phase sequence change detector

[Vis5254]

I was able to obtain a circuit stating it could be used for phase sequence detector . Does this circuit really do that ??

I'm really confused about the AND gate part dealing up with the R Y B phase signals and how does it really manage to keep its purpose ? Simulations are not working as expected .. Is this is a reliable circuit ??

 

[rjenkinsgb]

I think the logic part will work to some extent, though the LED connections have no chance or working - the output of D4 can never go low due to the base-emitter junction of the transistor and no logic family can give enough output current to hold that output less than 0.6V from positive while sourcing around 100mA to feed R7 & VL1.
(Which would blow most LEDs anyway).

The logic part looks as if it gives some indication; the D2 & D3 latch is set by D1 when both L2 & L3 are high (above zero) which is for 1/6th each AC cycle.

With "forward" rotation it is reset at the end of that time when L2 goes low, before L1 goes high so no output from D4.

With "reverse" rotation the reset is delayed and there is a 1/6th cycle overlap with L1, causing D4 output to pulse low with a 1/6th duty cycle.

If D4 was replaced with an edge-triggered latch (eg. a D type) then L1 used as clock and D3 out as data, it should give a continuous forward or reverse signal.

(All assuming my brain is working properly, before my first coffee of the day...)

 

[Diver300]

I agree that the logic works, but that the LED bit doesn't. When rjenkinsg says "no output from D4", he means that it stays high the whole time. If R7 and VL1 were removed, the circuit would work quite well, with VL2 lighting 1/6th of the time at mains frequency with one rotation, and not at all with the other rotation. The output of D4 would be kept at a high voltage all the time by transistor V, but that wouldn't actually matter. Transistor V would be turned on when the output of D4 tries to go low, and not when the output doesn't try to go low, so the circuit would work.

R7 and VL1 were probably an attempt to make the circuit have one light for one direction, and another for the other direction. However, that won't work as the output of D4 is high all the time with one direction, and high for 5/6 of the time with the other direction, and turning on a light when it is high will result in the light being nearly the same brightness whatever.

There are other things to be aware of if trying to make this circuit. The 12 V supply means that the logic gates need to be work at 12 V, so the obvious choice is a CD4011. https://www.ti.com/lit/ds/symlink/cd4011b.pdf

The negative of the supply has to be connected to the neutral of the incoming supply, so the whole circuit should be treated as being possibly at mains voltage for safety reasons. The zener diodes have to keep the voltage down to a voltage that the logic gates can survive, and the 68 kOhm resistors will produce about 0.5 W of heat each, on a 230 V supply, so you need resistors rated to 1 W or more, and they will get hot.

You could power the circuit from the supply that you are sensing. I've added some components to show how that can be done. It's still all possibly at mains voltage. VS1 - VS3 should be 10 or 12 V, the additional diodes should be the same types as VD1 - VD3 and the capacitor should be 16 V or more, and 100 uF or more. I suggest you increase R8 as I have put in the diagram.

 

[Vis5254]

Thanks for the help sir !

But I think the Transistor part might bring some complications I think . I came across almost a similar circuit online excluding the PNP part . The simulation was like LED was turned ON for always under RYB . And was blinking every 3 seconds on phase reversal !

 

[rjenkinsgb]

That still needs a resistor (or two) in the transistor base circuit!

 

[Diver300]

That still needs a resistor (or two) in the transistor base circuit!

I think that it would work without a resistor. A series resistor is to limit the current, but the IC will only sink about 25 mA so it won't damage things if the is nothing except the IC to limit the current. A base - emitter resistor isn't needed where the drive circuit always makes the base - emitter voltage less than about 0.5 V, which the IC would do.

 

[rjenkinsgb]

the IC will only sink about 25 mA so it won't damage things

That's roughly a quarter of a watt power dissipation in a single gate, it will shorten the life if not just cook it.

Also it's unnecessarily adding to the overall power requirements of the device...

It's just plain bad design practice. Two resistors vastly reduce both dissipation and consumption.

 

[Diver300]

That's roughly a quarter of a watt power dissipation in a single gate, it will shorten the life if not just cook it.

Also it's unnecessarily adding to the overall power requirements of the device...

It's just plain bad design practice. Two resistors vastly reduce both dissipation and consumption.

I agree that a resistor is good practice. I don't see why a second resistor would help.

However the dissipation will be 1/6 of the time, so about 50 mW average, and the package is rated to 500 mW.

The series resistors from each phase will pass around 5 mA each, and any current not needed will flow through the zener diodes, so if the device is powered like that, the overall consumption won't change if the IC takes more power