The following circuit schematic presents a two-pole Sallen-Key low pass filter.
Determine the value of resistor R1 to achieve the Butterworth frequency response.
Calculate the cut-off frequency of the filter.
RA = 10 kΩ
Rb = 10 kΩ
Ca = 100nF
Cb = 100nF
R2 = 10 kΩ
VCC = 12V
VEE = -12V
To achieve the Butterworth frequency response, we need to use the formula:
R1 = R2 / (√(2) - 1)
R1 = 10 kΩ / (√(2) - 1) ≈ 2.61 kΩ
To calculate the cut-off frequency, we use the formula:
fC = 1 / (2π√(R1R2C1C2))
fC = 1 / (2π√(2.61 kΩ × 10 kΩ × 100 nF × 100 nF))
fC ≈ 159.2 Hz
Therefore, the cut-off frequency of the filter is approximately 159.2 Hz.
To determine the value of resistor R1 to achieve the Butterworth frequency response, we need to use the Butterworth filter design equations.
The Butterworth filter is a type of low pass filter with a maximally flat frequency response in the passband. The cutoff frequency, also known as the 3dB frequency, is the frequency where the output voltage is reduced to 70.7% (1/sqrt(2)) of the input voltage.
The Butterworth filter design equations for the Sallen-Key topology are as follows:
Fc = 1 / (2 * pi * R * C)
Where Fc is the cutoff frequency, R is the resistance, and C is the capacitance.
In this circuit, we have Ca, Cb, and R2 already provided. We need to find R1 and calculate the cutoff frequency.
Let's first calculate the cutoff frequency:
Fc = 1 / (2 * pi * R2 * Cb) = 1 / (2 * pi * 10kΩ * 100nF) = 1 / (2 * pi * 10^4Ω * 10^(-7)F) = 1 / 6.283 * 10^(-3) = 15.92Hz
Now, let's determine the value of resistor R1:
To achieve the Butterworth frequency response, we need to make the transfer function of the filter have a Q of 0.7071, which is obtained by setting R1 to be equal to R2.
So, R1 = R2 = 10 kΩ
Hence, the value of resistor R1 to achieve the Butterworth frequency response is 10 kΩ, and the cutoff frequency of the filter is approximately 15.92 Hz.