Need help designing a basic LED light

[Mavruss]

Hello everyone!

Could someone give me a hand designing a circuit? I used to be able to do this back in the day (late 90s) but I forgot everything. I remember some things but the math is gone….

Anyway, I want to make an IR light for night vision. I purchased some LEDs online and I just need to develop the circuit.

The IR LED specs are:
• Voltage: 1.5V~1.6V
• Current: 140mA
• Peak(pulse): 700mA
• Power: 200mW

The light will consist of 12 IR LEDs, a switch, and it will be powered by a 9 volt rechargeable battery.

I wasn’t sure if series or parallel would be better for the array. Here is what I came up with so far. The current is too low I believe. Also, Im not sure how long the battery would last. Maybe a 6 LED version? Thoughts?

Thanks in advance,
Russ

I attached a screen shot. Please see below.

[Thomas W]

Hi Russ,

What's the purpose of R3 and R4?
You would want a current of 140 mA in XMM3.

Thomas

[Mavruss]

R3 and R4 is a voltage divider. I didnt want the full 9 volts hitting the LEDs.

Yes, I believe 140ma is the max on those LEDs.

-Russ

[Thomas W]

There will only be 1.5 V across each LED. With 2 x 6 LED's in series the total voltage across the LED's is 6 * 1.5 = 9 V.
You should probably only put in 2 x 5, and let the rest of the voltage fall across R5 (and R2). 5 * 1.5 = 7.5 V, which leaves 9 - 7.5 = 1.5 V across R5 (and R2).
Try connecting net 13 directly to net 12 and see what happens.

[Mavruss]

Well, here is what happened:

[Thomas W]

That's good.
The resistor values can be calculated like this:

R5 = R2 (They're in parallel so the same voltage (VR2) will be applied to each one.
IR2 = ILEDx = IR5
R2 = VR2 / IR2 = (VBATT - (5 * VLEDx)) / ILEDx = (9 - (5 * 1.5))/0.140 = 10.71 ohm

Power:
PR2 = PR5
PR2 = IR2^2 * R2 = ILEDx^2 * R2 = 0.140^2 * 10.71 = 0.2099 ~ 210 mW (1/4 watt resistor will do).

The LED's will load the battery with 2 * 140 mA = 280 mA.
I have a 9V rechargeable NMH battery that holds 175 mAh.
If the voltage is the same it will last 175 / 280 = 0.625 hours = 0.625 * 60 min. = 38 minutes
but check the datasheet for the battery if you want exact numbers.

Did you consider the safety of your eyes regarding IR light?... (I have no idea how much it will take to do any damage...)

[Mavruss]

Yikes. 38 minutes isnt very long. How many LEDs would you think it would take for the battery to last at least an hour? 8? 6?

I havent really tested the LEDs yet. Im trying to find the sweet spot of maximum light for long battery life. I want an hour + at the least. Hell, maybe 5 LEDs would do the trick.

[Thomas W]

If you put in an on/off switch for the R5 branch you can switch off 5 of the LED's and you'll the double battery life.
In order to calculate a solution you'll need to decide how much light you want measured in candela [cd] and then go backwards and find the best LED's / current / battery combination. This can be difficult so a field test with the mentioned switch in the circuit will give you something to work with in order to move forward with the project.

[Mavruss]

I really appreciate the help. I have been crazy busy at work and havent had the time to build this yet. Im hoping for either this weekend or next.

I will keep you posted on what happens.

Thanks again for the help!

-Russ

[Mavruss]

I built the circuit last night. it turned on but the LEDs were pretty dark when looking through the camera.

Anyway, I was thinking about ditching the battery option all together and go with a AC/DC transformer.

I have a 12V 200ma laying around that I could use. Do you foresee any issues?

[Thomas W]

You should limit the loading of the AC/DC adapter by using only one LED branch (140 mA), and redo the calculations with 12 V instead of 9 V.

Did you test it with 5 or 10 LED's?

Do you have access to a physical ampere meter so that you can check the actual LED current?

(Here's the schematic for unregistered forum readers:)
[img]http://diagramtips.com/images/2011/ir_light2.png[/img]

[Mavruss]

I built the circuit with the 10 LEDs and a 9V battery. I do have access to a multi meter.

I will fire it up when I get home and measure the current off the battery.

[Mavruss]

Ok, either the circuit is hosed or I'm not using my DMM correctly.

I set the DMM to DCma. It has three settings. 2, 20, and 200.

I selected 200 and wired the DMM in line of the power. The meter is reading 21.5

If I'm reading this right, the circuit only has a 21.5ma draw? No wonder it isnt very bright.

Suggestions?

[Thomas W]

Are you using the battery or the AC/DC adapter? What is the voltage output?
It should be close to 9 V or 12 V.

[Mavruss]

I'm at work right now so I cant do anything but when I took the measurement, it was on the 9V battery.