I am connecting 7 solar cells, the data or info shown below -

data1 solar2


simulate this circuit – Schematic created using CircuitLab

The above schematic is how I am rigging up my solar cells. There are 7 cells in series. I am assuming each gives out 2.740V(Voc). So 7 in series = 19.18V. I am making 2 different panels (each with 7 cells). They are connected in parallel to increase the amperage to the batteries via an MPPT charger. max current drawn is 0.5Amps per panel.

Now, my doubt can be stated in 2 phases -

1) Imagine there is full light incident on panel A and not so full on panel B. So, panel A gives out 0.5Amp, while panel B gives out a lower ampere, right ?

2) Panel A , suppose gives the max of 19.18V. So, in Node 1 we get approx that value. Panel B,due to lesser incident light , gives approx 15V,at node 2. What will be the voltage at node 3 ? Will any current be even drawn via node 2 ? because the cathode of D9 will be at higher voltage than anode of D9, so current will be forced in or rather the D9 will be reverse biased, thereby blocking any current from panel B.

Hope my reasoning and logic is correct. If this is so, any panel, connected in parallel will work only if same light is incident on all of them, which is a stupid thing, right ?

  • 2
    \$\begingroup\$ 1N4148 is completely inappropriate in this situation. Use a higher current Shottky diode. \$\endgroup\$ – Olin Lathrop Apr 26 '16 at 10:58
  • \$\begingroup\$ @OlinLathrop. How is it unappropriate ? \$\endgroup\$ – Board-Man Apr 26 '16 at 11:52
  • 1
    \$\begingroup\$ Did you even look at its datasheet? \$\endgroup\$ – Olin Lathrop Apr 26 '16 at 13:20
  • \$\begingroup\$ @OlinLathrop. I mistook the peak surge of 2Amps for the peak current. In fact peak continuous is 0.5Amps. My mistake. Thank you. \$\endgroup\$ – Board-Man Apr 26 '16 at 13:22
  • \$\begingroup\$ @OlinLathrop You may (or may not) be interested in my conclusion in my second (lonnng) answer - which contradicts popular wisdom. At least for the curves I used - and quite likely for many real ones, the power contribution of lower illuminated panels when paralled with fully illuminated panels is vey good % of what it would be if they are optimally loaded. NB: This is for the whole panel evenly illuminated. ie it applies to parallel connection and not to partial shading of series strings. | .... \$\endgroup\$ – Russell McMahon Apr 28 '16 at 0:05

Super summary:

  • This is not the result you thought you were expecting.
    Neither was I.

  • Identical panels that are each evenly illuminated but at different light levels per panel, their power outputs add when paralleled.

  • If one panel is illuminated at 'full sun' and the other at a level of say 25% or more of full sun, the power output of the less illuminated panel will be 80%-95% of its optimum power output at the insolation level concerned.

    Read the above carefully.
    Slightly more detailed summary at end of answer.


The following description shows what happens when you parallel evenly illuminated PV panels which are identical in characteristics but which have different light levels falling on each panel. Typically this will be a parallel combination of individual panels, typically with all cells in a given panel in series.

Note that the following "analysis" relates to WHOLE PANELS where any given panel has the same insolation (light energy) level across the whole panel.
This is not the place for a shaded cell discussion, but:

  • Where a panel of cells in series is shaded across part of the panel
    the Ipanel is ~= the current able to be made by the most shaded cell.
    As I_most-shaded drops
    V_shaded cell voltage drop from R_effective x I rises
    until the cell is reverse biased by the voltage from the rest of the cells -
    and the bypass diodes conduct.

The following process must be followed carefully. When following various load lines and similar it is easy to 'decide' to follow the wrong line for some reason. If (when) that happens the result will be incorrect.

Advice #1: If the result you get does not match the result given, go back and follow the instructions as written.

The results will vary with panels used. Voc, Isc, V, Vmp, I at various light levels will all vary according to what panel is actually used. It is assumed that the light level across a whole single panel is constant.



insolation - energy input to panel - usually "sunlight"
Insxx = Insolation xx% of max. mp maximum power point at max insolation
mpxx maximum power point for a given insolation level

V panel voltage
I panel current
Voc panel voltage at 100% insolation and no load
Isc panel current at 100% insolation and full short circuit
Vxx voltage at insolation xx% curve (in the context being discussed) Vmpxx voltage at maximum power point on the xx% insolation curve.

Similar terms that follow the above 'rules' may be used - eg Imp20 has the 'obvious' meaning.


The curves below show the performance of a panel at insolation levels from 10% to 100% in 5% steps.
The blue dots represent the maximum power point for that curve. At that point, for the insolation level involved the product of V x I = W is maximised. This is termed Wmp.
For eg 85% insolation the peak power will be termed Wmp85

Drawing a line vertically downward from a blue dot yields a voltage. This is Vmp for that curve.
Drawing a line horizontally to the left yields a current. This is Imp for that curve.

Drawing vertical and horizontal lines from ANY point on a curve yields a V and and I. For the given amount of insolation the V will always be that value if I is always that value and vice versa. For example, on the 100% insolation curve, at the Wmp point V is 43V and I is 5.1A. If V as Ins100 is ever 43V then we know that I = 5.1A.

As a panel is loaded with more load that at the Wmp point so that more current is taken, V will fall. If less current is taken V will rise.

The graph below of a PV panel performance with load was chosen as a good example as it has 5% insolation change lines (most have fewer). The curve shapes appear typical of most panels but differences may produce substantially different results for a given "real world" panel. Or not. .

Graph from here fron fig 4a/4b on this webpage - (irrelevant to use of curves here)

enter image description here

Let the fun begin.

The following will show how much power is contributed by and identical panel with a lower insolation level IF THE VOLTAGE DOES NOT CHANGE. This could be due to powering a battery voltage held constant by a DC-DC converter, ... . You can in fact use this method for any combination of panels and insolation levels but the problem becomes iterative and messy.

The curves below are a subset of those above. The X axis has been limited to a voltage range around those where the Vmps all lie. This is done entirely and only to increase clarity.

Note that this not at all hard - just able to be followed incorrectly if inattentive.

(1) Draw a line vertically downwards from the blue dot at mp100
(= existing vertical blue line) The voltage shown = Vmp = Vmp100 = 43V

(2) Draw a line horizontally to the left from the blue dot at mp100 (= existing orange horizontal line) The current shown = Imp = Imp100 = 5.1A

Wmp is not shown but is obviously 43 x 5.1 = 219.3 Watt

This is the output wit one panel optimally loaded in full sun.
These are the reference maximum output values.

(3) On the vertical blue line, note the intercepts with the other lower insolation curves. These are the operating points the other panels will be at for V is the same as Vmp for the initial panel.

(4) From the intercept points with the blue line and the X% insolation curves draw lines horizontally left to intersect the Y axis (green lines). Note that these are NOT drawn from the blue dots = mpxx points, they are drawn from the intercept points with the Vmp lime.

(5) For a give % insolation follow the corresponding horizontal line and read the current. This is the current the panel will deliver at the voltage at which V = Vmp for 1st panel.
eg the 80% insolation curve (the 5th red curve from top) intercepts the blue line at 4A. So when the 1st panel delivers 5.1A the 2nd panel will deliver 4A.
Note that 4/5.1 = 0.78 or 78% of maximum output. As the insolation is 80% of maximum, very little has been lost by operating the panel off its Vmpxx.
As a more severe case consider the 20% insolation curve.
At 43V (blue line) it will produce 0.7A.
At its Vmpxx (at blue dot on 20% curve) it would have produced about 0.95A (extend line from blue dot to Y axis).
So Iout is about 0.7/0.95 = about 74% of the current it would make at Vmp20.
Vmp20 can be seen to be in the 38 to 39V range, so the panel is being made to rise to a voltage (43 - 39) = 4V above its ideal mp voltage.
Note that the I20 curve has a maximum voltage of about 46V.
If another panel is operating at above 46 V no current will be output by an I20 panel.

enter image description here

How much current is lost by operating a panel off optimum can be seen by comparing the y axis intercepts of lines drawn horizontally from the Ixx curves blue dot (Wmp point) and the green lines of the corresponding Ixx curve.
OR - easier - see how far below the blue dot the green lines pass.

In the chart below I've rotated a subset of the charts above counter-clockwise by 90 degrees. SO either turn your head on its side (head on left shoulder :-) ) or mentally change the terms above such as "left" and "above" as required.

The short red lines from the blue dots (mpxx points) to the relevant green line shows the magnitude of the current drop caused by running the panel at a higher voltage than optimum. It's "higher than optimum" as when insolation falls Voptimum also falls.

Pseudo MPPT:

The above illustration of change in current as voltage is taken off-optimum nicely illustrates the effect of a pseudo MPPT method which comes close to MPPT results with much less effort. THe method is to predict an optimum panel voltage Vmp regardless of insolation and load the panel to that voltage during operation. As can be seen from the above, current is not greatly lower than optimum when V is set to 43V and, as V is now higher than Vmpxx, the power produced is not as greatly reduced as the current is.
In the extreme example above, at 20% insolation
...... mp = 0.95A, 39V say = 37 W
operating = 0.7A, 43V = 30 W
So current is 0.7/0.95 = 74% of optimum BUT
power is 30/37 = 81% of optimum.
At high insolation levels but under 100% the difference between optimum power and power at constant voltage will be very small.


Cells in series (say within a panel) should have the same light levels on them.
If light levels per cell varies in a series string then

    I_series_string ~= I of least illuminated panel. 

For parallel panels, with any one given panel illuminated reasonably evenly across the surface, the power made by individual panels will add.

When paralleled with fully illuminated panels, the power output from a less illuminated panels will be lower than it would at that illumination level BUT not vastly lower.
eg if a panels has insolation of say 50% or more then its actual output will be say 80%-95% of what it would have contributed at that light level if optimally loaded.

Made up example

2 x identical 500 Watt panels
Panel_1 full sun
Panel_2 70% sun

Output of Panel_1 optimally loaded = 500W
Output of Panel_2 optimally loaded at 70% sun = 350W

When parallelld at Vmp for panel 1

Pout Panel_1 = 500W

Piut Panel 2 = 350 W x (80 - 95%) = + 290 to 330 Watt range. Say 310W

Pout P1 + P2 = 810 W

enter image description here

  • \$\begingroup\$ This confirms that while MPPT is nice to have, it's far from essential in many applications, +1. The big gotcha is temperature variation, as that has a direct effect on voltage, making picking a fixed voltage more difficult. I know you did a lot of work already, but your point would be even more intuitively explained by showing a set of curves of power as a function of voltage. Then we can see the relatively flat region near MPP, and see that a little sliding in voltage near the peak makes little difference. You seem to not have the data in machine readable form, so I understand. \$\endgroup\$ – Olin Lathrop Apr 28 '16 at 11:19

You've pretty much got it, the output voltage of most solar cells is not that dependent on the amount of light (within reason). The open circuit voltage can be 10+ % higher than Max power voltage so in situation 1, panel A will be doing the bulk of the work, but B might be helping a little. In situation 2, the voltage at node 3 will be a diode drop below whatever panel A's voltage is. Panel B might be contributing a little current (Voc > Vmaxamps so it might have enough voltage to push some current), but node 3 will always be a diode drop down, if it was higher, no power would be able to get through the reverse biased diodes and the only way node 3 could be lower than 1 diode drop was if there was a problem with the diode. A solar cell is a voltage source and a voltage source will push power into a lower voltage node and draw power from a higher voltage node (unless there's blocking diodes)

  • \$\begingroup\$ I am confused here. Imagine node 1 is 19V. Then node 3 will be 18.4V(0.6V diode drop). Now, node 2 is 15V, This means the diode D9 is reverse biased, so, no current is contributed,right ? SO, panel B will be completely useless and contribute almost nothing, right ? \$\endgroup\$ – Board-Man Apr 26 '16 at 10:38
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    \$\begingroup\$ Correct, panel B isn't going to be doing anything unless it can forward bias D9 \$\endgroup\$ – Sam Apr 26 '16 at 10:40
  • \$\begingroup\$ So, the moral of the story is that, panels in parallel will be useful only if the incident light is same. \$\endgroup\$ – Board-Man Apr 26 '16 at 10:41
  • \$\begingroup\$ More or less, most panels have reasonably flat voltage vs light plot and their open circuit voltage can be 10+ % higher than their maximum power voltage so they can sort of share the load, a more heavily loaded cell will have a greater voltage drop, if that drop is big enough, the next panel can start to push a little power as well \$\endgroup\$ – Sam Apr 26 '16 at 10:45
  • \$\begingroup\$ Then why would anyone connect solar panels in parallel to get additive current/amperage ? alternative-energy-tutorials.com/energy-articles/… The link says otherwise. \$\endgroup\$ – Board-Man Apr 26 '16 at 10:49

The "problem" is that panels all need about the same light level to be usefully connected together - in parallel or series.

ie re

any panel, connected in parallel will work only if same light is incident on all of them, which is a stupid thing, right ?

The answer is NO! - it is how it must be done for full output.

However, the example you gave shows a misunderstanding.
Tom noted that panel voltage is relatively constant with light level and that loaded voltage is lower than unloaded voltage.
Look at the data sheet for your cells. Note the loaded voltage for your panels.
Now also look at Vmp (= at max power) and Voc. In your example, if panel A is in full sun, and panel B is in say 90% sun. Voc for both panels will be ABOUT the same. Vb will NOT be 90% of Va. If you load panel A to max efficiency and Vout is say 20V and load panel B to max efficiency then VB will be ABOUT the same as Va (slightly lower) as 90% sun means it can make about 90% of the current of panel A. If you instead connect panel B to the load as well then it's voltage is higher than its best efficiency point so it can't make 90% so it will reduce I somewhat. If you get to the point you descrive where VB is 15V and Va is 20V then light levels must be very different so you will get less poser.

If you place the panels in series and Pa can make say 10A and Pb can make 9A then you can get only 9A as Pb restricts max current possible.

In your diagram, your top diodes D3, D9 are not usually needed of panels are well matched and lit equally. If not you need the diodes But as you say, one panel will not do much.

  • \$\begingroup\$ Thank you for explaining. But, I am still confused. Sorry sir. Let me elaborate - 1) How can I fine Vmp ? Voc - max voltage (when not loaded). How did you get Vmp ? 2) You mention loading panels A and B at different instances. They are bothe being loaded at the same time. 3) The 90% example is understood. But, imagine my panes are at opposite ends. So, when one panel(say panel A) is being used, the other (panel B) is in a relative shadow. In such a case, the voltage from Panel A is almost max, while panel B is almost low. Right ? 4) Without the diodes, the current from A will enter B. \$\endgroup\$ – Board-Man Apr 26 '16 at 11:26
  • \$\begingroup\$ @Board-Man I had a closer look at my advice, and at what most people will tell you, and I decided it was misleading. My new advice is different enough that I've put it in a new answer. The results are "interesting". In summary. - Cells in series (say within a panel) should have the same light levels on them | - If light levels per cell varies then I_series_string ~= I of least illuminated panel. | - For parallel panels, with any one given panel illuminated reasonably evenly across the surface, The power made by individual panels will add. | .... \$\endgroup\$ – Russell McMahon Apr 27 '16 at 13:28
  • \$\begingroup\$ ... When paralleled with fully illuminated panels, the power output from a less illuminated panels will be lower than it would at that illumination level BUT not vastly lower. | eg if a panels has insolation of say 50% or more then its actual output will be say 80%-95% of what it would have contributed at that light level if optimally loaded. \$\endgroup\$ – Russell McMahon Apr 27 '16 at 13:28

A concise answer to this question is that you must consider the resistance of the load (battery charger) and remember that real voltage sources are not ideal.

When the resistance of the charger load is high, (battery full or missing), then what you described is correct -- the panel providing the higher voltage will provide all of the drawn current. The voltage of node3 will be that of node1 minus the forward voltage drop of the diode.

Now say the resistance of the charger load is low (battery charging -- load perhaps in the tens or hundreds of milliohms). You can no longer think of a single series panel as an ideal voltage source. It will have its own internal resistances at various points within the cell (near photosensitive junctions and in bus wires). These values will be comparable to that of the load, and that will mean in a noticeable drop in node1's voltage. When node1's voltage lowers to that of node2, then the second panel will start to contribute significant amounts of current. And so in a higher current situation, node3 will approach the voltage of node2.

"Higher current situation" here means "near the limits of the first panel's power output but still within reach of the output of panel 1 and panel 2 combined."

To really understand this thoroughly, you can imagine shorting node3 to ground. Now the load's resistance would be nearly negligible compared to that of the internal resistances of the panels and the latter are likely to be quite similar to each other. The voltage of node3 is now very close to ground, and if you could probe the insides of each paenl you would find that instead of a single +19V or +15V drop you'd find a series of smaller voltage down-gaps across the bus wires and up-gaps across the photosensitive junctions. The short you created would correspond to one of the down-gaps. Of course, all these up- and down- gaps would sum to zero in accordance with Kirchhoff.


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