I bought some LEDs for a school science lab, and they say they are for 1.5V throught 3V. I can’t get them to light with 2 AAA batteries. I’ve switched the input direction to account for the fact that they’re diodes, trying both directions. What am I missing? Here is the eBay listing. Pardon Our Interruption...
Link doesn’t work.
Did you just connect the LED leads to the battery? If so, they lit for a millisecond or 2 before burning out.
Loose LEDs need a current-limiting resistor in the curcuit somewhete.
URRRGGHH. The link works only for me, it seems.
OK, a resistor like what?
The strength of the resistor depends on the rest of the circuit you’re designing and what it’s for. This is very basic electronics, but if all that is new to you, you probably want to read or watch some electronics hobbyist stuff to get the beginnings of knowledge and understanding about all this stuff.
Here is an article which explains how to choose a current-limiting resistor for an LED:
https://eepower.com/resistor-guide/resistor-applications/resistor-for-led/#
Without any data on the actual LED (i.e., operating current) it will likely take some trial and error.
I see 30 mA. Thanks for the link. I can calculate the resistor needed. Then the challenge is finding one.
LEDs are “happiest” when they are driven by a constant current (CC) source, and unhappiest when they are driven by a constant voltage (CV) source. Connecting an LED directly to a battery is an example of the latter. The reason is due to the LED’s negative temperature coefficient. Powering it with a CV could cause it to go into thermal runaway.
A resistor in series with a battery is not a CC or CV. It is sort of “in between” a CC and CV. A low voltage in series with a low resistance is closer to a CV, and a high voltage in series with a high resistance is closer to a CC. So be careful if you’re using a low voltage in series with a low resistance, as it could sort of look like a CV and thus cause the LED to go into thermal runaway.
The best solution, IMO, is to use build a CC using an LM317 and one resistor. It’s just one more component (vs. a simple resistor), but offers much better performance and protection.
Ok, thanks. The thing is that voltage range is 1.5-3V and 30 mA. To put in 2V and keep it in the middle of the range, we use: 3V - 1V / .03 = 66.7 Ohms. Right?
Now I see that the setup linked by @Crafter_Man gives a max of 10 mA. I’m not experienced enough to go and devise something that is different. I’m a HS chem guy, and I can plan a buffer to get a 9.2 pH, but I haven’t done this kind of thing. Is asking for help on this too much? At some point there is a thing as asking for too much free work.
Yeah, no need to overcomplicate this for your presumably simple application. If these are loose LEDs with assorted colors, then the voltage range indicated might just be an approximation of the range of “forward voltages” of the LEDs in the pack. Different LEDs (particularly different colors) have different forward voltages. Assuming you grab an old-school red LED, 1.8 V would be a safe guess.
Anyway, take your applied voltage, subtract the LED’s forward voltage, divide by 0.02 A (a fine target current), and whatever resistance you get out, that’s your resistor. As noted upthread, you might opt for a higher voltage and higher resistance to ensure a more stable current (since variations in the LEDs forward voltage would matter less), but honestly these things are pretty robust as long as there’s some current limiting resistor in there.
So, a red LED with a 9 V battery would suggest a 360 ohm resistor, via (9 V - 1.8 V)/(0.02 A) = 360 ohm.
(As a side note, two AAA batteries won’t be enough to overcome the forward voltage of some LED types, and for those where it can, it leaves only a small excess voltage to judge the resistance (and thus regulate the current) by. You might be less frustrated with at least three AAA batteries, if not a different voltage source entirely.)