Hi everyone,
Can I connect 5.6v adapter to arduino (uno r3) usb connector and pls guide me how much volt arduino uno r3 can bear if I power from usb connector?
thanks
Why would you have a 5.6V adapter with a USB connector on it?
You can drop the extra voltage with a silicon power diode, but it will not be as well regulated.
thanks Paul__B for response.
Is 5.6v or 5.4v safe for arduino? if not, can I use resistor to drop extra voltage?
szacpp:
Is 5.6v or 5.4v safe for Arduino?
Just barely. See page 299 of the datasheet.
szacpp:
if not, can I use resistor to drop extra voltage?
No. See my answer above.
Why can't you just use a regular 'power wart' type of supply for your usb connector? 5.6v is pushing it.
ok 5.5v is maximum 
...could you tell me the number of diode to drop extra voltage?
ChrisTenone I made power supply by the below schematic it provides 4.9v so i changed regulator from 5 to 6 now I'm receiving 5.6v I used 6.8 resistor to drop voltage it provides 5.04v in arduino output pins but I test usb connector from multi meter it shows 5.6v.
ok 5.5v is maximum
You can use a small signal diode to drop the voltage by ~.7v.
The forward drop across any silicon diode is about 0.7V under "normal" conditions. With less current you can get less drop, but 0.5 - 0.7V is a reasonable approximation.
[u]1N400x[/u] diodes are common & cheap and I usually keep several on hand. They are good up to 1 Amp. The last digit indicates the voltage and I usually buy the 1N1004 (400V) or higher because they'll handle any voltage I'm likely to be using and the price doesn't vary much no matter which variation you buy.
I used 6.8 resistor to drop voltage it provides 5.04v in arduino output pins but I test usb connector from multi meter it shows 5.6v.
That's because with nothing connected, no current flows. With only the multimeter connected, very-very little current flows.
The voltage dropped across a resistor is proportional to the current ([u]Ohm's Law[/u]). i.e. If you double current, the voltage dropped across the resistor doubles. With no current, there is no voltage drop.
As a general rule, a resistor is a BAD way of reducing voltage from a power supply. A [u]voltage divider[/u] (2 resistors) can be used to get a lower voltage in certain low-current applications, but not for a power supply.
Hi,
Remove D1 and replace it with a shorting link.
4.9V would have been within spec for 5V USB with the original 5V regulator.
The Original Circuit
The original circuit that you show, has a 0.7V drop across the B-E jucntion of Q1, so 5V from the regulator would give
5 - 0.7 = 4.3V
To compensate for this D1 adds 0.7V to the output of the 5V regulator output.
However not all junctions are the same and after the compensation there is a 0.1V error, and thats fine.
The Modified Circuit
In this case you have placed a 6V regulator instead of the 5V, with D1, so the output should have gone up by 1V to 5.9V, however not all junctions are the same, especially under load so you got 5.6V.
Again your original circuit would have been okay.
Tom.... 
thanks all specially Paul__B, DVDdoug and zoomkat who help me.
I used 1N5819 diode it provides 5.1v
again thanks