Timeline for Is this an RC low pass filter?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Nov 5, 2021 at 20:17 | vote | accept | Less | ||
| Nov 5, 2021 at 19:13 | comment | added | Ben Voigt | The capacitor not only "debounces" the switch (eliminating repeated intermittent contacts) but also stretches out the pulse to a certain minimum width. | |
| Nov 5, 2021 at 16:19 | comment | added | Jay M | @Lundin: A contact bounce can never be a pure sine wave. At the most parasitic level it is a bunch of RLC circuits (charging parasitic L and C) which creates a complicated circuit with many resonances. Those freqs are triggered a step in current (the switch) with infinite harmonics. The parasitics make the ringing (sine waves) not the contact bounce. The purpose of the capacitor the OP is asking about is to over dominate those C parasitics, removing the harmonics that trigger ringing. If the charging current were limited then at best you would have a sawtooth due to C*dv/dt not a sine wave. | |
| Nov 5, 2021 at 14:57 | answer | added | TonyM | timeline score: 2 | |
| Nov 5, 2021 at 14:20 | comment | added | Less | t = 10 ms (datasheet) --> f = 100 Hz. | |
| Nov 5, 2021 at 14:12 | comment | added | Lundin | @JasonM A contact bounce is a sine wave. Also if you know the time something bounces no matter signal characteristics, then 1/t=f. | |
| Nov 5, 2021 at 13:21 | history | edited | Null♦ | CC BY-SA 4.0 |
added 16 characters in body; edited tags; edited title
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| Nov 5, 2021 at 12:48 | comment | added | Jay M | The equation 1/(2*piRC) is in the frequency domain and only meaningful for a sine wave, you want time domain. V=Vo x e^(-t/(R x C)) -> ln(V/Vo)=-t/(R x C) -> C= -t / (ln(V/Vo) x R) | |
| Nov 5, 2021 at 12:03 | comment | added | Lundin | The standard solution would be ground -> switch -> series resistor 1k -> pull down cap 100nF -> MCU input. The more professional solution would be to connect a bench supply to the raw unconnected switch, measure the contact bounce with a scope, determine a reasonable cut-off frequency, then calculate some suitable R and C. | |
| Nov 5, 2021 at 12:01 | comment | added | Less | But when the button ist pressed....there is a signal flow from ground to D6. That's why I drew the circuit so that the resistor is in front of the capacitor. | |
| Nov 5, 2021 at 11:57 | comment | added | Less | The purpose is to debounce the button | |
| Nov 5, 2021 at 11:49 | comment | added | Lundin | The simplest RC lowpass filter is a resistor in series and a cap to ground and that's it. So no, this schematic isn't a low pass filter, the cap should go to ground and you should place the filter between the switch and the MCU, since I'm guessing the purpose is hardware debouncing. | |
| Nov 5, 2021 at 9:55 | review | Close votes | |||
| Nov 5, 2021 at 23:03 | |||||
| Nov 5, 2021 at 9:24 | answer | added | Marko Buršič | timeline score: 1 | |
| Nov 5, 2021 at 9:09 | history | asked | Less | CC BY-SA 4.0 |