Schedule Date 3 Nov Day Mon Class No

Schedule… Date 3 Nov Day Mon Class No. 18 Title Operational Amplifiers 4 Nov Tue 5 Nov Wed 19 6 Nov Thu 7 Nov Fri Recitation 8 Nov Sat 9 Nov Sun 10 Nov Mon 20 12 Nov Tue Chapters Lab Due date LAB 6 13. 1 – 13. 2 HW 8 Exam Review LAB 7 ECEN 301 Exam 8. 4 Binary Numbers HW Due date Discussion #18 – Operational Amplifiers EXAM 2 1

Give to Receive Alma 34: 28 And now behold, my beloved brethren, I say unto you, do not suppose that this is all; for after ye have done all these things, if ye turn away the needy, and the naked, and visit not the sick and afflicted, and impart of your substance, if ye have, to those who stand in need—I say unto you, if ye do not any of these things, behold, your prayer is vain, and availeth you nothing, and ye are as hypocrites who do deny the faith. ECEN 301 Discussion #18 – Operational Amplifiers 2

Lecture 18 – Operational Amplifiers Answer questions from last lecture Continue with Different Op. Amp configurations ECEN 301 Discussion #18 – Operational Amplifiers 3

Op-Amps – Open-Loop Model 1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an op. Amp form a closed circuit when (i 1 = i 2 = 0) ? i 1 – + + v– vin + v+ i 1 – + i 2 v– io – + – Rin vo – – – v+ + Rout + + AOLvin– i 2 + vo – NB: op-amps have near-infinite input resistance (Rin) and very small output resistance (Rout) AOL – open-loop voltage gain ECEN 301 Discussion #17 – Operational Amplifiers 4

Op-Amps – Open-Loop Model 1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an op. Amp form a closed circuit when (i 1 = i 2 = 0) ? i 1 v– – – Ideally i 1 = i 2 = 0 (since Rin → ∞) Rin v+ + Rout + + AOLvin– i 2 + vo – What happens as AOL → ∞ ? → v– ≈ v+ ECEN 301 Discussion #17 – Operational Amplifiers 5

Op-Amps – Closed-Loop Mode 1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an op. Amp form a closed circuit when (i 1 = i 2 = 0) ? RF RS i. S v. S(t) + – v– i. F – i 1 v+ + + vo – ECEN 301 Discussion #17 – Operational Amplifiers 6

Op-Amps – Closed-Loop Mode 1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an op. Amp form a closed circuit when (i 1 = i 2 = 0) ? RF RS i. S v. S(t) + – v– i. F – NB: if AOL is very large these terms → 0 i 1 v+ + + vo – ECEN 301 NB: if AOL is NOT the same thing as ACL Discussion #17 – Operational Amplifiers 7

Op-Amps – Closed-Loop Mode 1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an op. Amp form a closed circuit when (i 1 = i 2 = 0) ? va R R – + R 1 R 2 vb R ECEN 301 R – + vo R – + NB: Current flows through R 1 and R 2 Discussion #17 – Operational Amplifiers 8

Op-Amps – Closed-Loop Mode 1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an op. Amp form a closed circuit when (i 1 = i 2 = 0) ? va R R – + R 1 R 2 vb R R – + vo R – + NB: Inverting amplifiers and (RS = RF) → vo = -vi ECEN 301 Discussion #17 – Operational Amplifiers 9

Op-Amps – Closed-Loop Mode 1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an op. Amp form a closed circuit when (i 1 = i 2 = 0) ? i. F R 1 -va i 1 R 2 R – + vo -vb i 2 ECEN 301 Discussion #17 – Operational Amplifiers 10

More Op. Amp Configurations ECEN 301 Discussion #18 – Operational Amplifiers 11

Op-Amps – Closed-Loop Mode The Differential Amplifier: the signal to be amplified is the difference of two signals RF i. F RS i. S v. S 1(t) + – v. S 2(t) + – ECEN 301 RS v– – i 1 v+ + i 1 RF + vo – Discussion #18 – Operational Amplifiers 12

Op-Amps – Closed-Loop Mode The Differential Amplifier: the signal to be amplified is the difference of two signals RF NB: an ideal op-amp with negative feedback has the properties i. F RS + – i. S v. S 1(t) v. S 2(t) + – ECEN 301 RS v– – i 1 v+ + i 2 RF + vo – Discussion #18 – Operational Amplifiers 13

Op-Amps – Closed-Loop Mode The Differential Amplifier: the signal to be amplified is the difference of two signals RF i. F RS + – i. S v. S 1(t) v. S 2(t) + – ECEN 301 RS v– – i 1 v+ + i 2 RF + vo – Discussion #18 – Operational Amplifiers 14

Op-Amps – Closed-Loop Mode The Differential Amplifier: the signal to be amplified is the difference of two signals RF i. F RS + – i. S v. S 1(t) v. S 2(t) + – ECEN 301 RS v– – i 1 v+ + i 2 RF + vo – Discussion #18 – Operational Amplifiers 15

Op-Amps – Level Shifter: can add or subtract a DC offset from a signal based on the values of RS and/or Vref RF Vsensor RS AC voltage with DC offset v– – v+ + Vref + – DC voltage ECEN 301 Discussion #18 – Operational Amplifiers + vo – 16

Op-Amps – Level Shifter Example 1: design a level shifter such that it can remove a 1. 8 V DC offset from the sensor signal (Find Vref) RS = 10 kΩ, RF = 220 kΩ, vs(t) = 1. 8+0. 1 cos(ωt) RF Vsensor RS v– – v+ + Vref ECEN 301 + – + vo – Discussion #18 – Operational Amplifiers 17

Op-Amps – Level Shifter Example 1: design a level shifter such that it can remove a 1. 8 V DC offset from the sensor signal (Find Vref) RS = 10 kΩ, RF = 220 kΩ, vs(t) = 1. 8+0. 1 cos(ωt) RF Vsensor RS – v+ + Vref ECEN 301 + – Find the Closed-Loop voltage gain by using the principle of superposition on each of the DC voltages v– + vo – Discussion #18 – Operational Amplifiers 18

Op-Amps – Level Shifter Example 1: design a level shifter such that it can remove a 1. 8 V DC offset from the sensor signal (Find Vref) RS = 10 kΩ, RF = 220 kΩ, vs(t) = 1. 8+0. 1 cos(ωt) RF RS Vsensor + – DC from sensor: Inverting amplifier v– – v+ + + vo – ECEN 301 Discussion #18 – Operational Amplifiers 19

Op-Amps – Level Shifter Example 1: design a level shifter such that it can remove a 1. 8 V DC offset from the sensor signal (Find Vref) RS = 10 kΩ, RF = 220 kΩ, vs(t) = 1. 8+0. 1 cos(ωt) RF DC from reference: Noninverting amplifier v– RS – v+ + Vref + – + vo – ECEN 301 Discussion #18 – Operational Amplifiers 20

Op-Amps – Level Shifter Example 1: design a level shifter such that it can remove a 1. 8 V DC offset from the sensor signal (Find Vref) RS = 10 kΩ, RF = 220 kΩ, vs(t) = 1. 8+0. 1 cos(ωt) RF Vsensor RS v– – v+ + Vref ECEN 301 + – + vo – Discussion #18 – Operational Amplifiers 21

Op-Amps – Level Shifter Example 1: design a level shifter such that it can remove a 1. 8 V DC offset from the sensor signal (Find Vref) RS = 10 kΩ, RF = 220 kΩ, vs(t) = 1. 8+0. 1 cos(ωt) RF Vsensor RS v– – v+ + Vref ECEN 301 + – Since the desire is to remove all DC from the output we require: + vo – Discussion #18 – Operational Amplifiers 22

Op-Amps – Ideal Integrator The Ideal Integrator: the output signal is the integral of the input signal (over a period of time) CF The input signal is AC, but not necessarily sinusoidal RS i. S(t) v. S(t) + – i. F(t) v– – i 1 v+ + + vo(t) – ECEN 301 Discussion #18 – Operational Amplifiers NB: Inverting amplifier setup with RF replaced with a capacitor 23

Op-Amps – Ideal Integrator The Ideal Integrator: the output signal is the integral of the input signal (over a period of time) CF RS i. S(t) v. S(t) + – i. F(t) v– – i 1 v+ + + vo(t) – ECEN 301 Discussion #18 – Operational Amplifiers 24

Op-Amps – Ideal Integrator Example 2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10 ms, CF = 1 u. F, RS = 10 kΩ CF RS i. S(t) v. S(t) + – A i. F(t) v– – T/2 i 1 v+ + T + vo(t) -A – ECEN 301 Discussion #18 – Operational Amplifiers 25

Op-Amps – Ideal Integrator Example 2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10 ms, CF = 1 u. F, RS = 10 kΩ CF RS i. S(t) v. S(t) + – i. F(t) v– – i 1 v+ + + vo(t) – ECEN 301 Discussion #18 – Operational Amplifiers 26

Op-Amps – Ideal Integrator Example 2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10 ms, CF = 1 u. F, RS = 10 kΩ NB: since the vs(t) is periodic, we can find vo(t) over a single period – and repeat CF RS i. S(t) v. S(t) + – i. F(t) v– – i 1 v+ + + vo(t) – ECEN 301 Discussion #18 – Operational Amplifiers 27

Op-Amps – Ideal Integrator Example 2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10 ms, CF = 1 u. F, RS = 10 kΩ NB: since the vs(t) is periodic, we can find vo(t) over a single period – and repeat CF RS i. S(t) v. S(t) + – i. F(t) v– – i 1 v+ + + vo(t) – ECEN 301 Discussion #18 – Operational Amplifiers 28

Op-Amps – Ideal Integrator Example 2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10 ms, CF = 1 u. F, RS = 10 kΩ CF T/2 RS i. S(t) v. S(t) + – T i. F(t) v– – i 1 v+ + + vo(t) -50 AT – ECEN 301 Discussion #18 – Operational Amplifiers 29

Op-Amps – Ideal Differentiator The Ideal Differentiator: the output signal is the derivative of the input signal (over a period of time) RF The input signal is AC, but not necessarily sinusoidal i. F(t) CS v– – i. S(t) v. S(t) + – i 1 v+ + + vo(t) – ECEN 301 Discussion #18 – Operational Amplifiers NB: Inverting amplifier setup with RS replaced with a capacitor 30

Op-Amps – Ideal Differentiator The Ideal Differentiator: the output signal is the derivative of the input signal (over a period of time) RF i. F(t) CS v– – i. S(t) v. S(t) + – i 1 v+ + + vo(t) – ECEN 301 NB: this type of differentiator is rarely used in practice since it amplifies noise Discussion #18 – Operational Amplifiers 31

Op-Amps – Closed-Loop Mode Circuit Diagram RS Inverting Amplifier + – v. S RS 1 + – Summing Amplifier + – RS 2 v. S 2 RF – + + vo – RF + vo – RSn + – ECEN 301 v. S 1 ACL v. Sn Discussion #18 – Operational Amplifiers 32

Op-Amps – Closed-Loop Mode Circuit Diagram RF RS Noninverting Amplifier Voltage Follower ECEN 301 – + R v. S + – – + + – ACL + vo – Discussion #18 – Operational Amplifiers 33

Op-Amps – Closed-Loop Mode Circuit Diagram RF RS Differential Amplifier ECEN 301 + – – + RS v 1 + – v 2 ACL RF + vo – Discussion #18 – Operational Amplifiers 34

Op-Amps – Closed-Loop Mode Circuit Diagram RS Ideal Integrator + – v. S ACL CF – + + vo(t) – RF CS – Ideal Differentiator ECEN 301 + – v. S + + vo(t) – Discussion #18 – Operational Amplifiers 35

Op-Amps Example 3: find an expression for the gain if vs(t) is sinusoidal CF = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, CS = 1/6 F CF vs(t) R 1 R 2 i 1(t) i 2(t) i. S(t) ECEN 301 i. F(t) CS v+ + i 1 v– vo(t) – Discussion #18 – Operational Amplifiers 36

Op-Amps Example 3: find an expression for the gain CF = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, CS = 1/6 F ZF=1/jωCF Node b Vs(jω) Z 1 I 1(jω) Node a Z 2 I 2(jω) ZS IF(jω) v+ + 1. 2. Transfer to frequency domain Apply KCL at nodes a and b Vo(jω) Iin v– IS(jω) – NB: v+ = v– = vo and Iin = 0 ECEN 301 Discussion #18 – Operational Amplifiers 37

Op-Amps Example 3: find an expression for the gain CF = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, CS = 1/6 F ZF=1/jωCF Node b Vs(jω) Z 1 I 1(jω) Node a ECEN 301 Z 2 I 2(jω) ZS IS(jω) IF(jω) v+ + 1. 2. Transfer to frequency domain Apply KCL at nodes a and b Vo(jω) Iin v– – Discussion #18 – Operational Amplifiers 38

Op-Amps Example 3: find an expression for the gain CF = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, CS = 1/6 F ZF=1/jωCF Vs(jω) Z 1 I 1(jω) IF(jω) Z 2 I 2(jω) ZS IS(jω) ECEN 301 v+ + 1. 2. 3. Transfer to frequency domain Apply KCL at nodes a and b Express Vo in terms of Vs Vo(jω) Iin v– – Discussion #18 – Operational Amplifiers 39

Op-Amps Example 3: find an expression for the gain CF = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, CS = 1/6 F ZF=1/jωCF Vs(jω) Z 1 I 1(jω) IF(jω) Z 2 I 2(jω) ZS IS(jω) ECEN 301 v+ + Vo(jω) 1. 2. 3. 4. Transfer to frequency domain Apply KCL at nodes a and b Express Vo in terms of VS Find the gain (Vo/VS) Iin v– – Discussion #18 – Operational Amplifiers 40