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This is a follow-up to Is this line detector safe?

I'm re-designing the front-end based on some great feedback from the community. I've decided to move to the following topology:

diagram

  • The phase of the source is set to 90º to model the worst case where a user closes a switch at the top of the line curve; in this case the caps would not yet be charged and R1 + R2 would have to absorb all surge energy
  • The surge current would be \$168.6 \text{V} / 1.6 \text{k} \Omega / 2 \approx ~ 53 \text{mA} \ll 150 \text{mA}\$, so the opto would be fine
  • During surge, the TVS would still not conduct, because \$(168.6\text{V}/2)+1.1\text{V}=85.4\text{V}\$, which is smaller than Vbrmin=86.5V
  • During surge, each resistor would see a peak power of ~4.5W. If I use the CF14JT1K60, based on its pulse derating curve, it would support 4W for ~0.1s, or 50W for ~1ms. Since we have 4W for less than 1ms, we should be quite OK.
  • D2 is shown disconnected - in reality it would be connected but I wanted to simulate the circuit as if the protection zener never quite conducts under typical operation. I intended for it to only conduct if there's a transient above the usual line voltage.
  • R1 absorbs energy in the event of a transient spike if D2 conducts.
  • C1 and C2 are chosen to relieve the burden on R1 and R2 so that their typical non-surge power dissipation is small, and decrease typical current to the opto while still keeping it on. The intent is to provide about 10mApk to the opto. At 60Hz, the total impedance is about

\$2 \left( 1.6\text{k}\Omega + \frac 1 {2\pi j \cdot 60 \text{Hz} \cdot 300 \text{nF}} \right) \approx 18.0 \text{k}\Omega \angle -79.4º \$

which allows ~9.4mApk.

  • This equates to ~6.7mA RMS; the resistors each then dissipate about 53mW RMS of power each. This is well below the 250mW for which they're rated.

Transient Simulation

Voltages:

voltages sim

Current:

current sim

Resistor peak power:

power sim

Have I missed anything? Is my method sound?

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  • \$\begingroup\$ Are you expecting this circuit to provide zero crossing information or is it simply to detect that AC is present? \$\endgroup\$ – Jack Creasey Sep 25 '18 at 4:54
  • \$\begingroup\$ @JackCreasy the latter, although I haven't shown the second half. \$\endgroup\$ – Reinderien Sep 25 '18 at 12:27
  • \$\begingroup\$ OK, if you are looking for power failure signals, you should be detecting the peak of each half sine. This way you can detect both brownout and power loss. Most reasonably highly loaded power supplies may have holdup times in the 16 - 300mS range so depending on what you need to do on power loss the timing may be quite critical. \$\endgroup\$ – Jack Creasey Sep 25 '18 at 15:51
  • \$\begingroup\$ @JackCreasey interesting - but I'm not looking for power failure, per se, simply the presence of any voltage on this particular mains line. \$\endgroup\$ – Reinderien Sep 25 '18 at 16:00
  • \$\begingroup\$ Curious, did you build this circuit? Is it working as you expected? I ask because I am endeavoring to do the same thing. \$\endgroup\$ – Alex Oct 6 '19 at 7:41
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You missed nothing important. I did this most every month for 15 years at a surge manufacture plant. While most people might cut corners and have a single RC based reactive power supply it is wise to double up on R and C at times, especially if Vin is prone to surges. Two 3 watt resistors spread the heat and generally had a breakdown voltage of 400 volts, so 2 in series made it 800 volts, with a little safety margin.

Also R dissipates the real current as heat, so even for 120 VAC we used 2 18K 3W resistors in series with an AC type optocoupler. For 600 VAC (Canadian products) we used 2 47K 3W resistors, always the flame-proof type. We only used capacitors (X2) when we needed actual power from one or more phases. It was no more than 60mA with a range of 50mA to 70mA, so capacitor tolerance could be 5%.

Due to UL and ISO demands above 277 VAC we HAD to use 2 capacitors rated 630 VDC to be sure shoot-through was not possible. Please see this link about the capacitor issue and why 2 is much better than one, even for the best grade of capacitors.

You seem to have a keen insight as to how to make reactive sensing or reactive power supplies safe, even during a surge event. By the way, if used at the main service entrance panel you are allowed to use ground as a return path for current, as the neutral to ground strap would be in that panel.For outlets or sub-panels neutral must be the return path.

From our point of view shoot-through meant trusting a single capacitor not to arc internally during a surge, which is why series resistors are mandatory for us. They limit surge currents and turn-on currents if the power is switched on during a cycle peak (90 \$^{o}\$). The resistors protect any down-stream diodes and opto-couplers by limiting peak possible currents to their safe region. A well built reactive power supply is supposed to be safe to short out, as the capacitors set the overall current and resistors limit surge currents. We also used 1 mH inductors to block RF or surges with sharp rise/fall times.

Remember the capacitors are transparent to RF and surges with sharp rise/fall times, so the resistors are an insurance policy that works well. Adding an inductor (as an RF choke) is a cost and space-allowed decision.

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  • \$\begingroup\$ For shoot through current definition I read this - maximintegrated.com/en/glossary/definitions.mvp/term/… so it basically means shorting the supply? \$\endgroup\$ – Reinderien Sep 25 '18 at 1:24
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    \$\begingroup\$ From our point of view it meant trusting a single capacitor not to arc internally during a surge, which is why series resistors are mandatory for us. They limit surge currents and turn-on currents if the power is switched on during a cycle peak. The resistors protect any down-stream diodes and opto-couplers by limiting peak possible currents to their safe region. A well built reactive power supply is supposed to be safe to short out, as the capacitors set the overall current and resistors limit surge currents. We also used 1 mH inductors to block RF or surges with sharp rise/fall times. \$\endgroup\$ – VTNCaGNtdDVNalUy Sep 25 '18 at 1:36
  • \$\begingroup\$ Remember the capacitors are transparent to RF and surges with sharp rise/fall times, so the resistors are an insurance policy that works well. Adding an inductor (as an RF choke) is a cost and space-allowed decision. \$\endgroup\$ – VTNCaGNtdDVNalUy Sep 25 '18 at 1:40
  • \$\begingroup\$ I put these comments in the answer so they are not deleted over time. \$\endgroup\$ – VTNCaGNtdDVNalUy Sep 25 '18 at 1:45
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    \$\begingroup\$ @Sparky: "I did this most every day for 15 years at a surge manufacture plant." So it was you generating all those surges!? \$\endgroup\$ – Transistor Sep 25 '18 at 17:44
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If all you are looking for is to detect that AC is present within reasonable limits (what are your limits??), I'd use a peak detector based on your TVS.

Something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

I've shown two schematics here, one with a 33V clamp and one with a 96V clamp, just to show options. Obviously you only need to select one of them. The circuit will run under the simulator to see the differences in clamp voltage.

enter image description here

The signal you get from the opto-isolator will occur at some unknown time between initial conduction of the TVS (with its variable threshold) and the CTR of the opto. However the signal will be symmetrical around the AC input sine wave peak at 90Deg which you wanted to achieve.
There will be some minor fluctuation due to the capacitance of the TVS of course, but this should not concern you for your particular application.

I'd suggest that the P6KE100CA would be a better choice of TVS, but either would work.

The issue of surge suppression I see as separate to your issue of sensing presence, and you should of course have surge suppression on the AC input. This would typically be a VDR surge suppressor close to your AC input.

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    \$\begingroup\$ Why remove the capacitors (your resistors now need to dissipate 0.6 and 1W)? And what is the purpose of the two branches - what kind of circuit would the two clamp voltages be fed to? \$\endgroup\$ – Reinderien Sep 25 '18 at 18:15
  • \$\begingroup\$ The two branches simply allow you to simulate both ….you would only need one of them of course. Getting rid of the capacitors gets rid of the uncertainty of pulse position (phase compared to AC input). You have to also include the power dissipated by the TVS of course, but I don't consider it excessive. In the end it's up to you, but you tried to achieve a 90deg sensing phase shift with capacitors and can see it is ineffective. At no time did you say absolute low power was the objective. There are ways to achieve very low power sensing, but it's more complicated. \$\endgroup\$ – Jack Creasey Sep 25 '18 at 19:00
  • \$\begingroup\$ The circuit is not intended for "90-sensing", simply to sense voltage. The 90 figure is just for the simulator, to start the line voltage at its peak to model a worst-case surge. Other than that, phase (and a few ms of delay) don't matter at all. \$\endgroup\$ – Reinderien Sep 25 '18 at 19:03
  • \$\begingroup\$ I'm aiming for lower power dissipation while still maintaining simplicity. I'd like to be able to run without a heatsink, or even in a plastic case with poor thermal characteristics, without having to worry. \$\endgroup\$ – Reinderien Sep 25 '18 at 19:04
  • \$\begingroup\$ @Reinderien Then why use resistors at all in your proposed circuit? You could simply use two (two for safety) 600nF mains rated capacitors and the opto-isolator and get about 14mA RMS or you 300nF for about 7mA RMS. A VDR would protect the input from surges. \$\endgroup\$ – Jack Creasey Sep 25 '18 at 19:43

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