Newbie needing help with DIY project...

[bfinleyui]

I am hoping to wire together some LEDs to a single 3v 2032 battery.

The most important thing is battery life, they're for my wedding reception, and will need to run for, at least, 6-8 hours.

The LEDs have various forward voltages, and it would be nice if i could mix them, but if need be, I can use the same color throughout.

The LEDs are 2v and 3v. Hoping to be able to drive at least 4 or 5 of them. Is this possible to drive 4 or 5 leds for that long on a coin cell?

Any help with regards to the layout of this project? The LEDs came with a bunch of 200ohm 1/4 watt resistors, will those work properly?

Sorry for the newbie questions, I'm good with electronics once they're manufactured, but before that I'm pretty lost.

Thanks.

 

[AnAppleSnail]

Sounds like you have:

3v (White/Green/Blue/Purple)
2v (Yellow/Red/Orange)

The 3v ones could run directly off the CR2032. Some little keychain lights work this way. You can actually just pinch these onto the battery in the right orientation and wrap it in tape. I suggest using a plastic "pull tab" between the wire and tape so that you can pre-assemble them all and pull the tab out to turn them on. You will have to check brightness, runtime, and if they burn out like this.

The 2v ones need to get rid of about 1v at 20 mA. This requires (V=IR, 1v = 0.02A * R) where R=50 ohms. 200 ohms is meant for about 6v, or much dimmer settings.

 

[Gunner12]

A simple thing to do would be to get a few spare batteries, a potentiometer, and hook the LED through the potentiometer to the battery and measure the current with a multimeter. Adjust the potentiometer until you get the output you want, and then measure the resistance and get a resistor with a similar value (if you end up with 183 ohm, just use those 200 ohm resistors). I could drive a LED for a year on one of those batteries, if I run the LED at a really low level (~20 µa).

I'm assuming these are the normal lower power LEDs (20-30 mA max) and you aren't looking to light up anything. According to wikipedia, the CR2032 battery has 190-220 mAh depending on chemistry, so you can estimate runtime from there. The rated discharge current is 3 mA, and at that current you are looking at ~60 hours of runtime and a decently lit LED.

This is a really basic circuit, with just a power source, a load, and maybe a resistor to limit the current. Remember that the battery has its own internal resistance, so the more current you draw, the lower the battery voltage is. But for your uses, you should have a problem with internal resistance. Those batteries can't supply high currents at all since they have a decent amount of internal resistance.

Even if you drive the LED at 20 mA, you should still get a good few hours of runtime, and a few more hours of decreasing output. You could just hook up a LED directly to the battery and test the runtime.

What is your end goal for this project? (what should it look like)

Congrats on the wedding!

:welcome:

 

[bfinleyui]

The 'inspiration' picture I've been given to work with. (Linked because it's a fairly large photo)

Basically, paper lantern, with a small vase inside, and I figured I could wire up 4-6 LEDs, wrap those around the vase, and all will be pretty and well.

I've got some 3-pin switches to control the on-off functions, just to make it idiot-proof for those who'll actually be turning these things on.

I'm still a little confused about how to estimate the time. All the articles and things I've found to try and figure out what is going on have gone way over my head (I was a humanities major)

The LEDS max out at 20 mA a piece, but it says they can drop to 10-15 without much noticeable drop in brightness.

What I'm lost on right now is:

How we get from a led that runs at 20 mA, to 60 hours of life? I may be dumb (may? likely am), but 200/20 = 10 hours for a single LED. Confused about "rated discharge current" vs the power of the LEDs.

Secondly, if the light maxes out at 20mA, how do we tell it to drop to 10-15, to use less of the battery?

Sorry for the dumb questions, but you guys have been super helpful already.


Edit: So after reading some more elsewhere, I think I've got this figured out, at least for a single LED. If i'm using a 3v battery and a 3v LED, which i would want to run at 15mA, i'd need to put a 200ohm resistor ahead of the LED, correct? Do resistors not drop the voltage, as well, though? Or is that not a concern? And assuming I do this all in parallel, the same 3v can supply everything, as long as they're wired in parallel, not series. I feel like I'm getting closer, it just doesn't seem to make sense that putting a resistor ahead of an LED would drop the amps, but not the volts...

 

[AnAppleSnail]

How we get from a led that runs at 20 mA, to 60 hours of life? I may be dumb (may? likely am), but 200/20 = 10 hours for a single LED. Confused about "rated discharge current" vs the power of the LEDs.

Batteries, honestly, get tired if you suck power out too fast. Every battery in the world has a rating called "Capacity," which is in Amp-hours. They are tested at some rate (amps) and will last for some time (hours) at that rate. If I have a battery that can give me 1 amp for 1 hour, I have a 1 amp*hour battery. I could take the same power out of the battery in half that time, or have 2 amps for half an hour.

However, if I take that battery and drain it dry at 20 amps, it won't last even 3 minutes (1/20th of an hour), because batteries get tired. If you drain a battery faster than its "rated current," you won't have as much power. Since these CR2032 cells have 200 mAh in them at 3 mA, you can run them for about 60 hours. Higher drain rates will reduce how much power you have. At 20 mA, I would conservatively estimate five to six hours runtime.

Secondly, if the light maxes out at 20mA, how do we tell it to drop to 10-15, to use less of the battery?

Sorry for the dumb questions, but you guys have been super helpful already.

Edit: The LEDs don't max out at 20 mA, really. They are "rated" for 20 mA, which means they last a reasonable while (Several hundred hours, we hope) without getting too much dimmer. Like all machines, you can "overdrive" the LED with a bit more current. But even doubling the current won't look very much brighter, but it greatly cuts battery life and LED life.

Using resistors, we play with voltage. Imagine the electrons in the circuit. If they went long enough, they would go in a loop from the (+) end of the battery, through the wires, LED, and resistor, to the (-) end of the battery, and move back "up" from 0 volts to about 3 volts. The voltage of the whole circuit as a loop amounts to 0, a closed loop.

If I had a resistor and a battery, the resistor would have the opposite voltage as the battery, and current would flow through the resistor following Ohm's Law: V=IR (Volts equals Amps (I) times Resistance in ohms). Resistors predictably change their voltage with current.

LEDs are tricky. Being semiconductors, they have a nonlinear response (A curve, not a line). In general, a white LED may light up very dimly (Low current) at 2.5v, be quite bright at 3.0v, be extremely bright and short-lived at 3.4v, and burn out much over that.

You're combining an LED and a resistor to get 3v at 20 mA. Technically, you are using the voltage of the LED at 20 mA V(led at 20 mA) plus V(resistor at 20 mA) equals 3v. As a rule of thumb, we like to use data-sheets for each LED, but general rules apply:

1.8-2v, Red, Yellow, Orange
2.5-3v, Green, Blue, Purple, White

LED color is based on the voltage needed by a semiconductor part. 'Higher Energy' photons take higher voltage, so we can guess what the LED's voltage is going to be at a reasonable current. The abstract, math-ey way is to plot the LED's voltage response to current and pick a resistor so it all adds up to 3v.

The engineer way is to borrow or use a multimeter on a "milli-amp (mA)" setting, and measure. Connect one leg of the LED to a resistor between 50 and 200 ohms, then touch the free legs to the battery. If it didn't light, turn the battery around and try again. Now tape one "free' leg to the battery, and connect the multimeter probes in. Touch the free leg with one, and the battery with the other. A number should pop up telling you the current draw. Larger resistors in series will decrease this number (Dimmer). Adding resistors in parallel will increase it (Brighter). The current draw will indicate battery life, but nothing replaces a runtime test. The light will gradually dim over time, eventually fading to nothing.

PS: Don't get the lithium coin cells wet! Metals in non-pure water can corrode very fast when they have voltage on them. Nothing exciting is likely to happen, but the lights won't light.

 

[bfinleyui]

Batteries, honestly, get tired if you suck power out too fast. Every battery in the world has a rating called "Capacity," which is in Amp-hours. They are tested at some rate (amps) and will last for some time (hours) at that rate. If I have a battery that can give me 1 amp for 1 hour, I have a 1 amp*hour battery. I could take the same power out of the battery in half that time, or have 2 amps for half an hour.

However, if I take that battery and drain it dry at 20 amps, it won't last even 3 minutes (1/20th of an hour), because batteries get tired. If you drain a battery faster than its "rated current," you won't have as much power. Since these CR2032 cells have 200 mAh in them at 3 mA, you can run them for about 60 hours. Higher drain rates will reduce how much power you have. At 20 mA, I would conservatively estimate five to six hours runtime.




Edit: The LEDs don't max out at 20 mA, really. They are "rated" for 20 mA, which means they last a reasonable while (Several hundred hours, we hope) without getting too much dimmer. Like all machines, you can "overdrive" the LED with a bit more current. But even doubling the current won't look very much brighter, but it greatly cuts battery life and LED life.

Using resistors, we play with voltage. Imagine the electrons in the circuit. If they went long enough, they would go in a loop from the (+) end of the battery, through the wires, LED, and resistor, to the (-) end of the battery, and move back "up" from 0 volts to about 3 volts. The voltage of the whole circuit as a loop amounts to 0, a closed loop.

If I had a resistor and a battery, the resistor would have the opposite voltage as the battery, and current would flow through the resistor following Ohm's Law: V=IR (Volts equals Amps (I) times Resistance in ohms). Resistors predictably change their voltage with current.

LEDs are tricky. Being semiconductors, they have a nonlinear response (A curve, not a line). In general, a white LED may light up very dimly (Low current) at 2.5v, be quite bright at 3.0v, be extremely bright and short-lived at 3.4v, and burn out much over that.

You're combining an LED and a resistor to get 3v at 20 mA. Technically, you are using the voltage of the LED at 20 mA V(led at 20 mA) plus V(resistor at 20 mA) equals 3v. As a rule of thumb, we like to use data-sheets for each LED, but general rules apply:

1.8-2v, Red, Yellow, Orange
2.5-3v, Green, Blue, Purple, White

LED color is based on the voltage needed by a semiconductor part. 'Higher Energy' photons take higher voltage, so we can guess what the LED's voltage is going to be at a reasonable current. The abstract, math-ey way is to plot the LED's voltage response to current and pick a resistor so it all adds up to 3v.

The engineer way is to borrow or use a multimeter on a "milli-amp (mA)" setting, and measure. Connect one leg of the LED to a resistor between 50 and 200 ohms, then touch the free legs to the battery. If it didn't light, turn the battery around and try again. Now tape one "free' leg to the battery, and connect the multimeter probes in. Touch the free leg with one, and the battery with the other. A number should pop up telling you the current draw. Larger resistors in series will decrease this number (Dimmer). Adding resistors in parallel will increase it (Brighter). The current draw will indicate battery life, but nothing replaces a runtime test. The light will gradually dim over time, eventually fading to nothing.

PS: Don't get the lithium coin cells wet! Metals in non-pure water can corrode very fast when they have voltage on them. Nothing exciting is likely to happen, but the lights won't light.

EXCELLENT explanation. So let me see if i've got this right...

I have a purple LED, it's 3v. If I put a 200ohm, 1/4v resistor before it, it's still going to burn purple, just less bright, using fewer mA, but lasting longer.

My point of confusion was that the 3v was an either/or situation. Like if it got 2.75, it wouldn't light up at all. In reality, it will just be dimmer. To test 'how low' i can go, i could put more resistors ahead of the LED, and it would get dimmer and dimmer with each one, until there's not enough power to light it at all.

Am I reading that right?

 

[AnAppleSnail]

EXCELLENT explanation. So let me see if i've got this right...

I have a purple LED, it's 3v. If I put a 200ohm, 1/4v resistor before it, it's still going to burn purple, just less bright, using fewer mA, but lasting longer.

My point of confusion was that the 3v was an either/or situation. Like if it got 2.75, it wouldn't light up at all. In reality, it will just be dimmer. To test 'how low' i can go, i could put more resistors ahead of the LED, and it would get dimmer and dimmer with each one, until there's not enough power to light it at all.

Am I reading that right?

Correct on all points. My experience is that at about 2.5v a blue LED will light about as bright as a star in the sky - it will appear as a glint of light if you look directly into it in a dark room. On the other hand,

You might wonder, "Why doesn't a 3v battery burn up the LED?" The exact reason is that batteries have internal resistance. For small batteries, this is high. Even with small batteries, lower-voltage LEDs (Red, yellow, orange especially) can burn out on a 3v cell. They need extra resistance to lower the voltage they see.

Always remember that LEDs are best controlled by controlling current supplied to them. In this case, we are lowering the voltage available to the LED so that it (hopefully!) can't get enough voltage to crisp. If you overdrive a small LED to destruction, the LED will usually turn off, turn black and smell 'burnt' for a bit.

 

[bfinleyui]

Wow, again, thank you *SO* much for your help. I am literally beaming ear-to-ear having figured this out.

Now I can't wait for the end of the workday to go get some alligator clips, tear apart an old network cable for some wire, and start trying things out!

Thank you thank you thank you. I'm blogging about my wedding experience, as a dude trying to figure out how to plan a wedding, and will definitely link to this thread and give shout outs. Absolutely great, thanks thanks.

 

[Gunner12]

Great explanations AnAppleSnail!

In the pictures you showed, it seems like the lanterns are decently lit up and lit up evenly, so you might want to diffuse the LEDs a bit for an even beam, and maybe use a different power source for a brighter light (brighter = less runtime), or use a couple lights for each lantern.

I like to test things, so I'd also suggest taking one of the white/blue LEDs, taping it to a CR2032 battery (positive of the LED to positive of the battery, negative to negative, no resistor), and seeing both how bright that is, and how long it lasts.

With the lower voltage LEDs, hook the resistor up and see how bright it is. If it is too dim, try a resistor with a lower resistance and keep lowering it until you get the output you want (50 ohm should give you 20 mA to the LED with a 3v battery).

Remember, V = I*R, V being voltage dropped, I being current, and R being resistance. So if I want to drop 1 volt with a current of 20 milliAmps (0.02 Amps), the equation would be R = V / I.

So: R = 1 v / 0.02 A, R = 50 Ohm.

Realistically, there will be some variation, the LED won't be exactly 2v, the battery might have a higher voltage when fresh, etc. But the equation is good enough for estimates, and most of the time, an estimate is all that you need.

So, with your 200 Ohm resistors, assuming a 1v drop, you should get: I = V/R -> I = 1/200 -> I = 0.005 A, or 5 mA.

If you have a potentiometer (adjustable resistor), that would be even better, since you could tweak the resistance until you get the output you want.

When I tried a single white LED with one CR2032 battery a couple years back, it was decently bright, didn't test the runtime though.

Sorry that I wasn't that clear in my previous post.

 

[bfinleyui]

As far as LEDs, I accidentally bought clear-housing LEDs for my first package, but some 600-grit sandpaper should fix that no problem. I also ordered some straw hat wide-throw LEDs, for a little less work. If the 2032's don't give enough life/brightness tradeoff, I'll change it out for a 9v and adjust the layout, which would still be fairly cheap (any recommendations on where to get reliable bulk batteries?).

Luckily i've got about 8 months to mess around with this before it's go-time.

As far as multimeter and potentiometers, sounds like i've got the beginnings of my christmas list.

 

[Gunner12]

Potentiometer can be pretty cheap (~$1 on eBay, ~$4 at radioshack). Radioshack also has multimeters for $30-$100.

I've got a small electronics kit from college, it's nice for tinkering and stuff.

For bulk batteries, eBay? Amazon? Straight from the battery company? I have seen boxes of 9v batteries, so I know they can be bought, I just don't know where the best place would be.

 

[bfinleyui]

Mmkay, thanks. Didn't know if there was some "secret" that only the lighting guys knew about that had quality batteries for cheaper.

Looks like I can get 100 2032's on amazon for about $30, so it might be cheaper to wire up more than one 2032, even if the same battery life could be achieved with one 9v. 9vs seem to be about a buck a piece.

 

[AnAppleSnail]

9vs seem to be about a buck a piece.

Yes, but they have twice the capacity and triple the voltage (Run 3 white LEDs in series off a 9v battery). 9v battery holders are cheaper to buy, but with CR2032s you don't necessarily need them.

 

[RoGuE_StreaK]

Looks like I can get 100 2032's on amazon for about $30

I just searched ebay and found multiple sellers with 100x for around $10-14 (including shipping). Also found 100x holders for the same kind of price; maybe forget the switches, just use a holder and instruct those setting up to insert the battery. QED.
NB: when searching ebay for stuff like this, always tick the "worldwide" location box. Oh, and always check out feedback ratings before choosing a seller, particularly what people are saying in negative feedback.

 

[bfinleyui]

After all this trouble, it looks like I'm going to end up having to go 1:1 batteries->lights to drive the purple LEDs at the brightness that's acceptable. Luckily, with volume purchasing (thanks to several tips here), that ends up at about 15 cents per LED, and 15 cents per battery, add in some electrical tape and cardstock on/off tabs, and i'm out the door for 30 cents a light, as opposed to the next highest power source option, a 9v, which would be about $1.15 just for the battery, plus $0.15 per LED, and it just didn't make sense.

I'm sure I'll be back for other projects though, now that I have a general understanding of the way an LED circuit works. Thanks!!!

 

[Gunner12]

Well, the ones in most flashlight are a bit more complex (they have to raise the voltage, or lower the voltage, transistors, inductors, etc :p). If you understand generally how the lights out just made work, then that should make a great starting point for electronics and stuff.

I'm glad the lights worked out for you, nice and simple.

A cool next project if you have time might be the Joule Thief. Personally I think they are pretty cool circuits and are fairly easy to built (runs a LED from a 1.5v battery and can run the battery down to pretty low voltages). It can take some time to understand how they work though.