Electronic – How to clamp a signal

A Thyristor or Triac drops ~1.2V because it needs some power to keep itself turned on. This property is also used to turn the Triac off at the zero crossing point (ie. when voltage drops below 1.2V).

You could replace a Triac with back-to-back MOSFETs which have very low voltage drop at 1A, but you will need extra circuitry to detect the trigger pulse and zero crossing, and to drive the FETs. Something like this:-

^{simulate this circuit – Schematic created using CircuitLab}

When a trigger pulse is fed into the R/S flip-flop formed by NOR gates NOR2 and NOR3, NOR2 toggles to a logic high output. The flip-flop will stay in this state even after the trigger pulse is removed. Gate driver U1 converts the 5V logic signal into a high voltage and current to turn the MOSFETs on quickly.

Op-amp OA1's output is high when the sawtooth wave is positive, while OA2's output is high when the wave is negative. R3 introduces a small bias voltage that causes both op-amp outputs to be low when the wave is close to zero Volts. When this happens NOR1 produces a logic high output, resetting the R/S flip-flop and turning the MOSFETs off.

But why do you need two FETs? MOSFETs have an inherent body diode that would conduct when the wave went negative, even when the FET was 'off'. With two FETs wired back to back the body diodes oppose each other, so one is always blocking this unwanted current path.

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.