Phase Sequence Indicators

[ed downey]

This is the same calculation only using Nodal analysis.

 

[ronaldrc]

Question

Now that we are on this subject again. I know it doesn't make any difference which leg it is A, B or C as long as they are the right sequence.

Since

ABC=CC
ACB=CCW
BCA=CC
BAC=CCW
CAB=CW
CBA=CCW

But is there anyway of knowing that A is A, B is B, and C is C ?

 

[ronaldrc]

Reposted to correct a mistake

Question

Now that we are on this subject again. I know it doesn't make any difference which leg it is A, B or C as long as they are the right sequence.


CW=clock wise
CCW=Counter clock Wise

Since

ABC=CW
ACB=CCW
BCA=CW
BAC=CCW
CAB=CW
CBA=CCW

But is there anyway of knowing that A is A, B is B, and C is C ?

 

[mikeames]

I don't see how this would work.

You start with a 60Hz signal, and use it to drive a frequency divider so as to pulse an LED, say at 1Hz.

Now you take 3 of these 60Hz signals, displaced in _time_ by about 5.5mS. You take each of these signals and run it through a circuit as described above. Won't the output be three LEDs, each pulsing 1x per second, displaced in time by 5.5mS?

-Jon

As each phase peaks it turns on the respective LED. So you get three LEDs blinking in the same sequence as each phase. It really does nothing but reduces the frequency to the point where you can visually see the phase rotation. No phases peaks at the at the same time as any other and thats what your following.

 

[winnie]

As each phase peaks it turns on the respective LED. So you get three LEDs blinking in the same sequence as each phase.

Okay. So if you have three 1Hz signals each 120 degrees apart and driving an LED, you would see a 'rotating' pattern of the blinking lights. Not only do I believe that; I've done that, generating rotating patterns of blinking LEDs.

What I don't get is this: If you have 3 60Hz signals with 120 degrees phase between them, and you run each of these signals through a frequency divider, what is the phase angle difference of the resulting outputs? I see nothing in frequency dividing process that holds the relative phase angles in a fashion that makes the lights blink in the right way.

-Jon

 

[weressl]

Here is a very simple crude frequency divider. If you take the source voltage and reduce it, you could use 3 of the following circuits to run 3 LED's. Then you could visually see the the three pase sequence. You would need to reduce the frequency more than what is shown by altering the values of R1 and C1.

I don't think this would work as you would see three short flickers one-next-to-the-other then a pause. You still would not be able to distinguish WHICH phase blinks first, second and third.

 

[mikeames]

Why a pause?

It is my understanding that this would be a crude way but the OP did ask for a DIY circuit.

I believe the phase angles would be roughly maintained. The blinking LED would correspond to the phase its connected to telling you when its peaked in relation to the other 2.

 

[weressl]

Why a pause?

It is my understanding that this would be a crude way but the OP did ask for a DIY circuit.

I believe the phase angles would be roughly maintained. The blinking LED would correspond to the phase its connected to telling you when its peaked in relation to the other 2.

This is a divider, so it will trigger on say every 20th cycle. The pause will be 20/60th or 1/3rd second. The three phases will still be 1/60th of a second apart from each other so unless you control the relationship between the three COUNTERS with a master clock and trigger each counter differently then you will not see anything else, but a burst of flicker then a pause.

 

[mikeames]

I understand what you are saying and I think the 555 circuit would do what you describe, however the 4017 is actually a decade counter. So by using only 1 of the LEDs as an output you will have effectively created a divide by 10 circuit. I don't believe The 4017 circuit would not have the issues you described with the 555.

The U2 is a 4017 decade counter.

 

[weressl]

I understand what you are saying and I think the 555 circuit would do what you describe, however the 4017 is actually a decade counter. So by using only 1 of the LEDs as an output you will have effectively created a divide by 10 circuit. I don't believe The 4017 circuit would not have the issues you described with the 555.

The U2 is a 4017 decade counter.

Where is the driver signal input on this circuit?

 

[mikeames]

Pin 14 on the 4017 is the clock or frequency input. Everything that is connected to pin 14 can go away and be replaced by one of the phases.

 

[weressl]

Pin 14 on the 4017 is the clock or frequency input. Everything that is connected to pin 14 can go away and be replaced by one of the phases.

So where are the other phases and how do we get a rotation indication?

 

[mikeames]

You would need three of these circuits. One for each phase. Your would have to figure the rotation out yourself by looking at which LED is on and what phase its connected to. I don't think it would pose the same problem that the 555 circuit would.

 

[weressl]

You would need three of these circuits. One for each phase. Your would have to figure the rotation out yourself by looking at which LED is on and what phase its connected to. I don't think it would pose the same problem that the 555 circuit would.

It still would not work without a master clock over the three chips so each will start the count at the same time.