0
\$\begingroup\$

Have a design that is taking an input from .549V-.838V out of a DAC and am trying to get it to amplify it into a range that would be more useful in an ADC, like 0-5V. So far I came up with this design: enter image description here

This gets me from .549-.838V to 1.921-2.933V, which is way better. I feel like I should be able to shift the output left so that the .549V translates closer to 0V or 1V, and I could amplify to 4+V for the high end. Thanks for any help or insight.

enter image description here

EDIT for new problem

So I am trying to apply this to a similar problem now with what you showed me to do. I have a thermistor as part of a voltage divider, and the output is currently going into an ADC. I am trying to use an Op-Amp to increase my resolution.

My current range would be 2.22V-2.73V, so a spread of .51V. I am trying to spread this out over 0-5V, and specifically I am trying to do this with my op-amp limited on the rails with 0-5V, because my current design(not shown) would overload my ADC with like 10V if I unplugged the thermistor.

So, I did 5V/.51=9.9. 8.9 for a non-inverting setup, and i used this for my gain resistor, R102, then I used a 1k resistor for R101.

enter image description here

I am modeling this graph to show the voltage as the Thermistor changes in values. 2.222 is at 8k, and 2.73k is at 12k.

enter image description here

Thanks again for all the help. Learning a lot here.

\$\endgroup\$

2 Answers 2

2
\$\begingroup\$

You have two parameters you need to tweak: gain and offset.

Your input swings .289V (.838 - .549), and you want an output swing of 5V (0 to 5.0).

So you need a gain of 5/.289 = 17.3.

The non-inverting amp has a gain of 1 + (R2/R1).

So R2/R1 = 16.3

schematic

simulate this circuit – Schematic created using CircuitLab

Now for the offset, you want your input of .549V to create an output of 0.0V.

We need to figure out the DC voltage V1 to make this happen with our chosen R1 and R2.

schematic

The same current flows through R1 and R2.

Since R2 is 16.3 times larger than R1 the voltage across R2 will be 16.3 larger than the voltage across R1.

The voltage across R2 is .549 - 0.0 = .549.

So the voltage across R1 is .549/16.3 = 0.034V.

Since the right side of R1 is .549V the left side is .549 + 0.034 = .583V = V1.

If you have a .583V power supply around, great. If not, we can create one with a resistor divider from +15V. The resistor divider will replace R1 so not only do we need to choose values R3 and R4 such that we get .583V, but also so that their Thevenin equivalent impedance is R1.

schematic

simulate this circuit

Now, just pick a convenient value for R1 and use the equations to calculate R2, R3, R4.

Check your work by verifying your high voltage input .838V yields a 5.0V output.

\$\endgroup\$
6
  • \$\begingroup\$ This was perfect and really showed me how to think about the problem in the correct way. Thanks for the help! \$\endgroup\$
    – tangomonky
    Oct 29, 2020 at 16:24
  • \$\begingroup\$ So I am still running into some issues here. I managed to get my design to work perfectly in the previous example, and I decided to try and apply that same methodology to another example. \$\endgroup\$
    – tangomonky
    Oct 29, 2020 at 19:03
  • 1
    \$\begingroup\$ What's the issue? Is it just a different input/output values, or an entirely different topology? \$\endgroup\$
    – td127
    Oct 29, 2020 at 19:20
  • \$\begingroup\$ Yeah I didn't mean to send that. Thanks for prompt reply, I am going to do a better job explaining here. \$\endgroup\$
    – tangomonky
    Oct 29, 2020 at 19:21
  • 1
    \$\begingroup\$ After your edit: You can see something's not kosher because + input on opamp is 2.379V and it should be the same as the - input, 2.222V. Why is this? Because the LF347 opamp is not a rail-to-rail opamp and can't run (properly) with 5V supply. It's going as low as it can, which is only 1.55V, and that's why the + input is pulled up to 2.379. The oapamp is saturating. You need to find a rail-to-rail opamp that can run at 5V, or limit ADC input with resistor from opamp and schottky diode to 5V, limiting to 5.3V. Another diode for negative excursions, although opamp shouldn't ever go negative. \$\endgroup\$
    – td127
    Oct 29, 2020 at 20:47
2
\$\begingroup\$

TI has a document called “Designing Gain And Offset in Thirty Seconds” which gives a procedure for designing op amp circuits to do this kind of level shifting. (It took me rather longer than thirty seconds to use it, personally, until I put the formulas into a spreadsheet.)

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Great resource thanks a lot for that. Going to help a lot for my future designs Russel! \$\endgroup\$
    – tangomonky
    Oct 29, 2020 at 16:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.