I will have to adjust the power supply for the Arduino Mega, since my voltage supply is 24 V. I'm thinking of using LM 317 voltage regulator. Any comment or advice very much appreciated!
LM317 is an adjustable regulator right? I'd use a 7805 for the sake of simplicity.
Does 7805 convert from 24V to 5 V?
Yes, but not for long without a good heatsink. Input voltage for 7805 should not exceed 35V DC. Difference beetween INPUT voltage and OUTPUT voltage becomes "HEAT" on a linear regulator like this.
Then it is probably better to use LM 317?
No, LM317 has similar issue since it is also linear. You won't be changing the scenario that much. Invest in heat sink.
If you plan on using an external power supply while the usb is connected, you should use Vin, not +5v. Read this thread
http://arduino.cc/forum/index.php/topic,82046.0.html
and especially this post from the Arduino crew.
The minimum voltage to prevent the failure is Vin >= 7 volts. Maybe use the LM317 set at 8 volts connected to Vin?
Yes on heatsink.
Since it will be surface mounted, is it enough to use PCB as a heat sink?
Dear SurferTim,
I will redesign the PCB of Arduino Mega and change the power supply part on it, since my external voltage supply is 24V and I have to convert it down to 5 V. So, basically I will make a new circuit or better to say replace the existing power supply part (with LM2734) on the PCB with a new one (maybe LM317 or something...). Therefor there is a lot of option now available and I would like to have the best one. You have some idea what would be the best to do?
Where do you plan on connecting the 5 volts? If you plan on having a usb cable connected to a computer while powered externally, you must use Vin, and a voltage at or above 7 volts. Otherwise, never connect the usb cable while powered externally.
Linear power supplies are pretty easy to calculate the heat produced.
P = I * E
If you use about 200ma for the current (I use an ethernet shield) and using the voltage drop to 8 volts
P = .2 * 16 = 3.2 watts
I advise you use a 7809 or LM317, either providing 9V output, and use the round power input plug on the Arduino. The arduino will bring that down to 5V, since it has its own voltage regulator.
This was, part of the heat will be dissipated by your regulator and part by the arduino regulator.
another point is that if your circuit outputs 5V, it cannot be used in the arduino round power plug.
You mean the round jack input?
borutg:
You mean the round jack input?
Yes. That is Vin.
Yes, I was thinking to replace this part with a new voltage regulator, since I will make a custom made PCB board for arduino. I don't want to have a separate PCB with voltage regulator and separate arduino board.
borutg:
Yes, I was thinking to replace this part with a new voltage regulator, since I will make a custom made PCB board for arduino. I don't want to have a separate PCB with voltage regulator and separate arduino board.
Since you are planning on a custom board, I would recommend the new voltage regulator circuit either being based on a switching regulator or a combination of a step-down converter and linear regulator (where the step-down circuitry drops the voltage level to just above the minimum for the linear regulator to function properly). Use whichever approach you find easiest to fit on the board while still being cost-effective.
From the spec sheet for Arduino Mega R3:
Input Voltage (recommended) 7-12V
Since the mega regulates whatever comes in from this down to 5V and this range acceptable, I would look for an unused 7-12V wall wart power supply that I could repurpose for the Mega. Everyone has at least 10 of these things, I must have 50. Look for one around the house.
The main reason they say MAX 12V... is because of the heatsink you DON'T see on the board. Supplying more than 12V will cause excessive heat on the part and the board.
Maybe convert 24vdc to your desired voltage using some of those adjustable LM2596 chained together? They are very cheap, you can get 5 of them for less than $9 (shipping included) 
I plan to use a pair of them to convert 12vdc to 9vdc for powering my Mega 2560 and then 9vdc to 5vdc for powering other things (to avoid "overloading" the arduino's 5V pin)
What do you think? You advice will also help me to decide if I should use this or not 
What is the maximum current needed for powering the Seeeduino Mega?
http://www.seeedstudio.com/wiki/index.php?title=Seeeduino_Mega#Key_Specification
Edit: Sorry you want to know the current... No idea, but I think there is no "maximum current needed", I think there is no problem if you give 5A and the seeduino Mega only require like 1A (suppose). That's what I've been told anyway.