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Op amp w/ varying DC offset

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apw5020

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Hello everyone, this is my first post so please take it easy on me! I am desperate right now. I have been trying to figure this problem out for hours to no avail :confused: ...someone please help!!

So here it is: I need to design a variable level shifting amplifier. The input into the inverting terminal is .2 Vpp and needs to be amplified to a maximum of -12 Vpp. The input signal ALSO has a varying DC offset between .5 and 2.5 V. I need to use a 20 kohm potentiometer (in combination with resistors) at the non inverting terminal to cancel out this DC offset. The extremes of the potentiometer must cancel out .5 and 2.5, respectively.

I know I need to use voltage division somehow, but I can't figure out the resistor values. I know the negative feedback needs a gain of -60. The resistor values I have right now to achieve this gain are 60 kohm and 1 kohm.

Can anyone help me figure this out tonight?

Thanks,
Aaron
 

MikeMl

Well-Known Member
Most Helpful Member
Is the opamp to be powered on split supplies? Do you have any specific opamp in mind?
 

apw5020

New Member
The op amp we are using is the LF412CN, and it is powered by Vcc and Vee of +/- 12V. If you want, I can post the schematic of what I have so far.
 

MikeMl

Well-Known Member
Most Helpful Member
Stare at this for a while. It is not an accident that this configuration subtracts out the offset automatically. Note that R2/R1 = R4/R3 = 60. Further note that R1 = R3 and R2 = R4.

Note that the supplies have to be more than +-12V.

Google "Difference Amplifier"
 

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apw5020

New Member
I like that design, but I need to use a potentiometer somewhere at the non inverting terminal. When the 20 kohm pot is at its max, it needs to subtract .5 V DC offset, and when it is at its min, it needs to subtract 2.5 V DC offset. Any ideas?
 

MikeMl

Well-Known Member
Most Helpful Member
I like that design, but I need to use a potentiometer somewhere at the non inverting terminal. When the 20 kohm pot is at its max, it needs to subtract .5 V DC offset, and when it is at its min, it needs to subtract 2.5 V DC offset. Any ideas?

Well, I was trying to get you to do some of the work. As the offset goes from 0.5 to 2.5V, what is happening at P1? Hint: It has something to do with R1 and R2.

If you configure a voltage divider consisting of fixed resistors and a pot which is capable of producing the identical voltages that appear at the non-inverting input (Node P1) in the circuit I posted, you will have the magic values that subract the offset to keep the output at the same level...
 

apw5020

New Member
The voltages I came up with earlier were .4918 V and 2.459 V. I am having trouble with the voltage division and picking the two resistor values. I did:

(offset * gain) / (1 + gain)
 

MikeMl

Well-Known Member
Most Helpful Member
Ok, one other question? What reference voltage are you starting from to create the two required voltages of 0.5*60/61 = 0.4918V and 2.5*60/61 = 2.459V ?

Do you know how to solve a system of two simultaneous equations in two unknowns?
 

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