# Help calculating the value of a smoothing capacitor

I prefix this with I know virtually zero about electronics, sorry.

Preface

In an attempt to try and gain some basic knowledge and learn a lower level programming language (usually work with more abstracted languages) I'm working on a project to create a power folding mirror module with an Arduino.

My current progress is located here if anyone is interested.

It currently works. I listen to door lock CAN message and simulate ignition through an Arduino pin. Using a couple of relays it controls the mirrors.

However I can't seem to decipher ignition signals of from CAN messages so I need to take a 12V feed and bring it down to the safe 5V level my Arduino can work with.

I have an of the shelf offering which I have opened up and am trying to use as a reference to some of the circuitry and design.

Question

Based on my off the shelf offering I have traced the following circuit.

simulate this circuit – Schematic created using CircuitLab

It is my understanding that R1 and R2 make a voltage divider and bring the voltage down to about 4.5v. It is also my understanding that R3 will act as current limiting. I assume not for the benefit of the Arduino but more to protect anything I may have taken the original 12V feed from.

The part I'm having trouble deciphering is C1. Based on my attempted research i think this is acting as a smoothing capacitor.

How can I work out its rating? I have no equipment capable of measuring it. Are there any sort of generic default values such a component could be in this context?

Iv'e tried looking at the rules for calculating it and the various equations I've seen are just a little mind boggling (maths is a sticking point for me, which is odd being a programmer writing HR and payroll software.)

Thanks for any advise. This is the first question I've written in this domain so hopefully it is acceptable.

• @winny and whover upvoted your comment: This is about a signal, not power. The Arduino needs to know when the ignition switch is turned off so it can activate a relay to make the rear view mirrors on a car fold in. The voltage divider is to provide a 5V signal from the ignition switch 12V.
– JRE
May 9, 2020 at 8:59
• I upvoted both now! May 9, 2020 at 9:26
• @JRE Sorry about that. May 9, 2020 at 10:06