I need assistance in refining my AC capacitor coupling circuit to effectively filter the DC component of my input signal.

The circuit consists of a current generator, a 1 Ω resistor, and an AC coupling circuit.

enter image description here

The input signal features four sinusoids with frequencies 13.9085, 16.9993, 20.0901, and 27.8171 Hz, a DC level of 12 mA, and a range of 4-20 mA.

Given the low frequencies of my signal, I've opted for higher capacitance and resistor values to achieve a lower cut-off frequency.

This is my input signal:

enter image description here

Despite setting capacitance and resistor values for a very low cut-off frequency, the results of my transient simulation are not as expected. For instance, in the attached picture, I anticipated a voltage difference of 1 Ω·12 mA = 12 mV, but the actual result differs.

enter image description here

Upon examining the FFT, I believe the amplitude of the green signal should be less than the one of the current signal, am I correct?

enter image description here


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1 Answer 1


The FFT is plotted in logarithmic scale, so a difference of let's say 2x won't be readily visible with 13 orders of magnitude on the vertical axis (!). Stop the plot at 100Hz, it's all noise past that.

Even if there was a small difference in amplitude on the FFT plot, you wouldn't be able to see it with the huge vertical axis range. A 2x amplitude difference is only a 6dB difference. But you have 260dB of vertical axis range...

Current and voltage have different units yet are on the same plot, but since a 1Ω sampling resistor is used, they happen to have the same numeric values. You seem to expect different numeric values - why?

the results of mu transient simulation are not as expected

The p-p current is 20mA-4mA=16mApp. The p-p voltage is 4mV-(-12V)=16mVpp. They agree, and are the same amplitude. I think you just misread the plots. Make sure you take the signs into account.

I believe the amplitude of the green signal should be less than the one of the current signal

It shouldn't be, because if you inspect the time-domain plots, they agree exactly as they should, due to the scaling constant being a numerical 1 (1Ω=1V/1A).

I've opted for higher capacitance and resistor values to achieve a lower cut-off frequency.

You've set the cut-off frequency to about 0.1Hz. The filter will need about a minute for the DC average to settle to zero.

This very low cut-off is unnecessary given that the frequencies of interest are all 2 orders of magnitude higher (≥10Hz vs 0.1Hz).

You should replace the 1MΩ resistors with 100kΩ resistors, for a 1Hz cutoff. The settling time will be a couple of seconds instead of a minute.

With a 16..24-bit ADC, the AC coupling trick won't be necessary at all. You can attach the ADC input directly to the 1-ohm resistor, and add overvoltage clamps to protect the ADC from transients. The ADC should be isolated from the rest of the circuit. Modern ADCs are very low power devices and a combined integrated signal+power isolator will do well for the ADC.

  • \$\begingroup\$ I expected a different offset value in the voltage result. I designed the current to produce a signal oscillating with an amplitude of 8mA and a DC level of 12mA. Consequently, I expected the resulting signal to have a p-p voltage of 16mV (and it works) with a DC level of 0V. For istance, at t=133ms, I1=4.5mA and VR=-11.7mV (I expected it to be 4.5m-12m=-7.5m). Based on the plot, it appears to have an offset of approximately -4mV, contrary to my expectation. \$\endgroup\$ Feb 25 at 16:00

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