Can I use a voltage divider on a 230 volt 60 Hz supply followed by a diode to trigger a 555 when applied to the #2 pin? The 555 will work like a Schmitt trigger to provide short square pulses at 60 Hz. Is there a simpler way of doing this?
R
My mistake. The supply is 50 Hz.
R
You can do that if you plan on doing a one shot for a narrow pulse.
You can also do that with a "DIAC" diode to discharge a cap that'll give you a short pulse.
But I guess it depends on what the application is that governs the selected components.
Yes, use a cmos ??hc14 schmitt-trigger chip. You get 5 unused extra gates as spareparts.
I once tried to get a Xilinx salesperson to lower the cost of one of their FPGAs if I promised not to use its embedded PPC core. He grinned.
Bob
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I do hope you realise that connecting your circuit direct to the 230V mains is very dangerous?
The usual method to convert mains to logic level is to take your AC input (from a safe isolation transformer) and use a schmitt inverter (e.g. 74HC14) and suitable voltage divider/diode combination as you mention.
Dave.
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What are you trying to achieve here? Are you just using the mains a a convenient 60 Hz reference to produce short square pulses or does it have to be the mains for some other reason?
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There devices made for this job.
It would work. It is not usually a good idea to feed mains voltage into a low volt DC circuit without isolation though.
Use a step down isolation transformer then square it with a 555 wired as a Schmidt trigger, or use a resistance divider to power the led in an opto isolator, or something along those lines to provide isolation.
Squaring is the easy part. Any op amp is likely to have a gain of
200,000 feed a 3 V p/p sine wave in and get a square out.
Yup, Step down isolation transformer is what I would suggest. Say could you use an opto-isolator with a suitablly large resistor on the input side to generate a half wave rectified, scaled down version of the power lines?
George Herold
What about the low PIV of the LED in the opto-isolator?
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Michael, Sorry I don't know what PIV stands for.
I was trying to think of a method of isolating the circuit from the line with out using a bulky transformer. If someone was really going to tyr this they'd want another diode across the input of the opto- isolator to take care of the reverse current and some sort of protection for the HV spikes on the power line. (Would a capacitor be enough?)
George H.
Peak Inverse Voltage, or the maximum reverse voltage a diode can withstand,, before failing.
A capacitor can make things worse, because the ESR goes down, as the frequency goes up, so a narrow spike will pass more current, than a clean sine wave.
The safest & most reliable is a transformer. You can salvage one out of an old wall Wart.
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Michael, In parallel with the diode in the opto-isolater I would put another diode but with reverse polarity and then a capacitor. Second diode is to take care of reverse voltage and capacitor to take care of voltage spikes. Oh and something like a 100k ohm resistor is series with the whole lot. It would certainly be smaller than a wall wart.
George H.
Well, in context of the post, I don't see it. He wants to use a
555 - that means a low voltage DC supply is required. Since he has mains power available, he would be really smart to use it to make his DC supply for the chip. That does not automatically mean a transformer - he could use a switcher. But given that he wants to use the 230 mains for triggering, he's likely best served by a wall wart with low voltage AC output. He can make that the input for the DC supply and the trigger circuit.The other issue is having 230 volts on the printed circuit board, assuming that one is used. Your idea will protect the opto, but if 230 is on the PCB, it won't protect the person who might handle the board.
Ed
I don't think a cap for spikes . . . no point in filtering when you are interested in preserving a sine wave there - and you might adversely affect the phase angle. A zener or MOV.
I would suggest a cap to drop voltage into the opto isolator's diode - but that won't fly if phase angle is important (but it is more efficient 10 ma of led current dropping nearly the full 240 volts is
2.4 watts of heat)He just says he wants to trigger it, at a 50 HZ rate. If the application is just a time base, phase angle probably doesn't matter.
He says "square" pulses - that suggests an op amp to square it rather than trying to set a 555 to hold a precise 50% duty cycle.
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Yup you are right... It's just a crazy idea.
George H.
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