FDP EX-CPL Installer Manual Download

Page: 88 / 91

Page 88 of 91

Rev. 4.20

Problem

Cause

Solution

System fault

It usually appears for the first
installation.

One of the two microprocessors
does not work properly.

Press the resetting key on the motherboard
and type in the requested level code. The
control panel will start up again without losing
the made programming. The loops are put in
out of service. If the correct level code is not
typed in, the system fault will continue to exist.

Control panel under repair.

LOOP OVERLOAD

Properly uninstalled points have been
inserted in the loop.

Remove the last installed points.

DOUBLE ADDRESS

There are two or more elements with
the same address in one loop.

Their elements flash. It is necessary to remove them
from the loop physically and such a point from the
control panel. Afterwards it is necessary to zero the
removed points (refer to the element manual for the
procedure of the address zeroing) and install them
again.

S.C.AFTER ISOLATOR
x-L

In the loop a short circuit and a cut off
on the line +LOOP or –LOOP is
present.

If the line +L is aborted, one of
the two cut off ends short-
circuits itself with the –L.

A loop thin cable might have
come off the clip and the thin
cable has short-circuited itself
with the other one.

The fault message points out the loop where the
problem has appeared (L = 1, .., 4) and the
isolator/side-control panel where there has been the
short circuit (x = A, B, 121, .., 127).
If x=A and L=2, the short circuit will be on the side a
of the loop 2.
If x=123 and L=1, the short circuit will be either
before or after the isolator with address 123 of the
loop 1.
Once the starting point from where to begin the fault
research has been identified, the loop will have to be
put in out of service and to be disconnected from the
control panel. Search for the short circuit along the
line with the tester and afterwards the line cut
resetting the fault.

LOOP SECT. S.C. x-y-
L

In the loop a short circuit is present
bL and –L in the section
pointed out by the line isolators x and
y.

The fault message points out the loop (L = 1, .., 4)
and the section where the short circuit (x = A, B, 121,
.., 127) is present.
If x=A, y=121 and L=2, the short circuit will be on the
side A of the loop 2 and the isolator 121.
If x=123, y=124 and L=1, the short circuit will be
between the isolator 123 and 124 of the loop 1.
Once the section interested by the short circuit has
been identified, the loop will have to be put in out of
service and to be disconnected from the control
panel. Search for the short circuit along the line with
the tester, resetting therefore the fault.

-L/+L SHORT CIRC. L

In the loop L (L = 1, .., 4) a short circuit
has taken place between the line and
the shielded generic mistake.

Put the loop in out of service (if it has not already set
up by the control panel) and disconnect the loop from
the control panel. Search for the short circuit along
the line with the tester, resetting therefore the fault.

±L/SHI SHORT CIR. L

In the loop L (L = 1, .., 4) a short circuit
between the line (+L or –L) and the
shielded has taken place.

In the loop L (L = 1, .., 4) an
interruption has taken place along the
positive line (+L) or negative (-L).

Put the loop in out of service (if it has not already set
up by the control panel) and disconnect it from the
control panel. Search for the interruption along the
line with the tester, resetting there fore the fault.

±L INTERRUPT. L

If –L, there could be an addressable
isolator not added from the control
panel.

To put in out of service the loop and to restart this
with research of new isolator. If the internal address
at this isolator is present in the control panel, to find
the isolator to install and remove the address.
Repeat the procedure to restart loop with research of
new isolator.

PART 18

HOW TO SOLVE THE PROBLEMS