ISL28022_15 Datasheet, PDF (27/32 Pages) Intersil Corporation – Precision Digital Power Monitor

ISL28022
TRACE WIDTH
Matching the current carrying density of a copper trace with the
maximum current that will pass through is critical in the
performance of the system. Neglecting the current carrying
capability of a trace will result in a large temperature rise in the
trace, and the loss in system efficiency due to the increase in
resistance of the copper trace. In extreme cases, the copper
trace could be severed because the trace could not pass the
current. The current carrying capability of a trace is calculated
using Equation 20:
1
Trace _width

Imax
0.725



kT^0.44


Trace _Thickness
(EQ. 20)
I_max is the largest current expected to pass through the trace. T
is the allowable temperature rise in Celsius when the maximum
current passes through the trace. Trace_Thickness is the thickness
of the trace specified to the PCB fabricator in mils. A typical
thickness for general current carrying applications (<100mA) is
0.5oz copper or 0.7mils. For larger currents, the trace thickness
should be greater than 1.0oz or 1.4mils. A balance between
thickness, width and cost needs to be achieved for each design.
The coefficient k in Equation 20 changes depending on the trace
location. For external traces, the value of k equals 0.048 while
for internal traces the value of k reduces to 0.024. The k values
and Equation 20 are stated per the ANSI IPC-2221(A) standards.
TRACE ROUTING
It is always advised to make the distance between voltage
source, sense resistor and load as close as possible. The longer
the trace length between components will result in voltage drops
between components. The additional resistance will reduce the
efficiency of a system.
The bulk resistance, , of copper is 0.67µΩ/in or 1.7µΩ/cm at
+25°C. The resistance of trace can be calculated from Equation 21:
R _trace
Trace _length
  _{Trace widthTrace thickness}
(EQ. 21)
Figure 42 illustrates each dimension of a trace.
For example, assume a trace has 2oz of copper or 2.8mil
thickness, a width of 100mil and a length of 0.5in. Using
Equation 21, the resistance of the trace is approximately 2mΩ.
Assume 1A of current is passing through the trace. A 2mV
voltage drop would result from trace routing.
Current flowing through a conductor will take the path of least
resistance. When routing a trace, avoid orthogonal connections
for current bearing traces.
FIGURE 43. AVOID ROUTING ORTHOGONAL CONNECTIONS FOR
TRACES THAT HAVE HIGH CURRENT FLOWS
Orthogonal routing for high current flow traces will result in
current crowding, localized heating of the trace and a change in
trace resistance (see Figure 43).
CURRENT ^FLOW
CURRENT FLOW
FIGURE 44. AVOID ROUTING ORTHOGONAL CONNECTIONS FOR
TRACES THAT HAVE HIGH CURRENT FLOWS
The utilization of arcs and 45° traces in routing large current flow
traces will maintain uniform current flow throughout the trace.
Figure 44 illustrates the routing technique.
TRACE
THICKNESS
TRACE
WIDTH
TLREANCGETH
FIGURE 42. ILLUSTRATION OF THE TRACE DIMENSIONS FOR A STRIP
LINE TRACE
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FN8386.7
October 2, 2015