Electronic – Solenoid capacitor circuit for paintball gun

The cap has little to do with battery life. Ultimately the current thru the solenoid comes from the battery.

A cap can momentarily supply a higher current, which is then backfilled a little later and more spread out in time from the battery. That can be useful for loads that draw currents in large spikes but with a low average. In that case, the cap can let the battery see more the average rather than the spikes. This can be good for a battery.

However, that is not the case here. A solenoid is a inductor with significant series resistance from the circuit's point of view. When switched on, the current thru the solenoid will be a exponential decay towards the steady state value. It won't ever exceed the steady state draw that the battery has to be able to handle anyway. In this case, there is little point in a reservoir cap to power a solenoid.

The only reason I can think of one being useful for solenoid power is if the connection back to the power supply (the battery in this case) had such high resistance that there is significant drop on the solenoid voltage in steady state. This might be the case, for example, if the solenoid and switch is at the end of a long cable. In that case, the cap provides a higher voltage for a short time when the solenoid is switched on. This is likely when it needs more force. Depending on the external mechanics, it may be good to have a initial higher pull, which can then be relaxed a bit once the solenoid has finished traveling. This is common in relays, for example, where the two are often referred to as the turn-on current and the holding current.

Adding the capacitor smooths the rectified AC voltage and not having one probably accounts for you measuring different voltages when you applied different loads. Pure half wave, as a waveform looks like this: -

If you add a capacitor the red waveform is transformed into the blue waveform for normal loads. This is the first benefit you get - the dc voltage remains largely at the peak value of the AC waveform and is therefore more predictable when you measure it with your meter.

When you connected DC motors this has a similar effect in that between half wave pulses, the back emf from the motor "free-running" holds the dc voltage and prevents it falling to zero (like in the red waveform).

The voltage you measured was 7.1V off-load and this largely corresponds to what a half-wave rectifier produces. Mathematically it is \$ \frac{V_{ACpeak}}{\Pi}\$ = \$ \frac{16.3\times\sqrt{2}}{\Pi}\$ = 7.34V.

A better option is to use a full bridge rectifier as this fills in the gaps on the negative cycle of the AC waveform: -

This has a better average dc voltage without a capacitor of twice what the half wave circuit gives i.e. about 14.7V. Also the smoothing capacitor can be smaller (when using one) because it gets charged-up to peak voltage twice as often.

There are a whole load of other things I could go into but this document from wiki says most of them and covers the basics above.