The RC delay element is a way to create a time delay in your circuit by connecting a resistor and a capacitor. It’s super simple. And very useful.

The ‘R’ is a resistor, and the ‘C’ is a capacitor. That’s where the ‘RC’ comes from. And here’s how you connect the two:

## How does it work?

A capacitor is kinda like a tiny little battery. You can charge it with a voltage. And you can use this voltage for a short time until the capacitor is discharged.

The time it takes for the voltage to rise across the capacitor becomes our time delay.

A capacitor with a higher Farad value can store more energy than one with a smaller value. Therefore it also takes more time to charge a high-value capacitor versus a small-value capacitor.

The “speed of the charging” is determined by how much current that flows through the capacitor.

The more current that flows, the faster it charges.

If we connect the capacitor directly to the battery, there is no restriction on the amount of current that flows through the capacitor (other than the batteries own maximum current capacity).

So a lot of current flows, the capacitor charges really quick, and the delay becomes very small.

That’s where the resistor comes into play.

The task of the resistor is to reduce the flow of current to the capacitor to slow down the time it takes to charge it.

## The RC Time Constant

You can calculate the delay time of your RC delay element with a simple formula:

That’s the RC time constant, also called *tau*, which is written like **τ**.

It gives you the time it takes for the voltage to rise from zero to approximately 63.2% of the voltage you apply.

If your RC delay element has a resistor of 10 kΩ and a capacitor of 100 µF, then your time delay becomes:

Put that into a calculator and you’ll get a time delay of 1 second.

Arturo says

Hello from arturo @ [email protected]

Just finished reading your whole program

admin says

Hey Arturo, I hope you enjoyed it!

Michael says

Hey there,

I am a new reader and have spent substantial time with your blog during Easter. I have almost read all articles and love your style.

I have some quite basic questions towards this schematic. I thought, voltage splits up equally in a parallel circuit, so why does (2) not immediately have full voltage? I would understand if the current starts low and rises high.

A follow-up question that I would love to hear an answer to is: How can I calculate current/voltage development over time for both capacitor and (2) during the charging process?

I hope, you will publish many more articles, I really enjoy reading them!

admin says

Hi Michael,

Glad you enjoy my articles!

“I thought, voltage splits up equally in a parallel circuit, so why does (2) not immediately have full voltage?”

The resistor and capacitor are not connected in parallel. To be connected in parallel, both side of both components must be connected. Also, I’m not sure what you mean by “voltage splits up equally in a parallel circuit”….

But maybe this will become clearer when you look at the calculation of voltage over time. Learn about that here:

https://www.electronics-tutorials.ws/rc/rc_1.html

Michael says

Oh sorry, I missed some crucial detail.

I mean, if I had some extension of the schema at (2), so that the current and voltage that arrive there actually arrive somewhere. I mean, the current/voltage that arrives at (2) should have some purpose. I could, for example, put another resistance there like in this schematic: https://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=2ahUKEwi1rZKBq-bhAhXRIVAKHaXvBcsQjRx6BAgBEAU&url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3D4I5hswA45CM&psig=AOvVaw3HDX8ZMLzu2xpvE8riLY2N&ust=1556112999420315

Now, the new resistor R2 and the capacitor are, per my understanding, connected in parallel, so that the capacitor and LED should have the same voltage. I mean that in the same sense as, e.g., wikipedia states it: https://en.wikipedia.org/wiki/Series_and_parallel_circuits#Voltage_2

The circuit in your link has a resistor and capacitor wired in series, this is not the case that I mean here.

Does my question get clearer? I only find material for RC circuits with AC instead of DC, or articles that do not explicitly describe how current/voltages behave during the charging process but only when the capacitor is fully charged.

Michael says

sorry, this is the link with the image: https://i.ytimg.com/vi/4I5hswA45CM/maxresdefault.jpg

admin says

Ok, I think I know what you’re asking. You’re saying that if we ignore C in the circuit you linked to (https://i.ytimg.com/vi/4I5hswA45CM/maxresdefault.jpg), then the voltage would split up equally, right? So why doesn’t it happen right away when the C is there?

First. the voltage doesn’t split equally, but it splits based on the voltage divider formula:

Vout = Vin * R2/(R1 + R2)

Only when the two resistors have the same value, the voltage is split equally.

Vout would be the same point as (2) in my circuit.

In the case where we also have the capacitor, R2 in the formula above would be replaced by the total resistance of R2 and C in parallel.

A capacitor also has a resistance. But it’s resistance varies depending on how “charged” it is.

When it’s completely discharged, its resistance is close to zero. When it is completely charged, its resistance is theoretically infinitely high.

So, when the capacitor is completely discharged:

R2 is in parallel with 0 Ohms. Which gives a total resistance of 0 Ohms. Which gives a voltage of 0 V for Vout – or (2).

When the capacitor is completely charged:

R2 is in parallel with a resistor of infinitely high resistance.. Which gives a total resistance equal to the value of R2. Which gives a voltage out that basically ignores that the capacitor is there.

Michelle says

Hi, I’m trying to use this circuit to create a 500ms delay on the output of an IC going to a clock for a flip flop, which is not working. My best guess is that because low for the IC is <1.2V not 0 the calculation above no longer applies? Wondering if you have any suggestions on how to create such a time delay?

admin says

The easiest would be to just increase the resistor and/or capacitor values. If you want to calculate exactly 500 ms, you can find more info on how the voltage across the capacitor increases with time:

https://www.electronics-tutorials.ws/rc/rc_1.html