# How long will this LED last with this battery?

1. May 9, 2012

### samak

I have an Led with the following specifications:
3-3.2 V
Size (mm) : 5mm
Lens Color : Water Clear
Reverse Current (uA) : <=30
Life Rating : 100,000 Hours
Viewing Angle : 140 Degrees
Absolute Maximum Ratings (Ta=25°C)
Max Power Dissipation : 80mw
Max Continuous Forward Current : 24mA
Max Peak Forward Current : 75mA
Reverse Voltage : 5~6V

IF i use two or 3 LR41 batteries connected to it, how long will the power last? Do I really need to use a resistor for this setup?

If I use two or 3 LR44 batteries, how long will the power last?

If I use one CR2032 battery, how long will the power?

How can I solve these problems?

Last edited: May 9, 2012
2. May 9, 2012

### cepheid

Staff Emeritus
An ideal diode acts like a short circuit in one direction (so it will conduct and have 0 V across it) and like an open circuit in the other direction (it will not conduct). So the typical application for a diode is to put it in series with some other component that you want current to be able to flow across in one direction but not the other. (I realize that you have an LED whose typical application is to light up...but I am imparting this information about general diodes to you because it is relevant here).

A *real* (as opposed to ideal) diode will not have a dead short (0 V) across its terminals in the forwards direction, but will typically develop some small voltage across it when conducting. A typical number is 0.7 V for small signal diodes, but it could be much larger for a diode designed for higher power.

So, the answer is, yeah, you do want to put a resistor in series with a diode when it conducts, because if you don't, and you put a voltage across it that is much larger than 0.7 V (e.g. like 5 V), then that's equivalent to taking a 4.3 V source and putting a short ( a very low resistance wire) across the terminals...a tremendous amount of current is going to flow between those terminals (and through the diode). Much more current than the diode is rated to handle. You will burn out the diode.

So, how do you decide how much resistance to put in series with the diode? Suppose your diode has a forward voltage drop of V_f (i.e. V_f is the voltage that develops across the diode when it is conducting). You have to figure out what V_f is for your LED. Suppose also that your supply voltage is V_s (e.g. maybe V_s = 5 V).

Your circuit is set up like this: positive terminal of supply ---> resistor ---> LED ---> negative terminal of supply.

Then the voltage across the resistor will be V_s - V_f. Therefore, Ohm's law says the current across the resistor (and hence the diode) will be given by:

I = (V_s - V_f) / R

Pick a value of R such that the current I is at a safe level (no larger than 24 mA in this case).

To figure out how long your battery will last, just look up how much charge these batteries store in ampere-hours (Ah). Then take this number and divide it by the current to get the amount of time required to use up all the charge.

For example, if I look up "button cell" on Wikipedia, I see numbers on the order of 200 mAh being quoted for the charge amount. Therefore, the time would be:

(200 mAh)/(24 mA) = 8.33 h

EDIT: I used the *max* continuous current, and so this is a lower limit on how long the battery will last. I hope this helps.

Last edited: May 9, 2012