Calculating Resistor

[brthegreat117]

Okay, I am not a fan of math, but I did my best and used Ohm's law to calculate what resistor I would need.
I will be using a Lithium Ion 3.7 Volt 18650 battery
The LED is B/B/W Tri-Cree Xp-E2
This is the info for the Blue led
Wavelength:   485 nm   
Luminous Intensity:   -   
Luminous Flux/Radiant Flux:   45.7 lm   
If - Forward Current:   350 mA   
Vf - Forward Voltage:   3.1 V   
Mounting Style:   SMD/SMT   
Power Rating:   3 W
Info for the white
info for the WHITES
Illumination Color:   Cool White   
Color Temperature:   10500 K   
Luminous Flux/Radiant Flux:   360 lm   
Color Rendering Index - CRI:   65   
Luminous Intensity:   -   
Viewing Angle:   125 deg   
If - Forward Current:   1050 mA   
Vf - Forward Voltage:   3..5 V


For the blue I calculated
A resistance of 1.71429 ohms. & a Led power of 1.085 watts. Would this mean that I would need to get a 2 ohm/1 watt resistor OR a 2.2 ohm/5 watt resistor.
For the white I calculated
A resistance of 0.19048 ohms & a Led power of 3.675 watts. I strongly believe this value is incorrect. I will be double checking and updating this post to fix that error.

[smaneesint]

I've been wanting to know this as well.

Waiting for someone to enlighten us...

[brthegreat117]

I found this on the TCSS forum
Ohm's Law - the down and dirty guide to "Doing the Math" yourself.

Ohm's Law comes in two parts, meaning there are two values you need to calculate. The "Ohms" value and the "Watts" value. The formulas for both are as follows:

R <Ohms> = (VfBattery - (VfLED1 + VfLED2)) / I

Where:
Vf Battery = Battery Voltage (usually either 3.7V or 7.4V)
Vf LED1 = Forward Voltage of LED #1
Vf LED2 = Forward Voltage of LED #2 (If you have a second in series)
I = (in Amps) Current you wish to run your LED at (Reds and Red-Oranges are usually 700 mA and Blues, Greens and Whites are usually 1A)

The second part of the formula is as follows:

P <Watts> = R * I^2

Where:
R = the resistance value from the first calculation
I = (in Amps) Current you wish to run your LED at (Reds and Red-Oranges are usually 700 mA and Blues, Greens and Whites are usually 1A) - This value needs to be squared in your calculation.

In the following example, I calculate a resistor value for a single Red LED
(Vf= 2.4V) running at 700mA running it off of a 3.7V battery.

R = (3.7 - 2.4) / .7 ==> 1.3/.7 ==> 1.86 ----> Rounded up to 2 Ohms

P = R * I^2 ==> 1.86 * (.7)^2 ==> .91W ---> Rounded up to 1W

Since there are only so many values that are commonly produced, it will be very likely that your will need to round UP to the next resistor size, to get the resistor that you would need (and as I did in the above example).

So, I will need a 2 Ohm, 1W resistor.

[Forgetful Jedi Knight]

Awwww, someone came across my tutorial.   

What current are you basing your calculations on?

[TheBaconWizard]

Resitance (ohms) = total voltage / current (amps)

Note: We are using milliamps not amps, so the result will need to be multiplied by 1000 afterwards.

Total voltage of the blue is 3.7v (battery) - 3.1v (forward voltage of LED) = 0.6v total

R= (0.6/350) x 1000

R = 1.71ohms. Always round up to the nearest available resistor value. So you need a 1.8ohm resistor.


Your calculation for the white is correct (I only checked R not watts output)


This is assuming all your LEDs are in parallel.

[Forgetful Jedi Knight]

@TheBaconWizard:  Running a LED at 350mA makes for a dull blade. The way I had the equation is accurate, and it's possible to express 350mA as .350 in the denominator of the equation.

P.S. The resistor you calculated is not the optimal resistor for Cree Blues.

[brthegreat117]

Hey all thanks for the help!

I have no clue as to how to do this. I know how to solder, but when it comes to math and resistors then my brain fries.
I am not sure about the current?
Someone on TCSS told me about having them all go at 1000mA, and connecting a resistor to each diode. I want the blues in parallel (I believe I am using the term correctly) and the white for FoC. I am not sure if my terminology is correct.
Would I need to only use one resistor for the blue dies or one each.

[Forgetful Jedi Knight]

I always recommend (and teach) one for each. a resistor is less than a buck, and a new Tri-Cree is over $20. Better safe than sorry. 

I saw the thread on TCSS.

[Darth Chasm]

Hey all thanks for the help!

I have no clue as to how to do this. I know how to solder, but when it comes to math and resistors then my brain fries.
I am not sure about the current?
Someone on TCSS told me about having them all go at 1000mA, and connecting a resistor to each diode. I want the blues in parallel (I believe I am using the term correctly) and the white for FoC. I am not sure if my terminology is correct.
Would I need to only use one resistor for the blue dies or one each.

Haha psst...FJK taught me ;)

[brthegreat117]

Then I will be using one for each :D. i am currently doing math. I thought that after my semester of college, I would never do math, but xxx lol.
I have my formula set up.
R=(3.7-(3.1-3.5))/I
I am not sure what the I is. Will it be the 350mA or 1000mA

I am subtracting the Blue from the White. If I do blues alone then I get 3.7/I

[Forgetful Jedi Knight]

Then I will be using one for each :D. i am currently doing math. I thought that after my semester of college, I would never do math, but xxx lol.
I have my formula set up.
R=(3.7-(3.1-3.5))/I
I am not sure what the I is. Will it be the 350mA or 1000mA

I am subtracting the Blue from the White. If I do blues alone then I get 3.7/I

You want to run everything at 1A (so you can see it). Your calculations are wrong either way, please go over the example I have in my tutorial, and you'll see how the numbers "relate". 

[TheBaconWizard]

@TheBaconWizard:  Running a LED at 350mA makes for a dull blade. The way I had the equation is accurate, and it's possible to express 350mA as .350 in the denominator of the equation.

P.S. The resistor you calculated is not the optimal resistor for Cree Blues.

Yes, if you wish to use a different current, it will obviously change the result of the calculations.

[brthegreat117]

I Feel like I am back in high-school asking my calculus teacher for help 

I want the desired current to run at 1000 mA (1A) to get an uber bright blue. So my value on the bottom should be 1A.
I am slowly unraveling the mystery that is Ohm's law.
Strangely I took Physics for 2 years in high school and still couldn't grasp the concept of Ohm's Law or Newton's 2nd Law.

[Forgetful Jedi Knight]

Take your time, and you'll "get it". 

[smaneesint]

So this is my understanding, please tell me if it's correct. I graduated in BA and this is like the first time since middle school doing math   

-----------------------------------------------------------
Here is my setup:

Tri-Cree Blue Green White

Blue Cree Part# XPEBBL-L1-0000-00301
45.7 Minimum Luminous Flux @ 350mA
Forward Voltage @1000mA = 3.47v

Green Cree Part# XPEBGR-L1-0000-00F01
122 Minimum Luminous Flux @ 350mA
Forward Voltage @1000mA = 3.58v

White Cree Part# XPEBWT-L1-0000-00F51
266 Minimum Luminous Flux @ 1000mA
Forward Voltage @1000mA = 3.15v
-----------------------------------------------------------

Blue:

Step 1: Resitance (ohms) = total voltage / current (amps)

-> R = (3.7-3.47)/1
-> R = 0.23

Step 2: Step 2: P (Watts) = R *( I^2)

-> P = 0.23*(1^2)
-> P = 0.23

Green:

Step 1: Resitance (ohms) = total voltage / current (amps)

-> R = (3.7-3.58)/1
-------------------
-> R = 0.12
===================

Step 2: Step 2: P (Watts) = R *( I^2)

-> P = 0.12*(1^2)
-------------------
-> P = 0.12
===================

And if I to use one resistor for a parallel Blue/Green I can just add their numbers up, am I correct?

R = 0.23 + 0.12 -> 0.35
P = 0.23 + 0.12 -> 0.35

This seems weird...now for the white

White:

Step 1: Resitance (ohms) = total voltage / current (amps)

-> R = (3.7-3.15)/1
-------------------
-> R = 0.55
===================

Step 2: Step 2: P (Watts) = R *( I^2)

-> P = 0.55*(1^2)
-------------------
-> P = 0.55
===================

Please comment on my calculation.