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I found this "Plant Grow LED light" circuit on the Internet and I wanted to know if: (Please don't feel pressured to answer all my questions. You can just answer a few only if you like.)

  1. this is correct?
  2. the best way to build this circuit?
  3. is the cooling fan necessary?
  4. what is the heat threshold necessary to implement a cooling fan?
  5. how can I determine the amps correctly of 12 volt power supply if the label on the power supply is wrong? I use a 120 volt outlet here in Toronto, Canada.
  6. I have old 12 volt power supply but it gets too hot to the touch. Should I discard that?
  7. Is R1 0.20 ohm resistor the same thing as a 20 ohm resistor?
  8. How can I add a switch with a potentiometer? What kind of ohms?
  9. Could a 555 timer be used to increase the amount of LEDs to double by alternating 2
    separate circuits back and forth - on and off? This would appear to be 20 led light on simultaneously but really a set of 10 go on while the 2nd set of 10 turn off and then in a split second later this reverses the 1st set of 10 turn off and the 2nd set turn on. And so forth.

I'm wanting to grow Orange Pixie Tomatoes.(I bought the seeds from Tomato Grower's Supply Company I'm just a customer, have no endeavor with them, just in case your curious about that tomato cultivar.)

The circuit to be built would hang 4 inches above the plant. I'm scared that if I purchase a 50 WATT(includes a cooling fan) grow light this would catch fire when I'm sleeping or away from home. Thus I thought a 12 volt circuit would be safe.

Later I would like to show my final draft which will include white LEDs to produce better nutrition in the tomato plant. Many grow lights now come with between 10.66% to 25% white lights to produce carotenoids in the plant as a health benefit I believe. So thats one mistake the circuit has.

This is my first time here in posting. Thank You for reading this.

enter image description here

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  • \$\begingroup\$ If supply is a steady 12V (as many modern regulated supplies are) then the voltage "headroom" allowed is on the VERY low side for the LH LED string at 0.2V and low ish for the rh string at 1.1V. LEDs are not as precise as shown if Vf ratings. If yout lh red LEDs ran at say 2.05V at 1A then ILED = 3 Amps and ... . Using one less LED per string and a larger resistor (more volt drop) OR a constant current driver would work better. You can buy CC LED drivers on ebay for very little. \$\endgroup\$ – Russell McMahon Jan 19 '15 at 10:05
  • \$\begingroup\$ My constant current driver for this answer would work but has about 1 V min drop. Transistors need to be rated correctly and resistor values and R1 wattage need to suit. This is not a fantastic CC cct but probably meets your need. Design for minimum acceptable driver V drop to minimise thermal losses. \$\endgroup\$ – Russell McMahon Jan 19 '15 at 10:07
  • \$\begingroup\$ Thank You Russell for the reminder of getting the right resistors. 0.2 ohms, 2W sounds very tight. VERY low side as you mentioned. A meter would help. \$\endgroup\$ – Fraser Dunn Jan 21 '15 at 7:20
  • \$\begingroup\$ it's not just 'tight' - it's 'ill conditioned' . Nice way os saying that it has no hope of working. SMALL changes in the LED Vf utterly swamp the adjustment made by resistor value. Use less LEDs and larger R OR my CCish cct. Cheap and easy. \$\endgroup\$ – Russell McMahon Jan 21 '15 at 8:01
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  1. Depends on what the LEDs are rated for. The left branch LEDs have a forward drop of 11.8V, and the right branch drops 10.8V. The limiting resistors at the bottom of each branch should restrict the current to 1A, so make sure your LEDs are able to handle that current.

  2. A better way to drive this would be with a constant current LED driver. A switching regulator will be far more efficient and you won't be burning as much power as heat (if that matters for your application). The method you have here will work fine as long as your supply stays fixed. If it sags at all your LEDs will dim proportionally.

  3. Depends on the LEDs. You need to read the specifications for whatever in particular you're using. See how much power the LEDs will dissipate thermally, and look at the environment you're running them in. If you're in a well ventilated space you could get away with just affixing them to a metal heat sink if the average temperature of your space isn't too high. If it's in a tight or enclosed space you might want to have the fan just to force some circulation over the LED heat sinks.

  4. Depends on the specs of your components, the environment you're operating in, and how effective your passive heat sinks are.

  5. Read the specifications of your supply. Generally you buy the supply to fit the loads. In this case you have 2A drawn by the LEDs (1 A in each of the parallel LED branches), plus whatever your fan draws which isn't indicated on the schematic.

  6. Might be a good idea, there's the potential the supply could catch fire if overheated significantly, however it's more likely it will just burn out. Are you overdrawing the intended output current from it? If that's case you can keep it for a lighter load. If it's just getting hot while unloaded or at normal load recycle it.

  7. No. There's a 100x difference between the two.

  8. Switch is not an issue, just make sure it's rated for the current. Keep in mind if you just put the potentiometer in series with the LED supply whatever power doesn't get used by the LEDs will get dissipated as heat in the potentiometer...which might hurt. It's not an efficient way to do it. Better option would be to have an adjustable constant current regulator. Or you could use a MOSFET and a 555 timer to switch the lights at a fixed frequency and adjust the duty cycle to control the brightness.

  9. No. You don't have the voltage to drive any more LEDs down either branch. I just saw the edit though. You could use the Q and /Q outputs of a 555 to drive some FETs to alternately run two separate circuits.

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  • \$\begingroup\$ Depends on what the LEDs are rated for. The left branch LEDs have a forward drop of 11.8V, and the right branch drops 10.8V. The limiting resistors at the bottom of each branch should restrict the current to 1A, so make sure your LEDs are able to handle that current. >>> 1. Thank You alphasierra I shall make sure the LEDs that I purchase have this restricted to a certain amperage such as 1A. \$\endgroup\$ – Fraser Dunn Jan 20 '15 at 5:51
  • \$\begingroup\$ 2. Thank You alphasierra I shall aim(plan) to add a constant current LED driver to the schematic. This will keep the LEDs at “proper” illumination and emitting colour. \$\endgroup\$ – Fraser Dunn Jan 20 '15 at 5:52
  • \$\begingroup\$ 3. I’m surprised to learn that even a 12 volt project may require a fan. I therefore presume that even 1.5 volt projects such as a wrist watch may need a fan. Must look a little further into this topic. \$\endgroup\$ – Fraser Dunn Jan 20 '15 at 5:53
  • \$\begingroup\$ 4. That looks to me like a compete answer thank you alphasierra: components specification(design), environment(hot, humid, wet, dusty, lack of air flow, electronic components near a hot electronic component, an earthquake 11.0, etcetera), heat sinks(wire grade, parts grade). \$\endgroup\$ – Fraser Dunn Jan 20 '15 at 5:54
  • \$\begingroup\$ 5. Thank You this is IMPORTANT! Amperage is increasing. Need the right power supply to fit the amp loads! \$\endgroup\$ – Fraser Dunn Jan 20 '15 at 5:55
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1) Maybe yes, maybe no. The left branch, in particular, assumes an unreasonable precision in specifying the LED voltages. I assume you have access to data sheets for the LEDs you're hoping to use, and that's where you got the numbers. Check them again. Those are probably typicals. If they are maximums, the design is in trouble. Let's say, for instance, that the forward drop for each LED is .05 volts less than you're showing. Then the total forward drop will be 11.55 volts, and the current for that branch alone will be 2.25 amps. To some degree the LEDs will compensate by requiring a greater voltage drop for this current, which will reduce the current level, but the end result will still be higher than you think, and this will get you in trouble with your power supply AND may be more than the LEDs are rated for.

2) Depends on what you mean by "best". If you've already got a 12 volt supply that will do the job, the additional resistor cost will be much less than getting a constant-current supply, and if cost is your metric, then your circuit is, indeed, best. If you want consistent light levels over time and temperature, then no, it's probably not too good, let alone best.

3) As alphasierra stated, it depends on your LEDs. It also depends on your mounting. If you put all 10 LEDs on 4 x 5 inch piece of plywood, a fan is not necessary since it won't work and your LEDs will die no matter what you do. If it's a piece of aluminum, the size and shape will matter a whole lot (thicker is better, for instance) and a fan may be necessary (to keep the LEDs from dying in a matter of minutes or hours), a good idea (to prolong LED lifetime) or unnecessary. It depends on how hot the LEDs get without a fan.

4) Like alphasierra says, it all depends. The must-have threshold occurs when, without a fan, the LEDs exceed maximum rated temperature.(Note - this is the LED temperature, not the mounting surface temperature. If you do a crappy mounting job the LEDs will bake while the heatsink stays cool.) And if you add a fan and the temperature is still over the rating, you need more fan or, more likely, a complete rethink about how you're mounting your LEDs. The cooler you keep the LEDs, the longer they will last. The data sheet is your friend.

5) Are you talking input or output error? If it's a 220 volt supply, and you plug into 120, the odds are very good that the output voltage will not reach rated value. If you're talking about output current, the only way to tell is to load the output with a known load and check the voltage. In your case, you need a 6 ohm resistor (at 25 watts minimum). This will draw 2 amps at 12 volts. If you connect it to the supply output and the output drops by more than a few 10s of millivolts, the supply is probably not happy.

6) Like alphasierra says - if it's getting hot with no load, don't use it.

7) No.

8) Don't do it. You will not easily find a variable resistor which will handle the current levels you're dealing with. (They're called rheostats.)

9) Absolutely. Of course, each bank of LEDs will only be on for half the time, so the total light will remain the same, so you might want to rethink this one.

And finally, fire concerns. Wattage alone is not the issue - it's how the power is handled. If you make a 50-watt unit which runs at 40 degrees C, and you don't install it so that airflow gets blocked, and you replace it with a 25-watt unit which runs at 70 degrees C, and are careless about your mounting, guess which one will be the bigger hazard?

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  • \$\begingroup\$ 1. The numbers are not mine but rather a schematic from the Internet. I believe they are typicals yes. “To some degree the LEDs will compensate by requiring a greater voltage drop for this current, which will reduce the current level…” So the compensation makes this even worse you are saying? Interesting! This is so important to know! I learned something vital! I Wonder how the LEDs compensate to make matters worse? How they electromechanically do that!? \$\endgroup\$ – Fraser Dunn Jan 21 '15 at 7:35
  • \$\begingroup\$ 2. Yes this is cost metric and as you say a voltage regulator is some item I wish to add to keep the "proper" intensity and colour for growing those tomatoes at home. \$\endgroup\$ – Fraser Dunn Jan 21 '15 at 7:38
  • \$\begingroup\$ 3. I wanted to build a stand alone cool temperature project without using a fan. \$\endgroup\$ – Fraser Dunn Jan 21 '15 at 7:44
  • \$\begingroup\$ 4. Exceeded temperature for the parts – check. Must mount good to keep components cool– check. Fan creates an ironic problem since adding one makes the load of the circuit greater and in turn generates more heat caused by the fan – check! THAT IS VITAL! Cost Metrics again, does a person design for intensity and low life LED or medium intensity and medium LED life or low intensity and long lasting LED life? This is gardening so looks like somewhere above medium. \$\endgroup\$ – Fraser Dunn Jan 21 '15 at 7:52
  • \$\begingroup\$ 5. Good point the error could be input or output. A good Main Fuse box looks after the input error. For the output error have some item that I know the actual load – check. Get a super resistor. I’ll have to research the idea of a 6 ohm resistor that can withstand 25 watts minimum. Have to get a book and learn or re-learn this matter and other matters. Thank You. \$\endgroup\$ – Fraser Dunn Jan 21 '15 at 7:59

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