Hi,

I have what is probably a simple question for the forum. I am making an IR-LED array of 3 to 6 IR-LEDs in series with a ballast resistor of whatever value is rendered by the online series LED calculator-wizards. I have also done the calculations by hand. Works fine. I want to add an additional LED to the circuit but visible red to indicate there is power as a safety precaution. How do I add in the visible LED to the circuit containing the series array of IR-LEDs? One assumption in all of the series LED arrays is that the LEDs are each electrically the same as the other.

Thank you.

 

What current do the IR LEDs run at and what current does the visible LED need?

 

What current do the IR LEDs run at and what current does the visible LED need?

IR LEDs I = 100ma
red led = forward voltage of 2.0V and a rated forward current of 20mA.

 

In that case it would be best to have a separate resistor for the visible LED connected to the same supply that feeds the IR LEDs.

 

That would work as an indication that voltage is being applied to the string of IR LEDS, but if you wanted to check that they were actually passing current then a power resistor could be connected in series with them and the voltage dropped across that resistor would be the supply for the indicator LED.

 

skip/delete

 

In that case it would be best to have a separate resistor for the visible LED connected to the same supply that feeds the IR LEDs.

Would I have to change anything else in the existing circuit or just add the red LED. My circuit is simple enough: I have 3 IR leds pulling 100 mas each with a single 100 ohm ballast resistor (calculated by a wizard at some site). Works great. So I add the red "power" LED to another circuit loop, not in the series.

 

Yes.

 

What is the power supply voltage?
What is the part number for the IR LEDs? Or vendor datasheet.

ak

 

PS is 15v
no datasheet (Ebay) but posted specs (below). Hope that is enough.

Package:
※1xPCS 50Pcs 5mm Infrared IR LED Night Vision 940nm Luminous Diode LED infrared Lamp.

Features:※ Small and light appearance, it is convenient to carry.※ Small volume, low power consumption, excellent directivity.※ With IR function, it is easy to use.※ It can be used in the field of monitoring, remote control, photoelectric control, target tracking etc.

Specifications:※ Quantity: 50pcs※ Diameter: 5mm / 0.2"※ Material: GaAsP※ Emitted color: Infrared ray※ Forward voltage: 1.4--1.65V※ Backward voltage: 5V※ Forward current: 100mA※ Wavelength: 940nm※ Angle: 90/60/45/30 degree※ Luminous power: 150mw※ Frequency characteristic: Low frequency.

The default angle is 90 degree,if you need other angle,pls kindly contact us when you place order!

 

3 IR leds pulling 100 mas each with a single 100 ohm ballast resistor

If I understand you, your circuit works but you want to know if it is on or off. Now, I assume those three LEDs are in series with each other and with your 100 ohm resistor. Then the voltage across the resistor must be 10 volts. Please measure it to be sure. If it is right or nearly right, just use your wizard to calculate the resistor needed to light your visible LED (probably 2.2 volts at 10 ma or about 390 ohms) and wire it across your 100 ohm resistor.

 

Thanks for your reply. Could you please expand on this part of it:

wire it across your 100 ohm

I understand the rest of the post. Wire it across - does end up in series?

 

Yes, series to the other LEDs but parallel to the resistor.

If you are already running your IR LEDs at 100mAs you may not want to place another across the resistor as this will increase the current through the IRs.

You might just want to pick up one of those ½ watt red LEDs and place it in series with the IRs and recalculate the resistor.

Just be aware those ½ watt LEDs are very bright, so you will need to diffuse it somehow.

If you haven’t bought the parts yet, then you are golden, and you can buy the proper resistor to allow a parallel low power LED across it and still maintain the 100mAs.

 

in series with the IRs and recalculate the resistor.

That's where I am stuck. All of the wizards solve for the same value LED not mixed.

 

If you know the total forward drop of the LEDs (simple addition) you don't need a wizard.

Or you can use this one...just place the total drop where it says "Voltage Drop Across LED".

https://www.quickar.com/starresistorcalc.htm

 

Awesome. I don't know anything! : )

Here it is: using quickar I enter the total voltage drop that includes the Red LED. I enter 100ma for forward current(ratings). Now my problem is (I think) I am subjecting the Red LED to 100ma in current I need for the IRs? This is where I get stuck because that sounds like a problem.

 

Yes, that is why I recommended using a 1/2 watt red LED.

You can't place a low power (20mA-30mA) LED in series with those IRs. (unless bypassed with the dropping resistor)

 

Your comments about diffusing the LEDs: is the brightness a health hazard -- I use them for IR beacons. Also will diffusing the LEDs increase the viewing angle? I am okay with where they are now (~90 degs), but would like to experiment with something more luminescent than focused. They need to be detectable at 9 to 12 feet hence the selection for 100ma. Would diffusing it increase the viewing angle?

>>Yes, that is why I recommended using a 1/2 watt red LED.
Found an 80ma one at Sparkfun - guesstimating it almost passes 100ma and is fit for use.

 

I wouldn't diffuse the IR LEDs unless you have a reason to...I also wouldn't look directly into them, because they are very bright, even though you can't see the light. And yes they can hurt your eyes, even though they are not lasers. I can't really speak to long term damage, all I know is if I catch a glance at the LEDs I use they hurt my eyes.

You should probably do more research if you intend to create powerful "beacons" that cannot be seen.

80mA is not close enough...sorry.

As far as diffusing the red LED, I used the term "diffuse" loosely, you will need to block most of the light, to use as an indicator.

 

Problem is user faces them, though not looking directly into the beacons . Will require more assessment. Lack of a blink response can result in "baking" the retinas under certain conditions if there exists a harmful condition.