CENG4480 Lecture 03: Operational Amplifier 2 Bei Yu - PowerPoint PPT Presentation

CENG4480 Lecture 03: Operational Amplifier – 2 Bei Yu byu@cse.cuhk.edu.hk (Latest update: August 19, 2020) Fall 2020 1 / 21

Overview Preliminaries Integrator & Differentiator Filters 2 / 21

Overview Preliminaries Integrator & Differentiator Filters 3 / 21

Euler’s Identity e j θ = cos θ + jsin θ ◮ real component ◮ imaginary component ◮ magnitude � | e j θ | = cos 2 θ + sin 2 θ = 1 3 / 21

Prove: 1 1 √ | 1 + ja | = 1 + a 2 4 / 21

Sinusoidal Signal x ( t ) = Acos ( ω t + φ ) ◮ Periodic signals ◮ A : amplitude ◮ ω : radian frequency ◮ φ : phase 5 / 21

Time Domain ◮ Voltage gain against time For sinusoidal signal: v ( t ) = Acos ( ω t + φ ) +1V Voltage 0 Time (second) -1V (usually linear scale) 6 / 21

Frequency Domain ◮ Voltage gain against frequency For sinusoidal signal: V ( j ω ) = Ae j φ = A ∠ φ = Acos φ + jAsin φ Power 0dB Gain (dB) -3dB Frequency (Hz) (can use log scale) 7 / 21

Impedance A complex resistance or frequency-dependent resistance . That is, as resistors whose resistance is a function of the frequency of the sinusoidal excitation. 8 / 21

Resistor Impedance Assume source voltage v = A cos( ω t ) , then ◮ V ( j ω ) = A ∠ 0 ◮ I ( j ω ) = A R ∠ 0 Impedance of A Resistor Z R ( j ω ) = V ( j ω ) I ( j ω ) = R ∠ 0 = R 9 / 21

Capacitor ABC Capacitance C A measure of how much charge a capacitor can hold. ◮ Amount of charge Q = C · V ◮ current is the rate of movement of charge: I = dQ dt = C · dV dt 10 / 21

Capacitor Impedance V ( j ω ) = A ∠ 0 I ( j ω ) = ω CA ∠ π/ 2 Impedance of A Capacitor Z C ( j ω ) = V ( j ω ) I ( j ω ) 11 / 21

Capacitor Impedance V ( j ω ) = A ∠ 0 I ( j ω ) = ω CA ∠ π/ 2 Impedance of A Capacitor Z C ( j ω ) = V ( j ω ) I ( j ω ) 11 / 21

1 Z C ( j ω ) = j ω C Capacitor Rule 1 Low Frequency ⇒ Open circuit Capacitor Rule 2 High Frequency ⇒ Short circuit 12 / 21

Overview Preliminaries Integrator & Differentiator Filters 13 / 21

Frequency Response of An Op-Amp 13 / 21

Frequency Response of An Op-Amp ◮ Inverting amplifier: V out ( j ω ) = − Z F V S Z S 13 / 21

Frequency Response of An Op-Amp ◮ Inverting amplifier: V out ( j ω ) = − Z F V S Z S ◮ Non-Inverting amplifier: V out ( j ω ) = 1 + Z F V S Z S 13 / 21

Integrator i S ( t ) = − i F ( t ) i S ( t ) = v S ( t ) R S i F ( t ) = C F · dv out ( t ) dt 14 / 21

Integrator i S ( t ) = − i F ( t ) i S ( t ) = v S ( t ) R S i F ( t ) = C F · dv out ( t ) dt Therefore: � t 1 v S ( t ′ ) dt ′ v out ( t ) = − R S C F −∞ 14 / 21

Differentiator i S ( t ) = C S · dv S ( t ) dt i F ( t ) = v out ( t ) R F 15 / 21

Differentiator i S ( t ) = C S · dv S ( t ) dt i F ( t ) = v out ( t ) R F Therefore: v out ( t ) = − R F C S · dv S ( t ) dt 15 / 21

Overview Preliminaries Integrator & Differentiator Filters 16 / 21

Low-Pass Filter A ( j ω ) = − Z F Z S 1 R F Z F = R F || = j ω C F 1 + j ω C F R F Z S = R S 16 / 21

Low-Pass Filter A ( j ω ) = − Z F Z S 1 R F Z F = R F || = j ω C F 1 + j ω C F R F Z S = R S ⇒ A ( j ω ) = − Z F R F / R S = − 1 + j ω C F R F Z S 16 / 21

Given: A ( j ω ) = − Z F R F / R S = − 1 + j ω C F R F Z S 1 w c = R F C F Prove: | A | = R F 1 · � R S 1 + ω 2 / w 2 c 17 / 21

Low-Pass Filter | A | = R F 1 · � R S 1 + ω 2 / w 2 c 1 ◮ w c = R F C F ◮ 3-dB frequency ◮ or cutoff frequency BTW, lim ω → 0 | A | = R F , lim ω →∞ | A | = 0 R S 18 / 21

High-Pass Filter A ( j ω ) = − Z F Z S 1 Z S = R S + j ω C S Z F = R F ⇒ j ω C S R F A ( j ω ) = − Z F = − 1 + j ω C S R S Z S 19 / 21

High-Pass Filter j ω C S R F A ( j ω ) = − Z F = − Z S 1 + j ω C S R S ω → 0 | A | = 0 lim ω →∞ | A | = R F lim R C High freq. cutoff unintentionally created by Op-amp 20 / 21

Band-Pass Filter 21 / 21