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Blog Gr0G - Follow up
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  • Author Author: amgalbu
  • Date Created: 31 Dec 2020 12:48 PM Date Created
  • Views 315 views
  • Likes 4 likes
  • Comments 0 comments
  • 1meterofpi
  • gr0g
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Gr0G - Follow up

amgalbu
amgalbu
31 Dec 2020

A very short update: there are more and more sprouts, however they are growing slower than I expected and there are a lot of sprouts on the wrong side of the vase. Need more light?

 

imageimage

YES, the problem was definitely that the amount of light provided by LED strips (mounted on the edges of the Gr0G cube) and by the high efficiency grow LEDs (mounted on a standpoint in the middle of the cube) is not enough

So the only quick-and-dirty solution I can apply right now is to provide more current to the LEDs strips. LEDs strips are currently powered by 5V: I could use a 12V power supply and add a current limiting resistor

 

Back to the basics

LEDs are typically 3-volt devices. What this means is that if a 3-volt differential is applied between the positive and negative ends of an LED, it will light up. The LED strips I am using is rated for 5V, this means I have a LED and a current limiting resistor connected as shown in figure below

 

image

 

This is the "building block" of the LED strip. Many of these building block are connected in parallel

 

image

 

If I want to draw more current, I can connect 12V instead of 5V and add a current limiting resistor to keep the current just a little above the 5 mA (which the typical value for the current for this kind of LEDs. This will obviously affect the duration of the LEDs, but, as I said, this is just a temporary solution)

The value of the current limiting resistor is 470 ohm, so the current through the LED is about 4 mA

If I increase a little bit the voltage applied to the parallel circuit, the current will be increased as a consequence.

 

I calculated the number of LEDs installed: there are 96 LEDs on all the 12 edges of the cube (8 LEDs on each edge). If I want the the current in each parallel circuit is 5mA, the current supplied by the power source is

5mA x 96 = 480 mA

 

This means that the I need to add a resistor to drop voltage from 12V to 5V (i.e. a voltage drop of 7V). According to Ohm law:

7V / 480mA = 15 Ohm

 

So if I connect a variable resistor that can vary from 0 to 100 Ohm, I can increase a little bit the current and thus the brightness of the LEDs

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