changing 0 to 12 volts to 0 to 5 volts

Other Hardware General Electronics
1

I am building a solar panel sun tracker using the output of the panel as a sensor of direction (light intensity).
Can anyone give me any hints on how to change the 0-12 V from the panel to a 0-5 V signal to be used on the analog in pins?
Use of a resistive voltage divider may be to much waste of current from the panel?
Possibly a voltage divider using high value resistors , in the hundreds of Mega ohms?
I am using an Arduino duemilanove.
Also I am having trouble finding the impedance value for the analog pins.
Mike

2

Also I am having trouble finding the impedance value for the analog pins.

I believe the analog input pins are designed to be driven by 10k ohms (or lower) of source impedance.

Lefty

3

Drawing a few microamps from a solder panel (using >100k resistors in a voltage divider) should not be a big deal. The problem is the larger the resistors the less accurate your voltage reading is. The AVR datasheet suggests less than 10k input impedance but that's for full 10-bit accuracy. As long as you're OK with "close enough" then resistors in the 100k-1M range should do.

The analog input impedance is very high (Table 28-16 in the ATmega328P datasheet), 100Mohms typical, but the voltage error is caused by the sampling capacitor that has to charge very quickly during a sample acquisition. The bigger the input resistance the slower it charges towards the "real" value.

--
The Aussie Shield: breakout all 28 pins to quick-connect terminals

4

Thanks for the quick responses.
I will try a voltage divider with high resistances.
My input can be very inaccurate , I am more interested in the relative change in intensity than a true value.
Mike

5

"Waste of Current"

With a 12 volt supply and, say, a 25k divider chain (15k + 10k), the "wast of current" will be less than 0.5mA. What kind of solar panel are you working with if a 0.5mA load is deemed detrimental to its output efficiency.

6

jackrae:
"Waste of Current"

With a 12 volt supply and, say, a 25k divider chain (15k + 10k), the "wast of current" will be less than 0.5mA. What kind of solar panel are you working with if a 0.5mA load is deemed detrimental to its output efficiency.

Only about 0.1 to 0.2 % of produced current. Point taken.