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Provided is the DC power supply block to the micro-controller development board from one of the leading MCU manufacturers. MCU multiple power sources

I need help to understand this arrangement. I will use two sources V5-9V_VIN and V5V_SDA (USB). The document says that multiple DC power sources can be provided simultaneously because protection is provided ( schottky diodes). BTW notice that the same diodes are used, hence the Vforward is the same for all diodes and the voltage supplied could be the same too (threshold turning on/off is the same).

But what happens when I feed, say, USB 5V and VIN=5v simultaneously? One diode turns on allowing +5V to the regulator input and not allowing the other diode to turn on because there is not enough forward potential difference? Can they turn on at the same time? If only one diode can turn on at a time, then which one? I need to make sure that VIN will be the source whenever/if I connect VIN to the board, otherwise USB supplies the power.

Please clarify.

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  • \$\begingroup\$ It is virtually impossible for two voltages be absolutely, exactly the same in real world. So, whichever is higher, however little it is, will reverse bias all other diodes. In the extreme case two or more diodes with very close voltages will be slightly open, but so what? The voltages are very close anyway. \$\endgroup\$ – Maple Jun 11 '18 at 21:39
  • \$\begingroup\$ It's not clear how you connect P5-9_VIN. IF it is with a jack the same as J24, you could use the ground switch in it to disable the USB ground. Then when there is no plug in the jack the USB ground will be connected to the circuit's ground and the USB will deliver power. If there is a plug, the USB's ground will be disconnected from the circuit and all power comes from the power supply connected through the jack. \$\endgroup\$ – HarryH Jun 12 '18 at 12:52
  • \$\begingroup\$ @HarryH not USB ground but USB +5V I will disconnect using jumper pins, whenever VIN is connected (either DC_JACK or VIN). I will still use USB when VIN is powering the board and USB signals are referenced to the ground. \$\endgroup\$ – VladBlanshey Jun 12 '18 at 18:30
  • \$\begingroup\$ Ok, it was not clear from your description that the output of your circuit goes to the USB plug. In that case indeed you can not use a jack with built-in switch disconnecting the USB. So yes, in that case you have to switch the power manually. \$\endgroup\$ – HarryH Jun 14 '18 at 4:58
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To understand this you need to read the IV curve for the Schottky diodes. The DFLS130L series datasheet Fig. 1 shows this.

enter image description here

Fig 1. IF versus VF curves. 25°C curve highlighted by me.

Let's say we have one source and our circuit is looking for 0.5 A. We can see from the curve that at 0.5 A (1) we can expect a voltage drop across the diode of about 0.28 V (2).

Now let's switch on a second supply of exactly the same voltage. (You could feed the same supply into two inputs to test this.) It should be clear from the graph that the action point of the second diode will start at 0 A but run up the yellow curve because of the voltage drop across the first diode. As the second diode starts to pass more current the first diode passes less and so it will move down from point (1) and the pair of them will stabilise somewhere around 0.25 A (3) with a resultant voltage drop of about 0.24 V (4).

But what happens when I feed, say, USB 5V and VIN=5v simultaneously? One diode turns on allowing +5V to the regulator input and not allowing the other diode to turn on because there is not enough forward potential difference?

No. Both will share the current.

Can they turn on at the same time?

Yes.

I need to make sure that VIN will be the source whenever/if I connect VIN to the board, otherwise USB supplies the power.

It's not at all clear why you would need to ensure this. If it is the case then you would need to sense VIN on the anode of the relevant diode and use that to somehow disable the other inputs.

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  • \$\begingroup\$ Thank you very much - this is the absolute and highly professional answer. I need to ensure Vin is the source of current because the board may draw more than max 300mA allowed from USB port. Even if they share current simultaneously, I am not sure if a draw from USB port will stay under the limit. The simplest way probably to install a jumper pins pair to disconnect USB +5V line when Vin is connected. Again, thank you for being here. \$\endgroup\$ – VladBlanshey Jun 12 '18 at 1:22

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