I will give a very specific example: Hardware:
- Iphone X device with 2716 mAH capacity battery, 3.81V
- Iphone "dongle" lightning to USB 3.0 port, bus-powered
- USB hub 3.0, bus-powered, 3V, 200 mA, four-ports
- USB flash drives 3.0, 900 mA and 5V +- 10% operating voltage bus-powered
Circuit setting and analysis:
Connect Iphone to dongle, to USB hub and plug in USB flash drives.For me, the circuit is in series for component 3 (Hub) and components 4 (the drives). The drives are in parallel inside their component.
Hence the hub receives 3.81/2 = 1.905 V and the drives 1.905 V. The flash drives receive x/n mA where n is the number of drives plugged in and x the current in amps release by the IPhone battery. Here I have a missing variable: maximum discharge current of the iPhone X battery, which may of may not be limited by software.
I have battery capacity at 2716mAH, and current requirements of 4*900mA = 3,600 mA so I can last up to 2716seconds on a full iPhone battery theoretically (if the battery had a 5V voltage).
Questions
- Is the above analysis correct?
- When an electrical device says 5V operating voltage, 900mA, does it have to be these precise conditions or can power the device with equivalent wattage i.e 1.905V and 2,362 mA?
- Assuming maximum discharge current is 3,600 mA, how many USB flash drives as above can I bus power?
Edits
- For 1.,it was commented that the circuit is in parallel, the USB hub is not in series.
- For 2., a link was provided by brhans Choosing power supply, how to get the voltage and current ratings? to a similar question which advises to keep voltage within recommended tolerance band and current amperage too. Hence, as user towe commented, there will be a problem connecting 5V appliances to 3.8 V battery. Maybe some USB hubs can do this (ex:)
- User Chris Patton has mentioned that connecting multiple flash drives to a phone is a bad idea. Why is that?