Lecture 6 - ShanghaiTechsist.shanghaitech.edu.cn/faculty/zhoupq/Teaching/Fall15/Lec6... · Lecture 6 -Operational ... extremely high differential voltage gain of the op-amp ... •

Electric Circuits (Fall 2015) Pingqiang Zhou

Lecture 6

- Operational Amplifiers

10/20/2015

Reading: Chapter 5

Lecture 6 1


Outline

• Operational amplifier (op amp) model

• Ideal op amp

• Inverting op amp

• Noninverting op amp

Voltage follower

• Difference amplifier

Instrumentation amplifiers

• Differentiator/Integrator

• Application: DAC

Lecture 6 2


The Operational Amplifier

• The operational amplifier (“op amp”) is a

basic building block used in analog circuits.

Its behavior is modeled as dependent source

(voltage-controlled voltage source).

When combined with resistors, capacitors, and inductors,

can perform various functions:

–amplification/scaling of an input signal

–sign changing (inversion) of input signal

–addition/subtraction/multiplication/

division of input signals

– integration (over time) of an input signal

–differentiation (w.r.t. time) of an input signal

–analog filtering

–nonlinear functions like exponential, log, sqrt, etc.

Isolate input from output

Lecture 6 3


What are Inside an Op Amp?

• Origins at Fairchild Semiconductor in 1968.

4

Source: http://www.ibiblio.org/kuphaldt/electricCircuits/Semi/SEMI_8.html#xtocid109742.

Lecture 6

http://www.ibiblio.org/kuphaldt/electricCircuits/Semi/SEMI_8.html#xtocid109742


Op Amp Terminals

• Five important terminals

The inverting input

The noninverting input

The output

The positive (+) power supply

The negative (-) power supply

• The rest three terminals

2 Offset Null (Balance)

–May used in auxiliary circuit to

compensate for performance

degradation due to aging etc.

1 No Connection (NC)

– Unused, not connected to the

amplifier circuit.

Lecture 6 5


Powering an Op Amp

• As an active element, the op-amp

requires a power source.

• Often in circuit diagrams the power

supply terminals are obscured (ignored).

It is taken for granted that they must be

connected.

The supply current cannot be overlooked.

• Most op-amps use two voltage sources,

with a ground reference between them.

This gives a positive and negative supply

voltage.

𝑖0 = 𝑖1 + 𝑖2 + 𝑖+ + 𝑖−

Lecture 6 6


Output Voltage

• The voltage output of an op-amp is

proportional to the difference between

the noninverting and inverting inputs

Here, A is called the open loop gain.

Ideally A is infinite.

In real devices, it is still high: 105 to 108.

2 1( )o dv Av A v v

Lecture 6 7

𝑣2

𝑣1


Voltage Saturation

• Is the output voltage unlimited?

One cannot expect the output to exceed the supply voltages. When

an output should exceed the possible voltage range, the output

remains at either the maximum or minimum supply voltage. This is

called saturation.

Outputs between limiting voltages are referred to as the linear region.

𝑣0 =

−𝑉𝑐𝑐 𝐴𝑣𝑑 < −𝑉𝑐𝑐𝐴𝑣𝑑 −𝑉𝑐𝑐 ≤ 𝐴𝑣𝑑 ≤ +𝑉𝑐𝑐+𝑉𝑐𝑐 𝐴𝑣𝑑 > +𝑉𝑐𝑐

How do we know the op-amp

is operating in linear region?

Lecture 6 8


Application – Comparator

• This comparator compares 𝑉𝑖𝑛 with 𝑉𝑟𝑒𝑓, the reference

voltage set by the potentiometer 𝑅1:

If 𝑉𝑖𝑛 drops below 𝑉𝑟𝑒𝑓, 𝑉𝑜𝑢𝑡 saturates to +𝑉, lighting the LED;

If 𝑉𝑖𝑛 is above 𝑉𝑟𝑒𝑓, the LED will remain off.

• If 𝑉𝑖𝑛 is a voltage signal produced by a measuring

instrument, this comparator can function as a “low” alarm.

9

𝑅1𝑉𝑟𝑒𝑓

𝑉𝑜𝑢𝑡

Source: http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/the-operational-amplifier/

Lecture 6

http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/the-operational-amplifier/


Application – Square-Wave Converter

• The input voltage 𝑉𝑖𝑛 is an AC sine wave.

It doesn’t have to be sine wave

10

𝑉𝑟𝑒𝑓

• Useful for creating square waves of varying duty cycle

(on/off time)

Referred to as pulse-width modulation (PWM).

𝑉𝑟𝑒𝑓 𝑉𝑟𝑒𝑓

Source: http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/the-operational-amplifier/

Lecture 6

http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/the-operational-amplifier/


Feedback

• Output of the op-amp is fed back into the inverting terminal.

Feedback to the inverting terminal is called “negative feedback”.

A self-stabilizing system (also true for any dynamic system in

general), giving the op-amp the capacity to work in its linear (active)

mode.

Op-amp gain A does not have to be precisely set by the

factory in order for the circuit designer to build an amplifier

circuit with precise gain.

𝑉𝑖𝑛 ↑ ⟹ voltage differential ↑ ⟹ 𝑉𝑜𝑢𝑡 ↑⟹ voltage differential ↓ ⟹ 𝑉𝑜𝑢𝑡 ↓⟹⋯⟹ 𝑉𝑜𝑢𝑡 → 𝑉𝑖𝑛 but small difference exists

Source: http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/negative-feedback/

Lecture 6 11

http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/negative-feedback/


Correct or Not?

12Lecture 6


Example

A 741 op amp has an open-loop voltage gain of 2 × 105, input

resistance of 2𝑀Ω, and output resistance of 50Ω. Find the

closed-loop gain 𝑣𝑜/𝑣𝑠. Determine current 𝑖 when 𝑣𝑠 = 2𝑉.

13

𝑣𝑠 − 𝑣110𝑘Ω

=𝑣1 − 𝑣𝑜20𝑘Ω

+𝑣12𝑀Ω

𝑣1 − 𝑣𝑜20𝑘Ω

=𝑣𝑜 − 𝐴𝑣𝑑50Ω

𝑣𝑜𝑣𝑠

= −1.9999699

𝑣1 = 20.066667𝜇𝑉

20.066667𝜇𝑉−3.9999398𝑉

Lecture 6


Ideal Op Amp

• Attributes of ideal op-amp:

infinite open-loop gain, 𝐴 ⋍ ∞– Implies that 𝑣2 = 𝑣1.

infinite resistance of the two

inputs, 𝑅𝑖 ⋍ ∞

–This means it will not affect any

node it is attached to

– Implies that 𝑖1 = 𝑖2 = 0.

zero output impedance, 𝑅𝑜 ⋍ 0–From Thevenin’s theorem one can

see that this means it is load

independent.

20.066667𝜇𝑉 −3.9999398𝑉

Lecture 6 14


Comments on Negative Feedback

• The basic principle of negative feedback is that the output

tends to drive in a direction that creates a condition of

equilibrium (balance).

With negative feedback, the circuit tends to prevent itself from

driving the output to a full saturation.

–Whether the output is directly fed back to the inverting (-) input or

coupled through a set of components, the effect is the same: the

extremely high differential voltage gain of the op-amp will be “tamed”

and the circuit will respond according to the dictates of the feedback

“loop” connecting output to inverting input.

15

Negative feedback

Source: http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/negative-feedback/

Lecture 6

http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/negative-feedback/


Positive Feedback

• The inverting terminal is free to receive an

external voltage.

• Let’s see what happens if we ground the

inverting input:

𝑉𝑜𝑢𝑡 will be dictated by the magnitude and

polarity of 𝑉𝑝𝑙𝑢𝑠

–Small +𝑉𝑝𝑙𝑢𝑠 ⟹ large +𝑉𝑜𝑢𝑡 ⟹ full positive

output saturation because of feedback.

–Small -𝑉𝑝𝑙𝑢𝑠 ⟹ large -𝑉𝑜𝑢𝑡 ⟹ full negative

output saturation because of feedback.

16

Sources:

• http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/positive-feedback/

• http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-

electronics-spring-2007/video-lectures/lecture-21/

Positive feedback

𝑉𝑝𝑙𝑢𝑠

▪ Bistable output: either saturated positive or saturated negative

▪ Known as hysteresis.

Lecture 6

http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/positive-feedback/

http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-21/


Application - Hysteresis

17

Glitch

Source: http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/positive-feedback/

• When 𝑉𝑜𝑢𝑡 is saturated positive,

𝑉𝑟𝑒𝑓 will be more positive than before.

• When 𝑉𝑜𝑢𝑡 is saturated negative, 𝑉𝑟𝑒𝑓will be more negative than before.

𝑉𝑟𝑒𝑓

𝑉𝑟𝑒𝑓

Lecture 6



Application - Oscillator

• Oscillator is a device that produces an alternating, or at

least pulsing, output voltage.

Known as astable device, no stable output state.

18

• When 𝑉𝑜𝑢𝑡 is saturated positive, 𝑉𝑟𝑒𝑓 is positive, capacitor 𝐶 charges up.

• When 𝑉𝑟𝑎𝑚𝑝 exceeds 𝑉𝑟𝑒𝑓 by the tiniest margin, 𝑉𝑜𝑢𝑡 becomes saturated

negative, and 𝐶 will charge in the opposite direction (polarity).

• Oscillation occurs because

positive feedback is instantaneous,

negative feedback is delayed by means of an RC time constant.

Source: http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/positive-feedback/Lecture 6



Ideal Op Amp

• Attributes of ideal op-amp:

infinite open-loop gain, 𝐴 ⋍ ∞– Implies that 𝑣2 = 𝑣1.

infinite resistance of the two

inputs, 𝑅𝑖 ⋍ ∞

–This means it will not affect any

node it is attached to

– Implies that 𝑖1 = 𝑖2 = 0.

zero output impedance, 𝑅𝑜 ⋍ 0–From Thevenin’s theorem one can

see that this means it is load

independent

Lecture 6 19


Ideal Op-Amp Analysis Technique

Assumption 1: The potential between the op-amp input terminals, v(+) –

v(-), equals zero.

R2R1

+

V0VIN

EXAMPLE

Assumption 2: The currents flowing into the op-amp’s two input

terminals both equal zero.

No Potential DifferenceNo Currents

Lecture 6 20


21Lecture 6

Ideal Op-Amp Analysis: Inverting Amplifier

R2

I2

VOUT-VR

R1

RLVIN

021

R

VV

R

VV OUTRINR

RinROUT VVR

RVV

1

2

Only two

currents

for KCL

When 𝑉𝑅 = 0, 𝑉𝑂𝑈𝑇 = −𝑅2

𝑅1𝑉𝐼𝑁

From this one can see that:• Closed-loop gain = ratio of feedback resistor 𝑅2 and 𝑅1.

• The polarity of output 𝑉𝑂𝑈𝑇 is the reverse of the input,

thus the name “inverting” amplifier.

• 𝑉𝑂𝑈𝑇 independent of 𝑅𝐿 - ideal voltage source

VR


22Lecture 6

Practice

• Determine 𝑣𝑜 in the circuit shown below


Equivalent Circuit of Inverting Amplifier

• The equivalent circuit

has a finite input resistance

a good candidate for making a

current-to-voltage converter

𝑣𝑜 = −𝑅𝑓

𝑅1𝑣𝑖

Lecture 6 23


Ideal Op-Analysis: Non-Inverting Amplifier

Assumption 1: The potential between the op-amp input terminals, v(+) –

v(-), equals zero.

R2R1

+

VoutVIN

Assumption 2: The currents flowing into the op-amp’s two input

terminals both equal zero.

021

R

vv

R

v outinin

inout vR

Rv

1

21

KCL with currents in only two branches

Non-inverting AmplifierVIN appears here

Lecture 6 24


25Lecture 6

Non-Inverting Amplifier

_

+

vin+-

R2

R1

RL

v2

vo

2

• Closed-loop gain

𝐴 =𝑣𝑜

𝑣𝑖𝑛= 1 +

𝑅2

𝑅1

• Input impedance = ∞

_

+

vin+- RL

v2

v0

• Voltage follower

𝑅2 = 0, 𝑅1 = ∞

𝑣𝑜 = 𝑣𝑖𝑛 (𝐴 = 1)

separating two circuits while

allowing a signal to pass

through.


Practice

• Find 𝑣𝑜

26Lecture 6


Summing Amplifier

• Aside from amplification, the op-amp can be made to do

addition very readily.

• If one takes the inverting amplifier and combines several

inputs each via their own resistor:

The current from each input will be proportional to the applied

voltage and the input resistance

1 2 3

1 2 3

1 2 3

a a av v v v v vi i i

R R R

𝑖 = 𝑖1 + 𝑖2 + 𝑖3 =𝑣𝑎 − 𝑣𝑜𝑅𝑓

, 𝑣𝑎 = 0


𝑅1𝑣1 +

𝑅𝑓

𝑅2𝑣2 +

𝑅𝑓

𝑅3𝑣3

Lecture 6 27


Practice

• Find 𝑣𝑜 and 𝑖𝑜 in the circuit shown below

28Lecture 6


Difference (or Differential) Amplifier

• Subtraction should come

naturally to the op-amp

since its output proportional

to the difference between

the two inputs.

• An op-amp with no feedback

is already a differential

amplifier, amplifying the

voltage difference between

the two inputs. However, its

gain cannot be controlled,

and it is generally too high to

be of any practical use.

2 21

1 1

1o a

R Rv v v

R R

42

3 4

b

Rv v

R R

2 1 2 2

2 1

1 3 4 1

1

1o

R R R Rv v v

R R R R

Lecture 6 29


Common Mode Rejection

• It is important that a difference amplifier reject any signal

that is common to the two inputs.

Which implies that when 𝑣1 = 𝑣2, 𝑣𝑜 = 0.

• For the given circuit, this is true if𝑅1

𝑅2=

𝑅3

𝑅4, then

2 1 2 2

2 1

1 3 4 1

1

1o

R R R Rv v v

R R R R

𝑣𝑜 =𝑅2𝑅1

𝑣2 − 𝑣1

Lecture 6 30


Example

• Design an op amp circuit with inputs 𝑣1 and 𝑣2 such that

𝑣𝑜 = −5𝑣1 + 3𝑣2.

31

2 1 2 2

2 1

1 3 4 1

1

1o

R R R Rv v v

R R R R

⟹ 𝑅2

𝑅1= 5,

𝑅2 1 + 𝑅1/𝑅2𝑅1 1 + 𝑅3/𝑅4

= 3

⟹ 𝑅3𝑅4

= 1

Lecture 6


Example – An Instrumental Amplifier

• Show that 𝑣0 =𝑅2

𝑅11 + 2

𝑅3

𝑅4𝑣2 − 𝑣1 .

32

𝑣0 =𝑅2𝑅1

𝑣02 − 𝑣01

𝑣01 − 𝑣02 = 𝑖 2𝑅3 + 𝑅4

𝑖 =𝑣1 − 𝑣2𝑅4

Lecture 6


Practice

• Obtain 𝑖𝑜 in the following instrumentation amplifier circuit

33Lecture 6


Limitation with the Difference Amplifier

• One significant drawback:

The input impedance is low.

Each input voltage source has to

drive current through a

resistance, which constitutes far

less impedance than the bare

input of an op-amp alone.

• By placing a noninverting

amplifier stage before the

difference amplifier this can

be resolved.

Source: http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/building-a-differential-amplifier/

Lecture 6 34

http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/building-a-differential-amplifier/


Instrumentation Amplifier

• It is beneficial to be able to adjust the gain of the amplifier

circuit without having to change more than one resistor

value

• Instrumentation amplifier has this capability

a buffered differential amplifier stage with three new resistors

𝑉𝑜𝑢𝑡 = 1 +2𝑅

𝑅𝑔𝑎𝑖𝑛𝑉2 − 𝑉1

Overall gain 𝐴𝑣 = 1 +2𝑅

𝑅𝑔𝑎𝑖𝑛

Source: http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/the-instrumentation-amplifier/

Lecture 6 35

http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/the-instrumentation-amplifier/


Instrumentation Amplifier

• Instrumentation amplifiers are so useful, they are often

packaged as a single chip with the only external

component being the gain resistor

• They are very effective at extracting a weak differential

signal out of a large common mode signal

In circuits exposed to external electrical noise, this is

important in order to maintain a high signal-to-noise ratio.

Lecture 6 36


Differentiator

• The differentiator (not to be confused with differential)

produces a voltage output proportional to the input

voltage’s rate of change.

• Want

• By introducing electrical reactance into the feedback loops

of op-amp amplifier circuits, we can cause the output to

respond to changes in the input voltage over time.

𝑣𝑜 = 𝐾 ∙𝑑𝑣𝑖𝑑𝑡

Source: http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/differentiator-integrator-circuits/

Lecture 6 37

http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/differentiator-integrator-circuits/


Differentiator

• We can build an op-amp circuit which measures change

in voltage by measuring current through a capacitor, and

outputs a voltage proportional to that current:

38

i

The RHS of the capacitor is held to

a voltage of 0 volts, due to the

“virtual ground”effect. Therefore,

current“through”the capacitor is

solely due to change in the input

voltage.

Capacitor current moves through the

feedback resistor, producing a drop

across it, which is the same as the

output voltage.


Lecture 6



Integrator

• The integrator produces a voltage

output proportional to the product

(multiplication) of the input voltage

and time.

• Want

o inv K v dt

R+

-C V0

+

-

vin

𝑣0 =1

𝑅𝐶

−∞

𝑡

𝑣𝑖𝑛 𝑑𝑡 =1

𝐶

−∞

𝑡𝑣𝑖𝑛𝑅𝑑𝑡


Integrator

• As before, the negative feedback of the

op-amp ensures that the inverting input

will be held at 0 (the virtual ground).

• if we apply a constant, positive voltage

to the input, the op-amp output will fall

negative at a linear rate, in an attempt

to produce the changing voltage across

the capacitor necessary to maintain the

current established by the voltage

difference across the resistor.

• Conversely, a constant, negative

voltage at the input results in a linear,

rising (positive) voltage at the output.

40


Lecture 6



Cascaded Op Amps

• It is common to use multiple op-amp stages chained

together.

• This head to tail configuration is called “cascading”.

• Each amplifier is then called a “stage”.

• The gain of a series of amplifiers is the product of the

individual gains:

𝐴 = 𝐴1 ∙ 𝐴2 ∙ 𝐴3

Lecture 6 41


Application - Digital to Analog Converter

• The function of a digital to analog converter (DAC) is to

take a series of binary values that represent a number

and convert it to an analog voltage

In a digital signal, the input voltage will be either zero which

represents ‘0’ or a non-zero voltage which represents ‘1’

• The summing amplifier can be used to create simple DAC

Recall that each input has its own multiplier resistor


𝑅1𝑣1 +

𝑅𝑓

𝑅2𝑣2 +

𝑅𝑓

𝑅3𝑣3

Lecture 6 42


Digital to Analog Converter

• By selecting the input resistors 𝑅𝑖s

such that each input will have a

weighting according to the

magnitude of their place value

Each lesser bit will have half the

weight of the next higher bit

• The feedback resistor 𝑅𝑓 provides

an overall scaling, allowing the

output to be adjusted according to

the desired range

−𝑉𝑜 =𝑅𝑓

𝑅1𝑉1 +

𝑅𝑓

𝑅2𝑉2 +

𝑅𝑓

𝑅3𝑉3 +

𝑅𝑓

𝑅4𝑉4

Lecture 6 43


A DAC can be used to convert the digital representation

of an audio signal into an analog voltage that is then

used to drive speakers -- so that you can hear it!

0 0 1 0 10 0 1 10 1 0 0

1.52

0 1 0 10 1 1 00 1 1 11 0 0 0

2.53

3.54

1 0 0 11 0 1 01 0 1 11 1 0 01 1 0 11 1 1 01 1 1 1

4.55

5.56

6.57

7.5

Binarynumber

Analogoutput(−𝑉𝑜)

0 0 0 00 0 0 1

0.5

MSB LSB

S1 closed if LSB =1

S2 " if next bit = 1

S3 " if " " = 1

S4 " if MSB = 1

4-Bit D/A

8V

+ V0

5K

80K

40K

20K

10K

S1

+-

S3

S2

S4

+

DAC Example

“Weighted-adder D/A converter”

(Transistors are used

as electronic switches)


Digital Input

0

1

2

3

4

5

6

7

8

0 2 4 6 8 10 12 14 16

An

alo

g O

utp

ut

(V)

111110000100

0000 0001

Characteristic of 4-Bit DAC


Summary

Lecture 6 46


To Read Further…

• Nonlinear Circuits

Online notes “Introduction to nonlinear circuit analysis” from

Berkeley.

Ch. 4 in reference: Anant Agarwal and Jeffrey Lang, Foundations of

Analog and Digital Electronic Circuits, Morgan Kaufmann, 2005.

• More about Op Amps

Online textbook http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/

Lecture 6 47

http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/

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Lecture 6 - ShanghaiTechsist.shanghaitech.edu.cn/faculty/zhoupq/Teaching/Fall15/Lec6... · Lecture 6 -Operational ... extremely high differential voltage gain of the op-amp ... •