I have a 100 watt solar panel with the specifications:

  • Optimum operating voltage = 18.1 V
  • Optimum operating current = 5.52 A
  • Voc = 22.1 V
  • Isc = 5.86 A.

I connected this solar panel directly to a 100 watt microinverter and connected a 33 watt lamp to the inverter.

The voltage from the solar panels dropped from 19 volts to 8 volts.

Did something go wrong with the solar panels?

  • \$\begingroup\$ What's the voltage rating of this lamp? \$\endgroup\$
    – Mitu Raj
    Nov 10, 2020 at 11:34
  • \$\begingroup\$ 220 Volt @Mitu Raj \$\endgroup\$ Nov 10, 2020 at 12:19
  • \$\begingroup\$ Did the lamp get on, or remained off? \$\endgroup\$
    – Mitu Raj
    Nov 10, 2020 at 13:35
  • \$\begingroup\$ Does the panel run the inverter even without the lamp? \$\endgroup\$ Nov 10, 2020 at 15:58
  • \$\begingroup\$ To debug this you need to remove some variables. I would suggest connecting a simple resistor to the solar panel. Something like 18.1V / 5.52 Amps = 3.3 Ohms. You can probably use a 25 or 50 Watt resistor as long as you don't leave it connected too long (like a couple of seconds). Measure the voltage across the resistor and calculate the power delivered to the resistor (P = V^2/R). Compare that with the power delivered to the micro-inverter. Maybe the MPPT algorithm of the microinverter is not working. Or maybe the microinverter doesn't work the way you think it does. \$\endgroup\$
    – mkeith
    Nov 10, 2020 at 22:04

3 Answers 3


Your panel is very probably fine (no guarantees, though).

A solar panel is roughly a current source over most of its V/I characteristic, not a voltage source.

So, the voltage you see across it depends on the impedance of the load that is connected (or the voltage of the battery that is connected); it isn't set by the solar panel itself.

The impedance of the load you have is pulling the solar panel's voltage down to 8 V, but the solar panel still delivers about 5 A under full sun, or about 40 W, which is all the power it needs to deliver for your lamp; it just isn't sitting at its Maximum Power Point, where it could and would deliver more (and, in this case, too much) power.

Here's a (measured) example of a 3 W load (a DC/DC converter with a loaded output) connected to a nominally 12 V, 10 W solar panel:

Solar panel characteristic
(Image source: me)

  • 1
    \$\begingroup\$ but the results of testing the current from the solar panel that I did was around 1 ampere instead of 5 ampere \$\endgroup\$ Nov 10, 2020 at 12:18
  • \$\begingroup\$ 1A at 8V is only 8W and isn't on a 100W panel's I/V characteristic under full sun; maybe at 10% of full sun you could get 1A at 8V (See centsys.com.au/pdf/Solar/… for an example of a 100W panel's characteristics). How did you do your measurement? \$\endgroup\$
    – ocrdu
    Nov 10, 2020 at 12:42
  • \$\begingroup\$ I don't think there is enough information to conclude that the panel is fine. But it might be. \$\endgroup\$
    – mkeith
    Nov 10, 2020 at 22:48

No, that just means the lamp is using more power than the panels are providing. You'll only get 100W out of the panels at "optimum" conditions.

  • 2
    \$\begingroup\$ the power of the lamp that I use is only 33 Watt \$\endgroup\$ Nov 10, 2020 at 9:52
  • \$\begingroup\$ And how much power is the sun providing? Are you at the equator at noon? How far north are you, are the panels oriented directly at the sun, and what time of day is it? \$\endgroup\$
    – pjc50
    Nov 10, 2020 at 9:54
  • \$\begingroup\$ See comment under JRE's answer. \$\endgroup\$
    – ocrdu
    Nov 10, 2020 at 10:29
  • \$\begingroup\$ It's not just about how much power the load requires. It's about the impedance match between the load and the panel. Look at the I-V plot in Ocrdu's answer. For some given level of illumination, the instantaneous current and voltage must lie somewhere on that green line. (For a different level of illumination, it would be a different line, but with a similar shape.) The load in Ocrdu's answer that is drawing only 0.57 Amps receives much more power from the panel than the load that draws 0.62 Amps because the panel voltage at 0.57 Amps is so much higher. \$\endgroup\$ Sep 21, 2021 at 14:45
  • \$\begingroup\$ Sophisticated solar installations use a so-called Maximum Power Point Tracker (MPPT) that automatically adjusts its input impedance to keep the panel operating at its maximum power point (again, see Ocrdu's answer) for any given level of illumination. \$\endgroup\$ Sep 21, 2021 at 14:49
  1. Your solar panels are rated to provide 100 watts under optimum conditions.
  2. Under optimum conditions and no load, your panels will have a voltage of 22.1 volts.
  3. With no load, you say the voltage is 19 volts - that means your solar panels are not getting full sunlight to produce 100 watts.
  4. The inverter will waste a good bit of power in converting the DC from the solar panels to AC. It would not be surprising if the inverter wasted as much power as it puts out - your 33 watt lamp would then require 66 watts from the solar panels.

Solar panels do not provide a fixed voltage and current. They convert a certain percentage of light to power. Less light means less power. Less power means lower voltage and current.

Your solar panels aren't getting enough light, and your load needs more power than you think it does.

  • 4
    \$\begingroup\$ What the OP is seeing would happen under full sun with a total load of 33W also. A solar panel is roughly a current source over most of its characteristic, and the impedance of the load is setting the operating point's voltage, which is much lower than the panel's voltage at its MPP. At its MPP, it would be delivering more power than is needed. \$\endgroup\$
    – ocrdu
    Nov 10, 2020 at 10:14
  • \$\begingroup\$ true, I did the test when the sun was shining brightly and was able to produce a voltage of up to 19 Volts but when I connected the lamp the voltage from the panel dropped @ocrdu \$\endgroup\$ Nov 10, 2020 at 12:13

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