<html><head><meta http-equiv="Content-Type" content="text/html; charset=utf-8"></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; line-break: after-white-space;" class=""><br class=""><div><blockquote type="cite" class=""><div class="">On 24 Feb 2023, at 15:22, Mike Bryant <<a href="mailto:mbryant@futurehorizons.com" class="">mbryant@futurehorizons.com</a>> wrote:</div><br class="Apple-interchange-newline"><div class=""><meta charset="UTF-8" class=""><div class="elementToProof" style="font-style: normal; font-variant-caps: normal; font-weight: normal; letter-spacing: normal; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration: none; font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; background-color: rgb(255, 255, 255);">If the original LED signal comes from a MCU and you have spare pins then another approach which uses less board area is to have two LEDs, one on the front panel and another hidden inside the module. Run a resistor from VCC to the two LEDs, then feed the other ends of the LEDs to two pins on the same MCU port. Writing to the port causes and instantaneous change from one LED to the other. However because the LEDs are in different locations, 100nF from the LED/resistor joint to ground may be needed to smooth out the last wrinkles.<br class=""></div></div></blockquote><div><br class=""></div>If the LED signal is coming from a cpu (with a push/pull output), wouldn’t something like this get you there 90+% of the way? (But as the VCC is probably noisy anyway, not sure how effective this would be.)</div><div><br class=""></div><div><img apple-inline="yes" id="40BD5024-0730-4E2A-9055-50D929BF32FA" width="460" height="822" src="cid:561776B5-5A71-4589-AC19-4A3D26B07503" class=""></div><div><br class=""></div><div>Ben</div><div><br class=""></div><br class=""></body></html>