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schematic

simulate this circuit – Schematic created using CircuitLab

I replaced the feedback resistor of an inverting configuration by a pot. I'm meant to vary from edge to edge the pot so I can get a variable gain within -100 < A < -10 (I mean, without getting out from that range not even on the edges). Therefore, I need to fix the gain when the pot is at 0 Ohm and so for its max Ohm to avoid getting it greater than 100 times or minor than 10 times.

My first attempts were to put a resistor in series with the pot (both as feedback), later I tried with placing a resistor in parallel with the pot (again as feedback); at any case, I couldn't get the gain to be limited at 10, it tends to zero invariably.

What can you kindly suggest to me?

EDIT: opamp polarized with +12 and -12 V

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  • 1
    \$\begingroup\$ One resistor in series and one in parallel? \$\endgroup\$
    – Dampmaskin
    Commented Mar 28, 2017 at 8:44
  • \$\begingroup\$ Where do I have to place them? \$\endgroup\$
    – Julio
    Commented Mar 28, 2017 at 8:48
  • \$\begingroup\$ The series resistor would fix that, if your schematics is OK. This is an inverting opamp, such will deliver negative voltage output. So, do you have dual power supply? If no, then your circuit is wrog. \$\endgroup\$
    – Marko Buršič
    Commented Mar 28, 2017 at 9:17

1 Answer 1

Reset to default
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That should fix your problem.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Yes, now it works! I forgot to substract a portion of the pot, which in your schematic is R4. Thank you. \$\endgroup\$
    – Julio
    Commented Mar 28, 2017 at 9:24
  • \$\begingroup\$ Since you are extremely unlikely to be able to buy a 90K pot, either shunt the rheostat with 900K and live with the slight nonlinearity or scale everything up by 100/90 and use 1.1 for R1, 11 for R4 and 100K for the pot. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Mar 28, 2017 at 9:25
  • \$\begingroup\$ Where can you get 90k pots? Surely 100k is the standard value. \$\endgroup\$
    – JIm Dearden
    Commented Mar 28, 2017 at 9:25
  • \$\begingroup\$ By using the 100k pot and R4=11k, the output voltage is very close to the expected (less than 5% error). \$\endgroup\$
    – Julio
    Commented Mar 28, 2017 at 9:47
  • \$\begingroup\$ You can't even get 100k pots actually: the variance is 20%; so you will need to trim according to the actual value. \$\endgroup\$
    – user207421
    Commented Jul 2, 2017 at 23:58

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