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'Zero cross over detector'
2000\02\24@102036
by
Andy Bassam
Hi all
I have written the coding to detect zero cross over, it does this by
using a square wave derived from a 60 V ac input. If anyone
could point me in the way of a suitable circuit to output this square
wave. I have a few circuits, but there wasn't a nice clean square
wave outputted.
Thanks
Andy B
2000\02\24@111858
by
M. Adam Davis
Use a small bridge rectifier, where the output goes through a resister into an
optocoupler. The LED in the optocoupler will be on for mist of the wave, and
only off just before each zero crossing, turning back on just after the zero
crossing.
-Adam
Andy Bassam wrote:
{Quote hidden}>
> Hi all
>
> I have written the coding to detect zero cross over, it does this by
> using a square wave derived from a 60 V ac input. If anyone
> could point me in the way of a suitable circuit to output this square
> wave. I have a few circuits, but there wasn't a nice clean square
> wave outputted.
>
> Thanks
>
> Andy B
2000\02\24@114237
by
Eisermann, Phil [Ridg/CO]
> Use a small bridge rectifier, where the output goes through a resister
> into an
> optocoupler. The LED in the optocoupler will be on for mist of the wave,
> and
> only off just before each zero crossing, turning back on just after the
> zero
> crossing.
>
or just an optocoupler like the H11AAx, which has two diodes, designed
for AC detection. But either way, you'll still have a somewhat rounded
signal when the input voltage is close to the threshold of the diode. If
you really want a nice square wave, try a comparator/logic gate/transistor
after the opto to clean up the signal. Although I don't see why it can't
work with just the opto coupler...
Phil Eisermann
H:(440) 284-3787 (spam_OUTmazerTakeThisOuT
ix.netcom.com)
O:(440) 329-4680 (.....peisermaKILLspam
@spam@ridgid.com)
2000\02\24@114649
by
Jose Souto
part 0 448 bytes
Andy Bassam wrote:
>
> Hi all
>
> I have written the coding to detect zero cross over, it does this by
> using a square wave derived from a 60 V ac input. If anyone
> could point me in the way of a suitable circuit to output this square
> wave. I have a few circuits, but there wasn't a nice clean square
> wave outputted.
>
> Thanks
>
> Andy B
Attachment converted: wonderland:60Hz.gif (GIFf/JVWR) (00018824)
2000\02\24@115907
by
Quitt, Walter
2000\02\24@164027
by
Wagner Lipnharski
|
part 0 2577 bytes content-type:text/plainI thought about something like Souto's circuit, but there is a problem.
The sine wave is small, lets say 8 to 12 Vac to a 5V power supply. It
means that the transistor will conduct when the sine wave is above 0.8
or 1.0 Volt, so it will exist a large gap meaning "zero crossing" and
not just a narrow pulse. 1V into a 10V sine wave means aprox 5.7¡ down
and up, or 11.4¡ per zero crossing, as it happens each 180¡, it means
6.33% of the time. It means the transistor will be off (collector up)
during 527.56µs from the 8.33ms period between zero crossings. The
zero crossing circuit should get a pulse as close as possible from the
real zero volts at the AC line. The above slack window will represent a
AC line window with a maximum of 10Vac (in a 120Vac line), what can not
be considered zero crossing. Of course that timing circuits of 263µs
could be implemented after the transistor collector goes up, but it will
always be ac voltage dependent, and never precise. The other problem at
your circuit, it gets only one side of the sine wave, it would require
another diode at least.
Using the opto-coupler idea, driven directly by the ac line, it means
that the opto emitter would start to lite around 2 to 3 Volts, so it
would represent 3/117, or 1.46¡ x 2 = 2.92¡, or just 1.62% of the time.
This ac line zero crossing slack would be exactly the necessary voltage
to lite the LED. Remember that the LED would be attached to the ac line
via a resistor, and if this resistor value is high, the LED would not
lite when ac line is only 3V requiring a higher voltage, if the resistor
is small enough to do it, when the ac line reaches high value will fry
the LED. Protecting the LED with zeners or something like that would
always require a low value resistor, so it will ask for a power resistor
(and zeners).
The big problem here, is that to sense the zero crossing we are trying
to measure "when it is not zero", what is wrong.
Now, if you double rectify a sinewave from the output of a power
transformer, you will have a double half sine wave, and exactly at the
zero crossing you will have a sharp edge, or... high frequency pulse.
Using a simple RC filter it is possible to recognize exactly when the
double sinewave reverts state, so it is exactly zero crossing, or in
this circuit you can call it zero peak.
I don't have the minimum idea how to calculate what will be the peak
pulse width, so its equivalent frequency.... anybody?
Wagner
Attachment converted: wonderland:hfp.gif (GIFf/JVWR) (00018872)
2000\02\24@165857
by
paulb
Wagner Lipnharski wrote:
> The above slack window will represent a AC line window with a maximum
> of 10Vac (in a 120Vac line), what can not be considered zero crossing.
> Of course that timing circuits of 263µs could be implemented after the
> transistor collector goes up, but it will always be ac voltage
> dependent, and never precise.
Considering the risetime at this point in the waveform, I don't think
the variation with voltage will be that much (in fact it will be closely
proportional).
The question to be asked is: What is this for? Usually for mains
phase control switching. The time delays can be corrected in software;
indeed I think a software PLL to correct for interference and general
jitter is the way to go (but haven't tried it yet).
And it is *not* a good idea to attempt to fire the Triac within 10V of
zero-crossing in either direction anyway. Part of the circuit design is
to avoid this very thing.
--
Cheers,
Paul B.
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