Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] Favorite low phase shift 240Vrms isolated zero cross detector

Status
Not open for further replies.
F

fourtytwo

Guest
Hi everybody I have a requirement for a very low phase shift galvanically isolated zero cross detector that will operate on UK-grid 240Vrms. What is your favorite trusted solution ?
I find transformers have to much phase shift and opto-couplers need to dissipate quite a lot of power in there droppers.
 

47K in L and N lines, bridge using 1N4148 diodes and a sensitive opto-coupler (I use 6N139). Runs virtually cold and has almost zero phase shift and dead zone.

Brian.
 
  • Like
Reactions: fourtytwo

    F

    Points: 2
    Helpful Answer Positive Rating
Hi Brian thank you for your reply, may I ask what your opto load circuit is like ?
I am using dual PC817D that I think have a similar sensitivity in inverse parallel in common collector configuration with a 33K load and 5V supply each connected to a PIC MPU pin (10pf).
If I increase RL the turn-off time gets to long but to get the deadtime down to a few hundred uS I have to use ~40K input resistors :(
So maybe the 6N139 is a lot better then the 817D I will investigate :)
 

The load is a 4.7K resistor to VDD feeding a PIC.

The biggest problem is the turn-on threshold of the LED in the optocoupler. The 6N139 has a high CTR (~400%) and switches when the LED current is >0.5mA whereas the PC817 has a low (~50%) CTR and needs around 5mA.

You can reduce the turn on requirement a little by using small Schottky diodes in the bridge, something like BAT85 which have a Vf of about 0.3V at 0.5mA. As long as you wire the opto directly across the bridge and put the resistors in the AC side they will never see high reverse voltages.

Don't forget that the delay between real zero crossing and the optocoupler output switching is constant so you can compensate for it in software to some extent.

Brian.
 
  • Like
Reactions: fourtytwo

    F

    Points: 2
    Helpful Answer Positive Rating
I am sorry I had not realized the 6N139 was a darlington till after I posted and I was not allowed to edit it! Normally I think of darlingtons as to slow but it seems this one is faster than most I have seen. I assume your VDD is 5V so IL is ~1mA and you make no connection to the base. You can help your turn on voltage by using two opto's in inverse parallel like me and eliminate the bridge. I also introduce phase lead by adding a capacitor across part of the dropper chain. It seems a spice model for the 6N139 is a long outstanding problem according to websearches I might try another post here in case someone knows of a similar device. Sadly I have a box full of 6N138's but no 6N139's so I wont be able to try this out till after the holiday. Thanks very much for your help
Roger
 

Well, at least I know who has the other half of the Worlds stock of 6N138s!

There is a trick you might wish to try but obviously no promises as to the outcome, if you add a resistor to ground from VB on the 6N139 it reduces the sensitivity a little but greatly speeds its switching speed. You might be able to recover a little more speed while still keeping it faster then a PC817. Try values between 10K to 20K and see what happens. I've got about 50 6N139s in stock here but a workbench on overload already so I can't help with experiments.

There is another completely different approach which might be worth looking at, use a small DC-DC converter to generate an isolated DC supply and use it to power a comparator driven from the AC line. The comparator output would drive an optocoupler LED to give the isolation you need.

Brian.
 
  • Like
Reactions: fourtytwo

    F

    Points: 2
    Helpful Answer Positive Rating
47K in L and N lines

Do you mean two 47K on both line and neutral sides? What is wrong with one 100K only on one side?
 

Correct - the value actually needed is about 100K but splitting it into two resistors overcomes the resistor voltage limitation. I'm not sure common mode noise is a problem as it feeds an optocoupler LED anyway but from a safety aspect, I would rather there was a resistor in each leg in case the line wires were reversed and it got touched.

Brian.
 

Well, at least I know who has the other half of the Worlds stock of 6N138s!

There is a trick you might wish to try but obviously no promises as to the outcome, if you add a resistor to ground from VB on the 6N139 it reduces the sensitivity a little but greatly speeds its switching speed. You might be able to recover a little more speed while still keeping it faster then a PC817. Try values between 10K to 20K and see what happens. I've got about 50 6N139s in stock here but a workbench on overload already so I can't help with experiments.

There is another completely different approach which might be worth looking at, use a small DC-DC converter to generate an isolated DC supply and use it to power a comparator driven from the AC line. The comparator output would drive an optocoupler LED to give the isolation you need.

Brian.

Hello Brian I always have lots of stock of things I don't use, so if you want anything useless I am your man :) Yes I think your right the base connection is very handy for tuning the tradeoff so now I can't wait to get some in the new year to try :)

Yes a very interesting idea about the op-amp & isolation barrier that would decouple the sensing current from the opto-current and allow a normal transistor coupler to be used without high power dissipation, of course a step up in complexity!

I have looked everywhere for other alternatives, logic/gate-drive opto's came to mind but they all have threshold currents >=5mA.

Ohh I almost forgot I also thought of using a constant current regulator on the hot side but it did not cut the dissipation by a huge amount and again circuit complexity is my enemy!
 
Last edited by a moderator:

Hello Brian I always have lots of stock of things I don't use, so if you want anything useless I am your man :) Yes I think your right the base connection is very handy for tuning the tradeoff so now I can't wait to get some in the new year to try :)

Yes a very interesting idea about the op-amp & isolation barrier that would decouple the sensing current from the opto-current and allow a normal transistor coupler to be used without high power dissipation, of course a step up in complexity!

I have looked everywhere for other alternatives, logic/gate-drive opto's came to mind but they all have threshold currents >=5mA.

Ohh I almost forgot I also thought of using a constant current regulator on the hot side but it did not cut the dissipation by a huge amount and again circuit complexity is my enemy!

A constant current source is a good idea that both limits power dissipation and should result in sharper turn-on with less variation.

Not complicated - A depletion fet and a resistor form a high voltage current source quite well (any depletion can replace this for a lower cost):
https://ixapps.ixys.com/DataSheet/DS98729A(IXCP-CY10M90S).pdf
 

quoting asdf44:
Not complicated - A depletion fet and a resistor form a high voltage current source quite well (any depletion can replace this for a lower cost):
I'm puzzled why you suggest that. The problem with opto-sensing the current is the need for the voltage to rise enough after crossing zero for the LED to light sufficiently. There will always be a short dead period between zero and LED Vf being reached. I'm not sure how a constant current source helps that, especially in an AC circuit. I agree the current waveform is unipolar sine and for most of the cycle it just wastes power but as long as peak cycle voltage doesn't result in excess LED current I don't see that as a problem. Rectifying first re-introduces the Vf of the bridge rectifiers again.

Brian.
 

Thank you for the link asdf44, I dismissed fet solutions due to the high AK voltage (~5) that unfortunately makes the deadzone around zero cross rather wide. I experimented with bipolar versions but there the bias current was high enough to dissipate quite a bit of power, high voltage bipolars being a bit short on current gain! Again part of the problem is the very wide compliance voltage range added to the minimum operating voltage being very low to maintain a narrow deadband. I hope I don't sound like I am dismissing alternatives its good to see them but I guess the application is somewhat fussy! For now I shall pin my hopes on the 6N139 unless something else pops out of the woodwork!
 

quoting asdf44:

I'm puzzled why you suggest that. The problem with opto-sensing the current is the need for the voltage to rise enough after crossing zero for the LED to light sufficiently. There will always be a short dead period between zero and LED Vf being reached. I'm not sure how a constant current source helps that, especially in an AC circuit. I agree the current waveform is unipolar sine and for most of the cycle it just wastes power but as long as peak cycle voltage doesn't result in excess LED current I don't see that as a problem. Rectifying first re-introduces the Vf of the bridge rectifiers again.

Brian.

That's my fault I suggested it as a potential solution to the power dissipation problem, however as you say most forms of current limit effectively substantially increase the diode vf and contribute to the deadzone. Unipolar/bipolar is easily dealt with by various circuit configurations.

I cannot reply in realtime due to moderation of my posts, sorry.
 

quoting asdf44:

I'm puzzled why you suggest that. The problem with opto-sensing the current is the need for the voltage to rise enough after crossing zero for the LED to light sufficiently. There will always be a short dead period between zero and LED Vf being reached. I'm not sure how a constant current source helps that, especially in an AC circuit. I agree the current waveform is unipolar sine and for most of the cycle it just wastes power but as long as peak cycle voltage doesn't result in excess LED current I don't see that as a problem. Rectifying first re-introduces the Vf of the bridge rectifiers again.

Brian.

Am I missing something? The 100k resistor proposed is going to have a similar dead zone because it needs voltage before it reaches the required forward current of the opto. Fourtytwo cites 5V dead zone as a problem yet a 100k resistor will provide only 50uA at 5V. No likely to be a guaranteed ON through the opto I wouldn't think.

Now consider the nature of the linked current topology. The depletion is 'normally on' with a resistance <100ohm and the source resistor will be on the order of 1k. Before the depletion starts clamping off it's just a 1k resistor and will turn on the opto when a 1k resistor would. This is surely an improvement over 100k I think?
 
  • Like
Reactions: fourtytwo

    F

    Points: 2
    Helpful Answer Positive Rating
Am I missing something? The 100k resistor proposed is going to have a similar dead zone because it needs voltage before it reaches the required forward current of the opto. Fourtytwo cites 5V dead zone as a problem yet a 100k resistor will provide only 50uA at 5V. No likely to be a guaranteed ON through the opto I wouldn't think.

Now consider the nature of the linked current topology. The depletion is 'normally on' with a resistance <100ohm and the source resistor will be on the order of 1k. Before the depletion starts clamping off it's just a 1k resistor and will turn on the opto when a 1k resistor would. This is surely an improvement over 100k I think?

Now I am feeling enlightened, I spend so much of my life with enhancement mosfets I think I got that completely muddled and I can see where you are coming from. Indeed even at 1mA opto current that is near where I am working atm your depletion fet solution would only add 1.1V to the threshold of the LED. Just to be awkward I am using a bipolar solution with two opto's so I would either need two fets or one and a rectifier (not good due to increased VF).

I reckon with the 817D's my threshold current is around 100uA at 5V grid and am using 40K in droppers for a dissipation of ~1.5W. Using a depletion fet assuming VGS ~5V then for 200uA R=25K and Pd is ~48mW a huge saving :) But there is a fly here and that is the increased value of Rs that also increases the switching voltage threshold so probably the actual power will have to be higher.....compromise again :)

Potentially however changing to a more sensitive opto may also allow a substantial power reduction provided I can keep the speed up so I will try that first, I simply did not know photo-darlingtons existed that were fast enough!

Thank you both for all your help and it seems the Xmas holiday's now intervene (no postage, no parts) with the serious stuff in life......merry Xmas to all.
 

Note that if you can tolerate ~1.5W/~3mA and set the depletion source to the same 3mA you get ~2k source resistor.

Don't you need at least one extra diode anyway since the optos surely can't block mains? So rectifier would be one more......


An alternative strategy would be to generate a traditional DC bus and use a proper comparator. For the same power you're currently dissipating (3mA/1.5W) you could easily power a comparator from the line. Or alternatively price out what it costs to use an isolated supply powered from your secondary. It may come out favorably considering the cost and size of the high voltage parts you're proposing (fets or R's).

These IC style isolators provide a small amount of power in a small package and signal isolation:
https://www.analog.com/en/parametricsearch/11036

And this is a cost effective isolated supply not much bigger than an opto
https://www.digikey.com/product-det...ions-inc/NXJ1S1205MC-R7/811-3095-2-ND/5697916
 

The constant current idea is perfectly valid but I'm still not convinced it will work in practice. If we assume the zero crossing has to be detected at each zero and not whether there is a (say) positive voltage or no positive voltage, detecting half cycles only, the constant current circuit has to work on both half cycles. I do not know of any simple AC constant current devices and adding a bridge rectifier to make it DC will negate the purpose of using it by adding 2xVf back into the equation.

Maybe the solution is half hardware and half software: only look for half cycles, whether with or without a CC circuit. Shunting the optocoupler LED with an external reverse wired signal diode will protect it from reverse polarity. That eliminates the bridge rectifier and associated voltage drop. Then in software, to look for the rising and falling edges of the optocoupler output instead of expecting a pulse from it.

Brian.
 

Hi,

To be true I didn'd go through each word in this thread.
What I am missing: There is a requirement for "low phase shift" ...but I don't see any specification.
And I can't find the information what the signal is used for...

Without a value you may discuss a long time and never get to an end.

****
Zero cross.
I'm designing high precision phase control for the industry. And I use zero cross circuits to align the firing angle oir the SCRs.
I found out, that the "phase shift" is a minor problem for a precise regulation.

There are other, like
* timing jitter caused by overtones / distortion of the input (mains) signal
* multiple zerocross trasitions
* voltage fluctuations of mains, causing output signal fluctuations (even with precise phase)

Klaus
 

What you say is quite true Klaus. Often overlooked it another problem, when trying to detect ever lower voltage thresholds to be nearer 'true' zero crossing the amount of noise becomes more problematic. Faster responding detectors are also more likely to give false outputs when seeing spikes on the lines. There is some merit in feeding the AC through a low pass filter to clean it up then compensating for the phase shift in software. At least the shift added by the filter is predictable and constant.

There may even be some sense in detecting rise/fall reversals at peak voltage rather than at zero so the noise becomes almost irrelevant compared the background. A comparator sensing the voltage with a time delay in one input will do that and again give predictable and removable phase shift.

Brian.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top