You really asked two questions:
- Can a stepper motor be used as a generator?
- Can a charged capacitor activate a solenoid?
Yes and yes.
Stepper motors almost always have permanent magnets, so just spinning the shaft will cause each winding to produce a voltage. Each winding should produce the same AC waveform, but at a different phase. Each would have to be rectified separately onto the DC bus. The stepper motor must also be turned at some minimum speed to get enough AC voltage out to overcome the diode forward drops. Lower than that speed, there will be no DC output. The unloaded AC output voltage will be proportional to speed, so at some level it should be possible to get a reasonable DC voltage. The details depend on the stepper motor, and can vary greatly. Note that while stepper motors are usually driven rather slowly in normal operation, there is no such restriction when running it backwards as a generator. You should therfore be able to get substantial voltage from such a stepper. The easiest way to get some idea what you can get out is to watch the voltage of one winding on a scope while twirling the shaft.
Obviously, the resulting DC voltage can be used to charge a cap.
If a capacitor holds enough energy at the right voltage, it can activate a solenoid for a short time. You need the solenoid particulars to compute the capacitance required to activate it. Find out how much current the solenoid draws at what voltage and for how long this needs to be held to activate it.
If you are asking about a trigger circuit to turn on the solenoid when the cap gets to a certain voltage, then that is possible too. The simplest, but not most efficient, way is to use the capacitor voltage to fire a SCR, which in turn discharges the cap onto the solenoid. You have to adjust the voltage divider from the cap to the SCR gate so that it fires as the desired cap voltage. This requires only two resistors and the SCR.
I would probably use a tiny micro like the PIC 10F204. It includes a fixed voltage reference, comparator, and internal oscillator. These are all the elements you need to determine if the capacitor is at high enough voltage, and then to time the turning on of a low side FET switch to discharge the cap onto the solenoid.
I think that the hold time for the MLCC can be calculated numerically as in the following example. The total charge in a linear capacitor Q is C times V. But MLCC is not a linear capacitor and therefore Q=f(V) (some function that we will assume known now).
At time 0, let be V=5V. At this voltage Q0=f(5)=240 uC.
After some unknown small time step, the voltage dropped to 4.9 V. The charge in the capacitor is now Q1=f(4.9)=237.65 uC. (for example).
Assuming a constant current sink I of 10 mA and remembering that I·(delta time)=delta Q. We can calculate delta time=(240-237.65 uC)/(10 mA)=0.235 ms. The first time step took 0.235 ms.
After the following time step, the voltage dropped to 4.8 V. The new charge will be Q2=f(4.8)=235.2 uC. This time step is then (237.65-235.2)/10 mA=0.245 ms.
If this is continued until the voltage arrived to the minimum allowable voltage for your circuit, you only need to add all the time steps to get the hold time.
I chose voltage steps of 0.1 V, but values smaller or bigger can be chosen to get more or less accuracy in the final result. The problem remains to find function f(V).
The capacitance values from the "Capacitance vs DC Bias" graph in the datasheet gives the the relationship between Delta_Q and Delta_V at every DC bias voltage; i.e. it gives the capacitance seen by a small signal.
I think that a good approximation of f(V) could be obtained doing Integral(from 0 to V, of C(V')·dV'). Where C(V') is read from the "Capacitance vs DC Bias" graph.
Finally there is a FAQ from Murata http://www.murata.com/en-global/support/faqs/products/capacitor/mlcc/char/… where the physics behind the capacitance change are explained:
Without a DC voltage, spontaneous polarization can happen freely. However, when a DC voltage is externally applied, spontaneous polarization is tied to the direction of the electric field in the dielectric, and independent reversal of spontaneous polarization is inhibited. As a result, the capacitance becomes lower than before applying the bias.
This explanation would also apply to decreasing DC voltages. If DC voltage slowly decreases (during capacitor discharge) the polarization won't be tied to a particular direction and then the capacitance will increase.
The calculation of the hold-up time, using this method, can be done rather easily with Excel. I attach a worksheet with real datasheet data for a given 47 uF MLCC capacitor and the necessary calculations:
Hold-up time comparison betwween given 47 uF MLCC capacitor and a 40 uF linear one
Best Answer
I would not suggest deliberately directly shorting anything much bigger than 1000uF/16V ~= 0.1J (1/50 or so of your example), though I sometimes do it when I'm in a hurry and "pretty sure" 99.5% it's discharged through circuitry or whatever but would rather risk taking a nick from a screwdriver tip than my finger or meter. You could damage the shorting element or the capacitor doing that if it's fully charged. I have some fat 5W and 10W wirewound cement resistors around to do it properly, when necessary. If you misjudge the surge energy badly compared to what the resistor is capable of it's possible to have them explode or silently open up leaving the capacitor with a charge on it.
Very large capacitors can hold a lot of energy and demand serious precautions. Sometimes just the dielectric absorption can result in a big enough charge to give a significant jolt. That happens when you properly discharge a large capacitor and then remove the resistor- the capacitor appears to partially re-charge itself due to the way the dielectric behaves.