What is the reason of using integrator with Instrumental Amplifier

Thread Starter

Hamza100

Joined Dec 18, 2021
12
Hi,
We are developing an EMG sensor as shown in the figure below. Often an integrator is used with IA (Instrumental Amplifier) to set its reference voltage. I have searched online to find how the integrator works in this case and I found that it is working as an inverting high pass filter and removes DC offset as well, however, I found it difficult to understand the exact working for the integrator.

https://e2e.ti.com/support/amplifie...tive-integrator-rather-than-passive-high-pass

My questions are:

1): The integrator can increase the DC offset with time and it is important to consider it, otherwise the amplifier can saturate. Often a feedback resistor is used to minimize this impact. For an A.C signal like EMG, can the integrator ever get saturated or it is only for a D.C signal?

2): Integrating is working as an inverting high-pass filter. The cut-off frequency is set at the lower limit of the signal range. My understanding is that no part of the actual signal is integrated as it is more than the cut-off frequency of the integrator and is simply inverted and given to the reference pin of the IA. Can the DC offset of the Operational Amplifier being used can shift the signal up and down or will it remain constant?


1700032628694.png
 

Ian0

Joined Aug 7, 2020
9,826
The circuit is known as a "servo".
The integrator adjusts the level on the Vref pin to make the average signal on the output zero*.
(The integrator is a low-pass filter)

By the way, 100k output impedance would seem inadvisable unless the connection to the next part of the circuit is very short.

*zero, in this case, is the half-supply reference.
 

Hymie

Joined Mar 30, 2018
1,284
The integrator is formed by the op-amp and components C2 and R3.

The non-inverting input to the integrator op-amp is set at ½ Vcc by resistors R4/R5 and the buffer op-amp.

When the AD620 output is greater than ½ Vcc, the integrator output will go low (at a rate determined by C2 and R3 and the voltage output of AD620 above ½ Vcc). The opposite will happen when the AD620 is less than ½ Vcc.

The mathematics of the integrator is not that simple, but in your circuit is given by the following formula (which can be transposed to suit).

Vin/R3 = -(dVout x C2)/dt

Where Vin is the difference between the AD620 output and ½ Vcc.
 
Last edited:
Top