user29562: I have a powerbank rated 5v. The battery is actually 3.6v, powering the built-in SMPS to give out 5v. I have stripped one end of the USB cable coming from the power bank, powering a motor driver and the Arduino in parallel. The positive end of the power bank, which is 5v, is going into the Arduino and the motor driver IC's Vcc.
I heard that 5v is not enough to drive the Arduino from the Vin pin. And there is only one power out from the powerbank, so directly connecting the Arduino through the USB would leave nothing to drive the motor driver IC.
A USB splitter would be the obvious solution, but is there anything else I can do? The power source has to be mobile. Can I power the Arduino through the Vin with those 9v batteries commonly used in electronics hobby projects?

user29692: I had this problem last year when I needed to power a prototype using a LiPo battery which supplies only 3,2 up to 3,7V (and up to 4,2V when charging). 
My way of solving the problem was to use a voltage boost module. Boost circuits are a kind of circuit where you the inertial properties of passive components to elevate (or attenuate) DC voltage levels. 
But you don't need to worry about the math behind use, there is modules like the MT3608 which will supply your circuit up to 28V starting from as little 2V, check the datasheet.

user10794: You have a powerbank which has a USB output, right? Why not just plug a cable from that into the Arduino USB port? That will connect the 5V directly into the 5V pin of the Arduino (probably via a switching MOSFET). Then you can connect to your motor driver from the 5V pin on the Arduino (and Gnd of course). There may be some limitations on how large a motor you can drive (based on the capacity of the power bank, and to an extent also the thickness of the traces on the Arduino board).
As others have commented, Vin is not intended for that purpose (ie. to supply 5V). You should supply 7V or more to Vin to force the switching MOSFET to disconnect the USB input. In your case you don't have 7V from the powerbank.

Here is how the Uno Rev3 power system works:

The input jack (circled in blue) goes through a diode (D1) to the Vin line.
Vin then goes through the 5V voltage regulator (U1) to provide +5V.
However on the top left you see 2 x 10K resistors (RN1A and RN1B) which make a voltage divider from Vin to Gnd. This is compared to the output of the 3.3V voltage regulator by U5A op-amp (circled in red). If higher, then T1 (circled in green) will turn off, disconnecting USBVCC from the +5V line. If lower, then USBVCC is connected to +5V. The nett effect is that if Vin in greater than 6.6V (because of the voltage divider) then the circuit chooses to use Vin, otherwises it uses USBVCC (USB power).
That's why they recommend to use 7V or more into Vin.
Note that you lose around 0.7V through the diode, which is why you need something like 7.3V at the input jack to allow for the loss of the 0.7V before the op-amp compares it to 6.6V.

29562: <p>I have a powerbank rated 5v. The battery is actually 3.6v, powering the built-in SMPS to give out 5v. I have stripped one end of the USB cable coming from the power bank, powering a motor driver and the Arduino in parallel. The positive end of the power bank, which is 5v, is going into the Arduino and the motor driver IC's Vcc.</p>

<p>I heard that 5v is not enough to drive the Arduino from the Vin pin. And there is only one power out from the powerbank, so directly connecting the Arduino through the USB would leave nothing to drive the motor driver IC.</p>

<p>A USB splitter would be the obvious solution, but is there anything else I can do? The power source has to be mobile. Can I power the Arduino through the Vin with those 9v batteries commonly used in electronics hobby projects?</p>
 29692: <p>I had this problem last year when I needed to power a prototype using a LiPo battery which supplies only 3,2 up to 3,7V (and up to 4,2V when charging). </p>

<p>My way of solving the problem was to use a voltage boost module. Boost circuits are a kind of circuit where you the inertial properties of passive components to elevate (or attenuate) DC voltage levels. </p>

<p>But you don't need to worry about the math behind use, there is modules like the MT3608 which will supply your circuit up to 28V starting from as little 2V, check the <a href="https://www.olimex.com/Products/Breadboarding/BB-PWR-3608/resources/MT3608.pdf" rel="nofollow noreferrer">datasheet</a>.</p>
 10794: <p>You have a powerbank which has a USB output, right? Why not just plug a cable from that into the Arduino USB port? That will connect the 5V directly into the 5V pin of the Arduino (probably via a switching MOSFET). Then you can connect to your motor driver from the 5V pin on the Arduino (and Gnd of course). There may be some limitations on how large a motor you can drive (based on the capacity of the power bank, and to an extent also the thickness of the traces on the Arduino board).</p>

<p>As others have commented, V<sub>in</sub> is not intended for that purpose (ie. to supply 5V). You should supply 7V or more to V<sub>in</sub> to force the switching MOSFET to disconnect the USB input. In your case you don't have 7V from the powerbank.</p>

<hr>

<p>Here is how the Uno Rev3 power system works:</p>

<p><a href="https://i.stack.imgur.com/ajmr5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ajmr5.png" alt="Uno power switching"></a></p>

<p>The input jack (circled in blue) goes through a diode (D1) to the V<sub>in</sub> line.</p>

<p>V<sub>in</sub> then goes through the 5V voltage regulator (U1) to provide +5V.</p>

<p>However on the top left you see 2 x 10K resistors (RN1A and RN1B) which make a voltage divider from V<sub>in</sub> to Gnd. This is compared to the output of the 3.3V voltage regulator by U5A op-amp (circled in red). If higher, then T1 (circled in green) will turn off, disconnecting USBVCC from the +5V line. If lower, then USBVCC is connected to +5V. The nett effect is that if V<sub>in</sub> in greater than 6.6V (because of the voltage divider) then the circuit chooses to use V<sub>in</sub>, otherwises it uses USBVCC (USB power).</p>

<p>That's why they recommend to use 7V or more into V<sub>in</sub>.</p>

<p>Note that you lose around 0.7V through the diode, which is why you need something like 7.3V at the input jack to allow for the loss of the 0.7V before the op-amp compares it to 6.6V.</p>
