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Is it possible to create a circuit with an op amp (single supply) that have a nonlinear gain (below 1) slope and a start offset?

See image Green = input Red = output

enter image description here

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  • \$\begingroup\$ Yes, it is possible. If you want an overall non-linear response then you add non-linear components to the circuit. You need to supply more details if you expect a more detailed answer. In particular, how precisely do you want to control the point where the gain changes? \$\endgroup\$ – Elliot Alderson Aug 1 '18 at 22:09
  • \$\begingroup\$ Thanks for the reply, I need to be able to adjust the gain breakpoint by 0.1V. The breakpoint should be at 1.5V and the input is between 0-5V DC. The output will end at 4V when input is at its maximum (5V). \$\endgroup\$ – Honken Aug 1 '18 at 23:19
  • \$\begingroup\$ What do you mean by "able to adjust the gain breakpoint"? Do you mean by redesigning the circuit and selecting different components or by twiddling a knob? \$\endgroup\$ – Elliot Alderson Aug 1 '18 at 23:36
  • \$\begingroup\$ Redesigning when doing simulations... \$\endgroup\$ – Honken Aug 1 '18 at 23:45
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You start with a non-inverting differential amplifier with 0V reference then increase the gain when Vout =Vth of transfer function and then increase the gain to linearize the input. Since the non-inverting gain is always 1+ Rf/Rin, you have two choices; { if Vout > x , increase Rf or if Vin >y decrease Rin.

For Rin, Depending on the voltage error tolerance of linearity you have options to use a precision switched resistance or a std Zener or better, a precision Zener IC with a series R, all in parallel with the fixed Rin to increase the gain.

When the threshold of the Zener is relatively low resistance , Zzt compared to the fixed Rin, the knee is as sharp as you need it to match the input response.

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A precision clamp circuit will do what you want. Consider this: -

enter image description here

U1 and U2 act as clamps. They clamp the signal at a positive level of V1 and a negative level of V2 (as shown in the little picture under the "clamped output").

If R1 becomes a pot then you basically have a new signal that is somewhere between fully clamped and not clamped at all. If the input doesn't go negative then you don't need the U2/V2 circuit.

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