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I just obtained o solar panel which shows a max output of about 18-19 votlts . I put it in partial shade to get an output of about 2.7 volts and connected it to an LED ensuring the correct positive and negative legs. But the LED failed to light. Other small appliances and toys using 3-6 volts also failed to work. Any suggestions ?

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    \$\begingroup\$ Solar panel data sheet? \$\endgroup\$ – Leon Heller May 18 '11 at 9:55
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Use a series resistor with the LED to limit the current (220R, say), you could damage it otherwise.

Measure the panel voltage when you have the item connected to it. You will probably find that it drops well below the 2.7V you were getting with no load. and is too low for it to work.

Assuming that the panel has enough output in full sunlight, you will need a suitable regulator to power low voltage items. Switchers are often used, because of their high efficiency.

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Make sure the panel is supplying enough current. Current is what determines if an LED lights up, not voltage. That is to say, even if you have 1.7V (enough volts) but are only able to supply 0.3mA your LED won't light if it needs 1mA.

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PV (photovoltaic) or solar panels are close to being constant current devices with current out being close to proportional to light levels.
The voltage output of a PV cell is close to constant across its usual working range.
Only by reducing light levels to a very ow percentage of full power can you get the voltage to drop substantially.

This means that if you shade an 18V panel to the extent that it is only providing 2.7V, the illumination level will be exceedingly low, so that the available current will be close to zero. LEDs driven with close to zero current produce close to zero light out :-).

If you MUST use am 18V panel to drive a 2.7v load, you can load it with a 2.7V voltage clamp (such as a zener diode or a number of silicon diodes in series, so that when the load conducts the voltage is clamped to their forward voltage at the panels operating current.
In full sunlight the current at 2.7V will be close to the panel's shortcircuit current which is about 10% to 20% higher than the panel's maximum power point current.

I max_power = Watts max power / Vmaxpower
eg If you have a 1 Watt panel then Imp ~= 1W/18V ~+ 55 mA.
So for a say 6W panel Imp = 6W/18V = 330 mA. Your load needs to be able to handle that much current and dissipate the power provide. (Whether an LED or motor or other device)..

Power = V x I = 2.7V x Imp x 110% or so in this case.

You can get 2V7 zeners diodes BUT a 3V3 would be better - allows most white LEDs to be driven. Or you could use about 5 x silicon diodes in series. 5 x 0.6V = 3V. As current rises Vf_diode rises and depending on diode and current may be between 0.6V amd 1V. (1V unusual). Using a string of eg 1N400x diodes allows you to clamp about 1A of continuous current..

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  • \$\begingroup\$ Correct for the current output, but for the LED what is important is not the voltage but the current flowing through. If you use a zener, a resistor in serie with the LED is still required to limit the current through the LED otherwise it will be destroyed. Also depending of how much current the panel will produce the Zener has to be sized accordingly is not really a device to dissipate power. A LDO or even better a current limiter would be more suited to that usage. \$\endgroup\$ – Damien Mar 2 '17 at 8:59
  • \$\begingroup\$ @Damien Your comment has merit but is too broad to be useful in this context. Your comment about LED current needing to be designed to be <= I_LED_max is correct, but that was not a detailed part of the discussion. While a series resistor is a common way to limit LED current that would be hard to design in Panel_Wmp was >> W_LED_max. Whereas eg an 18Vmp, 6W panel has Imp of 333 mA and Isc of typically about 370 mA so an eg 1W LED connected directly across the panel would draw a max of Isc ~= 370 mA and run at about 1W ) eg 3V x 0.37 = 1.1W. 3.3V x 0.37 = 1.2 W. ... \$\endgroup\$ – Russell McMahon Mar 3 '17 at 2:07
  • \$\begingroup\$ ... @Damien ... Whereas using a series R at say 300 mA would dissipate about (18-3.3)V x 0.3A = 4.4W for ~= 1W in the LED with no "wasted" power. Using a non-switching regulator or current limiter would still dissipate about 4W of wasted power. ... \$\endgroup\$ – Russell McMahon Mar 3 '17 at 2:07
  • \$\begingroup\$ If he / you work through what I said you will find that it is (1) correct on all key points (2) addresses the points you raise. | I noted that PV panels are essentially constant current devices in their normal operating range, that when Vout is 2.7V for an ~= Vmp 18V panel that Isc will be much lower than 2.7/17 x Imp. I noted that Imax in full sun will be about Imp x 1.1 and a clamp or load must handle this, that whatever load needs to be applied must be capable of accepting the delivered power, that a clamp load needs to be "suitable" (eg 5 x Si diodes (eh 1N400x) . \$\endgroup\$ – Russell McMahon Mar 3 '17 at 2:08
  • \$\begingroup\$ @Damien As a result of your input I went back and fixed up various typos and spelling errors and made minor changes to the post proper. The two sentences which end in .. rather than . are added to not change the meaning but to increase its understandablity in the area that you suggested. \$\endgroup\$ – Russell McMahon Mar 3 '17 at 2:13

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