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 ?
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.
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.
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..