# Most efficient way to drive an LED from 9v source?

I have a circuit that I am powering from a 9v battery. I bought little 3mm flat LEDs and I want to use one for a power indicator, but don't want to use a voltage divider that will consume too much power. The rated voltage is 3.2-3.4v and it came with a current limiting resistor of 430 ohms.

What would be the optimum method of reducing the voltage and limiting current such that I minimize power drain? I'm assuming it will be a voltage divider and/or a voltage divider, but what would be the optimum values?

(Update to question) Is it possible to use a single large resistor in series with the LED to both reduce the 9v down to 3.3v and provide the required current limiting function?

• Do you care enough about the efficiency to implement a switching supply? Jul 2, 2018 at 17:18
• No, I have major space constraints Jul 2, 2018 at 17:25
• There are endless series resistor calculators on the web ...here's one from Digikey: digikey.com/en/resources/conversion-calculators/… A single resistor is all you need to set the LED current. Jul 2, 2018 at 17:28
• "voltage divider and/or a voltage divider"
– Bort
Jul 2, 2018 at 17:30
• Jack, if you can tell me how to determine the forward current as an input to the calculator and create an answer from your comment, I'll accept it as the answer Jul 2, 2018 at 17:33

## 2 Answers

A voltage divider will waste power, as it will have current flowing through the lower resistor, as well as through the LED.

The 430 Ohm resistor will allow about 12 mA through the LED. A higher value series resistor will reduce the LED current, and decrease the brightness. You may find that a 1K or higher resistor will still allow the LED to produce sufficient light.

• Is it possible to use a single large resistor in series with the LED to both reduce the 9v down to 3.3v and provide the required current limiting function? Jul 2, 2018 at 17:20
• I believe that is what I sugggested in my answer. Jul 2, 2018 at 17:29
• The use of a single (series) resistor was the key to the answer I was looking for. While there are ways to calculate this, what I really needed was the minimum current draw with a viable brightness. I connected a 50K pot to the LED and find a high resistance that works with 9V. I settled on 15K, which works nicely at 9V and still illuminates with a 3.7V battery. Current drain is 423 microamps which by itself looks like about 1200 hours. Thanks also for your answer @Misunderstood, as that would have given a correct value, but simply dead reckoning seems to have worked fine here. Jul 3, 2018 at 2:10

Find an LED with the highest luminous flux output.
In a 3mm package this recently released green LED may fit the bill:
QT-BrightekQBL7IG30C

Keep in mind the discharge curve of a 9V alkaline drops to a cutoff of 5V. When selection a resistor use 7.5V as the supply voltage rather than 9V.

In the case of the LED referenced here the forward voltage below 5 ma is 2.85V.

With 18,000 mcd @ 20 mA you could use 1 mA (5KΩ, 7.5 mW) and still have 900 mcd.
You could likely get away with a 300 µA (15KΩ, 2.5 mW) and the 300 mcd may be sufficient.

Example for 1 mA @ 7.5V supply and 2.85 Vf.

NOTE: The graph for the forward current of a InGaN (green) begins a 5 mA but usually an LED will work well below 5mA.