I need to use OP184 opamp to scale a 0.5-4.5 Voltage range to -3.000 to +3.000 voltage range. The opamp will be powered by +/- 12V. I have another available 5V line i can use. please help.
Clarification - The input to the opamp is the output of an accelerometer so it will vary with rotation. The output of the acceleromer is a DC +0.5 to +4.5V. I need to use OP184 to turn this voltage to -3.000 to +3.000 V. Thanks.
The gain from the + input to the output is 1+R2/R1,
so 1.5 = 1 + R2/R1, or R2/R1 = 0.5.
The center of the input range is (0.5+4.5)/2 = 2.5V.
If we apply 2.5V to the + input, what voltage on the gain setting resistor connected to the - input would cause the output to go to zero? The gain from the inverting input to the output is R2/R1 = 0.5, so to subtract out the 2.5V offset, you must supply an offset of 2.5 + 2.5/0.5 = 7.5V to the resistor.
I would use 10.00K for R1, so that makes R2 5.00K.
Since you dont have a 7.5V source, and 5V is too low, you can transform the 7.5V and 10K input resistor to 12V and a two resistor voltage divider by using Thevenin's Theorem. Look at the sim.
The only problem now is that the accelerometer I will use is very sensitive to Voltage changes in the millivolt range. My +12 volt is not as stable as I would like. My 5 volt will be very precise +/- 1 to 2 millivolts. Would you recommend to use OP284 and use the second opamp to amplify the 5 Volts to 10 Volts or even 7.5V and use that voltage? BTW this project is very important to me, and I would like to offer you a small donation for your help. What is the best way we can do this?
Here is problem #2. The accelerometer outputs 1V/G. 1 G equals gravity which would really be my maximum reading needed. If I decrease my range to +/- 1.5 g, I should be able to take advantage of the 5V instead of needing a 7.5V right?
In short, take the differential amplifier transfer function
Vout = V1 * R2/(R1+R2)*(1+R4/R3) - V2*R4/R3
and write two equations: One for Vout = -3V, V1 = 0.5V and V2 = 5V, and another one for Vout = 3V, V1 = 4.5V and V2 = 5V. Solve this system with two unknowns, R2/R1 and R4/R3. Choose R1 and R3 and calculate R2 and R4. The result is R1 = 2k, R2 = 12K, R3 = 2k, R4 = 1.5k.