Go Down

Topic: Supply Due with 5V (Read 424 times) previous topic - next topic



I am currently thinking about powering the Due using a 5V power supply as it will be in the box together with the Due anyway and I would like to avoid putting a second power supply in there. Main idea is to keep all functionality like programming through USB without connecting or with connecting the 5V power supply.

Just connect the 5V Power supply to one of the 5V pins. Problem would be if I connect one of the USBs in if would crosslink the Power supplys which I think would be bad:

Connect my 5V Power Supply to a 5V Pin and the VIn pin. I would need to remove IC2 as the power would come directly through a 5V pin and I only use the VIn pin to cut of the power supply from the USBs. But as VIn an 5V are connected connecting a USB to power up the Due would trigger it to shut the power down. So not a good plan.

Connect my 5V power supply to a 5V pin and remove transistors T1 and T2. This should work but I would loose the possibility to power the Due from the USB ports in order to program it. Best plan yet, but not perfect.

Like plan 2 but use a diode in between 5V pin and my power supply (adjusting it in a way that I get 5V at the power pin). Connect the VIn pin directly with the 5V power supply. I am not sure that I would need to remove IC2 here (would need to check the data sheets) but I would waste a lot of the power on the diode (at least 0.3V at least 1A but its a 3A power supply so I should expect at least 1W power dissipation at the diode) which would heat the diode a lot. I might be able to use both, my 5V power supply and the power from the USB connector to program it.

Are there any other options. Currently I think I am going for Plan 3 as i fear the diode from Plan 4 might unsolder itself. Has anybody a briliant idea how to overcome this problem?

Thanks a lot in advance


Apr 27, 2015, 12:18 am Last Edit: Apr 27, 2015, 12:58 am by promacjoe
what problem. You really don't have a problem. The Arduino DUE has built-in power protection. it will accept a 5 V power supply to any of the 5 V pins and not interfere with the USB power. there is circuit protection built-in to keep any of your scenarios from happening. In fact some manufacturers for 7 inch TIF displays, Have a power supply built into the shield, To provide power to the screen and due board.

your only problem is making sure that you don't short out anything. You are bypassing all the fuse protection within the due.

use the Arduino dues power input jack as you prototype your boards. Only switch to the 5 V power supply when you're sure everything works correctly.

PS, transistors T1 and T2 Is used to isolate between the to power sources. Do not remove these transistors.


Thanks for the answer. I will need to have a look at the schematic again to understand this build in power protection. I might come back with a few questions if I got stuck.
As I am currently just trying out things I am not already using the 5V power supply but I will in a few weeks. But I like to be prepared regarding these things (as these are the things that might occupy you for a long time).


Ok, I tried to understand the circuit in between USB Power Supply and 5V on the DUE board. There are the transistors T1 and T2 in between (which are p-channel MOSFETs regarding to the data sheet), with source connected to +5V and drain connected to the USB power supply.
For T2 gate is connected to GND. If I apply 5V through any of the 5V pins the transistor gets conductive. As long as there is no voltage applied at the USB port (or it is lower then the 5V on the board) no current is flowing. But as soon as the voltage level at the USB port is higher then the 5V on the DUE port there will be a current from the USB port to the board. I don't see any curcuit that protect the differnt power sources from each other? Where am I wrong?
For T1 it is basically the same as for T2 except that the Gate voltage can switch between GND and +5V depeding on the inputs to the comparator IC1B where +5V would make T1 non conductive. But if VIn is not connected and as I understand it UOTGVBOF is normally low therefore we have the same conditions for T1 and T2.
I hope somebody can explains it to me.
Thanks in advance


okay, let's start with the T2 transistor. The gate is pulled to the Negative, turning off the transistor. This leaves the internal diode of T2 left to transmit power from the USB port. the Voltage drop will be about .3V depending on the type of diode used. it will allow approximately 1/2A to go through it without a problem. It will not allow the power to go in the other direction. The USB voltage will be 5V - .3V = 4.7V . The USB voltage should always be lower than the 5V onboard power supply. if the onboard power supply is not used, The USB power will be used. 4.7V is adequate to power the USB chip and a 3.3 V regulator. Total consumption should not exceed .5A .

Note: the MF-MSMF050-2 500mA "F1, F2" are electronic fuses. Limiting the current of the USB ports to .5A each.

the only difference between T2 and T1, is the UOTGVBOF circuit. When a UOTG cable is used,transistor T1 is turned on. allowing current to flow from the 5V onboard regulator to the USB port.

the onboard regulators are switching regulators . if 5V is detected by these regulators, They will turn off. They will not allow voltage to feedback through the regulator.

Hope this helps.



Can we connect 5v power directly at usb port?


Can we connect 5v power directly at usb port?
Yes, you can connect a "regulated" 5V power supply, directly to the USB port. however, it will be limited to 500 mA.

Go Up