5 second delay on circuit

[randyrogers04]

Hey all, i am new to electronics but love tinkering with them. I have built a few small circuits but have a very low knowledge base.

So to get started, the circuit i am trying to build is a 5 second ON delay. The idea behind this is: press momentary ON/OFF button, after 5 seconds LED(s) turn on for 10 seconds(this time can vary) then turn off. This whole thing would run off a 9v battery or similar DC power supply.

If you need any more info please let me know since im not entirely sure what other info you may need. Thank you in advance for any and all help building this circuit.

[KMoffett]

Two LMC555's or one LMC556 timer IC and a few resistors and capacitors. Does the "on/off" button also connect the battery or just start the timing circuit?

Ken

[randyrogers04]

The on/off would preferably connect the battery, however i am open to suggestions.

[KMoffett]

Adding power on/off adds another level of hardware complexity. What's the purpose of this circuit?

Ken

[randyrogers04]

The purpose of this circuit is to turn on X leds after Y seconds. Then preferably(but not necessarily) turn back off after Z seconds.

X= amount of leds
Y= delay ON time
Z= time leds are on

Also if it is easier to set the circuit up with the ON/OFF not connecting the power source that is perfectly fine. Since i am still very new, easier is better.

[pebe]

555 delayed start timer 3.GIF (5.7 KiB) Viewed 19522 times

Here is a circuit I made some time ago that will do what you want. It consists of a 555 timer for the delay followed by a second 555 for the ‘on’ period.

When you first switch on C2 is discharged so it triggers on IC1and ensures that IC2 starts up in reset mode. C1 will then start to charge via R1 until the voltage on pin 6 of IC1 reaches 2/3 of the supply voltage, then the 555 will reset. The falling voltage on IC1 pin3 now triggers on IC2 via C3; the relay switches on, and C4 starts to charge via R4. When pin6 reaches 2/3 of supply voltage the relay switches off. That ends the cycle until power is switched off then on again. In your case you can replace the relay with a number of LEDs in parallel in place of the relay.

The time period of the 555 delay is a function of R1 and C1.
R1 (in megohms) x C1 (in microfarads) is known as the ‘Time Constant’ (TC). In the circuit, TC = 0.470 x 10 = 4.7seconds. The time taken for C1 to reach 2/3 of the supply voltage is 0.72 x TC seconds. So the time will be 4.7 x 0.72 = 3.38seconds.

If you use 22µF and 330K the time will be 4.7secs, and if you use 25µF and 270K the time will be 4.8secs.

You can easily work out values for 5secs and 10secs by varying C1 and R1. There are two provisos:
1. Electrolytic capacitors have variable leakage currents, so if you want accurate repeatable times then use bead tantalum capacitors.
2. The value of R1 should never exceed 2megohms.

You can use the same calculations to get the relay-on time from C4 and R5.
R3 is there to discharge the capacitors when the supply is switched off.

I hope you find that useful and come back with any questions

[pebe]

Hi randy..
Was that information useful?
pebe