Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Help with LED's

Status
Not open for further replies.

PanicMode

New Member
Well im new to the forum, and i hope you guys can help me out since im a noob when it comes to LED's.

I need to connect 5 LED's in parallel to a 12v source.
The LED's specifications are as follows:

yellow forward voltage 2.1
20mAMP current
12 V source.

I bought 100oHm 1/4 watt resistors, but when i connected them in series the resistor got really hot.

I dont know what i did wrong or if this is normal for the resistor.

Any help is greatly appreciated.
 
12V source - 2.1 V forward voltage = 9.9V

I = 20 mA or .02 A

V = IR or R = V/I

R = 9.9/.02 = 495Ω

Replace the 100 Ω resistors with 500 Ω and they should not get hot.

When you had the 100 Ω resistors the current through them was:

V/R = I = 9.9/100 = .099 A or 99 mA.

P = IV so P = .099*9.9 = .9801 W That's almost 4 times what a 1/4 W resistor is rated for.

The 500 Ω resistor will dissipate only about .196 W
 
Last edited:
Well im new to the forum, and i hope you guys can help me out since im a noob when it comes to LED's.

I need to connect 5 LED's in parallel to a 12v source.
The LED's specifications are as follows:

yellow forward voltage 2.1
20mAMP current
12 V source.

You need to put the LEDs in SERIES (not parallel)!!!

you likely used the SERIES calculator on the web link you provided, which comes up with close to a single 100Ω resistor in SERIES with the five LEDs.

The power dissipation in the resistor will drop to (12-(5*2.1))*0.02 = 30mW, which is well within its 250mW rating. It won't even feel warm...
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top