555 timer

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Reza_liaghat

Joined Jun 17, 2016
12
Hi everybody
I want to make an circuit with 555 IC that connects a relay for 10 to 15 seconds and then disconnects for 2 to 3 seconds and repeats.
any one can help me ?
 

dl324

Joined Mar 30, 2015
12,871
Welcome to AAC!

Do you want to make an attempt on your own? If you do that, we can guide you past any obstacles you encounter and you'll learn more that way.
 

crutschow

Joined Mar 14, 2008
27,212
I want to make an circuit with 555 IC that connects a relay for 10 to 15 seconds and then disconnects for 2 to 3 seconds and repeats.
You want that cycle to continue indefinitely?
A 555 configured as an astable can do that.

What type of relay?

How will you turn the circuit on and off, by just applying and removing power?
 

dl324

Joined Mar 30, 2015
12,871
I try to do it myself, but I'm not familiar with electronics.
If you want to try on your own, this is all the information you need:
clipimage.jpgclipimage.jpg
You want t1 to be 10-15 seconds and t2 to be 2-3 seconds. Frequency and duty cycle calculations aren't relevant.

EDIT: Since you're learning, you should know that Nat Semi drew the schematic "backwards". Flow should be primarily left to right and top to bottom. Inputs should be primarily on the left side of the symbol and outputs should be on the right. There are exceptions, but their symbol is awful.

Like this:
clipimage.jpg
 
Last edited:

dl324

Joined Mar 30, 2015
12,871
@Reza_liaghat It's straightforward. You have two equations with 2 unknowns (you pick a value you like for C). Solve for t2, substitute values for RB and C into the t2 equation and solve for RA.

The only consideration is that larger electrolytic capacitors will be leaky and affect timing.
 

Tonyr1084

Joined Sep 24, 2015
5,976
@dl324 gave you a nice schematic. I've copied and pasted it below so you can follow what I'm telling you:

C1 charges through the total resistance of R1 + R2 and discharges only through R1. So using the formulas above R1 will be the lower value resistor. Since two resistors in series (R1 & R2) form a higher value resistance the cap (C1) will charge slowly through the resistor pair. During this time pin 3 will be active high, meaning it will actuate the relay. When C1 reaches 2/3 the supply voltage (2/3 of 5V) the 555 will toggle pin 3 to active low and pin 7 will begin discharging C1 While it discharges pin 3 will be low. If you want the opposite functioning then R1 should be the higher value of the two resistors.

Q1 (2n3906 PNP Transistor) will be ON when pin 3 is active LOW. For what you describe you won't need the RLOAD, R3, LED1, R4 and LED2. The relay takes the place of R4 & LED2. Since I haven't yet looked up the relay you say you're using I'll just say it needs to work on the supply voltage. Whether you use 5V or anything higher up to (I think) 15V (not sure, check the spec sheet) the relay will have to operate on that chosen voltage. The contact current handling needs to be rated at or above the load you're going to use it to switch. It's always better to go 33% over rating to 50% over rating. Some people may choose double the rating. For instance; should a load draw 10 amps, the contacts should be able to handle 13.3 amps using the 33% over rating. 15 amps would be 50% over rating and double would be 20 amps. It's up to you to decide what you need the circuit to do and how robust you want to make it. You don't want to go way over the amperage rating either. Doing so just means the contacts will get dirty and fail sooner than later.
1613855194123.png
 

BobaMosfet

Joined Jul 1, 2009
1,776
Hi everybody
I want to make an circuit with 555 IC that connects a relay for 10 to 15 seconds and then disconnects for 2 to 3 seconds and repeats.
any one can help me ?
Google: 555 timer tutorial

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3
 
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