Lecture 4: Feedback and Op-Amps• Last time, we discussed using transistors in small-signal

amplifiers– If we want a large signal, we’d need to chain several of these

small amplifiers together– There’s a problem, though: at each stage of the amplification,

distortion (noise) is introduced along with the signal we want– Eventually the amplified noise becomes comparable to or bigger

than the signal• This can be avoided by designing amplifier circuits with

feedback– that means that part of the output from the amplifier is routed back

to the input– Concept first developed by Bell Labs engineer Harold Black– This made long-distance calls (which must be amplified many

times en route) possible

Feedback example

The small-signalamp from lastlecture would gohere.From amp gain,we know thatout iv Av=

Negative output sentback to positive input– “negative feedback”

• We want to find the gain for the entire circuit (feedbackincluded)– i.e., want to know vout/vin

• We start with what we already know:

1

1

out i

i in f

f out

F

v Av

v v v

Rv v

R R

=

= !

=+

• From which we find:

• Note that the gain of the feedback circuit is less than thatof the main amplifier– Seems counterproductive!

1

1

1

1

1

1

1

1

1

1

i in out

F

out in out

F

out in

F

out in f in

F

Rv v v

R R

Rv A v v

R R

ARv Av

R R

Av v A v

AR

R R

= !+

" #= !$ %

+& '" #

+ =$ %+& '

= (+

+

Benefits to feedback• The general form of the equation from the previous slide is:

where A is the main amplifier gain (often called the open-circuit gain), β is the fraction of the output fed back to theinput, and the ± indicates whether the feedback is positive ornegative

• The quantity L = ± βA is called the loop gain for the circuit• Now let’s assume A = 100, and β = 0.25, and the feedback is

negative• Then

( )1 1f

AA

A!=

" ±

1003.8

1 0.25 100fA = =

+ !

• What happens if the main amp degrades such that A isreduced to 50?

• So a drastic change in A results in a tiny change in Af– As long as β is constant– But β can be determined by resistors (as in our example

circuit) and is therefore very stable• Thus we see that negative feedback improves the stability

of our amplifier

503.7

1 0.25 50fA! = =

+ "

Frequency response• Let’s assume that A is bandwidth-limited

– i.e., it depends on ω, with large frequencies amplified less:

• In a negative-feedback system, the overall gain is:

( ) 0

1

c

AA !

!

!

=

+

( )( )

( )

( )( )

( )

0

0

1 /

11

1 /

1

1 / 1 1 /

1 10

1 1 / 1 1 / 1

cf

c

o o o

c o o c o

of

o o c o c

A

AA

A A

A A A

A A A

AA

A A A

! ! !!

" !"

! !

"! ! " " ! ! "

" ! " ! ! " !

+= =

+ # $+ % &

+' (+

= = )+ + + + +

= ) =+ + + + +# $ # $' ( ' (

• So we see that the cutoff frequency is increased by a factorof 1+βAo– that’s the same factor by which the amplifier’s gain is

reduced by the feedback circuit– we’re trading reduced gain for increased bandwidth

Differential Amplifiers• So far we’ve talked about amplifying a voltage, but often

what we really want is to amplify the difference betweentwo voltages– Example: telephone carries your voice as a difference in

voltage between two wires– Electrical noise from outside sources will typically change

the voltage in both wires at the same time, in the samedirection

– We want to amplify the voice, but not the noise!

• A circuit that does this is called a differential amplifier,and the following is an example:

It’s basically twocommon-emitteramplifiers placedback-to-back

Point A

• Let’s calculate the differential gain for this circuit:

to calculate this, let’s see what happens when equal butopposite signals are input to V1 and V2:– note that VA is unchanged by this, since the two emitter

resistors have equal value

( )1 2

outdiff

VG

V V

!=! "

1 2

2

out C C E C

E E E

E C Cdiff

E E E E E

V I R I R

V V I R V

I R RG

I R I R R

! = "! # "!

! = ! = "! = !

"!= ="! " !

• We now compare this to the common mode gain thatoccurs when both V1 and V2 move in the same direction:

• Let’s see what happens when we increase both voltages bythe same amount:

( )1 2

out

CM

VG

V V

!=! +

( )

1 1 2

1 1

2

2 2 2 4

out C C E C

E E E E

E C C

CM

E E E

V I R I R

V V I R I R V

I R RG

I R R R R

! = "! # "!

! = ! = "! " ! = !

"!= =

! + +

• If we choose R1 >> RE, signal differences will be amplifiedmuch more than common signal changes

• We define the common mode rejection ratio (CMRR) as:

for our circuit, it’s:

CM

diff

GCMRR

G=

1 12 4 2

2

E E

E E

R R R RCMRR

R R

+ += =

Operational Amplifiers (Op-Amps)• An op-amp is an integrated circuit (chip) whose behavior

approximates an ideal amplifier:– high input impedance– low output impedance– large gain (factors of a million aren’t uncommon)

• These are used rather than transistors in circuits thatrequire an amplifier

• Internally, an op-amp might look something like this:

• We’re not really going to care too much about the innardsof the op-amp– we just need to be familiar with how it behaves

• An op-amp looks like:

• Requires external power input (typically ±12 or ±15V)– these connections often not shown on schematic

on a schematic:

in real life:

invertinginput

non-invertinginput

Op-amp rules• An ideal op-amp behaves as follows:1. Inputs draw no current (infinite input impedance)2. Output tries to adjust itself so that inputs are at the same

voltage• Real op-amps come pretty close to these ideals• Limitations appear as:

– nonzero input currents– finite slew rate for output– limitations in bandwidth– finite CMRR

Op-amp circuits• Inverting amplifier:

• The voltage at the (-) input is:

• Since the op-amp wants its inputs at the same voltage,

• Since the inputs draw no current, I1 = -I2

1 1 2 2i oV V R I V I R

!= ! = !

0V!=

Note the negativefeedback

• Putting it all together, we find:

• Note that the gain depends totally on the resistors, and notat all on the properties of the op-amp– except of course that the op-amp is assumed to behave

ideally!• Input impedance is just R1, so we can choose the value

– but raising input impedance lowers gain – not an optimalfeature

• Output impedance is small (<1Ω)

1

2

2

1

i

o

o

i

V R I

V R I

V R

V R

=

= !

=

Follower• We can remove the resistors from the previous circuit to

get:

• From op-amp rule #1, we see that Vo will adjust itself toequal Vi– the output “follows” the input

• This is an “amplifier” with a gain of 1– but does offer very high input impedance and low output

impedance

Comparator• Consider what happens with the following circuit:

• The voltage at the + input changes due to the variableresistance

• Since there’s no feedback, the op-amp can’t do anything tomake the inputs equal– any difference between Vi and V+ will be subject to the op-

amp gain of a million or so

• Does that mean that a difference of 1V at the inputs willresult in an output of ~1 million V?– No! The op-amp can’t exceed the voltage of its power

supply• What does happen is that the output swings all the way to

the limits (upper or lower) of the power supply wheneverthe inputs are different– in other words, the output tells us whether V+ is greater than

or less than Vi – it gives one bit of information about theanalog input voltages

• This circuit is at the interface between analog and digitalelectronics– We’ll go into the digital realm starting next lecture