How long will this LED last with this battery?

[samakbass]

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 one LR41 battery connected to it, how long will it last? Do I really need to use a resistor for this setup?


If I use one LR44 battery, how long will it last?

If I use one CR2032 battery, how long will it last?



How can I solve these problems?

 

[godfreyl]

LR41 and LR44 won't work. They are 1.5V batteries. The LED needs at least 3V. A CR2032 battery might work.

 

[nikhildascl]

Resistor is must so as to limit the current.

 

[enjunear]

First, you need a source with a voltage greater than the LEDs maximum forward bias voltage (>3.2V). Make it a volt or so extra, so you can use a reasonable value of current-limiting resistor (yes, you need one... you always need a current-limiting device of some kind with an LED!)

For run-time calculations, you need to look up/calculate the number of Amp-hours that the battery is rated for. For a small cell like the ones you mention, they might be given in mA-hrs. It should be pretty obvious how to estimate run time by looking at the units, but just as an example:
Your battery pack had a rating of 200 mA-hrs, and the LED is set up to draw 20 mA (stay a little below max continuous fwd current, for saftey). 200 mA-hrs / 20 mA = 10 hours. It's a rough estimate, but it's a place to get you started.

 

[tpetar]

This is discharging curve for CR2032 :

**broken link removed**

**broken link removed**


LR41 is 1,5V battery! You need 3V.

 

[FoxyRick]

A lot of pocket-sized and keyring-style LED flashlights have the LED directly driven from a small button cell, either lithium or two alkaline. Generally, the internal resistance of these small cells also helps limit current if the voltage is close to the recommended Vf for the LED.

It won't get the most light out of the LED, but it will light. Depends how much light is needed.

As to how long the LED will light when used like that, it's difficult to estimate because current drawn will drop as the batteries become exhausted (less so with lithium). Also, at what point to you consider the light to be no longer bright enough...?

 

[samakbass]

When you say "brightness" do you mean mcd? Or how is brightness measured when you talk about LED's? The LED's I used in the example have 20,000 mcd. So I would guess that 10,000 mcd will be the lowest acceptable brightness for my purposes. Maybe I should just get an led with a rating of 10,000 mcd to begin with?

 

[tpetar]

When you say "brightness" do you mean mcd? Or how is brightness measured when you talk about LED's? The LED's I used in the example have 20,000 mcd. So I would guess that 10,000 mcd will be the lowest acceptable brightness for my purposes. Maybe I should just get an led with a rating of 10,000 mcd to begin with?

Do you have datasheet of that led or model name ? You may be surprised with darknes... :-D

 

[FoxyRick]

Without getting into lux/candela/lumen comparisons, yes, by brightness I mean mcd output.

Personally I would go for the highest efficiency* LED, with at least or preferably more intensity (mcd) at the required beam divergence, that you can get.

It's always easy to lower the output (with a larger resistor for instance), and that will give longer running time too. By the way, the term for running an LED directly from a cell with no current limiting is termed 'direct drive' by flashlight geeks like me.


*Efficiency: for this purpose = mcd/watt (where watts = volts x amps)

 

[D.A.(Tony)Stewart]

In order to design this, one needs to fully understand the LED and battery specs,. Some LED specs are missing that are essential include the Vf at rated voltage. If we assume by convention, Vf is given at rated max steady current, Vf=3.0~3.2 @24mA Since the Battery is only rated at 3.0max @ no load. but the load drop of .1V at 0.19mA implies an ESR of 526Ω. Since the LED has an IR or ESR of < 20Ωcapacity.

Now consider what the LED voltage is a low current. 20Ω*24mA = 0.48V,
so the LED Vf@0mA ≈ (3.0-.48) to (3.2-.48) or 2.5V to 2.7V
Thus for a battery ESR of 526Ω connected direct to the LED,
we get a range of LED voltages with internal battery drop giving..of 526Ω
and ESR of LED much smaller @20Ω. (IR on graph)
Vbat/Rbat = (3.0-2.5)V / 526Ω ≈ 0.95 mA max
to (3.0-2.7)V / 526Ω ≈ 0.57 mA min

ANSWER:
However the battery with a lighter load of 0.19mA will last 1000hr to 2.7V and 1100hr to 2.5V.

So assuming a constant battery capacity of 0.19*1000 = 190mA-hr and using calculated currents, the battery life time will be 190/0.95 =200hr and 190mA-h/.57mA = 333hr
... with no added resistor,

but these are poor batteries and that is a very low current ...
...compared to rating of 25mA = 2% to 4% of rated brightness. (dim but visible)


However I happen to know where Vc values for LEDS binned to a lower voltage and batteries with a lower ESR and get much more candella's for the same time.

eeff I am correct then I been good student and score high? mebbe not batman.. you decide.