Active Low-pass Filter


The circuit configuration shown on Picture 1 is active low-pass filter. We assume here that the input signal is sinusoidal. Thus, we can work in the frequency domain and use the concepts of impedance and reactance. If we look at the circuit we can see that this is an inverting amplifier where the feedback resistor has been replaced with impedance Zf, which is the parallel combination of the resistor R2 and the capacitor C. Therefore, we can write:

Zf = ((1/sC)*R2)/(1/sC + R2) = R2/(1 + sCR2)


Now, if we write the expression for the output voltage of the inverting amplifier, we will have:

Vo(s) = -(Zf/R1)*Vi(s) = -R2/(R1*(1 + sCR2)) * Vi(s)

So, according to the last expression, the output voltage Vo is large for s small or is small for s large.


Picture 1: Active Low-pass Filter


The frequency-domain characteristic of this circuit configuration is shown on the Picture 2. The magnitude of the output voltage is 20 dB for frequencies of the input signal up to 1.5 kHz. The output voltage decreases for 3 dB at frequency of 1.6 kHz, which is actually the dominant frequency pole for this circuit configuration. So, this circuit pass through the low frequencies up to 1.6 kHz.



Picture 2: AC Analysis - output voltage [dB] and its phase [degrees] (frequency-domain)

2 comments:

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