# LED Current Drawing Question

1. Aug 14, 2008

Hello,

You've probably already seen sites about "LED throwies" which are little circuits made of

An LED
A small 3V watch battery
A Magnet
Some Tape

The idea is that you make these and then stick them to different things and the glow and are pretty.

Anyway, with that background out of the way, I'm confused about something and I haven't been satisfied by the answers I've gotten when discussing it with people.

Your standard LED throwie apparently stays lit for a about a week, but the battery is only rated for ~180mAhr. This doesn't make too much sense to me because I see over and over again that LED's draw 20mA. If that is true, wouldn't they only last 9 hours?

My thinking is that one of the following is true in some form, but I don't know which is the most correct.

1. The LED itself has some current limiting resistance.
2. Because the 3v of the battery barely covers the threshold voltage, the LED won't be really on and therefore only draw a small amount of current
3. The battery for physical reasons, someway, only outputs a small amount of current and doesn't short to a larger one.

I know this is probably a simple problem, but it is one that I'm really confused on. Any help would be greatly appreciated.

2. Aug 14, 2008

### Redbelly98

Staff Emeritus
LED's (or any diode) do have some internal resistance, but in this case it is the battery's internal resistance that limits the current.

LED's are often specified to run at 20mA. That doesn't mean they always draw 20mA, it can be more or less depending on the power source they are attached to. Even with just 1 or a few mA running through it, there will be a visible glow from the LED.

3. Aug 14, 2008

### Oberst Villa

You should google for "Low Current LED", as far as I remember they can give sufficient brightness down to 2mA.

Are you sure the only electronic parts are the battery and the LED ? Depending on the colour of the LED, the threshold voltage can be below 2V (for red LEDs). If we assume 2mA, the internal resistor of the battery would have to be 500 Ohm to drop 1V. Are they really that high ? Maybe, I don't know enough about these little watch batteries. Or is there an external resistor hidden somewhere ?

4. Aug 14, 2008

### Redbelly98

Staff Emeritus
I imagine the current is actually rather high at first, probably even exceeding the 20 mA spec. As the battery is depleted, the battery voltage and LED current would drop towards the low levels we are talking about.

I've measured 15 to 20 ohm resistance for the 2032 size 3V coin cells when new. Resistance typically increases as the battery ages.

5. Aug 15, 2008

### Oberst Villa

Thanks for the data. So maybe what I wrote in #3 about low current LED is a bad idea - might exceed the max. current rating. Still I think it would be best to include an external resistor, but OK, at least it's possible with just a LED and battery.

6. Aug 15, 2008

### cabraham

An LED must be driven with *current*, not voltage. The forward voltage drop of an LED varies with temperature and forward current. Also, the reverse saturation current Is, (I = Is * ((e^(V/Vt)) - 1)) is strongly temperature dependent with a positive coefficient. If a low impedance constant voltage source is connected across the LED, the power (current times voltage per previous equation) results in a temperature increase, which raises Is, which raises the current and the power, which further raises Is, further raising the current and power, Is increasing more etc. etc. This is thermal runaway.

With constant current drive V = Vt * (ln((I/Is)+1)). An increase in temp results in an increase in Is which results in a **decrease** in voltage. Hence thermal stability is immediately achieved.

But, constant current sources are not always available. If the primary power source is constant voltage, adding a resistor provides stability. Any increase in forward current results in an increase in the voltage across the resistor, producing a **decrease** in the LED forward voltage. Thermal stability is assured.

I recommend around 5 to 10 mA of forward current if you want long steady lamp life. Todays LED lamps are quite bright without the need for driving them hard. Just subtract the forward LED voltage from the supply voltage, and divide by the current, 5 to 10 mA, and that is the resistance value. If the supply is 12V, and the forward drop is 2.0V, and the current is set to 10 mA, then (12V - 2.0V)/(10mA) = 1.0 kohm. Does this help?

Claude

7. Aug 16, 2008

### Redbelly98

Staff Emeritus
Claude is entirely right about the "correct" way to power an LED.

But the fact remains that many cheap key-chain LED flashlights are sold commercially that are powered with coin cell batteries and no resistor. Coin cells are not "constant voltage" sources, so it works. Not the ideal way to run an LED, but they are cheap and do what they need to do.