# Circuits with different LEDs?

1. Apr 15, 2010

### Gersty

I know there are already a number of posts in this general subject area, but they all seem to involve circuits with one or more identical LEDs. I'm trying to come up with circuits containing red (approx 2.5 Vf) and blue (approx 3 Vf) LEDs that will run off a 2AA (3V) battery harness. I'm assuming that I'm going to have to combine both series and parallel wiring in order to supply each LED with the proper voltage at 20 mA, but I'm having trouble designing the circuits.

2. Apr 16, 2010

### svenkesd

A few things to consider here.

You should be able to power a red LED from 3V but maybe not a blue LED. It depends on the LED and the battery charge. Usually AA batteries start with a charge of about 1.6V per battery so you might be able to run a 3Vf LED from 2 of these for a limited time.

"I'm going to have to combine both series and parallel wiring in order to supply each LED with the proper voltage at 20 mA"

A proper LED circuit design does not care about the supply voltage, but is designed to source the appropriate current (20mA in your case) regardless of voltage. That being said, the voltage must be high enought to overcome the Vf of the LED.

Since 3V is not enough to power the two LEDs in series, I am assuming you will put them in parallel. The problem is that it does not work very well to put different LEDs in parallel with simple current limiting resistors. What is needed is a current limiter in series with each LED separately and it will likely have to involve a current sensing circuit.

That being said, I have doubts about how well an active current limiting circuit will work with an LED Vf that is very close to the supply voltage, you may have trouble finding components that can achieve this. Your other option is to use a boost converter to pump up the voltage of the batteries.

Hope that helps, good luck!

3. Apr 16, 2010

### Gersty

4. Apr 16, 2010

### Gersty

So a 3 V Blue LED with a 1 ohm resistor in series paralleled with a 2 V red LED with a 50 ohm resistor in series won't cut it?

5. Apr 16, 2010

### svenkesd

Technically that should work, you will get very uneven current distribution without a feedback mechanism, but it will light the LEDs.

The problem with what you are suggesting is that for the blue LED you are working with the steep part of the I-V curve of the diode. What that means is a tiny change in the voltage will cause a huge change in the current through the diode.

For example, your batteries will likely start out around 3.2V total for a pair of fully charged alkaline batteries. With a 1 ohm resistor as you suggest, and a 3V diode, the current will be (3.2V - 3V)/1ohm = 200mA. That will burn out a low power LED. Once the supply voltage drops to 3V, the current limiting will be dominated by the LEDs I-V curve and depending on the diode will probably be around 20mA. You can see in this example that a drop in the supply voltage of 0.2V will cause the current to change by a factor of 10.

If you have a steady supply voltage you can size the resistor correctly but even a tiny drop in the charge of the battery can have a large impact on the LED brightness, so not a very good design for anything but demonstration purposes.

What is the purpose of the circuit? Will it need to operate for extended periods of time?