I have a nokia phone charger which has the following specifications:
When I checked the output of the charger with a multimeter it shows 7.7 V. Why is it so? Is it because the charger voltage should be greater than battery voltage (for powering the load and charging battery) or any other reasons?
Some regulated supplies (cheap switchers) allow their outputs to drift high when completely unloaded. With a typical load they will work correctly. What's happening is short duration switching spikes containing very little energy are rectified and gradually charge up the smoothing capacitor. Because they have very little energy this only affects the output voltage when the supply is unloaded.
Re-measure with a minimal load (maybe 10ma, or a 500 ohm resistor) and the voltage may be much closer to nominal. If it's still over 7V you have an unregulated supply.
The designers assume that because no load is connected, this deviation from the specification doesn't matter. And for the original application, they are correct; however if you are using the supply for another purpose, beware ... or take appropriate precautions - like wasting 10ma in a resistor!
There are two kinds of DC transformers you can get, regulated and unregulated. The regulated kind have strict margins of error controlled by an internal voltage regulator, they are more expensive but are useful in certain applications. Unregulated power supplies are much more common (like the one in your phone charger), and the voltage they supply is dependent on the current draw of the load they are powering. So when your phone battery is charging and drawing however-many mA, the power supply is operating at 5v; but with the comparably minimal load of the multimeter it is throwing 7.7. This is normal. If you are trying to use that particular adapter with another device it's best to make sure that the current draw, as well as the voltage requirement, of the other device is as closely matched as possible.
