0
\$\begingroup\$

Ok so I pulled out some old electronics stuff I have laying around. I have two AA batteries in a holder, in series. I have common cathode RGB LEDs where you can find all their specs here.

So I plugged in the cathode in and started touching the other pins with a wire. Red works, blue works, and green works, rubbing blue and green is making teal, cool so how about white? all three connected just gives me red. After trying white green stopped working.. and eventually tapping blue does nothing or is very dim. I'm thinking maybe I did something wrong maybe too rough with the pins or something I pull out another and all are working again and eventually red's all thats lighting up. I start to think maybe I have the positive and negative thing backwards and switching them does the same thing. Is it normal that the red will work both ways? rubbing or tapping green or blue on and off will turn the color on for a millisecond. Seems to be a brighter flash of color when the electricity is jumping from the wire to the leg, but holding it on it wont do it. So holding it just far enough that the shock is constantly jumping seems to make blue/green flicker. Maybe too much information but I just think it's weird that it's doing that and doesnt seem to be burnt out or anything and that it works fine until I try to get white out of it. I was hoping that if I burnt something out it would be more obvious or instant instead of really having no sign of damage. So I'm thinking I'll need a resistor, I have a few in this box here. Batteries put 3.0v since in series, right? They're new from the store, Duracell MN1500 batteries. I'm now looking at the chart and it looks like I might need 3.2-3.5 volts to power green/blue? but it was working for a moment. If I'm trying to power white, do I need to add up the volts for each one or can I get all three up with the same voltage?

So my simple question is what circuit do I need to create here to get this thing to work for one night? I'm starting to think I need another battery in series with a resistor. If I jump up to 9v battery and resistor, will I achieve longer battery life?

  • Noob
\$\endgroup\$
  • \$\begingroup\$ You'll probably need to add resistors, depending on what the LED specifications are. If you've got a datasheet for the LED, it might already have resistors on the terminals but it will be rated for a certain voltage and if you're not using that you'll need to add more resistance to each leg to compensate. You'll need to supply 3.5v+ to allow proper forward conduction on the Blue LED as from memory that's the largest voltage drop of all the colours. \$\endgroup\$ – NathanielJPerkins Dec 22 '15 at 4:31
2
\$\begingroup\$

You need 3 batteries minimum, and 3 resistors. With 3 batteries, try

schematic

simulate this circuit – Schematic created using CircuitLab You have 2 problems. The really fundamental one is tieing the LEDs in parallel. They operate at different voltages. The red has the lowest voltage, and it's hogging the current from the others. You need a separate resistor for each LED. Red has the lowest voltage, green next, and blue the highest. The current (and brightness) is determined by the resistor, so for the same supply voltage you really ought to have different resistor values. This is not entirely true, since different LEDs can have different efficiencies, but it's a good starting point.

The other problem you have is using a 3 volt battery. This is just about the blue LED voltage. When you first connect the blue LED, the battery has just enough voltage to light the LED, but this quickly gets exhausted. When you disconnect the battery, it recovers and will briefly work again.

\$\endgroup\$
0
\$\begingroup\$

Human vision needs 3 different "stimulus" values to see white:

Blue 420–440 nm
Green: 530–540 nm
Red: 560–580

You'll have to read your datasheet, install the correct resistors to produce the correct Vf (forward voltage) & current on each diode, in order to produce the right combination of light from all 3 diodes to see "white light".

\$\endgroup\$
0
\$\begingroup\$

Having a look through your datasheet, you've got 3.5v drop, a 1.95v drop and a 3.2v drop, with a desired current of 20mA. You'll need at least 3 AA's to properly work the RGB LED, so assuming that, you've got a supply voltage of 4.5v. 3*1.5 for AA. To get the rated current for the device you need to add a resistor. Forgive me, I don't know how to insert proper equation symbols yet, but the math goes like this.

(V_supply - V_led)/R = I

I = 20mA. So you can rearrange and get that you need a 128 ohm resistor for the 1.95v LED, a 65ohm for the 3.2v and a 50ohm for the 3.5 You can get all three diodes up off the same voltage level, however with all three on the battery will be supplying 60mA.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • 1
    \$\begingroup\$ The circuit with a single resistor won't work, due to the different voltage drops of the three LEDs. The lowest voltage LED (red) will draw all the current, and keep the voltage across the other two LEDs too low for them to light. \$\endgroup\$ – Peter Bennett Dec 22 '15 at 5:05
  • \$\begingroup\$ Very true. Editing now \$\endgroup\$ – NathanielJPerkins Dec 22 '15 at 5:06
0
\$\begingroup\$

These answers so far are unfortunately off the mark.

You need to provide three separate resistors on the anodes to control the current to each LED to get your white balance. The reason you only see red is because the forward drop of the red LED is less than it is for blue or green, so ALL of the current will go through that one if it has the chance.

You have to keep it from getting that chance by using separate resistors on the three R G and B anodes.

The typical Vf values for each from the datasheet are

  • \$V_{f(red)}=1.95V\$
  • \$V_{f(green)}=3.2V\$
  • \$V_{f(blue)}=3.5V\$

To get the basic resistor value for each leg, use the formula $$R = (V_{bat} - V_f) \div 20mA$$

Now experiment with increasing various resistor values to get white output. You will probably start out with a pale green or blue.

Note that \$V_{bat}\$ will change as the charge goes down, which will change the quality of the white balance somewhat.

schematic

simulate this circuit – Schematic created using CircuitLab

And as @whatroughbeast points out, the Vf exceeds the battery that you've specified, so that's another problem.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.