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I'm considering purchasing a pre-made circuit assembly with the following associated schematic: enter image description here

Having looked at a typical 555 device datasheet, I'm unable to confirm the operating mode as the datasheet modes show the Discharge pin connected in some fashion.

What I would like the circuit to do is upon application of 5VDC as Vcc, the relay pulses once for the set time controlled by R1, and that nothing further happens. Can someone confirm that this is the operation of the relay or otherwise define what the operation would be?

BartmanEH
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  • What does the seller of this circuit claim it's able to do? – jwh20 Jan 28 '21 at 17:45
  • there is a serious language barrier: https://www.aliexpress.com/item/33045823748.html – BartmanEH Jan 28 '21 at 17:51
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    That circuit, as shown, will switch the relay on after a delay from when power is first applied. The relay will then stay powered. That is to say there is a switch on delay from power-up. To switch the relay on immediately the power is applied for a fixed time and then switch it off, the positions of the capacitor and variable resistor would need to be swapped over with each other. –  Jan 28 '21 at 18:00
  • In the image on aliexpress there is nothing connected to the outside edge of pin 7, but there could be something underneath the chip. From this schematic, it is a one time output from power up. – Aaron Jan 28 '21 at 18:00
  • [What to check for when buying an electronic component or module](https://electronics.stackexchange.com/questions/504044/what-to-check-for-when-buying-an-electronic-component-or-module). – Andy aka Jan 28 '21 at 18:09
  • If all loads share the same voltage, it is possible the relay will chatter turning off. So it would be better to define exactly what you expect rather than gues what could go wrong. As is , NG. – Tony Stewart EE75 Jan 28 '21 at 18:13
  • so this circuit is not quite what I want but if I swap C1 an pot R1 it is what I want? It'll energize the relay on power up and shut off relay after the timed delay? – BartmanEH Jan 28 '21 at 20:53
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    @BartmanEH Yes, the circuit should perform as you require if you swap the positions of C1 & R1. One small problem, if you power back-on straight-a-way after powering off then there will be no output pulse from the circuit. It is necessary to wait a few seconds before powering back-on to give the capacitor time to discharge. The way to get around this problem is to place a diode in parallel with the pot (R1) with its cathode connected to VCC. The added diode will then rapidly discharge C1 at power-down ready for the next power-up. The diode should preferably be schottky (low voltage drop). –  Jan 31 '21 at 08:11

1 Answers1

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It delays the relay energization after power is applied, because the capacitor is connected to Vcc. Adjustable via the "chronopotentiometry". After the delay the relay pulls in and never drops out. If you connect the normally closed contact in series with Vcc you might have what you want.

If you interrupt the power for a short period of time the next delay may not be as long as if the interruption was very long.

If you have a very heavy load (or accidentally short) Vcc to ground you might destroy the 555. If your power supply has a lot of current capability and the pot is turned close to minimum time the pot might be damaged. Both issues could have been avoided by a couple of almost-free resistors.

It's an "okay, not great" hobby circuit.

Edit: As suggested by @James, if you switch the positions of the pot and C1 it should do exactly as you say you want, with similar caveats as mentioned above.

Spehro Pefhany
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  • what I want is when Vcc is applied, that the relay energizes briefly for the timing period and then de-energizes. A previous comment suggests that swapping the capacitor and variable resistor (potentiometer) would accomplish this. I assume they mean C1 and R1. Do you agree? If so, please update your answer so that it addresses what I would like as explained in my post and I can accept this answer. – BartmanEH Jan 29 '21 at 16:24