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I want to use a sensor that delivers a DC voltage that varies from 0.174V to 1.174V depending on the temperature.

After processing the signal, I want to have a voltage between 1 and 9V (when Vin = 0.174, Vout = 9V and Vin = 1.174, Vout = 1V.)

Here is a circuit diagram that was given to me by an electronic engineer but I don't know how dimensionate component :

circuit

Here is my calculate :

In linear mod, E+ = E- because Espilon = 0. More over, I, in born "+" is equal to 0 so :

E+ = 5V

From milman :

E- = (UeR3+UsR4)/(R3+R4).

So 5 =(UeR3+UsR4)/(R3+R4) <=> 5(R3+R4) = UeR3+UsR4 <=> 5(R3+R4) - UeR3 = R4Us (5(R3+R4) - UeR3)/R4 = Us

I put my system from my point : Ue = 0.174V Us = 9V; Ue = 1.174V Us = 1V And here is my system :

9 = (5R3 + 5R4 - 0.174R3)/ R4 R4 = (4.872/4)R3 <=> 1 = (5R3 + 5R4 - 1.174R3)/ R4 1 = 3.872R3+(5(4.872/4)*R3)/4.872/4)*R3

So 1 = 9.962R3 / 1.218R3 <=> 1 = 9.962/1.218 so that's not good.

Somebody can help me to solve this please ?

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  • \$\begingroup\$ You forgot GND at R4. This circuit is called "non-inverting op amp with non-inverting reference voltage". It's not an integrator as bobflux stated. Capacitor is needed for stability only if you use high value R3. If you wrote your question correctly, this circuit will not give you what you want because it can't invert a signal. It can work if you don't want to invert but it would need a negative voltage as reference voltage \$\endgroup\$
    – Hedgehog
    Apr 21 at 10:40
  • \$\begingroup\$ @Hedgehog thanks for your answer. I talk with the electronic engineer and he did a mistake because he didnt look the datasheet of the sensor before. Now, the voltage source (Temp sensor equivalent) is at R4 and ground at R2. This circuit is an differential op amp that's it ? \$\endgroup\$
    – Benjamin M
    Apr 21 at 11:43
  • \$\begingroup\$ I modificated my post with my new equation ^^' \$\endgroup\$
    – Benjamin M
    Apr 21 at 13:38
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You don't need an integrator (unless you want to integrate).

when Vin = 0.6, Vout = 9V and Vin = 2.4, Vout = 1V

So it needs to invert (output goes down when input goes up).

enter image description here

(source)

Gain should be : (9-1)/(2.4-0.6) = 4.44 so that gives you the ratio of R2/R1.

In the inverting amplifier, the "+" input is usually connected to 0V but it doesn't have to. It you set it to a voltage V, then :

\$ V_{out}-V = R_2/R_1 (V_{in}-V) \$

Just solve it and you get the voltage you need to put on the "+" input, you can make it with a voltage divider from the supply voltage if its accuracy is good enough, and filter by a capacitor.

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  • \$\begingroup\$ Well, I did some mistakes in my post and there is some changes in my project. Firstly, now my sensor output start at 0.174V (-40°C) and go to 1.174V(120°C). Secondly, I have a schema which was proposed to me by an electronic engineer but I dont know what is this montage (I'll add the schema in my post) if you can tell me what is it to help me ^^ Thanks \$\endgroup\$
    – Benjamin M
    Apr 21 at 9:55
  • \$\begingroup\$ You can use the schematic from my answer, recalculate gain and just change the resistor values. The schematic in your post is not inverting, so it would be okay if you wanted output voltage to go up when input voltage goes up, but your question says you want the opposite polarity. \$\endgroup\$
    – bobflux
    Apr 21 at 10:10
  • \$\begingroup\$ I change my post, it is good now ? :) \$\endgroup\$
    – Benjamin M
    Apr 21 at 13:50
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R1 is not needed in your circuit.

enter image description here

Design Notes

  1. Use op amp linear output operating range. Usually specified under AOL test conditions.
  2. Amplifier common mode voltage is equal to the reference voltage.
  3. Vref can be created with a voltage divider.
  4. Input impedance of the circuit is equal to R2 .
  5. Choose low-value resistors to use in the feedback. It is recommended to use resistor values less than 100kΩ. Using high-value resistors can degrade the phase margin of the amplifier and introduce additional noise in the circuit.
  6. The cutoff frequency of the circuit is dependent on the gain bandwidth product (GBP) of the amplifier. Additional filtering can be accomplished by adding a capacitor in parallel to R1 . Adding a capacitor in parallel with R1 will also improve stability of the circuit, if high-value resistors are used.

enter image description here

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  • \$\begingroup\$ My problem is VRef isn't variable :/ \$\endgroup\$
    – Benjamin M
    Apr 21 at 14:03
  • \$\begingroup\$ You can not do this with 5V reference voltage, If it's a school assignment, it's wrong. \$\endgroup\$
    – Hedgehog
    Apr 21 at 14:06
  • \$\begingroup\$ Well, I didnt understand what the exercise wants. I finally find, and you are right, sry and thanks ! \$\endgroup\$
    – Benjamin M
    Apr 21 at 14:38
  • \$\begingroup\$ Please provide a citation or a link for the graphics that you copied into your answer. We want to be sure to give credit to the original creator. \$\endgroup\$ Apr 21 at 14:54
  • \$\begingroup\$ I would like to point out one thing, the OP is attempting to use a dual supply op amp on a single supply so initially, the output zero crossing is 1/2 of 15V (7.5V). \$\endgroup\$ Apr 21 at 16:36

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