Need help planning an LED indicator light

[IsaacT]

Hey guys,

I don't know enough about LED's to know what needs to be done. I am planning a build with an LED indicator light and I havent been able to find how people are driving the LED's. The build will be a single cell 488nm laser, so I guess the input voltage will be 3.7V. The DTR 488 module/driver combo states 2.5V-6V 1A+. A random 5mm LED I found shows a typical forward voltage of 2.0V and current of 20mA. I am assuming that forward voltage means the amount of voltage lost from one side of the component to the other.

1. To avoid issues with losing the necessary voltage across the Laser Diode, as 3.7V - 2.0V is only 1.7V, well below the stated input requirement, should I run the LED and the Laser Diode in Parallel to supply both with a voltage input of 3.7V?
1a. Alternatively, should I use two batteries for a total of 7.4V coming in to the circuit and have the LED in series with the Laser Diode such that the 7.4V - 2.0V will yield 5.4V which also fits into the stated input requirements for potentially longer runtimes?
-Caveat to this is would this be playing a dangerous game? I assume if the LED emitter dies, the component would no longer be soaking up that extra voltage and would give too much voltage to the Laser driver. Which, in turn, may kill the diode?

2. Do I need something to drive the LED emitter similar to how we use laser drivers to limit the current coming in?

Tentatively this is planned for a modified laser66 host so space concerns are very real.

Thank you for any advice you can provide!
-IsaacT

 

[Singlemode Laser]

You need to limit the current through the LED or you will kill it. Take a resistor between your voltage supply and the led so the current through the resistor is about 20mA.


Singlemode

 

[IsaacT]

You need to limit the current through the LED or you will kill it. Take a resistor between your voltage supply and the led so the current through the resistor is about 20mA.


Singlemode

Based on that it sounds like I absolutely need to run the LED parallel to the Laser Diode rather than in series, which knocks my other questions away nicely.

Using an Ohms law LED calculator, for a source Voltage of 3.7V with a voltage drop of 2.0V to get 20mA I would need a resistance of 85 ohms. Which is not a resistor value I can locate on any component sites.

If I understand correctly, resistors in series is equal to R=R1+R2+...., so if I took two 47ohm resistors in series it should equal 94ohms. Plugging 94 ohms in to the calculator instead of specifying current gives 18mA. Would that be sufficient to drive it?

EDIT: Or a 50 ohm plus a 33 ohm would give 83 ohm, which would give 20.48 mA. This is a bit more than specifications, but would such a small overage be a big deal?

 

[paul1598419]

An 82 ohm is a common value.

 

[IsaacT]

An 82 ohm is a common value.

Oh fantastic! I didn't see those when I was looking. Would the .72 mA over the rating for the LED have a negative impact on it or is that a small enough difference that it should function normally?

 

[paul1598419]

Three ohms difference is negligible in this case.

 

[Anthony P]

This is from "Engineer's Mini-Notebook, Basic Semiconductor Circuits" by Forrest Mims. There is a whole series of these short books that can be very helpful to someone learning electronics, or even someone experienced who just needs a quick reference.

 

[absolute]

Take into account that a fully charged Li-Ion cell is often at 4.2V and make sure the the higher current that will flow (for this example 4.2-2.0=2.2V 2.2/82=26.83mA) won't kill the LED. If it would, find a different resistor value by doing a calculation with the max current rating for the LED and the maximum voltage your battery will provide.

Also, you're talking about running it parallel to the laser diode but I assume you mean running it parallel to the laser driver i.e. connected straight to the battery. If you connect both the LED and LD to the output of the driver you'll have a different problem altogether

 

[Cyparagon]

If it is for indication, the full 20mA is not typically necessary. 1 or 2 mA is often bright enough and can save a bit a power. I would use a 1K resistor, but it really doesn't matter.