Find $R_1$ and $R_2$ in lossy integrator so that the peak gain is 20 dB and gain is 3dB down from its peak

when $ω$=10000 rad/s . Use capacitance of 0.01 MF Given: Max gain = 20 dB, 3dB frequency. $ω_s=10000 \frac{rad}{s} ,Cr=0.01Mf$


enter image description here

Component values:

$∴\text{DC gain} = 20 dB= 20 \log_{10⁡}|Av| \\ ∴|Av|=10 \\ \text{But} |Av|=\frac{R_2}{R_1} \\ ∴R_2=10 R_1 \\ \text{Also,} w_a=10000 \frac{rad}{\sec} \\ ∴f_a=\frac{10000}{2π}=1591.55 Hz \\ \text{But} f_a=\frac{1}{2πR_2 C_f} \\ ∴C_f=0.01 \text{MF given} \\ ∴R_2=\frac{1}{2πf_a cf}=\frac{1}{10000 \times 0.01 \times 10^{-6}} \\ ∴R_2=10 kΩ \\ ∴ \text{substituting,} 10 kΩ=10 R_1 \\ ∴R_1=1 kΩ$

Please log in to add an answer.

Next up

Read More Questions

If you are looking for answer to specific questions, you can search them here. We'll find the best answer for you.


Study Full Subject

If you are looking for good study material, you can checkout our subjects. Hundreds of important topics are covered in them.

Know More