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EDIT: Thanks everyone, I got it to work: enter image description here

Now if only I could accept multiple answers...

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Is the output of this op-amp circuit linear in input voltage? The last equation \$A = \frac{R_f}{R_1}\$ seems to imply the output is linear but when I simulated it in multisim with a sine wave input, the output was either a sine with an amplitude of 0.1V or a square wave with amplitude of 4V for small changes in input amplitude. How can I keep the output linear?

I am trying to amplify a 10µV to 100 µV differential signal to at least 1V using an analog amplifier so that I can measure it using an ADC such as the one on an arduino. Is there an easier way to do this with only a few parts?

Frequency range: 8Hz to 50Hz

The opamp I used in multisim is OP497 (this was the first one I found in the multisim catalog that had a DIP package)

edit: I flipped the op amp and now the output seems linear but the peak to peak output voltage is only 45mV. Shouldn't it be much higher than the input?

Input in simulator: 20Hz sine at 500mV peak, 500mV offset

Screenshot: enter image description here

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    \$\begingroup\$ I won't go into details, but in general, Op Amps are quite noisy. You may want to amplify your signal using discrete components such as transistors. \$\endgroup\$
    – Christoph B
    Commented Oct 28, 2012 at 9:38
  • \$\begingroup\$ The circuit should be linear. If you post a picture of the circuit you simulated we might be able to see why you got the strange results you did. What op-amp model did you use for the simuation, and what power supplies did you use, for example, might be involved. \$\endgroup\$
    – The Photon
    Commented Oct 28, 2012 at 15:48
  • \$\begingroup\$ Also, what is the frequency range of your signal. What is the maximum frequency you need to amplify, and do you need to dc couple or not? \$\endgroup\$
    – The Photon
    Commented Oct 28, 2012 at 15:57
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    \$\begingroup\$ @AlfredCentauri Agreed - looks like he has + and - mixed up. OP also needs some DC bias to not clip the negative part of the AC waveform. \$\endgroup\$
    – Adam Lawrence
    Commented Oct 28, 2012 at 20:12
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    \$\begingroup\$ Your revised circuit is no longer a differential amplifier. it is a inverter of gain -1... The (+) input is grounded. \$\endgroup\$
    – markrages
    Commented Oct 28, 2012 at 21:10

4 Answers 4

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I think you've got your opamp "upside down" in your schematic though it's hard to tell for sure due to the coarse resolution of the image. But it does look like your feedback divider is connected to the non-inverting input rather than to the inverting input.

If so, you've got something like a schmitt trigger circuit there rather than the difference amplifier you intended.

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    \$\begingroup\$ "Open picture in new tab" and you'll see you got it right. OP has the op-amp configured for positive feedback. \$\endgroup\$
    – The Photon
    Commented Oct 28, 2012 at 20:07
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Your circuit is wrong. You can't expect the output to swing negative when the supply is 0V and +5V. Opamps don't generate negative voltages on their own.

Option 1: Fix your supply. Supply the opamp with -5V and +5V.

Option 2: Arrange for the output to swing around +2.5V instead of ground. For example you could split R3 into two 2kΩ resistors. Connect one to ground and one to +5V. This generates a 2.5V supply with 1kΩ Thévenin impedance.

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Your real-world big challenge with such a small signal is going to be the opamp input offset voltage, which can be well into the millivolts for general-purpose opamps. Even super-accurate ones like the LT1001 can be into the tens of microvolts, which is a significant proportion of your input signal.

If your input signal isn't large enough to swamp the opamp input offset voltage, your gain formula must take into account the loss of signal caused by it.

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  • \$\begingroup\$ Another answer suggested the TC7650. Is 0.7μV small enough that I can assume it is 0? If not, how do I account for that? \$\endgroup\$
    – Navin
    Commented Oct 28, 2012 at 19:02
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    \$\begingroup\$ This looks like a nice part. Imagine the offset voltage as a DC source in series with one of the inputs, opposing your signal - your AC signal has to overcome that voltage level before the opamp 'sees' it. Multiply the input offset by the gain you've programmed and that's your expected output error due to the offset. \$\endgroup\$
    – Adam Lawrence
    Commented Oct 28, 2012 at 20:10
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You might want to consider an op amp such as the TC7652, available in a DIP package, which has a typical input offset voltage of 0.7 µV (max 5.0). If you can use a surface mount device (SMD), then the MAX4239 (which comes in a SOT23 package) has a typical input offset voltage of 0.1 µV (max 2.0).

For R1 and R2, and Rf and Rg, I would use matched resistors.

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  • \$\begingroup\$ I can only use through hole packages so the TC7650 seems good. How do I simulate this op amp? \$\endgroup\$
    – Navin
    Commented Oct 28, 2012 at 19:00
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    \$\begingroup\$ @Navin I haven't found any model for this op amp, so all I can suggest is you find one as close as possible in terms of gain and other characteristics, and assume an ideal device in terms of input offset since this one is so low. Note: while searching for a simulation model for the TC7650, I ran across the TC7652 which is identical to the TC7650 but has 10x lower noise, and is only slightly more expensive ($5.54 in single quantities). I have updated my answer with the better device. \$\endgroup\$
    – tcrosley
    Commented Oct 28, 2012 at 19:17

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