Should I use a resistor in this scenario?

I am building a IR pen and wondering if I need a resistor for this. Here is my specifications for my IR LED:

• Forward Voltage: 1.2 V
• Forward Current 100 mA

So on and so on. I am powering it with a AA battery. According to Ohm's law, I would only need a 3.3 ohm resistor, and that is when the battery is at 100%. That seems unnecessary, but I also don't want to damage my LED because I want it to last a long time (with a couple of battery replacements).

Link to calculator and specs (Not that most people would need the calculator but it has a diagram of my circuit).

P.S. Should I put a resistor on just so it doesn't draw too much current? What resistor would I need?

• – Chris Laplante Feb 2 '13 at 22:01
• This is NOT a duplicate of the above question. This deals with battery voltages with time, as well as LED driving. More to learn here. – Russell McMahon Feb 3 '13 at 1:12

Yes - you need a resistor.

Your understanding of batteries is wrong.
If that is an AA Alkaline or carbon-zinc battery it will start with Vout ~= 1.5V BUT this will fall to about 1.0V across the lifetime of the battery.

A good place to look for a wide range of battery information is Battery University - that link is to their page discussing AA Alkaline batteries.

The graph below is from the above page. IGNORE the time scale and stated load. The red curve shows the typical discharge voltage with time for a constant current load - which is what you want to have. As can be seen, the voltage varies widely as the battery discharges, so your resistor required would change , and at 1.2V you reach the LED NOMINAL VF with about 25% of battery capacity remaining.

One solution is to use two batteries with similar characteristics and a constant current regulator. Such a regulator can be very simple and cheap - two "jellybean" (= common. cheap, nothing special) transistors and a very few resistors.

A possible alternative (which needs an expensive battery) is to use an AA LiFeS2 "Lithium" Battery. This has a characteristic much closer to what you had in mind originally.

You need to say what LED you intend to use. Here is a datasheet for an Everlightht IR323_H0_A rated at 100 mA. In the Digikey brief description they say it has a typical Vf of 1.2 V
BUT in fig4 of the datasheet they say Vf is typically 1.5V at 100 mA.
AND On page 3 it says 1.4 typical to 1.8V max at 100 mA
You'd have ahard time designing for this LED, even with the Lithium Battery. The LED you intend to use is liable to have similar uncertainties in its spec. Using a higher voltage and a current regulator (or even a higher voltage and a resistor) will work much better. The Li battery is better than nothing.

http://www.everlight.com/datasheets/IR323_H0_A_datasheet.pdf

What if I used one Lithium battery?

IF Vf = 1.2V and if Vlithium = 1.45V (a compromise) then for 100 mA.
R = V / I
= (Vbat - V_LED) / 100 mA
= (1.45 - 1.2) 0.1 A = 0.25/ 0.1
= 2.5 Ohms.
There is so much uncertainty that a 2.7 Ohm or 2.2 Ohm would be OK enough. Even a 1.8 Ohm MAY suit.
A "bad" LED with high Vf would reduce the current greatly.