Simple Circuit for high power LED

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SUMMARY

This discussion focuses on designing a simple circuit for a 3V high power LED with a maximum current of 0.8 amps, powered by 4.5 to 9V batteries. Users recommend connecting the LED in series with a resistor calculated as (Battery Voltage - LED Voltage) / Current, specifically using a 7.5 Ohm resistor for a 9V battery and a 1.875 Ohm resistor for a 4.5V battery. It is crucial to select a resistor with a power rating of at least 10 watts to handle the heat generated. For improved efficiency, utilizing a buck switching regulator is advised to reduce heat and extend battery life.

PREREQUISITES
  • Understanding of Ohm's Law for resistor calculations
  • Familiarity with LED specifications and requirements
  • Knowledge of voltage regulator ICs, specifically LM317 and L7805
  • Basic electronics skills for circuit assembly and heat management
NEXT STEPS
  • Research "Buck Switching Regulators" for efficient LED driving
  • Learn about "Constant Current Configuration" for LED circuits
  • Explore "Thermal Management Techniques" for high power components
  • Investigate "LED Series Configuration" to optimize power usage
USEFUL FOR

Electronics enthusiasts, hobbyists designing LED circuits, and engineers looking to optimize high power LED applications.

selseg
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What simple circuit that I can design for a 3V high power LED with max current 0.8 amps , I think the component names would help me , I would like to power it on 4.5 to 9 V batteries .Preferably lower voltages. Thanks
 
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Connect the 9v battery to the led in series with a (9-3)/.8 Ohm resistor.

On a 4.5v battery use a (4.5-3)/.8 ohm resistor.
 
Antiphon said:
Connect the 9v battery to the led in series with a (9-3)/.8 Ohm resistor.

On a 4.5v battery use a (4.5-3)/.8 ohm resistor.

Just make sure the resistor has suitable power rating - in the above scenario, the resistor is dropping 6 volts at 0.8A, which is 4.8 watts. I'd use at least a 10 watt resistor in that case, and that is going to be a LOT of heat... it would need mounted on a pretty beefy heatsink. The LED will need mounted on a big heatsink too.

If the OP can do it, it would be better to have more LED's in series rather than burning all that power as heat, or if you want to get more advanced, a buck switching regulator will get you into the 90%+ efficiency range and make the battery last 3 times longer and massively reduce the heat output.
 
These are voltage regulator ICs that ensure the LED is constant brightness even as the battery voltage drops lower.
 
Last edited by a moderator:
You may want some size-D or lantern batteries for something that draws this much current. (3) size D alkalines have a nominal voltage of 4.5. That might get you there without wasting too much of your power as heat.
 
selseg said:
Thank you very much , I checked from the link below earlier on . Why do they include the LM317 in that circuit ? in another circuit i found online , they used l7805. Why do they include these ics and capacitor ?
http://www.instructables.com/id/Super-simple-high-power-LED-driver/?ALLSTEPS
They're using voltage regulator chips in a "constant current" configuration.
 

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