In this answer poster calculates a safety factor of 10 for the minimum base current in a transistor used as a switch. Seems like a lot to me, and for higher collector currents this may mean that a microcontroller won't be able to drive it anymore. Is a factor 2-3 not enough?
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Any fixed factor like that is mostly superstition. The point is to make sure the transistor saturates if you want low C-E voltage, which you usually do in a switching application. Generally you know or can specify the highest collector current the transistor must support and you know the minimum current gain from the datasheet. Dividing those two yields the minimum base current you must supply. You have to do some calculation to determine the base current you need, so you might as well do the right one. Using bad numbers doesn't make it easier, just less useful. It's not like doing this divide is any significant work or burden.
Once you know the minimum base current, you find the voltage accross the base resistor, and divide the two to get the maximum base resistance. This is the place to apply some judgement now that you have a meaningful value. Good engineers know that stuff happens and you don't want things to operate right on the edge of correctness. Once you have the maximum base resistor you can make it a bit or a lot lower depending on other issues in the design. Almost always I would go down at least a couple of standard values from the maximum, but there are times when every uA matters and you want to think about it carefully. A factor of 10 is gross overkill in most situations unless there are other constraints you haven't mentioned.
There is one possible reason you might want to saturate a bipolar transistor by a wide margin. The C-E voltage flattens out once the transistor saturates, but it still goes down a little with increased base current. If extra low C-E voltage is more important than higher base current, then it can be worth overdriving the transistor by 10x. Generally if on state voltage is so important you'd use a FET though.
One case I can imagine, admittedly a bit contrived, is you have enough current to switch so that a little lower C-E voltage matters for dissipation or voltage loss in the system and you are driving the switch from a low voltage circuit that has a good amount of current available. FETs don't work well with gate voltages as low as a B-E junction drop, so driving a FET in this situation would require extra parts and higher voltage. A overdriven bipolar transistor could be a good answer in this case.
When the factor of 10 results in a current that is perfectly acceptable I would stick to that, simply because it takes more effort to be more precise. Effort is best spent on those details in a design where it matters.
If you want to calculate the lowest base current that is acceptable, you will need to know the maximum CE current and the CE voltage that is acceptable to you, and consult the datasheet to find the corresponding worst-case minimum base current. No margin factor involved, but more work.
I'm the one who used that "10" factor.
In any case, I would in fact go for an NMOS. I gave him a BJT answer because he mentioned "to connect the base and the PWM output pin" in his question. And, for a BJT, I used a factor of "10", because:
I would probably work with lower margins (down to 1.5 or 2), but he will be safer with that one (if the MCU can provide it).