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Calculates minimum PCB trace width for current capacity considering voltage drop and temperature rise. Prevents overheating and signal degradation.
ସୂତ୍ର
Formula: width (mils) = current (A) ÷ (ampacity × thickness)
- A
- A value — Variable used in the calculation
ଷ୍ଟେପ୍-ଷ୍ଟେପ୍ ଗାଇଡ୍ |
- 1Current flowing through trace
- 2Trace length and thickness (oz/ft² of copper)
- 3Ampacity nomogram: 1 oz, 1 A/mil² roughly
- 4Formula: width (mils) = current (A) ÷ (ampacity × thickness)
- 5Account for temperature limits
ସମାଧାନ ହୋଇଥିବା ଉଦାହରଣ
ଇନପୁଟ୍
Current 2A, copper 1oz
ଫଳ
Width 20 mil
ଏଡ଼ାଇବା ଯୋଗ୍ୟ ସାଧାରଣ ଭୁଲ
- ✕Using minimum trace width to reduce design time
- ✕Not accounting for temperature derating in high-current boards
ବାରମ୍ବାର ଜିଜ୍ଞାସା
Why does trace width matter?
Thin traces carry less current, overheat, and can catch fire; wide traces prevent damage.
What's voltage drop in traces?
50-100 mV acceptable for power; too much dropout regulator can't maintain voltage.
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