Answer / prabhakaran

hi,

It depens upon the current rating used. i Dont know is
there any formula for it But there is an Thumb rule
followed that is for 1 Amps trace width should be 20 mills.

Answer / poornimashankara

1Amps is equal to 40 mils .This is the thumb rule.

Answer / gb

trace width calculation for what type of signal ?
if it is for controlled impedance traces, need to consider
the heigh (distance )from refference plane, just thickness
and width alone is not engouh, if it is for power, consider
the thickness of copper and wdith of trace.

Answer / anandhan k

Hi

This is depending on the copper wight, current rating and
the transmission line location (outer layer or inner layer)

Trace will reduce half when you increase the copper wight double

Anandhan K

Answer / sowmya

it based on the current and the maximum temperture rise are
litte comple,.for 1Amps,1ounes and 10degree C temp is equal
to 10 amp current..

Answer / jyothi

An approximate formula for the minimum trace
width required to carry a given current which is
accurate over a current range of 1 to 20A is

T =(2/CuWt)(-1.31 + 5.813I + 1.548 I2 - 0.052 I3)

where T = trace width in mils; I = current in
Amperes, and CuWt = copper weight in ounces.
The formula assumes that the current causes a temperature
rise of 10 degrees Centigrade in the traces.
*NOTE: In the above equation it is I sq. and Icube

Answer / harish

A 12 mils trace can carry 1A current easily.

Answer / kkr

1A=20MIL

Answer / prabhakaran

Actually the thing is for 1 oz copper we will follow 1 Amp
is equal to 40 mills and for 1/2 oz copper 1 Amp is equal
to 20 mills.

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