Is it some sort of "Latch" I need?

You could do something with good old-fashioned 4000-series CMOS - which Motorola (now ON-Semiconductor) sold as MC14000 parts.

Unfortunately, the "then comes back on permanently" part of the specification makes it an insoluble problem.

What you are implying is that the signal starts off low, goes high to 4 seconds, goes low for 300msec, than stays high forever.

In reality, it presumably eventually goes low again, so it can subsequently go high.

The simplest way of doing what you seem to want is to use a non-retriggerable monstable, set up for a period of slightly longer than 4.3 seconds, to control your light.

can be set up to be non-retriggerable.

The outputs clearly won't drive your LED directly but could drive the gate of MOSFET which could.

Is this your airbag warning light?

This should work:

I think he means the source is 12V or it's open, so it needs a pull-down resistor. You have pul-ups on the gates at the left.

This is a beautiful drawing. May I ask, what software do you use?

werner Dahn

I'm not sure your description is complete. Once your input pulse has gone high after the 300 ms drop out and your output is low (LED off), when do yo u want the LED to come on again? The circuit Pimpom provided will only let the LED light one time and never again until power is removed. How long d oes the input signal remain high?

Is the LED a typical small LED that uses no more than 20 mA?

Rick C.

The OP says he has to work with a "signal.... which is initially off, then on, then off ....."

"on for...4 seconds.....out for 300ms....then on permanently"

"after the 300ms of zero volts".

I take this to mean initially low, then high for 4 sec, then low for 0.3 sec, then continuously high for an indefinite period until the whole process is terminated.

This is represented in the lower right side of my diagram. Let's see what the OP has to say about it.

Yes, it is the "on permanently" part of the description that I find odd. I don't know of any input signal that is like that. After all, "permanent" is a long time.

How would the "process" be terminated? What exactly is your "process"?

Rick C.

Not *my* process. I meant until the OP stops doing whatever he's doing with the gadget - maybe switch the whole thing off.

Nice drawing! This would need an inverter gate at its output for my purposes, though. The output voltage I require therefore appears in your diagram labelled as Iled. Not at all sure how that confusion arose. :-/

Correct.

As I said earlier, for some reason you have the output inverted. Or I'm going mad. Or both!

That is what I want!

See my original post.

Yes, it is - and I should have stated that at the outset for which I apologise.

I thought of asking that first but decided to go with the assumption that it's only for an indication of the state.

The confusion, if it should be called that, arose because the only desired output you mentioned was about the state of the LED. Based on that, my circuit is still correct. The LED lights up or goes dark when you want it to.

If you need an inverted output to do some additional function, you can take U1D and use it to invert U1C's output instead of connecting the two in parallel.

BTW, after reading your original post, it took me a few seconds to get the idea in my head and a minute or two to work out the fine details. It took much longer to draw the diagram and prepare it in a format suitable for uploading to Dropbox.

CircuitMaker 2000. It's an old program as the name suggests. I do a virtual print of the schematic and save it as a pdf file, then capture a screenshot of the pdf display and save it as a bitmap image after cropping as desired.

Sometimes I capture the CAD display directly but it's not antialiased and raster "jaggies" are visible on curves and slanting lines.

It's not for everyone. Drawing the schematic and preparing it for uploading took much longer than designing the circuit in my head.

So it is driven high and low - not just open or closed to the +12V source to drive a lamp?

That's because his circuit drives an LED to +12V whereas you seem to have a lamp with one side grounded, or do you?

It's over-elaborate IMHO, though. If it were me, I'd just use a 555 in single-shot mode.

I think you would need two of them. Do you have a way of making the 555 trigger only once and never again? Or will you make the pulse longer than the input pulse masking the "glitch"? The spec is it only happens once I guess.

Rick C.

It's always interesting to see if the same function can be performed by a simpler circuit.

Apart from the LED and its series resistor which make up the load and are not a part of the solution, my design uses three resistors, two capacitors and a diode. One resistor (R3) can be eliminated if the input is permanently connected.

True, a basic 555 monostable needs only one resistor and one capacitor. But it needs a negative-going trigger while the OP's signal is positive-going. This requires some form of inverter at the input which adds complexity and increases parts count.

The OP specifies only approximate input pulse lengths. Some margin, maybe a lot, will have to be added to the monostable time constant to accommodate that uncertainty as well as tolerances in component values. The output pulse length will then not conform to the input pulse length.

Finally, there are other factors which may or may not be important depending on the application: My design draws practically no current in the idle state whereas a 555 uses ~100mW at 12V. It also produces huge current spikes while switching states. Both the idle and spike currents may be reduced by using the CMOS versions of 555, but they're still orders of magnitude greater than those associated with a 4011.