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I am trying to implement a power-on delay circuit using a 555 timer. I referred to the circuit shown below:

enter image description here

In my application:

  • A switch is used to apply power. When the switch is closed, the input supply goes instantly to 14V.

  • I need 2 power lines, one instant when the switch is closed and the other delayed, both nodes denoted by the 'out' label in the circuit linked below.

  • The delayed output should drive a relay coil, and the relay contacts will also be switching the same 14V supply that powers the coil.

Based on the above. I have created the following circuit.

My questions are:

  1. Will the circuit still work the same way when I use a switch to apply the input power (instead of a constant supply)?
  2. Is it okay that both the relay coil and the relay contacts are connected to the same 14 V input supply?
  3. How can I make sure the capacitor discharges fully when the switch is opened, so that if I press the switch again, the delay restarts correctly (rather than being shortened because the capacitor was still charged)?
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3 Answers 3

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It looks Ok, but if you are using the same voltage for relay and contacts, take care of decoupling the supply and control pin (5) with capacitors.

To help discharging the capacitor, use a diode to bypass the timing resistor, and if necessary, use the NC contact of relay to connect a resistor for the residual voltage. enter image description here

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1: this will probably work. 2: no problem 3: add a resistor from vcc to gnd

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By using a DPDT switch, you can discharge the capacitor every time you turn the power off. Additionally, you can use a SPST relay instead of SPDT by not depending on the other contact for discharging. It can be a less bulky, inexpensive reed-switch type depending on the application. Of course, the power switch becomes more complicated.

Also, it's usually best practice to place a reverse biased diode across the relay coil to clamp the energy when it denergizes, and to add a decoupling capacitor to voltage control (pin 5).

If the DPDT switch operation isn't clear, I added two images: one with the power off, and another with the power on.

When the power is off, the capacitor is shunted to ground. Once power is turned on, the 555 output stays low while the capacitor slowly charges through the resistor until the voltage across the resistor falls below the trigger voltage. That's when the output goes high, and ultimately the relay is energized.

enter image description here

enter image description here

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