Hi John:
> >I presume that the fault current path for the ungrounded
> >SELV pole must be routed through the SELV source to ground
> >as that path is the lowest impedance path to ground. It
> >seems to me that the fault current would have to pass
> >through the transformer winding. In some cases, the fault
> >current may have to pass backwards through the rectifier
> >to ground -- which is an impossibility.
>
> I don't understand that. If the fault current can't flow through the
> source, it will flow through the load. That is OK, if the product
> remains safe. It doesn't have to work after such a fault has occurred.
The load impedance of many SELV circuits is relatively
high. If a mains fault should occur to the ungrounded
SELV pole, then the impedance of the load must either:
1) carry the fault current until the overcurrent device
operates, or
2) present a sufficiently low impedance relative to the
fault impedance such that the voltage on the SELV
circuits does not exceed the SELV rating.
For example, consider a 240-volt mains, a 3-amp mains
fuse, and a 5-volt, 4-amp SELV. To operate the fuse
within 1 minute, the fault current must be at least 6
amps.
The aggregate 5-volt load impedance is E/I or 1.25 ohms.
When the fault occurs between the mains and the ungrounded
SELV pole, 240 volts will appear across the aggregate
1.25-ohm resistance. There is no single component with an
impedance of 1.25 ohms.
Clearly, the aggregate current is sufficiently low to
easily operate the 3-amp mains fuse, E/R or 192 amps.
However, not a wire or PWB trace could carry 192 amps.
So, this scenario is impossible.
We need 6 amps to operate the fuse. So, we need the
aggregate SELV impedance to be E/I, 240/6, or 40 ohms.
40-ohm devices in a 5-volt circuit are reasonable.
However, such devices would be rated no more than
1/2-watt (for the 5-volt circuit).
When the fault occurs, 240 volts is applied across a
40-ohm, 1/2 watt resistance. Since 240 volts, 40 ohms,
and 6 amps is 1440 watts, the question is: What opens
first, the resistor or the fuse?
If the resistor opens, then the SELV voltage goes
to 240 volts, and the scheme is unsafe.
So, in this scenario, the 5-volt circuit would need
to include a 40-ohm, 1400-watt (240 x 6) resistor
to carry the fault current until the fuse operates.
My conclusion is that relying on the SELV load to
operate the primary protective device is not
practicable.
Best regards,
Rich
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