# Licensed Electrical & Mechanical Engineer

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Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Engineering 43 Chp 14-1 Op Amp Circuits Bruce Mayer, PE Licensed Electrical & Mechanical Engineer

Ckts W/ Operational Amplifiers
OpAmp Utility OpAmps Are Very Useful Electronic Components We Have Already Developed The Tools To Analyze Practical OpAmp Circuits The Linear Models for OpAmps Include Dependent Sources A PRACTICAL Application of Dependent Srcs

Maxim (Sunnyvale, CA) Max4241 OpAmp
Real Op Amps LM324 DIP Physical Size Progression of OpAmps Over the Years LMC6294 MAX4240 Maxim (Sunnyvale, CA) Max4241 OpAmp

Apex PA03 HiPwr OpAmp Notice OutPut Rating PwrOut → 30A•75V → 2.25 kW!

OpAmp Symbol & Model The Circuit Symbol Is a Version of the Amplifier TRIANGLE The Linear Model Typical Values OUTPUT RESISTANCE INPUT RESISTANCE GAIN

OpAmp InPut Terminolgy
The Average of the two Input Voltages is called the Common-Mode Signal The Difference between the Inputs is called the Differential-Signal, vid

OpAmp Power Connections
BiPolar Power Supplies UniPolar Supply For Signal I/O Analysis the Supplies Need NOT be shown explicitly But they MUST physically be there to actually Power the Operational Amplifier

OpAmp Circuit Model DRIVING CIRCUIT LOAD OP-AMP

vi→vo Transfer Characteristics
Linear Region vo/vi = Const Saturation The OUTPUT Voltage Level can NOT exceed the SUPPLY the Level

Unity Gain Buffer (FeedBack)
FeedBack Loop Controlling Variable = Note that Ro is connected to v− Solve For Buffer Gain (I = Vin/Ri) Thus The Amplification

UGB Algebra

The Ideal OpAmp The IDEAL Characteristics The Consequences of Ideality
Ro = 0 Ri =  A =  (open loop gain) BW =  The Consequences of Ideality The Open Loop Gain is that found on the spec sheet, The textbook uses AOL

Summing Point Constraint
The MOST important aspects of Ideality: Ri = ∞ → i+ = i− = 0 The OpAmp INput looks like an OPEN Circuit A = ∞ → v+ = v− The OpAmp Input looks like a SHORT Circuit This simultaneous OPEN & SHORT Characteristic is called the → Looks Like an OPEN to Current Looks Like a SHORT to Voltage SUMMING POINT CONSTRAINT

Analyzing Ideal OpAmp Ckts
Verify the presence of NEGATIVE FeedBack Assume the Summing Point Constraint Applies in this fashion: i+ = i−= 0 (based on Ri = ∞) v+ − v− = 0 (based on AO = ∞) Use KVL, KCL, Ohm’s Law, and other linear ckt analysis techniques to determine quantities of interest OpAmps are effectively LINEAR Devices

Voltage Follower The Voltage Follower Usefulness of UGB
Also Called Unity Gain Buffer (UGB) from Before Usefulness of UGB Connection w/o Buffer Buffered Connection VS & RS are the Thevenin Equivalents of the signal source The SOURCE Supplies The Power The Source Supplies NO Power (the OpAmp does it)

Inverting OpAmp Ckt Determine Voltage Gain, G = Vout/Vin Start with Ao
Finally The Gain Now From Input R Apply KCL at v− Next: Examine Ckt w/o Ideality Assumption

Replace OpAmp w/ Linear Model
Consider Again the Inverting OpAmp Circuit Identify the Op Amp Nodes Draw the Linear Model

Drawing the OpAmp Linear Model
Redraw the circuit cutting out the Op Amp Draw components of linear OpAmp (on the circuit of step-2) i R

Drawing the OpAmp Linear Model
UNTANGLE as Needed Before ve = v+ − v− The BEFORE & AFTER After

NonIdeal Inverting Amp
Replace the OpAmp with the LINEAR Model Label Nodes for Tracking b - d b - a Draw The Linear Equivalent For Op-amp Note the External Component Branches

NonIdeal Inverting Amp cont.
On The LINEAR Model Connect The External Components ReDraw Ckt for Increased Clarity Now Must Sweat the Details

NonIdeal Inverting Amp cont.
Node Analysis Note GND Node 2 Controlling Variable In Terms Of Node Voltages The 2 Eqns in Matrix Form

Inverting Amp – Invert Matrix
Use Matrix Inversion to Solve 2 Eqns in 2 Unknowns Very Useful for 3 Eqn/Unknwn Systems as well e.g., The Matrix Determinant  Solve for vo

Inverting Amp – Invert Matrix cont
Then the System Gain Typical Practical Values for the Resistances R1 = 1 kΩ R2 = 5 kΩ Then the Real-World Gain Recall The Ideal Case for A→; Then The Eqn at top

Compare Ideal vs. NonIdeal
Ideal Assumptions Gain for Real Case Replace Op-amp By Linear Model, Solve The Resulting Circuit With Dep. Sources

Compare Ideal vs. NonIdeal cont.
The Ideal Op-amp Assumption Provides an Excellent Real-World Approximation. Unless Forced to do Otherwise We Will Always Use the IDEAL Model Ideal Case at Inverting Terminal Gain for NonIdeal Case

Example  Differential Amp
KCL At Inverting Term KCL at NONinverting Terminal Assume Ideal OpAmp Have Negative FeedBack, so summing point constraint applies By The KCLs A simple Voltage Divider

Example  Differential Amp cont
Then in The Ideal Case Now Set External R’s R4 = R2 R3 = R1 Subbing the R’s Into the vo Eqn

Ex. Precision Diff V-Gain Ckt
Find vo Assume Ideal OpAmps Which Voltages are Set? What Voltages Are Also Known Due To Infinite Gain Assumption? Now Use The Infinite Resistance Assumption Have negative feedback so summing point constraint applies CAUTION: There could be currents flowing INto or OUTof the OpAmps

Ex. Precision Diff V-Gain Ckt cont
The Ckt Reduces To Fig. at Right KCL at v1 KCL at v2 Eliminate va Using The above Eqns and Solve for vo in terms of v1 & v2 Note the increased Gain over Diff Amp OpAmp Current

NONinverting Amp - Ideal
Ideal Assumptions Infinite Gain Since i− = 0 Arrive at “Inverse Voltage Divider” Have negative feedback so summing point constraint applies Infinite Ri with v+ = v1

Example  Find Io for Ideal OpAmp
Ideal Assumptions NONinverting ckt KCL at v−

Example: TransResistance Ckt
The trans-resistance Amp circuit below performs Current to Voltage Conversion Find vo/iS Use the Summing Point Constraint → v− = 0V Now by KCL at v− Node with i− = 0 Notice that iS flows thru the 1Ω resistor Thus by Ohm’s Law Have negative feedback so summing point constraint applies

Key to OpAmp Ckt Analysis  IOA
Remember that the “Nose” of the OpAmp “Triangle” can SOURCE or SINK “Infinite” amounts of Current Can source/sink up to Isat IOA = ± ∞ |IOA,max| = Isat

Example: TransResistance Ckt
Notice that the OpAmp Absorbs ALL of the Source Current The (Ideal) OpAmp will Source or Sink Current Out-Of or In-To its “nose” to maintain the V-Constraint: v+ = v−

Example: TransConductance Ckt
The transonductance Amp circuit below performs Voltage to Current Conversion Find io/v1 Use the Summing Point Constraint → v− = v+ → i− = i+ = 0 Thus v− = v1 Then by KCL & Ohm Notice the Output is Insensitive to Load Resistance

Example: Summing Amp Circuit
Label Important Quantities By Virtual Short (v+=v−) Across OpAmp Inputs Also by Summing Point Contstraint (i+ = i− =0) KCL at vf node Have negative feedback so summing point constraint applies

Example: Summing Amp Circuit
Use Ohm for iA & iB Also by Ohm Thru Rf

Example: Summing Amp Circuit
Recall iA & iB Sub for iA & iB But by Virtual Short vf=0; Thus after ReArranging

Example: Summing Amp Circuit
Input Resistances: Similarly for RinB

Example: Summing Amp Circuit
For the OUTput Resistance as seen by the LOAD, put the Circuit in a “Black Box” Thévenizing the Black Box Recall vo

Example: Summing Amp Circuit
Thus Have From the LOAD Perspective Expect But the Previous analysis vo does NOT depend on RL at ALL If vL = vo in all cases then have Since vo is Not affected by RL, then Ro = 0

Example: Summing Summary

Example: I & V Inputs Set Voltages Infinite Gain Assumption Fixes v−
Use Infinite Input Resistance Assumption Apply KCL to Inverting Input Then Solving Required Find the expression for vo. Indicate where and how Ideal OpAmp assumptions Are Used Have negative feedback so summing point constraint applies

Example – Find G and Vo Solving Ideal Assumptions
Have negative feedback so summing point constraint applies Yields Inverse Divider

Example  OpAmp Based I-Mtr
Desired Transfer Characteristic = 10V/mA → Find R2 NON-INVERTING AMPLIFIER Have negative feedback so summing point constraint applies

Offset & Saturation A NonInverting Amp Start w/ KCL at v−
But How to Handle This??? Start w/ KCL at v− Assume Ideality Then at Node Between the 1k & 4k Resistors Have negative feedback so summing point constraint applies Notes on Output Eqn Slope = 1+(R2/R1) as Before Intercept = −(0.5V)x(4kΩ/1kΩ) Then the Output

Example – Offset & Saturation
Note how “Offset” Source Generates a NON-Zero Output When v1 = 0 The Transfer Characteristic for This Circuit IN LINEAR RANGE −2V Offset “Saturates” at “Rail” Potential

WhiteBoard Work Let’s Work a Unity Gain Buffer Problem
Vs = 60mV Rs = 29.4 kΩ RL = 600 Ω Find Load Power WITH and withOUT OpAmp UGB

What’s an OpAmp? All Done for Today

Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Engineering 43 Appendix Bruce Mayer, PE Registered Electrical & Mechanical Engineer

29.4 k 29.4

Unity Gain Buffer Source-Driven Circuit
Vs = 60mV Rs = 29.4 kΩ RL = 600 Ω

Key to OpAmp Ckt Analysis  IOA
Remember that the “Nose” of the OpAmp “Triangle” can SOURCE or SINK “Infinite” amounts of Current Can source/sink up to Isat IOA = ± ∞

Unity Gain Buffer Controlling Variable = Solve For Buffer Gain by KVL
Thus The Amplification

Comparator Ideal Comparator and Transfer Characteristic
“Zero-Cross” Detector → Heart of Solid State Relay Cnrtl

Find G & Rin for NonIdeal Case
The OpAmp Model Add Input Source-V Determine Equivalent Circuit Using Linear Model For Op-Amp

Example Required Draw The Linear Equivalent Circuit Locate Nodes
Write The Loop Equations + - o v R i O Locate Nodes Place the nodes in linear circuit model

Example cont Add Remaining Components to Complete Linear Model
Examine Circuit Two Loops One Current Source Use Meshes Mesh-1 DONE But Could Sub for (v+-v-) and solve for i2 Mesh-2 Controlling Variable

Find G & Rin for NonIdeal Case
Add The External Components The Equivalent Circuit for Mesh Analysis Now Re-draw Circuit To Enhance Clarity Be Faithful to the NODES There Are Only Two Loops

Find G & Rin for NonIdeal Case
Now The Mesh Eqns Mesh-1 2 1 Mesh-2 The Controlling Variable in Terms of Loop Currents Eliminating vi Continue Analysis on Next Slide

G & Rin for NonIdeal Case cont
The Math Model From Mesh Analysis The Input-R and Gain Then the Model in Matrix Form →

G & Rin for NonIdeal Case cont
The Matrix-Inversion Soln Invert Matrix as Before Find Determinant,  Adjoint Matrix Solving for Mesh Currents Then the Solution

G & Rin for NonIdeal Case cont
The (Long) Expression for The input Resistance By Mesh Currents This Looks Ugly How Can we Simplify? Recall A →  Ri → 

G & Rin for NonIdeal Case cont
Use A→ in Expression for  Now Since For Op-Amps Ri→ Also, then vo Finally then G: Infinite Input Resistance is GOOD And The Expression for Rin: