FEEDBACK
INTRODUCTION
Most physical systems incorporate some form of feedback. Feedback can
be either negative (degenerative) or positive (regenerative).
In amplifier design, negative feedback is applied to effect one or more of
the following properties:
1. Desensitize the gain; that is, make the value of the gain less sensitive
to variations in the value of circuit components.
2. Reduce nonlinear distortion; that is, make the output proportional to
the input (make the gain constant).
3. Reduce the effect of noise.
4. Control the input and output impedances.
5. Extend the bandwidth of the amplifier.
All of the above desirable properties are obtained at the expense of a
reduction in gain.
In short, the basic idea of negative feedback is to trade off gain for other
desirable properties.
The General Feedback Structure
Figure. General structure of the feedback amplifier. This is a signal-flow diagram,
and the quantities x represent either voltage or current signals.
The open-loop amplifier has a gain A; thus its output Xo is related to the
input Xi by
Xo = A Xi
(1)
The output Xo is fed to the load as well as to a feedback network, which
produces a sample of the output. This sample Xf is related to Xo by the
feedback factor B,
Xf = Β Xo
(2)
The feedback signal Xf is subtracted from the source signal Xs to
produce the signal Xi ,
Xi = Xs – Xf
(3)
So the negative feedback reduces the signal that appears at the input of
the basic amplifier. The gain of the feedback amplifier can be obtained
by from Eqs. 1-3 :
Af ≡
Xo
AXi
AXi
A
=
=
=
Xs Xi + Xf
Xi + ABXi 1 + AB
(4)
The quantity Af is called the loop gain.
The quantity 1 + AB is called the amount of feedback.
In many circuits, the loop gain AB is large, AB » 1, it follows
that Af ≈ 1 / B - The gain of the feedback amplifier is almost entirely
determined by the feedback network.
SOME PROPERTIES OF NEGATIVE FEEDBACK
1.Gain Desensitivity
Assume that B is constant. Taking differentials of both sides of Eq. (4)
results in
dAf =
dA
(1 + AB )2
(5)
Dividing Eq. (5) by Eq. (4) yields
dAf
1
dA
=
Af
(1 + AB ) A
(6)
The amount of feedback, 1 + AB, is also as the desensitivity factor.
2.Bandwidth Extension
Consider an amplifier whose high-frequency response is characterized by
a single pole. Its gain at mid and high frequencies
As =
AM
1 + s / ωH
(7)
where AM denotes the midband gain and ωH is the upper 3-dB frequency.
Application of negative feedback, B, results in a closed-loop gain :
Af ( s) =
AM / (1 + AM B )
A( s )
=
1 + BA( s ) 1 + s / ω H (1 + AM B )
Thus the feedback amplifier will have a midband gain of AM / (1 + AM B )
and an upper 3-dB frequency: ω Hf = ω H (1 + AM B )
Similar a lower 3-dB frequency is ω Lf = ω L / (1 + AM B )
THE FOUR BASIC FEEDBACK TOPOLOGIES
Based on the quantity to be amplified (voltage or current) and on the
desired form of output (voltage or current), amplifiers can be classified
into four categories.
Fig. The four basic feedback topologies: (a) voltage-sampling series-mixing (seriesshunt) topology; (b) current-sampling shunt-mixing (shunt-series) topology; (c)
current-sampling series-mixing (series-series) topology; (d) voltage-sampling shuntmixing (shunt-shunt) topology.
Voltage Amplifiers
Since the signal source is a voltage source, it is convenient to represent it
in terms of a Thevenin equivalent circuit. In a voltage amplifier the
output quantity of interest is the output voltage. It follows that the
feedback network should sample the output voltage. Also, because of the
Thevenin representation of the source, the feedback signal Xf should be a
voltage that can be mixed with the source voltage in series.
A suitable feedback topology for the voltage amplifier is the voltagesampling series-mixing one shown in Fig. (a). This feedback topology is
also known as series-shunt feedback, where "series" refers to the input
and "shunt" refers to the connection at the output.
Current Amplifiers
Here the input signal is essentially a current, and thus the signal source
is most conveniently represented by its Norton equivalent. The output
quantity of interest is current; hence the feedback network should sample
the output current.
The feedback signal should be in current form so that it may be mixed in
shunt with the source current.
Thus the feedback topology suitable for a current amplifier is the
current-sampling shunt-mixing topology, illustrated in Fig. (b).
This feedback topology is also known as shunt-series feedback.
Again, the first word in the name (shunt) refers to the connection at the
input, and the second word (series) refers to the connection at the output.
Transconductance Amplifiers
Here the input signal is a voltage and the output signal is a current. It
follows that the appropriate feedback topology is the current-sampling
series-mixing topology, illustrated in Fig. (c). It is also known as the
series-series feedback configuration.
Transresistance Amplifiers
Here the input signal is current and the output signal is voltage. It
follows that the appropriate feedback topology is of the voltagesampling shunt-mixing type, shown in Fig. (d). It is also known as
shunt-shunt feedback.
THE SERIES-SHUNT FEEDBACK AMPLIFIER
The Ideal Situation
The ideal structure of the series-shunt feedback amplifier is shown in Fig.
(a). It consists of a open-loop amplifier (the A circuit) and an ideal
voltage-sampling series-mixing feedback network (the β circuit). The A
circuit has an input resistance Ri a voltage gain A, and an output
resistance Ro.
Fig. The series-shunt feedback amplifier:
(a) ideal structure; (b) equivalent circuit.
This circuit of exactly follows the ideal feedback model of and
therefore the closed-loop voltage gain Af is given by
Af ≡
Vo
A
=
Vs 1 + AB
Note that A and B have reciprocal units. This in fact is always the case,
resulting in a dimensionless loop gain AB.
Input resistance
The equivalent circuit model of the series-shunt feedback amplifier is
shown in Fig. (b). Here Rif and Rof denote the input and output
resistances with feedback. The relationship between Rif and Ri, can be
established by considering the circuit in Fig. (a):
Rif ≡
Vs
Vs
Vs
Vi + BAVi
=
= Ri
= Ri
= Ri ⋅ (1 + AB )
Ii Vi / Ri
Vi
Vi
That is, in this case the negative feedback increases the input resistance
by a factor equal to the amount of feedback. The relationship between
Rif and Ri, is a function only of the method of mixing and does not
depend on the method of sampling.
This result is physically intuitive: Since the feedback voltage Vf
subtracts from Vs, the voltage that appears across the input becomes quite
small. Thus the input current becomes correspondingly small and the
resistance becomes large.
This relationship can be generalized to the form
Z if ( s ) ≡ Z i ( s ) ⋅ (1 + A( s ) B( s ) )
Output resistance
To find the output resistance, Rof, of the feedback amplifier in Fig. (a)
we reduce Vs to zero and apply a test voltage Vt, at the output, as shown
in Fig.
Fig. Measuring the output resistance of the feedback amplifier of Fig. (a):
Rof = Vt/I.
Rof ≡
Since Vs=0 , therefore
Thus
Vt
I
and I =
Vt − AVi
Ro
Vi = −Vf = − B ⋅ Vo = − B ⋅ Vt
I=
Vt + A ⋅ B ⋅ Vt
Ro
And we can write
Rof =
Ro
1 + AB
That is, the negative feedback in this case reduces the output resistance
by a factor equal to the amount of feedback. This relationship does not
depend on the method of mixing and depends only on the method of
sampling. Finally, we note that this can be generalized to
Z of ( s ) ≡ Z o ( s ) / (1 + A( s ) B ( s ) )