never really played with a PUT before (actually never heard of em) but i was interested and read the datasheet.
It looks like the current through the PUT is dependent on the resistance between gate and ground, which explains why when the cap is feeding the LED it doesn't get really mad about the LED not having a current limiting resistor. In this case the Rg gate resistance is your R3. My guess is that when you moved R3 up to 96k your limiting the current so much that your LED isn't getting to full brightness.
Additionally the low limit of this current combined with a really big cap means your capacitor discharges much slower. Combine this with the very small R1, which charges the cap quickly, and i'm betting you are getting some oscillation, but its happening very, very fast.
Try a larger R1, smaller R3 and whatever sized R2 you need to keep the divider ratio the same. Ideally track down a smaller cap, it would make finding the resistor sizes needed easier.
That article is crap, and you should probably forget you ever saw it. Besides the fact that it's totally wrong, as you've discovered, it's full of half-truths that cultivate an incorrect understanding of how electrical phenomena actually work.
The plate on the capacitor that attaches to the negative terminal of the battery accepts electrons that the battery is producing.
wrong. Batteries don't produce electrons. In fact, nothing in your circuit produces electrons. As far as physicists have been able to demonstrate, charge is never created nor destroyed. A physicist can tell you how to make an electron by assembling more fundamental particles, but unless your circuit includes a particle accelerator or operates inside a star, there won't be any relevant electron creation or destruction in your circuit. Batteries pump electrons. They don't produce them.
In the first few paragraphs, that article uses the word charge in several senses. An experienced engineer can distinguish the senses by context, but the novice is more likely to confuse them. Go read Bill Beaty to immunize yourself against this misconception.
The bulb will get progressively dimmer and finally go out once the capacitor reaches its capacity.
Not exactly true. The bulb will go out once the voltage across the capacitor is equal to the battery voltage. When this happens, there can be no voltage across the bulb, and thus no current, thus no light. I don't know what they mean by capacity in this sense, but it seems to me that they are phrasing it this way to avoid explaining how capacitors actually work. You might define a capacitor's "capacity" any number of ways, but they haven't defined it at all. This kind of half-thinking isn't going to help you understand once you figure out that the explanation is incomplete.
A capacitor's storage potential, or capacitance, is measured in units called farads.
If you take "capacity" and "storage potential" as synonymous, as anyone would naturally do, this is wrong and self-contradictory. If this were true, then we could re-write the previous statement as
The bulb will get progressively dimmer and finally go out once the capacitor reaches its capacitance
...which is totally bogus. Ordinary capacitors don't change capacitance under normal operation conditions. Again, the problem here is they haven't fully defined the underlying concepts. If capacitance is a capacitor's storage potential, then what is it storing?.
A 1-farad capacitor can store one coulomb (coo-lomb) of charge at 1 volt.
Here, they try to define the stuff that the capacitor is storing, but it hardly makes sense. The problem is if you use words like "capacity", this implies there is some sort of "full" or "at capacity". But, there is no concept for an ideal capacitor. If you push 1C of charge through a 1F capacitor, the capacitor will have a voltage of 1V. For 2C, you get 2V, and 1000C gets you 1000V. You can push 1000C through a 0.5F capacitor also, but it will then be at 2000V. An ideal capacitor is never full, and you can push charge through it forever and it will never be "full". That's not a capacity, it's the ratio of charge to voltage, which is evident in the definition of farad (a farad is a coulomb per volt):
$$ F = \frac{C}{V} $$
Real capacitors, incidentally, have a maximum voltage, which if exceeded will damage or destroy the capacitor. So in this sense, a capacitor can be "full" and can have a "capacity". But, this is not how this article is using the word.
I could probably write an entire article on the problems with that article, but hopefully you get the idea. Please wipe your memory of what that article has said, and seek a more sound explanation.
Best Answer
This is not a good debounce circuit.
One problem is that (at least ideally) the switch and its connecting wires have a resistance of zero. This means that the capacitor will instantaneously discharge when the switch is closed. (In practical terms, also, this rapid discharge could even be bad for the switch contacts or the wiring, if there is a high enough voltage on the capacitor and it has a high enough capacity.)
A capacitive switch debounce should slowly charge the capacitor when the switch is in one state, and slowly discharge it when it is in another state. The RC constant doesn't have to be the same, but it should be something nonzero. The circuit has resistors that control the charging of the capacitor; it just needs a resistor in the switch loop to discharge it gracefully.
Another problem with this circuit is that the LED is only off if the circuit has been on for some time, such as if the circuit has existed since the beginning of time with that same voltage source. But what if, at time \$t = 0\$, the voltage source has been 0V and suddenly jumps to its voltage? At that time, the capacitor, which must have been empty, begins charging. While it is charging, current flows and the LED will light up briefly and then go dark. (Well, maybe not, because your source has only 1V, but that's another story).
In CircuitLab, you can distinguish these two situations in the "Time Domain" simulation. You can either "Skip Initial" or not. The solver can either pretend that the circuit has existed in the given state for all eternity until the time \$t = 0\$, and start solving it from there. Or it can solve it from the point of view that the circuit just came into existence at \$t = 0\$ and the voltage sources spring into life, the capacitors are empty, and so on.
One final consideration here is that the circuit only lights a LED, so switch bounce is basically moot, unless the LED is shining on some optical detector where the switch bounce turns into a glitch in the signal. If the LED's job is to just provide a pretty light, then your eye won't even be fast enough to see the switch bounce.
Here is a time domain simulation of the circuit (after changing V1 to 3V). What is plotted is the LED current. Important: the Skip Initial parameter is set to Yes, so we can see what happens when the capacitor is initially empty and the voltage source energizes to 3V. This is all with the switch in an open state.
As you can see, current surges through the LED and then dies down. If your intent was that the LED is strictly controlled by the operator via the push button, then your design does not implement your intent one hundred percent.
With regard to the comment below, suppose that the aim is to actually drive a microcontroller pin (everything running at 5V). Firstly, we can do that without any capacitance and handle the debouncing in software by sampling the pin at a reasonably low rate.
simulate this circuit – Schematic created using CircuitLab
When the switch is open the output is pulled to 0V by the pull-down resistor. When we close the switch, the voltage at the top of the resistor rises to 5V. This output can be regarded as a signal. We are interested in the low frequency component of the signal: relatively slow switch presses. We want to reject high frequencies, like switch bounce. To that aim, we can add a passive, one-pole RC low-pass filter:
simulate this circuit
Now when the switch closes, the voltage rises gradually as the capacitor is charged. You can see this in the time domain simulation:
When the switch is opened, the capacitor will discharge through R1 and R1, gradually dropping the voltage back to zero. The capacitor basically follows the voltage of R1, but with lag due to having to charge through R1, and discharge through R1 and R2. (Note that the discharge is twice as slow as the charge!)
The microprocessor input senses the voltage with high impedance, so we can ignore its loading effect and not even show it on the diagram. We cannot do this in the case of the LED because it requires current which our circuit must supply. That current flows through our resistors and develops voltages that we must account for: in other words, it has "loading effects".
This type of circuit works even better if we feed the output to a Schmidt trigger. A Schmidt trigger is a kind of buffer for digital signals which shows hysteresis similar to a thermometer. Its output goes high when some high input threshold is exceeded, and falls low when a different low threshold is exceeded. For instance, it might go high when the input goes above 3.5 volts, and only go low when the input falls below 1.5.
So even if the capacitor allows through some noise that could still cause some small flipping back-and-forth near the crossing of an input's threshold, the Schmidt trigger will reject that.
Suppose we want to debounce the LED with a capacitor? The problem is that the resistances end up being too low due to the need to supply current to the LED. If we just use the same circuit and make the resistors smaller (and the capacitor larger by the same factor), we end up with something which wastes power. The way to do this is to use a small signal loop to handle the switch, and debounce it, and then use the voltage to control a transistor which dumps current into the LED.
Though debouncing a LED might be useless, if we make the resistors and/or capacitor large enough, we can obtain a nice behavior: that of the LED slowly fading on when the button is pressed and held, and fading out when it is released.
simulate this circuit
This is the same circuit as before: the "out to microcontroller" node now connects to the base of an n-channel MOSFET which drives current to the LED. The MOSFET "buffers" the debounce logic from the LED driving. The debounce circuit isn't disturbed by the low impedance of the LED, and the LED is not starved of current by the high impedances in the debounce circuit.