Questions on Isolation Panel

[ActionDave]

Nice .

Glad you approve.

 

[mbrooke]

That looks like the "product over sum" formula which is used to find the total resistance of 2 resistors in parallel. So the bottom number is the impedance of the capacitor (1,200,000 ohms) plus the resistor (1000 ohms).

I don't think its 100% correct to use it in this case since they have a capacitor in parallel with a resistor.

But since the resistance of the resistor is so much less than the capacitive reactance, the formula gives a pretty close answer. (The current through the capacitor will be much less than the current through the resistor.)

So their conclusion (which is valid) is that we can basically ignore the one capacitor. You can see in the next diagram (#7) that is what they did to simplify the circuit.

If I am correct, the impedance from each wire to ground is based on a parallel capacitor and resistor. The resistor is usually the smaller of the two values, while capacitance is the bulk of the ground leakage.


When a phase faults to ground, all of the combined capacitance and resistance from the other phase and transformer (all circuits, conductors, buss bars ect) govern the amount of fault current. If one faulted say "A" leg, every piece of conductor on "B" leg will contribute current. If one was to start opening breakers the fault current would start to drop for every circuit disconnected since every time a breaker is opened a "B" leg is disconnected, removing its resistance to ground and capacitance to ground from the system.

When the system is designed so that the fault current is below 5ma its considered safe.

 

[steve66]

If I am correct, the impedance from each wire to ground is based on a parallel capacitor and resistor. The resistor is usually the smaller of the two values, while capacitance is the bulk of the ground leakage.

Yes, the resistor is basically the resistance of the wire insulation, and its usually high enough to ignore. The capacitor is the capacitance between the wire and conduit. You want to keep it small - that's why they use wire with a certain type of insulation, avoid the use of wire lubricant, keep the runs short, and use oversized conduits among other things.

When a phase faults to ground, all of the combined capacitance and resistance from the other phase and transformer (all circuits, conductors, buss bars ect) govern the amount of fault current. If one faulted say "A" leg, every piece of conductor on "B" leg will contribute current. If one was to start opening breakers the fault current would start to drop for every circuit disconnected since every time a breaker is opened a "B" leg is disconnected, removing its resistance to ground and capacitance to ground from the system.

The "fault" is usually considered to be through a person (usually the "patient", although it could also be a doctor working in the OR). The point is the resistance of the patient is really small compared to the capacitance reactance mentioned above, so the size of the capacitance really determines the fault current.

When the system is designed so that the fault current is below 5ma its considered safe.

The fault current is usually called the "hazard current" since that is what could flow through a patient if they somehow became connected between the isolated 120 volts and ground.

This is what they are trying to show with Diagram 7 of the link in the original post.

 

[mbrooke]

Yes, the resistor is basically the resistance of the wire insulation, and its usually high enough to ignore. The capacitor is the capacitance between the wire and conduit. You want to keep it small - that's why they use wire with a certain type of insulation, avoid the use of wire lubricant, keep the runs short, and use oversized conduits among other things.




The "fault" is usually considered to be through a person (usually the "patient", although it could also be a doctor working in the OR). The point is the resistance of the patient is really small compared to the capacitance reactance mentioned above, so the size of the capacitance really determines the fault current.



The fault current is usually called the "hazard current" since that is what could flow through a patient if they somehow became connected between the isolated 120 volts and ground.

This is what they are trying to show with Diagram 7 of the link in the original post.

True, not disagreeing, with today equipment that is a much less frequent occurrence. But still in a wet location where GFCI trips cant be tolerated this really helps.

 

[Pharon]

Ironically, the problem with isolated power panels is that while they don't have 5mA nuisance trips, they tend to have a limited amount of breaker capacity, which can lead to overload tripping conditions, due to the migration toward a more heavily concentrated electric load in these rooms.

For this reason, in our hospitals we've gone in the other direction - namely back to regular 42 circuit panels feeding self-testing GFCI receptacles. This way, we can basically run dedicated circuits to each one and not worry what the BioMed folks plug into it for overload. We also have not heard any issues with nuisance tripping this far.