# Calculator's solar cell powered compass

Discussion in 'General Electronics Chat' started by LÃ®tus, Aug 23, 2011.

1. ### LÃ®tus Thread Starter New Member

Aug 23, 2011
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0
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. ### iONic AAC Fanatic!

Nov 16, 2007
1,422
68
I don't believe hooking up the solar cells to a coin cell or capacitor is going to help you.

You will first need to calculate you total power requirements of all components. It seems that you will need at least 3V, but I don't know what the voltage requirements of the LCD are as well. Solar cells are not a regulated power source and your devices may need a regulated source to prevent damage to one or more of the devices. In order to really help you out you will need to provide us with specifications of all devices involved.
One thing for sure, 2mA is not going to work!

For example the HMC6352 sensor Minimum Voltage is 2.7V, Typical is 3V, and Max is 5V. It's current needs are 1mA @ 3V and 2mA - 10mA @ 5V. I suspect its sensitivity/accuracy is enhanced at the greater voltage as I can see no other advantage of supplying it a greater voltage and current otherwise.

If you measured 2mA in bright sun that seems rather small. How did you make this measurement?

Last edited: Aug 23, 2011
3. ### LÃ®tus Thread Starter New Member

Aug 23, 2011
9
0
I was thinking in creating the same circuitry the calculator had. It had the little solar cell and a LR54 coin cell. When I replaced the coin cell the calculator still worked fine (I could only notice a little decrease in the LCD contrast). That's what I'd like to do: power the thing with a little solar cell while having a battery to be able to supply enough power when there's not enough light or no light at all (at night, for example). The LR54 battery is not rechargeable, but I'd like mine to be.

The 2 mA were measured during noon, without any clouds and pointing the cell orthogonally to the sun. Of course, it is the short-circuit current, not the real one. Remember it is a 2-calculator's solar cell!

As I have no components yet (except the magnetometer and solar cell) I can't really calculate the power requirements, but I can guess it:

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 (the Sparkfun one)
1 mA @ 3V (steady state)
1 uA @ 3V (sleep mode)

LCD: ??? (maybe the calculator's display?)

Thanks iONic!!!

4. ### iONic AAC Fanatic!

Nov 16, 2007
1,422
68
I have some 2032 coin cells, nominal 3.6V, 4.1V when charged. Can't vouch for their longevity, but you also need a charger for them and they only charge two at a time.

A option, maybe. Step-u regulator It's darn small. Just in case you need to go with a 5V solution.

5. ### LÃ®tus Thread Starter New Member

Aug 23, 2011
9
0
So do you recommend me to elevate the voltage? It might be enough with 3V (using NCP1402SN30T1) but, then what current can I expect at the output? Enough to drive the system or (in the same board) charge a battery?

6. ### iONic AAC Fanatic!

Nov 16, 2007
1,422
68
No, not really. Just thought I mention it if there was a need for more than 3V.

Here's the link to where I bought the 2032 Button Cells.
LIR2032

7. ### LÃ®tus Thread Starter New Member

Aug 23, 2011
9
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Thanks!

I bought a cheap solar light (3 LED and a solar cell) that had a more powerful solar cell and a LIR2032, so I'll use both of them to build my power supply. In the instructions paper of this solar light it is stated that 7 hours in sunlight will fully charge the battery.

I'll keep you updated with the results!