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
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.
Your could also use this program:
It has many different LED circuits for both AC and DC and it will calculate all components value for your.
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.
--- View in Courier:
Your circuit is:
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
...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 Vancouver Power Squadron: http://vancouver.powersquadron.ca
Yeah, I usually calculate a minimum resistor value, and another for
5mA, and use whatever's handy between those two.
-Bill
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