relationship between Voc, Isc and Wm2

[GPack]

If I have 16 panels, each with an Voc of 30V and Isc of 8.37A and the irradiance is at 500Wm2 what is the equation to make sure the measured array Voc and Isc are roughly in the right area?
Also when measuring the Isc of the array what is the best way to conduct this test. I have seen someone shorting the array at the DC isolator in the off position placing the clamp meter around then turning the isolator on but got the impression this was presenting a hazard. Can it be avoided?
Thanks for your help

 

[waterleat]

I was told to use 1 mV/W/m2 as Irradiance coeficient to calc voltage drop due to less light than STC.

Voltage drop = 1 X (1000-500) = 500mV = 0.5 V per panel

Assuming 16 panels in one string expect
Array Voc (measured) = 16 X (30 - 0.5) = 472 V

BUT you have not taken into account the drop due too the temperature of the cells not at STC

Temp Voltage drop = temp coeff X diff in temp
temp Vd = 100/1000 X (45 - 25) = 2V per panel -------assumed temp coeff = -100mV/C and 45C

16 panel X 2 V = 32V

expect voltage to be 16(30 - 0.5 - 2) = 440V

 

[waterleat]

Forgot to give short circuit current calc assuming current temp coeff is +0.05%/C and again temp is 45C. Note some panels datasheets give voltage coeff as %/C

expected Isc = 8.37 X (500/1000) X (1 + 0.05/100)
= 4.187 A

others reading this please correct these calcs if I have made any errors

 

[yellowvanman]

Not saying there is anything wrong with it but I've never seen the 1mV/W/m2 coefficient, and just looking at a Sharp module spec sheet it doesn't refer to a voltage to irradiance correlation.

Generally Voc is affected by temperature, and its the temperature of the module not the ambient that is required, so difficult to measure in practice.

As a rule of thumb Voc with reduce by 0.5% for every degree rise, and increase for every degree fall.

So if the modules are at 45C temp difference from STC is 20C
String Voc = 16 (30 - 0.5%*20) = 16 (30 - 10%) = 432V

Look at the module spec sheet for a more accurate temperature coefficient, as has been mentioned they are usually in %/C.

For current
Isc = Isc(Stc) x (measured irradiance/STC irradiance) = 8.37 * 500/1000 = 4.185A

The method is as stated, but you can buy (or make) a dedicated test box for doing the Voc and Isc measurements

 

[waterleat]

The cell temp however has significantly more effect than the irradiance coefficient but it was what we were told on a C&G 2377 course and was needed for one of the calculations that was in the exam. I have never seen this figure quoted anywhere else. Perhaps it was a City & Guilds thing that only occurs in their exams?

 

[GPack]

Thanks for your help and thorough explanation. I've definitely got it. Looking at the data sheet for the panels I can see that the .05% rule of thumb is pretty accurate. It actually works out at .035% for these panels. on the sheet it is down as as 104mV/degs C and this represents .035% drop per degree rise.
The Ioc makes sense as well. The figure I took on the day are starting to ring true.
Thanks again. It's great to understand it.
How do you guys measure the Ioc for the PV test sheet?

 

[yellowvanman]

There is no such thing as Ioc - if your talking about Isc then you include your current measurement and your irradiance measurement in the test sheet.