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- Dec 18, 2013
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Does the output of A and B repeat as long as you have a 50 Hz square wave? Or does it just happen once on the first pulse from the sensor?
Adam
Adam
Ok I think we are getting there. So the output of B is triggered by the high to low transition of the sensors output?
You could use ak's design he did for you once it's been modified. I could do it but it would be rude of me to do so. Lets wait and see what he thinks.
Adam
In the rest state, i.e sensor output low won't FET B be high? Is this correct and is this ok?This should do what you describe. Because C2 is so large, diode D1 is added to protect the U2C inputs. With every sensor input pulse, FET-A is the width of the sensor pulse, and FET-B turns on the instant FET-A turns off. This circuit uses a true monostable - the output pulse runs it's full time whether the input pulse is longer or shorter. If you are sure that the sensor input will not go high again before the 5-second output has finished, then we can go back to the 40106-based circuit. With that circuit, if the sensor input goes high in the middle of the FET-B output pulse, that pulse ends immediately. That's the difference between a pulse-stretcher or boxcar circuit and a true monostable.
ak
No and no. My bad. I rarely use the internal gate for the monostable output, but I didn't have an inverter for the normal output. Fingers got ahead of brain. Add 1 transistor inverter to U2C pin 8.
Another solution is to add an RC differentiator to U2D pin 13, but then it's just like a 555, and we can't have that.
ak
Oh dear, I am keeping quiet for once.