Driving a solenoid with low voltage source
I thought I would work out what is being said (started to being said) in the comments to your question, for the heck of it:

Basically a solenoid needs a magnetic field strong enough to move, so you need a high enough current through it's X windings. To get that current, you usually apply a voltage that induces that current in the given coil.

In the next bit of theorising I will use square-wave signal points to indicate uC outputs.

If you want that voltage without a huge booster built into your system and you can "stand" for the cap to be charging a while, it can be done very small:







simulate this circuit &ndash; Schematic created using CircuitLab

The finding of a diode of type "Good One" I leave out of this discussion.

In this, if your solenoid switching is irregular, you might want to make a protection, if you design this too well, eventually it could charge up beyond a break-down voltage somewhere.

You can add another well-configured transistor, maybe two, or a small sot23-5 op-amp configuration and tie it to another uC pin to indicate it should go from regulation mode to burst mode (&lt;0.05 duty cycle).

I have set C1 = 1mF as an absolute limit for myself while designing, because at 16V that is already not going to be very tiny at all.

To find L1, you need to consider uC_pin1's maximum on-time. In this I will assume that it is 100kHz and 50% duty at most, as drawn.

Let's say you have a 2N7002 that handles up to 150mA with the 3.3V (probably easy enough to find, because averaged over time it'll still only dissipate a maximum of about 130mW then)

To find L1, you can calculate the actual curve, that the best to do, but I am not going to, because if you know about those integrals, you already know how to do it and if you don't know about them, this is a much too small box to go into how those work. In stead, I am going to blatantly (and wrongly) assume the voltage is constant at about 2/3 of the actual input voltage. Then it becomes simply: I(max) = ( (2/3) * V1 * t ) / L where V1 is the 3.3V input and t is the time in seconds M1 is turned on.

Remember: You need to calculate I(max) at the maximum input voltage. Since my assumption is not very accurate, I would take 4.5V for a LiIon battery, just to be on the safe side.

The equation above can be re-written as:

L = ( (2/3) * V1 * t) / I ) which is: L = ( (2/3) * 3.3V * 10us ) / 150mA =~ 146uH

Of course, I made a risky assumption on, let's say gut-feeling, not knowing the ESR of the inductor you will choose, so I would say, for your first experiment round up very well: Start with 220uH. Unless you use a beefier MOST.

220uH with 150mA saturation may not be very large, but it itn't tiny either, so you might want to see if you can "up" it to 1MHz, or if you can allow a lower on-time and slower charging. That's down to your requirements.

Now, from fiddly complex maths and stuff I could get a very accurate representation of that current and of how much the voltage on the cap will increase each cycle, but in the real world ESRs and such will bug you anyway, so I'm just going to estimate the second part too. 

Again, if you know the equations, you are welcome to them, but I am a bit pressed for time and the most likely readers will not follow them easily through this medium. (Not saying anyone is stupid, just that there's digitally better equipped media or people for that ;-) I'd be spending a lot of time drawing and photographing where I am now )

If M2 is turned off and leakage is small, and the ESRs are average, I would expect the capacitor to rise with 4V in about 550 cycles of M1 at 100kHz at the maximum on-time of 50% with a decent cap of 1mF. But, as I said, the maths I used is shoddy and partly based on gut-feelings and they do sometimes turn out to be off by a factor of two to five. But it's a starting estimate for an experiment.

If that is too slow, you can decrease C1, but the required minimum size of C1 is dictated by the amount of energy the solenoid needs (which I do not know at this time) and all the ESRs in the C1-Solenoid-M2 path. So you could experiment with that value and when it doesn't switch on from the cap charged at 8V or 10V any more, double that value for safety.



An alternative you could try is without M2 and without the Capacitor and a small diode substitution:







simulate this circuit

I didn't look up the IRF530, but I believe it's a bit over-kill. The 9V zener can be various voltages, it depends a bit on the Solenoid to L1 ratio and whether this actually works depends on a few more details, but it is a fun experiment that might work or fail miserably.

In the place of the Zener you can also put other components with an even sharper break-down, such as diac-types, several kinds of TVS or even a gas-discharge tube (mind the MOST breakdown though).

If you shortly pulse M1, long enough to have L1 build up a decent amount of current (which depends on the value of L1 you end up needing, which depends on the Solenoid) it will build up that current, then if you release it, the current can't go anywhere until the zener breaks down. This creates a special case, where there COULD be enough "motivation" in the resulting signal on the solenoid to convince it to move. For one-off some experimentation required, for production some parametric testing with a broader range of values.

Also be aware that a zener for currents that are high will get hot, in which case a better device with a more awesome curve, as mentioned above, would be a better choice all around.