Hi,

I am designing a relay which is based on MOSFET, this relay has 14 output.
The relay is being operated on 5V dc supply.The supply is being provided to the board by using a MOSFET, the attachment shows the power supply section and one of the output ciruity from the available 14 output.


TPC8124, Now I want to calculate the total available power that can be obtained through this relay board:

As I am given the input 5v supply to this transistor (TPC8124) which is further supplying this 5v to the rest of the board, the datasheet of this transistor says that Drain Current (DC) is -12 A, as my input supply voltage is 5v so I calculate that total output power that be get through this board is 60W.
is this a right estimation or I am missing something ??

The datasheet is available on the following link
http://www.toshiba.com/taec/components2/Datasheet_Sync/200912/DST_TPC8124-TDE_EN_11751.pdf

with regards,
Muhammad Faisal.

 

is this a right estimation or I am missing something ??

Heat. The datasheet says the maximum Rds could be 10mΩ. At 12A current, that's 1.44W of heat that needs to be dissipated by the MOSFET. Without a heatsink or other measures, I think that would be very likely to fail.

Running any component up to its max rating risks imminent failure.

 

60W would be the power delivered to the load.

As wayneh noted, you shouldn't operate the device at it's Maximum Rating for best reliability. 75% of this or 9A maximum would be a good derated value.

At 9A the maximum dissipation when ON would be 810mW. To minimize the temperature rise you should mount each transistor on a one square-inch minimum PCB copper plane as shown in Note 2(a) on page 2 of the data sheet.

Edit: R51 should be a small value or zero ohms to insure the transistor is fully turned on.

What is the purpose of Q30 in the power supply section? It is wired so it is always ON.

 

What is the purpose of Q30 in the power supply section? It is wired so it is always ON.

Q30 is used to isolate the rest of the circuit in case the voltage supply is connected incorrectly. It will certainly heat up if all the current (14A) needs to go through it.