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<DIV dir=ltr align=left><SPAN class=441122723-04122020><FONT color=#0000ff
size=2 face=Arial>Also, most CV sources are coming directly from opamps, and
will therefore have very low output impedances. If not, then the CV input
should be buffered prior to any shifting circuit.</FONT></SPAN></DIV>
<DIV dir=ltr align=left><SPAN class=441122723-04122020><FONT color=#0000ff
size=2 face=Arial></FONT></SPAN> </DIV>
<DIV dir=ltr align=left><SPAN class=441122723-04122020><FONT color=#0000ff
size=2 face=Arial>Opamps are cheap.</FONT></SPAN></DIV><BR>
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<FONT size=2 face=Tahoma><B>From:</B> Synth-diy
[mailto:synth-diy-bounces@synth-diy.org] <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> synth-diy@synth-diy.org<BR><B>Subject:</B> Re: [sdiy] CV
input op-amp circuit<BR></FONT><BR></DIV>
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Non-UBC Email]</SPAN></SPAN></SPAN>
<DIV dir=ltr align=left><SPAN class=582121723-04122020><FONT color=#0000ff
size=2 face=Arial>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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<FONT size=2 face=Tahoma><B>From:</B> Ian Fritz [mailto:ijfritz@comcast.net]
<BR><B>Sent:</B> Friday, December 04, 2020 1:26 PM<BR><B>To:</B> David G
Dixon<BR><B>Cc:</B> synth-diy@synth-diy.org<BR><B>Subject:</B> Re: [sdiy] CV
input op-amp circuit<BR></FONT><BR></DIV>
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Non-UBC Email]</SPAN></SPAN></SPAN>
<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
<dixon@mail.ubc.ca> 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><FONT
size=2></FONT></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><FONT
size=2></FONT></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><FONT size=2></FONT></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><FONT size=4></FONT></SPAN> </DIV>
<DIV><SPAN class=945072518-04122020><FONT size=4>
<DIV><poop.png></DIV></FONT></SPAN></DIV>
<DIV><SPAN class=945072518-04122020><FONT size=4></FONT></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><FONT size=2></FONT></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><FONT size=2></FONT></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><FONT size=2></FONT></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>
<DIV dir=ltr align=left><FONT size=4></FONT><BR></DIV>
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<FONT size=2 face=Tahoma><B>From:</B> Synth-diy
[mailto:synth-diy-bounces@synth-diy.org] <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> synth-diy@synth-diy.org<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>
<DIV id=CanarySig>
<DIV>
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