help Using 2v 45mA mini solar panel as solar radiaton sensor

Hello to you that clicked in this topic by curiosity or interest.

i'm 19 years old currently studying renewable energies, i'm new to arduino world, i know couple basic stuff like voltages, current, resistance, and some laws and effects they have. Well, i have an arduino Sainsmart UNO R3 board, and i got hanging around 4 solar panels each one rated at 2v and maximum Short circuit current 0.045A, that gives a 0.09 power at perpendicular to sun radiation exposure, this is a polycristalline silicon panel, with an efficiency of 4.5%. All i want is to manage to use this as a radiation sensor measuring voltage or current, without burining my arduino out, i have already burned an original arduino before putting 12v from a gadget transformer in Vin, there i saw how easily arduino can damage. i am here posting asking help how should i connect this mini solar panel and use it as a sensor.

for those who are not aware of solar panels funcion, basically research the photovoltaic effect, and also the output power of the panel depends very much on the radiation and angle when its exposed. so lets say i put it flat on the groud and the sun is in winter at noon, where the sun is at the lowes angle in a noon time. the power output will be very very low.

having explained this, i want to manage this, in order to put it side by side or a industrial scale solar panel of 270 watts (also Poly-Si), and manage to use the arduino to record and power generated by the panels without using complex and expensive power meters, this way, paired up with a HC-05 bleutooth module, i can also have real-time data to my phone or tablet or computer.

if there is a post related to this, or the same as this, please apologise me, the internet is too big and we cannot always find everything just like that right? i apologise for that.

if not, i would be very very much apreciated if someone can help me wire this mini panel without damaging my Arduino.

Your sense panel generates 2Vdc Max. Hook it up to the arduino (- to GND, + to an analog input pin, with a series resistor , say,,,1k). modify one of the analog in examples and map the readings with various levels of light hitting the panel.

purtty much dirt simple........ Have fun Y'all!

Thanks for your help!! I will try that right now

i have already burned an original arduino before putting 12v from a gadget transformer in Vin, there i saw how easily arduino can damage.

A [u]Protection Circuit[/u] can protect against excessive or negative voltages.

I like the 2nd schematic with two "normal" silicon diodes (just because I always have diodes on-hand). And, I'd increase the resistance to 1K (just to reduce the current pulled from the source in case of an over-voltage).

However, the circuit with the 2 regular diodes relies on the presence of the 5V power supply. A Zener diode will provide protection even when the power is off.

NOTE - The Arduino actually has protection diodes built-in already, but you need to limit the current (with a series resistor), and the built-in diodes can only handle a few milliamps whereas something like the 1N4001 diode can handle 1 Amp.

so, assuming the battery represented in the picture, is the Photovoltaic mini solar panel, i wire it like this to make it a radiation sensor?

To use a photovoltaic panel as a radiation sensor, it has to have a "load". The current generated by the panel (not the voltage) is proportional to the light intensity.

For a 2V, 45 mA panel, a 47 ohm load resistor would be a good choice. The voltage across the load resistor (V = IR) is then proportional to the light intensity.

Connect the load resistor between the + and - terminals of the panel and then connect the + to an analog input and - to ground.

The 1K resistor shown in your diagram is not necessary, but can help protect the input pin from overvoltage.

will a 100 ohm resistor suffice?

Yes, but 47 ohms would work better.

jremington:
Yes, but 47 ohms would work better.

there are quite a few types of resistors, some are 0.25W, other 0.5W, some are 5W. what's the wattage you most recoment ?

and thank you a lot for the help!

The wattage is irrelevant as the power dissipation is very small.
V2/R = 4/47 = 0.085 watt. So, use a 1/8 or 1/4 watt resistor.