250.30(A)(2) Supply side bonding jumper

[unsaint34]

In a grounded, service-supplied system, the enclosure of an outside service equipment (meter socket or transformer) is bonded to the grounded conductor within the enclosure. In most cases, PVC conduit runs between the outside service equipment and the service disconnect, and no separate bonding jumper wire between them either(Just phase and grounded conductors). When a fault happens at the outside service equipment, the fault current will travel through the grounded conductor to energize OCPD.

Why doesn't this same configuration apply to a grounded, separately derived system? Why does NEC require SSBJ (supply side bonding jumper) to run from the outside equipment to the first disconnect if the SBJ (system bonding jumper) is installed in the disconnect?

 

[jumper]

I am confused on the question.

There is no grounded/neutral conductor ran to a SDS, so how would a fault clear without the EGC and SSBJ?

http://ecmweb.com/qampa/code-qa-104

 

[roger]

In a grounded, service-supplied system, the enclosure of an outside service equipment (meter socket or transformer) is bonded to the grounded conductor within the enclosure. In most cases, PVC conduit runs between the outside service equipment and the service disconnect, and no separate bonding jumper wire between them either(Just phase and grounded conductors). When a fault happens at the outside service equipment, the fault current will travel through the grounded conductor to energize OCPD.

Why doesn't this same configuration apply to a grounded, separately derived system? Why does NEC require SSBJ (supply side bonding jumper) to run from the outside equipment to the first disconnect if the SBJ (system bonding jumper) is installed in the disconnect?

From the service source the Grounded Conductor serves as both the neutral and the fault clearing conductor, on an SDS this is not the case and the SBJ or EGC is required to clear a fault.


Roger

 

[unsaint34]

That makes sense now, but only to a certain point due to my lack of basic electrical knowledge. I have to ask an apprentice question, please bear with me.

In order for fault current to trip a breaker, the fault current has to travel through the breaker.

In a SDS, a fault at the transformer will cause the fault current to travel through the transformer enclosure, supply side bonding jumper, system bonding jumper then to the grounded terminal of the transformer. So, how is that fault current take the path with the breaker?
If I illustrate the fault current path using the posted picture (the bottom one),... the X2 faults at the transformer, the fault current follows the supply side bonding jumper, then the disconnect enclosure, then the system bonding jumper, then back to the transformer's X0. The fault current cycles but does not flow through the path with the breaker.

 

[RUWired]

If a fault happens at the SDS, the primary fuses protecting the transformer will open.

 

[unsaint34]

If a fault happens at the SDS, the primary fuses protecting the transformer will open.

I see. But if the enclosure was directly bonded to the X0 at the transformer, wouldn't the fault at the enclosure trip the primary side fuse also?

 

[roger]

I think part of the problem is your terminology. What are you calling a "supply side bonding jumper"?

The illustrations properly show the "system bonding jumper". In the bottom diagram if X2 faulted to the transformer enclosure the current would flow through the transformer enclosure to the EGC to the panel enclosure and through the "system bonding jumper" to the gronded conductor (neutral) then back into the XO terminal.


Roger

 

[RUWired]

wouldn't the fault at the enclosure trip the primary side fuse also?

Any fault on the line side of the secondary conductor OCPD would open the primary transformer fuses.

 

[roger]

I see. But if the enclosure was directly bonded to the X0 at the transformer, wouldn't the fault at the enclosure trip the primary side fuse also?

Yes it would which is shown in the first diagram.

Roger

 

[unsaint34]

Any fault on the line side of the secondary conductor OCPD would open the primary transformer fuses.

I'm trying to figure out whether the bonding jumper between the transformer and disconnect is required for that to happen. If there was no such bonding jumper and the transformer enclosure was bonded to the X0, wouldn't the fault at the transformer enclosure open the primary transformer fuses?

 

[unsaint34]

Yes it would which is shown in the first diagram.

Roger

then what is the purpose of the green wire connection between the transformer and the disconnect?

 

[roger]

then what is the purpose of the green wire connection between the transformer and the disconnect?

With out it there would be no fault clearing path to XO from the panel. Follow the arrows showing the fault current flow.

Roger

 

[jap]

If there is a fault in the transformer or anywhere on the primary side of the Service Disconnect Switch the high fault current would seek its path back to its source which is the power company's high voltage line and hopefully blow the high voltage fuse (not shown in the picture) just ahead of the transformer. But not blow the fuses or trip the breaker in the Service Disconnect.

If there is a fault on the load side of the Service Disconnect Switch the fault current would still seek its path back to its source but would hopefully blow the fuse or trip the breaker in the Service Disconnect. but not affect the High Voltage fuse or any Branch Breakers in the Distribution Panel.

If there is a fault on the load side of the branch circuit fuses or breakers in the Distribution Panel the fault would still seeek its path back to its source but would hopefully blow the branch fuse or circuit breaker in the Distribution Panel and not affect the overcurrent device in the service disconnect or the primary high voltage fuse and so on..................

 

[unsaint34]

With out it there would be no fault clearing path to XO from the panel. Follow the arrows showing the fault current flow.

Roger

what if I get rid of the green wire and just bond the transformer enclosure to X0? Would that not open the transformer's primary fuses?

 

[unsaint34]

If there is a fault in the transformer or anywhere on the primary side of the Service Disconnect Switch the high fault current would seek its path back to its source which is the power company's high voltage line and hopefully blow the high voltage fuse (not shown in the picture) just ahead of the transformer. But not blow the fuses or trip the breaker in the Service Disconnect.

If there is a fault on the load side of the Service Disconnect Switch the fault current would still seek its path back to its source but would hopefully blow the fuse or trip the breaker in the Service Disconnect. but not affect the High Voltage fuse or any Branch Breakers in the Distribution Panel.

If there is a fault on the load side of the branch circuit fuses or breakers in the Distribution Panel the fault would still seeek its path back to its source but would hopefully blow the branch fuse or circuit breaker in the Distribution Panel and not affect the overcurrent device in the service disconnect or the primary high voltage fuse and so on..................

Thanks for the thorough explanation. Now I understand that an OCPD only works on the faults on the downstream side. But another question I had was this... NEC 2011 requires the transformer enclosure to be bonded to the disconnect by using supply bonding jumper. However, the requirement is only for a SDS (separately Derived System), That requirement does not apply to a service supplied system. So, I am trying to figure out why that is. In a service supplied system, the transformer enclosure is bonded to the grounded conductor. So, the fault at the transformer will open the utility fuse. By definition, a service supplied system can come from a transformer with the service point on the 2ndary side of a transformer. Now, if the service point happens to be on the primary side, the secondary side becomes a SDS, therefore NEC does not allow me to bond the enclosure to the grounded conductor of the secondary. By "grounded" conductor, I mean the conductor that was connected to the electrode and EGC at the disconnect.

 

[unsaint34]

Okay, now I can explain better. Please refer to the figures in the picture. Figure 1 is my rendering of a SDS (separately derived system). Figure2 is the SSBJ (supply side bonding jumper) required by 250.30(A)(2). I have listed the article below. So, ultimately my question is "why do we need this SSBJ?" In the figure3, X2 ground-faults. In the figure4, I showed the fault current path through the SSBJ. This fault current then opens the primary side fuses.
Now, in figure5, I got rid of the SSBJ and bonded the enclosure to X0. In figure6, X2 ground-faults. In the figure7, the fault current ends up at X0, opening the primary side fuses.

So, why do we need this SSBJ?

250.30(A)(2) Supply-Side Bonding Jumper. If the source of a separately derived system and the first disconnecting means are located in separate enclosures, a supply-side bonding jumper shall be installed with the circuit conductors from the source enclosure to the first disconnecting means.

 

[jap]

Well, what if you take a piece of alluminum flex or sealtite from the transformer to the 1st Means of disconnect shown in your picture #7,,,,,, then arent you back to figure 4 ?

 

[jap]

You either bond the neutral in the transformer or in the 1st means of disconnect but not both.

 

[unsaint34]

Well, what if you take a piece of alluminum flex or sealtite from the transformer to the 1st Means of disconnect shown in your picture #7,,,,,, then arent you back to figure 4 ?

The green wire in the drawing denotes SSBJ. It can be a form of a wire or nonflexible metalic raceway. So, when said I get rid of the SSBJ, I meant there is no reliable electrical connection between the transformer enclosure and the disconnect.

 

[unsaint34]

You either bond the neutral in the transformer or in the 1st means of disconnect but not both.

I understand that only one SBJ (One MBJ in a service-supplied system also) in a SDS according to the code. But I am trying to learn why. The reason for only one SBJ is to prevent a parallel fault current path. However, the figure 5 has no parallel current path either. So I am trying to learn why I cannot make a SDS like figure 5 when doing so does not create a parallel fault current path.