dhenry:
divide it by the minimum hfe(gain) of the transistor
hFE is a concept for linear amplification and has very little meaning for switching applications. Typically, you design a circuit assuming the switcher is driven to saturation: Ic / Ib < 10.
IC/IB < 10 is a just rule of thumb that should work for any transistor, but may not always be practical or the right thing to do. It's not a bad rule for most cases, but you can certainly do a lot better if you take the specifications of the transistor being used into account.
While hFE is a variable quality and is only valid during linear operation, it can be used to help find a more reasonable saturation bias to get the most out of your transistor. A transistor is in saturation when both the BE and BC junctions are forward biased. This will occur when:
VBE>VCE, which beings to occur when IBE x hFEmin > IC. Where the value for hFEmin is minimum for the expected IC.
In other words, saturation occurs when an increase in base current will no longer significantly increase collector current.
To insure adequate saturation we include a factor of 1.5 so that all we need is:
IBsat > (1.5 x IC) / hFEmin
If we use the transistor quoted above, the ZTX851, and we need to drive a 1000ma motor, we can get the hFEmin from the spec sheet. Then we have:
IC = 1000ma
hFEmin=100 (between 10ma and 2A)
Therefore IBsat > (1.5 x 1000) / 100
Or, a base current of just over 15ma, which is a whole lot better for MCU use than the 100ma as prescribed by the rule of thumb. In many cases you can get away with even less current, but this would involve testing each transistor for actual hFE at the required collector current.
BTW, this is the method and assumptions by which I arrived at the 240 ohm base resistor in my second schematic which is a conservative selection. 270 ohms would still get the job done.