AD8233 Question

Hi

I have a question AD8233.

AD8233 Evaluation board UG-1016 written,

A lower common-mode signal cutoff frequency improves RFI rejection; however, it can increase the risk of instability with a right leg drive feedback loop.

Content of consultation
· Please explain concretely the relation between the cutoff frequency and the feedback loop.
· Please tell us about the characteristics of the OP amplifier used for the integrator.
· Please tell me about calculation method of loop gain. ? (We are asking you because the notes on loop gain are stated in the data sheet of AD8233)

phenomenon
· The values of R36 and R42 were increased to improve the static tolerance.
Since it is on the order of kHz as the cutoff frequency, it is judged that the signal component is not affected
· In fact it is discovered that the waveform is affected

 

 

Waveform at 47 kohm (Raw filter)

 

Waveform at 180kohm (Raw filter)

 

Best Regards

HOD

  • 0
    •  Analog Employees 
    on Jun 12, 2018 4:08 PM

    Hi,

    I contacted the product expert for your query. We'll get back to you with the answers for your question.

    Regards,
    Goz

  • 0
    •  Analog Employees 
    on Jun 27, 2018 9:46 PM

    Hi HOD,

    Here are the answers to your questions:

    • Once all of the electrodes are connected to the patient, this completes the RLD feedback loop.  The RFI filter is inside of this loop, so you have to include its cutoff/poles in the stability analysis.
    • The RLD amplifier used as the integrator has specs listed in the table under "Right Leg Drive Amplifier (A2)" on page 4 and then the TPCs start on page 14.
    • Are you referring to the RLD amplifier gain? The gain of the RLD amplifier can be calculated using the GBP listed in the spec table, 20kHz, and then the crossover frequency. So with the 1nF cap and 150kOhm res example, you get a 1/(2*pi*1n*150k) = 1kHz crossover and 20kHz/1kHz = 20V/V (26dB) of gain.
    • Changing the resistors R36 and R42 changes the filter cutoff frequency.  Going from 180kOhms to 47kOhms you are pushing out the cutoff of that filter - signal components that were at higher frequencies are now going to be attenuated less which is why you see a difference.

    Regards,

    Jordyn