1. Operational Amplifiers
EE 174 Spring 2019
Operational Amplifiers

2. Table of Contents Introduction to Operational Amplifier (Op-Amp)
Brief of History
Op-Amp Internal Circuit
Op-Amp: Ideal, Typical Circuit and Characteristics Table
The Op-Amp Golden Rules
Op-Amp: Negative and Positive Feedback
Gain and Bandwidth (GBP)
Op-Amp: The Four Amplifier Types
Common Mode Rejection Ratio (CMRR)
Fundamentals of Op-Amps
Basic operation
Gain
Offset
CMRR
Applications

3. Introduction to Op-Amp
Operational Amplifier (a.k.a. Op-Amp) is used in circuit to perform mathematical operations such as add, subtract, filter, integrate, differentiate, etc.
Op-Amp is an active circuit element consisting of transistors, resistors, diodes and capacitors.
Op-Amps are two-port networks in which the output voltage or current is directly proportional to either input voltage or current. Four different type of amplifiers exits:
Voltage amplifier (VCVS): Av = Vo / Vi
Current amplifier (ICIS): Ai = Io / Ii
Transconductance amplifier (VCIS): Gm = Io / Vi (Siemens)
Transresistance amplifier (ICVS): Rm = Vo / Ii
Op-Amp is ideal device which takes a relatively weak signal (sensor) as an input and produce a much stronger signal as an output without affecting its other properties.
Op-Amps are commonly used for both linear and nonlinear applications: Inverting/Non-inverting Amplifiers, Variable Gains Amplifiers, Summers, Integrators/Differentiators, Filters and Schmitt trigger, Comparators, A/D converters.

4. Brief History of Op-Amp
Monolithic IC Op-Amp
First created in 1963 μA702 by Fairchild Semiconductor
μA741 created in 1968, became widely used due to its ease of use 8 pin, dual in-line package (DIP)
Further advancements include use of field effects transistors (FET), greater precision, faster response, and smaller packaging
Vacuum Tube Op-Amps (1930’s-1940’s)
Dual-supply voltage of +300/-300 V
Output swing +/- 50 volts
Open-loop voltage gain of 15,000 to 20,000,
Slew rate of +/- 12 volts/µsecond
Maximum output current of 1 mA
George Philbrick
Solid State Discrete Op-Amps (1960’s)
Dual-supply voltage of +15/-15 V
Output swing +/- 11 volts
Open-loop voltage gain of 40,000,
Slew rate of +/- 1.5 volts/µsecond
Maximum output current of 2.2 mA

5. Op-Amp Internal Circuit
Op-Amp 3-stages:
Differential Amplifier (Blue)
Gain Stage (Magneta)
Output Stage (Cyan)
Current Mirrors/Sources (Red)
Voltage Level Shifter (Green)

6. Op-Amp: Ideal, Typical Circuit and Characteristics Table
Parameter
Ideal
Op-Amp
Practical Op-Amp
Differential/Open Loop Gain (Ad) ꚙ
Common Mode Gain (AC)
10–5
Common Mode Rejection Ratio (CMRR)
> 60dB
Input Impedance (Ri)
Output Impedance (Ro)
100-1kΩ
Bandwidth
1-20 MHz
Slew rate (SR)
A few V/µs
Offset Voltage
< 10mV
Offset Current
< 50 nA

7. Op-Amp: Common Mode Rejection Ratio (CMRR)
Differential Mode Gain (Ad) is defined as:
Ad = VO Vd = VO V2 − V1  VO = Ad Vd where Vd = V2 − V1
Common Mode Gain (AC) is defined as:
AC = VOCM VCM Ideally, VOCM = 0, since V+ = V-
The op amp common-mode rejection ratio (CMRR) is the ratio of the common-mode gain to differential-mode gain.
CMRR = Ad ACM or CMRR = 20 log | Ad AC𝑀 | in dB
Example: Calculate the CMRR for the circuit with measurement values as follow:
VCM = 1mV, VOCM = 5mV and V1 = - 0.2mV, V2 = 0.8mV, VO = 10V
Solution:
AC = VOCM VCM = 5mV 1mV = 5, Ad = VO Vd = mV − (− 0.2mV) =  CMRR = Ad ACM = = 2000 or 20 log(2000) = 66dB

8. Op-Amp Offset
Another practical concern for op-amp performance is voltage offset.
The primary cause of the Op-Amp offset is a slight mismatch of the base-emitter voltages of the differential input stage of the Op-Amp.
For an ideal op-amp with both inputs shorted to ground or when there is no difference between its inputs would produce output to zero.
 v+ = v– = 0  VO = A(v+ – v– ) = 0.
However, for a real op-amp  An offset input can drive the output to either negative or positive saturated level even if the op-amp in question has zero common-mode gain (infinite CMRR). This deviation from zero is called offset.
The 741 OPAMP have offset voltage null capability by connecting 10kΩ potentiometer between pin 1 and pin 5. By varying the potentiometer, output offset voltage (with inputs grounded) can be reduced to zero volts. For the 741C the offset voltage adjustment range is ± 15 mV.
Effect of offset voltage on Op-Amp produces error at the output.
VO = – 𝑹𝟐 𝑹𝟏 Vin + (1 + 𝑅2 𝑅1 ) 𝑉𝑂𝑆+
Desired output Error

9. Op-Amp Offset Example
1) Determine the output voltage for the open loop differential amplifier Fig #1.
vOS- = 5 µVdc, vOS+ = –7 µVdc
vOS- = –10 mVdc, vOS+ = 0Vdc
Specifications of the Op-Amp are given below:  A = 200,000, Ri = 2 MΩ , RO = 75Ω, + VCC = + 15 V, - VEE = - 15 V, and output voltage swing = ± 14V.
Solution:
a) VO = A(vOS+– vOS-) = 200,000 (– 7 – 5) µV = 2.4 V
Remember that vo = 2.4 V dc with the assumption that the dc output voltage is zero when the input signals are zero.
b) VO = A(vOS+– vOS-) = 200,000 (0 – (– 10mV)) = 2000 V  Saturation VO = 14V
Thus the theoretical value of output voltage vo = 2000 V However, the OPAMP saturates at ± 14 V.
2) Given circuit in Fig #2 with Vin = 0.25V, R1 = 1kΩ , R2 = 4 kΩ Vos+ = 5mV. Determine the offset voltage error.
Use superposition to obtain:
VO = – 𝑅2 𝑅1 Vin + (1 + 𝑅2 𝑅1 ) 𝑉𝑂𝑆+ = – 4(0.25V) + 5(0.005V) = (– )V = – 0.975V
Fig #1
Fig #2

10. Op-Amp: Gain and Bandwidth
The Voltage Gain (A) of the operational amplifier can be found using the following formula:
In Decibels or (dB) is given as: V0 +  V+ V
Open loop gain:  Gain is measured when no feedback is applied to the op amp. Open-loop gain AOL is very high.
Closed loop gain:   Gain is measured when the feedback loop is closed and the overall gain of the circuit is much reduced.
Loop gain: The difference between the open-loop gain and the closed-loop gain. This is useful information because it gives you the amount of negative feedback that can apply to the amplifier system.
Gain versus Bandwidth: Applying feedback will reduce the gain but increase the bandwidth.
Gain-bandwidth (GBW) product is defined as the op-amp gain A multiplied by the bandwidth BW.
The GBW product can be used to calculate the closed-loop gain bandwidth.
GBW = A x BW where A is in ratio (not in dB)
Example: Closed loop gain 70 dB. What is closed-loop BW?
Gain: 20log = 70dB  logA = 70/20 = 3.5 or A = = 3162
 Closed-loop BW = 10 MHz / 3162 = 3163 Hz.
A = 105, BW = 10Hz, GBW = 105x10Hz = 1MHz V0 + 
VIN
A = 1, BW = 1 MHz, GBW = 1x1MHz = 1MHz

11. Op-Amp Slew Rate
The slew-rate (SR) is the maximum possible rate of change of the op amp output voltage.
Slew Rate (SR) = dVO dt max in V/µs
Square Wave: Applied a square wave input to the 741 Op-amp in Fig. 1, find the slew rate  SR = ΔVO/Δt = 2.06V / 3.16 µs = 0.63V/ µs (Ref Lab #1).
Sine Waves: The maximum rate of change for a sine wave occurs at the zero crossing and for a given vo = VP sin(2π f t), the slew rate SR is:
Slew Rate (SR) = dVO dt max = 2π fmax VP
SR = 0.63V/ µs.
Fig. 1
Example: Given Slew rate SR = 1 V/μs; input voltage vin (t) = 1 sin(2π × 105t);
closed loop gain AV =10.
Sketch the theoretically output and the actual output in the same graph.
What is the maximum frequency that will not violating the given slew rate?
What is the maximum gain A that will not violating the given slew rate?
What is the required SR to prevent distortion output?
Solutions:
See graph Fig. 2
fmax = SR / (A x 2π) = 1 x 106 / (10 x 2π) = 15.9 kHz
VPmax = SR / (f x 2π) = 1 x 106 / (105 x 2π) = 1.59 V  A = 1.59
For voltage gain A = 10  vo(t) = 10 sin(2π × 105t)  SR = 𝑑𝑣𝑜 𝑑𝑡 | = 2π x f x 10 = 2π × 105 x 10 = 6.28 V/μs
Fig. 2

12. Op-Amp: Negative and Positive Feedback
Negative feedback: Vout = A(V+ – V–) = A(Vin – Vout)
If Vout > Vin  Vout ↓ goes down toward Vin
If Vout < Vin  Vout ↑ goes up toward Vin
This drives Vout = Vin or V+ = V-
 Negative feedback Op-Amp can produce any voltage in the supply power range.
Positive feedback: Vout = A(V+ – V–) = A(Vout – Vin)
If Vout > Vin  Vout ↑ goes way up to + saturation
If Vout < Vin  Vout ↓ goes way down to – saturation
This drives Vout to either directions: + / – saturation
 Positive feedback Op-Amp can only produce maximum and minimum voltages of the range.
Positive feedback:
Negative feedback:

13. The Op-Amp Golden Rules
vd = v+ – v–
vo = Avd = A (v+ – v–)
i+ = i – = 0 (Since Ri = ꚙ. This is always true for any feedback configuration).
v+ = v– (In negative feedback, the output of the op amp will try to adjust its output so that the voltage difference between the + and − inputs is zero  V+ = V−).
Rule 2. is not applied if VO in saturation.
Note: The resistances used in an op-amp circuit must be much larger than Ro and much smaller than Ri in order for the ideal op-amp equations to be accurate.

14. Op-Amp: The Four Amplifier Types
Description
Gain Symbol
Transfer Function
Voltage Amplifier or
Voltage Controlled Voltage Source (VCVS) Av
Vout/Vin
Current Amplifier
Current Controlled Current Source (ICIS) Ai
Iout/Iin
Transconductance Amplifier
Voltage Controlled Current Source (VCIS) gm
(siemens)
Iout/Vin
Transresistance Amplifier
Current Controlled Voltage Source (ICVS) rm
(ohms)
Vout/Iin

15. Op-Amp Saturation
The op-amp output voltage is limited by the supply voltages.  In practice the limits are about 1.5 to 2 V below the value of the supply voltages.
The op amp has three distinct regions of operation:
Linear region: −VEE < Vo < VCC  Vo/Vi = A = constant
Positive saturation: Vo > VCC  Vo = VCC
Negative saturation: Vo < −VEE  Vo = –VEE
Note that the saturation voltage, in general, is not symmetric.
For an amplifier with a given gain, A, the above range of Vo translate into a certain range for Vi
− VEE < Vo < VCC or − VEE < A Vi < VCC
 − VEE / A < Vi < VCC / A
Any amplifier will enter its saturation region if Vi is raised above certain limit. The figure shows how the amplifier output clips when amplifier is not in the linear region.
Example: For A = 105, –VDD = –12V, VCC = +15V, find range of input Vi to prevent saturation.
Solution: –12V / 105 < Vi < 15V / 105 or – 0.12mV < Vi < 0.15mV

16. Op-Amp: The Four Amplifier Types

17. Op-Amp Two Basic Operations
The Inverting Op Amp
The Non-inverting Op Amp
In negative-feedback configuration, op-amp always “wants” both inputs (inverting and non-inverting) to be the same value.

18. Op-Amp Applications n = 3 For RI = nRO where n is # of inputs
(c) Voltage follower IO
(d) Scaling Summer
(d1) Voltage Adder
n = 3
(d2) Averager
For RI = nRO where n is # of inputs

19. Op-Amp Applications I1 = IO and I2 = I3 For R1 = R2 = RI and R3 = RO 
(e) Differential Input
(f) Adder-Subtracter Ef
I1 = IO and I2 = I3 Ef
For R1 = R2 = RI and R3 = RO 
As an adder-subtracter, unused inputs should be grounded.
R and RI not necessarily equal

20. Op-Amp Applications General Form of the Inverting Amplifier
One practical method of reducing noise and preventing instability is shown below. The transition “point” is the frequency at which
 XCI = RI
Differentiator (High pass filter): Bode plot is shown below.
When XC = 1/(2πfCI) = RO  Gain = 1
Since highest gain is encountered at high frequencies, this circuit is very susceptible to random noise.
Low frequency cutoff fO = 1 2π𝑅𝑂𝐶𝐼
High frequency cutoff fI = 1 2π𝑅𝐼𝐶𝐼
Gain: 𝐸𝑂 𝐸𝐼 = – 𝑅𝑂 𝑅𝐼 𝑅𝐼 2π𝑓𝐶𝐼 𝑅𝐼 2π𝑓𝐶𝐼+ 1

21. Op-Amp Applications
Integrator (Low pass filter): Bode plot is shown below.
Gain: 𝑉𝑂 𝑉𝐼 = – 𝑍𝐶 𝑅 = – 1 𝑅 2π𝑓𝐶
When ZC = 1/(2πfC) = R  Gain = 1
Integrator (Low pass filter with Gain): Bode plot is shown below.
DC Gain: 𝑉𝑂 𝑉𝐼 = – 𝑅2 𝑅1 AC gain: 𝑉𝑂 𝑉𝐼 = – 𝑅2 𝑅1 1 𝑅2 2π𝑓𝐶+1
Corner Frequency: f0 = 1 2π𝐶𝑅2
Example:
For LPF with gain, find suitable components to achieve a −3-dB frequency of 1 kHz with a dc gain of 20 dB and an input resistance of at least 10 kΩ
At what frequency does gain drop to 0 dB?

22. Op-Amp Applications (a) Find i, if and vo. i = 0 A
(b) For the ideal op amp shown below, what should be the value of resistor Rf to obtain a gain of 5?
Solution: Want Vo = 5Vi
For non-inverting:
For Input V+ = 2 3 Vi and Output VO = 5Vi  Gain = 𝑉𝑂 𝑉+ = 5 2/3 = 15 2
Gain = 1 + 𝑅𝑓 3𝑘Ω =  Rf = 19.5kΩ
Another way: i = 𝑉− 3𝑘Ω = 𝑉+ 3𝑘Ω = 2/3(𝑉𝑖) 3𝑘Ω = 2𝑉𝑖 9𝑘Ω =𝑖𝑓
Vo = (Rf + 3kΩ)i = (Rf + 3𝑘Ω) 2𝑉𝑖 9𝑘Ω = 5Vi Rf = 19.5kΩ V+

23. Op-Amp Applications
(a) Given an op-amp circuit below. If the power supplies for the op-amp are ± VCC = ± 12V. Determine:
Overall gain A = VOUT / VIN, VOUT and IOUT .
If op-amp in saturation, what is voltages at V+ and V-.
Solution:
(a) Vx = 1kΩ x 4V 1kΩ+1kΩ =2𝑉=V+ = V− =Vy
i = 2V 1kΩ =2 𝑚𝐴
Vz = 2kΩ x 2 mA=4𝑉
Vy = 1kΩ x 2 mA=2𝑉
IOUT = 2 mA + 2 mA = 4 mA
VOUT = (3kΩ x 4 mA)+ 4V = 16V  A = VOUT / VIN = 16V / 4V = 4
However, VOUT = 16V > Power supply 12V
Saturation  VOUT = 12V Clamp
For VOUT = 12V  IOUT = 12V / (3kΩ + 2kΩ//2kΩ) = 3 mA
i = 3 mA / 2 = 1.5 mA
Vz = 2kΩ x 1.5 mA=3𝑉
Vy = V- = 1kΩ x 1.5 mA = 1.5V
V+ = Vx = 2V
V+ ≠ V- when op-amp in saturation mode.

24. Op-Amp Applications
(a) Given gain A1 = RF/RS = 100, GBW product = A0 ω0 = K = 4π × Determine the overall 3-dB bandwidth of the cascade amplifier below:
Solutions: The 3-dB bandwidth for each amplifier is:
ω1 = K A1 = 4π × = 4π × 104 rad/s
 BW of the cascade = 4π × 104 & cascade gain A = A1A1= 102 × 102 = 104.
To achieve the same gain A = 104 with a single-stage amplifier having the same K  BW: ω2 = K A = 4π × = 4π × 102 rad/s << 4π × 104 rad/s of cascade
(b) Voltage follower application:
Solutions:
Assume Rs = 1 kΩ, RL = 100Ω  VL = 0.1 Vin
 Huge attenuation of the signal source.
If voltage follower (buffer amplifier) is used.
 VL = Vout = Vin

25. Op-Amp Applications (1) Find Vx , Vo (3) Find Vy , Vo Vx = 3.2V
Vo = 6Vi
Vx = 0.5Vo
Vo = 0.8Vi

26. Op-Amp Applications (5) Find Vo (7) Find Vout as a function of Vs.
Due to supply voltage limitations, 0V ≤ Vout ≤ 10V. Plot Vout vs. Vs.
(a) (Vs – Vout)/20k = (1 – Vs)/2k
Vout = 11Vs – 10
(b) See Plot
+10V
v+= v-= 6V
–10V
(6) Find Vo
v+= v-= 5V

27. Basic Comparators
A comparator is a device that compares two input voltages (inverting and noninverting input) and produces an output indicating the relationship between them. The inputs can be two signals or a signal and a fixed dc reference voltage. The output that usually swings from rail to rail.
A typical comparator has low offset, high gain, and high common-mode rejection.
Comparators are designed to work as open-loop systems, to drive logic circuits, and to work at high speed, even when overdriven.
Non-inverting comparator
Inverting comparator
Zero detection Vref = 0V

28. The Comparators with and without Hysteresis
Non-hysteresis uses a voltage divider (Rx and Ry) to set up the reference voltage. The comparator will compare the input signal (Vin) to the reference voltage (Vref). The comparator input signal is applied to the non-inverting input, so the output will have a non-inverted polarity
Hysteresis uses two different threshold voltages to avoid the multiple transitions introduced in the previous circuit. For this non-inverting comparator, the input signal must exceed the upper threshold (VH) to transition high or below the lower threshold (VL) to transition low.
Calculate VH and VL:
1kΩ // 5.76kΩ = 0.85k Ω
VH = 1kΩ x 5V 1.85kΩ = 2.7V
VL = 0.85kΩ x 5V 1.85kΩ = 2.3V

29. Design of Hysteresis Comparator
The design requirements are as follows:
Supply Voltage: +5 V
Input: 0V to 5V
VL (Lower Threshold) = 2.3V
VH (Upper Threshold) = 2.7V
VH - VL = 0.4V
Vth ± 0.2V = 2.5V ± 0.2V
Equations (1) and (2) can be used to select the resistors needed to set the hysteresis threshold voltages VH and VL.
One value (Rx) must be arbitrarily selected. In this example, Rx was set to 100kΩ to minimize current draw.
Rh was calculated to be 575kΩ, so the closest standard value 576kΩ was used

30. When To Use Op-amp Comparator Circuit???
Although op amps are not designed to be used as comparators, there are, nevertheless, many applications where the use of an op amp as a comparator is a proper engineering decision.
WHY USE AN OP AMP AS A COMPARATOR?
• Convenience
• Economy (e.g. use spare op-amps in quad package
and cheaper than comparator)
• Low bias current (IB)
• Low offset voltage (VOS)
Note: This is not a good design practice.
WHY NOT USE AN OP AMP AS A COMPARATOR?
• Speed
• Inconvenient input structures (Comparator is designed
for large differential input voltages)
• Inconvenient logic structures (Compartor TTL, CMOS)
• Stability/hysteresis (stability)

31. References:
Fundamentals of Electrical Engineering, Giorgio Rizzoni, McGraw-Hill Higher Education