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When 2 solar panels are connected in series, the output voltage is sum of both panels but the output current (measured by short circuiting) is the same as single panel.

What I dont understand is that according to ohms law, if volts increase, current also increase. But in solar panels case why is it the same?

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    \$\begingroup\$ Hint: a solar panel is not a resistor. If the solar panel was an ideal voltage source what current would you expect in a short circuit? \$\endgroup\$
    – Transistor
    Commented Apr 13, 2020 at 13:32
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    \$\begingroup\$ But then you are expecting double the voltage and double the current so four times the power? \$\endgroup\$
    – Transistor
    Commented Apr 13, 2020 at 13:43
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    \$\begingroup\$ A solar panel is not a resistor. Ohms law does not apply. \$\endgroup\$
    – winny
    Commented Apr 13, 2020 at 14:47
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    \$\begingroup\$ No you don't. If the first pump pumps 1 L/min and that feeds into the second which also pumps 1 L/min then the output is 1 L/min. They are in series. The current has to be the same. \$\endgroup\$
    – Transistor
    Commented Apr 13, 2020 at 15:02
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    \$\begingroup\$ You are confusing resistors and ideal voltage sources with solar panels. Pick one and understand how it works. Then move on to the other. \$\endgroup\$
    – winny
    Commented Apr 13, 2020 at 18:44

3 Answers 3

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Your confusion stems from how soar panels produce electric current and from mixing two different concepts.
They can produce no more than their rated current, even if you make a short circuit.
This has nothing to do with Ohm's law (except for the internal resistance of the panels), but I see that you lack a basic understanding of how two power sources interact when connected in series or parallel.
Ohm's law ignores the internal resistance and the maximum current capability of a power source like battery, solar panel, power supply for the sake of simplicity because it only focuses on the relationship between the voltage and current ACROSS and THROUGH the RESISTANCE/load (this is the key point here).
The tiny detail that confuses you is that you THINK you have DOUBLE the voltage across that resistance, but the voltage across that resistance will not increase when you add another panel in series if that resistance was ALREADY drawing the maximum amount of current a single panel can give.
Remember the basics about batteries: to increase voltage, you connect batteries in series; to increase CURRENT you connect batteries in parallel, but the voltage remains the same.

A solar cell can only push a certain number of electrons when exposed to sunlight. It can be considered like a "current source" with a limited voltage.
A current source will give you the same current whether you connect some resistance at its ends or if you short it out.
However, if that resistance/load limits the current through it because of insufficient voltage across it, you can double the current if you connect 2 panels in series to raise the voltage and push more current through that load/resistance, but, again, that current cannot go above the maximum (or short-circuit) rated current of the panel.

If you want more current, you add a panel in parallel.
If you want more voltage, you connect panels in series.
If you want both higher voltage and more current, you need to connect 2 panels in parallel with each other, and then in series with another 2 panels that are connected in parallel with each other.

So, once again, you need to learn some basics about series and parallel battery connections, and read carefully the specifications on the power sources like solar panels.
You are jumping all over the place without connecting the dots.

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  • \$\begingroup\$ So what i get from ur answer is that solar panels are limited in current when in series but batteries are not. Is that right? \$\endgroup\$
    – Asim
    Commented Apr 13, 2020 at 14:34
  • \$\begingroup\$ I understand some basics of the ohms law. Which states that if u increase the voltage on a constant resistance u would get increased current as well. \$\endgroup\$
    – Asim
    Commented Apr 13, 2020 at 14:43
  • \$\begingroup\$ Sort of. Batteries also have internal resistance which limits their output current, and the current is limited by the size of their plates (larger battery plate surface, more electrons, more current). It is just that batteries usually have a lower internal resistance than solar panels (but not all batteries) and that means they can give more current before hitting the limit. There is no such thing as an ideal source that can give you unlimited current. They are all limited to a certain maximum current. \$\endgroup\$ Commented Apr 13, 2020 at 14:44
  • \$\begingroup\$ The current would increase only if the source is capable of producing more current. If the resistance was already drawing the maximum current, increasing the voltage will not increase the current. Think of it in terms of water: amount of water in liters per second is like the amount of electrons (or current in amperes), the pressure under which water comes out is like the voltage in electric circuit. If your source gives you 1l of water per second, you can only draw up to that amount every second, regardless of whether it comes out in a slow wide flow or narrow high-pressure stream. \$\endgroup\$ Commented Apr 13, 2020 at 14:50
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    \$\begingroup\$ @Asim, no it's not just because of the series resistance, it's because of the constant current nature of the source. See my answer too. \$\endgroup\$
    – Transistor
    Commented Apr 13, 2020 at 15:30
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enter image description here

Figure 1. The equivalent circuit of a solar cell. From Wikipedia's Theory of solar cells.

To understand the electronic behavior of a solar cell, it is useful to create a model which is electrically equivalent, and is based on discrete ideal electrical components whose behavior is well defined. An ideal solar cell may be modelled by a current source in parallel with a diode; in practice no solar cell is ideal, so a shunt resistance and a series resistance component are added to the model. The resulting equivalent circuit of a solar cell is shown on the left. Also shown, on the right, is the schematic representation of a solar cell for use in circuit. [Emphasis mine.]

If the solar cell is behaving as a constant current source then it doesn't matter how many you put in series, you can only get IL from the combination. If, for example, IL = 1 A then two or more panels in series will give 1 A into a short circuit.


If you don't short-circuit the panels and allow them to work at their optimum point - maybe 12 V and 0.5 A, for example then the series connection will give out 24 V @ 0.5 A and the power into the load will be twice the power of one panel. Note that your load resistance will have to be double that of the 12 V load.


From the comments:

If we leave the short circuiting aside and assume a solar panel connected to 10 ohms draws 1 A . And if we connect another panel in series to same 10 ohms would it draw 2 A? Assuming both panels are capable of providing more than 1 A according to load.

enter image description here

Figure 2. Solar cell I-V curve. Modified from source: Alternative Energy Tutorials.

The I-V (current vs voltage) curve of the solar panel is required to answer this question. Starting at the Open Circuit point on the V-axis we can see from the blue I-V curve that as we increase the current drawn from 0 A the voltage starts to decrease. At the maximum power point the curve is turning almost horizontal and it has gone into current limit and the voltage collapses. The purple P-V curve shows us that maximum power will be obtained at MPP and that is where you try to operate for maximum efficiency.

  • If you were to run your 12 V panel into a 10 Ω load and it happened to be operating at point (3) then if we double the load (half the resistance - which is the same effect as your comment) the current would increase to that at (2) but the voltage would drop to maybe 90%. Your power out would be less than double: it would have gone from \$IV\$ to \$2I \times 0.9 V\$ or 80% extra. Note that you if you were operating at (1) then you were were not operating the panel efficiently.

  • If you were operating at (2) and tried to increase the current the voltage will collapse as shown at (1).

    • Relative to (1) current has doubled but voltage has collapsed to 50%. Now instead of \$ P_2 = VI \$ you have \$ P_1 = \frac {V_2} 2 2I_1 \$ which is the same power you started with.
    • Relative to (2), the optimum, I has gone up about 10% but V has collapsed to < 50%. Now \$ P_1 = 1.1 \times I \times \frac V 2 = 0.55 P_2 \$.
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    \$\begingroup\$ See the update to respond to your comment in Edin's answer. Solar cells do not behave like batteries. \$\endgroup\$
    – Transistor
    Commented Apr 13, 2020 at 16:19
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When 2 solar panels are connected in series.

  • If the data sheet for a single solar panel said it produces 12 volts (for example under certain lighting conditions) into an open circuit you probably wouldn't be surprised.

  • If the data sheet also told you that the maximum output current (short circuit) is 1 amp you probably wouldn't be surprised either.

  • The data sheet might also tell you that the maximum output power is at 6 volts and 0.5 amps. Would that be a surprise?

So, if you put two of these solar panels in series you would expect 12 volts at a current of 0.5 amps to achieve the maximum output power. In other words, the current doesn't change if you want maximum output power.

If you tried to get 1 amp from the dual panel you would have to short out the whole panel and that means zero power.

One solar panel: -

schematic

simulate this circuit – Schematic created using CircuitLab

Two solar panels will have an open circuit voltage and effective internal series resistance of 24 volts plus 24 ohms.

This means that the short circuit current is 24 volts / 24 ohms = 1 amp.

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  • \$\begingroup\$ My queation is for theoritical understanding and not a practical one. What make me confuse is the contradiction in ohms law. If we take two batteries connected in series and put them in constant resistant. We get the double amps from both because volts increase also the current. But why its not the case in solar panels \$\endgroup\$
    – Asim
    Commented Apr 13, 2020 at 13:45
  • \$\begingroup\$ There is no contradiction - if two series panels are shorted you can still only get the short circuit current of 1 panel. If the two panels were in parallel then you would get twice the current but only the same terminal voltage as 1 panel. \$\endgroup\$
    – Andy aka
    Commented Apr 13, 2020 at 13:46
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    \$\begingroup\$ You keep asking and I keep saying "no". Also when shorted, they cannot put out any power because v x i will be zero due to v being zero. \$\endgroup\$
    – Andy aka
    Commented Apr 13, 2020 at 13:59
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    \$\begingroup\$ YOUR QUESTION STATES: \$\color{RED}{\text{but the output current (measured by short circuiting) is the same as single panel.}}\$. Your question IS ABOUT short circuit current. Sure, if you are talking about two batteries not being run close to their short circuit current then you get twice the current (as you would when you used half the resistance across one battery). \$\endgroup\$
    – Andy aka
    Commented Apr 13, 2020 at 14:17
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    \$\begingroup\$ You won't get double the short circuit current and that is what your question is about. \$\endgroup\$
    – Andy aka
    Commented Apr 13, 2020 at 14:31

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