Is this an RC low pass filter?

Question

The D6 is a digital pin (input) of the Arduino Nano.

Function of the circuit: If the button is not pressed the internal pull up pulls the input D6 up to 5 V (high). If the button is pressed the input D6 is low (0 V). The capacitor and resistor are used for debouncing the button.

My Question: Is the wiring concept of the capacitor and resistor a low pass? And can I calculate the capacitor with $$f=\frac{1}{2\pi R C}$$?

Answer 1

You can just connect a capacitor from I/O pin D6 to GND and connect the switch in parallel with it.

^{simulate this circuit – Schematic created using CircuitLab}

The Arduino Nano uses the ATmega328P which has Schmitt trigger inputs for all its GPIO pins. This ensures that the relatively slow rise voltage on the capacitor will not cause input 'chatter' as it passes through the uncertainty voltage range found in logic signals.

When the switch is pressed, it will short-circuit the capacitor, drawing an instantaneously-large discharge current through itself. This arrangement of switch and debounce capacitor is very common and has been used in millions of products and equipment around us for decades. Few or none of these switches or capacitors have been damaged by this.

You need to know the switch bounce period. You should take this from the datasheet. If you do not have this, most switches are 3..10 ms so you can use a working guess value of 10..20 ms. But look the value up if at all possible.

To calculate C, you need the following values:

• Switch debounce period (or use 15 ms) tdb
• Schmitt input trigger voltage (min.) Vtmin
• Pull-up resistor value (min.) Rpumin
• Supply rail voltage (min.) Vddmin

Note that the minimums are used. The maximum for any of these values will make the RC charging or detection time longer, which means the circuit will debounce its input for longer. If it debounced for a shorter time when the RC tolerances changed, the switch might still be bouncing and those might get detected, so this goes in the safe direction.

Some of these can be found in the ATmega328P datasheet:

• Vt = approx. 2.6 V (from Pin Threshold graph)
• Rpu = 30..60 kohm, so Rpumin = 30000 ohms

For the Arduino Nano supply rail, I'll assume (a terrible word) a 5 V +/- 0.25 V regulator. So Vddmin = 4.75 V.

Transposing the capacitor charging equation $$v=V(1-e^\frac{-t}{RC})$$

for C gives $$C=\frac{-t}{R*ln(1-\frac{v}{V})}$$

So $$C=\frac{-0.15}{30000*ln(1-\frac{2.6}{4.75})}=630 nF$$

Use 680 nF.

Answer 2

^{simulate this circuit – Schematic created using CircuitLab}

This is more correct, no internal pull-ups.

R2 limits the current through the switch. \$I_{SW}=\dfrac{5V}{1000\Omega}= 5mA\$. The \$\tau_c=R_3C=10k\cdot C\$ for charging the capacitor, while it is being discharged with \$\tau_d=(R_3 \parallel R_2) C\$. The remaining capacitor voltage is \$V_c=5V\dfrac{R_2}{R_2+R_3}=0.45V\$