So i have to design a board around a dialog SOC (DA14681). The dialog part supports li-ion batteries, it has a charger and fuel gauge and USB detection (to power device in case both usb and dc voltage/li-ion connected at the same time). As part of the requirements, my board should be compatible with an arduino shield (so i need headers that are identical to arduino uno).

This then means i need to add the Vin pin present on an arduino. I will also need to generate the 3v3 and 5V rails. The dialog part generates a 3v3 rail but this is limited to 100mA so will have to generate the rails externally to the chip.

Could i boost the li-ion battery to 9V and treat it as the DC Jack voltage (connected to Vin through a diode?). Leave the rest of the circuitry as would be present on an arduino? Would this circuit work? My concern is that if li-ion is turned on, it will disconnect the USB voltage to the rails (the pfet) but the VBUS voltage will still be present. So the dialog part will be turned on by VBUS (USB) whilst the rails will be turned on by the li-ion.

Would i need protection circuitry for the li-ion (considering there is protection in the dialog part)?

Is there an easier way of doing this?

Sorry for the long question, i will be so grateful for the help or push in the right direction.

Having three voltage sources (li-ion, usb and Vin) complicates things here.enter image description here

  • \$\begingroup\$ Why are you powering Vin? A normal arduino does not power Vin, it takes power from Vin (if theres a voltage source connected to it). \$\endgroup\$
    – BeB00
    Jul 5 '18 at 8:33
  • \$\begingroup\$ On a normal arduino, you can power the board up using Vin can't you? It can act as an input and an output(vout) i thought? \$\endgroup\$
    – Hasman404
    Jul 5 '18 at 8:35
  • 1
    \$\begingroup\$ It can act as an input, I haven't seen it act as an output. Why would you boost your battery to 9V? You're wasting loads of energy. Your board should be able to receive power from Vin, but shouldn't be putting any power out into vin. Also, Vin is certainly not guaranteed to be 9V. \$\endgroup\$
    – BeB00
    Jul 5 '18 at 8:39
  • \$\begingroup\$ I was trying to replicate an arduino circuit. The boost in my case helps me mimic the dc jack voltage which is around 9V. Doesn't the DC jack put power into Vin? (When you probe the Vin with DC jack powered on you can see the voltage minus a diode drop?) If i don't boost the battery, how else would i plug it into Vin? (Considering arduino shields would expect Vin to be 7-14V?) \$\endgroup\$
    – Hasman404
    Jul 5 '18 at 8:54
  • 1
    \$\begingroup\$ Most shields don't use Vin, and those that do know it's not guaranteed to be 9V. As far as I'm aware, the only shields that do use it are high power ones like motor drivers. If you try to drive them from your boosted li-ion battery, youre probably going to run out of power very quickly, and might break your battery. You'll certainly need a very powerful regulator. Why not just boost to 6V through a diode into an LDO, and put Vin through another diode into the same 5V LDO? Obviously this will not be a very power efficient design, but it will work. \$\endgroup\$
    – BeB00
    Jul 5 '18 at 9:03

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