EE 42/100 Fall 2005Week 8, Prof. White1 Lectures Week 8 The operational amplifier (“op amp”) Feedback Comparator circuits Ideal op amp Unity-gain voltage

EE 42/100 Fall 2005 Week 8, Prof. White 1

Lectures Week 8

• The operational amplifier (“op amp”)• Feedback• Comparator circuits• Ideal op amp• Unity-gain voltage follower circuit

ReadingCh. 14, Secs. 14.1-14.3, 14.5-14.7 lightly,

14.8-14.9, 14.10 lightly


The Operational Amplifier• The operational amplifier (“op amp”) is a basic building block used in

analog circuits.

– Its behavior is modeled using a dependent source.

– When combined with resistors, capacitors, and inductors, it can perform various useful functions:

• amplification/scaling of an input signal• sign changing (inversion) of an input signal• addition of multiple input signals• subtraction of one input signal from another• integration (over time) of an input signal• differentiation (with respect to time) of an input signal• analog filtering• nonlinear functions like exponential, log, sqrt, etc.


Op Amp Circuit Symbol and Terminals

+

–

V +

V –

non-inverting input

inverting input

positive power supply

negative power supply

output

The output voltage can range from V – to V + (“rails”) The positive and negative power supply voltages do not have to be equal in magnitude (example: 0V and +3V DC supplies)


Op Amp Terminal Voltages and Currents

+

–

V +

V –

• All voltages are referenced to a common node.• Current reference directions are into the op amp.

–

Vcc

+

+

Vcc

–

+

vn

–

+

vp

–

common node(external to the op amp)

ip

in

io

+

vo

–

ic+

ic-


Op-Amp Voltage Transfer Characteristic

In the linear region, vo = A (vp – vn) = A vid

where A is the open-loop gain

Typically, Vcc 20 V and A > 104

linear range: -2 mV vid = (vp – vn) 2 mV

Thus, for an op-amp to operate in the linear region,

vp vn

(i.e., there is a “virtual short” between the input terminals.)

vo

vid = vp–vn

The op-amp is basically adifferentiating amplifier:

“linear” “positive saturation”“negative saturation”

Vcc

-Vcc

Regions of operation:

slope = A >>1


Achieving a “Virtual Short”

Recall the voltage transfer characteristic of an op-amp:

Q: How does a circuit maintain a virtual short at the input of an op-amp, to ensure operation in the linear region?

A: By using negative feedback. A signal is fed back from the output to the inverting input terminal, effecting a stable circuit connection. Operation in the linear region enforces the virtual short circuit.

vo

vp–vn

Vcc

-Vcc

slope = A >>1

~1 mV

~10 V

vo

vp–vn

Vcc

-Vcc

slope = A >>1

Plotted using different scales for vo and vp–vn

~10 V

~10 V

Plotted using similar scales for vo and vp–vn


Familiar examples of negative feedback:

• Thermostat controlling room temperature

• Driver controlling direction of automobile

• Pupil diameter adjustment to light intensity

Familiar examples of positive feedback:

• Microphone “squawk” in sound system

• Mechanical bi-stability in light switches

Negative vs. Positive Feedback

Fundamentallypushes toward

stability

Fundamentallypushes toward

instability orbi-stability


Op Amp Operation without Negative Feedback(Comparator Circuits for Analog-to-Digital Signal Conversion)

1. Simple comparator with 1 Volt threshold: V– is set to 0 Volts (logic “0”) V+ is set to 2 Volts (logic “1”) A = 100

2. Simple inverter with 1 Volt threshold: V– is set to 0 Volts (logic “0”) V+ is set to 2 Volts (logic “1”) A = 100

If VIN > 1.01 V,V0 = 2V = Logic “1”

If VIN < 0.99 V,V0 = 0V = Logic “0”

V0

VIN1 20

1

2

+ V0

VIN

+1V VIN1 20

1

2V0

If VIN > 1.01 V,V0 = 0V = Logic “0”

If VIN < 0.99 V,V0 = 2V = Logic “1”

+ V0

VIN

+1V


Op Amp Circuits with Negative Feedback

Q: How do we know whether an op-amp is operating in the linear region?

A: We don’t, a priori.

• Assume that the op-amp is operating in the linear region and solve for vo in the op-amp circuit.

– If the calculated value of vo is within the range from -Vcc to +Vcc, then the assumption of linear operation might be valid. We also need stability – usually assumed for negative feedback.

– If the calculated value of vo is greater than Vcc, then the assumption of linear operation was invalid, and the op-amp output voltage is saturated at Vcc.

– If the calculated value of vo is less than -Vcc, then the assumption of linear operation was invalid, and the op-amp output voltage is saturated at -Vcc.


Op Amp Circuit Model (Linear Region)

vp

vn

Ri

+–

Ro

vo

A(vp–vn)

Ri is the equivalent resistance “seen” at the input terminals, typically very large (>1M, as large as 1012 for FET input op-amps), so that the input current is usually very small:

ip = –in 0

Note that significant output current (io)

can flow when ip and in are negligible!

)(

0

cco

cconp

iii

iiiii

ip

in

io


Ideal Op-Amp

• Assumptions:

– Ri is large (105 )

– A is large (104)

– Ro is small (<100 )

• Simplified circuit symbol:

– power-supply terminals and dc power supplies not shown

+

–+

vn

–

+

vp

–

ip

in

io

+

vo

–

ip = –in= 0

vp = vn

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.


Unity-Gain Voltage-Follower Circuit

V0

+

VIN

V0(V)

VIN(V)1 2

1

2

Note that the analysis of this simple (but important) circuit required only one of the ideal op-amp rules.

Q: Why is this circuit important (i.e., what is it good for)?

A: A “weak” source can drive a “heavy” load; in other words, the source VIN only needs to supply a little power (since IIN = 0), whereas the output can drive a power-hungry load (with the op-amp providing the power).

vp = vn V0 = VIN

( valid as long as V – V0 V + )

vn

vp

IIN


What’s Inside an Op-Amp?


Lecture Week 8 (continued)

• Op-Amp circuits continued: examples:• Inverting amplifier circuit

• Summing amplifier circuit

• Non-inverting amplifier circuit

• Differential amplifier circuit

• Current-to-voltage converter circuit

Reading

(Note: amplifiers are discussed

in great detail in Ch. 11)


Negative feedback is used to “linearize” a high-gain differential amplifier.

V0

+

V+

V

V0(V)

5

μV50

5

VV

510A

Without feedback

V0

+

VIN 510A

VIN1 2 3

V0

123

4 5

45

Review: Negative Feedback

With feedback


Gain vs. Frequency of the Basic Op-Amp without andwith Feedback (Hambley, Sec. 14.5)

Facts:

1. The open-loop gain of an op--amp (no feedback from output to inverting input) is constant from DC to a frequency fBOL, after which the open-loop gain drops at a rate of-20dB/decade of frequency (Fig. 14.20 – next slide).

2. As the negative feedback is made stronger, the gain decreases but the bandwidth of the amplifier increases. (Bandwidth of an amplifier is the frequency range over which it amplifies.)

3. One can show (Hambley) that the product of the dc gain and the bandwidth is a constant that is independent of the amount of negative feedback. This is called the gain-bandwidth product for the op-amp. Example: The op-amp you’ll use in the lab(National Semiconductor LMC6482) is rather like that shown in Hambley Fig. 14.22, with a DC open-loop gain of 100 dB (voltage amplification of 105) out to about only40 Hz! With negative feedback, however, the amplifier has an increasing bandwidth but with decreasing gain. The unity gain (0 dB) bandwidth is 4 MHz; one can get40 dB of gain (Vout/Vin = 100) from DC to about 40 kHz.


Op-Amp Frequency Response with and without Negative Feedback


Application of Voltage Follower: Sample and Hold Circuit


Inverting Amplifier Circuit

+

–

+vo

––+

vs

Rs

Rfin

in = 0 is = -if

vp = 0 vn = 0

+vp

–

+

vn

–

is

if

ss

fo v

R

Rv


Analysis using Realistic Op-Amp Model

• In the analysis on the previous slide, the op-amp was assumed to be ideal, i.e.

Ri = ; A = ; Ro = 0

• In reality, an op-amp has finite Ri, finite A, non-zero Ro, and usually is loaded at its output terminals with a load resistance RL.


Summing Amplifier Circuit

+

–

+vo–

–+

vc

Rc

Rfin

+vp

–

+

vn

–

ic

if

ib

ia

–+

vb

–+

va

Rb

Ra

c

c

fb

b

fa

a

fo v

R

Rv

R

Rv

R

Rv

in = 0 ia + ib + ic = -if

vp = 0 vn = 0

superposition !


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(volts)

0 0 0 00 0 0 1

0.5

MSB LSB

S1 closed if LSB =1S2 " if next bit = 1S3 " if " " = 1S4 " if MSB = 1

4-Bit D/A

8V+ V0

5K80K

40K

20K

10K

S1

+ -

S3

S2

S4+

Application: Digital-to-Analog Conversion

“Weighted-adder D/A converter”

(Transistors are used as electronic switches)

EE 42/100 Fall 2005 Week 8, Prof. White 23Digital Input

012345678

0 2 4 6 8 10 12 14 16

An

alo

g O

utp

ut

(V)

111110000100

00000001

Characteristic of 4-Bit DAC


Noninverting Amplifier Circuit

+

–

+vo

––+

vg

Rg

Rf

ip = 0 vp = vg vn = vg

in = 0 Rs & Rf form a voltage divider

+g

s

fso v

R

RRv

vp

–

Rs

+vn

–

in

ip


Differential Amplifier Circuit

0

b

on

a

an

R

vv

R

vv

nbdc

dp vv

RR

Rv

aa

bb

dca

bado v

R

Rv

RRR

RRRv

)(

)(

+

–

+vo

––+

Rbin

+

vp

–

+

vn

–

vb

–+

va Rc

Ra

in = 0

ip = 0

Rd

ip

aba

bo

d

c

b

a vvR

Rv

R

R

R

R then , If


More usual version of differential amplifier (Horowitz & Hill, “Art of Electronics”, 2nd Ed., p. 184-5 (part of their“smorgasbord” of op-amp circuits):

Let Ra = Rc = R1

and Rb = Rd = R2

Then v0 = (R2 / R1)(vb – va)

To get good “common-mode rejection” (next slide)you need well-matched resistors (Ra and Rc, Rb and Rd), such as 100k 0.01% resistors

Differential Amplifier (cont’d)


Differential Amplifier – Another Perspective

Redefine the inputs in terms of two other voltages:

1. differential mode input vdm vb – va

2. common mode input vcm (va + vb)/2

so that

va = vcm – (vdm/2) and vb = vcm + (vdm/2)

Then it can be shown that

)(2

)()(

)(

dca

dcbbaddm

dca

cbdacm

dmdmcmcmo

RRR

RRRRRRA

RRR

RRRRA

vAvAv

and where

“common mode gain” “differential mode gain”


Differential Amplifier (cont’d)

• An ideal differential amplifier amplifies only the differential mode portion of the input voltage, and eliminates the common mode portion.– provides immunity to noise (common to both inputs)

• If the resistors are not perfectly matched, the common mode rejection ratio (CMRR) is finite:

a

bdmcm

d

c

b

a

R

Rvv

R

R

R

R and then , If 0

d

c

b

aab

cm

dm

R

R

R

RRR

A

ACMRR )1(

/1

if


Op-Amp Current-to-Voltage Converter


Summary• Voltage transfer characteristic of op-amp:

• A feedback path between an op-amp’s output and its inverting input can force the op-amp to operate in its linear region, where vo = A (vp – vn)

• An ideal op amp has infinite input resistance Ri, infinite open-loop gain A, and zero output resistance Ro. As a result, the input voltages and currents are constrained:

vp = vn and ip = -in = 0

vo

vp–vn

Vcc

-Vcc

slope = A >>1

~1 mV

~10 V





Differentiator (from Horowitz & Hill)


Op-Amp ImperfectionsDiscussed in Hambley, pp. 651-665 (of interest if you need touse an op-amp in a practical application!)

Linear range of operation: Input and output impedances notideal values; gain and bandwidth limitations

(including gain-bandwidth product);

Nonlinear limitations: Output voltage swing (limited by “rails” and more); output current limits; slew-rate limited (how fast can the output voltage change: 0.5V/s for the 741 op-amp, to 6000V/s for high slew-rate op-amp); full-power bandwidth

DC imperfections: Offset current (input currents don’t sumexactly to zero); offset voltage; There are pins

(denoted offsets) for inputs to help control this.

Documents

EE 42/100 Fall 2005Week 8, Prof. White1 Lectures Week 8 The operational amplifier (“op amp”) Feedback Comparator circuits Ideal op amp Unity-gain voltage