Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10EE201 Lecture 27 P. 1Op Amp Basics—Ideal Op AmpsRevisit Op Amp with negative feedback+_v-v+Riv+ - v-)voRfRsRoFeatures: (v+ - v-) is small; A is large (104)Ro is as small as possibleVoltage input (not current)+_EE201 Lecture 27 P. 2Consider modifications to simplify analysis v+ = v-Implication of this assumption: voltage transfer characteristic is modified as shown below,(v+ - v-) voVoltage gain becomes infiniteVirtual short conditions: v+ = v-A = ∞EE201 Lecture 27 P. 3Output resistance Ro = 0 ΩImplication of this assumption: no voltage drop after dependent source (ideal output behavior)Voltage input (minimal currents into op amp) Ri = ∞ Implication of this assumption: i+ = i- = 0Performing KCL on op amp (see p. 3, lect. 26) i+ + i- + io + ic+ + ic- = 0Using above reduces this to, io = -(ic+ + ic-)EE201 Lecture 27 P. 4Ideal Op Amp conditionsRi = A = Ro = 0+_v-v+i-voi+v+ - v-)i- = 0 i+ = 0v- = v+Open circuit at input+_Now, revisit inverting and noninverting op amps.Ri = ∞Ro = 0EE201 Lecture 27 P. 5Inverting Amplifier+_25k 50mV100k+_ voifis+_Example: Find vo assuming ideal behavior.Solution:1) Compute v+ and v-. v+ = 0 (grounded)Using ideal op amp model, v+ = v-: v- = 0 2) Compute is. From Ohm’s law, is = vs/Rs is = 50 mV / 25kV = 2 AEE201 Lecture 27 P. 63) Compute if. From Ohm’s law, if = vo / Rfif = vo / 100k4) Relate is to if. From KCL, i- = is + if is = - if Because i- = 0 Therefore, is = vs / Rs = - if = - vo / Rfvo = - (Rf / Rs) vs vo = - (100k / 25k) 50 mV = -200mVThis is called an inverting amplifier because the output and input voltage polarity are opposite. One of the input terminals in this example is grounded (single-ended mode), which means the other is virtually grounded, because of the constraints of the virtual short circuit model (i.e. v- = v+).EE201 Lecture 27 P. 7Noninverting AmplifierGgGf+_ voifig+_Example: Find the voltage gain.Solution:1) Compute v+ and v-. v+ = vsUsing ideal op amp model, v+ = v-: v- = vs 2) KCL@A: i- = ig + if ig = - if +_Avsi-i+EE201 Lecture 27 P. 83) Express if and iin in terms of voltages. From Ohm’s law: ig = Gg (-VA) = - Gg vsif = Gf (vo - vs)4) Use current relation to find voltage gain. Gf (vo - vs) = Gg vs Voltage gain = (vo / vs) = (Gf + Gg) / Gf (vo / vs) = 1 + (Gg / Gf)Note that gain is positive. If noninverting terminal is more positive than inverting terminal, the output voltage will have the same polarity as the input voltage (amplifier is noninverting). If inverting terminal is more positive than noninverting terminal, the output voltage will have the opposite polarity as the input voltage (amplifier is inverting).EE201 Lecture 27 P. 9Summing AmplifierGf+_ vo+_+_vsG1G2G3G4v1v2v3v4Analyze circuit to show properties of this amplifier.GsEE201 Lecture 27 P. 10Approach:1) KCL@v- node:G1 (v1 - v-) + G2 (v2 - v-) = Gf (v- - vo)v- = (v1G1 + v2G2 + voGf) / (G1 + G2 + Gf) 2) KCL@v+ node:G3 (v3 - v+) + G4 (v4 - v+) = Gi (v+ - vs)v+ = (v3G3 + v4G4 + vsGs) / (G3 + G4 + Gs) 3) Equate V+ to V- (virtual ground):[Gf /(G1 + G2 + Gf )] vo = -[(G1v1+G2v2) / (G1+G2+Gf)]+[(G3v3+G4v4+Gsvs) / (G3+G4+Gs)]Output voltage is weighted sum of input voltages. See Examples 4.5 and 4.6 in
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