Will this schematic work - basic LEDs + strips

Question

This is my first LED project for my art bike and after a good chunk of reading, I wanted to know if the circuit would run properly.

LED are green 5mm from AliExpress. Forward voltage 3V-3.2V, 20mA

There is also a green LED strip in the design from AliExpress. 2835 SMD green lrd strip - use about 15 LEDs length. I'm guessing I won't need a resistor since it's 12v.

The battery will be 8xAA batteries. If my math is somewhat accurate about 24 hours run time - 16ah / 0.62ah -i.e (2000mah x 8 AA Battery) / (20mah x 31 leds)

I'm planning on using 22AWG if that makes any difference or should be advised on another size.

^ Here is an edited version #2 with a voltage converter... and 3.6 battery

Answer 1

A couple things to touch on here.

• first, your battery life calculation. Stacking batteries in parallel adds current (and current capacity). Stacking in series to get 12V like you've done will not increase capacity beyond 2Ah. If your draw is about 600mA, that'd only be 3 or 4 hours

• next, I would try to put 3 LEDs in series where possible so you only have to drop ~2 or 3 volts over the resistor. 150 is the correct resistor for 3 LEDs if you want 20mA per group of 3, though if you stay with the group of 2, that resistance should be higher (300 or 330)

• you're correct, no external resistor needed for an LED strip (almost always).

Other than that I think this would work just fine. Make sure that switch (and batterry!) is rated for whatever the total current draw is. Not sure what the current draw is for your length but I'd check it out.

Also your schematic has the battery flipped upside down, just a heads up!

Answer 2

LED Diodes are Green 5mm from AliExpress Forward Voltage / Current: 3V-3.2 V | 20 mA.

This seems very high for green LEDs and I would want to check the datasheet. "No datasheet? No sale."

There is also a green LED strip in the design from AliExpress. 2835 SMD Green Led Strip - use about 15 LEDs length. I'm guessing I won't need a resistor since it's 12 V?

Figure 1. Source: Flexible LED strip light.

If they're the standard type then they are wired as shown in Figure 1 and have the resistors built-in so can be driven at 12 V. See the linked article where I've written a little more on the topic.

The battery will be 8 x AA batteries... which if my math is somewhat accurate about 24 hours run time - 16 Ah / 0.62 Ah -i.e (2000 mAh x 8 AA battery) / (20 mAh x 31 LEDs).

Nope:

• 8 x 2000 mAh, 1.5 V batteries in parallel gives 16 Ah at 1.5 V.
• 8 x 2000 mAh, 1.5 V batteries in series gives 2000 mAh at 12 V.

Green LEDs usually drop about 2.2 V or so at 20 mA - unless they're something special. That would allow you to put four in each string and still be able to control the current reasonably well with the resistor. That would reduce your overall current to 20 mA x 32 LEDs / 4 per string = 160 mA. The Energizer E91 datasheet reckons you'd get 2000 mAh at that discharge rate so $$ \frac {2000 \text {(mAh)}}{160 \text {(mA)}} = 12.5 \ \text h$$

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Note that you can check your calcuations by watching how the units divide out. You divided Ah by Ah in your question and still got 'hours' in the answer. Another tip: SI units named after a person have their symbols capitalised and are lowercase when spelled out. 'A' for ampere, 'V' for volt, 'K' for kelvin, etc.

Answer 3

My recommendation would be to power each LED individually. I'd use an LDO adjustable voltage regulator at around 2.7 V then try powering the LED directly without a resistor and adjust (likely lower) the voltage to the desired intensity. I would power it with a single $4-5 ≈3000 mAh Samsung 18650 Li-ion battery. You would need to be careful not to discharge the Li-ion much below 3V.

I have tested powering, without a resistor, some Samsung mid power (35 lm, 2.75V, 65 mA) LM301B LEDs at between 2.4 V (very visible) and 2.7 V (very bright).

Use Cree 5mm Green C503B-GAS-CB0F0792 and you can lower the V_f and current significantly and get the more luminous intensity as the cheap ones you choose. Lower the current to about 5 mA and the V_f will be 2.8 V or less. The extra cost for the higher efficacy LEDs (64,600 mcd) will more than pay for themselves in cost of batteries. You may even be able to go under 1 mA.

This Cree LED should draw less than 10 mA @ 2.7 V. This would still give you over 25,000 mcd.

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RE: Your schematic:
If you want the eyes to be at the same 20 mA intensity, you should use a 300 Ω resistor. Better would be to combine the antenna and eyes in to a string of three LED or better yet use strings of 4 LEDs.

You could try six batteries and three LEDs in series with no resistor.

Understand that voltage of alkaline batteries will drop rather quickly. The typical alkaline AA drops from 1.5 V to 0.8V (20 mA drops to 10 mA) over its lifespan.

The correct way to power LEDs with alkaline batteries is to use a step-up/step down switching voltage regulator.

If you look at an alkaline AA datasheet you will see how quickly the voltage drops which mean the intensity will drop that fast. This is another reason to use high efficacy LEDs and a voltage controller.

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You will not get the 2000 mAh because that is if the batteries are discharged to 0.8 V which would be insufficient to power three cheap LEDs in series. The Cree LEDs would still be very visible but with much less intensity than with fresh batteries.

Whereas if you use Lithium AA you will get more capacity (3500 mAh) and the voltage discharge is much flatter than alkaline so the luminous intensity will hold until near the end of its lifespan.

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Your best bet would be to use two Li-ion batteries with a buck step down voltage regulator or a 3.3 V linear regulator and power each LED with its own resistor.
If you dropped the LDO voltage to about 2.7 V (for a 3 V LED) you could eliminate the resistors and use a single Li-ion cell.

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A simple adjustable LDO regulator LM317

You also could use a fixed output voltage LDO in the range of 2.8-3.0 V with a resistor (≈150Ω-200Ω) on each LED for a 2 mA current giving you about a very bright 5,000 mcd. The circuit for a fixed LDO would look the same without the 2 resistors.