<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="">Hi David,<div class=""><br class=""></div><div class="">It will, but if it’s for a microcontroller input, it doesn’t matter. It’s simpler to flip the ADC value in the firmware than add another op-amp to every CV input.</div><div class=""><br class=""></div><div class="">So, yes..but no!</div><div class=""><br class=""></div><div class="">Tom</div><div class=""><br class=""><div class="">
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<div><br class=""><blockquote type="cite" class=""><div class="">On 4 Dec 2020, at 19:01, David G Dixon <<a href="mailto:dixon@mail.ubc.ca" class="">dixon@mail.ubc.ca</a>> wrote:</div><br class="Apple-interchange-newline"><div class=""><span style="caret-color: rgb(0, 0, 0); font-family: Helvetica; font-size: small; font-style: normal; font-variant-caps: normal; font-weight: normal; letter-spacing: normal; text-align: -webkit-left; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration: none; float: none; display: inline !important;" class="">It seems to me that your circuit will invert the CV, which is not what you want.</span></div></blockquote></div><br class=""></div></body></html>