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--></style></head><body lang="EN-US" link="blue" vlink="purple"><div class="WordSection1"><p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1f497d">Hi Wrenches,</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1f497d"> </span></p><p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1f497d">Please see question below forwarded from one of my fellow engineers. Thanks, August</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1f497d"> </span></p><p class="MsoNormal"><b><span style="font-size:10.0pt;font-family:"Tahoma","sans-serif"">From:</span></b><span style="font-size:10.0pt;font-family:"Tahoma","sans-serif""> Eric Schoonbaert [mailto:<a href="mailto:eric@luminalt.com">eric@luminalt.com</a>] <br>
<b>Sent:</b> Monday, May 07, 2012 9:38 AM<br><b>To:</b> August Goers<br><b>Subject:</b> 120% Rule</span></p><p class="MsoNormal"> </p><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif"">The 120% rule, and how it is applied to panel board bus size has been widely discussed. There is one part of the rule (2008 and 2011 NEC quoted below) that gets much less attention and is the subject of this email. That is, how and when is the rule applied to a conductor? The heading and text both clearly say bus or <i>conductor </i>rating [emphasis added].</span></p>
</div><div><p class="MsoNormal"><b><span style="font-family:"Tahoma","sans-serif""> </span></b></p></div><p class="MsoNormal"><b><span style="font-family:"Tahoma","sans-serif"">2011 NEC 705.12 Point of Connection</span></b><span style="font-family:"Tahoma","sans-serif""> <b>(D)</b> <b>(2)</b> <b>Bus or Conductor Rating.</b> The sum of the ampere ratings of overcurrent devices in circuits supplying power to a busbar or conductor shall not exceed 120 percent of the rating of the busbar or conductor. </span></p>
<div><p class="MsoNormal"> </p></div><div><p class="MsoNormal"><b><span style="font-family:"Tahoma","sans-serif"">2008 690.64 Point of Connection (B) (2)</span></b><span style="font-family:"Tahoma","sans-serif""> <b>Bus or Conductor Rating.</b> The sum of the ampere ratings of overcurrent devices in circuits supplying power to a busbar or conductor shall not exceed 120 percent of the rating of the busbar or conductor. In systems with panelboards connected in series, the rating of the first overcurrent device directly connected to the output of a utility-interactive inverter(s) shall be used in the calculations for all busbars and conductors </span></p>
</div><div><p class="MsoNormal"> </p></div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif"">In Homepower 140 (December 2010 & January 2011), John Wiles has several examples of the application of the 120% rule in his code corner article. I have included one below. </span></p>
</div><div><p class="MsoNormal"> </p></div><blockquote style="margin-left:30.0pt;margin-right:0in"><blockquote style="margin-left:30.0pt;margin-right:0in"><div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif"">For example: Two inverters each require a 50 A backfed breaker in a main lug inverter combining load center to meet 690.8 requirements. A supply-side connection is going to be made with a 100 A fused disconnect. The rating of the combining load center and the ampacity of the conductor to the 100 A fused disconnect must follow the 690.64(B)(2) requirements. As noted, even with a supply-side connection, as soon as the circuit passes through the service entrance disconnect/overcurrent device, all load-side requirements apply, because the PC circuit is now on the load side of the service disconnect.</span></p>
</div></div></blockquote><blockquote style="margin-left:30.0pt;margin-right:0in"><div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif"">(50 A + 50 A +100 A) ÷ 1.2 = 200 A ÷ 1.2 = 166.7 A </span></p>
</div></div></blockquote><blockquote style="margin-left:30.0pt;margin-right:0in"><div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif""> </span></p></div></div></blockquote><blockquote style="margin-left:30.0pt;margin-right:0in">
<div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif"">The numbers indicate that a 200 A inverter load center/panel would be needed because there is no 175 A option available. Assuming a 75°C rated conductor, a 2/0 AWG conductor should be used between that panel and the 100 A fused disconnect</span></p>
</div></div></blockquote></blockquote><div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif""> </span></p></div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif"">My particular question concerns the application of this rule to the ampacity of the conductor. This is the only example I have come across where the rule is applied to a conductor in this way. Is it the intention of the code language to size the conductor between a main lug sub panel used to combine solar circuits and the OCPD for the panel using the 120% rule? Or is the inclusion of the word <i>conductor</i> to ensure that load centers that have conductors integral to them are covered?</span></p>
</div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif""> </span></p></div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif"">What about other equipment between along the conductor such as a utility of fire department disconnect? In the John Wiles example above, if an utility disconnect was to be located along the conductor, would 200 A rated disconnect be necessary?</span></p>
</div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif""> </span></p></div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif"">Applying this rule to the conductor does not seem to follow the same logic (at least as I understand it) as a panel board bus bar which has spots for additional circuits between the OCPD's supplying power to it while a conductor is typically does not. </span></p>
</div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif""> </span></p></div><div><p class="MsoNormal"><span style="font-family:"Tahoma","sans-serif""> </span></p>
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