Question

Problem 1 (25 Pts) Consider the OP amp circuit shown below with R = 100kN and C = 1(10)-5 F: Part a) 10 pts Find the complex transfer function for the circuit in terms of frequency f (Hz). Vi Part b) 5 pts Compute the gain of at as a function of frequency f(Hz). Part c) 10 pts Compute the corresponding gains at 100, 1000, 10000 Hz. Vo R 3R 5R с Vů V.

Answer 1

R 38 io HH 1/sc تل Vi tu 5R А Vo VA=OV TT

To solve this we assume that opamp is ideal. For an ideal opamp, there is a virtual short between the input terminal. As a result, no current goes inside opamp from input terminals.

Apply KCL to node A

lj = 2

Vi - 0 5R 0 – Vo R/SC 3R+ R+1/C
             
R| SC R/SC R+1/SC

Rearranging

$\frac{V_0}{V_i}=-\frac{3R+\frac{R/s C}{R+1/sC}}{5R}$

$\frac{V_0}{V_i}=-\frac{3}{5}-\frac{1}{5} \frac{1}{sRC+1}$

Put s=jw, RC=1

$\frac{V_0}{V_i}=-\frac{3}{5}-\frac{1}{5} \frac{1}{j\omega+1}$

put $\omega=2\pi f$

VO Vi 3 55j2nf +1

b)

To find the gain(magnitude) of this function we should do some modifications first

$\frac{V_0}{V_i}=-\frac{3}{5}-\frac{1}{5} \frac{(j2\pi f-1)}{(j2\pi f+1)(j2\pi f-1)}$

$\frac{V_0}{V_i}=-\frac{3}{5}+\frac{1}{5} \frac{(j2\pi f-1)}{(4\pi^2 f^2+1)}$      ($(a+b)(a-b)=a^2-b^2$)

$\frac{V_0}{V_i}=+\frac{1}{5} \frac{(-12\pi^2f^2-3+j2\pi f-1)}{(4\pi^2 f^2+1)}$

$\frac{V_0}{V_i}=+\frac{1}{5} \frac{(-12\pi^2f^2-4+j2\pi f)}{(4\pi^2 f^2+1)}$

We can take magnitude now

$|\frac{V_0}{V_i}|=+\frac{1}{5} \frac{\sqrt{(-12\pi^2f^2-4)^2+(2\pi f)^2}}{(4\pi^2 f^2+1)}$     $|x+iy|=\sqrt{x^2+y^2}$

$|\frac{V_0}{V_i}|=+\frac{1}{5} \frac{\sqrt{144\pi^4f^4+100\pi^2f^2+16}}{(4\pi^2 f^2+1)}$

This is our gain

c)

Put $f=100$

we get

$|\frac{V_0}{V_i}|=0.600000591$

Put $f=1000$

we get

$|\frac{V_0}{V_i}|=0.600000$

Put $f=10000$

we get

$|\frac{V_0}{V_i}|=0.600000$

Which is almost same

This makes sense also because at high-frequency capacitor act as a short circuit. So basically the gain converges to

$V_o/V_i=-\frac{3R}{5R}=-0.6$