Drop down DC voltage solution for low voltage LEDs

Status
Not open for further replies.

Hawker Pilot

Newbie level 3
I am building a test box that has a circuit in it with 28 to 24 vdc. I want to tap off it soley to power several low voltage LEDs for panel indication only. The LEDs can be from 1.8 up 12 volt DC. The circuit or device will decide that. What is the best way (some kind of circuit or voltage regulator)? I want this to be as small as possible.

betwixt

Super Moderator
Staff member
Re: Drop down DC voltage

The LEDs are constant voltage devices, all you need to do is add a resistor in series with them.

If you need to maintain constant brightness or cater for more than one LED in series, the simplest solution is to use a constant current generator to feed them. Within sensible limits, it will then work with a wide range of input voltages and any number of LEDs in series.

Brian.

Hawker Pilot

Newbie level 3
Re: Drop down DC voltage

I thought it would be something as simple as using resistors...

Where can I find information telling me what values of resistors to use to accomplish this task?

Lets say input voltage is 26vdc and the LED is a 2vdc unit.

betwixt

Super Moderator
Staff member
Re: Drop down DC voltage

Very, very easy...

The resistor value is based on the amount of voltage you want to drop and the amount of current you want to flow through it. The same current flows through the resistor as the LED so check the LED manufacturers data to find the optimum current. Usually, for panel indicators you will need about 10mA.

The formula is "voltage drop divided by the current", where the current is in Amps.

So using your figures as an example, the voltage drop is 26V - 2V = 24V.
If we assume 10mA is the current to use, the resistor value is (26 - 2) / 0.01 which gives a result of 2400Ω

Brian.

Hawker Pilot

Points: 2

Hawker Pilot

Newbie level 3
Re: Drop down DC voltage

Thanks,
This seems to be just what I am looking for. I'm just a dumb pilot but I think I can make that work.

Audioguru

Drop down DC voltage

You also must calculate the power dissipated by the resistor. Its dissipation is 24V x 10mA= 240mW so a 1/4W resistor will burn you and will burn the pcb. Use a 1/2W resistor.

Status
Not open for further replies.

Replies
2
Views
2K
Replies
10
Views
1K
Replies
9
Views
1K
Replies
3
Views
766
Replies
2
Views
1K