# Edge triggering circuit: how does it work?

This part is supposed to create a very short 0V signal to trigger 555; that is, even if I keep the button pressed, the 0V signal just lasts less than a second. Ok, it works, fine. But how does that work? I mean, how does that happen? What's the path of the current? I can understand it intuitively, but I cannot understand the precise path.

C1 is at VCC on both sides when the switch is open. When the switch is closed the left side goes to ground and right side pulled to ground. Then C1 starts charging through R2. Is that enough?

When the button is up, C1 is discharged through R1-R2. When you press the button, C1 is charged through R2. During this time, it appears to be a short to ground from the perspective of the trigger pin. After the time period set by R2*C1, 0.1ms in this case, the trigger pin returns to a high state. When the button is released, C1 discharges through R1 and the circuit is ready for the next cycle.

The trigger pin only responds to a low-going value. All others are ignored. That is what edge trigger means. Specifically, this is a falling edge trigger.

Did I just do your homework for you?

Thank you guys! I’ll read everything slowly…

salventre:
Thank you guys! I'll read everything slowly...

If you have (access to) an oscilloscope, it's very instructional to watch an RC divider in action...

blomcrestlight:
If you have (access to) an oscilloscope, it's very instructional to watch an RC divider in action...

Tell me more... It seems interesting... I don't know what an RC divider is...

salventre:
I don't know what an RC divider is...

RC series circuit

SteveMann:
Did I just do your homework for you?

Thank you, but I'm 37!

wolframore:
and right side pulled to ground.

Probably I cannot understand this part...

I cannot understand how the triggering circuit causes pin 2 going down just for very short time...

You say

wolframore:
and right side pulled to ground.

but very soon after that right side is 5v again, am I correct? I cannot understand the steps of this process...

When the switch is open they are both sides of the cap is 5V... this is simple.

When the switch is closed - both sides get pulled to ground... because the switch puts 0V on the side that connected to it... because there's no charge in the cap (since it was both at 5V earlier) so it pulls both sides to 0V immediately.

As it both goes to 0V the capacitor starts to charge through R2 since this side is still pulled up to 5V...

make any more sense?

It's like... when the switch is closed... I can easily understand the red path that makes the left side of C1 goes down, but I cannot understand the green path through which the right side is discharged... And why this discharge just lasts for a very very short time...

Before the switch is closed there is no voltage across C1. For there to be voltage across C1 it has to charge, which take time. When the switch is closed the end connected to the switch goes to 0V through the switch. The other end also goes to 0V because the capacitor is not charged and there is no voltage across it, so if the voltage on one end changes suddenly the voltage on the other end must follow it. After a short time the capacitor charges through R2 and the voltage on pin 2 rises back to +5V.

the green path doesn't happen...

so think of it like this:

anytime you have the same potential (voltage) at both leads of a capacitor it has ZERO charge and it's empty.

So if you put both sides to 5V it's empty

if you put zero on one side but have 5V on the other (this happens at the moment when switch is closed)

since it's empty and one side is 0 the other side is zero also... at exact time it's switched (because zero charge)

Next it starts to charge and goes up to 5V on one side...

wolframore:
if you put zero on one side but have 5V on the other (this happens at the moment when switch is closed)

since it's empty and one side is 0 the other side is zero also... at exact time it's switched (because zero charge)

Ok, but how you go from "one side 0V, the other 5V" to "the other side is zero also"?

PerryBebbington:
the voltage on the other end must follow it.

Maybe that's what I don't understand... How does it follow? I thought the two sides of a capacitor were separate...

they are both at the same potential when both sides were at 5V - therefore empty…

When one side goes to 0V (just at that instance) it’s still empty and capacitor have a charge time represented by the relationship: dv/dt - sooo since it’s empty it’s gotta be zero at both sides until it starts to charge

The two sides of a capacitor are separate, but they are coupled by
electrostatic force. If one side is given more electrons, they repel
electrons on the other side. It looks like a changing current flows
through the cap.

salventre:
Maybe that's what I don't understand... How does it follow? I thought the two sides of a capacitor were separate...

Define separate....!
The are electrically insulated from each other, but given their close proximity inside a tiny package I'm not sure they are separate in any meaningful sense. There will be micrometres between them.

Capacitor

PerryBebbington:
I'm not sure they are separate in any meaningful sense.

... especially since the wikipedia link reminds us they are separated by a dielectric medium, which is no "ordinary" insulator but

an electrical insulator that can be polarized by an applied electric field

So they are inextricably "linked" (for want of a better word).

I think any insulator can be a dielectric, some materials perform differently than others, hence the different type of capacitors with different dielectrics materials. I was going to type a brief description of how capacitors work then I thought a Wikipedia article would be far more informative and accurate than anything I could say.