U2481B
| Model | U2481B |
| Description | Automotive Lamp Outage Monitor |
| PDF file | Total 11 pages (File size: 153K) |
| Chip Manufacturer | ETC |
TELEFUNKEN Semiconductors
U2481B/ U2482B
Two equations for calculation of the shunt resistance:
R
shunt
= R
A
<L/W
cn
(with R
A
= Cu sheet resistance)
R
shunt
= V
Th
/ 1/2 I
LAMP
Thus the length of copper shunt is calculated as:
L = 2<V
Th<
W
cn
/ R
A<
I
LAMP
L = 215 mm
For a reasonable pc-layout a meander-shaped shunt
resistor is recommended. The high lamp currents may
cause hot spots at sharp edges of the copper shunts. That
may deteriorate accuracy of the measurement. Therefore
it is recommended to layout the copper shunts with
smoothed curves.
In accordance to figure...the meander may be formed by
4 straight tracks (length L
S
each) and 3 connecting 180°
ares (length L are each). If the mean are radius is selected
to r = W
cn
the are lenght becomes L
are
=
p
W
cn
.
Thus the total lenght is
L = 4 L
S
+ 3 L
are
= 4 L
S
+ 3
p
W
cn
With L = 215 mm the track length becomes
L
S
+
1 (L
*
3
4
L
S
= 47.8 mm
p
W
cn
)
Application Hints
Layout Recommendations for Copper
Layer Shunts
Lamp outage monitor systems can be produced most cost-
efficiently if stamped shunt resistors are replaced by
copper layer shunts which are generated with the pc board
layout.
The U2481B and the U2482B are suitable for this
application because of their comparator thresholds,
which are compensated in reference to the temperature
characteristic of copper.
A constant lamp current, I
LAMP
= V
Th
/R
sh
with threshold
voltage V
Th
= f(T) and shunt resistor R
sh
= f(T), is
achieved if the comparator threshold and the shunt
resistor have identical temperature characteristics. With
the temperature coefficient of copper a
cu
= 3.9<10
–3
1/K, a copper shunt changes its nominal value by 52% if
the automotive ambient temperature range of t
amb
= –40
to +95°C is taken into consideration.
Examples for sheet resistances of copper shunts
(T
amb
= 25°C):
R
A
= 0.5 mW/square (35.1
mm
layer thickness)
R
A
= 0.25 mW/square (70
mm
layer thickness)
How to Lay Out Copper Shunts (figure 3)
The width of the copper trace has to be selected in
reference to a low current-effected temperature increase.
The copper trace must be capable of peak currents which
do not blow the fuse. The peak currents are specified by
the car manufacturers.
Example:
A 7.5 A fuse allows a peak current of 26 A (1 s), 15 A
(10 s) or 10 A (60 s).
The copper shunt length has to be calculated between the
two sense connections to the comparator. The connection
of the common reference input of double and triple
comparators has to be considered carefully.
Calculation example for a copper shunt used with a 4 W
bulb.
I
LAMP
= 0.325 A
Failure criterion: I = 1/2<I
LAMP
10 A fuse is capable of I = 13.5 A
Copper layer thickness: 70
mm
Comparator threshold voltage (U2481B, U2482B):
V
Th
= 3.5 mV
Assumed copper width for temperature increase
DT
cu
< 50°C:
W
cu
= 2.5 mm (13.5 A, 70
mm)
Using Fuses as Shunt Resistors
This cost saving method can be used if the following
assumptions are fulfilled:
–
–
Each lamp needs a dedicated fuse
The fuse socket is mounted within the monitor
module to connect the sense linesare suitable for
this monitor
task because of their comparator thresholds are
compensated for the fuse temperature coefficient
of a
F
= 4.1<10
–3
1/K
–
Calculation Example
for a 55 W Bulb and a
7.5 A fuse:
Measured voltage drop across the fuse: V
F
= 52 mV
Measured current: I
LAMP
= 4.2 A
Calculated resistance: R
F
= 12.4 mW
Selected comparatore threshold:
V
Th
= 1/4 V
F
= 13 mV (typically);
With a lamp current I
LAMP
< V
Th
/R
F
=< 1.05 A
the comparator detects a blown fuse.
Rev. A1: 21.08.1995
Preliminary Information
7 (11)
