Question on zero-crossing circuit

Hmm, We'll have to wait for Jim, 'cause it's not obvious to me how it works.

Why not just a transformer and then a comparator with some hysteresis. You still would want to capacitively couple it I think.... let the average voltage be your "zero". And that looks to be what C1 is doing.

George H.

It's only complicated-looking because it's drawn badly, with the power supplies mixed up with the signal wiring.

The top and bottom rails (after R2 and R3) are the power supplies, Cap C1 turns the AC mains into essentially a regulated AC current source. R1 limits the inrush to an amp or so so you don't pop the rectifiers, and the two zeners clip off the tops to keep the voltage regulated. R2/C3 and R3/C4 are the supply filters. (The positive supply lead on the LM339 isn't shown, which adds to the confusion.)

So apart from whatever incidental voltage there is on the neutral, you can think of the junction of D1 and D2 as sitting still at ground potential. The other ends of D1 and D2 sit at +- a diode drop from there.

AC current is applied to the junction of D4 and D5 via R4 (plus a bit of despiking from C5, to get rid of switching hash and so on). On the positive half cycle, D4 conducts and the D3/D4 junction goes to +2 diode drops, so U1 pulls low.

On the negative half cycle, D3 conducts in the same way, and U2 pulls low.

Within a diode drop of the zero crossings, neither D3 nor D4 conducts, and both U1 and U2 are high. So the result is a brief pulse on each zero crossing.

It's not a circuit you'd want to do anything much with, except drive an optocoupler, because there are safety issues.

Cheers

Phil Hobbs

The impedance of C1 is ~12K @60Hz, limiting the recirculating current. Transformers are expensive, and bulky.

The object of that path is to provide the power supply current for the

Contrary to Phil, it is not drawn badly... it's quite easy to follow :-)

And... there is a whole world that doesn't need opto-couplers... think the appliance control industry where everything is operated directly off-line. ...Jim Thompson

That's a lot of parts for a simple function. It's from an old s.e.d. thread.

Here are a few that I suggested:

That last one powers the gate from the line input, via the esd diodes.

(Beeson. I knew a Dr Beeson, a physicist, in New Orleans.)

C1 is mostly a current limiter.

Three interesting things:

C1 and C2 need to be "XY" type caps, safe to put across the AC line

C4 appears to be not connected

It has 20 parts and its output is still not isolated from the AC line.

A transformer, or a wall-wart equivalent, doesn't need a cap.

There'd be something seriously wrong if the guy who drew it couldn't follow it. I'd much prefer flags on the PS leads (and it would be easier for beginners if VCC were shown on the 339s.

But if you're going to go to all that trouble, why put up with a floating output? You aren't going to drive much off those rails, not with 330 ohms in series.

And a 6N138, one BR, one resistor, and one cap will do the whole shebang and give you isolation besides. (Or a 2-channel opto, as John suggested.)

Cheers

Phil Hobbs

It's not meant for use as a PCB netlist. The 339 is totally behavioral (model is on the Device Models & Subcircuits page of my website), thus needs only IN+, IN-, OUT and a GND reference to function.

(I'm actually working up a nice model of (OpAmp & Comparator) output stages that properly reflects VDD and VEE currents with load, but have been too busy to put the final touches on it.)

Huh? The output is at the 339, with a 10K pull-up. A _logic_ signal, and it worked great in the washing machine it went into.

Whoopee doo ;-) Everything to the right of the bridge rectifier (except capacitors) went onto a single chip along with a ton of other goodies, timers, OpAmps, and a 3.3V output to power a micro.

And lots of things don't need much power... I recently designed a 1kHz oscillator (for a medical application) that consumes only 6nA! And a

...Jim Thompson

So you had a completely separate power supply just for the zero crossing detector? Why? There must have been circuitry on the other side, which would have needed its own power.

Yeah, I suspected it was just a doodle. Parts are free in ICs, which (like a trust fund) leads to lots of bad habits if you don't fight it.

Cheers

Phil Hobbs

SOP for Hobbs and Larkin. Snarky for snarkiness sake. ...Jim Thompson

Are you just looking at it to understand it, or are you looking for a Zero Crossing Detector to actually use?

I have one that I have used for many moons that is a heck of a lot less convoluted that this one and has proven to be quite reliable.

Sorry Jim, I have always found your schematics difficult. I figured it was just some standard that you needed in the IC industry. I would even prefer to decipher Jan's hand drawn light pencil marks. To end on a positive note, I'm sure you know what your doing. Mikek

It was conceptual. I suppose I could add dummy pins for the .... >:-}

My clients love them, particularly when I clean up their drawings.

Yes >:-} ...Jim Thompson

John, those first three won't work. If the LED's pass enough current so it detects 0 volts somewhere near 0 volts then they burn out at the top of the mains cycle.

Yeah, I also noticed there isn't any (+) rail powering up the LM339 cheap, so don't worry about isolation, it's only connected to the noodle! :)

Jamie

This one might be better for those who are afraid of "hot" circuits...

(Next to last listing on the S.E.D/Schematics page of my website.) ...Jim Thompson

Dr. Hobbs, Thank you for your very informative reply. In Jim's defense, I found the rest of the circuit to be pretty easy to understand. :-) doug

Ok, got it, although I am amused to see that allaboutcircuits.com expressly forbids discussions of mains-source current limiting circuits that use caps

So the 180 ohm 2W resistor basically provides protection from in-rush current?

Thanks, Jim. Personally, I like the stuff you draw and appreciate the help you give to beginners like me.