PDA

View Full Version : LED Resistors and Ohm's Law



RevengeoftheSeth
08-05-2010, 08:53 PM
So I've been studying Ohm's law at the suggestion of some of the sticky topics on these forums and I think I have a pretty firm grasp of the concept, now. Basically I=Current in Amps, V=Voltage, and R=Resistance in Ohms.

The formula is R=V/I.

I do have a question on how I figure out what resistor I need for the correct LED.

I will use a blue Lux III for example.

Forward Voltage = 3.9 V
Current = 1 amp
Battery Voltage = 6 V

My question is which voltage do I formulate off of - the battery or the forward voltage of the Lux III? And what do the Watts correlate to on the resistors?

Thanks!

EDIT: RAWR! I figured it out. I scowered the boards some more and found out that V is not just one voltage. Its the difference of the supply voltage and the forward voltage of the LED. That made a lot more sense. Now I just need to find out how many watts I need on the resistor. Can someone throw me a bone on that one?

cannibal869
08-05-2010, 11:42 PM
is this an academic pursuit - i.e. are you asking this question to try and understand better the mathematics behind the calculation?

I only ask cause there is an LED resistor chart link on the front page of the store and forums... and there are plenty of online resistor calculators available...

-C

Don Se Wion
08-06-2010, 02:20 AM
Watts can be calcutaled with the formula

W = V x I

In your case you should have about 2.1V across the resistor with 1A passing through, so the resistor will need to dissipate 2,1 watts.

In theory you could probably go for a 2W resistor but I would suggest to get a 5W.
They are more bulky but this way you can stay on the safe side cause they can get pretty hot.

Rhyen Skytracker
08-06-2010, 04:51 AM
Here is the formula to find the resistance you need for a LED.

(V source - V forward) / LED current = R

Say you had a 6 Volt power source and you had an accent LED that has a forward voltage of 3 volts and has a current of 20 mA or .02 A. Here is the formula with values in it.

(6V - 3V) = 3V
3V / .02 = 150

So you would need a 150 ohm resistor. To find the wattage of the resister you would multiply the LED voltage by the LED Current. Keep in mind that this formula will give you the exact resistance needed and resistors only come in certain sizes. If you want to overdrive a LED use a slightly lower ohm resistor and if you want to dim down a accent LED a bit then use a larger ohm resistor.

3V x .02A = 0.06 W

That means a 1/8 watt resistor will be fine. (If you need to convert the fraction to decimal divide 1/8 which = 0.125) since 0.06 W is less than 0.125 watt you can use the 1/8 resistor. You can always use a larger wattage resistor but don't use smaller wattage resistors.

RevengeoftheSeth
08-06-2010, 07:42 AM
is this an academic pursuit - i.e. are you asking this question to try and understand better the mathematics behind the calculation?

I only ask cause there is an LED resistor chart link on the front page of the store and forums... and there are plenty of online resistor calculators available...

-C

Yes, this is an academic pursuit. I am well aware of the resistor chart. I don't want to have to refer to it each time I change out LEDs, though.

RevengeoftheSeth
08-06-2010, 07:44 AM
Here is the formula to find the resistance you need for a LED.

(V source - V forward) / LED current = R

Say you had a 6 Volt power source and you had an accent LED that has a forward voltage of 3 volts and has a current of 20 mA or .02 A. Here is the formula with values in it.

(6V - 3V) = 3V
3V / .02 = 150

So you would need a 150 ohm resistor. To find the wattage of the resister you would multiply the LED voltage by the LED Current. Keep in mind that this formula will give you the exact resistance needed and resistors only come in certain sizes. If you want to overdrive a LED use a slightly lower ohm resistor and if you want to dim down a accent LED a bit then use a larger ohm resistor.

3V x .02A = 0.06 W

That means a 1/8 watt resistor will be fine. (If you need to convert the fraction to decimal divide 1/8 which = 0.125) since 0.06 W is less than 0.125 watt you can use the 1/8 resistor. You can always use a larger wattage resistor but don't use smaller wattage resistors.

Your assistance is much appreciated. That is exactly what I was looking for. I think I now have a full grasp on the relations of Watts, Current, Voltage and Resistance. This forum has been an excellent resource for learning.

Thanks to everyone else for their help, as well.

Jay-gon Jinn
08-06-2010, 09:43 AM
Here's an entire topic devoted to Ohm's Law:
http://forums.thecustomsabershop.com/showthread.php?t=9360
Sorry it took so long to post it, but I had to remember who posted it to find it. ;) Looks like your question has been answered already, though.

There was a similar post over a FX-Sabers also, here's the info from that one:
Calculating an LED resistor value

LED resistor circuit

An LED must have a resistor connected in series to limit the current through the LED, otherwise it will burn out almost instantly.

The resistor value, R is given by:

R = (VS - VL) / I

VS = supply voltage
VL = LED voltage (usually 2V, but 4V for blue and white LEDs)
I = LED current (e.g. 20mA), this must be less than the maximum permitted

If the calculated value is not available choose the nearest standard resistor value which is greater, so that the current will be a little less than you chose. In fact you may wish to choose a greater resistor value to reduce the current (to increase battery life for example) but this will make the LED less bright.

For example
If the supply voltage VS = 9V, and you have a red LED (VL = 2V), requiring a current I = 20mA = 0.020A,
R = (9V - 2V) / 0.02A = 350ohm, so choose 390ohm (the nearest standard value which is greater).

Ohm's Law

Next Page: Power and Energy
Also See: Voltage and Current | Resistance | Resistors

To make a current flow through a resistance there must be a voltage across that resistance. Ohm's Law shows the relationship between the voltage (V), current (I) and resistance (R). It can be written in three ways:

V = I × R

or

I = V / R

or

R = V / I

where:

V = voltage in volts (V)
I = current in amps (A)
R = resistance in ohms (ohm) or: V = voltage in volts (V)
I = current in milliamps (mA)
R = resistance in kilohms (kohm)

For most electronic circuits the amp is too large and the ohm is too small, so we often measure current in milliamps (mA) and resistance in kilohms (kohm). 1 mA = 0.001 A and 1 kohm = 1000 ohm.

The Ohm's Law equations work if you use V, A and ohm, or if you use V, mA and kohm. You must not mix these sets of units in the equations so you may need to convert between mA and A or kohm and ohm.

The VIR triangle

V

I R

Ohm's Law
triangle

You can use the VIR triangle to help you remember the three versions of Ohm's Law.
Write down V, I and R in a triangle like the one in the yellow box on the right.

* To calculate voltage, V: put your finger over V,
this leaves you with I R, so the equation is V = I × R
* To calculate current, I: put your finger over I,
this leaves you with V over R, so the equation is I = V/R
* To calculate resistance, R: put your finger over R,
this leaves you with V over I, so the equation is R = V/I

Ohm's Law Calculations
Use this method to guide you through calculations:

V

I R

1. Write down the Values, converting units if necessary.
2. Select the Equation you need (use the VIR triangle).
3. Put the Numbers into the equation and calculate the answer.

It should be Very Easy Now!

* 3 V is applied across a 6 ohm resistor, what is the current?
o Values: V = 3 V, I = ?, R = 6 ohm
o Equation: I = V/R
o Numbers: Current, I = 3/6 = 0.5 A


* A lamp connected to a 6 V battery passes a current of 60 mA, what is the lamp's resistance?
o Values: V = 6 V, I = 60 mA, R = ?
o Equation: R = V/I
o Numbers: Resistance, R = 6/60 = 0.1 kohm = 100 ohm
(using mA for current means the calculation gives the resistance in kohm)


* A 1.2 kohm resistor passes a current of 0.2 A, what is the voltage across it?
o Values: V = ?, I = 0.2 A, R = 1.2 kohm = 1200 ohm
(1.2 kohm is converted to 1200 ohm because A and kohm must not be used together)
o Equation: V = I × R
o Numbers: V = 0.2 × 1200 = 240 V

Rian Jardrit
08-15-2010, 10:55 PM
ah this will help me out as well in picking the right ones! :D This really was a case of "search the forums and you will find out" moment! :P

RevengeoftheSeth
08-16-2010, 05:18 PM
ah this will help me out as well in picking the right ones! :D This really was a case of "search the forums and you will find out" moment! :P

Yeah, the resistor chart on the store is very helpful, but it is a little out of date and it's great to be able to do the calculations on your own in the event you run into something that is outside of the norm.

morpheus1977
08-26-2010, 10:20 AM
How let me ask you this, how would this work if you want to chain lets say 8 accent LED's directly driven off of 1 power source?

Arkhan
08-26-2010, 10:50 AM
you would have to come up with a compatible array of LEDs. If you wire them all in series, you can't deal with the forward voltage requirement.

If you wire them all in parallel, you end up putting a resistor on each of them, or a very large resistor in front of them.

If you create an array, mixing parallel and series, you can keep the forward voltage reqs down into manageable range, while reducing the # and wattage of resistors necessary to protect them from over current.

If you google for LED parallel/series calculator, there is one I have used that will take the info about your LEDs, your power source, how many of them you have and create a workable array for you.

morpheus1977
08-26-2010, 11:15 AM
so if I want to wire up let say six of these
http://www.radioshack.com/product/index.jsp?productId=3097460
using 3 AA
this is how I would want to wire it up?

Then I would just place a mom switch on the battery neg wire.
after looking at the radio shack website it looks like they dont have exactly what the calculator recomended.
It suggested.
1/4 w
56 ohm

the closest I could find was
1/2 w 47 ohm. so it looks like they will be overdriven. a little.

RevengeoftheSeth
08-26-2010, 11:32 AM
Doesn't Makoto just wire his all in series? The voltage stays the same, but the mA stacks per LED.

For instance, you have 10x 3V 20mA LEDs which requires 3V 200mA to power them sufficiently.

Rhyen Skytracker
08-26-2010, 12:37 PM
In series the voltage will split and the current stay the same, it is parallel where the current will split and the voltage stay the same.

Morph,
You can use down to a 25 ohm resistor. Those LEDs have a max recommended voltage of 4 V so:

(4.5V source - 4 V forward) / .02A = 25 ohm resistor.

morpheus1977
08-26-2010, 12:48 PM
Thanks Rhyen
the lowest I could find is 33 so I will use that. I want these nice and bright. I will be using diffusers with them.

But this is going to be ugly wired up.(yea it will be in the lancer but still ugly)

quick question instead of having to buy a three AA battery pack. I still have the battery sled from my MR Vader saber can I use that?

Rhyen Skytracker
08-26-2010, 01:04 PM
Yes, you can use it if it will fit.

morpheus1977
08-26-2010, 01:19 PM
ok got it can somebody tell me which is pos and which is neg?

*edit* never mind figured it out LOL