FUSE TIME/CURRENT CURVE
CURRENT IN AMPERES X 10 AT 13800 VOLTS
.5 .6
.8
1
2
3
4
5
6 7 8 9 10
2
3
4
5
2
3
4
5
6 7 8 9 1000
2
3
4
5
6 7 8 9 10000
1000
900
800
700
600
500
400
400
300
300
200
200
FS-1
S&C
SMU
SMU-40
100E
100
90
80
70
60
50
40
20
10
9
8
7
6
5
40
30
20
10
9
8
7
6
5
4
4
3
3
2
2
1
.9
.8
.7
.6
.5
.3
.2
.4
.3
.2
.1
.09
.08
.07
.06
.05
.1
.09
.08
.07
.06
.05
.04
.04
.03
.03
.02
.02
.01
.5 .6
.8
1
2
3
4
5
6 7 8 9 10
2
3
4
Within this area
fuse will be
damaged but
current may or
may not be
interrupted.
1
.9
.8
.7
.6
.5
This side of curve:
Fuse element intact,
no fuse damage
.4
Note long time
melting point
100
90
80
70
60
50
This side of curve:
Fuse element melts,
current flow
interrupted.
30
TIME IN SECONDS
6 7 8 9 100
TIME IN SECONDS
1000
900
800
700
600
500
5
6 7 8 9 100
2
3
4
5
6 7 8 9 1000
2
3
4
5
Wide plot in
instantaneous
range shows
fuse to be noncurrent limiting.
.01
6 7 8 9 10000
CURRENT IN AMPERES X 10 AT 13800 VOLTS
1-16
©Quadrelec Engineering Corp.-2004
The left side of this curve is the minimum melt curve and the right side the maximum clearing curve. The solid band is actually shaded between these two
curves. If this fuse was rated as a current-limiting type fuse, then the right side of the curve would extend down below the plot instead of tending to flatten out
as the one depicted. Published fuse curves are only valid for a fuse that has no load prior to injection of the test current and a certain ambient air
temperature (25-30 degC )is present. If fuse is fully loaded or if the air temperature is any higher, the whole curve would shift to the left. A full loaded fuse
curve would shift about 20% just due to being full loaded. The actual shape of the curve would change due to loading and temperature. In actuality the entire
curve would not shift the same amount according to IEEE C37.48. More shifting would actually occur in the long time range. It is nevertheless the practice of
most coordination engineers to shift the entire minimum melt side of the curve 20% to the left to compensate for a fully loaded condition. In the absence of
other data IEEE C37.48 states that open power fuses and cutouts should be derated an additional 0.4% per degC above 25 degC. It is always best to get the
manufacturers data for actual derating factors for both preloading and temperature. Why is all of this important? Remember these effects when we get to the
discussion of coordination time intervals.
Note that the 100 A fuse depicted in this plot can handle almost 200 A continuously. This gives us a rule we can use for setting adjustable trip breakers.
How? If a fuse properly protects a cable having a certain ampacity, a breaker mimicking the fuse operation should also provide adequate protection. If you
had an adjustable trip device such as a relay mimicking this fuse curve, what set point value could you use as a pickup point to protect a 100A
rated ampacity cable?
Copyright © Quadrelec Engineering Corp., Tulsa, OK
1-16
PLOTTING DEVICE CURVES VS TRANSFORMER
CONNECTIONS
Transformer winding connections are also a factor when plotting log-log
curves.
TX-1
BUS-2
TX-1
0.00
0.87
0.87
1.00
0.00
0.00
1.00
1.00
1.00
BUS-2
0.
1
kV
TX-1
BUS-1
0.50
0.50
1.00
0.58
0.58
0.00
Phase
A,B,C Ipu
TX-1
1 MVA
0.1 - 0.1 kV
10%
BUS-1
L-L
fault
0.50
0.50
1.00
BUS-1
1.00
1.00
1.00
0.
1
BUS-1
L-N
fault
0.58
0.58
0.00
3-phase
fault
kV
System
Vpri = Vsec
1.00
1.00
1.00
•
BUS-2
BUS-2
1-33
©Quadrelec Engineering Corp.-2004
If transformers are connected Y-Y, Y-YG, YG-YG, or delta-delta, all secondary faults will reflected into the primary by the transformation ratio. If a delta-wye
or delta-zigzag connection is used, we must also compensate the log-log plot by an additional factor to handle the ratio of primary to secondary current. In
the above example I have made the transformation ratio 1:1 so that you can see the effect of the transformer connection for different faults. In each diagram
from left to right are shown the A,B, and C phase fault current values in per-unit. Notice for 3-phase faults no correction is necessary. The primary and
secondary current are the same. For L-N faults the primary currents will only be 58% of the secondary currents. From the secondary current standpoint this
makes the primary device appear 1/.58 or 1.732 times larger. For the L-L fault the ratio of largest phase current to secondary current is 87%. This has the
effect of making the primary side device 87% smaller.
Copyright © Quadrelec Engineering Corp., Tulsa, OK
1-33