(Note: I asked this on https://engineering.stackexchange.com/ but I was told I should ask this here, so this is what I'm doing.)

I'm installing solar panels (not connected to the grid) in units out in a field for research we're doing, that looks like this (this is an outdated image, but I can't find the updated one unfortunately):

Outdated model of the system

I have to safely measure power output from each unit (and there are 4 units).

Each unit has 9 panels with the following specs:

  • Maximum Power: 100W
  • Maximum System Voltage: 600V DC (UL) Optimum
  • Operating Voltage (Vmp): 18.6V
  • Open-Circuit Voltage (Voc): 22.3V
  • Optimum Operating Current (Imp): 5.38A
  • Short-Circuit Current (Isc): 5.86A
  • Operating Temperature:-40°F to 176°F
  • Output Cables: 14 AWG (2 ft long)
  • Maximum Series Fuse Rating: 15A

(The panels are from Amazon so I'm not sure how accurate the specs are. The link to reference them is at the bottom.)

These panels are going into an agricultural field, so my primary concerns are: 1) Not electrocuting anyone, 2) not starting a fire, 3) being able to accurately measure the power coming from each unit.

This is my main plan so far, which I'd like feedback on:

  1. Wire all 9 panels in series to minimize amperage
  2. Use 14 AWG wire to take the power out of the field (2-12 meters of distance from the field to the load box) to some sort of resistive load to use the power
  3. The unit will be grounded using grounding rods
  4. On the circuit, there will be a 10A circuit breaker, a manual on/off switch
  5. The resistive load will be connected to a GFCI
  6. Power will be measured using some sort of Arduino based module powered by it's own battery

Some notes:

  1. I'm not sure what the best, safest (and if possible, cheapest) way to burn the power from the solar panels is. I was thinking using a space heater since each unit is allegedly only putting out a maximum of 900W and commercial space heaters are rated for 1875W. I know I could ground each unit and not have a load and the power will just discharge, but my understanding is I can't measure that.
  2. On the subject of measurements: Are any of the micro controller compatible power meters I find on Amazon going to be sufficient for this task, or do I need to look for something specific?
  3. I'm a graduate student who has to get this done but I lack expertise in electrical engineering to do this properly.

This has to be done ideally ASAP, but definitely before the end of May. Although, again, my main concern is doing this safely first and foremost. I am willing to hold off on measuring the power as long as there's as the primary concern is discharging power safely then figuring out how to measure power. I imagine that would involve routing power to a grounding rod, but I'm not sure.

Any advice and feedback is greatly appreciated, and I am happy to answer any questions.

Solar panel link: https://www.amazon.com/dp/B07JXYTFF7

  • \$\begingroup\$ You could use hot water tank heating elements from the hardware store and submerge them in big buckets of water (will need to top off ocasionally, you can reduce evaporation with vented lids). Safer than a space heater I think. But if your goal is to measure power output isn't just using a fixed load going to mess with things a bit? \$\endgroup\$
    – DKNguyen
    Commented Apr 22, 2021 at 19:17
  • \$\begingroup\$ I had not thought about submerged heating elements. That does sound safer. As far as measuring power from a fixed load, I have thought about this, but I'm on the fence if using a variable load is better or not. I have old drone motors which are rated for this level of power which could be variable loads, but within the context of this experiment: we want to know if we can link the power usage to something usable like irrigation pumps (which I don't have access to right now), and they're generally fixed loads which only operate at one speed. \$\endgroup\$
    – Omar
    Commented Apr 22, 2021 at 19:25
  • \$\begingroup\$ I was mainly referring to the job of a solar panel controller, and if you were planning on using battery reservoirs. \$\endgroup\$
    – DKNguyen
    Commented Apr 22, 2021 at 19:27
  • \$\begingroup\$ I'm not intending to use batteries because we're not needing any energy reserves. I just want to measure the power from the panels and a battery seems like unnecessary expense because we'll still have to shed the power somehow as we're not connecting the panels to anything in particular. \$\endgroup\$
    – Omar
    Commented Apr 22, 2021 at 19:32
  • \$\begingroup\$ Air is much less messy than water, use the convector space heaters you first thought of. They won't need visiting to top up with water either. Do you need to adjust the load to MPPT these things? What is the question that this experiment is intended to answer? The precise question you're trying to answer may help us help you finish your experiment design. \$\endgroup\$
    – Neil_UK
    Commented Apr 22, 2021 at 20:08

2 Answers 2


This is my main plan so far, which I'd like feedback on:

  1. Wire all 9 panels in series to minimize amperage

Putting the panels in series will boost the nominal voltage to 200 volts. Frankly, I'd recommend 3 sets of 3 panels, rather than 1 of 9. This way the nominal voltage will be about 55 volts.

  1. Use 14 AWG wire to take the power out of the field to some sort of resistive load to use the power

14 ga wire has a nominal resistance of 2.5 mOhm per foot. This will give a nominal voltage drop at 5.38 amps of 27 mV/foot (over both wires). You don't say how long a separation you want between the panels and the load.

  1. The unit will be grounded using grounding rods


  1. On the circuit, there will be a 10A circuit breaker, a manual on/off switch

OK, but keep in mind that you'll need to specify DC circuit breakers. AC circuit breakers count on the fact that the voltage across the contacts falls to zero 100 or 120 times/second, and this reduces the arc length required for quenching once the breaker trips.

  1. The resistive load will be connected to a GFCI

GFCIs don't work on DC. Plus, it's not clear why you would need one in this application.

  1. Power will be measured using some sort of Arduino based module

How do you propose to power the measurement box? Will you run an AC line to it? Will you attempt to power it from the solar cells? Will you provide it with its own solar power system?

As for your load, I'd just go with some big-ass power resistors. If you use 3 circuits of 3 panels in series, your nominal load for maximum power will be 10.37 ohms at 300 watts. I'd be inclined to get a total of 15 ea, 50 ohms at 100 watts, such as these https://www.vishay.com/docs/31838/fvt.pdf They are in stock at Digikey, and would cost a total of about $180. Mounting is simple, and you could make a rack of them with a small roof overhead to keep the rain off. For each circuit you'd put 5 in parallel, and you'd have a 500-watt, 10 ohm resistor, so you'd have plenty of margin in terms of power dissipation. You'd expect something like 1 - 2% resistance variation over temperature, so you wouldn't need current meters. Just measure the voltage across the load.

Look, for measuring solar power, don't try for super-high precision. A few percent is all that makes sense. You can't control the amount of cloud cover, and it will vary from month to month and year to year. So don't overthink your data requirements. Keep it simple.

  • \$\begingroup\$ Thanks for your feedback. To answer the comments (and I will update the question): 3 sets of 3 was the other idea I had but I'm not sure if they should be recombined into parallel for taking the power out of the field. As far as taking them out of the field, there's 4 units between 2-12 meters inside the field from where the load will be. We don't want any heating occurring over the crops which is why we're wiring them out of the field. Another, but less important concern here, is the cost of that much wire being used. \$\endgroup\$
    – Omar
    Commented Apr 22, 2021 at 21:19
  • \$\begingroup\$ The grounding and the GFCI is being done because the safety people on campus want me to ground the panels somehow, but if you're saying the DC nature of this means it won't work that way, then maybe I'll have to try explaining to them I can't do that. I'm a grad student and I have to get this done but this stuff is beyond my expertise which is why I'm asking here. \$\endgroup\$
    – Omar
    Commented Apr 22, 2021 at 21:21
  • \$\begingroup\$ As far as the measurement device, I was going to use an Arduino that's powered with a battery pack. \$\endgroup\$
    – Omar
    Commented Apr 22, 2021 at 21:22
  • \$\begingroup\$ @Omar - Going to 3 separate circuits will also mean that you'll have redundancy. If one solar cell or its wiring or load fails, you'll still have 2 other circuits active. As for heating, put a sheet of foam under the resistors. Most of the heat will be dissipated via convection, and the foam will insulate the area under the resistors. If you look at the data sheet for the resistors I pointed at, you'll see that the size of the array will be small compared to the solar cells. So you might as well just mount the resistors as part of the solar cells, and use almost no wire to connect them. \$\endgroup\$ Commented Apr 22, 2021 at 21:30
  • \$\begingroup\$ @Omar - Actually, you're arguably better off putting the resistors near the solar cells. When the energy converted to electricity is exported to another area, this will have the effect of cooling the area near the cells - disregarding all sorts of complicated issues concerning the relative albedo of the solar cells vs the ground which they shade. Dissipating the electrical energy near the cells at least keeps the total energy budget for the field the same. ignoring heat distribution issues, of course. \$\endgroup\$ Commented Apr 22, 2021 at 21:35

Fixed resistors will not extract as much power as an MPPT based controller. But fixed resistors will be BY FAR the simplest possible way to get this up and running quickly. You don't need circuit breakers for a single string of panels (industry term for panel is "module"). The current is intrinsically limited to less than 6 Amps. There would be no point in putting a 10 Amp fuse on a single string of panels. The fuse can never blow unless the sun somehow gets a whole lot brighter. It would be prudent to have a DC rated disconnect switch between the panels and the resistor.

Ground the frames of the modules. That is standard practice. You can, but don't have to, bond (wire) the negative output of the solar array with the grounded frame. Put the fixed resistors in a large well ventilated metal box to make sure nobody can touch the conductors. Ground the box to the module frames. Use a resistor rated to 2000 or 5000 Watts. Use a resistor that is designed to dissipate into the air (not a heatsink mount resistor).

The only think you have to measure with your arduino is the DC current flowing into the resistors. Then you can calculate power using P = I^2 * R. You can measure current pretty safely using a hall effect DC current sensor.

This setup will not extract maximum energy from the array. An array with MPPT will get quite a bit more energy from the panels. But if you log current vs time and then do some math afterwards, you can estimate how much energy you COULD have extracted if you had used MPPT.

The resistor value should be roughly equal to Vmpp/Impp. Well, actually it would be better to use the real-world voltage and current (Vmpp and Impp under NOCT conditions).

Mpp is "maximum power point." Vmpp is customarily measured at a specific solar cell temperature (25 C) and with a specific level of illumination (1000 W/sq. meter). NOCT is "normal operating condition and temperature." I think 800 W/sq. meter and 40 C are ballpark or typical NOCT conditions. For your panel it looks like Vmpp is 18.6 V and Impp is 5.38 A. Under NOCT, Vmpp and Impp will be less.

As a guess, maybe around 17 V and 4.3 A. So R = 17 V * 9 / 4.3 A. Let's round that to 36 ohms. For each unit (9 modules/unit) you want a 36 ohm, 2500 W resistor. If it is feasible to forcibly ventilate the box the resistor is in you should do that. The box should be NEMA 3R if you are in the USA. This just means it is outdoor rated and designed to shed rain. If the box is not forcibly ventilated, it needs to have a LOT of holes or slots and be very big. Much bigger than the resistor itself. Putting several hundred Watts in a box is going to make the inside of the box hot.

  • \$\begingroup\$ I'm not familiar with Vmpp/Impp. Is this the theoretical maximum voltage and current and I'm selecting the resistor to match ohms law? And by NOCT conditions, these are the factory tested conditions? \$\endgroup\$
    – Omar
    Commented Apr 22, 2021 at 22:59
  • \$\begingroup\$ Oh, sorry. Mpp is maximum power point. Vmpp is customarily measured at a specific solar cell temperature (25C) and with a specific level of illumination (1000 W/sq. meter). NOCT is normal operating condition and temperature. I think 800 W/sq. meter and 40C are ballpark or typical NOCT conditions. For your panel it looks like Vmpp is 18.6 and Impp is 5.38. Under NOCT, Vmpp and Impp will be less. Not sure exactly how much less. Maybe around 17V. And the reduction in incident power will drop Impp to maybe 4.3 A. So 17 V * 9 / 4.3 A = 36 Ohms. So 36 ohms, 2 kW resistor in a box. \$\endgroup\$
    – user57037
    Commented Apr 22, 2021 at 23:11
  • \$\begingroup\$ Here is a 33 Ohm 2500 watt resistor. digikey.com/en/products/detail/te-connectivity-passive-product/…. 36 ohm may not be so common as 33 or 39 ohm. \$\endgroup\$
    – user57037
    Commented Apr 22, 2021 at 23:21
  • \$\begingroup\$ The reason I am using such a high resistor power rating is because you really need to put the resistor in a box for touch safety (shock hazard). This will cause the resistor to run pretty hot. I am compensating for that (somewhat) by using a high power rating. If you can mount the resistor inside a box with a ventilation fan, that would be ideal for keeping the temperature down while still being touch safe. \$\endgroup\$
    – user57037
    Commented Apr 22, 2021 at 23:29
  • \$\begingroup\$ @mkeith - Avoiding the shock issue is why I originally recommended going to a 3 x 3 array, rather than 9 x 1. A single circuit makes a great deal of sense when you're driving a single load, such as a power net, but it's worth keeping in mind that this system does not obviously have such a constraint. I do appreciate your overrated power resistor approach, though. Lots of people don't have a proper appreciation of the need for margin. \$\endgroup\$ Commented Apr 25, 2021 at 14:01

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