0
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ 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. \$\endgroup\$ – Transistor Dec 22 '18 at 12:03
  • \$\begingroup\$ Edited, Hope it makes my question a bit clear.. \$\endgroup\$ – BetaEngineer Dec 22 '18 at 12:09
  • 3
    \$\begingroup\$ Maybe a one shot monostable transistor multivibrator \$\endgroup\$ – cm64 Dec 22 '18 at 12:12
5
\$\begingroup\$

This will provide sharp trigger pulse with hysteresis; you can finetune the time by Rp:

enter image description here

At time zero the supply turns ON and provides a 5V two seconds single trigger pulse output.

\$\endgroup\$
3
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Simple RC time delay.

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%.

I asked about the threshold voltage for a '1' on EN because this determines what point on the discharge curve the device will switch. I've aimed for a time constant of 4.7 s so that you have a good chance that it will be above the '1' threshold for 2 s. You'll have to try it to see if it works.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.