# Quick question, how does LED power consumption work?

Discussion in 'The Projects Forum' started by Imdsm, Feb 15, 2011.

1. ### Imdsm Thread Starter Member

Feb 11, 2011
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Just a quick question that I'm sure one of the veterans here will be able to answer in their sleep. I've just had a chat with a friend of mine who used to do electronics as a hobby a while ago and he's said something that has confused me - either my understanding is wrong or he's rusty.

Now what he says makes sense kinda, but the way I see it, he's got the wrong end of the wire.

So, the question: if you have two circuits both with 9v batteries, and you put three LEDs onto the first circuit with a resistor there too, drawing 20 mA across each of the LEDs, and on the second circuit you have an individual resistor for each LED, which will drain the battery quicker?

The way I see it is, volts and amps are interchangeable, you can have 9v at 100mA or turn that into 18v at 50mA, or thereabouts, so when you put three LEDs in at 2v each, you're squeezing 6v out of the battery for the LEDs alone, and the resistor only squeezes 3v out of it. The current drain would only be 20mA. But with having three separate LEDs and resistors, you are using only 2v per LED and the resistors will be using 7v each. Each node will then use 20mA each, totalling 60mA - so the second battery will drain three times as fast as the first one.

Am I right in my reasoning?

2. ### SgtWookie Expert

Jul 17, 2007
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You're close enough to right, for now.

Batteries have internal resistance. They're rated in mAh, and the mAh rating is for a constant current over a 20 hour period until the voltage drops below a threshold value. So, a 500mAh battery would sustain a 25mA load for 20 hours before falling below the threshold value.

If you draw more current than that, the power consumption in the battery itself becomes excessive. You might think since it's rated for 500mAh, you could draw 500mA for an hour instead of 25mA for 20 hours, but it doesn't work like that; you'd probably have a dead battery inside of 20 minutes.

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3. ### Imdsm Thread Starter Member

Feb 11, 2011
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Thanks, that was how I understood batteries to work and I'm glad to have that part clarified.

But just to clarify on the LED side though, do three LED's in a series with a single resistor use less power* than three leds in parallel each with a resistor of it's own?

* is this the right term to use when talking about consumption?

4. ### SgtWookie Expert

Jul 17, 2007
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If you have three LEDs in series with a suitable resistor to limit current to 20mA, then you have a single circuit that will consume 9v*20mA = 180mW.

If you have three separate LEDs with suitable resistors in series to limit each LEDs' current to 20mA, then you will have 9v*20mA*3 = 540mW power dissipation in the resistors and LEDs.

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5. ### Imdsm Thread Starter Member

Feb 11, 2011
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Ah, that's exactly the response I was looking for.

Thanks a lot SgtWookie!

+1

Dec 26, 2010
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The set-up using 3 LEDs may or may not be practicable however. Depending on the LED voltage requirement, this arrangement may not be able to work, and continue working down to to a reasonably low voltage as the battery discharges. This depends on how much the voltage for three LEDs adds up to, chiefly a function of the LED colour. For instance, three red LEDs at say 1.8V each would give 5.4V, which could work. Three white LEDs needing 3.3V each gives 9.9V total, and that would not do.

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7. ### Imdsm Thread Starter Member

Feb 11, 2011
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Indeed Adjuster, the LED's I use (one red one blue one ultraviolet) add up to 8.8v, which works very nicely, at 20mA it only requires a 10 Ohm resistor.

The main thing I was interested in is the economy of squeezing every little bit out of the voltage. Rather than wasting power on larger resistors, the more I can cram on the better for efficiency.

Cheers

8. ### SgtWookie Expert

Jul 17, 2007
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It's not going to work well like that; as soon as your battery drains a bit, your LED current will drop a good bit.

You need more "margin" on top. I usually start off with 1v or ~10%, whichever is larger.

Another option you might consider is using a boost converter to keep the current through your LEDs more consistent.

I don't have the time to go into that at the moment, tho.

Dec 26, 2010
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But how much of the battery's milliamp-hour capacity can be used if your total LED voltage comes to 8.8V? Normally batteries are used until the voltage has fallen considerably below the nominal value.

With a 9V alkaline battery, the voltage will fall towards 8.8V relatively quickly and the LEDs will grow dim. What type of battery are you using?

10. ### wayneh Expert

Sep 9, 2010
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If that's voltage drop (min voltage, not max), your LEDs will go off very shortly after you turn them on, as soon as the battery voltage drops even a tiny amount.

11. ### Imdsm Thread Starter Member

Feb 11, 2011
39
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The batteries I use to test are duracell procell 9v's, they're supposed to last longer than standard retail batteries.

The point you make about not leaving much margin makes sense, but apart from testing some circuits with a battery, for any long term running I will be using a power supply (you'd have to be crazy not to really).

Would a 'boost converter' be required/handy even when using power supplies?