Question

Design an op amp-based low-pass filter with a cut- off frequency of 500 Hz and a passband gain of 8 using a 5 uF capacitor. Draw your circuit, labeling the component val- ues and output voltage

Answer 1

To design a low-pass filter with a cutoff frequency of 500 Hz and a passband gain of 8 using a 5 µF capacitor, an operational amplifier (op-amp) in an inverting configuration with a resistor and capacitor in the feedback loop can be employed. The circuit design comprises a resistor R1 = 10 kΩ and a capacitor C1 = 5 µF in the feedback loop of the op-amp, with the output voltage labeled as Vout.

Step-by-step explanation:

The cutoff frequency of a low-pass filter can be determined using the formula:
`\(f_c = (1)/(2\pi R_1 C_1)\)`, where
`\(f_c\)` is the cutoff frequency,
`\(R_1\)` is the resistor in the feedback loop, and
`\(C_1\)` is the capacitor in the feedback loop. Rearranging the formula to solve for the resistor value yields
`\(R_1 = (1)/(2\pi f_c C_1)\)`. Given the cutoff frequency
`\(f_c = 500\)` Hz and
`\(C_1 = 5 \mu F\)`, the resistor value is calculated as
`\(R_1 = (1)/(2\pi * 500 * 5 * 10^(-6))\), resulting in \(R_1 \approx 10 \, \text{k}\Omega\)`.

For the op-amp to provide the desired passband gain of 8, it should be set up in an inverting configuration with the gain equation
`\(G = -(R_f)/(R_i)\),` where
`\(R_f\)` is the feedback resistor and
`\(R_i\)` is the input resistor. Here, to achieve a gain of 8, the ratio of
`\(R_f\)` to
`\(R_i\)` should be 8. Choosing a convenient value for
`\(R_i\)` (let's say
`\(R_i = 1 \, \text{k}\Omega\)`, c can be calculated as
`\(R_f = 8 * R_i = 8 * 1 \, \text{k}\Omega = 8 \, \text{k}\Omega\)`.

With the calculated values, the circuit consists of an operational amplifier configured in an inverting mode, using a feedback resistor
`\(R_f = 8 \, \text{k}\Omega\)` and a capacitor
`\(C_1 = 5 \mu F\)`in the feedback loop. The input resistor
`\(R_i\)` can be selected as
`\(1 \, \text{k}\Omega\)`, providing a passband gain of 8. The output voltage, labeled as Vout, represents the filtered signal with frequencies below the cutoff frequency of 500 Hz passed through the circuit while attenuating higher frequencies.

Answer 2

The filter will attenuate frequencies above the cut-off frequency ( and pass frequencies below the cut-off frequency where the passband gain of the filter is 8.

The Component Values are highlighted below:

C1 = 5 μF (Capacitor)

R1 = 1/2πfC = 1/2π * 500 Hz * 5 μF

= 636.6 Ω (Resistor)

R2 = R1 * (Gain - 1) = 636.6 Ω * (8 - 1)

= 4453 Ω (Resistor)

The output voltage (Vout) of the low-pass filter can be calculated using the following formula:

Vout = Vin * Gain / (1 + jωRC)

Gain = 8 (passband gain)

ω = 2πf (angular frequency)

f = 500 Hz (cut-off frequency)

R = R1 (feedback resistor)

C = C1 (capacitor)

Frequency Response: