parallel led array

[lightman]

you mean one of these would this work with the led ? and the 3.7 battery ? http://www.ebay.co.uk/itm/2-7A-Step...hash=item4ae78ef9a6:m:m7Xj-YlSJHJE9CVsnRGxMCg

 

[Gryd3]

you mean one of these would this work with the led ? and the 3.7 battery ? http://www.ebay.co.uk/itm/2-7A-Step...hash=item4ae78ef9a6:m:m7Xj-YlSJHJE9CVsnRGxMCg

Good find.
Pretty much... but, the input range for that particular model is higher than 3.7V
You can do this one of two ways:
> Buck Converter. Lowers voltage.
> Boost Converter. Increases Voltage.

So.. the 3.7V battery, would probably be best to find a 'step-up' or boost converter.
This will output more voltage than 3.7V, but will need to draw more current from the battery to compensate. This will let you add more than 2 LEDs in series

Otherwise, you should get two 3.7V batteries, and connect them in series to produce 7.4V that you can use with a 'step-down' or buck converter. Because this device lowers the voltage, it compensates by pulling less current from the battery as well.

Buck converters are nicer on the battery, but you have to start with a higher voltage than what you need. Boost converters are not as nice on batteries, but give you a little more freedom with the output by being able to get 10s of volts out of a 3.7V battery. Remember this though. If you use a boost converter to pull out twice the voltage, the converter will pull a little over twice the current from the battery!

 

[lightman]

that's great two 3.7 battery's would be ok so that would be 7.4 v going in so three led at 1.4 will be 4.2v so I set the out volt to 4.2 is that right if a little more I would need a resistor to take up the extra power going out is that right buddy would this be a better way to power my led for brightness and battery life ? thank you so much for your time and help and to everyone else this is a great site thank you

 

[BobK]

No, you don't want to set the output voltage, you want to set the output current. The 1.4V is nominal, it will vary from one LED to another. If you happened to have 3 LEDs that ran a little low, setting the voltage to 4,.2V will overdrive them.

Most of the cheap DC to DC converters out there have 2 controls, a voltage setting and a current limit. To drive the LEDs with a constant current, you set the current limit, then set the voltage control to something higher than is needed. The current limit function will lower the voltage output so as not to exceed the set current.

Bob

 

[lightman]

so if I supply 7.2 v to in put what would I set the current to for three leds and what would I set the volt to sorry for being so dumb its just so confusing to me I just don't want to spend £80 on an ir illuminator when I know I can build one cheaper

 

[BobK]

You will set the current for whatever the forward current of the LEDs is. You have shown 2 different specs, the first one indicated 1400mA and the second one 1050mA. That is what you want for the current. You set the voltage output to a little above the expected voltage, in this case, maybe 5V. The way a current limiter works is by setting the voltage to whatever is needed to draw the required current, but not to exceed the voltage setting.

Since the LEDs are in series, they all receive the same current, so you do not multiply the current by the number of LEDs.

Assuming you have a multimeter, here is how I would proceed.

1, With just the voltmeter connected to the output, adjust the voltage output to 5V.

2., Set the current limit to minimum. Verify it by connecting an ammeter in series with a resistor and LED, with the resistor being enough to limit the current well below the desired current, say 100Ω. The current should read near zero at the minimum setting, and go up as you turn it up.

3. Now connect your 3 LEDs in series and in series with the ammeter to the output, and adjust the current limit up until you reach the desired current (either 1400mA or 1050mA depending on which posting of the specs was the correct one.)

By the way, the stars the LED are mounted on are not enough for heat sinking. These are intended to be mounted to a larger heat sink. If you do not, they will get very hot and possibly the LEDs be ruined. That said, I have run up to about 600mA to 1W LED on one of these stars without damaging it.

Bob

 

[lightman]

so if each led is 1050ma x3 = 3150mm is that what I set current then I set the volt till the red light comes on : the 1400ma is the max for led I think and will this run the best it can be for the led many thanks

 

[lightman]

if I follow what this fella is doing will I be on the right track my voltage will be 7,2 but will not run it at that but maybe 5v like you say

 

[lightman]

is this ok to use http://www.ebay.co.uk/itm/DC-Buck-S...504675?hash=item2ee3edbf63:g:QTkAAOSw2VJVfzTO

 

[dorke]

lightman,
These DC-DC down converters are fine(based on the LM2596 Ic etc).
But,
You will never get to a very good efficiency(over 90%) with a low output voltage of 5V and under.
depending on the actual output voltage you work with you should expect something in the range of 65-75% efficiency,
look at the graph from the LM2596

 

[lightman]

ok many thanks for your help

 

[BobK]

Just for comparison, two cells giving 7.4V and 4 LEDs in series, using 5.6V will give you 76% efficiency. This should be fine. Just calculate the resistor as I have shown you in previous posts.

Bob

 

[lightman]

thanks I will do ATB Phil.

 

[lightman]

would I need a 1.8 ohm resistor ?

 

[BobK]

A 1.8Ω resistor would give you about 1A.

Bob

 

[lightman]

told you I was dumb lol to you fellas this is easy to me its just numbers and letters if some one can draw on paper the parts I need and were they go then I could build it with no problem practical I can do but thinking out I am not so clever I guess I need around 2A so maybe a 2.2 ohm would work ?

 

[BobK]

Where did you get 2A? If you put 2A through either of the LEDs you mentioned you will toast them.


Bob

 

[lightman]

sorry my mistake 7.4 v main 1,4 forward 1400 ma 3 led at 3w = 1.5 resistor ? wired in parallel

7.4 v main 1.4 forward 1400ma 3 led 3w = 2.2 resistor wired in series

 

[BobK]

2.3Ω for 3 in series.

But I would not run that at the full 1400 mA because the battery will start out at 8.2V and you would get 1900mA through them. Also additional heat sinking is required if you run them at full current. I would use a 3.3Ω to get about 1A,

And, by the way, they are not 3W. 1.4A * 1.4V = 2W.

Bob

 

[lightman]

cheers so they must be bull shitting me as there ad says 3w yes I have some 3,3 resistors so I can now try making one thank you all for your help this may sound daft but could I use a DC motor speed controller as a dimmer its range is from 1.8v to around 30v