# Help with designing an LED circuit

I am working on an electrical project and I need some advice on resistors and batteries. I have little electrical engineering education, I learned a little bit about circuits in high school a few years ago. The LED lights I’m going to be using are 3.0-3.2V, 20mA current, and 0.06 wattage. My plan is to use 8 AA batteries to power 88 of these LED lights. I am wondering what resistor I am going to need. If you have any better suggestions for the circuit I am all ears. Also, how would I make it so the lights blink. I’m pretty sure I need a transistor and a capacitor but I’m not entirely sure how to get it all working.

Edit: So I decided to switch to C batteries as they have a much higher mAh (8,000) so I think battery life will be fine now. I found a holder that can support 4 batteries so the new voltage will be 6V. What resistor/other components do I need?

Edit: So would I need to put every LED on it's own resistor with a parallel circuit or even every 2 LEDs?

• Have you worked out what 88 x 20 mA is and how that compares to the maximum current you can draw from AA batteries? Commented Sep 27, 2017 at 6:38
• 88 x 20 is 1,936 mAh and for the current that I can draw, they each can give 2650 mAh and there are 8 of them so do I use the 2650 mAh or the 21,200 combined? Commented Sep 27, 2017 at 7:03
• Neither, mAh and mA are 2 separate things. 88*20mA is actually 1.76A. Before working out your resistor, I would quickly have a google of the difference between mAh/Ah and mA/A
– MCG
Commented Sep 27, 2017 at 7:07
• Also, alkaline batteries (whether AA or C) don't like providing almost 2 amps, so you'll need to deal with that. How did you intend to connect them all? In parallel or in series? Or a combination? Commented Sep 27, 2017 at 17:37

The first thing I would do if I were you is consider using a different power supply. 8xAA batteries is alright, but your LEDs are going to drain them sure quick. You mention the battery capacity is 2650mAh. This means it can power a circuit of 2650mA for 1 hour, or 1325mA for 2 hours etc etc. Your circuit is 1.75A which means assuming perfect batteries and perfect conditions with no loss of power during discharge, you still only get 1.5h of battery life. This seems like a complete waste of 8 batteries. Hence, I would say change your power supply.

If you insist on using these, consider lowering your current. Get some low current LEDs or limit these ones to around 5 or 10mA (if that is bright enough) to lengthen the time your circuit works. You will also be needing a voltage regulator. Your batteries are going to drain, and as the voltage gets lower, your LEDs will dim. You will need a regulator to stabilise the voltage.

Now to tackle making them blink. If you want to use the transistor/capacitor way of doing things, google astable multivibrator circuit. Read up on how they work and understand it before building it.

• With a 12V supply (i.e. 8xAA cells) a series/parallel LED arrangement would allow 3 LEDs in series (+ series resistor) giving a total current consumption of about 600 mA (allowing for 30 parallel connections each taking the maximum 20 mA) Commented Sep 27, 2017 at 14:59

You can use ohms law (V=IR) to calculate resistance required with known voltages. When powering the LEDs, assuming they will all be in parallel, your total current is going to be 88 X 20ma= 1.76A, which is quite a lot for 3 AA batteries to handle, if even possible. As for the resistor, assuming you have sorted out the power supply issue, you simply use ohms law to figure out the resistance required. R=V/I, where V = supply voltage - led voltage drop and I= desired/specified current through LED. In your case 3 AA batteries giving 4.5V with 20ma leds at 3 v. R=(4.5-3)/0.02 R= 75 so a 75 ohm resistor would be required for 1 LED.

• The OP said 8xAA batteries, not 3. And with that you would need some sort of regulator to keep the voltage steady. 8xAA would give you about 12V, but the LEDs would dim as that voltage dropped. If each battery was considered 'dead' at 1.1V, then that would be a voltage of 8.8V. That's a 3.2V drop from 'full' to 'flat' which will definitely change the brightness of the LEDs
– MCG
Commented Sep 27, 2017 at 8:09
• Plus at 1.75A you have the issue that those batteries are not gonna last long. OP mentions 2650mAh battery capacity. That means the circuit will only be working for approx 1.5h. And that is not taking into account battery drain and other factors too.
– MCG
Commented Sep 27, 2017 at 8:13
• @MCG The OP originally said 3 AAs Commented Sep 27, 2017 at 8:34
• Oh yeah whoops, I read 3 instead of 8 Commented Sep 27, 2017 at 8:42
• @HandyHowie oh yeah, just checked the edit log of the question... My bad, I only saw it as 8!
– MCG
Commented Sep 27, 2017 at 10:57

Considerring that you are using now a new type of batteries I suggest you to calculate againg the necesary R. As MCG suggested it's not a good idea to make them work at the limit, it's better to work with lover Current. (6-3)/0.01 = R.

For blinking you can do in many ways. MCG suggested a great sollution. Also you can use a manual switch a transistor and generate a Square signal with NE555 and feed it to transistor.

PS: If you use the regular batteries I suggest you to connect them not all 8 in serie but 4 in serie together with another 4 in paralel, this way instead of having 12 V and 2650mAh you get 6V and 5300mAh.

I would recommend three things:

1. Instead of having a resistor on each and every LED, split them into small groups of 2-4. Then, for every group, give it its own resistor. If you are using RGB LED's then remember the R LED tends to require 2.0-2.2v (check your datasheet), so they will require their own resistor. Use Ohm's law to determine resistance.
2. Making them blink would by far be the best with a board as small as an arduino mini. You might need a bigger size board depending on the number of inputs/outputs you have.
3. Use a voltage regulator (3.3v is probably best for you).

The new C size alkaline cells are 1.6V when new and soon drop to 1V each. Their capacity is 8000mAh when the current is low and is less than 1500mAh with the very high current of 88 LEDs. The battery will be 4V after about 45 minutes.

Then you calculate each resistor with 20mA, a 3.0V LED and a 6.4V battery which is (6.4V - 3.0V)/20mA= 170 ohms which is not a standard resistance, use 160 ohms for each LED. The power in each resistor will be 0.072W. The LEDs that are 3.2V will be a little dimmed. All the LEDs will dim more and more as the battery voltage quickly runs down.