As part of a pumpkin carving contest, I have been put in charge of lighting up our pumpkin using an Arduino and small household batteries. Our pumpkin will have 4 cutouts with the top of each being an arch leaving 4 columns of lights plus the arches at the top. The lights will be sequenced in rows, so the first light in all 4 columns will come on, then off, then the 2nd light in all 4 columns comes on… etc, until the top where more lights comes on as the tops of the arches are lit up. The following image can be used as an example of of how the pumpkin will be carved:
I will be using 44 LEDs in total in my design and most of the time only 4 LEDs will be on at a time, but I may want to have all 44 flashing at some point. I believe I understand the math involved to figure out the voltage required (i.e. 44 * 3.3v = 145 volts) if I were to connect all 44 LEDs in series, but as I need to turn on and off rows of lights, the concept is confusing to me. To complicate matters, the LEDs must be powered by small household batteries so I was thinking of connecting as many 9v (or other) in series to get the voltage that I will need. A large battery will not work for design and portability reasons and the LEDs only need to work for approx 30 minutes.
Should I connect each of the rows of lights in series, and then connect each of the series in parallel?
If I do connect in series and parallel, does the math change for the voltage requirements?
If I do connect in series and parallel, can I control each series with a separate pin on the Ardunio?
Sorry if these aren’t vary intelligent questions… searches aren’t leading me to the right answers or I don’t know how to intepret the postings I’m finding.
All of the above run off a 5v supply (3A max, = 44 LEDs on full white) and would allow you to control each LED individually. You could use a wallwart but if battery power is needed, I suspect a single 4 pack or AA, or even AAA, NiMH batteries will work.
I noticed the image didn’t post, however if you copy the image url into a new tab in your browser, it seems to load.
The LED’s will always be in in sets of 4 or 8… never a single one by itself. I should have also mentioned that these are just standard clear white LEDs, not RGB and unfortunately, I don’t have enough time to get anything shipped to me now.
There is no image link, it just says “Image.” You can attach an image to your post. Under the editor text box there is a tab that says “Upload Attachments.”
meierk:
The LED’s will always be in in sets of 4 or 8… never a single one by itself. I should have also mentioned that these are just standard clear white LEDs, not RGB and unfortunately, I don’t have enough time to get anything shipped to me now.
If you have a handful of LEDs what I'd do is put a resistor in series w/each LED and group 4 pairs (as shown) or even 8 pairs (to save wiring, not shown) to be switched by a single transistor. Put another resistor between the Arduino pin and the base of the transistor. Run all 11 such circuit blocks (in yellow) and the Arduino off a single 4 pack of AA batteries. 9v batteries don't provide much current capacity and the number needed would be larger than the aforementioned.
If you want you can use a common driver IC in place of the transistors and base resistors. The come 8 channels to an IC, so unless you combine some 4-pairs into 8-pairs, you’ll need 2 ICs.
I have an Uno and was planning on using a 2N2222 transistor so great to see the confirmation since I am very green when it comes to this stuff.
I didn’t think the AA’s would provide enough voltage for either the 8 LED series or 4 LED series as the forward voltage of each is approx 3v (using clear white) so 4 LED’s alone require 12V to operate and the most you can get from 4 AA is 6V if connected in series… or am I not missing something?
I guess using this logic, even the 8 LED series wouldn’t operate using the 18V I was intending to supply using 2 9V batteries so I’m really hoping I’m wrong, but every LED Resistor calculator I found was giving me the same results.
I didn’t think the AA’s would provide enough voltage for either the 8 LED series or 4 LED series as the forward voltage of each is approx 3v (using clear white) so 4 LED’s alone require 12V to operate and the most you can get from 4 AA is 6V if connected in series… or am I not missing something?
I guess using this logic, even the 8 LED series wouldn’t operate using the 18V I was intending to supply using 2 9V batteries so I’m really hoping I’m wrong, but every LED Resistor calculator I found was giving me the same results.
Series vs. Parallel circuit.
If you can’t put the leds in series because it requires too much voltage, you have to place them in parallel (as Mee_n_mac showed) and make sure the sum of their currents can be supplied by the battery. (and pass through the transistor) The current issue is usually more a problem of how long the batteries should last. Thus battery capacity. In parallel circuits the voltage across each led and resistor is the same. But at the junctions before and after them the currents divide and later add up.
meierk:
I didn’t think the AA’s would provide enough voltage for either the 8 LED series or 4 LED series as the forward voltage of each is approx 3v (using clear white) so 4 LED’s alone require 12V to operate and the most you can get from 4 AA is 6V if connected in series… or am I not missing something?
You're correct as far as you go and that's why I have all the resistor+LED pairs in parallel w/each other. Now doing that means you need less voltage but more current. The problem w/using 9v batteries is that they aren't really 9v under load and after some discharge. Look at the curves here...
At 500 mA it’s more like 6-7 v after some discharge and that’s not really high enough for 2 white LEDs in series. To run this type of ‘quad LED’ circuit (shown below), you need to keep the current draw to 100 mA or less. Then the voltage is 7.5-8.0 v for the initial part of the discharge. Each quad of LEDs will draw 2 x 20 = 40 mA when on, meaning you can only run 2-3 quads (12 LEDs max) and that means at least 3 - 4 (9v) batteries, which takes as much and perhaps more space than 4 AAs (or AAAs) and won’t last as long.
(click on to open)
And then there’s the difference in the initial battery voltage vs the voltage 30 mins later, as well as variation under load. In the circuit above the LEDs get 20 mA w/8.0v and the 68 ohm resistors. It’s only 15 mA at 7.5v and it’s 25 mA if the battery is 8.5v (ie: new and only 4 LEDs on). So you can use 9v batteries but the results will be less predictable and it’ll take up more space. Look at it this way … each LED draws 20 mA (max) and 2 in series (as above) use the same current as a single LED. So 44 LEDs, arranged as above will draw 22 x 20 mA = 440 mA max when all are flashed on. That would degrade 1 battery to a voltage close to the 500 mA curve in the link. That’s not going to work. Splitting the LEDs (plus Arduino) across 2 batteries halves the current per battery (and might, maybe work) but 3 batteries is the min number I’d see as usable. Then again perhaps you’ll run less than 20 mA/LED ?