Hi y’all. For a Christmas present for my sister-in-law I’m making a diorama of a scene from one of her favorite TV shows. It’s a street scene and I’m making it so the street lamp turns on when you rotate the manhole cover. The batteries, two 3-volt CR-2016 button type batteries, will be directly under the removable manhole cover. I’ve carved out the battery bay and put all the wires and fittings in and everything is set in plaster and fits nicely and all electrical connections are tight.
:smack:
It all fits and works but now I’ve found I’ve been using too many volts for the LED they are powering. The LED I want to use is rated at 3 volts and so two 3-volt batteries is too much. There is no space for a resistor. So…
Q1. Can I simply discharge one of the batteries, and glue it in there permanently. It will just fill in the space that needs to be filled and provide an electrical connection for the single other battery.
Q2. The battery is 3-volt and the LED is 3-volt. Do I absolutely need a resistor or can I do without one?
Can you rewire them so that they are in parallel instead of series. That is, connect them so that the two positives and the two negatives are linked directly, rather than p>n>p>n.
This will give you 3 volts for twice as long instead of 6 volts.
Q1. Can I simply discharge one of the batteries, and glue it in there permanently. It will just fill in the space that needs to be filled and provide an electrical connection for the single other battery.
No. The good battery will try to charge the dead battery. You can however use a metal spacer of some sort, like maybe a coin.
Be careful with gluing. If you don’t end up with metal on metal contact (because of the layer of glue) then you won’t have a working circuit.
**Q2. The battery is 3-volt and the LED is 3-volt. Do I absolutely need a resistor or can I do without one? **
It’s usually a bad idea to drive LEDs from a voltage source. LEDs go from too dim to see to burning out because they were over-driven over a very narrow voltage range. You can look up the specs for the LED that you are using and check its voltage and current curve against your battery’s actual voltage.
Even if you luck out and the battery voltage is within range of your LED’s operating voltage, two batteries (6 volts) plus a resistor to limit the current will give a more consistent light output over the life of the batteries. LEDs go from very dim to very bright over a fairly narrow voltage range and the small voltage change as your battery discharges may make a very noticeable difference in the LED’s output.
They make LEDs with built-in circuitry so that they can be easily driven from a voltage source. They cost a bit more but if you are trying to reduce the total parts count (for size reasons or whatever) they may be worth it to you.
ETA:
It’s usually a bad idea to parallel batteries. If you do this, make sure you use exactly the same type of battery (same brand, exact same type, preferably out of the same package) so that they will be equally charged and will discharge roughly equally. Mismatched batteries will end up having the stronger battery trying to charge the weaker battery. Some battery chemistries aren’t meant to be recharged and the batteries can get very unhappy when this happens.
Wow, lucky I asked. I was pretty sure a discharged battery would have been fine, but I asked here just to be safe. And, yeah, come to think of it, sometimes I’ve tried to just replace one battery in my TV remote controls and it never worked. :smack:
OK So I guess I should do 2 x 3-volt batteries (as it is currently set up) and make a space for a resistor, somewhere. It doesn’t matter if the resistor is positioned between the positive or negative terminal and the LED. does it?
You might be able to get away with the two cells. Some of those little keychain lights that use 2016s and a 5mm LED use exactly that 2 cell setup without any extra resistance.
The secret is that those cells aren’t a constant voltage source. Their relatively high internal resistance means they are awful at supplying high rates of current. Their voltage will sag as soon as you switch on the circuit. Here’s a good discussion of the effects of their internal resistance. You’ll do better with alkaline chemistry 2016s than lithium because of more voltage sag. You likely don’t kill the LED either way.
Following DinoR’s link, I found a PDF spec sheet for the lithium CR2016 battery. If I’m reading that bottom right plot correctly, the internal resistance of a single cell is about 16 Ohms when fresh, rising to 20 Ohms after 40 m Ah has been used, and getting larger as you use more. So whatever resistance you calculate you need, you could reduce by 30 Ohms or so.
Not only that, but there’s the issue with temperature coefficient. An LED has a negative temperature coefficient, which (as you already know) means the IV curve shifts to the left as temperature goes up. If the LED is driven by a regulated voltage source, the LED could burn up due to thermal runaway. This is also the reason why it’s not advisable to put LEDs in parallel; they won’t have identical IV characteristics over temperature, which means one of the LEDs will dissipate more power than the other.
As you and others have stated, the best approach to powering LEDs is to control the current through each LED. This can be done using a current regulator (best) or with a series resistor (which may - or may not - be adequate, depending on the design).
It’s most convenient to put all the LEDs in series. That way, you only need one current regulator. But the required power supply voltage may be too high when all the LEDs are in series. If this is the case, the design should incorporate multiple strings of LEDs connected in parallel. Each string consists of multiple LEDs connected in series, and each string has its own current regulator.
One of the best introductory tutorial about LEDs I’ve seen is at Adafruit: https://learn.adafruit.com/all-about-leds . They are dealers in Arduino hobbyist boards, so perhaps after your current project here’s another direction to go…