How to short a dc Motor with arduino

You say you want to control a motor that requires from 180 to 250 mA at no load from a 3.3 V digital output of questionable current sourcing capability.

First, the no load current is pretty useless. The important parameter is what the maximum load will ever be. That is what you need to design to. If the circuit can handle the maximum current the motor will draw, it will certainly be able to handle a lower amount. The maximum current, also called the "stall current" is probably 500 mA at least. Let's figure 1 A max.

Here is a circuit that should work nicely:

This FET has a guaranteed maximum on resistance of 80 mΩ at 2.5 V gate drive. At 1 A it will only drop 80 mV and dissipate 80 mW, which is fine for a SOT-23 package. You will notice it getting a little warm, but it will be well within its limits.

A FET is a better choice here than a bipolar like your 2N3904 for several reasons. First, it will drop less voltage and therefore take less from the motor. At such a low drive voltage, even a few 100 mV can become significant. Second, it will drop less voltage and therefore dissipate less power and not heat up as much. Figure a bipolar would drop at least 200 mV although probably more at 1 A. Even just 200 mV at 1 A is 200 mW, which would make a SOT-23 pretty toasty. Third, it won't load the digital output, which you say has low current sourcing capability. The FET gate will only look like a capacitor to the digital output. That will slow down its edges, but they will still be instantaneous on the scale of anything the motor will react to.

A 2N3904 is a pretty crappy transistor for more than just signals anyway. I ususally use 2N4401/4403 for jellybean bipolars. They have more current capability and are generally more robust, but still have good gain and are also cheap and widely available. But I wouldn't use even the 2N4001 in this case. The FET is a better choice.

You pretty much have the right idea about how the circuit works. When you pull Q10 down, you'll end up with Vcc across Vbe_12+Vbe_10 + resistors. Vbe's are generally around 0.7 volts, so you get Vcc-1.4V = Vresistors. Knowing the voltage over the resistors, you can calculate the current going through that path from VCC through the base of Q10. Once you know the current going through Q10_base with a known Hfe, you'll know how much current flows into the base of Q14. That current into Q14 will turn it on and its Hfe should allow plenty of current for the motor to flow through it from collector to emitter.

Raising the input high on Q10 causes the voltage drop on Vbe+Vbe+resistors to now be 0. Since there's no voltage drop, the PNP's are pinched off and no current flows through that path. Since no current flows through that path, no current is allowed to flow from the emitter to the collector of the PNP Q10. No current through that path prevents any current flowing into the base of Q14. Without any current flowing into the base of Q14, it prevents any current flowing from the motor to ground from that path.

Your third question falls somewhat in gray space. If you have matched transistors, then you can expect each complement to allow the same amount of current, you can expect the rise and fall times to be approximately equal and you can expect the ON-resistance of the transistor to be about the same. If you have a mismatch in this pair, you need to ensure that you're not compromising on any of those points. If one turns on faster than the other has time to shut off, you'll end up with longer times when both Q11 and Q14 are on at the same time. This causes a direct short often resulting in burned chips. If one transistor can't handle the amount of current that the other can supply, it will either prevent your motor from running optimally, or could cause your under-sized transistor to fail early. Basically, if you spec the non-complementary transistor correctly (ideally, overspec it a bit), then you shouldn't run into any issues.