Datasheet

LT6110
26
6110fa
For more information www.linear.com/LT6110
applicaTions inForMaTion
temperature then the wire’s resistance vs temperature is:
R
HIGH
= R
LOW
• (1 + α • (T
HIGH
– T
LOW
)).
An approximation to the temperature rise in a wire due
to current can be derived from the wire’s resistance vs
temperature equation using the wire’s resistance increase
vs safe operating current. If R
LOW
is the wire resistance at
a low current and R
HIGH
is the wire resistance at a higher
current and T
RISE
is equal to T
HIGH
– T
LOW
then the tem-
perature rise in a wire is:
T
RISE
(°C) = 256.4 • (R
HIGH
/R
LOW
– 1).
Table 4 is a list of measured copper wire resistance ver-
sus current
at 20°C for an arbitrary group of 18AWG
to
30AWG wires.
Example: Find the wire temperature rise for 3A flowing in
a 28AWG wire. The 28AWG wire on Table 4 has 62.31mΩ/
ft R
LOW
resistance at 1A and 65.23mΩ/ft R
HIGH
resistance
at 3A.
T
RISE
for 3A is equal to 256.4 • (65.23/62.31 – 1) = 12°C.
An LT6110 wire drop compensation design requires reli-
able information
of wire resistance and current capacity.
Published
copper wire tables are a convenient quick-start
guide to copper wire information. However accurate copper
wire data is obtained by actual
measurements of samples
of copper wire to be used from a reputable manufacturer.
A statistically small sample of copper wire is sufficient for
measurements (the average measured mass resistivity
deviation of a large sample of copper wire is only ±0.26%).
The International Annealed Copper Standard of mass
resistivity is:
153.28 • 10
–6
(Ω-kg)/m
2
in Metric and
31.39 • 10
–6
(Ω-lb)/ft
2
in English units.
Mass resistivity is the product of Resistance/Length and
Mass/Length and is useful for estimating the weight of
copper wire required and its cost (the cost of copper wire
depends on its weight and the price fluctuation of copper
in the commodities market).
The weight of copper wire is:
153.28 • 10
–6
(Length in m
2
)/(Resistance in Ω) in kilograms
or 31.39 • 10
–6
(Length in ft
2
)/(Resistance in Ω) in pounds.
Example: Find the weight of one hundred thousand feet of
18AWG wire and compare it to the weight of a 24AWG wire:
Table 4 shows 6.5mΩ/ft for 18AWG and 22.43mΩ/ft for
24AWG.
The weight of the 18AWG wire is:
(31.39 • 10
–6
) • [(100000)
2
/(6.5 • 10
–3
• 100000)] =
483 pounds.
The weight of the 24
AWG wire is:
(31.39 • 10
–6
) • [(100000)
2
/ (22.43 • 10
–3
• 100000)]
= 141 pounds.
The weight of the 18AWG is 3.4× the weight of the 24AWG
.
Using
an LT6110 simplifies wire drop compensation and
provides the option to specify the smallest size and lowest
cost of copper wire.
The US Department of Commerce, National Bureau of
Standards Handbook 100 is a comprehensive source of
copper wire information.
Power Dissipation
The LT6110 power dissipation is at a minimum for I
+IN
100µA or less. If the I
+IN
current is at its specified maximum
of 1mA or greater then the maximum power dissipation and
operating temperature must be considered. The LT6110
power dissipation is the sum of three components:
V
IOUT
• I
IOUT
,
V
REG
• (I
+IN
+ I
SUPPLY
) and
I
LOAD
2
• R
SENSE
(if the internal R
SENSE
is used)
Example of an extreme power dissipation case:
V
REG
= 50V, I
+IN
= 1mA.
V
IOUT
= 36V, I
IOUT
= 1mA,
I
SUPPLY
= 2.7mA (I
SUPPLY
is a function of I
+IN
. See the I
SUPPLY
vs I
+IN
plot under Typical Performance Characteristics).
I
LOAD
= 2A and R
SENSE
= 20mΩ
Calculate LT6110 power dissipation:
Power = 36 • 0.001 + 50 • (0.001 + 0.0027) + 2
2
• 0.02
Power = 0.301 Watts