Question

How do you analyze a simple active low-pass filter circuit?

Answer 1

Analyzing a simple active low-pass filter circuit involves understanding its components, analyzing the transfer function, and examining its frequency response. Let's walk through the steps to analyze the circuit:

Step 1: Circuit Components
A basic active low-pass filter consists of an operational amplifier (op-amp) and a few passive components: resistors and capacitors. The most common configuration is the "inverting" active low-pass filter, where the output is fed back to the inverting input of the op-amp. The basic schematic of the circuit is as follows:

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       R1
  Vin ---\/\/\---+----------------- Vout
                |     
               --- C1
               ---
                |     
                |
                |
               ===
                |
               GND


Where:

Vin is the input voltage
Vout is the output voltage
R1 is the feedback resistor
C1 is the capacitor

Step 2: Transfer Function
To analyze the circuit, you need to find the transfer function, which relates the input voltage to the output voltage as a function of frequency.

The transfer function for an active low-pass filter is given by the following equation:

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Vout(s)/Vin(s) = -R1 * C1 * s / (1 + R1 * C1 * s)


Where:

Vout(s) and Vin(s) are the Laplace transforms of Vout and Vin, respectively.
's' is the complex frequency variable (s = jω, where 'j' is the imaginary unit and 'ω' is the angular frequency in rad/s).

Step 3: Frequency Response
The frequency response of the filter describes how the output voltage varies with the input frequency. To obtain the frequency response, we replace 's' with 'jω' in the transfer function:

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Vout(jω)/Vin(jω) = -R1 * C1 * jω / (1 + R1 * C1 * jω)


The magnitude of the frequency response (|H(jω)|) is given by:

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|H(jω)| = |Vout(jω)/Vin(jω)| = R1 / √(1 + (R1 * C1 * ω)^2)


The phase shift (φ) of the frequency response is given by:

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φ = atan(-1 / (R1 * C1 * ω))


Step 4: Cutoff Frequency
The cutoff frequency (f_c) is the frequency at which the magnitude of the frequency response is attenuated by 3 dB. At this frequency, the output voltage is reduced to approximately 70.7% of the maximum (ideal) value. The cutoff frequency (f_c) is calculated using the following formula:

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f_c = 1 / (2 * π * R1 * C1)


Step 5: Bode Plot
A Bode plot is a graphical representation of the frequency response of the filter. It shows the magnitude and phase shift of the output voltage as a function of frequency. To create a Bode plot, you can plot the magnitude and phase shift equations as functions of frequency (ω).

By following these steps, you can analyze a simple active low-pass filter circuit and understand its behavior at different frequencies. This analysis is essential for designing and optimizing the filter for specific applications.