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Calculator's solar cell powered compass

Discussion in 'General Electronics Discussion' started by pelouse, Aug 23, 2011.

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  1. pelouse

    pelouse

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    Aug 23, 2011
    Hello,

    I'm planning to build a digital compass using an HMC6352 magnetometer and an XLP PIC. My idea is to display the sensor value in degrees using a simple numeral LCD.
    I though that it'd be fun to try to play with low-power stuff and so I decided to power the whole system using 2 or 4 calculator's solar cells. I bought a 2€ calculator that has a solar cell on it. Measuring the open circuit voltage at full sun it gave me 2.38V (which is enough for my PIC but not for my sensor, it needs 2.7V at least), and the shortcircuit current was 2,03 mA (0,15 mA indoors)

    Which is the best way to interconnect those solar cells with a coin battery? Or a capacitor? What's better? Or maybe it is impossible to power up a system with an LCD, a PIC and a sensor only with a few solar cells like those?
    Does anyone have experience in using this solar cells? All kind of advice will be gratefully welcomed!


    Regards,
    Lîtus

    PS: Excuse my grammar errors if any, I'm not English! :(
     
  2. davenn

    davenn Moderator

    13,865
    1,956
    Sep 5, 2009
    Hi Litus
    Welcome to the forums :)

    the 2.38V open cct voltage wont be enough for your PIC
    with a load on that solar cell the voltage is going to drop probably at least by 1/2 the value. just for a test put a 1k Ohm resistor across the output
    of the cell and measure the voltage again. You may well find you only have ~ 1 - 1.5V
    You PIC chip and associated circuitry may even load the cell more than that and you end up with say... 0.5 - 1V

    cheers
    Dave
     
  3. pelouse

    pelouse

    5
    0
    Aug 23, 2011
    Ok! I was looking for problems like that before start wiring things up. I'm thinking in using 4 solar cells: two parallel branches of two cells in series. Thus I'll get around 4V (~2V when I hook it up?) and a few milliamps (up to 4 mA). Still the problem is supplying enough power to the sensor. Reading the datasheet I found that the magnetometer can only run from 2.7V (!!) to 5.2V. It consumes 1 mA @3V (steady state) and 1 uA @3V (sleep mode) so I guess I'm having, at least, 1 mA of consumption to consider.

    My PIC can certainly run at low voltage, but will I be able to find a 2-3V LCD?

    Since this is a bit of a problem, I'd like to consider adding some sort of accumulator (say a high capacitance capacitor; little lithium battery; coin cell...) Does anyone have an idea of how to interface those power supplies to work properly and to be able to have enough power even to drive the system with no light at all?

    Thanks!!
     
  4. pelouse

    pelouse

    5
    0
    Aug 23, 2011
    The power requirements are pretty much the following:

    Microcontroller: PIC16LF1933 (eXtreme Low Power)
    7.0uA @ 32 kHz, 1.8V, typical (PIC16LF1933)
    I choose this one because it has a LCD driver module

    Sensor: HMC6352
    1 mA @ 3V (steady state)
    1 uA @ 3V (sleep mode)

    LCD: ??? (maybe the calculator's display?)
     
  5. davenn

    davenn Moderator

    13,865
    1,956
    Sep 5, 2009
    Yes as many calculator LCD's run off ~1- 2 V, that may be worth an investigation a bit of experimentation and you could figure out the pin outs for a display. Not something I have done.

    have fun :) and let us all know how you get on with the project

    Dave
     
  6. pelouse

    pelouse

    5
    0
    Aug 23, 2011
    I will!

    I'm now considering adding a step-up DC-DC converter to have 3 to 5V in my output. I've been reading about this one, from Onsemi: NCP1402SN30T1
    Do you think I'll be able to charge a battery with it?

    Thanks!
     
  7. davenn

    davenn Moderator

    13,865
    1,956
    Sep 5, 2009
    probably not, you are limited by what the solar cell/s can produce plus there will be losses in the DC-DC converter eg. say your cells produce 50mA after DC-DC conversion you may only have 25 - 30mA available. You would be able to do measurements across a load placed across the output of the DC-DC conv.
    IE you can measure the voltage drop across a resistor load, you will know the resistor value, so therefore you can calc the current.

    Dave
     
    Last edited: Aug 27, 2011
  8. pelouse

    pelouse

    5
    0
    Aug 23, 2011
    I bought a cheap LED-light that used a solar cell and a LIR2032 button battery. The solar cell is more powerful than the calculator's one and I'm using it to charge the rechargeable 2032 battery (that I'm using too). If anyone needs a cheap solar cell and battery, buy one of those Made-in-China lights!

    I'll keep you updated with final results!
     
    Last edited: Aug 26, 2011
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