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I am trying to power a high power LED from a 9V battery. I have 3 10k 1/2 W resistors in series to throttle the current. I most likely will run it at 200-300mA. Is there a better way to do this instead of using resistors? I am new and tinkering around, but could I use a n mosfet and voltage divider with the same 9v battery to throttle the current that way? How does would a non-newb approach this to not waste the battery with heat.

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    \$\begingroup\$ @Tony DC to DC converter \$\endgroup\$ – Voltage Spike May 22 '17 at 20:52
  • \$\begingroup\$ Do you have a nine-volt pitcher to go with that nine-volt batter? More seriously though, try a switching converter with current feedback. \$\endgroup\$ – Hearth May 22 '17 at 20:56
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    \$\begingroup\$ A 9 volt battery will not provide 300 ma for very long. You are better off using enough C or D batteries in series to provide a volt or two above the requirement of the LED and then use a series resistor to limit the current. This will be more efficient and run much longer. \$\endgroup\$ – Barry May 22 '17 at 21:11
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    \$\begingroup\$ A 9 Volt battery won't provide 300mA at all if it's got three 10k resistors in series with the load. \$\endgroup\$ – Finbarr May 22 '17 at 21:44
  • \$\begingroup\$ You should be using a battery more suited to moderate to high current loads, and a current-mode regulator switching regulator. Realistically, just buy an off the shelf flashlight, they are extremely cheap for the parts required. \$\endgroup\$ – Chris Stratton May 22 '17 at 22:08
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9V batteries are meant for low current. They have high internal resistance and low capacity. They will not power your LED for long at all.

It is much better to use either:

  • A single 18650 LiIon cell (but you will have to purchase a suitable charger)
  • Or several AAs in series. This is much safer for a beginner (no Lithium fire) and I'm pretty sure you got NiMH rechargeables in the house already, so it's cheap.

A constant current DC-DC converter will have much higher efficiency than a resistor. You can get those at places like dealextreme, aliexpress or ebay... but it would be simpler to just buy a flashlight.

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  • \$\begingroup\$ I wasn't was aware that 9V batteries were an issue with current draw. I switched to 4 AAs which is a cheaper source and added another LED in series to help so I am burning less. Thanks for the inputs. \$\endgroup\$ – Tony Chains May 27 '17 at 18:07
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The best voltage to use is one only a little higher (~1 volt) than that required to light the LED. If it is a white LED, about 3 Volts is fine. Other LEDs work with even lower voltages. Once you have a good voltage level, because a LED is a DIODE, you need a way to limit the current. A dropping resistor works fine, but really only works properly with the one voltage it is designed for. It is possible to construct a current source which will provides a fairly constant current for a wider range of voltages. A fairly good one can be constructed with two transistors and three resistors. See 'Currrent-Limiting' wikipedia topic. Remember, the one transistors will be dissipating the excess power from your voltage supply so with a lot of excess voltage, they can drop quite a bit of power and may need heat sinks.

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  • \$\begingroup\$ Indeed, linear regulators waste power (ie, limited battery capacity) as heat, which is why most solutions today use switching regulation. \$\endgroup\$ – Chris Stratton May 23 '17 at 3:56

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