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Help understanding solar lamp circuit

Discussion in 'General Electronics Discussion' started by ***Ed***, Aug 27, 2014.

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  1. ***Ed***

    ***Ed***

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    Aug 27, 2014
    Hi,

    My garden solar lamps failed so I drew the circuit to help my fault finding. It looks simple but I just can't understand its function.Can anyone please help?

    Thanks, Ed
     

    Attached Files:

  2. Gryd3

    Gryd3

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    Jun 25, 2014
    Something looks wrong... Check Q3 again.
    What part are you having trouble with? Appears as though the solar cell is being used to switch the unit on/off
     
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  3. ***Ed***

    ***Ed***

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    Aug 27, 2014
    The problem I have is that the conditions for the LEDs to be on appears to only occur during daytime which is the opposite of what should (and did) happen.

    When Q3 is not conducting the LEDs will be on ie at nighttime.

    Q3 will not conduct when its base emitter junction voltage is zero. This occurs when Q2 (and Q1 which appears to be always conducting due to negatively biased emitter base junction) is conducting. For Q2 to conduct there needs to be a voltage output from the solar panel ie it is daytime.
     
  4. Gryd3

    Gryd3

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    Jun 25, 2014
    The reason Q3 looks odd to me in your drawing is that it shorts out the LED when it conducts...
    In doing so it has essentially shorted out the supply through L1. I wanted to confirm your drawing as accurate
     
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  5. ***Ed***

    ***Ed***

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    Aug 27, 2014
    I can confirm that is the way it is. Also, the battery is actually 1.2V not 1.6V.
     
  6. BobK

    BobK

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    That is a boost circuit. When Q3 is on, L1 is building up current, when Q3 is off, L1 produces a higher voltage across the LED to light it. Not sure how the oscillator works, though, or what keeps in off in daytime and on at night.

    The configuration of Q3 and the LED are the same as in the Joule thief.

    Bob
     
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  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Q1 and Q3 are a two-transistor relaxation oscillator. Q2 kills the oscillation when the solar panel voltage is high.

    It would help a lot if you rotated the whole circuit 180 degrees, so the negative (0V) rail is along the bottom and the battery is on the left.

    Can you redraw the circuit like that, and mark the voltages on the collector and base of Q2 in darkness (with the battery charged) and under a bright light, measured relative to the 0V rail (Q2 emitter, etc). Also upload a photo of the board.
     
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  8. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Bobs right!
    They often do this to run LEDs from 1.2 Volts. The upside down circuit takes some getting used to. Could someone turn it around?
    Adam
     
  9. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Beat me again Kris :)
     
  10. davenn

    davenn Moderator

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    Sep 5, 2009
    OK

    [​IMG]


    still think as some one else said ... the LED shouldn't be wired like that

    Dave
     

    Attached Files:

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  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    No, that circuit is OK. The LED lights up each time Q3 turns OFF, due to the back EMF from L1. Here's a similar circuit I use for running a white LED from a AA or AAA cell for a miniature torch (flashlight). It does the same thing.

    WHITELED.GIF
     
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  12. davenn

    davenn Moderator

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    ahhh OK Kris ... just looked weird to me with t he LED across the transistor

    shows my lack of circuit analysis haha

    D
     
  13. ***Ed***

    ***Ed***

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    Aug 27, 2014
    I've redrawn the circuit and measured the Q2 base emitter voltage in darkness at 0.73V and the Q2 collector emitter voltage as 0.04V. However I've found a problem as the solar panel is not producing any voltage.

    Circuit photos uploaded for your info. Looks a bit rough as has been in my garden for 4 years and also I've fiddled a bit and bypassed the on off switch.

    I wonder what is wrong with the panel - again it looks fairly simple, I'll see whether it can be fixed... Solar Lamp w voltage.png

    IMG_4671.JPG IMG_4672.JPG
     
    Last edited: Aug 28, 2014
  14. ***Ed***

    ***Ed***

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    Aug 27, 2014
    I've fixed a loose connection on the solar panel, it generates 2V.

    In circuit in 'daylight' Q2 ce voltage is 0.05V and Q2 be voltage is 0.67V.
     
  15. BobK

    BobK

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    So is it working now?

    Bob
     
  16. ***Ed***

    ***Ed***

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    Aug 27, 2014
    Yes it is.

    I'd still like to understand how the circuit works so will try to digest all that's been written. Thanks for all your replies.

    The solar panel repair involved putting two big blobs of solder in place of the previous two blobs, one of which presumably was no longer in contact although I couldn't see what it was that either was connected to - nothing obvious.
     
  17. bigone5500

    bigone5500

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    I have found that the cheap $1 garden lights at walmart usually stop working because of this. But for some reason I continue to purchase them...
     
  18. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    When the solar panel is illuminated by sunlight, the voltage on the base of Q2 is about 0.7v as it cannot rise any higher than this voltage. This turns the transistor ON and the voltage on the base of the output transistor is about 0.3v and it is kept in an OFF condition.
    When the solar panel does not receive any illumination Q2 is not turned ON any more and you can consider it to be removed from the circuit.
    C1 now charges via the 20k resistor and when the voltage across it is about 1.2v, Q1 is turned ON and this turns on Q3 via the 330R resistor.
    Q3 pulls the right side of the 100 micro Henry inductor towards the 0v rail and it pulls the top of C1 down.

    This causes the energy in C1 to be delivered to the PNP Darlington transistor and has the effect of turning it on to its maximum capability. This turns on Q3 to its maximum.
    Both transistors are now saturated and when all the energy from capacitor C1 has been delivered to Q1, the transistor starts to turn OFF a small amount.
    This turns OFF Q3 a small amount and the voltage on the right lead of L1 rises. This action is transferred through C1 to the base of Q1 to turn it OFF more.
    Very quickly both transistors get turned off and you can consider them to be removed from the circuit.
    The current flowing through the 100uH choke ceases and magnetic flux produced by the current collapses and cuts the turns of the inductor and produces a voltage in the opposite direction. The voltage emerging from the right-hand end of the inductor is a POSITIVE voltage and it adds to the voltage supplied by the battery.
    This voltage can be considerably higher than the applied voltage and when the two are added together, they cause the LED to illuminate.
    This voltage is only a spike and when it ceases, the LED turns off.
    C1 then starts to be charged via the 20k to start the next cycle.
     
    Last edited by a moderator: Aug 31, 2014
  19. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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