Can you debounce a tactile switch using a capacitor? Well, let’s search around. Now here are some stachoverflow folks who don’t know what they are talking about.
20200312/DuckDuckGo capacitor button debouncing
20200312/https://electronics.stackexchange.com/questions/64770/is-it-possible-to-use-just-a-capacitor-to-debounce-a-button#64771
Okay, let’s cut through that nonsense and I’ll provide my own accurate explanation. Yes, you can debounce a tactile switch using a capacitor, and you can also debounce a selector switch using two capacitors. Here’s how.
A resistor together with a capacitor between your GPIO input and power/ground acts as a low-pass filter. It wil work with both pull-up and pull-down circuits if wired correctly. Here’s how it works. Either the capacitor is empty and it will need to be charged up, or it is “full” (charged up to same voltage) and needs to be discharged. Not until the capacitor charge and voltage changes will the voltage to the GPIO input change.
For a pull-up input, wire the capacitor between the input and positive voltage supply. For a pull-down input, wire the capacitor between the input and ground. Although, if you are using ceramic capacitors, it doesn’t really matter which way you do it since the polarity is invertable and actually both ways work.
One problem with the Wikipedia articles on low-pass filters, high-pass filters, and RC filters is that they don’t really explain the technical circuit aspects for having an operating RC filter very well. Upon careful thought and closer examination of other circuits such as the prominent RC filter circuits for Raspberry Pi audio, it occurred to me that the common knowledge does not have a completely accurate understanding of the implications of the design of RC filter circuits.
So, let me put forth my explanation about the operational circuit requirements to have an actual functioning RC filter, starting with explaining a low-pass filter since it is easier to explain. (Refer to the Wikipedia article for the diagram in the following discussion.)
20200325/https://en.wikipedia.org/wiki/Low-pass_filter
A low-pass filter, at the outset, seems easy to understand. You have your digital logic output, that’s connected immediately to a resistor, then you have a capacitor connected between that resistor’s output and ground. The rate at which the capacitor charges and discharges is controlled by the resistor and the capacitor’s capacitance, and those together determine the cut-off frequency.
However, there are several assumptions that must be fulfilled in order
for Wikipedia’s simple calculation of the half-amplitude cut-off
frequency calculation of 1 / (2 * pi * R * C) to hold.
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The digital logic output must be “fully buffered” and not an “open collector” design: a logic one is presented as the positive voltage source, a logic zero is represented as a path to ground.
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The digital logic output must be “low impedance,” i.e. it is capable of supplying a large amount of current without a voltage sag. At least relative to the subsequent analog input, that is.
20200325/https://en.wikipedia.org/wiki/Line_level#Line_out
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The analog input must be “high impedance,” i.e. it does not draw very much current, much like a load with a very high resistance would.
If these assumptions are not met, the low-pass RC filter DAC will not function correctly. Why not?
A capacitor only charges and discharges when there is a path for current to flow. If the output is an open collector output, charge will flow into the capacitor at logic level one, but charge will not flow out of the capacitor (at a reasonable pace) at logic level zero because there is no path for the current to flow out of the capacitor. What will happen is that the PWM signal with push the capacitor to a higher and higher level until its voltage is nearly equal to that of the positive supply voltage, and then it will be stuck at that logic level one voltage. Therefore, a fully buffered digital signal is required for the capacitor to have a symmetric charging and discharging profile.
The purpose of the resistor in the low-pass filter is to deliberately take the low-impedance output and pin it to known impedance value. Without a clearly known source impedance, you cannot calculate the charging of the capacitor. More importantly, if the source impedance is too low, then the capacitor will have no effect on smoothing out the waveform into an analog curve: since the source impedance is so low, the voltage supplied to the analog input would be virtually equal to the digital output. The purpose of the resistor and capacitor together is to deliberately cause a “voltage sag” condition when the capacitor is empty of charge and is being charged with high current draw, and not until the capacitor is “fully charged” does the analog input receive the same voltage as the digital output.
You might have noted that some of the problems I’ve mentioned could be solved if you had a low-impedance analog input. So why, then, did I specify a high-impedance analog input as a required assumption? What this comes down to is mainly the ability to calculate the specific operation of the circuit, rather than the fundamental low-pass filtering characteristics itself. If you have additional low-impedance current flow paths to discharge the capacitor, you must take all of those together in your calculations in order to properly calculate the cut-off frequency of your RC filter. In the case of a low-impedance analog input, this would require you to calculate the resistance of resistors in parallel, which decreases the resistance smaller than the smallest resistor in your circuit, and that’s generally not very intuitive when you are trying to target a specific resistance in circuit design. Adding the resistance of resistors in series is much easier and more intuitive.
So, let’s return to our discussion of debouncing a tactile switch using a capacitor. At the outset, something sounds wrong about this. A tactile switch conducts when it is pressed, and does not conduct when it is released. How, then, can you use a capacitor as a low-pass filter with a tactile switch?
The answer here, is a rather tricky one. You cannot get an ideal, symmetric low-pass filter using a capacitor with a tactile switch, but you can get an asymmetric filter. And, as it turns out, an asymmetric low-pass filter is all you need to eliminate key bounce. Key bounce is caused when there are both high-frequency presses and releases. If you have high-frequency presses, but low-frequency releases, you’ve still eliminated key bounce, provided that your low frequency releases are not so low that the capacitor does not discharge fast enough.
The trick, here, comes down to the use of pull-up or pull-down resistors. Let’s consider the case of a pull-down resistor. When the button is pressed, the capacitor charges at an non-deterministically fast rate. But, when the button is released, the pull-down resistor’s path to ground provides the other required leg of an RC filter. The rate at which the capacitor discharges can be calculated using the RC filter equations. Essentially, what you’re doing is your taking a square waveform and converting it into a sawtooth waveform.
Of course, the implications of this particular circuit presents a little bit of a dilemma. With a pull-down resistor, you want the resistance value to be as high as possible to minimize the leakage current consumption. However, with an RC filter, you want to use a relatively low value resistor so that your filtered output still has a low impedance relative to your input. How do you know what resistance value to pick?
The ideal solution involves inserting a buffer IC chip between your tactile switch output and your RC filter input. This allows you to use two separate resistors, each with ideal resistance values for the primary purpose of the particular circuit. Of course, that circuit design costs more money, and the reason why you’re asking this question was probably quite deliberately because you were trying to create a cheaper circuit that does not require a microcontroller for debouncing.
So, I’ll make my calls on this point about a good compromise. For pull-down resistors, 4.7K ohm is a good starting point, and 50K is more ideal. For RC filters, 100 ohm is ideal, but 220 ohm works alright too. So, if you make a compromise between those two ends, you can end up trying out the use of a 1K resistor. This should still provide a low enough output impedance for most digital logic inputs, while still minimizing leakage current as much as possible.
Now, high-pass filters, that’s a little bit trickier to explain, but when you build an explanation based off of the previous knowledge I covered in explaining low-pass filters, then it is fairly easy to understand. Again, the same assumptions must be withheld:
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The digital logic output must be “fully buffered” and not an “open collector” design: a logic one is presented as the positive voltage source, a logic zero is represented as a path to ground.
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The digital logic output must be “low impedance,” i.e. it is capable of supplying a large amount of current without a voltage sag. At least relative to the subsequent analog input, that is.
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The analog input must be “high impedance,” i.e. it does not draw very much current, much like a load with a very high resistance would.
So, with those assumptions in place, we can present the circuit design. The capacitor blocks the path from digital output to analog input, and on the output side of the capacitor, the resitor provides a path to ground. (Refer to the Wikipedia article for the diagram in the following discussion.)
20200325/https://en.wikipedia.org/wiki/High-pass_filter
So, let’s explain how this works. Of course because we do have a capacitor on the path between output and input, a steady voltage will not allow current to flow for long: the capacitor will get charged and prevent any more current from flowing. The resistor to ground provides a current path with a clearly known resistance which (1) controls the rate of charging/discharging of the capacitor and (2) is selected such that only a negligible amount of current is consumed by the analog input compared to the operation of the RC filter circuit. Again, the fully buffered input is the key to deterministic operation of the low-pass filter RC circuit: without the ability of the digital input to provide a path to ground, the capacitor would never be able to discharge properly. When a path to ground is provided, current can discharge by looping through ground and through the resistor.
Now, the real challenges and pitfalls come into explaining the combination of both a high-pass filter and a low-pass filter. This is a stumbling point for many, simply because you cannot compute the values for the two separate modules and plug them together. Rather, the modules are, in essence, intertwined.
Take for example this misguided explanation of the Raspberry Pi’s audio on Hackaday. (Please also refer to the article for the diagrams in the following discussion.)
20200325/https://hackaday.com/2018/07/13/behind-the-pin-how-the-raspberry-pi-gets-its-audio/
The numbers derived for the low-pass filter… they are almost accurate, only off by a factor of two when discharging. But the high-pass filter numbers, they are way off.
So, how do you properly understand this circuit? Start with the low-pass filter. Take a good careful look at all current paths you can trace from one side of the capacitor to the other. You’ll notice that there are in effect two parallel resistor paths… so for the path of discharing that resistor, you must compute the effective resistor value from those resistors in parallel. Then you’ll know the roll-off freqeuncy for the discharge side of the waveform. What about the charging side of the waveform?
Well, this is a bit tricky to understand, you must take into consideration all of the ways in which the circuit setup could be interpreted.
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The two resistors form a voltage divider across the capacitor. The capacitor, therefore, can only be charged to a fraction of the digital output.
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The parallel resistor with the capacitor is a type of bleeder resistor. It causes the capacitor to discharge more quickly than it otherwise would if left on its own.
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The parallel resistor and capacitor together is a type of low-pass filter. However, this type of low-pass filter does not affect the output voltage, it only has a low-pass filter effect if fed by a current source. This is explained in the RC circuit Wikipedia article:
20200325/https://en.wikipedia.org/wiki/RC_circuit
Altogether, this is tough stuff to understand. With so many potential interpretations at play, how does the final charging frequency filtering behavior look like? I’ll make my call. The parallel resistor causes no interaction in changing the charging frequency filtering. It does, of course, act as a voltage divider that limits the maximum voltage the capacitor can get charged to, but because RC filters are defined independent of voltage (and we are using a voltage source rather than a current source), this does not affect the cut-off frequency when charging. The statement of the parallel resistor behaving as a “bleeder resistor” is indeed correct with our previous analysis noting that it causes an effective lower resistance during discharge.
Now, let’s analyze the touted “high-pass filter” section of the circuit. Now, this is where the explanation about what’s really going on can get really convoluted. Particularly it’s the most convoluted with the original Raspberry Pi, so let’s start there.
The first problem you see here… there’s no resistor on the “back-end” of the high-pass filter capacitor. How then does the high-pass filter capacitor charge and discharge? The answer lies in one of the assumptions I’ve warned not to make… the assumed impedance of the audio input connection provides a path for the current to flow. Typically this will be 10K ohms, but you can never be sure that this resistance will be the same across all kinds of different audio devices that the user may plug in, it could get as low as 600 ohms. So, the high-pass filter does not have a clear filter frequency, but rather an approximate range of filter frequencies.
What then does the other resistor do that is between the low-pass filter and high-pass filter capacitors? That has somewhat of the same effect on the high-pass filter as it does on the low-pass filter… it increases the discharge rate but it doesn’t affect the charge rate. For discharging the high-pass filter capacitor, there are two parallel resistor paths to take, the 150 ohm path to ground and the 270 ohm path to the digital output. Now, we can assume the digital output will only be a path to ground say 50% of the time, so effectively as far as current draw is concerned, we can double that resistance value to 540 ohms. So, what is the parallel resistance in front of that capacitor?
R = 1 / (1 / 540 + 1 / 150) = 117.391
Okay, 117 ohms, not too far off from the Hackaday article on that resistance leg. But, the larger resistance leg they failed to account for was the current path through the audio output. This is effectively a resistor in series, so you must add this resistance, 117 + 10000 = 10117. Worst case impedance, 600 + 117 = 717. In any case, the resistance along that current path is dominated by the output impedance.
So, what is the cut-off frequency range of the high-pass filter?
min = 1 / (2 * pi * 0.000010 * 10117) = 1.573 Hz
max = 1 / (2 * pi * 0.000010 * 717) = 22.197 Hz
Not too bad… it’s just the uncertainty of not knowing for sure what the cut-off frequency of the high-pass filter may be.
Now, the Raspberry Pi 3 improves upon the old circuit design by adding a 1.8K ohm resistor to ground on the output side of the high-pass filter capacitor and in parallel with the analog output. This allows us to pull the minimum cut-off frequency up higher to a known value in the case of a high-impedance output, but in the low-impedance output case, the minimum cut-off frequency is pretty much the same.
front resistance = 1 / (1 / 440 + 1 / 100) = 81.481
min rear resistance = 1 / (1 / 1800 + 1 / 600) = 450
max rear resistance = 1 / (1 / 1800 + 1 / 10000) = 1525.424
min cut-off = 1 / (2 * pi * 0.000047 * 1606) = 2.109 Hz
max cut-off = 1 / (2 * pi * 0.000047 * 531) = 6.377 Hz
Now, that’s much better. I mean, the difference between the minimum and maximum frequency cut-offs is much smaller. That means the high-pass filter’s behavior is much more deterministic and independent of the specific impedance of the connected audio device, especially compared to the old design.
So, that pretty much concludes are discussion and analysis about combined low-pass + high-pass filters. As you can see, it is quite complicated to analyze and understand the operation of low-pass + high-pass filters. One more key that must be understood: the low-pass filter section must always be in front of the high-pass filter section, otherwise your circuit will not work as intended.