# Trigger Pulse for IC555 using Inverted Square Pulse

#### Arouse1973

Dec 18, 2013
5,178
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?

#### Varun Dua

Jun 18, 2015
24
Sorry made a mistake in the post.. have corrected it as well. It isn't 50Hz. It is dependent on the Voltage Sensing section.. as in it variable on the input that is provided to the sensing section. If it is greater than 60% of 230V AC then we get a positive in the square wave otherwise if lesser than 60% of 230V AC then 0V. So we dont get a perfect square wave but more like a rectangular wave.

Dec 18, 2013
5,178

#### Varun Dua

Jun 18, 2015
24
It repeats as per the highs and lows on the rectangular wave.

Dec 18, 2013
5,178
Thanks

#### Arouse1973

Dec 18, 2013
5,178
Ok I think we are getting there. So the output of B is triggered by the high to low transition of the sensors output?

#### Varun Dua

Jun 18, 2015
24
Ok I think we are getting there. So the output of B is triggered by the high to low transition of the sensors output?

Yeah from High to Low transition

#### Arouse1973

Dec 18, 2013
5,178
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.

#### Varun Dua

Jun 18, 2015
24
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.

Yeah sure

#### AnalogKid

Jun 10, 2015
2,810
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

#### Attachments

• FET-Driver-2-c.pdf
10.5 KB · Views: 99

#### Varun Dua

Jun 18, 2015
24
So basically this is being achieved using a 2-input NAND Schmitt Trigger.

Can this be implemented without the use of IC's? As in 555 as the only IC?

#### Arouse1973

Dec 18, 2013
5,178
Oh dear, I am keeping quiet for once.

#### Arouse1973

Dec 18, 2013
5,178
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
In the rest state, i.e sensor output low won't FET B be high? Is this correct and is this ok?

#### AnalogKid

Jun 10, 2015
2,810
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

#### Arouse1973

Dec 18, 2013
5,178
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

LOL, no we can't have that. Just thought I would check

#### Varun Dua

Jun 18, 2015
24
I'm sorry I kinda didn't understand your solution

Slight changes to the requirement. The 230V AC Signal is applied to a bridge rectifier which converts it to 326V DC. The whole system is isolated.

Now if due to voltage fluctuations if the applied voltage goes below 60% of 326V, then a LOW signal is created otherwise it's high if above the 60% marks. No such particular frequency of the rectangular wave being created.

What I basically need is how do I create the falling edge triggered monostable multivibrator.

And how do I produce 2 similar rectangular waves of amplitude 12V from 1 single 12V Rectangular wave that comes out of the voltage sensing section.. basically the buffer section I'm talking about

#### AnalogKid

Jun 10, 2015
2,810
1. A CD4093 is an IC. An LMC555 is an IC. Why is one allowed and the other not allowed?

2. After the bridge rectifier, is there a filter capacitor to smooth the fullwave rectified AC into a DC voltage? Then what, a zener diode and an optocoupler, or something more precise like an analog comparator?

3. What is the output stage of the voltage sensor? If it is a reasonably low impedance like an opamp or even an optocoupler, then one output can drive two circuits like the FET-A gate and a delay circuit. Why do you think you need a buffer section? From your descriptions, it looks like the voltage sensor output could drive FET-A directly, and it's negative edge could trigger a 555, so I without more information I don't see a need for a buffer, and without more information I can't design one anyway.

4. A falling edge triggered monostable can be done with two transistors, or an opamp, or logic gates, or a 555. Without knowing the output characteristics of the voltage sensor, it is difficult to determine the correct input characteristics of the delay circuit.

ak

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,509
The other thing which strikes me is that there is no hysteresis in the measurement of the voltage.

#### Arouse1973

Dec 18, 2013
5,178
He is another variation that should work, if your interested.R1 controls the pulse duration and is approx. 3 seconds at the moment. V2 is the pulse from the sensor.

Jun 10, 2015
2,810

Replies
6
Views
8K
Replies
0
Views
4K
Replies
1
Views
2K
P
Replies
17
Views
2K
Pimpom
P
Replies
2
Views
1K