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अवयव के लिए आवश्यक हीट सिंक का आकार निकालें। शक्ति क्षय दर्ज करें।

A
°C
oz

Trace Width (external)

53.8 mils

Trace Width (mm)1.37 mm
अनुप्रस्थ काट क्षेत्रफल74.2 mils²

Trace Width (external) vs Current

सूत्र

## PCB Trace Width (IPC-2221) ### Formula **Area = (I / (k x dT^0.44))^(1/0.725)** **Width = Area / Thickness** For external layers k = 0.048. Copper thickness at 1 oz = 1.378 mils. This is for external (outer) layers; internal layers need wider traces due to less cooling.

हल किया गया उदाहरण

3 A current, 10C rise, 1 oz copper.

  1. 01Thickness = 1 x 1.378 = 1.378 mils
  2. 02Area = (3 / (0.048 x 10^0.44))^(1/0.725) = ~79.6 mils^2
  3. 03Width = 79.6 / 1.378 = ~57.8 mils = ~1.47 mm

अक्सर पूछे जाने वाले प्रश्न

External vs internal layers?

Internal traces have less cooling (no airflow). Use k = 0.024 for internal layers, giving roughly double the width.

What is 1 oz copper?

1 oz/ft2 copper foil is 1.378 mils (35 um) thick. 2 oz doubles both thickness and current capacity.

Does this include safety margin?

IPC-2221 values are nominal. Add 20-50% margin for production variation and derating.

सीखें

Ohm's Law Guide

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