Generally, a resistance less than 1 Ohm is not significant in the signal path. If the line width of \u00bd-ounce copper is 5 mils, how long could a trace be before its DC resistance is > 1 Ohm?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve this problem, we need to calculate the length of a \u00bd-ounce copper<\/strong> PCB trace that would result in a DC resistance greater than 1 Ohm<\/strong>, given a trace width of 5 mils (0.005 inches)<\/strong>.<\/p>\n\n\n\n The formula for resistance of a conductor:<\/p>\n\n\n\n $$ Where:<\/p>\n\n\n\n $$ $$ $$ In printed circuit board (PCB) design, signal integrity and voltage drop are directly influenced by the resistance of copper traces. Resistance becomes especially relevant in power and analog signal paths where even small voltage drops can cause performance issues. In this problem, we aim to determine the maximum length of a 5-mil wide PCB trace made from \u00bd-ounce copper that would produce a DC resistance of more than 1 Ohm.<\/p>\n\n\n\n Copper’s resistivity and the physical dimensions of the trace are key to this calculation. For standard \u00bd-ounce copper, the thickness is 0.7 mils or 0.0007 inches. Given a trace width of 5 mils (0.005 inches), the trace\u2019s cross-sectional area is very small\u2014just 3.5 \u00d7 10\u207b\u2076 square inches. Copper\u2019s resistivity in inches is approximately 6.78 \u00d7 10\u207b\u2077 ohm-inches, a constant used for calculating the resistance of copper in imperial units.<\/p>\n\n\n\n Using the resistance formula $R = \\frac{\\rho L}{A}$, we solve for $L$ by substituting 1 ohm for $R$. Rearranging the equation and plugging in the values gives a result of approximately 5.16 inches. This means if the trace is longer than 5.16 inches, its resistance will exceed 1 Ohm, which may no longer be negligible in sensitive circuits.<\/p>\n\n\n\n This type of analysis is essential when designing power delivery networks or high-speed digital traces, where excessive resistance could lead to voltage drops, heating, or timing issues. In such cases, designers either increase trace width or use thicker copper (e.g., 1 oz or 2 oz) to reduce resistance and maintain reliable performance.<\/p>\n","protected":false},"excerpt":{"rendered":" Generally, a resistance less than 1 Ohm is not significant in the signal path. If the line width of \u00bd-ounce copper is 5 mils, how long could a trace be before its DC resistance is > 1 Ohm? The correct answer and explanation is : To solve this problem, we need to calculate the length […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-357","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/357","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=357"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/357\/revisions"}],"predecessor-version":[{"id":358,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/357\/revisions\/358"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=357"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=357"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=357"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 1: Understand Copper Thickness for \u00bd-Ounce Copper<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nStep 2: Use the Formula for Resistance<\/strong><\/h3>\n\n\n\n
R = \\frac{\\rho \\cdot L}{A}
$$<\/p>\n\n\n\n\n
\n\n\n\nStep 3: Calculate Cross-Sectional Area<\/strong><\/h3>\n\n\n\n
\n
A = 0.005 \\times 0.0007 = 3.5 \\times 10^{-6} \\text{ in}^2
$$<\/p>\n\n\n\n
\n\n\n\nStep 4: Solve for Length When R = 1 Ohm<\/strong><\/h3>\n\n\n\n
1 = \\frac{6.78 \\times 10^{-7} \\cdot L}{3.5 \\times 10^{-6}}
$$<\/p>\n\n\n\n
L = \\frac{1 \\cdot 3.5 \\times 10^{-6}}{6.78 \\times 10^{-7}} \\approx 5.16 \\text{ inches}
$$<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer: Approximately 5.16 inches<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation (300+ words)<\/strong><\/h3>\n\n\n\n