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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

  1. Is the above analysis correct?
  2. 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?
  3. Assuming maximum discharge current is 3,600 mA, how many USB flash drives as above can I bus power?

Edits

  1. For 1.,it was commented that the circuit is in parallel, the USB hub is not in series.
  2. 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:)
  3. User Chris Patton has mentioned that connecting multiple flash drives to a phone is a bad idea. Why is that?
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    \$\begingroup\$ 900 mA seems very excessive for a simple USB flash drive. Around 0.2 W or 40 mA seem to be more common. \$\endgroup\$
    – towe
    Aug 10 '20 at 8:27
  • \$\begingroup\$ I have similar 2.0 drives with 200 mA. I just want to get the reasoning right. \$\endgroup\$
    – sovann
    Aug 10 '20 at 8:42
  • \$\begingroup\$ Since you'll likely be limited by the 3.7V -> 5V internal step up, this might be relevant: apple.stackexchange.com/questions/119707/… The battery itself could probably handle quite a bit more. \$\endgroup\$
    – towe
    Aug 10 '20 at 9:39
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    \$\begingroup\$ Nothing is in series. Even though the drives are connected through the hub, the +5V passes through the hub to the drives. Everything gets 5 volts. \$\endgroup\$
    – DoxyLover
    Aug 10 '20 at 9:54
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    \$\begingroup\$ Does this answer your question? Choosing power supply, how to get the voltage and current ratings? \$\endgroup\$
    – brhans
    Aug 10 '20 at 13:26
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Apple doesn't say outright how much power you can draw via a lightning to usb host cable or adapter but its about 200 mA before the phone alerts you. So unless you have an externally powered hub this isn't working at all.

The current ratings for your devices should be the maximum they draw. That may be the operating current or spikes. It may also be what they request from the usb host, as the usb drives should be asking for permission to draw a lot. When in sleep mode, like when a flash drive is not being read or written to, the draw will be lower. Reading is also less power hungry than writing. So its hard to say what your power draw is with specifics. If we assume full power, plus conversion from 3.x to 5V of 80% efficiency it would drain the battery quickly at over 1A draw for a single drive.

Since everything is in parallel power wise they all get 5V and share the available current. Since they vastly outpower the available current supply that can lead to over drawing and damaging the source aka the iPhone. Luckily there is power limiting inside. So nothing will work but you may still damage the iPhone without a powered hub.

Finally no you can't just provide the equivalent wattage to a device. They will need 5V 900mA (or so) to work. Giving it 2V 2.3A wouldn't work. Current is drawn not pushed.

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