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I am looking to build a analog high pass filter with a cutoff frequency of 5.3 kHz. I have found a design which suits my need but I would like to adapt it from the 120V AC, 60Hz and a cutoff frequency of 50kHz it was designed for to the UK 240/230V AC, 50 Hz standard, with a cutoff frequency of 5.3kHz. Any ideas on how to do it to maintain a safe circuit and a good signal ?

I also would like to double check it is safe enough. Should I add a 1:1 transformer to isolate the mains ?

The overall procedure I hope to follow is :

"To measure the EMI on the power line, we used the same hardware used in LightWave [4]. An analog high-pass filter (HPF) with a 3 dB corner frequency of 5.3 kHz is used to reject the strong 60 Hz component. The output of the HPF is sampled at 1 MS/s using a 12-bit analog-to-digital (ADC) converter in the USRP (Universal Software Radio Peripheral) followed by transforming the signal into frequency domain using a 32,768-point fast Fourier transform (FFT), yielding a frequency resolution (or bin size) of 30.5 Hz. The signal from the USRP is then fed into the computer for data analysis. It should be noted that a USRP was used in this prototype simply for convenience." from UTouch Paper

I hope that I didn't say anything too stupid. I'm still a novice, but willing to learn ;)

High Pass Filter Schematic

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  • \$\begingroup\$ Besides I have no idea where the 5.3kHz here come from, all you have to do is make sure it can stand the higher voltage. If you want isolation, add isolation, if you dont want it, dont do it. \$\endgroup\$
    – PlasmaHH
    Aug 18, 2015 at 15:54

2 Answers 2

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The "safety" of the circuit depends on the voltage rating of the capacitors. 450 V should be high enough for either 120 VAC or 250 VAC.

However, the corner frequency of the filter with the component values shown is 50 kHz, not 5.3 kHz.

If you use an additional isolation transformer, it will have a strong low-pass effect that may wipe out the signal you're looking for.

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  • \$\begingroup\$ Thank you for the comment. Sorry if I didn't express my question clearly enough, English is not my first language. I really want to know how to go from 50kHz to 5.3kHz, I understand I could just change the resistance or the capacitors of the circuit but I don't know what makes the most sense and will maintain a better signal, while remaining safe. [I edited the main post] \$\endgroup\$
    – Sorade
    Aug 19, 2015 at 11:21
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    \$\begingroup\$ If you multiply all of the resistor values by 10, you'll get a cutoff frequency of 5 kHz, which might be close enough. If you really care about the difference between 5.0 and 5.3 kHz, change R1 to 910 ohms instead of 1000 ohms. \$\endgroup\$
    – Dave Tweed
    Aug 19, 2015 at 11:43
  • \$\begingroup\$ Thanks Dave. Purely out of curiosity, why only change the resitor values rather than the capacitors or both ? \$\endgroup\$
    – Sorade
    Aug 19, 2015 at 12:02
  • \$\begingroup\$ Because resistors are cheaper than capacitors? You really didn't put very many constraints on the design, so I chose the easiest path. \$\endgroup\$
    – Dave Tweed
    Aug 19, 2015 at 12:11
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@ Dave Tweed.

To answer your last comment. The only information I have on the protocol to follow is : "To measure the EMI on the power line, we used the same hardware used in LightWave [4]. An analog high-pass filter (HPF) with a 3 dB corner frequency of 5.3 kHz is used to reject the strong 60 Hz component. The output of the HPF is sampled at 1 MS/s using a 12-bit analog-to-digital (ADC) converter in the USRP (Universal Software Radio Peripheral) followed by transforming the signal into frequency domain using a 32,768-point fast Fourier transform (FFT), yielding a frequency resolution (or bin size) of 30.5 Hz. The signal from the USRP is then fed into the computer for data analysis. It should be noted that a USRP was used in this prototype simply for convenience."

So I don't know if there is some unspoken implication this might apply on the circuit ?

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