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<div class="moz-cite-prefix">On 03.03.2025 06:16, David G Dixon via
Synth-diy wrote:<br>
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cite="mid:212F4569F56F46CC93F343EC3E9B6FCC@david78c70950b">
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cite="mid:212F4569F56F46CC93F343EC3E9B6FCC@david78c70950b">
<div><span class="083460805-03032025">My question is this: given
that the 2N390X transistors have higher transition frequencies
and lower collector capacitances than the old Japanese
transistors (according to All Transistors), why should the
latter give significantly faster saw reset times? What is the
actual figure of merit for this?</span></div>
<div><span class="083460805-03032025"></span> <br>
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This is not caused by the capacitances and inductivities alone.
<p>There is a time required to remove all the charges in the base
area when you turn it off. (mainly due to recombination IIRC)<br>
</p>
<p>This creates a turn off delay in a bipolar transistor called the
storage time. <br>
</p>
<p><br>
</p>
<p>The question is how accurately this is modeled in simulations. </p>
<p><br>
</p>
<p>My preference for this style of VCO is to use switching
transistors. <br>
</p>
<p>These have low Vce(sat) and are optimized for (drum roll)
switching.<br>
</p>
<br>
<p>Best,</p>
<p> René<br>
</p>
<br>
<pre class="moz-signature" cols="72">--
<a class="moz-txt-link-abbreviated" href="mailto:synth@schmitzbits.de">synth@schmitzbits.de</a>
<a class="moz-txt-link-freetext" href="http://schmitzbits.de">http://schmitzbits.de</a></pre>
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