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Message: <blockquote data-quote="mcglsr2" data-source="post: 1392898" data-attributes="member: 15705">Lol I posted and then totally forgot to vote.  One of those days <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite2" alt=";)" title="Wink    ;)" loading="lazy" data-shortname=";)" />And yes, I sure do.  Assuming you are running a light to warn when there is an issue, the resistor goes between the two wires for the light. It's the easiest place to access.  You want a 510 Ohm resistor.  Check out this post <a href="http://www.sn95forums.com/showthread.php?68658-Instrument-Cluster-Gauge-Swap-How-To-(with-some-other-stuff)&p=1334748&viewfull=1#post1334748" target="_blank">here</a>, scroll down to the "Pd w/ Gauge" diagram, and look over at the right side at the "BATT bulb" - that's exactly what you need to do. You will see the resistor there in between the two wires.  Let me know if you have any more questions about it <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile    :)" loading="lazy" data-shortname=":)" />Edit: I used a regular bulb.  If you want to use an LED, then you need to measure the resistance across the normal bulb, and then measure the resistance across the LED, and then add that difference into the resistor.</blockquote>