Electronic – Art of Electronics zero-crossing circuit question

The speed of the phototransistor optocoupler is possibly a problem, as Andy has already flagged.

Also consider what threshold you are trying to achieve. If you want a zero-crossing pulse centered about the zero crossing and turning off at (say) +/-10V then you need to supply sufficient current to the LED to make that happen. Say that's 5mA. You now need to limit the current at the AC peaks to some reasonable value, and with a resistor it will be ~33 times higher, of course (\$\frac {230\sqrt{2}}{ 10}\$).

To get the width (ignoring the speed of response of the circuit) we know that the voltage is about 325 sin(\$\omega\$t) where \$\omega\$ is the angular frequency of \$2\pi 50\$ radians/second.

So the width of the pulse will be pw = (2/\$\omega\$)\$sin^{-1}\$(Vzc/325),

or, in general

pw = (\$1\over\pi f\$)\$ sin^{-1}(\frac{Vzc}{Vline \sqrt 2})\$

For example, if the voltage is +/-10V, then the pw will be about 200usec. For small voltages it will be approximately proportional since the rate of rise of a sine wave is almost constant near the zero crossing. So, a +/-5V voltage will give you a 100usec pulse, ideally preceding the zero crossing by 50usec and following the zero crossing by 50usec.

To make a precision zero crossing detector, you would be best to derive a power supply from the mains (maybe the TNY has some voltage you could borrow) and use a comparator of some kind (I've often used discretes because they're more immune to nasty stuff on the mains) to actuate a high speed logic-output coupler, so the LED in the opto sees only 5mA or 0mA.

using an op-amp as a comparitor often ends badly, some op-amps (not OPA340) can be damaged by this. and performance almost always suffers.

no you won't have 0V out the negative half cycle if you are comparing with automobile ground, only while that phase is less than automobile ground (1/6 cycle) that's probably enough.

the simplest solution is a 100K resistor to a the logic input (or Schmitt gate) and just rely on the protection diodes to clip the over voltage - 150uA is most unlikely to damage anything, de-glitch in software

that will give you about 1/6th of the cycle low and the rest high.

^{simulate this circuit – Schematic created using CircuitLab}

if you add a series capacitor will auto bias via the protection diodes and give about 50% duty cycle

this seems good, but the automibile is a surprisingly rough environment for electronics, load dumps and altenator faults can prioduce spikes of hundereds of volts

so add a transistor to buffer the current instead of letting it reach your logic a small transistor can handle tens of milliamperes base current so fault voltages over one kilovolt will be safely handled with this setup.

^{simulate this circuit}

the diode handles the current for the negstive half cycle and the capacitor allows the bias to self-center giving approximately 50 percent duty cycle. at the output.