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Looking for two-input comparator with hysteresis

Discussion in 'Electronic Design' started by Keith, May 23, 2006.

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  1. Keith

    Keith Guest

    Use an optical isolator to connect to the line, even if there is an
    isolation transformer inbetween.

    Depending on your clock circuit, it may be eaiser to do the
    filtering in the digital domain. After the isolator turns on wait
    1/119th of a second to look for it to go off, then 1/119th of a
    second for it to turn back on...
     
  2. Hi all,

    I'm building a digital clock circuit which will use the AC mains
    frequency for time keeping. I need a circuit (preferably a single IC)
    which can be connected via a transformer to the mains supply, and will
    send one output pulse to my TTL for each complete supply cycle. I think
    it would be good to include some hysteresis in case the AC mains
    waveform is noisy. I'm concerned about using a simple Schmitt trigger,
    where the input is taken relative to the lower power rail, in case I
    create a loop which causes a large current to flow back through the
    linear power supply. So I'm looking for a comparator with two high
    impedance inputs and hysteresis. Does such a thing exist as a single IC?

    Any suggestions would be much appreciated.

    Best wishes,

    Chris Tidy
     
  3. Joerg

    Joerg Guest

    Hello Chris,
    Huh? Which current? Which not just take a 74HC14 and use a nice large
    resistor value in series with its input if you are afraid to fry
    something? Although I am still puzzled about that frying process you
    described ;-)

    The main issue here is a lowpass. After all, you don't want the spikes
    from a starting vacuum cleaner to create a dozen pulses within one 60Hz
    cycle.

    Regards, Joerg
     
  4. Joerg

    Joerg Guest

    Hello Keith,
    That only works for a nearly perfect since wave, like if you'd tap into
    the output of one of the generators at Hoover dam. It could fail in very
    common situations where the since wave is distorted. That is common
    these days because of all the switch mode supplies and dimmers in
    households. A brief distortion can also be caused by a large AC motor
    spooling up (air conditioner, pool pump etc.).

    I'd go with a nice analog filter. Being an analog guy I am, of course, a
    bit biased here :)

    Regards, Joerg
     
  5. John Larkin

    John Larkin Guest


    Do your transformer thing into a CMOS schmitt trigger, but insert a
    100K resistor in series, then a 22 nF cap to logic ground. That will
    both limit schmitt input current and filter spikes.

    John
     
  6. Keith

    Keith Guest

    No, it simply works. Once an endge is sensed one ignores any more
    for some time. If there are multiple edges, no one cares. If the
    waveform is *really* distorted where the sensing is done may shift,
    but it'll soon find an "edge", rather like a phase detector.
    If it's a digital problem, stay digital. Unlike analog, digital
    transistors are free.
     
  7. Keith

    Keith Guest

    Both. But when people ask these sorts of questions I'm primarily
    worried about them, rather than some magic smoke.
    You can do that because it is isolated (sometimes it's unclear
    exactly what someone is intending). I'd still vote for the
    suspenders to go along with the belt. Optical isolators are cheap
    and may not add anything to the BOM.
    Yep. You know you're not going to see a negative transition for a
    half-cycle after a positive transition. If there are several
    transitions together you ignore 'em. If you delay a little too
    long, you'll still see the negative level and restart your counter
    waiting for the positive level.

    Depending on the accuracy of the oscillator the delays can be tuned
    close to the anticipated half-cycle the line. Then if the power
    is lost (assuming backup) the clock will still run, though drift
    off somewhat.
     
  8. The transformer has a tapped winding. The first part of the winding is
    connected to a linear power supply which provides the 5 V for the TTL.
    The second part provides the 50 Hz signal which is to be counted. I'm
    proposing to connect one end of the second part of the winding to the
    Schmitt trigger input, and the other end to the lower TTL power rail.
    Current could flow from the first part of the winding, through the
    linear power supply, through the lower power rail, and back into the
    second part of the winding. I think the circuit might still work if I
    just connected one end of the second part of the transformer winding to
    the Schmitt trigger input, and left the other end unconnected, but this
    doesn't seem quite right as one end is left floating. Taking the signal
    from one end of the winding only doesn't seem quite right.
    This is exactly what I'm concerned about.

    Many thanks,

    Chris
     
  9. Is this for safety reasons or to protect the logic? I just opened my
    cheap Morphy Richards digital alarm clock and it seems to feed the 50 Hz
    signal from the transformer directly into an LM8562 chip.
    I think I see what you mean. Once it has turned on, you wait a while
    because you know that signals received during this period are irrelevant?

    Chris
     
  10. Joerg

    Joerg Guest

    Hello John,
    That's pretty much how I have seen it done. Also, it is not necessary to
    provide a "personal winding" to the Schmitt input (although it might
    make the Schmitt feel really important, if Schmitts have feelings...).
    Most of the time both sides of the transformer are used for power, for a
    more efficient rectification. Then the 100K just goes to one winding
    before the diode.

    Regards, Joerg
     
  11. Joerg

    Joerg Guest

    Hello Keith,
    If you'd run it in a PLL fashion it can work. You'd have to because a
    spooling motor might cause a phase that wanders a bit for quite a few
    cycles. But wait until uncle Leroy pushes the button on his vintage
    margarita blender, the one from the days when the letters EMI were
    recognized as a record label :)
    Hmm, haven't seen many uCs under 20c and those were 4-bitters. Plus a
    crystal, plus some house-keeping parts.

    Let's see, in the analog world we'd need one Schmitt inverter. That's
    about 1.5c. If you can't rent out the other five it'll be 9c. Then we
    need a resistor to the tune of 1c and a cap for another cent.

    Ok, the Schmitt is semi-digital unless you take a vacant opamp or
    comparator.

    Regards, Joerg
     
  12. The really cheap way is to use a CT xfmr and biplex the display using
    two 1N4001s as digit drivers. Then the 60Hz input serves two purposes.


    Best regards,
    Spehro Pefhany
     
  13. Keith

    Keith Guest

    If it can be done in the analog domain it can also be done in the
    digital domain, given a few billion transistors. ;-)
    The OP is building a _digital_ clock. He's already got the
    *DIGITAL* part. Any additional transistors needed for a
    (*digital*) filter are now free.
    Yow does your Schmitt inverter tell time? ;-)
    It's still digital. ;-)
     
  14. Show us a schematic of your power supply and we can show you how to do
    it.


    Best regards,
    Spehro Pefhany
     
  15. John Larkin

    John Larkin Guest

    Like you said, to the logic common.

    Yes, and it gets trickier. Use cmos to avoid the complications.

    John
     
  16. Joerg

    Joerg Guest

    Hello Keith,
    You just hook up a piezo and count the clicks :)))

    But seriously, if his clock has a uC it might already contain Schmitt
    inputs. Or at least a comparator that can be "schmitted". All it then
    takes is an RC to get rid of the bulk of the noise issues.

    Regards, Joerg
     
  17. I guess what's really puzzling me here is that I can't connect one end
    of the transformer winding to logic ground (or at least I don't think I
    can). So I need some kind of differential to single-ended convertor, but
    with a Schmitt trigger too, so I can't just use a 741 or LM139. I'm
    surprised there isn't a single IC for this purpose.

    I'm thinking through your ideas - many thanks.

    Chris
     
  18. Thanks for the suggestion, John. This is a potential divider which acts
    as a low pass filter, is that right? In this case, where should I
    connect the second end of the transformer winding? Do the component
    values change if I use a TTL Schmitt trigger instead of CMOS?

    Many thanks,

    Chris
     
  19. Here's a schematic of the power supply I intend to use:
    http://www.mythic-beasts.com/~cdt22/clock_circuit1.jpg

    Note that it will be powering a mechanical clock dial which moves each
    minute, not an LED display (perhaps I should have explained this at the
    start). I'm concerned about connecting the lower power rail back to the
    transformer winding - surely they won't be at the same voltage because
    of the rectifier voltage drop?

    Best wishes,

    Chris
     
  20. Connect either of the ends of the transformer winding to a circut as
    John Larkin suggested. You will get a 50Hz output (not 100Hz as you
    might expect) from the ST. It should work okay as I said, but one 10K
    resistor from each end of the transformer secondary to ground (the
    common in your circuit, not earth) might be prudent. Each end of the
    winding (when lightly loaded) has 1/2 cycle of the mains voltage
    followed by 1/2 cycle of slight negative voltage relative to your
    common.



    Best regards,
    Spehro Pefhany
     
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