.

 

[02GF74]

Thanks 02GF...this maybe touches on a point i'd been struggling with.
So...can you supply one LED (3.7 to 5 volt Vf) with a battery of say 6 volts, so long as it's through (in my case) a current control driver.

let's say the Vf of your LEd is 3.5 V and you want to connect to a 6 V battery. You need to "lose" 2.5 V and this is done using a resistor, RB1.

The value of the resitor is 2.5 / 600 mA (from the graph) = 4.2 ohm

As mentioned, it is not efficient way if you are using high currents.