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<p>Yes, the FET reverse diode will go into conduction at about
-0.5-0.6V. So maybe 1 to 1.2Vpp. </p>
<p>If what is driving the headphone amp is 2Vpp, and the output
was open (or light load), it could clip. <br>
</p>
<p>The 22R resistors form a voltage divider together with the
other 22R and the load. <br>
</p>
<p>So the actual clipping level depends on the headphones
connected. </p>
<p>Assuming you don't run this full steam it's probably good
enough. <br>
</p>
<p>Not a great solution, but it works aparrently. <br>
</p>
<p><br>
</p>
<p>I would have used two NMOS in anti-series instead.<br>
</p>
<p>No need for exotic components or PMOS (which would just flip
the sign) as discussed.<br>
</p>
<p><br>
</p>
<p>Best,</p>
<p> René<br>
</p>
</div>
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<div class="moz-cite-prefix">On 24.04.2024 22:14, Chris McDowell via
Synth-diy wrote:<br>
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<div>hello list, I've been looking at minibrute schematics today
and I'm puzzled by the anti pop circuitry in the headphone
driver. it -seems- like the body diode of the mosfets would
conduct when the signal went below 0.6ish volts (something I've
run into this using mosfets for dc signal muting).</div>
<div><br>
</div>
<div> From the vca page, that signal can be up to 2Vpp. <a
href="https://hackabrute.yusynth.net/MINIBRUTE/analog-board/schematics/MiniBrute-08-VCA.pdf"
moz-do-not-send="true" class="moz-txt-link-freetext">https://hackabrute.yusynth.net/MINIBRUTE/analog-board/schematics/MiniBrute-08-VCA.pdf</a></div>
<div><br>
</div>
<div>What am I missing here? Is 2Vpp just fine in this case? Now
reading the 2n7002 datasheet I see the forward voltage of the
source-drain diode is typ 0.88V, 1.5V max. Did I just answer my
own question? 😜</div>
</blockquote>
<br>
<p><br>
</p>
<blockquote type="cite"
cite="mid:58FD7D2B-88E6-4342-9F46-5CDDBEA0DA95@gmail.com"><br
id="lineBreakAtBeginningOfSignature">
<div dir="ltr">Cheers,
<div>Chris McDowell</div>
</div>
<br>
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