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90 LED Infrared Illuminator - Total Draw???

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uspilot

New Member
Hi Everyone,
I'm not only new to this forum, but new to the electronics field.
I recently built a 90 LED Infrared Illuminator, and had a question about the power supply I was going to use...

Power Supply:
12vDC 500mA

Individual LED Info:
Voltage Drop = 1.5v
Current = 100mA

Project Info:
15 banks of 6 IR LED's, with one 33ohm 1/4watt resistor per bank.
The 6 LED's are wired in series, and the banks themselves are wired in parallel.

----------------------------------------------------------

R = V / I

R = (12v - 6*1.5v) / 0.1A
R = 3V / 0.1A
R = 30 ohm resistor (used 33 ohm resistor)

----------------------------------------------------------

I = V / R

I = (12v - 6*1.5v) / 33 ohms
I = 3v / 33
I = .0909A or 90.9mA

----------------------------------------------------------

So, 90.9mA is close enough to the required 100mA per LED.

Now, here's the question...
With the banks of 6 LED's themselves being wired in parallel in the circuit, will the 500mA power supply cover the total draw?

:oops: I don't know enough about the electrical end of it to know what the total draw on a DC parallel circuit would be.

Thank you to anyone that reponds!
Mike
 

uspilot

New Member
Wow, thanks for the quick response!

Okay, I guess I don't need the power supply I bought then. :oops:

I really appreciate the help! :D
Mike
 

uspilot

New Member
Hi fat-tony,
It's being used for my IR CCD security camera.
And by the way...a big thanks to Russlk!

I used a 12vDC 1500mA adapter, and this thing put's out so much light it looks like a 100,000 candle power flashlight to the IR camera! It has about a 6-8 ft. wide beam at about 8 ft. When I tested it, I was in my living room and I shined it into the hallway and bathroom (which was about 15 ft. away) and it was still as bright as can be.

When I turned it towards the camera, it drowned it in what looked like sunlight. It's a great feeling knowing I built this thing from scratch, and the only assistance I had was from Russlk on the power supply.

Mike
p.s. It was running a bit hot, so I had to throw in a cooling fan.
 

heathnelson

New Member
ir illuminator

Hello i was wondering if you could send me some pics and a schematic on your ir illuminator you built i am to new to this and wanting to build some for out sony night shot cams.
thanks
heath.

heathnelson@yahoo.com
 

audioguru

Well-Known Member
Most Helpful Member
100mA isn't the recommended operating current for the IR LEDs. It is their absolute maximum allowed current when all listed thermal conditions are not exceeded. Surely the LEDs are close enough together for tha ones near the center of the array to concentrate their combined heat to exceed the rating. Therefore I don't think they would last very long.

Since the very high illumination is not required, I suggest reducing the current maybe to half of max, by increasing the value of the resistors to 62 ohms.
 

drrogla

Member
yeah i agree 100mA does seam a bit too much...
It must be hot for those poor LEDS :?

if the output isnt as good with the 62 ohms try 47ohms instead
 

pittuck

New Member
whats the FPS of the camera? i think i am right in saying you can flash at 2x the fps and wont notice too much problems BUT will have some off time.

My IR LEDs work at 100mA for 20ms, i flash them for 20ms at the same time i start the sampling for the detector (Its a distance sensor). I have a cap in parallel to the phototransistor, i put the I/O high for the sensor output to 'charge' the cap and then after 20ms high / illumination i put the pin to ADC mode and take a reading at an exact time.

Thanks to duncan at MINOS 05 for that one!
 

electropierre

New Member
Hi Mike,

I'm also trying to find a plan to build an IR Illuminator to work with my Sony nightshot. I purchased a lot of 72 ir LEDs. I would be so thankful if you could send me a copy of your schematics. I've been on the Net all afternoon looking for one... The LEDs I have aren't the same as yours, I will have to compensate to meet the specs... My e-mail address is u4pierre@yahoo.com

Thank you so much...

Pierre
 

audioguru

Well-Known Member
Most Helpful Member
A schematic isn't needed for the simple array.
It is strings of 6 IR LEDs in series at 1.5V for each LED so the total voltage drop is 9V. He used a 12V supply so the current-limiting resistor in series with each string has 3V across it. Ohm's Law determines the value of the resistor: R= 3V/desired current. You have 72 LEDs so you would use 12 strings of 6 LEDs in series with a resistor.
 

freeskier89

New Member
I have a quick and very simple question. What is the advantage of putting strings of 6 in parallel over just putting them all in parallel? Does it have to do with distributing the power to be dissipated by the resistors? I know this is a really really simple question, but I just can't seem to find an advantage to either one of the configurations.
 

electropierre

New Member
Thank you so much Uncle $crooge. You made it very simple... I appreciate... And to answer the question of freeskier89, you need to put 6 LED in series with a resistor in order to drop the voltage to 1.5 V for each components. Remember, you want 1.5 volt, not 12... And you can add as many strings in parallel as your power supply can handle...
 

audioguru

Well-Known Member
Most Helpful Member
freeskier89 said:
What is the advantage of putting strings of 6 in parallel over just putting them all in parallel?
LEDs all have a slightly different voltage drop, so having many in parallel will result in the single one with the lowest voltage drop to hog ALL the current and blow up. Then the LED with the second lowest voltage will hog ALL the current and blow up, etc. Each LED or series string of LEDs needs its own current-limiting resistor.
Also as said by Electropierre, 1.5V or 2.0V is not available but 12V is available for a few LEDs to be in series.
 
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