Analysis of circuits with
negative feedback
Electronic Circuit Design AA 2011-12
A.Castoldi
v1.5
A mathematical introduction to feedback
•
Feedback is an important concept in electronics. Feedback is covered also in courses
on control theory but the complexity of circuit topologies require to develop a specific
approach in order to fully analyze a real circuit within the framework of feedback
theory. New concepts and terminology need to be developed (open and closed-loop
parameters, loop gain, ideal gain, direct feed-through, etc.). It must be stressed that
feedback is virtually present in all analog circuits, therefore it is important that
students develop a good understanding of feedback concepts in circuits and learn a
sound analysis method.
•
In most circuit design books, feedback analysis is often presented with reference to
the classical block diagram (*) – consisting in a unilateral forward amplifier (forward
gain, G) and a unilateral feedback network (feedback factor, H) - where sin and sout
can each be either a current or a voltage.
•
The closed-loop gain (sout/sin) is easily obtained from the following passages:
sout=Gse=G(sin-sf)=G(sin-Hsout) and, solving for (sout/sin), you get
Gclosed loop 
•
sout
G

sin 1  GH
As we will see, the definition of blocks G, H is not very useful for circuit analysis. In
the following we’ll work it out to get to a more specific approach for circuits.
(*) other possible block schemes are also employed but all lead to the same general properties (slides 4 and 5)
2
Loop gain
•
Loop gain (GLOOP): it is the distinctive feature of the feedback system. It is the
amplification of a test signal after “one lap around the feedback loop”. It can be
measured by inserting a test signal into the loop, with the input signal set to zero (it
must be stressed that the insertion of the the test source must be such that it does
not alter the loop gain). If GLOOP is equal to zero there is no feedback in the circuit.
The computation of the loop gain in the classic block diagram is shown below.
G LOOP 
•
•
•
•
•
u out
 GH
uin
The “ideal” condition for the feedback loop is to have infinite G and hence
infinite GLOOP.
The loop gain can be either positive or negative. When the loop gain is negative, if
you apply a positive step at the input, the loop returns a positive feedback signal sf,
which is subtracted from the input so that the error signal s is smaller (s=si-sf<si)
It will be clear after studying the stability of feedback systems (e.g. see Bode
criterion for stability) that systems with positive feedback (GLOOP >0) tend to be
unstable, i.e. with poles having positive real part, while systems with negative
feedback (GLOOP <0) can be designed to be stable.
Here we will restrict to the analysis of negative feedback circuits, which are
commonly used to design analog amplifiers with superior performances.
The inherent instability of positive feedback circuits is exploited in special circuits
(comparators, oscillators, etc.).
3
General properties of negative feedback systems
•
The main variables (sout , sf , s) of a negative feedback system, in response to the
input signal (sin), are summarized below. The asymptotic behavior for G∞ is also
shown.
sout
G
  1 

G
 
sin 1  GH
H 
sf
sin

GH

G
1
1  GH
s
1


G
 0
sin 1  GH
•
•
•
•
It is interesting to note that for G∞ the closed-loop gain does not depend on
G any more but it depends only on the feedback factor H. In this limiting
condition, the feedback signal (sf ) is forced to be exactly equal to the input signal, so
that the error signal (s) tends to zero.
The asymptotic (or ideal) closed-loop gain, GID, equals (1/H). Therefore in
order to have closed-loop gain with magnitude >1 the feedback factor H is an
attenuator.
It is normally easier to make a precise attenuator (e.g. the attenuation of a resistive
divider is related to the ratio of resistances, which can be made very precisely), than
to make an amplifying stage with precise gain because the parameters of any active
device depend on several variables (time, environmental conditions, aging, etc.).
This means that using the concept of feedback we can design an amplifier with
very precise gain (more precise than a single amplifying stage) by using a
precise attenuator and an (intrinsically less precise) amplifying stage,
4
provided it has very large gain.
Application of feedback to real circuits
•
•
The classical feedback diagram is very useful as it can explain all the general
properties of feedback. However it is an idealized picture that does not fit all real
circuits (see Notes below).
Alternatively we can re-express the closed-loop gain in terms of two other relevant
quantities: the Loop Gain (GLOOP) and the Ideal Gain (GID):
Gclosed loop 
•
•
sout
 G LOOP
G
 1  GH

 
 G ID
1  G LOOP
sin 1  GH  H  1  GH
GID corresponds to the asymptotic gain of the system, which is typically the designed
gain and it is a meaningful quantity by itself. By computing the actual value of GLOOP
one obtains the exact closed-loop gain.
The main advantage of this formulation is that both quantities can be derived
directly from the circuit, with no previous knowledge of G and H and of the block
diagram.
Notes:
o
The classical feedback diagram assumes unilateral blocks (while it is intuitive that even simple resistive networks
have bilateral transfer), moreover in electronic circuits the analysis is complicated by the interaction of the
feedback network with the forward amplifier (i.e. the gain of a stage depends on the ratio of its output resistance to
the load resistance, the well-known “load effect”).
o
In order to apply it to a real feedback circuit, we have to manipulate the circuit to fit the idealized block diagram,
i.e. we have to define the blocks G, H in terms of unilateral two-port networks. Then one can use the known
expressions to compute the variables (sout , sf , s). This procedure is tedious and not always feasible due to the
5
above mentioned assumptions.
Computation of the ideal gain in real circuits
•
Let us analyze the following feedback circuit in a qualitative way:
•
The input voltage vin (1) is applied to the positive input of the opamp (2) (there is no current on R)
and produces a positive step at the output of Vout (3) and therefore a positive step vf is fed back
to the negative opamp input (4). Therefore , due to the KVL at the vin-v-vf loop, the differential
voltage v=v+-v- which drives the opamp is reduced by the action of negative feedback.
•
If the loop gain is high, V tends to zero (i.e. V- tends to be closer to V+) and, asymptotically for
A∞, you have V-=Vin. The current on R1 (=Vin/R1) flows in R2 so that the output voltage
Vout=Vin+Vin(R2/R1)=Vin(1+R2/R1). This represents the “ideal” gain of the circuit.
More formally:
•
We refer to the same dependent source in the feedback loop used for the computation of
GLOOP.
•
Set the controlling quantity (in the example v=v+-v-) to zero (v is like the error signal in the
classical block diagram)
•
Find the output variable
•
GID = (Vout/Vin)v=0
6
Computation of the ideal gain in real circuits (2)
•
Let’s see a second example.
•
The positive step Vin, applied to the negative branch of the opamp, forces a current in R1 and a
fraction of it will go into Rid. This fraction produces a positive step on V- and, being V+ at ground,
a negative differential voltage V=(V+-V-) which drives the opamp output. The negative output
voltage increases the voltage drop on R2 and therefore the current fraction flowing into R2
increases, thus reducing the current fraction going into Rid.
•
If the loop gain is high, the current in Rid and V- tends to be closer to 0 and, asymptotically for
A∞, you have V-=0. The current on R1 (=Vin/R1) flows in R2 (as Rid can not take current) so
that the output voltage is given by the drop on R2: Vout=-(Vin/R1)*R2=Vin(-R2/R1), which shows
the ideal gain of this circuit.
7
Computation of the loop gain in real circuits
•
We need to break the loop without altering the signal transfer along the loop. To this end we
identify the dependent source which has generally the role to provide gain in the
feedback loop. This is straightforward as a dependent source is present in the equivalent model of
any active device (op-amp or transistor).
•
•
•
Then:
Set all independent sources to zero
Break the connection between the dependent source and the rest of the circuit and restore the
original impedance (Z).
Drive the circuit at the break with an independent source of same type (sT) to probe the loop gain
Find the output of the dependent source (sT0)
GLOOP = sT0/sT
•
•
•
G LOOP 
sT 0
( R1 // Rid )
 A
sT
( R1 // Rid )  R2
8
Computation of the loop gain in real circuits (2)
•
Breaking the loop may alter the impedance level at the break point and therefore it may lead to a
wrong value of GLOOP, i.e. different from the loop gain in closed-loop condition. Breaking at the
dependent source is a smart way to avoid the problem, as the output of the dependent source does
not depend on the value of the load impedance (Z) which need not be computed.
•
Breaking at a different point along the loop is also possible but - as the break point is no more
generally driven by an ideal source – in order to obtain the correct loop gain one must compute
the original load impedance Z at the break as it now affects the result. The computation of the loop
gain is shown in the following scheme. The result is of course the same as before.
G LOOP 
sT 0
( R1 // Rid )
 A
sT
( R1 // Rid )  R2
9
Frequency response of the closed loop gain:
graphical method
•
The closed-loop gain, i.e. Vo/Vin, is related to GID and GLOOP according to the equation below (see
slide 5). The Bode diagram of |Vo/Vin| can be therefore approximated by its two asymptotic
curves, corresponding to the limiting cases |GLOOP|>>1 or |GLOOP |<<1.
 G ID  j   G LOOP  j 
vout
 j   GID  j   GLOOP  j 
vin
1  G LOOP  j 
G ID  j 
for G LOOP  1
for G LOOP  1
•
It is useful to remind that this is the same level of approximation of the usual straight-line
approximation of the Bode plot of a real pole or zero: in the denominator we neglect 1 when
|GLOOP|>>1 or we neglect GLOOP when |GLOOP |<<1.
•
The frequency response of the closed-loop gain of the previous circuit example (slides 7-8-9) is
shown below (yellow line). It follows the asymptotic curve |GID| when |GLOOP|>>1 while is follows
the curve |GID GLOOP| when |GLOOP|<<1. The frequency at which |GLOOP|=1 (i.e. the crossing point
between the two asymptotic curves) corresponds to the closed-loop pole (CL). The expression of
Vo/Vin(s) can be easily deduced from the diagram.
A0 ( R1 // Rid ) R2
( R1 // Rid )  R2 R1
computation of CL :
G ID G LOOP
A ( R // R ) R2
R2
CL  0 1 id
0
R1
( R1 // Rid )  R2 R1
G ID
R2 / R1
0
•
CL

vout
 R2 R1

1  s / CL
vin
 CL   0
s  j
A0 ( R1 // Rid )
( R1 // Rid )  R2
Note: this graphical method is rigorous only for first-order systems. However, due to its
simplicity, it is often used in circuit design for first-order estimation of the dominant pole of a
feedback circuit.
10
Direct feed-forward
•
Let us consider the example of the inverting amplifier when the opamp has a finite output
resistance r0. In this case, if we set the gain of the dependent source (A) to zero, we obtain a nonzero output voltage.
•
But setting A=0 also means zero loop gain and, if we recall the general expression of the closedloop gain derived before, we should have ended up with Vout=0 (?!).
Gclosed loop 
•
vout
vin
 GLOOP
1  GLOOP
A 0
(GLOOP  0 )
0
The contradiction is due to the fact that the given expression of the closed-loop gain was
incomplete. The complete expression, valid for all feedback circuit, is the following:
Gclosed loop 
•
A 0
( GLOOP  0)
 G ID
G0
vout
G LOOP

 G ID
1  G LOOP 1  G LOOP
vin
G0 is called the direct feed-forward gain and it corresponds to the closed-loop gain when A=0
(more generally, when the output of the dependent source is set to zero).
11
Direct feed-forward (2)
•
•
Although the direct feed-forward is small and usually neglected when the loop gain is high, it
becomes more important at high frequency, where the loop gain generally tends to zero. Let us
study the frequency response of the previous circuit, including also the direct feed-forward term.
We will adopt the usual asymptotic approximations for the Bode plots.
The 3 ingredients of the frequency response are GID, GLOOP and G0:
G ID  
•
R2
R1
G LOOP   A( s )
A0
R1
R1

R1  R2
1  s  0 R1  R2
G0 
R0
R1  R2  R0
The asymptotic values of the closed-loop gain for |GLOOP|<<1 and >>1 are the following:
G ID G LOOP
vout
G LOOP
 G ID
vin
1  G LOOP

G ID
G0
( for G LOOP  1)
G0
1  G LOOP
G0
( for G LOOP  1)
 G LOOP
•The final amplitude plot (dashed
red) is obtained by selecting the
dominant term at each frequency,
as we usually do in the straightline approximation of Bode plots.
•Here the direct feed-forward
term is responsible for a zero at
high frequency (z).
 z   A00 
R1  R2  R0 
R2
R1  R2 
R0
CL   A00  R1 R1  R2 
12
Effect of negative feedback
on equivalent impedances
•
•
•
A general property of circuits with negative feedback is to stabilize all the voltages
of the nodes belonging to the loop and all the currents of the branches of
the feedback loop. Here to stabilize means to minimize the change in the
voltages/currents of the feedback loop in response to an external source of
perturbation. In particular, for infinite loop gain (i.e. “ideal” feedback) the change in
the voltages/currents of the loop tends to zero.
For instance, if one injects a test current into a node belonging to the loop, the action
of negative feedback tends to minimize the voltage change of the considered node.
The same happens if one inserts a test voltage source in a branch of the loop: the
current change in that branch is minimized by negative feedback. When the loop gain
is infinite, the response to the external perturbation tends to zero.
The same concept can be also stated in terms of equivalent impedance: for infinite
loop gain, the impedance “seen” by a test current source, perturbing the
voltage of a loop node, tends to zero (i.e. equivalent to Vnode0); the
impedance “seen” by a test voltage source perturbing the current of a loop
branch tends to infinity (i.e. equivalent to Ibranch0);
Note:
o
The “stabilization” effect of negative feedback can be explained in terms of transfer functions of the feedback
circuit, which have the general expression of the closed-loop gain given before (slide 10).
o
In case the output variable is the voltage of a loop node, it can be shown that the ideal transfer (Vnode/Isource)ID is
zero and the transfer function reduces to the direct feed-forward term (Vnode/Itest) = (Vnode/Itest)0 /(1-GLOOP). This
transfer function is the impedance seen by the current source which vanishes for infinite loop gain.
o
The same applies when the output is the current of a loop branch and the transfer function is reduced to
(Ibranch/Vtest) = (Ibranch/Vtest)0 /(1-GLOOP). This transfer function is the admittance seen by the voltage source, which
vanishes for infinite loop gain (or alternatively the equivalent impedance goes to infinity).
13
Qualitative analysis
•
To compute the output impedance Zout of this amplifier, we set the input to zero and we choose a
source to probe the output impedance. Here we choose a current source because if we had chosen
a voltage source the loop gain of the circuit would be zero (and the concepts of feedback could not
have been used).
•
Sourcing a positive current iT, the output node tends to rise and also V-. This produces a negative
differential voltage that tries to reduce the initial rise of the output voltage v. The effect of reducing
the output voltage stimulated by the input current is equivalent to decreasing the equivalent
impedance seen by the source. For infinite loop gain, v=0 and therefore the asymptotic value of
Zout is v/iT=0.
After this qualitative analysis it is clear that in this case we were perturbing the voltage of a loop 14
node and therefore the use of a current test source was the appropriate choice.
•
Computation of equivalent impedances
•
In general, we need to understand whether we are perturbing a node (voltage) or a branch
(current) of the loop and then use the appropriate excitation source.
•
This can be understood by computing the asymptotic value of the impedance Z seen from the
considered test source when the loop gain goes to infinity, i.e. when the controlling variable of the
dependence source is set to zero (i.e. v=v+-v-=0), as we do for the computation of the ideal value
of any closed-loop gain in feedback circuits.
•
Let us check in the previous example that the asymptotic value of Zout is in agreement with the
qualitative analysis of the circuit discussed before:
•
if we apply v-=0 for infinite loop gain, the current in R1 and in R2 is forced to be zero, therefore
v=0. This means that the feedback loop stabilizes the voltage of the output node and therefore we
must use a current source. The impedance Zout tends to zero (as we saw before).
15
Computation of equivalent impedances (2)
In general, the asymptotic value of the impedance Z will be either 0 or infinity:
–
–
•
•
Z0 is the open-loop impedance, i.e. the impedance seen when the output of the
dependent source is set to zero, that is Z0=Z|A=0.
If we fall in one of the 2 above cases, we just have to compute Z0 and GLOOP and
insert in the proper expression to obtain Z. Returning to our example, let us compute
Zout|0 and GLOOP:
•
•
If Z0: the excitation source must be a test current perturbing the voltage (v) of a node of
the loop stabilized by negative feedback and therefore Z=v/itest0/itest=0. For finite GLOOP the
following expression holds: Z=Z0/(1-GLOOP)
If Z∞: the excitation source must be a test voltage perturbing the current (i) of a branch
stabilized by negative feedback and therefore Z=vtest/ivtest/0=infinity. For finite GLOOP the
following expression holds: Z=Z0*(1-GLOOP)
Open-loop impedance Zout|0
•
Loop gain
G LOOP 
 AR1
R1  R2  R0
As we found Zout0 for GLOOP∞, we use the expression Zout=Zout|0/(1-GLOOP):
Z out  R0 // R1  R2 
1
1  A R1 R1  R2  R0 
16
Computation of equivalent impedances (3)
•
This example shows the case in which the current of a branch is “stabilized” by negative feedback.
•
Asymptotic impedance (GLOOP infinity)
•
As Zin∞ for GLOOP∞, we use the expression Zin=Zin|0*(1-GLOOP):
•
Open-loop impedance (A=0)
17
Computation of equivalent impedances (4)
•
Finally, consider this case:
•
when we apply v-=0 to compute the asymptotic value of Rin, we get RinR1, i.e. neither zero
nor infinite. This means that we are neither perturbing a voltage nor a current of the
feedback loop, i.e. it is a node/branch of the circuit “outside” of the loop.
We try to recover the standard situation by moving closer to the loop i.e. by splitting the
impedance computation in two steps:
•
•
We compute the new feedback impedance R*in, hoping that moving beyond R1 would bring us on
the loop and therefore fall into one of the two main cases Rin0 or Rin∞. (Note: splitting
impedance computation in smaller problems is often advisable, even if not needed in principle,
because it leads to easier computations).
18
Computation of equivalent impedances (5)
•
To compute the equivalent impedance Z*in seen from the new standpoint we apply the same
procedure to the new sub-circuit. From the figure we see that now we are actually on a node of the
loop, as from the asymptotic analysis we get v=0 and therefore the impedance tends to zero.
•
We can now compute R*in|0 and G*LOOP. It must be remembered that by splitting the circuit one
also changes the loop gain, which must be re-computed for this special sub-circuit.
•
The total impedance is therefore:
Rin  R1 
R2
1 A
19