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Current at Star Point on WYE generator(4)

The star point on a Wye generator connects to Phases A, B, C, and ground through a nuetral transformer. In a perfectly balanced system, is there current at the star point? 

(2) peterb (Electrical) 
2 Dec 01 18:49 
No current in the star point connection for a balanced system. Zero sequence or triplen harmonic currents can flow in this neutral connection if it is grounded, the value of the current depending on the neutral grounding resistance. 

Normally in a balance load there is no flow of current in the star connected Neutral point. If neutral is solidly grounded then there is a possibility of flow of current for outside faults through the neutral. 

230842 (Electrical) 
3 Dec 01 11:03 

OK, so far I've been told that there is no current at the star point. But I'm still not convinced.
My reasoning is that, according to Kirchoff(?), Current at any point in a circuit is identical to any other point in the same circuit. So how can full current be coming out of the high voltage side of the generator if there is no voltage at the star point (low voltage side of the generator)?
Can anyone clear this one up for me? 

peterb (Electrical) 
3 Dec 01 20:13 
Of course there is current flowing in the phase leads on the neutral end of the generator windings, before the star point connection. The currents sum vectorially at the star point. The sum of the balanced 3phase vectors is zero  I ang(0) + I ang(120) + I ang(240) = 0 Draw the vectors and add them & you will see right away that the balanced current triangle is closed, with a resultant current of zero flowing in the star point neutral connection. Unbalanced conditions are another matter. 

SteveKW (Electrical) 
3 Dec 01 20:46 
I have to disagree. If there is no current flowing through the "star point" even in a balanced system then you could OPEN the star IN THE GENERATOR and the generator would still work! Sorry...but that just wont fly! Your vector sums may add to zero ON THE NEUTRAL, but as Dakota stated you have Kirchoffs law of current ... and if A phase has 10 amps flowing out of it, it has to flow back into it at the STAR POINT to complete the circuit. Don't confuse the NEUTRAL with the STAR POINT ... they are NOT the same. 

peterb (Electrical) 
4 Dec 01 7:15 
What I said was that the currents flow on the phase leads on both sides of the winding. The 10 A current in the A phase flows out of the generator winding on the line side lead and flows into the generator winding on the neutral side lead. The A phase current on the neutral side is the resultant of the vector addition of the (B + C) phase currents at the star point, for a balanced system, which is what is under discussion. The resultant neutral current is zero. Certainly the generator wouldn't work with the star point open  that is self evident. 

I think we both see two different original questions. His question dealt with the STAR CONNECTION on a perfectly balanced generator. So there is no current on the neutral. No argument there.
For me, as a motor rewinder and electrician, the "star" is physically in the generator and the neutral is outside of the generator to carry unbalanced current back to the generator.
Let's remove the "neutral" from the question and the generator, which I think is what Dakota wanted to do. Is there current through the "star" on a loaded generator when there is NO neutral?
Looking at Dakota's second response, he thinks he is being told NO! I think the answer is YES! Which is it? 

230842 (Electrical) 
4 Dec 01 11:20 
MotorMan1K and Peterb, you both are right: A star (wye) connection is achieved when three branches (in that case the three windings of the generator) are interconected in such a way that one termination of every branch converge in a common node called "star point" while the other three terminations belong to external three phase line. Star point can or cannot conected outside to the neutral line. Under fully balanced conditions, the three currents come out through the line terminations and no current goes in or out the star center through neutral line. Julian 

peterb (Electrical) 
4 Dec 01 11:43 
Dakotahawk  Sorry if there was any misunderstanding of the original question and answer. Just to make the point absolutely clear   The generator is equipped with six (6) leads  Three (3) leads are connected to one end of each phase winding and to the generator output terminals  call these A, B, C  The other three (3) leads are connected to the other end of each phase of the generator winding  call these X, Y, Z  The load is connected to leads A, B, C  Leads X, Y, Z are connected together to form the generator star point  The star point is connected to ground through a neutral grounding transformer  Whatever current flows in lead A also flows in lead X, and correspondingly for B and Y, C and Z (this is the principle of generator differential protection)  At the star point, the currents in leads X, Y and Z are added vectorially. If the system is balanced, the vector sum is zero, if the system is unbalanced or if there are triplen harmonics present, the zero sequence or summated harmonic current will flow in the neutral connection 

alex444 (Electrical) 
4 Dec 01 11:54 
Hiya all,
just my 2 cents worth .... it may just be a question of semantics/perspective ...
There are 3 currents flowing through the STAR point, it happens that  and this is the beauty of the Tesla 3 phase design  at the STAR point all 3phase currents CANCEL out to zero ONLY ... as everyone has pointed out, in a balanced system. So therefore you get zero volts  no current ...
I hope this helps


SteveKW (Electrical) 
4 Dec 01 20:01 
Just to clarify, I posted as Motorman earlier from work.
At one particular instant in a generators cycle one phase can be ZERO which means no current is passing through it. That leaves two phases now in series with one another. Those two voltages add (Vectorally) and ... with a load (lets use a 10 amp load) across them, those two phases now push current through the load.
In a series circuit all currents are EQUAL. That means that through the "star point" you have 10 amps flowing through it. So the answer is YES. Current does pass through the star point regardless of weather or not the generator is balanced.
I may be rusty on my theory and say this wrong (but that's never stopped me before) but here is what your "vector sum" is telling you... When you add them up so you get ZERO you are (for a lack of me to explain it better) doing the same as calculating power factor. When you have a power factor of ONE you are in unity (everything is in phase).... When you have a vector sum of zero in a generator you also have "unity". It does not mean the current is zero...it means that there is no unbalance outside of the generator and its all in phase with each other. That's all the vector sum is telling you ... it all adds up to zero at your X Y coordinate system. That means all the current is staying "inside" that XY point and not wandering outside of it.
Again... All a vector sum of zero means is that its "all in phase". It does not mean there is no current passing through the center of XY or the star point.
So Dakota, the answer is yes. You cannot escape the fact that current is flowing and that in order for that to happen you have to have a complete circuit... a series loop. So YES ... current will ALWAYS be flowing through the "star point" whenever there is a load.
I hope that makes it a little more clear. 

peterb (Electrical) 
4 Dec 01 21:07 
SteveKW  Sorry, but the vector sum is in no way analogous to power factor  there is absolutely no correlation. What the vector sum means, in practical terms, is that if you put a clamp on ammeter over the three phase leads (on either side of the generator), the reading will be zero for a balanced loading condition. The vector sum does NOT mean that all is in phase  it means that when you add the balanced outofphase currents together the net result is zero. Taking your case of say the A phase being at its instantaneous zero value, the B and C phases at that instant are equal and opposite  what instantaneously flows in on B flows out on C and the net result is still zero.


SteveKW (Electrical) 
4 Dec 01 23:09 
Pterb, I disagree... PF in my example is analogous of you vector sums! Well, at least I think so. Let me quote you... "What the vector sum means, in practical terms, is that if you put a clamp on ammeter over the three phase leads (on either side of the generator), the reading will be zero for a balanced loading condition." I agree100% ...the vector sums are zero ... no current flowing through your NEUTRAL ... the currents cancel each other out so the ammeter sees nothing ... that does not mean there is no current flow ... there is. Put an equally balanced three phase load on the generator drawing 10 amps and your ammeter will still show zero while clamped on all three leads. "The vector sum does NOT mean that all is in phase  it means that when you add the balanced outofphase currents together the net result is zero." Hmm...I a agree... Here's my simple definition. The vector sum of "zero" does mean all is in phase and canceling each other out ... your proof was in the fact that when the vector sum is zero the ammeter will show zero when clamped on all three legs ... and that's still with a 10 amp load. "Taking your case of say the A phase being at its instantaneous zero value, the B and C phases at that instant are equal and opposite  what instantaneously flows in on B flows out on C and the net result is still zero." B & C are 120 degrees apart, not 180! Or am I missing something there? Are you saying the current is zero? Maybe we arguing both sides of the same coin.... Won't be the first time for me... Let's go back to Dakotas original question, "In a perfectly balanced system, is there current at the star point?" Do you still say "no"? 

Wow! I'm learning a lot from this! Mostly that I don't know as much as I'd like to know.
This question stemmed from a conversation I had at work. I am a HydroElectric Operator for Seattle City Light. While an electrician and an apprentice were working on CTs at the star point of one of our generators, I asked if there was current flow at that point. The electrician said yes, the apprentice said no.
However, it's been pointed out to me by another operator at work, Current cannot be measured at a point. It can only be measured between two points. With that in mind, it makes the question about current "at the star point" unanswerable (I think).
Maybe it would be more correct to say that we're measureing current flow "passing the star point". If that's the case, then do we go to the vector sum addition of currents? Aaarrrgggghhhh!


230842 (Electrical) 
5 Dec 01 5:00 
Peterb continues being right. I feell, SteveKW, you have to take into consideration the qualitative differences between instantaneous current values and phasors. These latter are mathematical "images" of time dependant sinosoidal variables whose algebraic performance allows an eassy calculations in frecuency domain and not in time domain. Phasor do not take zeros unless current is zero all time along. Star point in a node where are connected three windings and, in some cases, a fourth wire, the so called neutral. Kirchhoff laws state that for every instant, the sum of current values flowing in (or out) a node is zero. In a balanced three phase system, the three generator windig currents are sinusoidal time functions: Ia = Ip x sin(wt) Ip = current peak Ib = Ip x sin(wt  2pi/3) Ic = Ip x sin(wt  4pi/3) It is eassy to verify that these three currents sum zero in every instant time. If so, the fourth wire (neutral) current must be zero and can be suppressed with no effect for the rest of the system. Julian 

Let's back up to Peterbs earlier post. " The generator is equipped with six (6) leads  Three (3) leads are connected to one end of each phase winding and to the generator output terminals  call these A, B, C  The other three (3) leads are connected to the other end of each phase of the generator winding  call these X, Y, Z  The load is connected to leads A, B, C  Leads X, Y, Z are connected together to form the generator star point  The star point is connected to ground through a neutral grounding transformer  Whatever current flows in lead A also flows in lead X, and correspondingly for B and Y, C and Z (this is the principle of generator differential protection)  At the star point, the currents in leads X, Y and Z are added vectorially. If the system is balanced, the vector sum is zero, if the system is unbalanced or if there are triplen harmonics present, the zero sequence or summated harmonic current will flow in the neutral connection "
Let's take an instantaneous look at B & C while A has gone to zero volts. Lets make B flow to C. We now have a path that goes B to the LOAD and back to C. Then C comes out at Z which is connected to Y and completes our little loop back to B. That's a series loop and a complete circuit. If 10 amps goes through the load and ... in a series circuit ALL the currents are EQUAL ... then that means 10 amps is flowing through Z & Y. Peterbs points out that Z &Y are part of the star point. So the answer is YES, current does go through the star point.
Show me how that answer is "no".
I don't want to hear theory that the exact calculated phaser center (your node) happens to be one molecule of wire in the star point and no current flows in that one molecule. Dakota just asked the simple question is there any current going through the star point. And when he asked that he was looking at wires ... not a single point calculated down to a point impossible to physically measure.
Now that I have said that I can see what your point is when you say no current is flowing. But, again put an ammeter on any of the three wires going to the star point and you will measure a current.
Dakota, maybe the correct answer is both. There is NO current through the EXACT calculated center of the star point but every other molecule of wire in the star point carries the rest of the current!
How does that sound?
SteveKW. 

peterb (Electrical) 
5 Dec 01 9:40 
OK, now it seems that Dakotahawk really wanted to know whether current flows in the leads on the neutral side of the generator winding. The answer is of course YES (see my first post of Dec 4). My original point was that there is no resultant current at the star point. Yes, current actually flows between two points  however, we measure current flow at a particular point with a current transformer. CTs located in the generator neutral side leads are used for differential protection as well as backup protection of the generator  this location provides protection whether the generator is on line or not, where backup protection connected to the line side CTs will only provide protection when the unit is on line. SteveKW, I agree that we are on the same coin, but we're going to have to agree to disagree on your PF analogy.
Trying to sum the whole thing up   Current flows in the phase leads on the neutral side of the generator, as well as on the line side phase leads  For balanced loads, there is no resultant current flow in the neutral lead connected to the generator star point 

Boy do I love these forums. I hope I came across clearly and others got as much out of this as I did.
I am glad we are on the same coin now. As for my PF analogy, we can save that for another day!
SteveKW. 

Just for fun, lets unbalance the load. The current at the star point is still zero. We know that the sum of currents entering any point is always zero. 

peterb (Electrical) 
6 Dec 01 15:52 
Yes, but now the zero sequence (unbalanced) current will flow in the neutral  see above! 

SteveKW (Electrical) 
6 Dec 01 18:32 
I have a simple question...
If I make a light bulb out of three filaments and "star" connect them in the bulb and then bring out the three leads so they can make the "star" of my generator, will the bulb burn when the generator load is "balanced" or will it only burn when the generator is "unbalanced"?


peterb (Electrical) 
7 Dec 01 7:50 
A simple answer  it is abundantly clear from everything that has been said above that load current flows at all times in the phase leads connected to the star point. Current will only flow in the neutral lead connected to the star point when there is an unbalance condition involving ground or when triplen harmonic currents are present. 

The argument that current flow can only be measured between two points does not negate the fact that current is defined as the flow of electrical charge carriers. They are measured in amperes which are defined as 6.24 X 10 24th power electrons moving past a specific point in one second. In a balanced three phase generating system, the current flowing in, out, around or through a star point has to occur or the generator would not work. The argument has been made that when added by vectors, the phase currents sum to zero and they do, however they still flow. There are electrons flowing in, out, around and through the star point. They are equal to the rms value of single phase peak current and they are constant. They have four possible routes of travel and as long as everything is in balance, they will use three.The answer as I understand your question is yes. The voltages when added in vectors equal zero and that is why no current flows through the neutral lead to ground. The star point and ground are at equal potential constantly. 

Thanks Dipstick! I think you summed up quite clearly the "current flows" side of the argument. PeterB gets the attaboy on the "no current flow" side of the argument. It's interesting that you can argue either side of this question and make complete sense. This has turned into quite a discussion onthejob also. 

jbartos (Electrical) 
9 Dec 01 17:24 
Suggestions: 1. Differentialmode currents cancel in a balanced voltage supply and load system, i.e. Iad+Ibd+Igd=0 2. Commonmode currents can flow through the neutral and ground even for the balanced supply and load. (Iad+Igcm)+(Ibd+Igcm)+(Icd+Igcm)=3Igcm where 3Igcm=commonmode currents, which may be caused by various causes, e.g. by a dc component. Hypothetically, one can connect a battery between the star neutral point and ground to generate the commonmode currents. 

Guest (Visitor) 
9 Jan 02 17:24 
I have recently come across a similar issue in dealing with medium voltage drives. Does anyone have any comments on the commonmode voltage problems (voltages that cause shift from the neutral point in medium voltage motors) associated with retrofitting medium voltage motor with new medium voltage VFDs?
The waveform created by the switching algorithm is Additive PWM and 9 step, in cases where the gated switches simultaneously connect to the + or the  busway, a more severe neutral point shift is created with respect to voltage across teh motor windings. This is not a big deal with LVM because of the inverter wire that is so widely used (spike resistant), as well as the fact that the voltages are not so high. However, with 4160V this can be a real problem, especially with retrofits (old insulation sytems). It has been determined that this problem can be resolved by employing the combination pf an input isolation xfmr and a complicated control scheme that involves a Neutral Point Clamp. the switches are 3300V IGBTs. Can anyone explain how the two work together to solve this problem? I can sort of visualize it, but its complex as hell!
Input xfmr has 12 taps, 22.5, 7.5, 7.5, and 22.5, (so it provides 24 pulses to the rectifiers) 

TheDOG (Electrical) 
9 Jan 02 22:29 
tuffone, Differential mode currents are caused by VSD's because of the switching of the IGBT's  voltages at the motor terminals are not a sine wave and therefore the vector sum of the three phases does not equal 0. See jbartos' post above. Is the motor connected in star or delta? And what is "complicated control scheme that involves a Neutral point Clamp."?




