Effect of wire resistance on MPPT charge controller

Michal.7 · Oct 11, 2022

I want to ask about the cables connecting panels with the charge controller - will the effective power loss be the same as calculated from the resistance of the cable or is the loss non linear if I use an mppt cc? I don't know if my question is clear so let me explain it with an example. My question arose from my difficulty to figure out what the voltage will be at the panels and what it will be at the charge controller knowing that a voltage drop will occur between them.

For example, I have a panel with 18.7 Vmp yielding 100 watts so if I use a theoretical perfect 0 ohm cable the cc will work at 18.7 volts since this panel works best at this voltage (let's assume this works according to specs). But then, let's suppose I put a cable between the panel and the cc, which at 100 watts introduces 1V drop. What voltage will the cc work at? I'm thinking of two extreme possibilities:

1. Panel voltage stays at 18.7 and the cc establishes 17.7 as the mpp.
2. The cc still finds 18.7V to be best to work at and so the voltage at the panel will be 19.7.
3. Maybe the cc will work somewhere between the two?....

The problem for me is that in this setup the cc will not see the real panel voltage but the voltage lowered with the cable resistance so I'm not sure it will be able to find the best mpp.

The real question is: with an mppt cc will the effective power loss be the same as the theoretical power loss that I calculate from cable resistance? For example, I use a voltage drop calculator for my cable and find out that for the amperage at 100 watts the resistive loss at the cable will be 10W. Does this mean that I will effectively get 10W less from the cc with this cable?

RCinFLA · Oct 11, 2022

For panels, you just have to worry about wire resistance. Breakers resistance matter if PV Vmp is low. It is breaker's short circuit solenoid coil that adds most of their resistance. For a 20 amp 500vdc breaker they are 10-20 milliohms.

On battery side, there is additional resistance in BMS due to BMS series MOSFET's. As the MOSFET's get hot their resistance increases, like 150% just before BMS does thermal shutdown.

Typical Chinese spec'g is about 20watts of BMS heating for their rated current at 25 degs C temp. So if you have a 100A spec'd BMS, its series resistance will be about 20 watts / (100A squared) = 2 milliohms at 25 degs C and 1.5 x 2 milliohm = 3 milliohm when hot.

On this example, with 100 amp current, you have about 5 minutes before BMS shuts down for thermal overload.

If you want continuous current you need to get BMS dissipation down to 5-6 watts. In this example, square root of current = 5 watts / 2.5 milliohm = square root (2000) = 45 amps.

This is where the derate BMS amperage spec by half rule of thumb guideline comes from.

--------------------------

Also, DC breakers have a lot of series resistance. It is common for a 200 amp breaker to have 3 to 5 milliohms of series resistance. You cannot afford much voltage drop for a 12v DC input inverter, so better to use fuse instead of breaker for 12v high power inverters.

Michal.7 · Oct 12, 2022

Thanks, but I don't see how your answer relates to my question. I was asking about how an mppt cc will behave with voltage drop on the cable from the panels and you are talking about breakers, BMS and the inverter - I don't see a connection...

mikefitz · Oct 12, 2022

The MPPT will find max power from the panel and x amps will flow into the controller , this current determines the cable loss. Loss is current squared times cable resistance.
Example, 5 amps current, 4mm2 cable with a resistance of 0.00471 ohms/ meter, 10 meters cable run.
Cable total resistance , 0.00471 x 10 x 2 = 0.0942 Ohms.
Power loss, 5 x 5 x 0.0942 = 2.355 watts

Mike

RCinFLA · Oct 13, 2022

Michal.7 said:
Thanks, but I don't see how your answer relates to my question. I was asking about how an mppt cc will behave with voltage drop on the cable from the panels and you are talking about breakers, BMS and the inverter - I don't see a connection...

The simple answer is the net power loss will be up to twice the power loss of wire. I'm sure that statement will get folks attention.

The complete answer to that question is very complicated with several interacting pieces.

When you talk about multiple parallel arrays combined near MPPT controller input this can get very complicated as different wiring losses to the parallel arrays can upset the optimum MPPT loading on all parallel wired arrays resulting in even more output power reduction.

The wiring power loss is current squared times resistance of wire. If controller lightens panel loading, the result of this is slightly greater panel voltage and panels output current will decrease slightly as panel operating point moves slightly toward Voc.

This results in lower output power from panel but if the new higher voltage operating point of panel voltage times panel's decreased output current equals the wiring I2R loss the maximum extractable power at the MPPT charge controller is achieved.

With the wiring IR voltage drop, the voltage at MPPT controller input will be less than the slightly higher panel operating voltage.

So net result at controller will be both a current drop and voltage drop at MPPT controller input and nearly twice the I2R power loss of wire. The reason for nearly 2x loss is there is I2R loss in wire plus a nearly equal panel output power decrease. By MPPT controller reducing its panel loading it reduces panels operating output current and increases panel's operating voltage, so it is minimizing the wire power loss by the reduced current.

The reason for the 'nearly' twice statement is the additional loss depends on internal series resistance of panels compared to wire resistance. Twice the power loss of wire would be the case if panel's internal series resistance is very low compared to wire resistance. It is likely the panel's internal resistance is not insignificant compared to wire resistance so the net power loss due to wiring will be less than 2x the wire loss alone. In fact, the internal series resistance of panel is likely greater than wiring resistance, but I had to keep your attention to read a complicated reply.

This link has a chart with a slider you can play with series resistance to get a feel for what happens at MPPT controller input.

Series Resistance | PVEducation

www.pveducation.org

Doggydogworld · Oct 13, 2022

3. Maybe the cc will work somewhere between the two?....

Basically. The controller doesn't know anything about panel specs or actual panel voltage. It just perturbs panel load a little and observes the result. See this web page about MPPT algorithm basics. Starting with your 100W panel, but using round numbers so my head won't hurt:

At panel MPP:
Vmp = 20V
Imp = 5A
Wire resistance = 0.2Ω
Voltage drop = 5A * 0.2Ω = 1V
Wire power loss = 5A * 5A * 0.2Ω = 5W
Controller sees 19V @ 5A = 95W

Let's say the MPPT controller first reduces reduces panel voltage slightly from 20V, increasing current a bit. Panel output will drop and wire loss will increase. Both are in the wrong direction, so the controller will abandon that path and try increasing voltage slightly above 20V. Panel output and wire loss will both decrease. If wire loss drops more than panel output, the controller will increase panel voltage again. And so on until net power the controller sees maxes out.

So let's pretend the above panel ends up at 21V * 4.75A = 99.75W
Wire voltage drop = 4.75A * 0.2Ω = 0.95V
Wire power loss = 4.75A * 4.75A * 0.2Ω = 4.5W (rounded)
Controller sees 20.05V * 4.75A = 95.25W (rounded)

This is an exaggerated case, with pretty high wire resistance and very little panel output drop. If you look at the curve for a real 100W panel you'll see 1V above Vmp reduces current more than my 0.25A. Looks more like 0.4A, which ruins my whole example. In the real world voltage settles much closer to Vmp unless you have very high wire resistance.

RCinFLA said:
The simple answer is the net power loss will be up to twice the power loss of wire. I'm sure that statement will get folks attention.

I can't see this except with crazy high wire resistance, e.g. a 2Ω wire in the above example that burns 50W at Vmp (half of panel output!). Say the controller moves the panel to 21.5V @ 4A. The panel now outputs 86W, but you only lose 32W in the wire for 54W net output. The 14W panel loss is still less than the 32W wire loss, but it's getting in the same ballpark.

And even in extreme cases we're talking about panel loss approaching the reduced wire loss. It will never come close to the original wire loss at Vmp.

Michal.7 · Oct 13, 2022

Thanks guys for your replies, I see this can get complicated once we want to take all the factors into consideration. @RCinFLA, this was an interesting read about series resistance, I didn't know about this concept in solar panels.

RCinFLA said:
The simple answer is the net power loss will be up to twice the power loss of wire. I'm sure that statement will get folks attention.

Like @Doggydogworld, I am skeptical about this, however I don't have enough knowledge to prove or disprove this. You say the panel will decrease its output power the same amount as the wire - I don't understand why it would do so. If the panels voltage will increase due to lower current caused by wire resistance it should compensate for the current loss and so the power loss should not be as high. But this is just my guess, I don't know to what extent these two opposing symptoms will work together. I will probably have to read again your explanation after some time to try to understand all the nuances better.

@Doggydogworld, thanks for the calculations, I think I understand them for the most part and seem logical.

Doggydogworld said:
I can't see this except with crazy high wire resistance, e.g. a 2Ω wire in the above example that burns 50W at Vmp (half of panel output!).

One more thing comes to my mind, which is an extreme case occurring even with not so extreme wire resistance in cases where the panel voltage is low and close to the minimum working voltage of the mppt controller. For example, suppose we have a panel with 17Vmp and a controller that needs at least 16.5V to work - if we use a wire that would drop 1V at full load then this 1V drop would not happen because the controller would not be able to find an mpp below 16.5V so it will stay at 16.5V reducing available power because it will not be able to find the real mpp. I don't know how big the loss would be but I suspect it would be much higher than the loss occurring at voltages within the controllers working range. Am I thinking correctly?

Apart from learning all the theory and maths behind this I'm also interested in practical application: if I use a voltage drop calculator like this one for the wires coming from the panels to the mppt controller will the power loss on the controller's output be roughly the same as the loss I got for the wires with the calculator? This is important, because if what @RCinFLA says is true then we need to account for the double power loss and choose much thicker cables than what the calculator might suggest.

Doggydogworld · Oct 13, 2022

I guess you're right the losses could be higher near the MPPT's minimum operating range, though it seems the bigger risk there is the MPPT might not work at all on hotter days.

Seems to me you can use the voltage drop calculator at Vmp to estimate maximum power loss. The MPPT will either operate the panel at that point or find a point where the total reduction (panel output decrease minus wire loss) will be less.

RCinFLA · Oct 13, 2022

Doggydogworld said:
I can't see this except with crazy high wire resistance, e.g. a 2Ω wire in the above example that burns 50W at Vmp (half of panel output!). Say the controller moves the panel to 21.5V @ 4A. The panel now outputs 86W, but you only lose 32W in the wire for 54W net output. The 14W panel loss is still less than the 32W wire loss, but it's getting in the same ballpark.

And even in extreme cases we're talking about panel loss approaching the reduced wire loss. It will never come close to the original wire loss at Vmp.

It can be crazy high wire resistance or very low panel internal series resistance. Its the ratio between the two.

A panel can have a few tenths of an ohm series resistance for high quality panels. A 100 ft pair of #10 wire has about 0.2 ohms wire resistance.

The primary point is everything effects everything else and the more wire resistance the more 'off' the panel will operate from is inherent isolated MPP point adding more power loss than just the contribution of wire power loss.

Darren67-72 · Oct 13, 2022

RCinFLA said:
The simple answer is the net power loss will be up to twice the power loss of wire. I'm sure that statement will get folks attention.

The complete answer to that question is very complicated with several interacting pieces.

When you talk about multiple parallel arrays combined near MPPT controller input this can get very complicated as different wiring losses to the parallel arrays can upset the optimum MPPT loading on all parallel wired arrays resulting in even more output power reduction.

The wiring power loss is current squared times resistance of wire. If controller lightens panel loading, the result of this is slightly greater panel voltage and panels output current will decrease slightly as panel operating point moves slightly toward Voc.

This results in lower output power from panel but if the new higher voltage operating point of panel voltage times panel's decreased output current equals the wiring I2R loss the maximum extractable power at the MPPT charge controller is achieved.

With the wiring IR voltage drop, the voltage at MPPT controller input will be less than the slightly higher panel operating voltage.

So net result at controller will be both a current drop and voltage drop at MPPT controller input and nearly twice the I2R power loss of wire. The reason for nearly 2x loss is there is I2R loss in wire plus a nearly equal panel output power decrease. By MPPT controller reducing its panel loading it reduces panels operating output current and increases panel's operating voltage, so it is minimizing the wire power loss by the reduced current.

The reason for the 'nearly' twice statement is the additional loss depends on internal series resistance of panels compared to wire resistance. Twice the power loss of wire would be the case if panel's internal series resistance is very low compared to wire resistance. It is likely the panel's internal resistance is not insignificant compared to wire resistance so the net power loss due to wiring will be less than 2x the wire loss alone. In fact, the internal series resistance of panel is likely greater than wiring resistance, but I had to keep your attention to read a complicated reply.

This link has a chart with a slider you can play with series resistance to get a feel for what happens at MPPT controller input.

Series Resistance | PVEducation

www.pveducation.org

I wish I'd talked to you before I installed my system. Because your exactly right. About twice the loss.?

Michal.7 · Oct 14, 2022

Darren67-72 said:
I wish I'd talked to you before I installed my system. Because your exactly right. About twice the loss.?

Did you measure it in your installation and found it to be true? I'm tempted to test it when I get all the parts form my solar system, I'll use a purposefully long cable and see how much power will be really lost.

Another interesting question is how does the cable power loss influence the output of a pwm controller? It's weird but to my mind in most cases it will not introduce any loss at all. Suppose we have a 20Vmp panel connected to a 12V battery via a pwm. When the battery voltage is less than the maximum charge voltage (like 13V) the pwm controller simply connects the panel to the battery. So if the current at 13V is, say, 5A then if we add a wire with 1V drop the current will also be 5A because at 14V the current from the panel will be the same. The should be no power loss up to 7V being lost on the wire - up until the panel reaches 20V. When the battery voltage reaches the maximum charge voltage there should also be no difference as the controller would adjust its pwm signal to set proper voltage even if there is cable with 1V drop before it. Of course, with lower panel output or larger load eventually the wire loss will influence the output power but before that happens there will be no loss at all.

Is my reasoning correct? If it is then the conclusion is what I wouldn't have expected: with a pwm controller the loss is none or proportional to wire resistance while for mppt it is double the wire loss. Therefore, we need A LOT thicker cables for use with an mppt cc than with a pwm - is this right?

Darren67-72 · Oct 14, 2022

Michal.7 said:
Did you measure it in your installation and found it to be true? I'm tempted to test it when I get all the parts form my solar system, I'll use a purposefully long cable and see how much power will be really lost.

Another interesting question is how does the cable power loss influence the output of a pwm controller? It's weird but to my mind in most cases it will not introduce any loss at all. Suppose we have a 20Vmp panel connected to a 12V battery via a pwm. When the battery voltage is less than the maximum charge voltage (like 13V) the pwm controller simply connects the panel to the battery. So if the current at 13V is, say, 5A then if we add a wire with 1V drop the current will also be 5A because at 14V the current from the panel will be the same. The should be no power loss up to 7V being lost on the wire - up until the panel reaches 20V. When the battery voltage reaches the maximum charge voltage there should also be no difference as the controller would adjust its pwm signal to set proper voltage even if there is cable with 1V drop before it. Of course, with lower panel output or larger load eventually the wire loss will influence the output power but before that happens there will be no loss at all.

Is my reasoning correct? If it is then the conclusion is what I wouldn't have expected: with a pwm controller the loss is none or proportional to wire resistance while for mppt it is double the wire loss. Therefore, we need A LOT thicker cables for use with an mppt cc than with a pwm - is this right?

Short story... I spent months planning and doing the numbers. (Like three or four times) including wire size and resistance.
After my full installation I was coming up short on input power. Again, I did the real world figures. Not the rated power or our hopeful power.
I then measured voltage and amperage both under full load and open, at the panels and at the inverter. I did this 3 consecutive days and I came up with a little less than twice my predetermined loss, charge at the inverter (mppt).
I don't know if people would notice on a short run. Even though the same affect applies.
I also don't think this 2x loss would apply to a pwm.
It's not a big loss, but I'm considering if the mppt could be reconfigured to account for this, although it would be specific to the actual distance loss (or loss due to wire size) and could not be predetermined.... unless there was a variable adjustment.
Sorry, I get carried away, sometimes.

Doggydogworld · Oct 15, 2022

Darren67-72 said:
I then measured voltage and amperage both under full load and open, at the panels and at the inverter. I did this 3 consecutive days and I came up with a little less than twice my predetermined loss, charge at the inverter (mppt).

I don't understand. What does open circuit Volts and Amps tell you about loss? Especially Amps, which I hope was zero.

If you see more voltage drop than calculated for the round trip distance between your panels and MPPT controller then something in between (wires, connections, combiner box, etc.) has higher resistance than claimed. That's a different loss than what @Michal.7 is talking about.

The only way I can see to measure this "MPPT effect" is to measure output with the MPPT connected to the panels with the shortest, fattest wires you can find then again with a very long run of the wires you plan to use. According to his theory, the net loss in the 2nd case will be almost twice the measured loss in the long run of normal wires.

Pappion · Oct 15, 2022

The panels have a internal resistance higher than the typical wire run to cc. Wire W(loss) = Current*Current*Ohms.
Both wire lengths (round trip) count in the Ohm calculation. So a 25ft run is 50ft of wire.

Wire voltage drop is more critical for CC output. Typically CC measures/regulate voltage at it's output. Except for my Victron Smart Solar CC which can get voltage data from the Smart Shunt or a remote Voltage/Temperature sensor.

Michal.7 · Oct 18, 2022

Thanks guys for your answers. It happens sometimes that the more you ask the less you know because there are contradictory answers. But I understand it is not so easy to calculate the actual loss because it depends on many factors. A specific mppt controller, solar panel, its Vmp, Imp, internal resistance, type of battery, method of connecting panels into array, etc. can all make a difference. Well, I think I'll need to do some live tests when I find some time for playing with it.

Darren67-72 · Oct 18, 2022

Doggydogworld said:
I don't understand. What does open circuit Volts and Amps tell you about loss? Especially Amps, which I hope was zero.

If you see more voltage drop than calculated for the round trip distance between your panels and MPPT controller then something in between (wires, connections, combiner box, etc.) has higher resistance than claimed. That's a different loss than what @Michal.7 is talking about.

The only way I can see to measure this "MPPT effect" is to measure output with the MPPT connected to the panels with the shortest, fattest wires you can find then again with a very long run of the wires you plan to use. According to his theory, the net loss in the 2nd case will be almost twice the measured loss in the long run of normal wires.

Well, it is something a lot of people do not understand, and not really any reason to.
Open circuit voltage and a minuscule amperage would be a starting point without losses. It's testing data.
Most equipment has a (+-) differential, even cheaper asilasope's will give misreadings due to the resistances in the scopes wire and even if you leave a loop in the test lead wire during the test, among a hundred other things. This is why it's not advantageous for most people to test these suggestions (because it would cost you more to do a proper test than you will ever save) and why I responded the original statement of the almost 2x loss.
No need to take our word for it. The loss is minimal, but the information is there if you want to use it.

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