# Connect different "blinking LEDs" in parallel

I have a simple motor and two LED toy with different color leds:

LED D1 is blue blinking listed as 30mA/3V/0.09W (osub5x31a), LED D2 is red blinking listed as 20mA/3V/0.06W (OSHR5X31A). Battery is 2xAAA.

These are numbers the seller gives in homepage, but product detail sheets seem to have different voltage, so I'm a bit lost with that.

https://data.oomipood.ee/kasutusjuhend/oshr5x31a.pdf

https://data.oomipood.ee/kasutusjuhend/osub5x31a.pdf

Motor is rated as 1.5V-3V, I don't know other parameters.

I cannot figure what values R1 and R2 should have, or how to calculate them.

I initially made it without any resistors, but then the red LED barely works (does not blink,) blue does not light at all and motor works fine. References I found talk about 'voltage drop' but I'm not sure what it means, the LEDs I have don't list such property.

Update: hen I just connect the LEDs to the 3V source in parallel, no motor and no resistors, then they both seem to blink fine. Maybe they don't really need resistors at all, or is this not safe?

• Welcome to EE.SE! That's because the red LED will clamp down your battery if you are not using a series resistor, and possibly damage the red LED. There are plenty of LED resistor calculators online. I=(Vbatt-Vf)/R. With your blue LED having the same forward voltage as your battery, brightness will depends heavily on the battery voltage. Commented May 12, 2022 at 15:15
• thanks @winny I did not find online calculator for such case, only for simpler cases with single LED. Is motor element here irrelevant really? And what would be "Vf" in my case - thats also something I could not put to the online calculator Commented May 12, 2022 at 15:30
• Does this answer your question? How do I calculate the resistor value for a simple LED circuit? Commented May 12, 2022 at 15:57
• There are many duplicates and near-duplicates of this: electronics.stackexchange.com/q/148854/2028, electronics.stackexchange.com/q/22291/2028, etc. Commented May 12, 2022 at 16:01
• @JYelton not really, I'm mostly blocked in "check the datasheet or measure it to know how much voltage drops over your LED." part. See my linked datasheets, I cant find Vf or voltage drop from there, and dont know how to measure that either :( Commented May 12, 2022 at 16:03

Those are not simply LEDs. They are LEDs with an integrated circuit in an LED housing.

They are meant to operate on 5V. Your 3V power supply is not enough.

Without the motor, the battery voltage is (just barely) enough that the LEDs can operate

When you connect the motor, the battery voltage drops a bit and the LEDs quit working.

You need a higher voltage power supply, and it probably needs to be able to supply more current than AAA cells can deliver.

• I tried finally with both 3V and 4.5V (3*AAA) power, no resistors, and no visible difference: without motor LEDs blinks fine, with added motor motor itself works, blue LED is blind totally and red LED shows very little light. Commented May 12, 2022 at 19:06
• Like I said, the AAA batteries can't supply enough current to the motor. The motor draws so much current from the batteries that the voltage drops to something lower than the LEDs need.
– JRE
Commented May 13, 2022 at 9:44

These are blinking LEDs, they contain the circuitry to make them blink and current-limiting circuitry. So they don't need (or want) series resistors.

The bad news is that you must give them a voltage within the recommended range to have them reliably work. 4.5 minimum to 6V maximum. So they will work reliably from a 5V regulated supply. In parallel, no resistors.

The following in not exact but will get you close enough. The voltage drop of LEDs varies by chemistry which controls color. The following is in order from low to high voltage drop. Laser diode: ~1.5V IR LED ~ 1.5V Red: ~2V Amber: ~2V Yellow: ~2V Green: ~2.5V Blue: ~3.5V White: ~3.5V You need to know what current you will drive the LED at, I use 10 mA as a default value, it is generally safe for most LEDs.

For example let's use a Yellow LED at 10 mA. At 5 Volts source we subtract the forward voltage in this case 2V from 5V leaving a value of 3 volts. Using Ohm's law with 3V and 10Ma we get a resistor value of 300 Ohms. If we put two in series we get 5 - (2+2) = 1V. Using OHM's for 10mA and 1V we get 100 ohms. Same thing with 12V. 12V - 2V = 10V. 10V at 10mA = 1K or 1000 ohms. You can use the Ohm's Law calculator at: https://ohmslawcalculator.com/ohms-law-calculator. Lets try three in series different colors. Blue = 3.5V, Green = 2.5V and Amber = 2V again at 10mA. 3.5 + 2.5 + 2 = 8V. 12V - 8V = 4V at 10mA we need a 400 Ohm resistor. The reason this works LEDs are current NOT voltage devices. These values are per chip, some high power devices have two or more chips in series.

In summary you add the forward voltages together and if it is less than your source voltage you are good to go, if not you will need to do a parallel series combination. By subtracting the forward voltages from your source you have the voltage drop across the resistor. Using Ohm's you calculate the resistor resistance required to pass the current you want. Be sure to check the data sheet for speciatility LED's.

• It is worth taking a look at the datasheets. These are not LEDs, as implied in the question. Commented May 12, 2022 at 17:17