Charge a 7.2 V battery with a 5 V solar panel

Discussion in 'Power Electronics' started by DavidB3, Jan 13, 2019.

  1. DavidB3

    Thread Starter New Member

    Nov 28, 2018
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    Hi.

    I am trying to "convince" a 5 V/140 mA solar panel to charge a 7.2 V 1300 mAh NiMh battery, even in poor light.
    I think the best way is to convert the output of the solar panel to a constant current of ~ 0.2C = 260 mA.
    Perhaps by storing the energy in a capacitor; when the voltage hits a certain level, it is discharged into the battery.

    The circuit have to be very power efficient, even when the solar panel gets poor light.

    This is what I've done so far:

    schematic.jpg

    B1 and R8 simulate the solar panel. C1 is the "buffer" (multi-layer ceramic). R2 is the internal resistance of L1. B2 is the battery to be charged.
    So far the energy is transmitted in pulses/bursts.
    But the battery docs say "constant current".
    So the battery might not "like" it...

    charge_current.jpg

    Any idea on how to improve it?

    Thank you.
     
  2. wayneh

    Expert

    Sep 9, 2010
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    You need a boost circuit to get the voltage up efficiently. DC-DC converters that can do this are widely and cheaply available on ebay. I wouldn't bother to build one.
     
  3. dl324

    AAC Fanatic!

    Mar 30, 2015
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  4. DavidB3

    Thread Starter New Member

    Nov 28, 2018
    21
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    @wayneh

    Thank you.
    You could be right, although most of them are not optimized for low power.
    Also they aren't so power efficient (below 85%) - but I need minimum 90 (over 95% recommended).

    For example I tried with MT3608

    s-l300.jpg

    The energy on output was a lot lower than on input (less than 30% efficiency). The main problem was most of the input energy was used to power the MT3608 ic...

    There are specialized low power ics but couldn't find one for this situation: either the input was for > 5 V, the output was for a <= 5 V battery or the maximum current was for less than 140 mA and so on...

    @dl324

    Thank you for the information and the link.
    But I already know about this problem and the solution would be: even with the most efficient converters, the (average) charge current would be around 0.05C so the danger of overcharge is remote.
    Anyway I can add a protection - after a specific voltage it should not charge.

    Best regards,
    David
     
  5. wayneh

    Expert

    Sep 9, 2010
    16,018
    6,114
    Well you're like Sisyphus pushing the boulder up the mountain. Your panel at maximum power can make something less than 700mW. Your battery has a capacity of 9,360 mW•hr. That means over 13 hours of peak panel time at 100% efficiency. When you throw in real performance, various inefficiencies and such, and you're looking at 20 hours or more charge time. That could be an entire week depending on the weather where your panel is operating.

    Why not go the easier route and add another panel? That would boost the voltage and you might not need anything other than a blocking diode.
     
  6. DavidB3

    Thread Starter New Member

    Nov 28, 2018
    21
    0
    Yes, you're right, it could take a lot of time to charge but I will use this method of charging only if I can't use a normal charger.
    And, unfortunately, in my country, these situations are not so rare...

    And yes, I could add another panel, in fact I could add another 3 (in a 2 x 2 matrix):
    panels.jpg
    But the problem is how to install them on this:

    top img 1.jpg
    top img 2.jpg

    It's not easy, especially if the led lamp would be sometimes used by elderly/kids...

    Late edit:
    Forgot to mention other 2 things:
    - when 2 panels are in series, the one in the shade will lower the power generated by the other panel too (because of the high electrical resistance of the cells when generating low power);
    - even with 2 in series a low power MPPT circuit would help boost the efficiency.
     
    Last edited: Jan 13, 2019
  7. DavidB3

    Thread Starter New Member

    Nov 28, 2018
    21
    0
    Well, meantime I improved the circuit a bit:
    - replaced the coil with a bigger toroid ferite core and a smaller resistance;
    - increased the capacitor to 448 uF;
    - added the other comparator from LM293P ic to keep an eye on the current through Q2;
    if it's too high it closes Q2; this way the charge pulse current is limited to ~0.52A (0.4C).
    - removed the diode in parallel with the capacitor since is kinda useless now (the voltage on capacitor will never be negative);

    schematic.jpg

    It's much stable now.

    Any tips on how to calculate (fast if possible) the power efficiency of the circuit?
    This would help me because I can tweak the values of the components until I'll reach maximum possible.
     
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