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<p>Yep. I like to put the current limit R inside the feedback loop
to avoid loading errors. A couple examples from a MIDI to CV I'm
working on:<br>
</p>
<div class="moz-cite-prefix">On 12/4/20 7:11 PM, David G Dixon
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:616EE625354A40278DB89A14EEF650F7@david78c70950b">
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<div dir="ltr" align="left"><span class="517095900-05122020"><font
size="2" face="Arial" color="#0000ff">If I'm generating CVs
that are meant to be precise (as, for example, VCO
pitch-controlling voltages from a keyboard or sequencer),
then I take the opamp output from the other side of
the current-limiting resistor. I call this resistor
an "innie" and it doesn't affect the output voltage of the
circuit (the opamp compensates for it). I don't want that
damned 1k resistor dividing my carefully calibrated
voltage. </font></span></div>
<br>
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<font size="2" face="Tahoma"><b>From:</b> Ian Fritz
[<a class="moz-txt-link-freetext" href="mailto:ijfritz@comcast.net">mailto:ijfritz@comcast.net</a>] <br>
<b>Sent:</b> Friday, December 04, 2020 3:47 PM<br>
<b>To:</b> David G Dixon<br>
<b>Cc:</b> <a class="moz-txt-link-abbreviated" href="mailto:synth-diy@synth-diy.org">synth-diy@synth-diy.org</a><br>
<b>Subject:</b> Re: [sdiy] CV input op-amp circuit<br>
</font><br>
</div>
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<div dir="ltr">Most of mine come from 1k output resistors.</div>
<div dir="ltr"><br>
<blockquote type="cite">On Dec 4, 2020, at 4:29 PM, David G
Dixon <a class="moz-txt-link-rfc2396E" href="mailto:dixon@mail.ubc.ca"><dixon@mail.ubc.ca></a> wrote:<br>
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<div dir="ltr" align="left"><span class="441122723-04122020"><font
size="2" face="Arial" color="#0000ff">Also, most CV
sources are coming directly from opamps, </font></span></div>
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<hr tabindex="-1"> <font size="2" face="Tahoma"><b>From:</b>
Synth-diy [<a class="moz-txt-link-freetext" href="mailto:synth-diy-bounces@synth-diy.org">mailto:synth-diy-bounces@synth-diy.org</a>] <b>On
Behalf Of </b>David G Dixon<br>
<b>Sent:</b> Friday, December 04, 2020 3:24 PM<br>
<b>To:</b> 'Ian Fritz'<br>
<b>Cc:</b> <a class="moz-txt-link-abbreviated" href="mailto:synth-diy@synth-diy.org">synth-diy@synth-diy.org</a><br>
<b>Subject:</b> Re: [sdiy] CV input op-amp circuit<br>
</font><br>
</div>
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FONT-SIZE: 12px; FONT-WEIGHT: normal; PADDING-TOP: 3px">[<strong>CAUTION:</strong>
Non-UBC Email]</span></span></span>
<div dir="ltr" align="left"><span class="582121723-04122020"><font
size="2" face="Arial" color="#0000ff">Just to clarify,
my design did not require an inverting opamp. The
operation was all done in a single opamp. I just showed
two opamps in the picture because one was processing -5V
and other +7V, to show that the proper voltages were
obtained.</font></span></div>
<br>
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<hr tabindex="-1"> <font size="2" face="Tahoma"><b>From:</b>
Ian Fritz [<a class="moz-txt-link-freetext" href="mailto:ijfritz@comcast.net">mailto:ijfritz@comcast.net</a>] <br>
<b>Sent:</b> Friday, December 04, 2020 1:26 PM<br>
<b>To:</b> David G Dixon<br>
<b>Cc:</b> <a class="moz-txt-link-abbreviated" href="mailto:synth-diy@synth-diy.org">synth-diy@synth-diy.org</a><br>
<b>Subject:</b> Re: [sdiy] CV input op-amp circuit<br>
</font><br>
</div>
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<div dir="ltr">IMO, everyone should at some point go through
the derivation of the equations for the generalized opamp
summer. This often makes it easy to avoid using unneeded
inverting stages. One thing to watch, though, is that the
resulting equations assume zero impedance voltage sources
for inputs. Usually you have to take source impedances into
account.</div>
<div dir="ltr"><br>
</div>
<div dir="ltr">Ian</div>
<div dir="ltr"><br>
<blockquote type="cite">On Dec 4, 2020, at 12:03 PM, David G
Dixon <a class="moz-txt-link-rfc2396E" href="mailto:dixon@mail.ubc.ca"><dixon@mail.ubc.ca></a> wrote:<br>
<br>
</blockquote>
</div>
<blockquote type="cite">
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<div dir="ltr" align="left"><span
class="945072518-04122020"><font size="2">Hello
Christian,</font></span></div>
<div dir="ltr" align="left"><span
class="945072518-04122020"></span> </div>
<div dir="ltr" align="left"><span
class="945072518-04122020"><font size="2">It seems to
me that your circuit will invert the CV, which is
not what you want.</font></span></div>
<div dir="ltr" align="left"><span
class="945072518-04122020"></span> </div>
<div dir="ltr" align="left"><span
class="945072518-04122020"><font size="2">Here's how I
would do it: First, I calculated that the range of
-5V to +7V is 12V, and the range of 0 to 3V is 3V,
so you need a gain of 25%. This alone would change
the range to -1.25V to +1.75V. Hence, this needs to
be shifted by +1.25V. So, you need a circuit that
will apply a gain of 25% and a shift of +1.25V. I
am going to assume that you have a -5V reference
source available (or an inverted +5V reference).
So, the -5V reference requires a gain of -25%. So,
what circuit will apply a (non-inverting) gain of
25% to one input, and an (inverting) gain of -25% to
another input? This one, with 5% resistors:</font></span></div>
<div><span class="945072518-04122020"></span> </div>
<div><span class="945072518-04122020">
<div><CVShifter.png></div>
<font size="4"><br>
</font></span></div>
<div><span class="945072518-04122020"><font size="4"><font
size="2">Or, a slightly more accurate version with
1% resistors:</font></font></span></div>
<div><span class="945072518-04122020"></span> </div>
<div><span class="945072518-04122020"><font size="4">
<div><poop.png></div>
</font></span></div>
<div><span class="945072518-04122020"></span> </div>
<div><span class="945072518-04122020"><font size="2">The
CV comes into the + input through a 4:1 voltage
divider which applies a gain of 20%. However,
the 1:4 ratio of feedback to inverting input
resistors applies a gain of 125% to the
non-inverting input, and (125%)(20%) = 25%.</font></span></div>
<div><span class="945072518-04122020"></span> </div>
<div><span class="945072518-04122020"><font size="2">The
-5V reference comes into the - input
through feedback/input resistor ratio of 1:4, which
applies an inverting gain of -25% to that voltage,
creating a level shift of +1.25V.</font></span></div>
<div><span class="945072518-04122020"></span> </div>
<div><span class="945072518-04122020"><font size="2">The
convenient aspect of this is that both pairs of
resistors have a 4:1 ratio. The closest 5% standard
values are 33k and 8.2k. The closest 1% values are
102k and 25.5k.</font></span></div>
<div><span class="945072518-04122020"></span> </div>
<div><span class="945072518-04122020"><font size="2">Cheers,</font></span></div>
<div><span class="945072518-04122020"><font size="2">Doc
Sketchy</font></span></div>
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<hr tabindex="-1"> <font size="2" face="Tahoma"><b>From:</b>
Synth-diy [<a class="moz-txt-link-freetext" href="mailto:synth-diy-bounces@synth-diy.org">mailto:synth-diy-bounces@synth-diy.org</a>] <b>On
Behalf Of </b>Christian Maniewski via Synth-diy<br>
<b>Sent:</b> Friday, December 04, 2020 5:31 AM<br>
<b>To:</b> <a class="moz-txt-link-abbreviated" href="mailto:synth-diy@synth-diy.org">synth-diy@synth-diy.org</a><br>
<b>Subject:</b> [sdiy] CV input op-amp circuit<br>
</font><br>
</div>
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<div>Hi all!</div>
<div><br>
</div>
<div>I’m trying to come up with an op-amp design for a
CV input. I want to transform a signal ranging from
-5V to +7V to a more MCU digestable 0-3.3V. I came up
with the circuit you’ll find attached.</div>
<div><br>
</div>
<div>I have seen other approaches, where an offset
reference is injected in the feedback loop, while the
positive op-amp input is grounded. Are there any
disadvantages to my approach or is it also valid?</div>
<div><br>
</div>
<div>Thank you so much!</div>
<div><br>
</div>
<div>I’ve been following this email list for some time
now. This is my first question and first email
entirely. Please bear with me.</div>
<div><br>
</div>
<div><br>
</div>
<div>Chris</div>
<div><br>
</div>
</div>
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