loading

How exactly should an LED circuit be properly set up?

I know LEDs need resistors or else they'll die out really fast.
The problem is, I'm having trouble figuring out how this needs to be done.


Lets say I take this LED:


White 70° Diffused 5mm LED:
Luminous Intensity: 690mcd typ. @ 20mA
Max Forward Current: 30mA
Pulse Current: 100mA for <= 10ms, duty <= 1/10
Forward Voltage: 3.6V typ. 4.0V max @ 20mA
Max Reverse Voltage: 5V
Power Dissipation: 120mW
Operating Temp: -30 to +85 C
Soldering Temp: 265 C for 10 secs
Max Reverse Current: 50uA @ 5V


ok so it has a current of 20mA and a forward voltage of 3.6V.
Now, If I wanted to power 6 of these with AA or AAA batteries How many would I need and what ohm resistors would I need?

It's been a long time since I took a class on electrical currents, and it was pretty basic.
Also, looking at the Reverse Voltage, what is that? Is that if I put the batteries in the opposite direction, i.e. Send the voltage INTO the - and out the + ends?

I've been researching this but its hard to find the right answer.

Lets say I'm making a basic serial circuit with the batteries here (lined up).


seandogue6 years ago
I'll start with the reverse voltage

LEDs are diodes, which, for all intents and purposes (or nearly all) are one way valves.

The reverse voltage is the maximum "pressure" in the reverse direction (against the blocking "flap" of the diode, that the diode can take before the "flap" breaks.

How you hook them up kinda depends.

Let's say you just want to connect one LED using AA or AAA batteries, or for that matter, using any 1.5V per cell battery as the "cell" basis.

Since the forward voltage of the led is 3.6 V, you'll need 3 batteries in series to exceed the necessary forward voltage, for an expected source voltage of 4.5VDC.

to absorb the excess voltage, you'll need to "drop" 4.5-3.6 V = 0.9 V

Since the expected load current is 20mA,

And since the required drop voltage is 0.9V,

BY Kirchoff's law (V = I x R)

you'll need V/ I = R or

0.9V/0.020A = 45 ohm.

51 ohm is a standard value. I'd use a slightly higher value than the exact or lower value, just to give yourself a bit of comfort zone, and to ensure that you only need a single resistor. Doing a calculation and coming up with "I need a 48.34 ohm resistor" over complicates circuit design in most cases...

Finally one can find the required Wattage of the load resistor.

Wattage is power . The governing equation is

P = Vload x Iload

Since we know that Vload = 0.9V (for the resistor, rememebr)

and it's Iload = 20mA

Then P = 0.9V x 0.020A = 0.018 W

You can safely use any 1/4 W resistor.

In fact, if you have them, you can use a 1/8 W resistor to save space.

This is the basic methodology. I'm going to leave the rest to you and other respondents.

One last thing. I don't talk like this in general (flaps, etc.) , really, except to folks who don't understand, so please take that stuff for what it is, illustrative in "common speak"

 
One last thing. Use your brain and skip the online calculators if you can. They're mindnumbing and do little for your long term understanding. Learn the fundamentals and you'll never need to use their crutch again. I'm sure the suggestion to use them will come within the day.

For multiples run in parallel using the same basic configuration duplicated, or increase the battery pack voltage by adding batteries to run leds in series., or do a combination approach of three pairs of leds, where each pair is run in series (with a load resistor) then the individual units in parallel... Try to figure the unit again, with two trios (and each having a load resistor)

And skip the damned calculator. This stuff is highschool level math..barely..I was doing more complicated math in 7th grade.
ONe more last last thing.

I sure do wish instructable s would make it so we could edit our answers already. I can see things I'd like to change in my first bit, but I'm just not going back and doing it all over or cutting and pasting the entire thing

Things to review

"drop" => it's actually dissipate. resistors dissipate energy by conversion to heat. they're interchangeable but dissipate is the correct term.

Kirchoff's law. We often pay simple mathematical games with the equation

V = I x R is equivalent to
V/ I = R is equivalent to
V/R = I

I could have done what's known as "worst case analysis" based on the typical and max values for the forward voltage. look it up. neat stuff and can benefit your designs if you keep going beyond this needed led circuit.

sompin ta think about.