The HPF attenuates frequencies which are below the cutoff frequency. But, in many cases a HPF is used to remove the DC level from any voltage.

For example, in this case, the IC AD5933 supplies an output voltage of 2V DC with a frequency sweep from 50 kHz to 50.5 kHz. In step 3 is used a 10 nF capacitor and 100 kOhm resistor to make the HPF, the cutoff frequency is approximately 160 Hz.

enter image description here

The output at the end of the HPF will be 1Vpp with the same frequency.

In this case, the purpose of attenuating frequencies below the cutoff frequency does not matter. Because the filter is only used to remove the DC component. It does not matter if the cutoff frequency is 160 Hz, 10K HZ or 20K Hz.

I have two questions.

This change of purpose confuses me. Used this way. Is the circuit a high pass filter?

How exactly does this circuit work for this purpose to remove the DC component? The value of the resistor does not make much difference, what really matters is the value of the capacitor, which should be 10n F, correct?

  • 2
    \$\begingroup\$ A non-polar capacitor in the direct path of a signal should remove the DC component regardless of it being in an RC filter configuration, or the presence of the opamp. There's a question about DC blocking caps on here already which may be of some help. \$\endgroup\$
    – Polynomial
    Jan 22 '17 at 2:16
  • 2
    \$\begingroup\$ DC is the lowest frequency of all: zero Hertz. Any high-pass filter will remove it by definition, otherwise it isn't a high-pass filter at all. \$\endgroup\$
    – user207421
    Jan 22 '17 at 2:37
  • 1
    \$\begingroup\$ The lower you make the pass frequency the larger your capacitor will grow. With an infinite capacitor even the DC will get through. A compromise is to pick the filter frequency that will be far from the unwanted and wanted frequencies calculated on amplitudes, signal power, end result, noise margis etc. \$\endgroup\$
    – KalleMP
    Jan 22 '17 at 21:41

That circuit is a high pass filter (HPF). Specifically a single pole RC, buffered HPF.

If you want to remove DC (0Hz) from your 50kHz to 50.5kHz frequency sweep, then indeed it doesn't matter to you whether the cutoff frequency of the HPF is 160Hz, 10kHz or 20kHz. But if you had a 5kHz to 50.5kHz sweep, then only the 160Hz version would work.

What matters is the RC time constant, where R is not only that actual R that's drawn in the schematic, but also any component that's connected across it. As it happens, the TL072 opamp is a FET input type, and doesn't load that resistor by any significant amount. However, if the TL072 wasn't there, and you wanted to drive a low impedance load, then the load would have to be taken into account to get the effective value of R. That's the benefit of buffering the output with a unity gain amplifier, any load does not affect the filter.

The 10nF + 100kohm values you mention have a 1mS time constant, giving a cutoff frequency of 1000 radians/s or about 160Hz. You could get the same filtering action with 100nF and 10kohm, or 1nF and 1Mohm. If you used 10uF and 1kohm, it would take more current to drive the input. If you went as far as 1pF and 1Gohm, then you would need to start worrying about the loading of the opamp and noise, but the RC time constant would still be correct.

  • \$\begingroup\$ Unfortunately the less-experienced does not appreciate the false assumption in this question. "How exactly does this circuit work ...?" The fact is it does not. for too many reasons ... SNR, results of data, lack of noise filtering, no impedance bridge or reference impedance \$\endgroup\$ Nov 3 '19 at 1:21

The frequency response can also be seen using KVL or KCL rules to the loop or node and the result is an impedance ratio that goes to zero with frequency. (i.e. a voltage divider)

i.e. the cap blocks DC with a half power point when impedances are equal at 100 Hz when circuit only uses 50.0kHz~50.5kHz is a curious error.

What is more concerning is the poor quality of signal and results from this well documented but poorly designed circuit. The results tell me they are bogus with lots of random results with AC hum. Averaging does help but since there was no analysis on the results, I would have little confidence in using that design as is. Lack of attention to layout, Common Mode gain rejection, Shielding, hum rejection , random reactance with only one significant figure. etc. enter image description here

Conclusion: printed Body fat result shown as 17.81% is bogus.

Beware that random web designs may have flaws, even apparently if well documented

  • \$\begingroup\$ There are plenty of published papers using the AD5933 for BIS/EIT. Instructables would be the last place I would look for a BIS circuit to be truthful. As far as the cap goes, this looks more like a repurposed circuit. With a decent analog front end and knowing the AD5933 quirks, it can easily scan 200Hz to 100Khz with an impedance range of about 1 Ohm to 500K Ohms. And with that instructable you basically have a random chance to be able to measure body impedance even semi-accurately, the AD5933 requires the cal resistor to be in the ball park, and they just say to use a known resistor.:) \$\endgroup\$
    – GB - AE7OO
    Nov 2 '19 at 8:01
  • \$\begingroup\$ @GB-AE7OO Glad you agree. After reading a proper design, tinyurl.com/y4btqnbd , I think I was being far too kind to say these results were just "bogus" They use a 2R1C model as a parallel bridge reference model. also the Dk of water =80 and Fat<4 \$\endgroup\$ Nov 3 '19 at 1:19
  • \$\begingroup\$ I think so too... That paper is not bad, but there are others that go into a lot more detail of the design stage(I'm talking 10 or 20 pages) plus error details and how to correct them. When your talking about injecting a human with any type of signal, you had better have a decent fail safe option, or Murphy will grab you and not let go. \$\endgroup\$
    – GB - AE7OO
    Nov 4 '19 at 13:29
  • \$\begingroup\$ Oh, as a side note, if you use a 49 ohm resistor(avg human impedance or somewhere in that ballpark) for calibration, the AD5933 eval board(you need this or another AFE to get rid of the AD5933's minimum output impedance of 200 ohms.), and a circuit to keep current less than let's say 300uA, you can get a general(meaning +/- 10%) reading of body impedance. Oh and make sure you have either saline soaked pads or use those little tags for the skin interface or you'll get burns... ask me how I know...:) \$\endgroup\$
    – GB - AE7OO
    Nov 4 '19 at 14:00
  • \$\begingroup\$ The same authors wrote a more detailed report 2 years prior Ref 3. tinyurl.com/y2bf7m3p 49 ohms is not the "avg human impedance" 2 electrode measurements are very inaccurate vs 4. Electrode impedance and 2R1C network is also critical. \$\endgroup\$ Nov 4 '19 at 19:49

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