Zero detector for medium voltage line

Hi all,
I need a zero detector for a Medium-voltage line that has capability of determining the polarity of following half-cycle.
I am thinking of using AC optocoupler for zero detecting along with a simple circuit consist of (diode+DC Optocoupler+RC circuit at the other side of the optocoupler) which determine the polarity of the previous half cycle.
Do you have any suggestion or better circuit in mind?
I would be really grateful if you share your thoughts.

thanks

In usual electrical engineering terms "medium voltage line" designates 1 to 30 kV AC. Might be difficult with standard optocouplers.

I would prefer a dual DC optocoupler topology to get polarity and zero crossing.

Hi,

For isolationg at those high voltages I prefer POF.
Like Broadcom`s HFBR type.

Klaus

Klaus;

What is the meaning of the acronym "POF"?

Alternative method:

Use an ordinary opto-coupler (if voltage and isolation permit).
Detect the rising edge of the voltage (LED turning on) then after a short delay, check if it is still on.
Then detect the falling edge (LED going out).

The edges give you the zero crossing points and sensing after the delay tells you the polarity. The delay should be less than half a cycle in duration.

Brian.

Thank you all for sharing your opinions and comments.
The voltage is 1.2kV for now.

I would prefer a dual DC optocoupler topology to get polarity and zero crossing.

Click to expand...

I think low voltage side would be complicated, in this case. As Betwixt has stated before, we need to detect rising edge, falling edge, etc.

For isolation at those high voltages I prefer POF.
Like Broadcom`s HFBR type.

Click to expand...

I am not sure what POF stands for. I would be grateful if you give me more details.

Please see attached preliminary sketch and share your opinion.



Thanks

Hi,

POF = Plastic optical fiber
https://en.m.wikipedia.org/wiki/Plastic_optical_fiber

Broadcom HFBR:
.

Klaus

You need to protect the lower optocoupler better. A parallel diode wired to conduct when the LED is reverse biased will be safer and give a better zero crossing indication as it lets the LED conduct without the forward voltage drop of the existing diode. In both cases you have to allow for the voltage rise delay before enough LED current flows to operate the sensor side. With ~1.2KV you also have to take into account the LED curent at voltage peaks and the voltage ratings of the series resistors.

You could consider a different logic gate arrangement: If you use a single opto-coupler (the lower one in your diagram) and split the output three ways, one is the polarity sense, one goes directly to one input of an XOR gate and the other to the other XOR input through an RC delay network, it will produce an XOR output pulse on both rising and falling edges which coincide with the zero crossings. It uses fewer components to give the same result.

Brian.

Betwixt, Thank you for your response.
actually, since I have to detect the existence of the voltage I might use AC optocoupler ( the circuit should differentiate between the power outage and negative half-wave).
About diode, doesn't utilizing of parallel diode instead series increase series resistor rating power? I admit that there is a voltage drop over resistor which introduce slight error but the resistor power will reduce considerabily. According my calculation I already need 25W resistors.

Adding a parallel diode does double the current but it ensures the LED PIV is never exceeded. Consider that in reverse polarity, the 1.2KV blocking is shared between the diode, LED and resistor but the proportion each sees depends on leakage current and it may still be excessive at the optocoupler. You should at least add a resistor across it to bleed a little current away.

You have other options, a capacitive dropper for example, it does introduce a phase shift but the amount is predictable and constant so you can take it into account later in your circuit. A transformer based isolation circuit may be an option and it would make the optocoupler redundant as well.

I predict the problem you will have is a poorly defined zero crossing point anyway. Consider that an optocoupler LED needs a certain current to be able to activate its sensor side. If you fix the dropper resistor so it passes say 10mA at voltage peaks, it is likely the current will be too low well before the fall or rise from the zero point. You could be over 100V into the cycle before it registers a change in level. You can fix that fairly easily with some active circuitry on the AC side, either by using a comparator or even a constant current generator to the LED so it turns on at much lower voltage without being overloaded at voltage peaks.

Brian.

HI Brian,
First of all, thank you for your response and comments. I really appreciate that.

You can fix that fairly easily with some active circuitry on the AC side, either by using a comparator or even a constant current generator to the LED so it turns on at much lower voltage without being overloaded at voltage peaks.

Click to expand...

you mean I need to design driver for the diode of the optocoupler? So I need a low power supply related to neutral point of power side, transistor, etc doesn't it add to system complexity? do you have simpler way in mind? I would be grateful if you share it.

Thanks.

The problem with using a simple resistor is the amount of voltage it has to drop. As you pointed out the resistor needs to be large and will run very hot (~12W dissipation).

The bigger problem though is that the way the zero crossing is detected is to expect the optocoupler output to go high when the LED is not lit up but go low again as soon as the voltage rises above zero. Unfortunately, that wont happen, the reason being the LED needs a few mA before it will emit enough light for the sensor to see it.

For example, suppose the peak current through the LED is 10mA when 1.2KV is present and the LED needs say 5mA to operate the internal phototransistor: The series resistor should be, R=(V-Vf)/I which gives a value of 119.84K and a power dissipation of: W=V*I = 11.98W. I'm assuming the voltage is RMS as it is obviously AC.

For the LED to pass 5mA, the voltage has to be 599.2V. So the optocoupler thinks you are at zero crossing until the voltage has reached around 600V. (not very accurate!!)

Your options are to drop the voltage using a transformer to something much lower, maybe 6V or less then using a smaller value resistor. This poses two problems, one is that you hint the voltage may be variable which makes transformer choice difficult, the other is that the transformer will introduce some phase shift. The phase shift will be constant so it is easy to 'subtract' it afterwards.

The other option is to use a lower value resistor but prevent the LED current becoming too high by feeding it through a contant current generator. It will still go out when there is no voltage but you can make it light up at much lower voltage without damaging it as the voltage increases. The drawback to that aproach is the CC generator also dissipates quite a lot of the heat.

You could consider a capacitive voltage dropper, that removes the heat problem but unless you use it in conjunction with a CC generator it still gives a wide 'Zero' zone. If you want to try that method, pick a capacitor value that has Xc a little lower than 120K and add a fixed series resistor to make up the difference. For example if you are using 50Hz AC, you could replace 119.8K with a 27nF capacitor and a 2.2K resistor in series (effectively 120K) and the heat dissipated drops from ~12W to 0.22W. Make sure you use a suitable rated capacitor and a fuse in case it fails short circuit!

The most practical solution that gives high accuracy it probably to use a differential amplifier powered from the low voltage (logic circuit) side of the circuit and monitor the AC line through very high value resistors. The circuitry is more complicated but you get most precise zero crossing detection and you can monitor the polarity as well.

Brian.

Hi Brian,
Thank you for your help and patient to type. I really appreciate it.
I have tried Iso OPAMPs for low voltage(600V). About 25$ per circuit, power resistors extra. I will read your post again carefully, see if I can come to an innovative solution to solve shortcoming of the methods or not.

Thank you again.