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I've been using the AD8236 to make an EMG amplifier. I connected a second order low pass MFB butterworth filter at @20Hz cutoff , which filters the AD8236 output and feeds it to the reference pin. I am not pleased with the results. while the DC correction seems to work well, at lower frequencies the signal is not attenuated much, and instead is kind of a square wave output even at low amplitude input.

I was wondering, does the ref pin operate with a varying signal? say, if i input a DC shifted sine wave at the Vref, will the output contain the same signal added to it?

  • Hello Usama,

    I'm not sure I understand the behavior you are observing. There are a several things that can make the circuit not operate correctly. Using a servo loop to eliminate the DC component is a common application for instrumentation amplifiers. However, there are a few rules that must be observed:

    1.  Operating voltage range: the reference pin must remain inside its specified voltage range, or in this case, inside the supplies. This gets more difficult with high gains, since everything is amplified before it can be filtered, including DC. A good rule of thumb is that the maximum DC component expected at the input times the gain must be less than half the supply range.

    2. The circuit must meet DC and AC stability conditions. Because you are closing the loop around the amplifier, you must make sure that the circuit is stable.

    a) If you go around the loop and verify that when the inamp's output voltage level increases, the output of the amplifier driving the refernce pin goes down, the circuit meets the DC stability condition.

    b) The AC stability condition is a little more complicated. You need to make sure that you don't apply positive feedback at any frequency before the loop gain drops below 1. This will be a little more difficult if you have a higher order filter, such as a second-order MFB. And even if you meet this criteria, you still want to have enough phase margin. There is no single answer, as it will depend on the gain of your filter, the inamp and where are the poles and zeroes (if any) located. You have to perform the analysis. Sometimes is easier to use a circuit simulator for this purpose.

    You can read more about the topic on any control theory book. You may also find the following links useful:

    By the way, all the observations above assume your power supplies are properly decoupled and that you have implemented appropiate housekeeping techniques such as good grounding and RFI filtering (especially if you have long leads coming into the instrumentation amplifier). Let me know if you have any further questions.



  • Usama,

    Of course, an easier way to solve the AC stability problem is to add an integrator to drive the reference pin of the inamp. This way you only get a high-pass of 20dB per decade but you can cascade a second filter stage to obtain the overall DC rejection that you are targeting.  Please look at page 18 on the AD8236 data sheet.

    Another alternative you could consider is AD8420. Because it uses a different circuit topology, you will be less restricted by the voltage range of the reference pin; you can eliminate the DC component before applying gain. Figure 69 on the data sheet shows an example of how to implement the DC servo for this amplifier (in particular for an ECG front-end). The tradeoff is that you'll need a higher value time constant to place the cutoff frequency at the same location as you would with the classic 3-opamp topology, since it is divided by the gain of the inamp.



  • Hi Gustavo,

    Thank you so much for your quick response and for your help.

    Actually this is the related part of my EMG Circuit. I'm using a single ended supply. Basically it's supplied with a 3.7V Battery. I'm using a linear voltage regulator to get a 3V output. The other end of the battery is used as GND. The mid reference voltage, Vref of around 1.5V is being generated with a Shunt regulator.

    What happens is, if suppose I apply a differential input voltage of say 1Hz, I expect the output to be nullified. I expect that my output will be Vref. However, I still get a distorted 1 Hz signal on the output. I can see that if I apply a DC offset to the input, the servo mechanism brings it back to the Vref level. So far, it works for the DC removal. But not very well for the AC removal below 20Hz.

    I believe, in that sense the Ref pin does remain within the voltage range and according to what you said, I think it meets the DC stability condition.

    But I think there may indeed be an AC stability problem.

    Yes, the power supplies are properly decoupled and I've taken precautions against RFI etc.

    I am tempted to explore the AD8420 option. Though I prefer the AD8235 because of its WLCSP package as well.

    I tried some rudimentary calculations, and I'm not sure if I am correct here...

    I tried a simulation of the same circuit in Texas Instruments' Simulation program TINA. Accordingly, I used the somewhat equivalent INA333 and OPA2333 ICs. I believe, in theory, the AD8235 and AD4505-4 can replace these and have the same circuit (except for the gain resistor). I observed the following scenario;

    Single Sided supply, +3V. Reference voltage (virtual ground) = 1.5V

    For an Input differential voltage 2mVp-p 200Hz signal, with a positive DC offset of 1mV. With a gain of 500 at the Instrumentation Amplifier, I had an output signal of 1Vp-p. The Ref pin was connected to the Virtual ground 1.5V. Hence the output centered at 2V (1.5V Ref + 1mVx500). My Filter output (unconnected) dropped from 1.5Vdc to 1Vdc as desired.

    When I connected my 2nd order butterworth low pass MFB filter @ 20Hz to the ref pin, The output seemed to average at 1.75 V. I think, in this case, considering only the DC Aspect, we can consider that:

    The output of Instrumentation Amplifier Vo=AVdc + Vref

    (Where A is gain, Vdc is input offset voltage and Vref is the ref pin signal)

    Vref = 1.5 - Vo (At DC level)

    So, Vo= ( AVdc + 1.5)/2

    So with a gain of 500, Vdc of 1mV, and the initial value of 1.5V, We get (1.5+0.5+1.5)/2 = 3.5/2 = 1.75V

    Because of this, at a larger input offset, the output DC point will shift till it goes out of range.

    Hence it is not DC Stable?

    On the other hand, if we use a simple integrator, then

    Vo=AVdc + Vref

    Vref = (1.5-Vo)

    Vo = AVdc + 1.5t - Vot

    Vo = (AVdc+1.5t)/(1+t)

    Taking limit t-> infinity,

    We get (0+1.5)/(0+1) = 1.5

    hence with an integrator, the output DC should come down to virtual ground over time.

    What is your opinion?

  • Hello Usama,

    Thanks for sharing your schematic, that  makes it easier to understand the problem.

    Now the issue is evident. You don't have enough gain on your filter to eliminate the additional offset. That's the problem. If you remove R9, you'll get enough loop gain to bring your output back to 1.5V. Once you do this, you'll have a simple low-pass filter followed by an integrator. As you noted, the integrator does exactly the same, because it has plenty of loop gain at DC.

    If you wish to perform a simulation with our models, you can download the free Analog Devices edition of Multisim from our website. It is available here:

    The models for AD8235 and ADA4505 are already available in the library, you don't need to download them separately.  I already built the schematic so you can play with the values (attached to the end of this message).

    Let me know if you have any questions.