How to generate a single pulse using RC circuit from a fixed DC supply without any controller, timer or delay IC?

I have a DC supply (5 V) to turn some circuit on and in that circuit there is a module which uses a specific mechanism (i.e. a 2 seconds width high pulse) to turn on (even if a supply is connected, that module will turn on only by this mechanism i.e 2 seconds pulse).

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

I know we can do this using timer ICs like 555 timer, or any microcontroller, or using gates (but this uses IC, RC components etc,) but I was wondering if I can make a circuit using only R and C as components.

Problem: When DC supply (5V) is turned ON, the circuit gives '1' (5 V) for 2 seconds and then give '0' (0 V) for the rest of the time supply is connected.

Question: 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.

Edit:

This pulse is used as a trigger to turn on the module, there is no load attached to that pulse. Currently, there is a push button that does this job. I have to press that button for 2 seconds to turn that module on. I want to remove that button and want to automatically turn the module on when supply is connected.

• We need a few more details. What is the load on this pulse? What is the threshold voltage for '1'? Hit the edit link under your question. Dec 22, 2018 at 12:03
• Edited, Hope it makes my question a bit clear.. Dec 22, 2018 at 12:09
• Maybe a one shot monostable transistor multivibrator
– cm64
Dec 22, 2018 at 12:12

You can try the circuit of Figure 1. C1 is initially discharged so when the power is switched on both sides of C1 jump to +5 V pulling the enable, EN, high. The voltage on the lower terminal of C1 then falls exponentially towards 0 V taking one time constant $$\ \tau = RC = 470 \mu \cdot 10k = 4.7\ \text s \$$ to fall by 63%.