Thanks for these lectures .. but this one has a bit of confusion,, the talk at 35:00, you said the current will flow through Rp and will have a voltage of Ipv*Rp ... #That is not right I think ... Rp is very high and will not make Ipv pass. Also if it allows it will be Ipv-Isc ... The thing here is the current will not significantly pass through Rp , the evidence is that the output current is not Ipv .... Now, for the hot spot, it is not exactly as you said ... A negative voltage will be applied on the shaded cell as the difference between the voltage of the healthy ones and the load voltage .. that voltage will have reverse polarity and across the shaded one ... when multiply that voltage with the Isc of that cell the power will be negative that means it is a load and here is why hot spot
Thank your for this comment. Rp is high but it does not imply that it is open circuit. The argument that you have put forth is valid in case the shaded cell is producing some amount of current. At 33:40 i have mentioned that some of the current is diverted into the Rp. Later on, the case which i have discussed is assuming to have a substantially strong shade such that Isc=0. This evaporates the confusion you have mentioned.
Regarding the Hot spot my theory is intact and does not contradict the theory explained in your comment. As i haev said this is also a method to explain the hot spot mechanism. It will not be the Isc of the shaded cell but the string current.
Perfect
Thanks for these lectures .. but this one has a bit of confusion,, the talk at 35:00, you said the current will flow through Rp and will have a voltage of Ipv*Rp ... #That is not right I think ... Rp is very high and will not make Ipv pass. Also if it allows it will be Ipv-Isc ... The thing here is the current will not significantly pass through Rp , the evidence is that the output current is not Ipv ....
Now, for the hot spot, it is not exactly as you said ... A negative voltage will be applied on the shaded cell as the difference between the voltage of the healthy ones and the load voltage .. that voltage will have reverse polarity and across the shaded one ... when multiply that voltage with the Isc of that cell the power will be negative that means it is a load and here is why hot spot
Thank your for this comment. Rp is high but it does not imply that it is open circuit. The argument that you have put forth is valid in case the shaded cell is producing some amount of current. At 33:40 i have mentioned that some of the current is diverted into the Rp. Later on, the case which i have discussed is assuming to have a substantially strong shade such that Isc=0. This evaporates the confusion you have mentioned.
Regarding the Hot spot my theory is intact and does not contradict the theory explained in your comment. As i haev said this is also a method to explain the hot spot mechanism. It will not be the Isc of the shaded cell but the string current.
@@HadeedSherThank you
I remember the same voice 15 yeas back in UCP and now after some time seen the lectures
Isn't this amazing :)