From a physics point of view, it makes perfect sense. In the ideal world, a solar panel sends one electron around the circuit for each photon absorbed.
In the real world, the short circuit current is proportional to the number of photons received per second, which is the illumination intensity. That represents the maximum power available. The actual power delivered is some fraction of that, depending on the load.
jremington:
From a physics point of view, it makes perfect sense. In the ideal world, a solar panel sends one electron around the circuit for each photon absorbed.
In the real world, the short circuit current is proportional to the number of photons received per second, which is the illumination intensity. That represents the maximum power available. The actual power delivered is some fraction of that, depending on the load.
I had not been thinking of it like that.
Does it follow then (and leaving aside the complexities of how it might be achieved) that the maximum power is harvested when the load is chosen to get the maximum voltage at full current?
TempleofTyler:
Unfortunately, bylaw states within our district that all renewable energy sources (Wind/Solar) are forbidden. This is 100% a political issue, as our government decided on a multi-billion dollar hydro plant to supply power, and since the population is going to be paying for it for generations to come, they outlawed renewables to decrease the time before it's paid off!
So if you build a house, the law says you have to connect to the grid?
If you stay off the grid, what is to stop you going solar/wind.
As explained, the max power and that point depends on the level of solar radiation.
So at anytime to measure the max power point you could connect the PV up to a variable load and monitor voltage and current through the load to calculate power.
Tom..
TomGeorge:
So at anytime to measure the max power point you could connect the PV up to a variable load and monitor voltage and current through the load to calculate power.
If my understanding is correct both @wawa and @jremington are saying that is not necessary if all you want to do is measure the available power?