I'm planning on adding leds to a small quadcopter which runs on a 1s lipo battery. I am planning on attaching the leds straight to the battery rather than the regulated voltage the main circuit board uses. The battery when fully charged has a voltage of 4.2 volts and a nominal voltage of 3.3.

Assuming an led with a 2.5v forward voltage and a 20Ma forward current. the calculations at either range result in either a 100ohm resistor (for 4.2 volts) or 47ohm resistor (for 3.3 volts). What size resistor should I use? should I use the 100 ohm resistor and let the brightness dim towards the end of the battery life?

In general what is the best way of selecting components where the system is not complex enough to warrant using a voltage regulator?


3 Answers 3


An extremely simple and low cost constantish current source is shown in the circuit below.

Q1 is initially turned on by R2.

I_LED flowing in R1 produces a voltage per V = IR
of V = I_LED x R1
When V_R1 reaches about 0.6V Q2 is turned on and shunts base drive to Q1, thereby limiting any current increase.

ILED max ~= I_R2 x Beta_Q1 where Beta = current gain.
For say 5V and R2=1k IR2~= (5-0.6)/1k = 4.4 mA.
For Q2 Beta = 100 Imax = 440 mA - ample for most purposes.
This circuit drops a minimum of ABOUT 0.6V

Cost, size and complexity are minimal.
Results are far better than using just a series resistor. ie Current is not "constant" but is far closer than with a resistor alone.


simulate this circuit – Schematic created using CircuitLab


enter image description here


  • 2
    \$\begingroup\$ "Constantish" :D \$\endgroup\$ Jan 18, 2015 at 12:26

If you use \$47\Omega\$ resistor, then current will be 17mA if battery voltage is 3.3V.
But at 4.2V, current will be = 36mA. But this exceeds your maximum allowable current through the LED. So choose a resistance value greater than \$\dfrac{4.2-2.5}{I_{max}}\$. Considering \$I_{max} = 20mA\$, you have to use \$100\Omega\$ here.


Use the 100 Ohm resistor - you don't want to exceed the LED's recommended maximum current.

The LED current will reduce, dimming the LED, as the battery voltage drops. However, the reduction in light output may not be obvious to the eye - modern LEDs are very efficient, and can be quite bright at low currents. (I'm using a green LED with only 1 mA, and it is much brighter than I expected - I haven't tried it at 20 mA!)

  • \$\begingroup\$ Is there a regulated voltage available? If so there is a simple, one-transistor current source you could use. It relies on the regulated voltage for a reference, but still draws LED current from the battery. You can also use a simple current source without any external reference (uses two diodes and a transistor). This will eliminate the problem of the LED's getting dimmer as the battery goes down. If you are interested I will happily sketch it. Just let me know. \$\endgroup\$
    – user57037
    Jan 18, 2015 at 1:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.