Provided is the DC power supply block to the micro-controller development board from one of the leading MCU manufacturers.
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.
Best Answer
To understand this you need to read the IV curve for the Schottky diodes. The DFLS130L series datasheet Fig. 1 shows this.
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).
No. Both will share the current.
Yes.
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.