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How to correctly calculate the Rfb value in AD5933

Thread Summary

The user asked about calculating Rfb and Zmin for the AD5933 impedance analyzer, particularly for a bio-impedance measurement setup. The final answer clarified that the Rfb formula applies with or without an AFE, but the VDCOFFSET value differs. The user's derived formula is correct, and the negative sign in the datasheet formula is an error. For the user's setup, Zmin should be calculated using VDCOFFSET - Vpk, and Rfb should be 15-20% higher than Zmin to avoid saturation. The AFE's VDCOFFSET is always Vdd/2, simplifying the formula.
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I was reading the datasheet of AD5933, and I did not understand a few things about the calculation for the Rfb value and other things.

This is the Rfb equation:

How do I find the value of Zmin?

I intend to power the AD5933 with 5V, use the 1st Output Range (3Vpp in this case) and PGA gain = 1.
With a frequency sweep of 1K Hz up to 100K Hz:
Internal oscillator: MCLK = 16.776 MHz
fSTART = 1000 Hz
∆f = 1000 Hz
Increments = 99

In addition, I will use a high pass filter with a cutoff frequency of 100Hz, a circuit with a current source controlled by voltage (1mA), and an instrumentation amplifier INA118 (Gain = 2). This circuit will use 4 Electrodes. Similar to this:

In AN-1252 APPLICATION NOTE, page 3 of 12, it is said that for range 1 (Adding external op amp) Zout is equal to "> 100 Ohms". And on page 5, The maximum ratio, ZMAX / ZMIN, for my case is x45. Right? If it is correct. How to calculate Zmax and Zmin for "> 100"?

Furthermore. What is "RCal" for? If I do not use it in any equation.

In the example on page 7, the Zmin value is 4.7K and the Zmax value is 47K. Why? I dont understand.

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  • Thank you again, Miguel.


    Regarding circuit power, before the current source, I am using a high pass filter to remove the DC component of the signal, I am sending an AC signal to the electrodes.
    Regardless of the type of current source used, I believe that a in amp with a single supply, can not amplify negative signals because the power is not symmetrical. Right?

Reply
  • Thank you again, Miguel.


    Regarding circuit power, before the current source, I am using a high pass filter to remove the DC component of the signal, I am sending an AC signal to the electrodes.
    Regardless of the type of current source used, I believe that a in amp with a single supply, can not amplify negative signals because the power is not symmetrical. Right?

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