Contact Bounce: Why Capacitors Don’t Fix It

As part of our discussion around those Hall effect switches, I cautioned our Larval Engineer that she can’t use capacitors to “smooth out” mechanical switch bounce, even though all of her cronies and (most likely) her profs will advocate doing exactly that. The subject also came up at the local hackerspace when she showed off her project, so I should explain why capacitors don’t solve the problem.

Here’s some switch contact bounce:

Switch bounce - black panel-mount
Switch bounce – black panel-mount

Another push on the button, just to show how unpredictable the bounces can be:

Switch bounce - black panel-mount - 2
Switch bounce – black panel-mount – 2

Note the horizontal scale: 10 ms/div. The smaller glitches appear only by courtesy of the scope’s glitch-catching mode; they’re down around a few microseconds.

Now, let’s add the canonical 100 nF “debounce” capacitor in parallel with the contacts and record another set of bounces:

Switch bounce - black panel-mount - 100 nF cap
Switch bounce – black panel-mount – 100 nF cap

Notice that the switch contacts bounce in a completely unpredictable manner.

The pullup resistor is a rather stiff 1 kΩ, so the RC time constant is τ = 1 kΩ × 100 nF = 100 μs, but that applies only to the rising edges of the waveform as the switch opens. You can, indeed, see a slight rounding of those corners: the voltage requires about 5τ = 500 μs to reach 99% of the final voltage.

The capacitor also forms an LC tank circuit with the usual parasitic wiring inductance, producing spikes that exceed the supply voltage: that’s the first half-cycle of the tank oscillation as the switch opens. The Q is fairly low due to the relatively high resistance, so the oscillations die out quickly. If this were feeding a microcontroller’s input pin, its input protection diodes would clamp the spikes to one diode drop above the supply voltage and below 0 V, but that’s an entirely different study.

It should be obvious that adding the cap hasn’t done diddly squat to debounce the switch transition.

Increasing the pullup resistor to the usual 10 kΩ will increase the time constant to τ = 1 ms, round off the leading edges a bit more, and further reduce the Q. It won’t debounce the longest transitions, which are on the order of 20 ms for this particular switch. You can’t increase the pullup too much, because you want enough current through it to ensure a valid logic level despite external noise (which is also an entirely different study); 100 kΩ may be as much as you can stand.

But that’s just for glitches due to the switch opening. The closed switch puts a dead short across the capacitor, so the cap provides no filtering as the switch closes: the microcontroller will see every single low-going bounce. The photos show only bounces during the open→closed switch transition, but the closed→open transition can be equally ugly: yes, switches bounce closed as they open.

That means the microcontroller will see glitches as the switch opens.

So let’s increase the capacitor enough that the voltage can’t rise beyond the logic threshold until the switch stops bouncing. Ignoring LC tank effects, the voltage rises as 1 – e-t/τ, so we need that value to be less than 0.25 (for a bit of margin) of the supply voltage after the longest possible bounce as the switch opens. Let’s assume the switch has a single closed (low) glitch after a long time being open (high), at which time the voltage must still be under the logic threshold to prevent a false input. The datasheets only give the maximum bounce duration, if they give any bounce time at all, so let’s assume the longest bounce will be 60 ms.

That says τ = -60 ms / ln(0.75) = 210 ms. Given a 10 kΩ pullup, that’s C = 210 ms / 10 Ω = 21 μF.

No problem, right? Let’s just put a 22 μF electrolytic cap across every switch and be done with it!

Well, except for the fact that most pushbutton switches can’t tolerate that much energy through their contacts. Assuming a 100 mΩ resistance and ignoring stray inductance, the initial current will be 5 V / 100 mΩ = 50 A with a time constant of τ = 22 μF × 100 mΩ = 2 μs. At the usual 5 V logic supply, the cap stores 22 μF × (5 V)2 = 550 μJ of energy, so we’re now burning the switch contacts with a 250 W pulse. Some switches have a maximum energy rating to deter exactly this design blunder, but you should not assume the lack of such a rating means the switch can handle anything you throw at it.

No problem, let’s just put a resistor in series with the switch to reduce the initial current.

I think you can see where this is going, though, so I’ll leave all that as an exercise for the student.

Moral of the story: you must do debouncing in software by filtering the raw switch input. The trick will be to get that code right, which isn’t nearly as simple as you might think. In fact, the first half-dozen techniques you come up with won’t work, so use a dependable library and test the results… which is an entirely different study, too.

If it’s any consolation, I didn’t know this stuff when I was a Larval Engineer, either. In fact, I didn’t learn much of it until after I made all the usual mistakes…