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I was looking for a similar solution to this question:

I want to generate a pulse (high) of 2 seconds when that DC supply is turned ON.

Is there a solution where we can implement this without using any microcontroller, timer IC or any delay IC? The circuit can contain resistors, capacitors, transistors or diodes etc.

The last answer was my approach, but this does not discharge the circuit when the power supply is taken away. So it only works once.

Picture stolen from last answer

Is there an easy way to have the capacitor discharged (in less than 1 second) when the 'POWER' switch is opened again?

And with easy I mean with less components than this answer, which uses 8 resistors, a single capacitor and 4 transistors.

EDIT

The goal is to power a DC motor for maximum 2 - 10 seconds (adjustable with potmeter) every time the power is turned on. If the power gets turned off the motor will also stop running.

The actual load will be a 5W DC motor, but this will be done by using another transistor or mosfet.

Timing diagram:

                __________       __    ______
Power      ____|          |_____|  |__|      |_______
                ____             __    ____  
Motor      ____|    |___________|  |__|    |_________
                <--> = 2 s

EDIT

I edited the restrictions in the title because timer IC is being considered.

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    \$\begingroup\$ The many-components answer by Transistor is quite good; component count is good for what it does. If you want low partcount, why are you restricting timer ICs, they are perfect for these kinds of tasks? \$\endgroup\$
    – anrieff
    Apr 2, 2020 at 16:33
  • \$\begingroup\$ Is there something wrong with adding a resistor in parallel to your C1? That would discharge the capacitor over time, a time which you can set by selecting an appropriate resistor. \$\endgroup\$ Apr 2, 2020 at 16:39
  • \$\begingroup\$ I have very little experience with electronics so i was looking for a circuit i understand. But maybe i should look into this timer IC solution. Thanks for your comment. @anrieff \$\endgroup\$
    – Egon
    Apr 2, 2020 at 16:40
  • \$\begingroup\$ @Marcus Müller Would this not change the charging time and voltage left over R1? \$\endgroup\$
    – Egon
    Apr 2, 2020 at 16:43
  • \$\begingroup\$ yes, but you're in charge of R1, too :) \$\endgroup\$ Apr 2, 2020 at 16:43

2 Answers 2

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One way to reset the capacitor is to use a double throw switch.

schematic

simulate this circuit – Schematic created using CircuitLab

This circuit has a feature that may be problematic, depending upon what the +5V and EN nodes are connected to. C1 will hold its charge for a short time. If the +5V node falls much more rapidly than C1 discharges, the EN node could go negative. Therefore, it may be useful to add a clamping diode between the EN node and ground, to keep the node from going too negative.

schematic

simulate this circuit

I used a Schottky diode for a low forward voltage. However, Schottky diodes tend to have significant reverse leakage, and this will affect the timing of the RC circuit. Choose a diode that fits your needs.

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A faster discharge can be obtained with D1.

The true answer depends on what the load is. If, for example, your circuit draws 1 mA then the capacitor will discharge at a rate given by \$ \frac {dV}{dt}= \frac I C = \frac {1m}{470\mu} = 2 \ \text {V/s} \$.

If you give more details in your question we can help further.

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  • \$\begingroup\$ i added some details to the question. will your answer (adding a diode) do the trick? \$\endgroup\$
    – Egon
    Apr 2, 2020 at 17:14
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    \$\begingroup\$ You've made it into an entirely different question. The one you linked to was to apply a very low-current pulse to a microcontroller. (1) What What parts have you got to hand? (2) What is the motor voltage? (3) I've added a timing diagram to your question. Please check that you understand it and that it's correct. \$\endgroup\$
    – Transistor
    Apr 2, 2020 at 17:28
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    \$\begingroup\$ Are you assuming a load on the +5V line? Otherwise D1 does nothing. \$\endgroup\$ Apr 2, 2020 at 17:50
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    \$\begingroup\$ @Spehro, yes. The question has evolved since originally posted. Ho-hum. \$\endgroup\$
    – Transistor
    Apr 2, 2020 at 18:02
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    \$\begingroup\$ @Transistor Shouldn't D1 be across R1, anode to ground rather than across C1. \$\endgroup\$
    – user173271
    Dec 24, 2020 at 6:28

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