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I'm trying to add a DC offset level to a sine signal through an instrumentation amplifier (as the figure below shows.)

The problem is that, in the simulation, everything works perfectly, however, in reality I can't achieve the DC displacement of the signal. The input signal parameters are: 0.7Vpp and 100Hz

enter image description here

The image shows the wanted behavior of the circuit, however, I'm not able to obtain that in real life.

What could be the problem? Maybe the AD620 is not the best option to do that?

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The AD620 negative supply (-Vs) is at 0 volts but if you do this, you need to respect the input common mode range and that is specified as: -

−Vs + 1.9 to +Vs − 1.2 

enter image description here

In other words, tying pin 2 (one of your inputs) to ground is an illegal condition and all bets are off. The lowest voltage allowed on your inputs is 1.9 volts.

Just in case it's important to your design, you should also consider that the output signal is limited to this following range: -

−Vs + 1.1 to +Vs − 1.2

So, with a single 5 volt supply, the output can typically be between 1.1 volts and 3.8 volts but, over the full temperature range this might reduce to 2.1 volts to 3.6 volts.

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  • \$\begingroup\$ So, if I understood clearly, I could not inject a purely sine signal (with positive and negative components) at the inputs of the AD620 (taking into account that I'm using a positive supply). I can only use signals that oscillate between 1.9V and 3.8V in the inputs of the instrumentation op amp, is that right? \$\endgroup\$ Jan 3, 2020 at 14:56
  • \$\begingroup\$ That is correct @SantiagoNoriega \$\endgroup\$
    – Andy aka
    Jan 3, 2020 at 15:03
  • \$\begingroup\$ Thank you so much @Andyaka \$\endgroup\$ Jan 3, 2020 at 15:05

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