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I'm trying to learn some basics of electronics and electricity with a simple solution to a problem.

Problem: I have an African Violet in a terrarium that I want to help bloom, which means it needs more light.

Solution: Create a simple circuit with 2 high power grow LEDs connected to a battery (so it's portable) that I can place over the plant to extend the time and intensity of light it receives.

I did some math, came up with the resistors I needed and wired it together and it works. Great brightness, the resistors are handling the power and not burning up, etc. After some time, 30 min or maybe an hour, you can tell the LEDs are dimmer. I turn the battery off, wait a little, turn it back on and the lights are bright again.

What is happening? Is the current draw to big on the 9v?

circuit

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    \$\begingroup\$ 9V batteries are not known for their energy density compared to, say 1.5V batteries. See this link for an example: baldengineer.com/9v-battery-energy-density.html \$\endgroup\$
    – Phil N DeBlanc
    Nov 8, 2017 at 20:31
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    \$\begingroup\$ You are likely right, the battery is being drained too hard. Use a multimeter to measure the voltage when the LEDs are dim. You should either switch batteries or consider using many in parallel (preferably same brand/model and new). \$\endgroup\$
    – Wesley Lee
    Nov 8, 2017 at 20:33
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    \$\begingroup\$ how long are the leds bright again? i guess far shorter than the first time? \$\endgroup\$
    – Jogitech
    Nov 8, 2017 at 20:44
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    \$\begingroup\$ Is the LED heatsunk properly? Is there a noticeable difference in its temperature when it's off compared with after it's been running for a while? \$\endgroup\$
    – kjgregory
    Nov 8, 2017 at 20:49
  • \$\begingroup\$ @PhilNDeBlanc Good read, maybe I will look in to trying 6xAA. \$\endgroup\$
    – ndyr
    Nov 8, 2017 at 20:49

3 Answers 3

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It's a combination of a few things. 9v batteries are low capacity batteries designed for small current draws over a long time. Due to their chemistry, typically multiple 1.5V alkaline cells or piles, they have a high internal resistance, and as more current is drawn, the bigger the voltage drop across the battery's internal resistance. Just like the leds and the resistors, this internal resistance varies based on the temperature it is at, and high current draw will cause it to vary causing your device/leds to see different currents too.

Add to this, the voltage recovery effect seen in batteries, where energy in a battery that has experienced a high discharge takes a while to distribute within the battery chemistry. At the discharge rates I calculated (~300 mA) it will only last about 2 hours anyway, regardless of recovery. As the voltage drops, the discharge rate changes, but the useful life is low for what you want to do.

In short, high current draw is not good on 9v batteries and the recovery effect you see is a side effect of it.

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Your circuit is highly inefficient and it's not a good project to run off a battery.

The math you missed is the battery discharge rate. You also underestimated the amount of light required. You will need up to 10 watts.

Use a Mean Well HLG-40H-48 Type A LED driver to drive a string of 16 LEDs.

Two LEDs will not generate enough PPFD for the plants. You may need a PPFD of 150-300 µmol/m²/s

If you have to have battery wire them in parallel using a CCR (e.g. On-Semi NSI50150ADT4G or Microchip MIC4802) for each LED rather than resistors. Use a lower voltage like a Li-ion or Li-po 3.6V battery.

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The "performance" that you are describing, is typical of most batteries. As a battery is "used up," its output voltage drops. This is caused by an internal chemical process. Breaking the connection allows the battery to recover, so its voltage goes back up (although not to 100%), and the process repeats, until the battery is discharged to the point the light will not light.

I recommend using rechargeable 18650 Li batteries (in parallel), connected directly to the LEDs (no resistors required).

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