Question

Assuming an ideal op-amp find the 3dB frequency of the circuit, if R1=2.4 KS2, R2=13.4 K2, R3=20 KS2, and C=5 nF. HH C R2 R1 + Vi(jw) Vo R3 IH Answer: rad/s

Answer 1

1/sc HE R2 From the concept of virtual ground and considering op-amp to be ideal. Vi V+ = V- = OV Ri V- • Vo Rz V+ + EO Applying KCL at point A we get :- Vq-o Vo-o + Vo-o RU R2 1/sc Vi =-vol14 SR2C Ri R2 Re → Vo R2 Vi RI I+SR₂C = 1 - Vol kat sc) (it sokac) Magnitude at 3 dB frequency should be 1/2 Gitme aasta autor) - > 2 13.4K 2.4K 1 1+ (w X134x106x5) 7.896 1 * w2x(4.489x109) 61-3472 Solving for w, we get w=116.9Kvad/s or w= 116.9x103 rad/s So , 348 frequency is 116.9 x103 rad /s

Vo Vi wrac) • From equation D the transfer function is given by :- - R2 Ri It jwR 2C (here, saju) DC gain of the circuit is given at w=0 rad/sec. Thus, putting w=0 rad's we will get the DC gain. =-13.4R2 - 5.583 Ri 2.4K2 * DC gain - -R2 IDC gain = - 5.583 From equation 0 at frequency 2KHz :- R2 L-tan"WR2C RI Vo Vi 1+ WR2C Angle. Magnitude Thus angle of bransfes function = L-tan' 2n6R2C L-tañ' (211x2x10°x 795-74X100X109 L-tanil = -45 Angle of transfer function → Angle of transfer function

· From equation o 1 Vo Vi - R2 Ri It j w Rac At w=0 rad/s . Vo Vi Ra Ri ention At W= (infinity) rad/s ; Vo Vi So this filter passes signals with Low frequency and attenuates signals with high frequency. Thus, this filter circuit is a low Frequency pass Filtes So option '3' is correct Request Kindly upvote if you liked by answer. Thank you !