LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Nothing that I can see. Are you sure it's not only two phases rising?
Neutral Current on a 3-wire system
- Thread starter philly
- Start date
- Status
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
Where were you measuring the voltage?
310V sounds like a possible voltage across the neutral grounding resistor.
Yes we do see all three phases rise.
We are measuring voltage between Phase buses and ground bus in MCC.
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Okay, now that we know this is the VFD output, I can't make any meaningful contribution. Sorry.
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
Effectively the VFD output is a separately derived system (AC->DC->AC). Off the top of my head I do not how a L-G fault on the output of the VFD would impact the HRG feeding it.
Are you using some type of phase monitor on the MCC so you can see all 3 voltages at once? Do you have a voltmeter across your HRG?
Are you using some type of phase monitor on the MCC so you can see all 3 voltages at once? Do you have a voltmeter across your HRG?
Yes we do have a fluke 1735 meter where we can see all these voltages at onece. We have each of the voltage phase leads on the 480V bus and the neutral phase lead connected to the ground bus.
I belive when we read across the HRG at one point we read about 154V. Could this be due to the fact that the output of the drive is not 480V but maybe some lower number depending on frequency setpoint?
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
The output of the drive is probably a separate system and should not be impacting your HRG (assuming it is not a regenerative drive).
A voltage across the HRG exists because there is current flow through the resistor (which is probably about 55ohms). A value of 154V would tell me you have an accumulation of about 2.8A of current flowing on your 'ground' grid. the presence of current does not always indicate a fault condition as every system has some amount of normal leakage current.
You need to make sure that your Fluke is not reading the coupling capacitance of your circuit. You could read 310V if you meter input resistance is 10,000 ohms and you have 0.0310A of leakage. I would have to look into how harmonics and EMI/RFI might affect the amount of leakage due to coupling capacitance.
AEMC meters and IPC resistors both have some good 1-2 page articles on measuring coupling capacitance.
A voltage across the HRG exists because there is current flow through the resistor (which is probably about 55ohms). A value of 154V would tell me you have an accumulation of about 2.8A of current flowing on your 'ground' grid. the presence of current does not always indicate a fault condition as every system has some amount of normal leakage current.
You need to make sure that your Fluke is not reading the coupling capacitance of your circuit. You could read 310V if you meter input resistance is 10,000 ohms and you have 0.0310A of leakage. I would have to look into how harmonics and EMI/RFI might affect the amount of leakage due to coupling capacitance.
AEMC meters and IPC resistors both have some good 1-2 page articles on measuring coupling capacitance.
Yes 2.8A of current is just about what we are seeing with a clamp meter when measuring current through the HRG resistors and neutral of the transformer. This current disapears when when motor is shut off. Why would be only be dropping 154V across the HRG? Maybe the output of the drive as mentioned above, or maybe some impedence of the fault itself.
We did remove the motor, and megger it to indeed determine it was shorted to ground, so a ground fault did indeed exist.
Why would the readings change from 277V to 310V on all three phases when drive and motor are turned on? Would the leakage current increase when motor is turned on.
We have since changed out the motor and this issue has disapeared. When the drive an new motor start all L-G voltages stay at 277V and do not increase. So it is obvious that the increase we were seeing was a result of a fault. It doesn't make sense to me though why all three phases would increase.
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
The drive is the source of current for the motor so it is possible that the fault path was motor->ground->resistor->neutral->phase conductor->drive->motor.
Honestly, this is the first time I have had so much information about a VFD supplied motor causing this much current to flow on an HRG.
Honestly, this is the first time I have had so much information about a VFD supplied motor causing this much current to flow on an HRG.
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
I have to disagree with one of the points above. A VFD is _not_ a separately derived system. The key feature of an SDS is galvanic isolation; if you make a connection to any single output leg of the SDS you will not see fault current flow because there is no complete closed circuit.
The output of a VFD is electrically connected to the input, and if you were to connect the output of a VFD to bonded metal, you would be able to follow a continuous closed circuit for fault current.
The thing that makes a VFD 'neither fish nor fowl' is that the connection from input to output is via the rectifiers and switching transistors. This means that a fault on a single output phase of the VFD will cause current flow at different times on _all_ of the input phases. The fault current path will be via the input rectifier, the DC rail, and the output transistors, and depending upon the state of each of these switching elements, a different phase will be connected to the fault.
With a high resistance grounded system, a fault on the VFD output means that each phase will be grounded at different times, rather than having a single phase grounded continuously. As described above, the grounding path and which phase is grounded will be changing moment to moment. The phase-ground voltage will be a composite of the various different waveforms, and probably looks quite interesting on a scope.
-Jon
The output of a VFD is electrically connected to the input, and if you were to connect the output of a VFD to bonded metal, you would be able to follow a continuous closed circuit for fault current.
The thing that makes a VFD 'neither fish nor fowl' is that the connection from input to output is via the rectifiers and switching transistors. This means that a fault on a single output phase of the VFD will cause current flow at different times on _all_ of the input phases. The fault current path will be via the input rectifier, the DC rail, and the output transistors, and depending upon the state of each of these switching elements, a different phase will be connected to the fault.
With a high resistance grounded system, a fault on the VFD output means that each phase will be grounded at different times, rather than having a single phase grounded continuously. As described above, the grounding path and which phase is grounded will be changing moment to moment. The phase-ground voltage will be a composite of the various different waveforms, and probably looks quite interesting on a scope.
-Jon
Winnie this is interesting.
Can you explain more what you mean by different phases being grounded at different times?
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
First, I have to apologize, because re-reading what I wrote, I have to say I was too loose with my language.
Where I said 'different phases being grounded at different times', I should have said 'each phase is _faulted_ to ground at different times'.
In a high resistance grounded system, you have a system node (the transformer neutral) intentionally connected to ground via the grounding resistor. Any other connection between any circuit conductor and grounded metal is a ground fault.
You can similarly describe solidly grounded systems an ungrounded systems. In general you have at most a single intentional connection between the system and ground; everything else is a ground fault.
Now consider a VFD with a ground fault on its output. Which 'input phase' is connected to ground? Which 'input phase' has the ground fault.
Let's start with an analogy: instead of a VFD we have a simple reversing manual starting switch, such as you would find on a small machine tool with a three phase motor. Imagine a connection to ground at the motor. Depending upon the FWD, OFF, REV position of the switch, that ground fault might show up on one or another phase, or might not show up at all (if the switch is open).
A VFD is really just a set of switches, continuously going from one state to another in a well defined electronically controlled pattern. Based upon which switches are closed and which are open, each input phase will in turn get connected to each output phase. Thus a ground fault on an output phase means that each input phase in turn has a ground fault.
In most common VFDs, there are two sets of switches: the input rectifier and the output bridge.
The input rectifier is quite dumb; at any given instant in time, the phase that is most positive is connected to the positive DC bus, and the phase that is most negative is connected to the negative DC bus. As the input goes through its 60Hz repetition, each phase in turn gets connected to each DC bus rail. If the DC rail voltage is high enough, there may be periods when there isn't a complete circuit between supply conductors and the DC bus.
The output bridge is controlled to generate the desired voltage and frequency. Each output terminal is alternately connected either to the positive or negative DC bus.
Each input phase is _sometimes_ connected to each DC rail by a closed switch, and each output terminal is _sometimes_ connected to each DC rail by a closed switch. Thus if there is a ground fault on the output of a VFD, that fault will _sometimes_ be connected by a chain of closed switches to each input phase.
Fault current flowing on the output of a VFD flows from all of the input phases, each at a different time depending upon which switches are closed.
I hope that my babbling on this topic is somewhat more clarifying than confusing
-Jon
ATSman
ATSman
- Location
- San Francisco Bay Area
- Occupation
- Electrical Engineer/ Electrical Testing & Controls
Can someone tell me how to attach a file on this forum? Would like to attach a report on a problem we encountered on a people mover system that I think backs up what Winnie is trying to explain here. It has to do with the "Striking Ground Phenomenon" on an ungrounded or HRG system where a striking (intermittent) ground fault on one phase causes the voltage on the other phases to increase sharply puncturing insulation throughout the dist. system. In our case it burnt up all the ground detection light control transformers in all 5 substations simultaneously setting off the smoke alarms. The problem was caused by one of the train car propulsion motor's brushes coming loose from the holder and hitting the frame.First, I have to apologize, because re-reading what I wrote, I have to say I was too loose with my language.
Where I said 'different phases being grounded at different times', I should have said 'each phase is _faulted_ to ground at different times'.
In a high resistance grounded system, you have a system node (the transformer neutral) intentionally connected to ground via the grounding resistor. Any other connection between any circuit conductor and grounded metal is a ground fault.
You can similarly describe solidly grounded systems an ungrounded systems. In general you have at most a single intentional connection between the system and ground; everything else is a ground fault.
Now consider a VFD with a ground fault on its output. Which 'input phase' is connected to ground? Which 'input phase' has the ground fault.
Let's start with an analogy: instead of a VFD we have a simple reversing manual starting switch, such as you would find on a small machine tool with a three phase motor. Imagine a connection to ground at the motor. Depending upon the FWD, OFF, REV position of the switch, that ground fault might show up on one or another phase, or might not show up at all (if the switch is open).
A VFD is really just a set of switches, continuously going from one state to another in a well defined electronically controlled pattern. Based upon which switches are closed and which are open, each input phase will in turn get connected to each output phase. Thus a ground fault on an output phase means that each input phase in turn has a ground fault.
In most common VFDs, there are two sets of switches: the input rectifier and the output bridge.
The input rectifier is quite dumb; at any given instant in time, the phase that is most positive is connected to the positive DC bus, and the phase that is most negative is connected to the negative DC bus. As the input goes through its 60Hz repetition, each phase in turn gets connected to each DC bus rail. If the DC rail voltage is high enough, there may be periods when there isn't a complete circuit between supply conductors and the DC bus.
The output bridge is controlled to generate the desired voltage and frequency. Each output terminal is alternately connected either to the positive or negative DC bus.
Each input phase is _sometimes_ connected to each DC rail by a closed switch, and each output terminal is _sometimes_ connected to each DC rail by a closed switch. Thus if there is a ground fault on the output of a VFD, that fault will _sometimes_ be connected by a chain of closed switches to each input phase.
Fault current flowing on the output of a VFD flows from all of the input phases, each at a different time depending upon which switches are closed.
I hope that my babbling on this topic is somewhat more clarifying than confusing
-Jon
TT
Tonytonon
When you click on "quote" or "post reply" scroll down after you type your response and you will see an area called "additional Options". Use the manage attachements feature to add an attachement.
ATSman
ATSman
- Location
- San Francisco Bay Area
- Occupation
- Electrical Engineer/ Electrical Testing & Controls
tonytonon
I do not understand the purpose of the test button on your attached schematic. It appears the primary of these PT's are connected in a wye arrangement and the test button is connected between the star point and ground. By operating this test button you will only be disconnecting the transformer bond to ground.
I do not see what this is supposed to do?
ATSman
ATSman
- Location
- San Francisco Bay Area
- Occupation
- Electrical Engineer/ Electrical Testing & Controls
That's a good question. Off hand I am not sure how it is suppose to work but it does light the 3 lights when pressed. This is a standard scheme used by Cutler-Hammer (Eaton) It's been 6 years since I worked on this project and I will have to research this. Can anyone explain this so I don't have to research?? :grin:
physis
Senior Member
Nope, it's not.
I can't agree with that. I would rather say that a seperately derived system is a second current source aside from a utilility service.
I'd even go further to say that if you had no utility connection and ran a building from a wind mill, it would still be called an SDS. Even though it's hard to imagine a single power sourse as being "seperate". But don't blame me , it's Article 100 that does that.
And worse than that, the output current of the VFD comes from the service you're trying say it's seperately derived from. You can't have a seperately derived system from a single service. Although the code's a tad bit unclear about it.
- Location
- Illinois
- Occupation
- retired electrician
Are you saying that a transformer is not SDS?
- Status