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-signal
amp from last
lecture would go
here.
From amp gain,
we know that
vout = Avi
Negative output sent
back to positive input
– “negative feedback”
• We want to find the gain for the entire circuit (feedback
included)
– i.e., want to know vout/vin
• We start with what we already know:
vout = Avi
vi = vin ! v f
R1
vf =
vout
R1 + RF
• From which we find:
vi = vin !
vout
R1
vout
R1 + RF
"
#
R1
= A $ vin !
vout %
R1 + RF
&
'
"
AR1 #
vout $ 1 +
= Avin
%
R1 + RF '
&
A
vout =
vin ( Af vin
AR1
1+
R1 + RF
• Note that the gain of the feedback circuit is less than that
of the main amplifier
– Seems counterproductive!
Benefits to feedback
• The general form of the equation from the previous slide is:
A
Af =
1 " (±1) ! A
where A is the main amplifier gain (often called the opencircuit gain), β is the fraction of the output fed back to the
input, and the ± indicates whether the feedback is positive or
negative
• 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
100
Af =
= 3.8
1 + 0.25 ! 100
• What happens if the main amp degrades such that A is
reduced to 50?
50
A!f =
= 3.7
1 + 0.25 " 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
Frequency response
• Let’s assume that A is bandwidth-limited
– i.e., it depends on ω, with large frequencies amplified less:
A0
A (! ) =
!
1+
!c
• In a negative-feedback system, the overall gain is:
A0
A (! )
1 + ! / !c
Af (! ) =
=
1 + " A (! )
#
$
A0
1+ " %
&
1
+
!
/
!
c (
'
Ao
Ao
1 + " Ao
=
=
)
1 + ! / ! c + " Ao 1 + " Ao 1 + ! / ! c + " Ao
Ao
1
1
=
)
= Af (0 )
1 + " Ao 1 + ! / #'(1 + " Ao )! c $(
1 + ! / #'(1 + " Ao )! c $(
• So we see that the cutoff frequency is increased by a factor
of 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 between
two 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 same
direction
– 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 two
common-emitter
amplifiers placed
back-to-back
Point A
• Let’s calculate the differential gain for this circuit:
Gdiff
!Vout
=
! (V1 " V2 )
to calculate this, let’s see what happens when equal but
opposite signals are input to V1 and V2:
– note that VA is unchanged by this, since the two emitter
resistors have equal value
!Vout = "!I C RC # "!I E RC
!V1 = !VE = "!I E RE = !V2
Gdiff
"!I E RC
RC
=
=
"!I E RE " !I E RE 2 RE
• We now compare this to the common mode gain that
occurs when both V1 and V2 move in the same direction:
GCM
!Vout
=
! (V1 + V2 )
• Let’s see what happens when we increase both voltages by
the same amount:
!Vout = "!I C RC # "!I E RC
!V1 = !VE = "!I E RE " 2!I E R1 = !V2
GCM
"!I E RC
RC
=
=
2!I E (RE + 2 R1 ) 2 RE + 4 R1
• If we choose R1 >> RE, signal differences will be amplified
much more than common signal changes
• We define the common mode rejection ratio (CMRR) as:
CMRR =
for our circuit, it’s:
GCM
Gdiff
2 RE + 4 R1 RE + 2 R1
CMRR =
=
2 RE
RE
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 that
require an amplifier
• Internally, an op-amp might look something like this:
• We’re not really going to care too much about the innards
of the op-amp
– we just need to be familiar with how it behaves
• An op-amp looks like:
on a schematic:
inverting
input
non-inverting
input
in real life:
• Requires external power input (typically ±12 or ±15V)
– these connections often not shown on schematic
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:
Note the negative
feedback
• The voltage at the (-) input is:
V! = Vi ! R1 I1 = Vo ! I 2 R2
• Since the op-amp wants its inputs at the same voltage,
V! = 0
• Since the inputs draw no current, I1 = -I2
• Putting it all together, we find:
Vi = R1 I
Vo = ! R2 I
Vo R2
=
Vi R1
• Note that the gain depends totally on the resistors, and not
at 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 optimal
feature
• Output impedance is small (<1Ω)
Follower
• We can remove the resistors from the previous circuit to
get:
• From op-amp rule #1, we see that Vo will adjust itself to
equal 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 variable
resistance
• Since there’s no feedback, the op-amp can’t do anything to
make the inputs equal
– any difference between Vi and V+ will be subject to the opamp gain of a million or so
• Does that mean that a difference of 1V at the inputs will
result 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 whenever
the 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 the
analog input voltages
• This circuit is at the interface between analog and digital
electronics
– We’ll go into the digital realm starting next lecture