Low-complexity generation of single pulse using RC circuit from a fixed DC supply without the use of a controller

Question

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.

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.

Answer 1

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

^{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.

^{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.

Answer 2

^{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.