{"id":26380,"date":"2025-06-05T03:43:10","date_gmt":"2025-06-05T03:43:10","guid":{"rendered":"https:\/\/wattsemi.com\/?p=26380"},"modified":"2025-09-22T18:51:02","modified_gmt":"2025-09-22T18:51:02","slug":"26380","status":"publish","type":"post","link":"https:\/\/wattsemi.com\/?p=26380","title":{"rendered":"Calculating power strap WIDTH during Power planning\u00a0"},"content":{"rendered":"\t\t
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During Physical Design power grid planning, how do we calculate the WIDTH of the power stripes in order to meet the required IR drop limit?<\/p>\n

Calculating the width of power straps during ASIC physical design involves ensuring that the IR drop<\/strong> along the power delivery network (PDN) stays within the acceptable limits defined by the design specifications. Here’s a step-by-step approach:<\/p>\n

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1. Understand the Power Strap Design Requirements<\/strong><\/h3>\n
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    IR Drop Budget (Vdrop<\/span>):<\/strong><\/p>\n

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      The allowable voltage drop from the power supply (e.g., VDD) to the furthest point of the PDN.<\/p>\n<\/li>\n<\/ul>\n<\/li>\n

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      Maximum Current (Imax<\/span>):<\/strong><\/p>\n

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        The total current drawn by the block or design in the worst-case scenario.<\/p>\n<\/li>\n<\/ul>\n<\/li>\n

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        Resistivity (\u03c1<\/span>):<\/strong><\/p>\n

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          Material resistivity (e.g., copper or aluminum).<\/p>\n<\/li>\n<\/ul>\n<\/li>\n

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          Current Density Limit (Jmax<\/span>):<\/strong><\/p>\n

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            The maximum current density allowed for the metal layer to avoid electromigration issues.<\/p>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n

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            2. Formulate the Power Strap Width Calculation<\/strong><\/h3>\n

            The IR drop is a function of resistance (R<\/span>) and the current (I<\/span>) flowing through the strap:<\/p>\n

            Vdrop=I*R<\/span><\/pre>\n
            Resistance of the Power Strap:<\/h4>\n
            The resistance of a metal strap is given by:<\/p>\n
            R=\u03c1 ( L) \/ (<\/span>W * t)<\/span><\/pre>\n
            Where:<\/p>\n
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              \u03c1<\/span>: Resistivity of the metal (Ohm\u00b7cm).<\/p>\n<\/li>\n
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              L<\/span>: Length of the strap (cm).<\/p>\n<\/li>\n
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              W<\/span>: Width of the strap (cm).<\/p>\n<\/li>\n
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              t<\/span>: Thickness of the strap (cm).<\/p>\n<\/li>\n<\/ul>\n
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              Required Strap Width (W<\/span>):<\/h4>\n
              Rearranging to find the strap width:<\/p>\n
              W=\u03c1\u22c5I\u22c5L \/ (t\u22c5Vdrop)<\/span><\/pre>\n
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              3. Consider Parallel Power Straps<\/strong><\/h3>\n
              If multiple straps are used, the effective resistance decreases as they are in parallel. The equivalent width becomes:<\/p>\n
              Wtotal=\u03c1*I*L \/ (t*Vdrop*N)<\/span><\/pre>\n
              Where N<\/span> is the number of parallel straps.<\/p>\n
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              4. Ensure Compliance with Current Density Limits<\/strong><\/h3>\n
              Check that the strap width also satisfies the current density constraint:<\/p>\n
              W\u2265It\u22c5Jmax<\/span><\/pre>\n
              This ensures the strap can carry the required current without violating electromigration rules.<\/p>\n
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              5. Additional Factors to Consider<\/strong><\/h3>\n
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                Layer Resistance:<\/strong><\/p>\n
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                  Resistance varies by metal layer (e.g., M1, M2, etc.).<\/p>\n<\/li>\n
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                  Higher metal layers generally have lower resistivity and larger thickness.<\/p>\n<\/li>\n<\/ul>\n<\/li>\n
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                  Grid Density:<\/strong><\/p>\n
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                    More straps or a denser power grid can reduce the width requirement per strap.<\/p>\n<\/li>\n<\/ul>\n<\/li>\n
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                    Via Resistance:<\/strong><\/p>\n
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                      Account for resistance due to vias connecting straps across layers.<\/p>\n<\/li>\n<\/ul>\n<\/li>\n
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                      Dynamic and Static Current:<\/strong><\/p>\n
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                        Dynamic IR drop is caused by switching currents, while static IR drop is caused by leakage. Consider both in the calculation.<\/p>\n<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n
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                        Example Calculation<\/strong><\/h3>\n
                        Given:<\/h4>\n
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                          I=10\u2009A<\/span> (current)<\/p>\n<\/li>\n
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                          \u03c1=2.2\u00d710\u22126\u2009\u03a9 \\cdotpcm<\/span> (copper resistivity)<\/p>\n<\/li>\n
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                          L=1\u2009cm<\/span> (strap length)<\/p>\n<\/li>\n
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                          t=0.35\u2009\u03bcm<\/span> (metal thickness)<\/p>\n<\/li>\n
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                          Vdrop=50\u2009mV<\/span><\/p>\n<\/li>\n
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                          Jmax=1\u00d7106\u2009A\/cm2<\/span><\/p>\n<\/li>\n<\/ul>\n
                          Strap Width:<\/h4>\n
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                            Calculate based on IR drop:<\/p>\n
                            W=\u03c1\u22c5I\u22c5Lt\u22c5Vdrop<\/span><\/p>\n
                            Substituting values:<\/p>\n
                            W=(2.2\u00d710\u22126)\u22c510\u22c51(0.35\u00d710\u22124)\u22c50.05\u22481.26\u2009cm<\/span><\/p><\/li>\n
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                            Validate against current density:<\/p>\n
                            W\u2265It\u22c5Jmax<\/span><\/p>\n
                            Substituting:<\/p>\n
                            W\u226510(0.35\u00d710\u22124)\u22c5(1\u00d7106)\u22482.86\u2009cm<\/span><\/p>\n
                            The current density constraint dominates, so W=2.86\u2009cm<\/span>.<\/p>\n<\/li>\n<\/ol>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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