# LED' s resistor formula

Discussion in 'Electronic Basics' started by Lee, Aug 3, 2006.

1. ### LeeGuest

I am looking fot the math formula to figure out the voltage drop i need
and the associated resistor. such as the led uses 3 volts i have 12 i
need a resistor that will drop 9 volts. also how to figure out the size
resistor (watts).
thanks
Lee

Lee, Aug 3, 2006

2. ### Tom BiasiGuest

"Lee" <> wrote in message
news:...
>I am looking fot the math formula to figure out the voltage drop i need
> and the associated resistor. such as the led uses 3 volts i have 12 i
> need a resistor that will drop 9 volts. also how to figure out the size
> resistor (watts).
> thanks
> Lee
>

You need the forward voltage drop of the LED.
The desired operating current.
The source voltage.

How much does the resistor drop and what power is it dissipating?
E=IR, P=EI
Hint: the resistor only drops what the LED doesn'nt.
Can you use basic algebra to manipulate these formulas?
Give it a shot.
Tom

Tom Biasi, Aug 3, 2006

3. ### Noway2Guest

Lee wrote:
> I am looking fot the math formula to figure out the voltage drop i need
> and the associated resistor. such as the led uses 3 volts i have 12 i
> need a resistor that will drop 9 volts. also how to figure out the size
> resistor (watts).
> thanks
> Lee

First, determine the amount of current you wish to have flow through
the LED. You will need to get this information from the datasheet.

Once you have this piece of information, you can apply ohms law to
determine the needed value of resistor. Take the supply voltage minus
the voltage drop of the LED at the desired current and divide by the
desired current. This will give you the needed resitance. Typically,
I use 10ma for a general LED and using that value as an example with
your numbers: 12V - 3V = 9V / 10ma = 900 ohms. You can then use this
figure to pick the a standard resistor that is near this value. Once
you have the resistor picked, double check what current values this
will produce and verify that it is within acceptible range.

Noway2, Aug 3, 2006
4. ### HKJGuest

Lee wrote:
> I am looking fot the math formula to figure out the voltage drop i need
> and the associated resistor. such as the led uses 3 volts i have 12 i
> need a resistor that will drop 9 volts. also how to figure out the size
> resistor (watts).
> thanks

Your could also use this program:
http://www.miscel.dk/MiscEl/miscelLeds.html

It has many different LED circuits for both AC and DC and it will
calculate all components value for your.

HKJ, Aug 3, 2006
5. ### DJ DelorieGuest

R = (Vcc - Vf) / If

Vcc = supply voltage
Vf = LED's forward voltage
If = desired forward current

Note that "Vf" needs to include Vol of your output logic, transistor,
whatever - i.e. you need to calculate the actual voltage drop across
the resistor in your specific case, so you can calculate the ohms
needed to pass the desired current.

DJ Delorie, Aug 3, 2006
6. ### John FieldsGuest

On 3 Aug 2006 07:17:01 -0700, "Lee" <> wrote:

>I am looking fot the math formula to figure out the voltage drop i need
>and the associated resistor. such as the led uses 3 volts i have 12 i
>need a resistor that will drop 9 volts. also how to figure out the size
>resistor (watts).

---
View in Courier:

E1 E2
/ /
Vcc---[Rs]---[LED>]--GND
It-->

The current limiting resistor required is:

E1 - E2
Rs = ---------
It

The resistor's dissipation will be:

Pd(Rs) = (E1 - E2) * It

Per your example, and assuming an LED current of 20mA :

E1 - E2 12V - 3V
Rs = --------- = ---------- = 450 ohms
It 0.02A

The closest standard 5% resistor which will keep the current at less
than 20mA is 470 ohms, so the new current in the circuit will be:

E1 - E2 9V
It = --------- = ------ = 0.019A
Rs 470R

and the rsistor will dissipate:

Pd(Rs) = (E1 - E2) * It = 9V * 0.019A = 0.172 watts,

so a standard 450 ohm +/- 5%, 1/4 watt carbon film resistor would be
OK to use.

--
John Fields
Professional Circuit Designer

John Fields, Aug 3, 2006
7. ### Peter BennettGuest

On 03 Aug 2006 11:43:20 -0400, DJ Delorie <> wrote:

>
>R = (Vcc - Vf) / If
>
>Vcc = supply voltage
>Vf = LED's forward voltage
>If = desired forward current
>
>Note that "Vf" needs to include Vol of your output logic, transistor,
>whatever - i.e. you need to calculate the actual voltage drop across
>the resistor in your specific case, so you can calculate the ohms
>needed to pass the desired current.

....however, LEDs will work fine over a fairly wide range of current,
so you don't need to get _too_ scientific in your calculations.

The current rating given on a LED data sheet is frequently the maximum
recommended current, so you should plan on operating the LED at a
somewhat lower current, unless you really need maximum brightness.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter

Peter Bennett, Aug 3, 2006
8. ### DJ DelorieGuest

Peter Bennett <> writes:
> ...however, LEDs will work fine over a fairly wide range of current,
> so you don't need to get _too_ scientific in your calculations.

Yeah, I usually calculate a minimum resistor value, and another for
5mA, and use whatever's handy between those two.

DJ Delorie, Aug 3, 2006
9. ### Bill BowdenGuest

Lee wrote:
> I am looking fot the math formula to figure out the voltage drop i need
> and the associated resistor. such as the led uses 3 volts i have 12 i
> need a resistor that will drop 9 volts. also how to figure out the size
> resistor (watts).
> thanks
> Lee

1. Subtract LED voltage from battery voltage. (12-3=9)
2. Divide result (1) by desired current to obtain resistor value.
(9/.02 = 450)
3. Square current and multiply by resistor to obtain resistor power.
(.02^2 * 450=180mW)
4. Use calculator to check answers.

http://ourworld.compuserve.com/homepages/Bill_Bowden/led.htm

-Bill

Bill Bowden, Aug 3, 2006