# Calculating resistance - no idea where to start

Discussion in 'General Electronics Discussion' started by lynkynpark86, Apr 1, 2012.

1. ### lynkynpark86

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Oct 22, 2011
How do I calculate the resistance I need fro a circuit? I have 2 bright LEDs I want to power with 2 "AA" batteries. Here's the info on the LEDs:
FW current 25mA; FW supply 3.3 (typical), 3.6V (maximum)
Would this need resistance, and if so, how much for both in series (nothing else in circuit)?

7,682
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Jan 5, 2010
3. ### Harald KappModeratorModerator

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Nov 17, 2011
Look for "joule thief" for an inexpensice method to power 3.3 V LEDs from 3 V. However, a joule thief cannot ascertain any defined current through the LEDs.

A robust solution would be to use a specialized step-up LED driver. Here http://www.linear.com/products/step-up_(boost)_led_drivers you can find more info.

Harald

4. ### timothy48342

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Nov 28, 2011
mmm... I think Harold and Bob are both correct. No disagreement here, but I was thinking....

While 3V isn't going to bring a 3.3V LED to full brightness, it might be close. you could hook it up and see. You won't need a resistor, as has been said, so there's no danger of destroying it. If it is bright enough for your application, then your good to go. The only thing is that as soon as the batteries even start to loose their full charge, you'll start to see it dimming. My thoughts on that, though, is that your drawing so little from the batteries, it might take quite a while for that to happen.

I think, experiment a bit and see.

--tim

5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
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Jan 21, 2010

Say the batteries are new and have a terminal voltage neat 1.7 volts. Two of them will give 3.4 volts. This will easily forward bias the LED and may cause enough current to pass through it to destroy it. If it doesn't do it immediately, thermal runaway will be an issue -- it may get brighter and brighter until it goes dark.

As the batteries go flat, the LED will begin to dim very quickly (assuming it hasn't burnt out) and it will go out almost completely at (say) 3.2 volts (it may be a little lower than that). The batteries will still be almost fully charged at this point.

Three cells and a resistor would be the simplest solution.

6. ### timothy48342

218
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Nov 28, 2011
I been trying to figure how to defend myself, here. But.... Not happening.

I guess I was imagining that the internal resistance of the AAA's would protect the LED from a very small amount of overvoltage. But in looking into it, it won't.

The internal resitance of (double) AA's can be well under an ohm when warm. AAA's might be similar. It varies with temperture and from one type of battery to the next. In general it's not reliable. I can't even tell if 1/2 an ohm per battery would be a conservative lower limit.

Even using 1 ohm and assuming that the battery voltage might excede the voltage drop by only a 2 tenths of a volt...
.2V / 1ohm = 200mv
...or about 8 times the upper limit of the LED. Ha!