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### Topic: RGB LED voltage drop (Read 5973 times)previous topic - next topic

#### katherinel

##### Jul 30, 2012, 02:27 amLast Edit: Jul 31, 2012, 12:39 am by CrossRoads Reason: 1
I have a few basic questions. I'm thinking about ordering these: https://www.sparkfun.com/products/9264 (2nd pass made to link fixed, moderator)

First, what is forward current (20 mA for each color) vs peak forward current (30 mA)? On the datasheet they appear under "absolute maximum ratings", but I'm assuming 20 mA is what I should be aiming for?

Second, it says it has a voltage drop of 2.0, 3.2, 3.2 V respectively across R,G, & B. How do I take this into account when designing a circuit? Say I have a 5 V power supply, and want to connect 3 of these in parallel (so technically 9 total, since each RGB LED acts as 3 LEDs...?). Does it change the V=IR equation?

Assuming they should each run at 20 mA (if I understood the datasheet correctly like I asked above), does that mean the resistance within each diode would be calculated by:

red: 2 V = 0.02 * R
R = 100 ohms

green & blue (each): 3.2 V = 0.02 * R
R = 160 ohms

Finally, if that IS the case, do I need to take into account those resistance values in the circuit? So:

red: 5V = 0.02 * (R1 + 100)
R1 = 150 ohm resistor is needed

green & blue: 5V = 0.02 * (R2 + 160)
R2 = 90 ohm resistor is needed for each

#### baselsw

#1
##### Jul 30, 2012, 02:38 amLast Edit: Jul 30, 2012, 02:48 am by baselsw Reason: 1

I have a few basic questions. I'm thinking about ordering these: https://www.sparkfun.com/products/9264

First, what is forward current (20 mA for each color) vs peak forward current (30 mA)? On the datasheet they appear under "absolute maximum ratings", but I'm assuming 20 mA is what I should be aiming for?

Second, it says it has a voltage drop of 2.0, 3.2, 3.2 V respectively across R,G, & B. How do I take this into account when designing a circuit? Say I have a 5 V power supply, and want to connect 3 of these in parallel (so technically 9 total, since each RGB LED acts as 3 LEDs...?). Does it change the V=IR equation?

Assuming they should each run at 20 mA (if I understood the datasheet correctly like I asked above), does that mean the resistance within each diode would be calculated by:

red: 2 V = 0.02 * R
R = 100 ohms

green & blue (each): 3.2 V = 0.02 * R
R = 160 ohms

Finally, if that IS the case, do I need to take into account those resistance values in the circuit? So:

red: 5V = 0.02 * (R1 + 100)
R1 = 150 ohm resistor is needed

green & blue: 5V = 0.02 * (R2 + 160)
R2 = 90 ohm resistor is needed for each

It kinda depends on how you want to connect these leds to the Arduino. Each arduino pin is capable of handling a maximum of 40mA. So let us say that you want to connect three leds to one arduino pin. Each led will draw 20mA, if they are connected in parallel then this will result in a total of 60mA! I can help you with the calculations if you specify on how you are planing to connect these leds in detail please! A schematic would be more than awesome!!

Edit: Here is a link that will help you with calculations and will also show you a schematic of the chosen setup:
http://led.linear1.org/led.wiz

//Basel

#### Magician

#2
##### Jul 30, 2012, 04:22 am
Quote
red: 2 V = 0.02 * R
R = 100 ohms

No, there is correct version:  R = Vr / I ;  Vr = Vp -- V led = 5V - 2V = 3V - voltage accross resistor. Than R = 3 / 0.02 = 150 OHm

#### Grumpy_Mike

#3
##### Jul 30, 2012, 11:11 pm
Quote

That link is to a Ceramic Resonator and the question is about RGB LEDs so it is not very fixed.

So tell us if you have a common anode or common cathode LED.

If you put LEDs in parallel they each need their own resistor, calculates as if it was on its own. So for each RGB LED you need three resistors.

You subtract the forward voltage from the supply voltage to get the voltage to use when you want to work out the resistor for a specific current.

Quote
what is forward current (20 mA for each color) vs peak forward current (30 mA)?

The first is the current you can send through the LED continuously, the second is if you pulse the current, normally with a 10 to 1 pulse ratio.
If 20mA is shown as absolute maximum you should aim to use a current of 80% of this, that is 16mA.

#4
##### Jul 31, 2012, 12:40 am
Link is really fixed now - moderator.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

#### katherinel

#5
##### Jul 31, 2012, 03:18 am
Thanks, everyone. Assume it's going to be a simple parallel circuit of 16 RGB LEDs (= 48 diodes). It has to be able to light up (but not set on fire) whether one or all of them are on at once. (Is that possible? I could potentially use a decoder).

Next question: What effect does the rating for current of the power supply have on the circuit? If it's rated at 1200 mA, but my LEDs say "maximum 20 mA", and I have only one diode on at a time, does that mean it will burn out? Or do I ignore that when designing the circuit? I only understand how to use V = IR, so I'm confused when the power supply has both a V and an I rating...

#6
##### Jul 31, 2012, 04:45 am
If you were to actually them all on at once, then you'd need some transistors or some decent chips to sink all that current.
48 * .02 = 0.96A.
If you use 6 TPIC6B595's, you can sink each LED individually, and be able to have 16 different groups of color.

The rating the power supply only comes into play if you have a direct short. The LEDs will only draw as much as their current limit resistors will allow.

I have a '328 board that you can populate with up to 12 TPIC6B595s to drive plenty of LEDs.
Its available as a bare board that you can populate as you need for \$6.
Scroll all the way down. I haven't added PL & schematic yet, but its all just basic stuff.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

#### Grumpy_Mike

#7
##### Jul 31, 2012, 07:47 am
Quote
I only understand how to use V = IR, so I'm confused when the power supply has both a V and an I rating.

V is the voltage it WILL supply and I is the current it CAN supply.
See this:-
http://www.thebox.myzen.co.uk/Tutorial/Power_Supplies.html

Normally when you drive an RGB LED you want to control the brightness of each LED to allow you to mix colours. The TLC5940 is a good chip for that, see:-
http://arduino.cc/playground/Learning/TLC5940
You can chain them together, three will control your 16 LEDs. Arrange them so that one chip controls all the reds, one for the greens and one for the blues. In that way you can adjust the current drive of each chip so that you get a balanced white when they are all fully on.

#### funkyguy4000

#8
##### Aug 02, 2012, 02:03 am
As great as the sparkfun support is, I don't believe it to be worth the \$2 charge for each RGB LED.  You can find much cheaper ones in the U.S. on eBay.

In reference to your questions.  The equation doesn't change, in fact, the power behind RGB LEDs is governed by that very equation.  If you want to supply the Red led in the RGB led, you will have to have a different resistance to safely keep that led going at 20 mA if you so desire.

Since the Green and Blue leds have the same voltage drop, you'll use the same resistor value for those. What I mean is not the same resistor, but two resistors that are the same.

Since the Red led is lower, than you'll need a higher resistance to make sure it doesn't burn out.
Accelerate to 88 miles per hour.

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