🚀 go-pugleaf

In article <2006@lll-lcc.llnl.gov> rzh@lll-lcc.llnl.gov (Roger Hanscom) writes:
>...I'm building a TTL digital counter, and I need a
>1Hz pulse train to feed it.  I'd like to use the 60Hz signal on the
>power transformer, and a MINIMUM of components to condition it and
>divide it down.  The transformer I'm using is nominally a 9v
>wall-plug type.  How can I convert that to some thing I can work
>with (read reduce the ~20v p-p to something like 5v p-p)...

What I'd probably try is like this:

                                     |\
        ---->|-R--o-----o------------|+\
                  |     v            |  \
                  R     |      +5V   |   >-------------
                  |     v       R    |  /
                  o-----o       o----|-/
                  |             v    |/
              	ground----------o

where the "v"s and the thing on the input are diodes (1N4148), the Rs are
resistors of maybe, oh, 10k, and the op amp is a 3130.  The first diode
rectifies the input, the first R current-limits it, the pair of diodes
clips it to 1.2V, and the parallel R pulls it down to zero when the input
diode is in cutoff on the reverse cycle.  The op-amp is run as a comparator,
comparing a voltage fixed at one diode drop above ground (on the - input)
to one that's either two diodes above ground or zero.  Presto, a CMOS-level
square wave.  Note also that the clipping diodes protect things against
spikes and such -- the AC line is a noisy place and the transformer will
pass much of the crud through.  If you want a TTL square wave, use a
different op-amp, maybe an LM324.  Either way, you should probably run
the signal to a Schmitt trigger, since the rise and fall times will
probably be a bit slow for normal digital inputs.

There are probably more elegant ways.  I'm not really at home with this
analog stuff...  Improvements and corrections welcome.
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