parallel led array

[dorke]

that's cool Colin I will forget the diode

Don't listen to Colin's B.S !
Definitely use the diodes!

 

[dorke]

so was colin wrong dorky?

Don't listen to Colin's B.S.
Definitely use the diodes!!!

 

[lightman]

thank you kind sir I have been interested in light making for a bit now made a few I make all my own ghost hunting kit as its very expensive to buy and need all the help I can get are they easy to solder ?

 

[lightman]

some of the items I have made rem-pod em pump shadow detector

 

[dorke]

thank you kind sir I have been interested in light making for a bit now made a few I make all my own ghost hunting kit as its very expensive to buy and need all the help I can get are they easy to solder ?

Yes, the SMD diodes are easy to solder ,not a problem at all.

 

[bigone5500]

some of the items I have made rem-pod em pump shadow detector

What is the apparatus with the antenna?

 

[lightman]

Hi fella its a Rem -pod it works on electrical static energy they say that's what ghosts are made from so anything electrical or static will set a buzzer and the lights light up just like this one

 

[BobK]

The way you are doing this is the most wasteful way you could come up with. You are using only 38% of the power from the battery to power your LEDs. The rest is going into heat in the resistors.

Want 2X the battery life? Put 2 LEDs in series. The voltage is 1.4V so 2 is 2.8V, which is fine for use with a 3.7V battery. You will now be drawing exactly half the current and your batteries will last twice as long. Of course this does not work with 3 LEDs so use either 2 or 4.

Or, better yet, use a DC to DC boost converter in constant current mode to drive all 3 in series. This will give you nearly 3X the battery life.

Colin was not wrong, just unclear. Change "cannot" to "should not" and he is correct.

Bob

 

[BobK]

Also, how did you calculate your 2.7Ω resistor. I come up with 1.6Ω to run a 1400mA 1.4V LED from a 3.7V battery.

(3.7 - 1.4) / 1.4 = 1.64

With a 2.7Ω resistor you are running them at about half power, which is perhaps intended.

Also, 1.4V * 1.4A is less than 2W, not 5W as advertised. If these were white LEDs running at 1.4A and 3.6V they would be 5W. I think you (or more likely the seller) has mixed up the specifications.

Bob

 

[lightman]

hi the led I am using at the moment are 3watt each I am using three I have wired them in parallel so that each led gets the same power I have used a 2.7 resistor on each led so the led don't burn out I am new to this that's why I am asking for help I was told not to go much lower than the 2.7 I know there is a 2.2 which I could use but un sure

 

[lightman]

the led are infrared three led = 9watt the other light I want to make will have three 5watt so total 15watt trying to keep ever thing compact as they are fitted to the camcorder when walking around I am using that type of battery as they are just right size I was using one battery at first and was only getting around 1 and a half hours constant light but using two of the 3.7 I am now getting three so that will only be one battery change in the night plenty for an investigation but always looking to better them that's why I came to this great forum in the hope of bettering what I have thank you ATB Phil.

 

[dorke]

ligtman,

BobK is absolutely correct:
The way to drive a power LED(s) is with constant current.

It will be the most efficient way and will yield longer possible battery time.
It is more complicated to implement and involves constant current high frequency dc-dc converters. such as the LM3410

If you are willing and able to go this way it will be the best.
You can get help here for that as well.

Otherwise,
Bobs advice to put 2 leds in series is a very good one.

As for connecting batterys in parallel ,
always use the Schottky diodes

 

[lightman]

it seems to be getting more complicated lol ok what I am after is to run ok two three or four 5w IR led the brightest I can all the same without burning them out if I can get two to three hours continuous use then that would work fine so what would be the best way to do it remember I am not like you fellas I don't no much as just starting out so would need to be as simple as poss thank you all so very much for your time and help

 

[lightman]

if I run the led with a 2.2 ohm resistor would that be better than the 2.7 ohm ? looking for brightness to thanks

 

[BobK]

The brightness of an LED is roughly proportional to the current, though it falls short of this linear relationship as the current approaches the maximum. Or, to put it another way, LEDs are slightly more efficient when run at maybe 1/2 the max current, where they will put out more than 1/2 the light that they would at full power. Of course that is a tradeoff with the cost of the LEDs.

The simplest way to get more battery life for your project is to run 2 LEDs in series. Here is the difference. If you run 2 LEDs in parallel, they will draw twice the current of 1 LED. If you run them in series they will draw the same current as 1 LED while producing twice as much light.

Your current situation is as follows:

3 LEDs run in parallel
3.7V battery
2.7Ω resistor for each LED

Since this will not run them at full current, the voltage will be somewhat less than the 1.4V they drop at 1.4A, let's estimate this at 1,2V

So:

V = I * R

(3.7 - 1.2) = I * 2.7

(3.7 - 1.2) / 2.7 = I = 0.92 A

So you the power you are putting into the LED is:

1.2 * 0.92 * 3 = 3.3W

The power you are drawing from the battery is:

3.7 * 0.92 * 3 = 10.2W

So you are getting approximately 33% efficiency with your circuit.

Let's see what would happen to get the same light with 2 LEDs in series:

The current would have to be 3/2 of the current you are supplying to the 3 LEDs, so:

I = 3/2 * 0.92 = 1.38A

How clever, that is almost exactly the 1.4A that your LEDs are rated at!

So now we have:

V = 1.4 V
I = 1.38 A

We can calculate the resistor as:

V = I * R

(3.7 - 1.4) = 1.38 * R

R = (3.7 - 2.8) / 1.38 = 0.65Ω


So, now, the power you are putting into the LEDs is:

2.8 * 1.38 = 3.9 W

And the power drawn from the battery is:

3.7 * 1.38 = 5.1 W

So you are getting about twice the light for 1/2 the power or about 76% efficiency, which is really not that bad.

I would actually halve the current and use two chains of 2 LEDs in series / parallel, which would be a little more efficient in light output because LEDs are more efficient at lower currents. Plus it spreads out the heat making it easier to heat sink them. (You did know that you have to have a heat sink for these LEDs, right?)

Compare to a switch mode converter that can be up to 90% efficient, but is much more complex.

Bob

 

[Martaine2005]

I think I prefer Gomez'. He gets Morticia to type a comprehensive answer!!
Excellent post @BobK

Martin

 

[BobK]

Morticia can type faster and more accurately than I, but she lacks the electronics knowledge that I acquired while studying with the good Dr. Frankenstein.

Bob

 

[lightman]

hi Bobk I would sooner have the led running at the brightest I can more light through the ir led the better I would sooner have that than battery life if I can get an hour and half out of the battery then that would be ok I was told to wire in parallel so I would get the same power going to each led not sure if this is making any sense to you yes the led are mounted on star heat sinks

 

[lightman]

I just want a simple way to wire them to get the best light I can from them I was told if I wire them in series they would not be as bright I really thank you for all your time but I am not the brightest in the bunch so not sure of the parts I would need to run it your way or were to solder each bit I can not read diagrams so not sure of what the simbles mean

 

[lightman]

this is how I have done it if I use a lower ohm like a 2.2 would that make anything better ?