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Questions about solar cells

Discussion in 'Electronic Basics' started by royalmp2001, Mar 4, 2005.

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  1. royalmp2001

    royalmp2001 Guest

    I am considering a solar cell powered project but I have some

    What happens during low light levels. Does the rated volatage decrease
    or only the current, or both. If I want to implement a battery backup
    what kind of circuit will switch between the two. Are solar cells
    shortcircuit proof?
  2. Both the voltage and the available current go down. The voltage
    roughly proportional to the log of intensity (so small decrease) and
    the current in proportion to intensity, if you use power at the
    optimum rate.
    Usually, the circuit just runs from the battery, and the solar cells
    connect to the battery through an anti reverse diode, or regulator.
    Sure. They just get as warm as if they were laying in the sun with
    nothing connected. This allows them to be regulated with a shunt
    dummy load that soaks up all the excess current once the battery
    reaches full charge. For a reliable operation over many months, the
    cells have to provide a full day's charge in just a few hours, to
    allow for a run of rainy weather.
  3. mike

    mike Guest

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  4. royalmp2001

    royalmp2001 Guest

    Is an anti-reverse diode a special type of diode. Can I use a IN4148?
    Any advantage of using a schottky?
  5. It is nothing special. It must stand the battery voltage in reverse
    (when the cells are in the dark) and carry the solar cell current in
    the forward direction. The schottky type wastes about half as much
    voltage compared to a junction diode with the same current rating, but
    will also leak more current when the solar cells are in the dark. A
    1N4148 is fine for up to about .1 amp maximum cell current.

    You may not need an anti reverse diode at all, if your cell array
    leaks a small enough current when it is in the dark, and still
    connected to the battery. Put an ammeter in series and measure the
    leakage or put a small resistance in series, measure the voltage drop
    and calculate the current with ohm's law. For example, if you put a
    1k resistor in series and had 0.1 volt drop across it when the cells
    were dark, that would mean that the cells leaked about .1/1000=.0001
    ampere (100 microamps).

    If that current times the number of hours of darkness is a small
    fraction (say, .01) of the forward current times the number of hours
    of daylight, then the leakage will not use a significant fraction of
    the charge energy. Adding the anti reverse diode would lower the dark
    leakage, but would also lower the daylight charging current, so it
    might be a net loss to have it. Measure the two cases and find out.
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