AC-coupling of instrumentation amplifier

[sudil99]

I was trying to simulate ac-coupled instrumentation amplifier. The inputs have 300mV of differential dc offset and 5 mV signal. So, I was expecting output as :
Vout = gain * 5mV
without the dc offset in output, as the integrator feedback acts as a high pass filter.

But when I set gain of 12, I get the output of just 12 mV sine wave with no dc offset.
And when I set gain of 20, I get 100 mV sine wave with 2.20 dc offset. How is this dc offset coming here?

Can someone describe what is wrong with this circuit. I want to have maximum gain of 5mV sine wave.

 

[keith1200rs]

The whole point of an instrumentation amplifier is to amplify differential signals not common mode. You cannot apply 300mV differential signal with high gain and expect it to do anything other than saturate. The results are therefore not surprising.

Keith.

 

[sudil99]

Sorry, but I thought ac-coupling of instrumentation amplifier are used for this purpose (to remove the dc offset in the output).
Also, why do they use integrator as feedback to inst. amp. instead of using just a RC high pass filter in the output.

Thanx

 

[keith1200rs]

I guess you saw the circuit in their data sheet. They don't give any idea of the size of the offset it can remove or any equations. It can only remove an offset which stays within the operating range of the amplifier after it has amplified it.

The reason they do that is if you put input coupling capacitors you then need to add some resistors to bias the inputs to somewhere. That loses you the very high input impedance.

Keith

 

[sudil99]

Thank you for the explanation.

But I tried to test the circuit with only 30 mV of offset and gain of 10. The output has no offset but the gain is reduced by 5 times i.e. now I get gain of 2 only.

 

[keith1200rs]

You need to show your simulation stimulus. I have just tried it and it works fine - see attached. My guess is the frequencies you are using are incorrect. If your input frequency is too low the integrator will track it and remove the signal!

Keith.

 

[crutschow]

If the integrator frequency response is the problem, just increase the value of R12.

 

[sudil99]

ya, that was a silly mistake. i wanted to have cutoff at 0.5 Hz

I increased the gain to 26, I get
Vout = 26*5 mV + 3.7 V dc offset
where, 26 is set gain and 5 mV is signal

So, unexpectedly, it works for higher gain too. Can somebody explain this?

 

[keith1200rs]

I have just tried those values at 1kHz and my simulation starts to fail above 150mV offset = 3.9V. In other words the output is not centred around zero when the input offset is above 150mV. At 150mV offset the REF becomes -3.85V. With the offset above 150mV, REF hits -4V and cannot go any further so the output starts to shift. You still get an output, but not centred on zero, until the offset is above 300mV when output starts to hit the positive rail.

Keith.