Converting a pulse train to a single pulse

Question

Posting my question with a schematic

I want to detect AC line voltage using an optocoupler. When i feed the optocoupler input with an AC line, i get a 50 Hz pulse train at the output. Usually this scheme is used for a zero cross detection. But my idea is to use this pulses and convert them to a steady DC pulse so that i can feed this pulse at the input of a PCF8574. In that way i would know the status of PCF8574 port pins and hence the presence of an AC voltage as shown in the schematic below

When i close switch SW, a pulse train is generated at output of opto-coupler. I want the "some circuit" block to generate a steady pulse from the opto-coupler output. When the switch SW is open, there is absence of pulse train and want the "some circuit' block to output a zero until the switch SW is turned ON again.

Now my question is:

How can i convert a pulse train to a steady DC? I want the "some circuit" block in above schematic to do the job, so what circuit can replace the "some circuit" block

Also, is it the correct way to interface an opto-coupler with "some circuit" block ? I am getting a pulse train at emitter of the optotransistor Please correct me if i am wrong

Please share any circuits or ideas

Thanking you,

Gaurav

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Basically i want to know whether 230V AC Mains is present or not, so if i use HCPL3700 opto-coupler as shown below, will it do the job for me?

I just want a single steady pulse, so i assume if the AC signal crosses a certain threshold, the HCPL3700 will output a HIGH and when AC signal is absent it will output a LOW.

Am i right ? Please suggest

Answer 1

Using a retriggerable one shot like the 74ls123 with an on time longer than one full AC cycle will be enough to make your circuit work.These "digital monostables" are not stunningly accurate because the time constant relies on logic gate thresholds and analog components .This means that your proposed production circuit should have a RC constant of say 40ms .

Answer 2

The filter will give you a signal: AC present/AC not present, but I would let the software check the status unless your INT would interupt the controller in all cases. If you are unlucky, the controller might miss a flank and be in a wrong state.

Answer 3

One way to do this is to place a filtering circuit between the output of the opto-coupler and the input of the PCF8574.

A circuit that could so this for you would look like this:

Whenever the 50Hz signal is present the high times of the signal will allow R1 to charge capacitor C1 through D1. Select values so that after a pulse or two of the 50Hz the C1 is charged to a high level to be seen at the PCF8574.

Resistor R2 is in place so that the capacitor will discharge back to a low level when the 50Hz pulses stop coming.

Last thing to consider is that depending on the input signal characteristics of the PCF8574 you may want to buffer the signal from the circuit below through a Schmidt trigger type gate to square up its rise/fall times.

Answer 4

It is impossible to convert a pulse train into a single DC pulse as shown in your diagram becuase it would require that your circuit be able to see into the future and detect events that have not yet occured.

Imagine that it is currently at the moment of the falling edge of the 3rd and final input pulse. In order to know that it is time to lower the output single, you would need to know that there was not another input pulse coming.

The best you can hope for is to do something like "output a single pulse that rises when the input rises and falls after the input has been low for at least n amount of time".

Note that the n can be dependant on timings of the input pulse, but only things that could be measured up until that moment in time.

Answer 5

Neon-LDR optocoupler

I offer this as an incomplete answer as I have not tested it.

^{simulate this circuit – Schematic created using CircuitLab}

All of the LED optocouplers suffer from high dissipation in the mains voltage dropping resistors and generate some heat on the board. The LED current varies during the mains cycle and so trade-offs have to be made. It is possible to drop voltage using a capacitor but additional components are required to handle transients on power-up.

Neons are relatively low-power and a 1/4 W 220 kΩ resistor is all that's required to light it from 230 V mains. An LDR (light dependent resistor) is used on the low-voltage side. In the schematic above I have added a 1 uF capacitor to maintain the 'mains present' signal during the zero-cross.

Neon optocouplers were used by Fender guitar amplifier vibrato circuits and are still available. A web search for 'neon optocoupler' shows up some more recent implementations for modern micros for 'mains presence' rather than the usual zero-cross applications.

Neon optocoupler for Fender vibrato. Note use of opaque heatshrink to prevent ingress of stray light. The neon appears to be on the left with bulb point visible pointing to the LDR on right.

Given that four circuits are required in the question, I suggest that this idea is worth consideration. The neon optocoupler can be home-made using a regular neon, an LDR and black heatshrink.

Answer 6

I wanted just the same solution when working with an LED indicator running at 50-Hz mains frequency and convert that to steady single pulse while the LED was "on". The retriggerable 74HC123 works really well as mentioned before (using 20-mS retrigger to keep it sharp), but eventually used software on Arduino, quite simple, where the incoming pulse (from a TSL257 optical sensor) kept resetting the 20-mS timer - when no pulse was received, the timer (actually a mS counter) timed out and alarmed or whatever. It's a really useful way to monitor mains. The beauty of the software approach is that you can make the retrigger time as long as you like. I also use the HCPL3700 instead of relays to monitor pumps running etc.