7 Light Activated led driver (1.5V joule thief)

This is another LED driver circuit powered by joule thief circuitry using 1.5V cell. The only difference is it has light detection that automatically turns on the led when the environment is dark.
schematic diagram

Part list:
R1 – any light dependent resistor (LDR) or photoresistor
R2 - 100 kilo ohms potentiometer
R3 – 220 ohms 1/4W
R4 - 120 kilo ohms 1/4W
R5 - 2 kilo ohms 1/4W
Q1 - 9012, 2N4403, or any similar PNP transistor
Q2, Q3 - 2N2222, 9013, or any similar NPN transistor
C - 22pF to 47pF capacitor
L - any 100uH to 1mH inductor
LED - any type of LED  (see LED pinout in case you forgot)

Part list diagrams: Click to diagrams to enlarge
PNP transistor pins
NPN transistor pins

You can adjust sensitivity of the light detector by varying the value of potentiometer R2. Full value of R2 means highest sensitivity.

Note:  see Sir Acme Fixer recommendation on how to construct the joule thief in the comments section below.


  1. The collector of Q2 must go to the bottom of R5, not the top. It will not work with the connection the way it is.

  2. Thanks for your input, I appreciate your idea. Actually I have seen other versions of this circuit that is working fine. Maybe one working versions is the one you are trying to point out.
    There is a chance that the circuit will not work if you will connect the collector of Q2 is in the bottom part of R5. The design above is ok and working fine, but if you have tried building it and it’s not working kindly report it so I can reconstruct and re design it for you.

  3. I assembled it and it works, kind of. When I adjust the pot, it doesn't act the way I expected it to act. I'll have to take some measurements later. It's getting late.

  4. Thanks for your info
    This is how the circuit work:

    Case 1: When surrounding is dark
    LDR resistance is much greater than the sum of resistances of R2 andR3, and then circuit is ON

    Case 2: During the day (well lighted surrounding)
    LDR resistance is much lower than the sum of resistances of R2 andR3, and then circuit is OFF

    If pot is not working as the way it is expected, try paralleling 120 kilo-ohms resistor across the LDR, or try increasing the values of R2 and R3 to satisfy the 2 cases above.

  5. Excellent publish. I'm dealing with several these troubles.

  6. I'm no longer posting anonymously because someone else has now begun to also post anonymously. I don't want to be confused with the other poster.

    I've had a chance to do some experimenting and measurements on what I built. Here are some of my results and conclusion.

    When I adjusted the R2 100k pot, I expected it to only adjust the sensitivity of the CdS light sensor. Instead, it also adjusted the brightness of the LEDs. I temporarily put a 1 ohm resistor in series with the LEDs to monitor the LED current. With R2 in the maximum position the current was about 8 milliamps. With R2 in the minimum position, the LED current was zero (the LED was dark). When I adjusted R2 away from the minimum the LED got bright and its current peaked at over 10 mA.

    My conclusion:
    Q2 acts as a switch to turn the base of Q3 on and off. But to do this, Q2's collector must have a load resistor. The only thing between Q2's collector and the + supply is the collector of Q1. The resistance of Q1's collector has to act as the load resistance for Q2. But Q1 is really supposed to act as a switch and turn fully on which supplies current to the bases of Q2 and Q3. When it's dark, R1 CdS cell is open, almost infinite resistance. R2 varies the mount of current that can go to the base of Q1, which in turn varies the collector current of Q1 and hence its resistance. That in turn varies the amount of current going to both Q2 and Q3, and the brightness of the LEDs. If I turn the R2 to its lowest point, the amount of current going to Q1 is so high that the collector looks like a short to the collector of Q2, and it can't divert enough current from Q3 to turn it off. That causes the LEDs go dark.

    I hope someone else builds the circuit as I have done, because I believe the unexpected results I'm getting are typical for the circuit as shown. I thank you for encouraging me to build it, because I've learned a bit more about this circuit even though I've built similar circuits in the past.

  7. To Acme Fixer
    Big thanks for sharing your experience in building the circuit. Information you’ve shared is a great help to other readers,hobbyist, and especially to Simple Electronics.