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Automatic Solar Light using a Relay Changeover

This automatic solar light circuit using relay changeover is designed for charging a battery during day time or as long as the solar panel is generating electricity, and for illuminating a connected LED while the panel is not active.

Upgrading to a Relay Changeover

In one of my previous article which explained a simple solar garden light circuit, we employed a single transistor for the switching operation,

One disadvantage of the earlier circuit is, it does not provide a regulated charging for the battery, although it not might be strictly essential since the battery is never charged to its full potential, this aspect might require an improvement.

Another associated disadvantage of the earlier circuit is its low power spec which restricts it from using high power batteries and LEDs.

The following circuit effectively solves both the above two issues, with the help of a relay and a emitter follower transistor stage.

Circuit Diagram

Automatic Solar Light using a Relay Changeover

How it Works

During optimal sun shine, the relay gets sufficient power from the panel and remains switched ON with its N/O contacts activated.

This enables the battery to get the charging voltage through a transistor emitter follower voltage regulator.

The emitter follower design is configured using a TIP122, a resistor and a zener diode. The resistor provides the necessary biasing for the transistor to conduct, while the zener diode value clamps the emitter voltage is controlled at just below the zener voltage value.

The zener value is therefore appropriately chosen to match the charging voltage of the connected battery.

For a 6V battery the zener voltage could be selected as 7.5V, for 12V battery the zener voltage could be around 15V and so on.

The emitter follower also makes sure that the battery is never allowed to get overcharged above the allocated charging limit.

During evening, when a substantial drop in sunlight is detected, the relay is inhibited from the required minimum holding voltage, causing it to shift from its N/O to N/C contact.

The above relay changeover instantly reverts the battery from charging mode to the LED mode, illuminating the LED through the battery voltage.

Parts list for a 6V/4AH automatic solar light circuit using a relay changeover

  1. Solar Panel = 9V, 1amp

  2. Relay = 6V/200mA

  3. Rx = 10 ohm/2 watt

  4. zener diode = 7.5V, 1/2 watt

Need Help? Please leave a comment, I'll get back soon with a reply!


  1. I have 6v solar panel.which changes can i make in the parts list?

  2. Dear sir, thanks for sharing this design. what are the changes to be done for illuminating 5 LEDs with 1 watt power each so that total LED output is 5 watts. Thanks

  3. 6V panel will not work for the above application, you can try the following instead:

  4. Dear Gopal, you can simply use a 18V/ 1amp panel, 12V/400mA relay and a 12V/7AH battery in the above application for illuminating 5 nos of 1 watt LEDs in parallel, just make sure each LED includes a series 10 ohm 2watt resistor.

  5. How to calculate the biasing resistor?

  6. you can learn it here:

  7. But here tip122 is configured as a transistor regulator, unlike the relay driver in your link. Here base resistor is calculated on which formula?

  8. Google "emitter follower calculation" you will find the required formulas.

  9. Hi sir, I have implemented the design and here are the queries/findings/doubts.

    1. I did setup Panel 20V/20W, 12V relay, 15V zener, 12V 9AH battery for lighting six 1 watt white LEDs in parallel with 10 ohms/2 watt resistor in series with each LED. When solar input is given (> 18 V), the relay works perfectly fine and latches from NC to NO and charges the battery (voltage is clamped by the zener voltage close to 15V). But when we remove the solar input from the circuit, the relay is not returning back to NC from NO. The reason which I noticed is the battery voltage of 12V is getting applied to the zener diode, diode and hence to the relay. When we disconnect the battery once then the relay moves to NC. Do let me know what mistake I am doing. Also whats the solution to correct it?

    2. One more thing, when connected with 12V 9AH battery, the voltage drops across each LEDs and 10 ohm/2 watt resistors are 5.7V and 6.2V respectively and current flowing is 630mA. So should I put 30 ohms / 2watt resistor instead of 10 ohms to limit the excess current flow? What is the allowable current and voltage for the 1 watt high power white LEDs?

    3. What is the calculation by which you arrived at 10 ohm / 2 watt resistor? It would be great if you can explain that. My calculation was: R = (12V - 3.6V)/.35A = 24 ohms where 3.6V and 350mA are the allowed voltage drop and current of the 1 watt white LED. Is this correct?

    4. Also, charging time of the 12V 9AH battery would be 9AH/1A = 9 hours (ideally on full sunlight continuosly). Discharge time (12V * 9AH)/(3.6V * .35A * 6) = 14 hours (on full charge). Please confirm these as well.

    Thanks a lot for your time.

  10. Thanks VJ, but that's quite strange, because to keep the relay latched ON even after the panel is removed the TIP122 emitter must be leaking current in reverse path which cannot happen according to me unless the transistor is faulty or connected wrongly.
    anyway try adding a diode between the TIP122 emitter and relay N/O, and check the results (anode to emitter and cathode to N/O)

    10 ohms was recommended for 6V supply for 12V it should be:

    12 - 3.3 / .3 = 29 ohms, that's right around 30 ohms.

  11. Hi sir, I re-verified the TIP122 pinout (1)base,(2)collector,(3)emitter. Also tried with 3 other ones. All are giving the same result. Anyways, connected the diode between emitter and N/O and now it works perfectly fine. Thanks a lot for clarifying it.

  12. Hi VJ, the internal structure of a BJT or similar devices can be strange and difficult to understand, anyway I am glad that the problem could be solved with the diode inclusion, however this will cause around 0.7V less to reach the to counter this you can either add a balancing diode in series with the 15V zener diode (with opposite polarity)....or simply replace the 15V zener with a 10 k preset and adjust it such that the emitter output becomes exactly 14.4V for enabling optimal charging conditions for the battery.

  13. Yes you are absolutely right. Will do it and confirm. The battery charging voltage was 13.8 max which I made a note during daytime. Meanwhile, When the LEDs light up at night, the battery voltage got discharged from 13.8V to 7.5V in 6 hours. Is this OK? What is the lower cutoff voltage for a 12V 9AH battery? As per my calculations specified in my previous question - charging time of the 12V 9AH battery would be 9AH/1A = 9 hours (ideally on full sunlight continuosly). Discharge time (12V * 9AH)/(3.6V * .35A * 6) = 14 hours (on full charge). Looking at the present discharge rate, I don't think it will last for 14 hours. Please confirm this as well. Thanks a lot for your time.

  14. You should never allow a 12V battery to discharge below 11V under any circumstances, otherwise it will get damaged quickly.

    your discharge rate calculation is correct 9AH/1A = 9 hour, but that may be true under extremely ideal situations and at 100% efficiency which is not possible even with a new battery.

    so practically you can assume the back up time to be 60% of the calculated value that is around 6 hours (for 1amp discharge rate), that too if the battery is fully charged.

    so for 0.35amp it should be three times more that is around 18 hours

    In order to ensure a 95% charge make sure the charging voltage is around 14.5V and the current is at 1amp, you can easily confirm these with a digital multimeter

  15. Thanks for the details. I have implemented the battery low voltage cut-off circuit and now it cuts off at 11V. But the actual discharge time from 13V to 11V is only 4 hours (max) for four 1 watt LEDs. Current consumed by 1 LED is 0.26A @ 3.95V. So total current is 0.26A X 4 = 1.04A. Is there any way to improve the time duration of LEDs? Also, will PWM input to the LEDs would help in achieving such improvement? Do let me know. Thanks

  16. please check the charging current, it should be around 1amp for the mentioned 9AH battery, and the charging period should be enabled for 10 hours at this rate.

    The LEDs should not be operated above 3.4V preferably.

    PWM won't help because it will lower the consumption but will also lower the illumination.

  17. Hi Swagatam, Can this circuit be used for running a 12V BLDC fan operated using solar energy directly during day time and using battery (charged from solar in the daytime) during the night time? Pl let me know the modifications. Thanks

  18. Hi VJ, Yes it can be possible, but only if the solar panel is adequately rated to simultaneously operate the motor as well as charge the battery during daytime.

  19. ...the panel rating will depend on the amp rating of the motor...

  20. Dear sir,
    How can i modify the same circuit foe 12v 40Ah battery and 75Watt solar panel please help me....

  21. Dear Bhanu, just replace TIP122 with TIP147 and use a 6 amp rated solar panel

  22. correction, use TIP142, not TIP147

  23. Dear sir,
    I have 100w 5.1A panel and battery of 14.5v 470wh.pls sir help me with the circuit that can charge laptop of input 18.5v 3.5A & run dc fan of 18w even at the night time.
    Thank you

  24. Abubakar, you can use any IC555 based DC to DC boost converter for converting 14V to 18V and use it for charging your laptop battery and operating the fan.
    here's one example

  25. "Sir,my requiment is to use solar panel to the circuit during the day time,but the panel is 19v &battry is 14.5v all they exceed the circuit input.any idea to modifier the circuit.
    2,suggest me a circuit that serve as a controller to charge the above battry using 19v panle.
    Thank you!

  26. Abubakar, you can simply use an LM338 circuit for this purpose:


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