94Views16Replies

Author Options:

What resistor (or do I need one)? Answered

Picture of

I'm designing a simple prop that requires 2 LEDs. I figure a 2AA batteries connected to a switch that splits into 2 resistors and then into 2 LEDs then back to the batteries. See diagram below. The LEDs should take 20mA at 1.9-2.0V. So what resistor should I put in between the switch and LEDs.

17 Replies

user
JM1999 (author)2014-12-07

I am completely not used to electronics like this, can someone please enlighten me as to why you even need resistors?

Select as Best AnswerUndo Best Answer

user
iceng (author)JM19992014-12-07

Sure, the author has a 20ma 2v LED which is the same as the yellow LED curve below and the 100 ohm Resistor straight line !

If you raise the voltage to 3v the Resistor current will increase to 30ma while the LED current may Exceed 100ma and destroy the LED.

And the 3 volt supply with a 56 ohm resistor draws only 20ma....

Does that make you en lightened ?

Select as Best AnswerUndo Best Answer

user
JM1999 (author)iceng2014-12-07

So it basically keeps the voltage at a level that the LED can handle?

Select as Best AnswerUndo Best Answer

user
iceng (author)JM19992014-12-07

It keeps the current at a level the LED likes.

LEDs are diodes and small voltage changes cause Huge changes in current. It is current that can overheat the diode junction and destroy the LED..

Someone asked if one could really control that voltage would it work too ?

NO because a small room temperature change will push the LED into another part of that curve that could help or hurt it !

LEDs Are Current Devices..

Select as Best AnswerUndo Best Answer

user
JM1999 (author)iceng2014-12-07

Wow, so complex!

Thanks for explaining this, I will look further into making something with LEDs before I start! (I don't have resistors)

Select as Best AnswerUndo Best Answer

user
iceng (author)JM19992014-12-08

PM me with your address and let me send you some, if you like.

I presume you have LEDs and a soldering iron ?

Select as Best AnswerUndo Best Answer

user
iceng (author)2014-12-07

R = V / I = ( 3V - 1.9V ) / o.02 = 1.1 / o.02 = 56 ohms

55 is not available...

Select as Best AnswerUndo Best Answer

user
iceng (author)iceng2014-12-07

Your original schematic is correct

Select as Best AnswerUndo Best Answer

user
lucek (author)2014-12-07

OK so reading what people have said and following links I think I've got it now.

2AA>switch>1 33ohm resistor>2 LEDs in perralel(fig 1)

Or

3AA>switch>1 39ohm resistor>2 LEDs in series(fig 2)

When I enter the 3v of 2AA batteries into the calculator from petercd it gives me an error.

Select as Best AnswerUndo Best Answer

user
iceng (author)lucek2014-12-07

You never run LEDs in parallel some will be brighter.

Select as Best AnswerUndo Best Answer

user
steveastrouk (author)lucek2014-12-07

You CAN use 2 x aa with two chains of one diode and one resistor 47 Ohm

Select as Best AnswerUndo Best Answer

user
petercd (author)2014-12-07

What is the reason for 2 resistors, are the leds physically separated, as in far apart?

Normally we put the 2 leds in series and use one resistor.

You havent mentioned the supply voltage, so kinda hard figuring what the resistor value should be.

Use the calc here http://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator, its much quicker than waiting for an answer.

Select as Best AnswerUndo Best Answer

user
petercd (author)petercd2014-12-07

use one resistor of 25 ohms serving the 2 paralleled leds as in the last calc in the link.

Select as Best AnswerUndo Best Answer

user
iceng (author)petercd2014-12-07
user
petercd (author)iceng2014-12-07
user
Moonlight Maker (author)2014-12-07

In the circuit you've shown, the LEDs are in parallel. This means you can analyze it as two different "current loops". Using Kirchhoff's loop rule, we know that the sum of the voltage drops in any current loop is zero. This tells us that 3V (from the batteries) - I1 (the current through R1 and D1) * R1 = 0. For the current through D1 to be 20mA, R1 needs to be 150 ohms. The same goes for R2. The problem with this analysis is that I believe your LEDs will receive 3V which is greater than the 2V maximum that you listed. They will probably still function alright, but they might not last as long. Additionally, you should take this response with a grain of salt. I am not an electrical engineer. For more information on the subject, you might want to check out this instructable.

Select as Best AnswerUndo Best Answer

user

After looking at an online calculator, I realized my error. I neglected the voltage drop of the LED (2V). The calculation then becomes 3V - 0.02A*R1 - 2V = 0. This gives R1 to be 50 ohms like the calculator says. I would also agree with petercd that 2 resistors are probably unnecessary. You could use one 25 ohm resistor instead of 2 50 ohm ones.

Select as Best AnswerUndo Best Answer