So I bought this 5V 1A solar panel that can be used e.g. to charge your smartphone accroding to the product description.

When I measure the voltage using a multimeter I measured 6.9 V. A little bit higher than my expectations. However, when I measured the Amps it was ~0.3 A, less than I expected at blazing sun.

I decided to add a 6 ohm resistor and measure V and A again. To my surprise I measured to voltage to just 1.6 V and the current to 0.2 A. Later I read that I/V ratio for solar panels is not trivial...

My question: How should this solar panel be suitable to be used for charging a smartphone when a smartphone clearly needs 5V?

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    \$\begingroup\$ Are you sure Amps is not Isc. with Voc=6.9 , Optimal V is usually somewhere inside 80% +/-5% of Voc for PMT. It seems specs are incorrect. Normally given in Watt max.output. \$\endgroup\$ Sep 11, 2016 at 19:13
  • \$\begingroup\$ Were you indoors when you made the measurements? Was the panel horizontal or oriented directly towards the sun? \$\endgroup\$ Sep 11, 2016 at 19:31
  • \$\begingroup\$ outdoor, horizontal oriented yes. \$\endgroup\$
    – Chris
    Sep 11, 2016 at 19:43
  • \$\begingroup\$ @Chris - "I measured 6.9 V" - Did you measure that voltage at its cable-mounted USB socket? Or was that measured on the back of the "bare" solar panel before that small, black box attached to the USB socket? \$\endgroup\$
    – SamGibson
    Sep 11, 2016 at 19:54
  • \$\begingroup\$ 6.9V was measure directly at the cable-mounted usb socket, yes (the two outer pins of the usb female cable). \$\endgroup\$
    – Chris
    Sep 11, 2016 at 19:58

2 Answers 2


If you open the black box you will probably find that each of the Data+ and Data- lines are each connected to ground with a 15 kilo-ohm resistor. The phone is supposed to detect these resistors, know that it is connected to a "standard downstream port" (SDP) USB port, and configure itself to draw at most 100 milliamps of current.

If you measure the IV curve of the solar panel at 100 milliamps current, try a variable resistor or about 50 ohms fixed resistor, you will maybe measure 5 volts. Thus allowing the manufacturer to claim that the solar panel will charge a mobile phone (even though it will take about 50 hours of full sun, or about 12 days allowing for 4 hours a day of noonday sun).


This solar panel has a size of 165 * 210 mm. If we calculate the power for 1 kW per square meter solar input and 10 % efficiency, the result is 3.4 W only. The result is valid only for full sunlight from a clear sky and best orientation of the solar panel to the sun. The efficiency of 10 % was only a guess, the actual value might be lower.

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    \$\begingroup\$ Your guess is not that far from the truth. Typical polycrystalline panels have efficiency around 15-16%, but since not 100% of the surface is covered in actual photovoltaic elements, and you'll have unaccounted losses here and there, I'd say 10% estimation is very reasonable. \$\endgroup\$ Sep 15, 2016 at 13:54

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