I bought a Tecknet Powerbank 9000mAh in order to power my arduino Leonardo that I will use to record values on a scooter.
My problem is that a "powerbank" is mainly designed to charge a cellular phone and will detect when the phone is disconnected. The arduino draws very little current, so the powerbank auto shut off 30 seconds after powering.
I found this article on internet :
The guy who wrote the article have a R100 (0.1 ohm) resistor (current sensor) and replaced it with a 10 Ohms.
In mine, I found a big R050 resistor (0,05 ohms ?) near the 1A output plug. I replaced it with 6 ohms, 10, 20 and 30 ohms resistors, but nothing works, it still power off.
Here are two pictures of the powerbank. The resistor is near the usb port, just near the pen/sensor.
Does anybody have an idea for solving this problem ?
I was thinking about that, but those 3 batteries are 1.5 V each. So, I don't have enough Volts to put on the arduino ? I think there is something on the circuit that convert the voltage to an higher value..
Do not step up to 7 to 9 volts and apply to the dc power jack, that will waste battery energy. You want a +5 volt output that goes into the USB input for maximum efficiency.
Old post, but maybe helps someone else.
Probably your point wrong resistor.
The correct one must be the R100 in the first picture next to the battery B- connection.
If the power bank switch off by low load you must change this resistor with a bigger one.
As understand this resistor is in series with the load and the processor translate the drop of voltage in this resistor to load current.
I read in some posts to change this resistor with 10Ω resistor. I believe this can work with very low loads (big load resistance) but if by mistake use your power bank to charge a phone or other big load the resistor will blow off.
The best approach is to find the lower resistor keeps the power bank alive.
Remember smd R100= 0.1Ω.
I try to power up an andruino mini with xiomi 10000 mA power bank and 1Ω resistor keeps the power bank alive.
