Sorry if this topic has already been touched on, but I have been searching and searching and cannot find the information I am looking for. I have this project where we are trying to measure the power generated by our solar array. What our array consists of is two sets of different panels.
Panel array #1: single panel generating 195W, 45v, 5.8amps
Panel array #2: series of multiple small panels generating 128V, .11amps, and 14.08W
(these specs are the rated peak values)
So what I do know. We need to use a voltage divider to drop the volts down to <5V so as not to overload the arduino. We will be measuring the power separately in each array and then combining the power measurements to get total power generated.
What I am confused on is that we have 195W, does this mean we need 195W rated resistors for the voltage divider circuit? If so, is there another way to drop the power down enough using more common resistors?
I have basically just got myself incredibly confused in reading all the different articles on this topic.
We need to use a voltage divider to drop the volts down to <5V so as not to overload the arduino. We will be measuring the power separately in each array
That's correct to measure voltage but voltage alone is not power, power = voltage x current. So you have to measure the current as well.
does this mean we need 195W rated resistors for the voltage divider circuit?
No, there is no direct relationship between the two. The resistor power rating is based on the voltage across it and the value of the resistor. For your purposes you hardly have to think about it, any PTH resistors in the ~10k range will have heaps of headroom.
For example 12v across a 10k resistor only needs to dissipate 14.4mW of power. Any resistor can handle that.
I don't know where your solar panels are located but in the UK the output from my solar panels can change by a huge amount in the blink of eye (clouds) and it would be necessary to take samples of voltage and current very frequently to get an accurate total power output.
thank you all for your replies, we have a current sensor to measure the current and within the program it will do the multiplication p=I*V to get power. Our program is basically taking voltage and current readings in real time to create a sizeable data log. I can go on and on how that data is being compared to a range of other sensors to get efficiency and so on. The math and readings aren't the problem. Measuring 128volts is. What are the power rating on resistors for? Our max output for the large panel is 195w at 45 volts, so even a 1/4 Watt resistor can handle that much power flowing through it?
The simple way to calculate the power dissipation is V2/R. That's P = I * V replacing I with V/R.
If you're using a .25W resistor then the least resistance you can use is
.25 = 452/R or R = 452/.25.
That's 8100 ohms. But you want to give yourself a margin of safety by using a higher resistance or higher wattage resistor.
Our max output for the large panel is 195w at 45 volts, so even a 1/4 Watt resistor can handle that much power flowing through it?
What the panel CAN output is irrelevant if you use high(ish) value resistors. The power calculation is based on the voltage across the resistor (45v in this case) and the current flowing through the resistor, that figure is up to you, just chose a resistor with a value that doesn't cause too much current to flow.
So following on from Changrin's example, let's say you hang a 10k resistor from your 45v to 0v, that's 45v over 10,000 or 202mW, so a 1/4W resistor will do just fine although 1/2 would provide a good margin.
If you have a voltage divider from 128v (are the panels in series?) you could say use 82k and 3k3, that's a total of 85300R and 128v, or 192mW. Once again 1/4W resistors will be OK.
Now if you chose 82R and 3R3 resistors the divider ratio is the same but the power dissipated is 192W, not only would you need some serious resistors but there'd be no power left for the load/batteries
Thanks Chargrin and Graynomad, your responses helped a lot. yes, the small panels are in series. After doing some measurements on a multimeter we are thinking about reconfiguring them to raise the current to get better readings. We don't necessarily need to have to power anything or charge batteries. Our experiment is purely just measuring the output of the panels.
Thanks again for your help, hopefully we are on the right track now!
Are you trying to measure whether your particular solar panels perform as well as the manufacturer claims or are you trying to measure how much solar energy you can capture at your location.
You will almost certainly need to design different experiments depending on which you are doing.
Are you familiar with the fact that solar panels produce maximum power when the load is varied as the available sunlight varies. MPPT controllers are used to try to achieve this, but I doubt if your experiment would need an MPPT.
Another interesting experiment would be to find the fixed load that gives you the most energy at your location and how much energy is lost by not using MPPT.
