EE help needed: newbie LED question

[Dougmeister]

Trying to wire two (2) LEDs (2.1v, 25 mA, 6.3mcd, Radio Shack #276-0022) to a single 3V button cell battery (CR2032). I know it's only 3V, but it worked fine putting the battery straight to the LEDs.

Tonight I tried putting maybe 6" to 8" of wire in-between and it works for a few times. Turn it off and back on, then off again, and the battery is "dead". Take the battery out, give it a break, try it again in a few minutes, same thing.

The wire I used was from a SATA cable I ripped apart. My limited electrical background is telling me that either:

1) the wire I chose had too high of a resistance,
2) I need to use 2 batteries,
3) I need to use some resistors, or

a combination of the above.

It has to be done by tomorrow night, so I appreciate any time you can give to this. I'll only have a few hours to work on it, so I'm hoping it's either a simple fix or else I go back to the "direct LED to the battery" trick.

Thanks for any help you can provide.

P.S. I'm using a mini toggle switch from Radio Shack, too. Not sure if that is important. It has 2 terminals. I wired one to the + of the battery holder (from Radio Shack as well), and one to the - terminal. Then the same to the LEDs (+ to +, - to -, and each of the leads from the LED are twisted together)...

4) Oh wait. Parallel vs series. Is that it?

 

[esun]

3) You need to use resistors. Remember, a diode works like this (let V_D be the voltage across the diode): if V_D < 0.7 V, the diode is basically off and conducts very little current. If V_D > 0.7 V, the diode is like a short and conducts a ton of current.

Without a resistor to regulate that current, you're just going to drain the battery. Thus, you need a resistor in series with the diodes. When the battery is applied, you'll get a voltage drop of about 0.7 V across each diode, then the rest of the voltage will drop across the resistor (for a 3 V battery, that'd be 3 - 1.4 = 1.6 V across the resistor). Knowing that you have 1.6 V across the resistor, you can pick an appropriate value to get a reasonable amount of current. For example, a 1 k resistor would give 1.6 mA of current (from the energy ratings of the battery you can figure out how long it will last with that constant current draw).

 

[Dougmeister]

Thanks for the quick reply.

I have no idea what the energy ratings are for the CR2032 button cell batteries that I bought. I would assume that they are all pretty standard?

1) Can I use 1k resistors for this simple circuit?
2) If so, I would need one for each LED?
3) Do I wire the LEDs together at the end? Positive to Positive, Negative to Negative, then connect the wires coming from the resistors (and the battery) to them?
4) Would two (2) 3V CR2032 batteries help it to last longer?

Thanks for your help.

 

[esun]

Sorry, my previous reply was off. I was thinking silicon diodes, not LEDs, and I missed the specifications you posted.

So the LEDs you have are rated at 2.1 V. That means you cannot put them in series with only one battery, since you'd need a battery with at least 4.2 V to get them working.

You have multiple options. You can connect two batteries in series and then connect the diodes in series to the batteries. Series would be positive to negative, so something like this:

+-------|>|--------|>|------- connect to negative end *
-
|
|
+
-*

Where the vertical +- indicate batteries positive and negative terminals and |>| represents a diode with the negative terminal on the right.

If you did this, you'd have a 6 V supply and you'd have a 4.2 V drop across the diodes. Now the current rating of your diodes comes into play. You want them to have a current of 25 mA (according to their specifications). Thus, you'd need to add a resistor of size R = (6 V - 4.2 V) / (25 mA) = 72 Ohms. If you used a 100 Ohm resistor it would work fine.

Now, you could also connect the diodes in parallel and then to just a single battery (though that battery would run out twice as fast due to needing to deliver twice as much current, 25 mA to each diode). In this case, you'd need a resistor of size (3 V - 2.1 V) / (50 mA) = 18 Ohms.

Here are the answers to your specific questions:

1) No. They're too large. You need to pick a resistor so that your diode gets the amount of current it needs. Something on the order of 50-100 Ohms seems to be what you'd need. If you're not sure, err on the large side so you don't damage anything.
2) No. You can do it with just one resistor.
3) Depends on what you want to do. You can wire them in series (positive to negative) or in parallel (positive to positive, negative to negative).
4) Yes. I'd recommend putting the batteries in series as I mentioned to generate a 6 V supply. Then connect the diodes in series with that supply. This means each battery will only be supplying 25 mA, helping them last longer. If you put the diodes in parallel, you will inevitably need 50 mA of current, which will cause the batteries to not last as long. You could also make two separate circuits, one each with one battery, one diode, and one resistor. This means you'll need to use an extra resistor (two total, one for each circuit). However the amount of time the batteries last would theoretically be the same.

 

[Dougmeister]

Again, thanks.

But does that explain why the "circuit" would work when I directly attached the LEDs to the single 3V battery? It would light up both 2.1V diodes...

 

[Modelworks]

The CR2032 batteries are usually around 200mah . So if you use two led then you are using 50mah. The problem I think you are seeing is that while the battery may have 200mah capacity , that doesn't mean it can supply that all at one time. That battery will never power those LED.

Specs on that battery type:
Type: CR2032
Performance Data: (23?±3?)
Nominal Voltage: 3.0V
Nominal Capacity: 220mAh(discharge current: 0.2mA,end-point voltage: 2.0v)
Max. constant current: 4 mA
Max. pulse current: 12mA
Max. diameter: 20.0mm
Max. height: 3.2mm
Reference Weight: 3.2g

 

[Dougmeister]

According to ledcalc, using the input parameters of:

Supply voltage: 3V
Voltage drop across LED: 2.2V (my diodes are green)
Desired LED current: 25 milliamps (? I used the value on the package of the LEDs I bought)
How many LEDs connected: 2

I get:

Exact calculated resistance: 32 ohms
Nearest higher rated resistor: 33 Ohm

1) So should I try to buy as close to that (33 Ohm) without going lower?
2) It looks like that site it telling me to use two (2) resistors, not one (1)
3) It also looks like (correct me if I'm wrong), that that site is telling me to wire them in parallel and not in series

4) Lastly, are any of the other numbers significant?

Wattage recommendation for the resistor: 1/8 WATTS
Circuit's total current consumption 48.5 mA
Actual Single LED Current 24.2 mA
Power dissipated by the LED 53 mW
Power dissipated by the Resistor 19 mW

 

[Mark R]

Easiest way to do what you want is to connect the 2 LEDs in parallel direct to the coin cell. You don't need any resistors if you do it this way (CR2032 cells are very weak, and act as if they have an internal resistor which just so happens to be exactly right for this situation).

Anyway, CR2032s aren't designed for high power drain like LEDs - 2 LEDs is about 10x the maximum recommended load, which will make the battery go dead very quickly.

The other thing is that connecting LEDs in series makes the circuit very sensitive to battery voltage - if you're killing the battery with a crazy load, then its voltage will drop - and with a pair of LEDs in series, it could well drop it below the voltage the LEDs require to light. A parallel circuit is much less sensitive.

 

[Dougmeister]

Hmmm. Interesting.

I was really hoping to utilize a real switch. This is the one I bought at Radio Shack:

SPST Micromini Toggle Switch

But if I go directly to the button cell, I can't really use the switch.

Decisions, decisions...

(Edit: I had a Macguyver-esque "switch" by shoving a piece of thin cardboard in-between the contacts and the battery...)

 

[mindless1]

I think you may be misunderstanding the significance of what the replies are saying. It's not the wire or switch that is the issue, it's that your battery is now drained more than it was prior to all these test runs. You can use a button cell holder and switch with no problem. Because of the low current capability of the coin cell, that it will suffer voltage depression so far, I recommend you run these LEDs at significantly less than 25mA each. How about using two 3V batteries in series, the two LED in series, and a 120 Ohm resistor between them. That will give you 13mA with a fresh pair of batteries, then quickly less as the batteries drain but it should take multiple times as long for the batteries to drain below the critical 2.2V per LED (summed when in series) forward voltage.

Edit: This assumed fresh batteries, the lower the voltage of those you've already used, the more problematic it will be to plan for current consumption as they approach the forward drop of the LEDs. Ideally for anything you want to run for a length of time a higher capacity battery is needed.

 

[Dougmeister]

Gotcha. Will do some soldering. Thanks!