Thursday, October 18, 2018

Next Obscure Electrical Question

Are there LEDs that will only illuminate at some threshold  volume, such as 2 VDC.  My reading says that 2.1 VDC is needed to illuminate a green LED.  Here's the application: I really want an LED that tells me if there is enough power from the two AAA batteries that I do not need to change them.  I would prefer to know this before trying to observe for the night.  At the same time, I do not want too much power draw that drains the batteries too quickly.

I can almost visualize a use of XOR and OR gates that would produce a 1 output if the battery pack voltage is below 2 VDC.  But I would expect LEDs, or logic combos with an LED, that already have this capability to light up a low power LED.

Just to clarify since some of the answers seem to show that I failed to communicate my hopes:  I want an illuminated LED when voltage is >2 VDC or thereabouts, and dark below that level.  No intermediate brightness.  I want a "change the batteries LED."

This looks useful; 51 ohms with a 3 VDC battery pack for a yellow LED will draw little current from the battery pack. Now that Radio Shack is no more, where do you buy one 51 ohm resister and one LED?  Here for the resistor.  Here for the yellow LED.  Red would seem the best choice: forward voltage 1.8 V to 2.2 V.   75 ohm resister.  Red LED with resisters for 6 V-13 V (100 units).  I am guessing the included resisters won't be of any use.  20 mA draw at 3VDC would be .006 watts.  That does not sound like a fast drain on two AAA batteries.  This says a AAA holds 750 mA-hours.  Even if I had the power switch on four hours a night, that should be 125,000 hours consumption for the LED, far more than the motor will draw.   Not sure what to do with the other 99 LEDs.  (Make a red night light to avoid destroying my night vision when doing astronomy?)  Am I on the right path here?

Wayne Johnson said...

Pretty much any LED has a minimum forward voltage. You also want to include a series resistor, as otherwise the current drawn by the LED can climb very rapidly with voltage. Possibly tie it with a momentary push button so it only lights when you push the button. If you look on mouser, newark or digikey, you can usually search by forward voltage.

Jim Horn said...

LEDs work on quantum electro-optical principles. As such, they have a minimum voltage below which they do not emit light at all. The minimum depends on the LED material and the color of light it emits. Longer wavelengths (such as red) are lower energy per photon and need less voltage to light up. That's why blue LEDs may need over 3V while red ones do well with 1.7 or so.

That said, they don't have a sharp turn-on at that voltage so I wouldn't rely on it being very precise.

One simple way to keep battery drain to a minimum is to enable the LED circuit with a momentary action pushbutton switch ("press to test").

StormCchaser said...

Yes. LED's are diodes, and they only start significant conduction (pulling enough current to emit much light) at a critical voltage.

You cannot judge voltage well using the brightness of an LED. As the voltage goes down, the brightness will change very little, until it suddenly drops rapidly to nothing.

If you need to judge brightness, you have to put a resistor in series. BUT... that only works for voltages above the LED's threshold voltage.

NOTE: that threshold is not perfectly sudden, but the current is an exponential function of voltage that does a pretty good job of simulating a perfect threshold.

Zendo Deb said...

https://www.digikey.com/ should have everything you need, but you might not be able to buy just one.

Then there is Parts Express https://www.parts-express.com/cat/electronic-parts/10

There are more choices. Depends on the part and the order quantity.

But wouldn't your needs be serviced best by a battery tester?

RS said...

Not going to work. Lithium batteries under zero load have very little voltage drop until they are almost all the way discharged. Lead acid batteries have a nice voltage/charge curve, but li, nope.

Wayne Johnson said...

If you are going to get additional resistors, get larger ones, not smaller ones. LEDs will glow on remarkably low current. That LED is capable of tolerating 20 mA. It will probably glow visibly on closer to 2 mA. I would try the included resistors before going to a smaller resistor.

Best of luck!
Wayne

Clayton Cramer said...

RS: There are lithium AAA batteries?

Clayton Cramer said...

ZendoDeb: Battery tester requires removing the end cap and pulling the battery pack.

Clayton Cramer said...

Jim: precision not needed. As long as the voltage is >= 2 or thereabouts.

Clayton Cramer said...

StormCchasr: Don't need variable brightness. On/off at 2 VDC is all I need.

StormCchaser said...

I could design a circuit to do that to whatever precision was needed, but it's probably more than you want to do.

With no resistor, or with a resistor, the LED won't go from off to full brightness at one voltage - there will be a range of a few tenths of a volt over which that takes place. The second link has a table giving the "threshold" voltage for given LED colors - but with the caveat that the threshold isn't perfectly sudden.

You might find parts of the page at the link below to be useful.

And... there are Lithium and also NiMh batteries. I use rechargeable NiMh in my Mac's Bluetooth keyboard.

http://www.gizmology.net/LEDs.htm

Clayton Cramer said...

StormCchasr: Over several tenths of a volt is no problem. Can I leave the resistor out, and just pick an LED that lights up about in the range I want?

Clayton Cramer said...

So from reading https://www.radio-electronics.com/info/data/semicond/leds-light-emitting-diodes/characteristics.php, it appears that a red LED with leads across the contacts on the battery pack will light up in the 1.6-2.0 VDC range. No resistor needed? That is all that I need.

StormCchaser said...

I would strongly urge you to put in a resistor, to protect the LED. The LED will have a maximum current rating, so you could calculate the resistor value to allow up to that rating for a drop of, say, one volt, and you should be safe.

The resistor value comes from Ohms Law: R = E/I: resistance in ohms = voltage (1V) divided by maximum allowed current in Amperes (not milli-Ampteres). So if the maximum current was, say, 20 mA, use 1/.02 = 50 ohms.

John

StormCchaser said...

One more thought... Since you are using this to tell battery state, you obviously don't want to deplete the battery too much from this circuit. The way to avoid that is to use a higher resistance, which will result in less current through the LED, but also a less bright LED. If you are doing this on your astronomy stuff, where you would be in the dark, a very dim LED (high resistance value) might be best. I'd fiddle with it - get a few resistors of various sizes, or a variable resistor, and see what works best.