I’m beginner in electronics and i have project for grow LED light and i want to connect LED with timer. My inquire how I protect my circuit. My circuit

Adapter 12V 1.2A

Timer (5-20v) up to 5A

2 * LED CHIP (6.6V 0.300A) 2W One led (3.3V 0.300A 1W) i will conect series.

I want be sure 6.6 V will be arrive to LED chip not more to damage my LED I’m waiting for your recommendation.

Thank you

Is this circuit correct ?


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ With LEDs you should not look at the voltage so much, instead you want the current to be correct. So you do not want 6.6 V, you want 300 mA. Browse this site and search for "LED resistor" to learn how to calculate the proper series resistor. Put the specifications of the power supply and LEDs on separate lines because as it is now it is confusing. Also include a schematic of how you intend to connect everything. \$\endgroup\$ – Bimpelrekkie Jun 13 '18 at 12:12
  • \$\begingroup\$ You can use a zener diode in parallel to your input. Maybe if you show your schematics it would be clearer to indicate where \$\endgroup\$ – PDuarte Jun 13 '18 at 12:14
  • \$\begingroup\$ I concur with Bimpelrekkie with the added suggestion that you drive the LEDs with current sources (LEDs are current controlled and their forward voltage can vary quite a lot over manufacturing variations quite apart from temperature). \$\endgroup\$ – Peter Smith Jun 13 '18 at 14:04

Simple Ohm’s Law solution duplicated many times in this EE forum.

(V drop) / If = Rs

(12 - 9.9 )/0.3A= 7 Ohms

Pd=V drop * If = 2.1* 0.3A = 0.63W so to reduce 100’C temp rise use 2W part or two 1W 14 ohm parts in parallel or 8 * 1/4W resistors in parallel , each 8x 7 Ohms =56 Ohms 1/4W


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Must be the guy with a grudge again - the silent critic -1 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 13 '18 at 14:23
  • \$\begingroup\$ Thank you for reply. I will contact the LED by series way as i know the total for 2 LED will be 6.6v 0.3A. \$\endgroup\$ – Abdulaziz Jun 14 '18 at 8:49
  • \$\begingroup\$ R=(12-3.3-3.3)/0.3=18 ohms, should be use 18R 2W \$\endgroup\$ – Abdulaziz Jun 14 '18 at 12:18
  • \$\begingroup\$ Sorry I misunderstood your configuration. Yes correct but inefficient. If the R is rated for 2W at 25’C and dumps 5.4^2/18=1.6W what temp rise above ambient do you expect ? 1.6/2*100? Are you cool with that? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 14 '18 at 17:10
  • \$\begingroup\$ No, I will use it in room temperature. Are you recomend to conected in parallel as you mentioned. \$\endgroup\$ – Abdulaziz Jun 16 '18 at 11:46

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