[sdiy] CV input op-amp circuit

David G Dixon dixon at mail.ubc.ca
Fri Dec 4 20:01:34 CET 2020

Hello Christian,

It seems to me that your circuit will invert the CV, which is not what you
want.

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:

Or, a slightly more accurate version with 1% resistors:

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%.

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.

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.

Cheers,
Doc Sketchy

_____

From: Synth-diy [mailto:synth-diy-bounces at synth-diy.org] On Behalf Of
Christian Maniewski via Synth-diy
Sent: Friday, December 04, 2020 5:31 AM
To: synth-diy at synth-diy.org
Subject: [sdiy] CV input op-amp circuit

71C7_75E3E7C7EDA0AF42FB4C4E434016155C.png>
Hi all!

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.

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?

Thank you so much!

I've been following this email list for some time now. This is my first
question and first email entirely. Please bear with me.

Chris

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