Is this a spike or is it no different than a short, within the specs of transistor?
You got ahead of me with already talking transistor, I was still at understanding what happens with the button scenario.
But ok:
16VDC source--normally open momentary button--1uf cap--GND
Pushing button completes circuit, cap opens it again soon after but is now polarized. Now what? What goes where?
The cap charges slowly while the switch is open. This simulates a longer press of the button, until the upper voltage threshold of the input is reached. Once loaded to its full capacity, the energy waits for a closed switch or transistor, searching for the least resistance and thus maximum current. Even a small (10R) resistor can limit the current to tolerable values, while it doesn't prevent the cap from fully discharging. Note that 16V on 10R result in a peak current of 1.6A, or 16A on 1R, and even more on less resistance.
You can find out yourself, what happens if you charge a big cap (>100µF) to say 16V, then connect its legs. Most probably you'll see a small spark and hear a bang. The same will happen inside a mechanical switch, but switches can be constructed to tolerate much more than that, like for switching motors at 230V AC. Transistors can suffer from multiple effects, depending on their construction. In a metal case tiny wires connect the pins to the crystal, and these wires can be burnt by a cap discharge. Next comes the thin metal film, distributing the current over the chip surface. Finally the current traverses the crystal, where it can cause further damage.
Yeah, I'm still trying to make sense of capacitor behavior in a DC circuit. From the pretty animations I've seen, no electrons pass the gap, they just build up and have voltage potential. The more it 'charges', the higher its resistance. Except this implies that if it's not charged, it somehow leaks electrons across the gap until it gets enough charge to not leak anything.
So if all the electrons are coming from GND but the switch is opening the circuit, there's still no voltage potential, right? And pressing the switch then has whatever electrons are on that side spring back from the cap because that's when charge in the cap increases an polarizes.
Cap charges slowly while switch is open. Does this mean taking a cap, and sticking only the negative lead to the negative end of a battery, will case the cap to charge?
The switch is connected to a pullup resistor, which provides the current to charge the cap.
A balloon is a physical model of a cap, which doesn't let pass air, but can contain air (charge, electrons). The pressure inside the balloon is equivalent to the cap voltage.
Ahhh the pullup resistor makes it make sense. Thanks, doc.