How to power up rfduino and fona from the same power supply

arduinopower supply

I am building a simple sensor device that's using Adafruit Fona to be connected to the Internet.
I want to power up RFduino from the Fona USB 5v power supply. There is a pin on Fona that supplies 5 voltage as far as I understand and I am using voltage regulator to step it down to 3.3v
There is a schematic:
RFDuino with Fona

The problem is:
When Fona starts sending data over gprs RFduino gets rebooted. I guess there is a voltage drop or current spike or something. I've tried to put 100uF capacitors around the regulator, but it didn't really help. Although sometimes it works fine… but in most of the cases it does not.

How to fix the circuit that it works and stable?

Thanks

Best Answer

The FONA does not have a 5v output - there is no boost converter to provide one.

I see two possibilities which would be worth contemplating.

  1. The SIM800 module has an internal regulator providing a 2.8v output on VDD_EXT (Adafruit seems to simply label the pin 2.8v), which is within the allowable range of the RFduino. The data sheet specifies that this should be good for up to 50mA, which is more than the claimed transmit power draw of the RFduino. In theory it seems like this should be a workable power source, but you will have to check out all of the of the details. You will also have to see if the 2.8v supply is active at the times you want it to be.

  2. Since the RFduino claims to work down to 1.9v, you could presumably use a small low dropout regulator configured for a bit over 2v to derive a power supply directly from the lithium battery which the FONA uses. Select your regulator for low quiescent current and an enable if needed, but mostly make sure that the combination of your configured voltage and the regulator's dropout offset add up to something less than the minimum cell voltage you want to work at.

You could also consider a small switching regulator for efficiency, or even a boost converter if you feel you need a higher voltage, but it doesn't seem clear that this would be required.