How to turn a pulsed signal into a continuous one?

Question

I have a pulse signal at 1 hz which goes high +5V and low 0V. How do i turn it into a continuous one? +5V. i have searched in different places and what i found says that there are different methods such as using an op amp as an integrator, then i also found that it can be used a diode, resistor and a capacitor, others suggest that a sample and hold circuit can also work. However none of these answers seem to be clear to me. i am still a novice in this matter. Would somebody help me with an answer that has a schematic i could test?. I have also found that a retriggerable monostable can work but i am not sure if this is the right approach, as i need the "continuous signal" to go low immediately after the input pulse is interrupted. If possible i would like to do this using logic gates such as the ones found within 4000 cmos series. Thanks for taking your time reading my doubt.

Answer 1

You are failing to see the error in your belief.

If the pulse is high (as normal) and then goes low (as normal) and while low it is "interrupted", the signal will still be low and you won't have known the signal was interrupted until you realize that it has failed to go back high some time later.

It's like when your phone rings - it does so in bursts - you go to pick it up when the bell falls silence but you don't actually know that the person ringing you has hung up during that silence.

So forget about "immediately" and use a re-triggerable monostable. Here's the waveforms: -

Here's a retriggerable monostable circuit: -

Picture stolen from here. Basically, a re-occuring pulse fed to the input (the little circle to the left in front of the two resistors feeding a BJT) keeps the 555 from timing out.

Answer 2

If you have a 1 Hz square wave (high 500 ms, low 500 ms), then it takes at least 500 ms to tell that the pulses have stopped. You can't tell "immediately" that the signal has stopped.

When the signal goes low, for example, you have no way of knowing the difference between the signal having gone away, or just entering the low phase. You can only tell after 500 ms because then the real signal will go high, but if the signal went away then the voltage will stay low.

In any case, to turn a pulse train into a continuous signal (within the fundamental constraint described above), you can use a retriggerable monostable multivibrator, more commonly referred to as a retriggerable one-shot. When this device is hit with a pulse, its output goes on. With no other input, it stays on for some time, then turns off after the delay time has elapsed. The "retriggerable" part means that if it is hit with another pulse while on, the delay until off time is reset and the output stays on longer.

In your case, you need to set the on-time at least as long as the maximum time between pulses. If the device only triggers on a particular edge, then that time needs to be at least one second. Of course you want to make it a little longer for some margin, like 1.2 seconds or 1.5 seconds.

Answer 3

" I would like to do this using logic gates such as the ones found within 4000 cmos series."

Using and EXOR gate (4070) and an RC delay you can separate the incoming 1 Hz clock into its edges to give very narrow pulses at twice the frequency (2Hz).

Then a JK flip flop (4027) can recombine the two pulses back to the original 1Hz clock.

Now you have access to the SET and RESET inputs of the JK flip flop wich will allow you to set the output HIGH or LOW and monitor these control signals as the RS inputs override the incoming clock.

Answer 4

Can't a simple grounded capacitor of 1 uF or 0.1 uF between the input line and ground do the trick? It would cause a small delay thought. Not exactely "immediately". The higher the value of the capacitor, the longer the delay.