Hello,

More cost effective solution would be __any__ two (current + voltage) digital multimeters equipped with a digital interface. The ranges of such meters should meet mesuring PV system constraints. The cost of such measurment system development would be in this case moved towards a time spent to manage the process of acqisition (and storge) the measurements results onto a PC via a digital interface.

I hope you will consider this direction useful.

Adam.

I recommend this work, it can help you. It is a simple, functional and inexpensive alternative.

Article Low-cost virtual instrumentation of PV panel characteristics...

Dear Muhammad, for make an answer useful for you, I have some questions for you, like asks below:

Which parameter do you want obtained of the PV module?

What application do you need or do you want to make?

What is the main parameters of the of the PV module? (usually appear in the backside)

How many measurements do you think is the best for the application?

Dear Santso,

Thanks for your complete consideration.

1. PV output (Watt) is the main parameter to log.

2. I just need it to log and compare it to the another PV panel that contains PCM

3. You can find the PV module characteristics in the attached PDF file.

4. I couldn't get your last question but I need to log the data for at least 3 months.

Best,

Mohammad

Dear Silva,

Very kind of you to answer.

I'll look at that and back to you if I have further questions.

Best,

Mohammad

This one may be helpful to you

http://digitalcommons.calpoly.edu/cgi/viewcontent.cgi?article=2017&context=theses

Very appreciated Dr.Kumar

Dear Mohammad, the solution presented by Silva is really interesting, but I would like told you about one instrument that you can use for the investigation. The instrument is called PV210 the Seaward. This instrument is used for test PV Power Plants and tracing a lot characteristic of this systems. you can review the link below. I think is the best solution by you.

http://www.seaward-groupusa.com/userfiles/pv210-pv-installation-tester.php

Dear Santos,

The solution you have presented is really good but device is so expensive (2290$).

This is a college experiment and should be cost-effective.

However, I really appreciated for your support.

Best,

Mohammad

Hello,

More cost effective solution would be __any__ two (current + voltage) digital multimeters equipped with a digital interface. The ranges of such meters should meet mesuring PV system constraints. The cost of such measurment system development would be in this case moved towards a time spent to manage the process of acqisition (and storge) the measurements results onto a PC via a digital interface.

I hope you will consider this direction useful.

Adam.

Dear Tomaszuk,

Thank you very much for your complete response. First off, the availability of each measurements in Iran is a huge obstacle. Then, the cost of them is the second constraint.

However, I will put this useful addressing to the work and evaluate the different aspects for better data logging.

Best,

Mohammad

hi,

The easiest way is by using Arduino, https://www.arduino.cc/en/Tutorial/Datalogger. I have try it many times.

best regard

Ikhsan

Dear Muhammad,

I have plan to do the same but I have no sense of the procedure. I hope I read the tutorial and do it well.

Best,

Mohammad

Well the solution presented by Silva is the best for your purpose I believe or two instruments more sheep

Mohammad,

HANTEK offers inexpensive data loggers that read voltage or amps. They cost less than $150 each and can be purchased on Amazon.com. They comm with a PC via USB. The software is included. You'd need 2 units to measure voltage and amperage and then compute power in an excel spreadsheet. We use them at Johnson to measure battery performance.

Cheers,

Richard

I am agree with Richard Moscatiello, but also think you need a resistor for closed the measurement circuit, named, charge resistor as drawn in the followed image.

Thank you all for your comprehensive answers.

I am in agreement with all the answers. In my opinion though the answer by Silva is the best solution for a college project. I am glad that you are utilizing the power of IoT in Chemical Engineering. You could replace the F031-06 sensor with CR2550-R sensor to improve on accuracy of the current measured.

Good luck though on your project.

Mohammad,

welcome,

Adding to the colleagues above you can use data acquisition and control cards DAQ from national instruments Ni. You can use them to log the I-V data of your panels for long times as they are connected to Lap tops or PCs. This solution is not expensive.The programming is simple by using LAB VIEW from NI.

For the complete set up for measuring the panels please refer to the paper in the link: Conference Paper Experimental Performance Characterization of Photovoltaic Mo...

If you have any question concerning the code contact the first author on the researchgate.

Best wishes

Thank you Mr.Shitote and Professor Zekry for your interesting solutions.

Best,

Mohammad

Considering the different constraints you have mentioned I dare suggest - on my humble opinion - the simplest way to get at the lowest cost your data.

Your PV is a DC source with an internal resistance. The simplest way is to measure voltage drops on resistors. If you know the resistor value and the voltage drop you know the current flowing through.

I would suggest following circuit between the red and black connections you pace in series two known resistors. The total value has to be such that for the maximal generated voltage the current will not be over the PV limits in order to avoid a too high load on the internal PV resistor. You have to measure 2 voltages: between thee 2 poles and across one of the above mentioned resistors.

With those 2 voltage values and the values of the resistors you can compute as follows:

Let us name Ri the internal PV resistance and R1 , R2 the 2 external resistors.

The voltage between poles let it be Up and the one across R1 or R2 (whose values you know) U1 and U2.

The current will be i= U1/R1 or U2/R2.

The power will be P= i*(R1+R2)

If for a given light conditions you disconnect the resistance the voltage at the poles is the PV voltage U0 without current. The voltage Up you will measure for same light conditions will be less since there are losses due to the internal resistance Ri knowing the current from above computation you can determine the value of the internal resistor as Ri=(U0-Up)/i.

If you have questions do not hesitate to ask. the only condition is to have a good cooling for the resistances so that their temperature will not vary too much as function of current.