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This is only a preview of the July 1988 issue of Silicon Chip. You can view 42 of the 96 pages in the full issue and the advertisments. For full access, purchase the issue for $10.00 or subscribe for access to the latest issues. Articles in this series:
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Pl.3: Negative feedback and frequency response
WHAT IS NEG~I'IVE
This month we look into why open loop
amplifiers suffer poor frequency response
and we investigate how negative
feedback increases the bandwidth.
By BRYAN MAHER
Analog amplifiers form such an
important part of electronic circuitry, that we find them in almost
every piece of equipment. Though
many signals may be digitized these
days, you will always find linear
amplification being performed. For
example, the ultimate hifi
capabilities of CD players can only
be .realised by using high quality
amplifiers in the final stages.
Though feedback concepts can
be applied to almost anything electrical, mechanical, hydraulic or
others  our interest here is in
electronic systems only. So let's
start at the beginning, with an
(allegedly) linear electronic
amplifier without any feedback at
all.
Frequency dependence
Last month we made the bald
+15V
statement that the open loop gain of
an amplifier is different at differing
frequencies. Why is that so?
All open loop amplifiers have
reduced gain at the high end of the
frequency range and possibly at the
low end too. First, let's look at the
reasons for such loss of gain, then
we can investigate how negative
feedback improves the situation.
Fig.1 shows a simple "linear"
amplifier stage. We call it "linear"
simply to distinguish it from
deliberately nonlinear digital or
logic circuits.
By "linear circuits" we mean
those with an output voltage which
is supposed to be an enlarged but
otherwise identical copy .of the input voltage waveform.
Strictly we ought to say " roughly
linear" because we know that all
simple amplifiers have output
voltages which are a distorted
larger version of their input
voltages.
Last month, we told the story of
an enthusiastic young lady called
Krystie who had built a simple transistor amplifier to play her
favourite music. But the results
were disappointing. The amplifier
not only exhibited excessive distortion but the frequency range was
also much less than expected. In
fact, the high and low frequency
notes were so weak they could
barely be heard.
At this point we can find out how
negative feedback can be an almost
magical cure for all amplifier ills.
Looking again at Fig.1, let us consider in detail what should be happening at each point in the circuit
and what is actually happening.
First, the signal source (CD
player, tape deck or whatever)
generates some signal V(in), which
we apply to the input, and we expect that same signal waveform
V(in) to appear at the base of the
transistor. And it does too  more
.or less.
But why isn't the signal on the
base identical to the input? Well,
we observe that the signal V(in) has
POSITIVE RAIL
hreor==.:::r
1
hreo3dBr~
.Flg.1: this simple commonemitter amplifier
bas no negative feedback .and suffers
from harmonic distortion and deficient
low and ·high (requency response.
70
SILICON CHIP
h r e = 1             +      
l<
LOG FREQUENCY
Fig.2: all transistors suffer from loss of gain at high frequencies.
At a really high frequency, FT, the gain drops to unity.
FEEDBACK?
to pass through Cl which has an appreciable impedance, so there is inevitably some signal loss in Cl. The
effect is worse at low frequencies.
But by making Cl sufficiently large,
up to 1.0µF, very little low frequency response is lost.
DC coupling?
Why not just DCcouple the whole
thing? DC coupling means using no
coupling capacitors at all, and indeed that is the ideal. Circuit gain
remains undiminished no matter
how low the frequency, right down
to OHz (ie, DC).
But there are drawbacks to DCcoupling right from the input terminals, so common practice is to
employ only one coupling capacitor,
Cl right at the input. Thereafter,
most feedback amplifiers are DCcoupled throughout.
Now what about high notes? Why
is the simple amplifier also deficient in the top frequencies? There
are two main causes and again
Fig.1 gives us some answers.
Loss of hre
Last month we made some comments about the way the hre or current gain of any transistor chang~s
at different collector currents and
at different temperatures. But current gain is also different at various
frequencies.
At a constant collector current
and room temperature, the hfe
holds constant right down to DC. (It
may even be a little higher at DC for
the secondary reason that large
steady DC currents tend to heat the
transistor junction more than AC
currents of the same peak value).
However, at high frequencies the
value of hre decreases as shown in
Fig.2, and at some very high frequency called fT the value of hre
has dropped to 1.0. On the curve
Fig.2 we define another point fhfe,
the frequency at which the hre has
fallen to 0. 707 of the DC value.
If the frequencies of interest extend up to fhfe or beyond, the current gain of the transistors will
reduce, hence the circuit gain must
be less than it is at low frequencies.
In practice, we never use transistors anywhere near their upper
frequency fT; that frequency is
quoted in transistor data sheets only to enable us to draw the curve
Fig.2.
True, we could go out and buy
different transistors having higher
values of fhfe and fT, and that is a
good idea for the small "front end"
transistors in an amplifier. But high
frequency large power transistors
cost a fortune, so we have to make
do with lower frequency types in
power transistors.
Shunt capacitances
Though Fig.1 is all you see if you
look at the physical circuit of such
an amplifier first stage, there are
many "stray capacitances" all
around the circuit, which we show
dotted. These are due to the natural
effects of capacitance which exists
between all separated conductors
and semiconductors.
As well as the stray capacitance
Cs from wiring and components to
ground, there is the baseemitter
capacitance Cbe in the transistor
itself. Also there is a capacitance
Cm from the transistor's collector
back to the base. This Cm is called
the "mutual" or the "Miller"
capacitance.
Cm is the most important of the
stray capacitances. It is due to the.
collectorbase capacitance Ccb, but
as the voltage across it is the input
voltage plus the stage output
voltage, the capacitive current
flowing is equivalent to it being
much larger capacitance Cm,
where Cm = Ccb(l + A), where A is
the stage gain. The total input
capacitance Cin is the sum: Cin =
Cm + Cs + Che·
The source resistance (in parallel
with Rl and R2) forms a lowpass
filter with Cin, reducing the
amplitude of high frequency
signals. This happens in every gain
stage of an amplifier.
But there is more. At the output
side of the transistor stage we have
a number of parallel paths:
(1). The equivalent collector output
resistance.
(2). The collectoremitter capacitance.
(3). The collector load resistor.
(4). The input resistance and input
capacitance of the following stage.
(5). The wiring stray capacitance.
Those capacitances tend to
reduce the high frequency response
of the stage, but the lower the
parallel resultant of the resistances
mentioned the less this reduction.
Therefore, while higher collector
a
r+15V
y.
STAGE 2
STAGE 3
STAGE 4
a+ OUTPUT
RA
FB
Fig.3: this fourstage amplifier, with all stages identical to Ql, has
negative feedback applied via RA from the output back to the emitter
of the first stage.
JULY 1988
71
       
T = 100        i
ERROR SIGNAL E
Vln
1DmV
X
DIFFERENCING ACTION
BETWEEN BASE ANO
EMITTER VOLTAGES
K ,___o_F _sT_AG_
E 1_ _
r
•G = 10000•1
\E ...,
1
__,____ _ _ STAGE 1 
GAIN OF
STAGE 1
FB

y
STAGES 2,3,4 i+ouwT
r
:7
I
I
I
Kl
I
I
: RB
tH
R
Some names
= 1/101
L _.,._I
Fig.4: this is a block diagram of the fourstage amplifier shown in Fig.3.
RA and Re set the proportion of the output voltage fed back to
input K.
load resistors may result in higher
lowfrequency gain, they also cause
the gain to drop off badly at higher
frequencies.
As most amplifiers have two or
more stages, the reduction of gain
at higher frequencies will occur in
every gain stage. The Miller
capacitance will have its greatest
effect in the front end highgain
stage (because stage gain A is high),
while the frequency dependence of
hre will be responsible for the poor
high frequency response in the final
highpower stage (because fT is
low).
Enter negative feedback
Fig.3 shows the outline of a fourstage amplifier where Ql and its
sundry components form stage 1.
Each stage is inverting, meaning
that positivegoing inputs produce
negativegoing outputs, but four
stages of phaseinversion result in
overall noninversion; ie positivegoing signals at X result in a
positivegoing output at Y.
In previous episodes we showed
some negative feedback block
diagrams. Now in Fig.3 we show
one possible way to actually apply
the feedback voltage from the output back to the input stage. Input
signals from the voltage source applied at X naturally increase the
output voltage.
But you will recall that the
negative feedback must be applied
In applying negative feedback to
our whole amplifier we have reduced its voltage gain. But as we will
see, we have improved the
amplifier's characteristics in about
the same proportion as the reduction in gain.
to the front stage in such a way that
the feedback voltage reduces the
output.
As you probably know, a positivegoing signal at X causes an increase in Ql 's collector current,
which in turn produces greater
voltage drop across R3, hence
greater stage output at the collector. In Fig.3, resistors RA and RB
form a voltage divider across the
stage 4 output, from Y to ground.
Do not be confused by the
presence of R4, as it is bypassed by
C2. The impedance of C2 (at all
audio frequencies) is much lower
than the resistance of R4, hence
there is little or no signal volt~ge
drop across R4. As far as AC signal
voltages are concerned, point W
(junction of RB and R4) is at ground
or zero potential).
Hence at the emitter K of transistor Ql we have applied some
feedback called "FB", a signal
which is some fraction of the output
voltage. Let us call that fraction
"H". So the fraction H will be given
approximately by:
H = RB + (RA + RB)
K is the point where we have applied feedback signals to the emitter of Ql, and you will observe that
we now have a "closed loop
system" from Ql collector C,
through stages 2, 3 and 4, to Y,
through RA, back to K, through
transistor Ql to C.
(1). The gain of the whole amplifier
(measured from the input terminal
X to the output terminal Y) before
any feedback is applied is called
the "Open Loop Gain". This is given
by the symbol "G".
(2). The gain of the whole amplifier
(measured from X to Y) with feedback applied is called the "Closed
Loop Gain". This has the symbol
''T''.
(3). Because of negative feedback, T
is always smaller than G.
(4). The fraction of output used as
feedback is called "H" In Fig.3, H
= RB + (RA +, RB).
(5). The feedback signal derived
from the feedback voltage divider H
is (naturally enough) called "FB".
FB is simply equal to H x Output.
(6). The gain around the loop (from
C through stages 2, 3, & 4 to Y,
through the voltage divider to K,
and through the transistor gain
back to C) is called the "Loop
Gain".
As the Loop Gain is clearly the
product of G and H, we simply refer
to the Loop Gain as "GH".
(7). The input signal at X is to be
amplified. So we would like the output signal at Y to be an enlarged
replica of the signal at X.
(8). As the feedback signal at K is a
smaller copy of the output signal at
Y, by (6) above we would want the
signal at K to be exactly like (but a
little smaller than) the input signal
at X. If it is, we have succeeded. If
not, then we ask the circuit to take
corrective action.
(9). The difference between the
signals at X and at K is called the
"Error Signal". This is given the
symbol "E".
(10). It is the error signal E which is
amplified by the amplifier.
(11 ). As the error signal is small, we
The difference between the signals at X and K is the error signal
and it is this signal that is amplified by the amplifier.
72
SILICON CHIP
ET,
T = 99.99
Vln
10mV
T = 99       ,
6 = 10000 OU11'11T
E = Yin  FB
999.9mV
E = 0.09999mV _ _ __,
K
rRA 1
I
FB
FB
= Y/101
= 9.90000991nV
10k
E
K
I
I
I
1
r,.H = 1/101
6 = 5000
'
E = 10mV  9.80199mV
E = 0.198mV
r;
10k
I
1
I
I
11~!0
L_..:...J
will need considerable gain in the
amplifier.
Fig.4 is a block diagram of the
circuit of Fig.3. In Fig.4 we have
split the action of the first stage Ql
into its two functions:
(a) The differencing action between
the input signal at X (the base) and
the feedback signal FB at K (the
emitter); and
(b) The action of transistor Ql
which (as in all transistors)
amplifies the difference between
the signals at base and emitter.
Also we have labelled Fig.4 to
point up three facts:
(1). It is the error voltage E (not the
input voltage at X) which is
multiplied by the amplifier gain G.
(2). H is simply a fraction, set by the
voltage divider ratio.
(3). FB is a signal which is some set
fraction of the output signal.
So just how does all that jazz
cure amplifier ills?
One thing at a time, please. Consider first the poor response to high
frequencies we noted earlier.
If, for any reason, the output is
not be as high as it should be, then
the feedback will be smaller. This
means that less will be subtracted
from the input signal in forming the
error signal E, so E will be a bit
larger. This will increase the output
to (nearly) the right signal level.
Keeping the same amplifier circuit, we redraw Fig.3 and Fig.4,
grouping parts of similar function,
giving us the simpler Fig.5, where
we have written a possible set of
conditions; ie resistances of RA and
RB, together with gains, and
voltages which might be measured
990mV
FB
I
FB = 990mV/101 = 9.801999mV I
I
·Fig.5: the signal E, which is the difference signal
applied between X and K, is multiplied by 10,000 to
give the output signal. Note that Vin is relatively
large but signal E is extremely small.
____
YOUTPUT
I
I
I
I 188
RB
Yin
10mv"'
I
I
1H = 11101
L_...:_I
Fig.6: even though the openloop gain of the amplifier
has dropped to 5000, the negative feedback ensures
that the output signal is still very close to 1 volt by
making the difference signal E larger.
in a typical fourstage amplifier at
low and middle frequencies.
Low and middle frequencies
We have chosen the amplifier
open loop gain G = 10,000, as four
stages each having various gains
could easily multiply up to that
figure. We choose the following
values as typical for such an
amplifier: input voltage V(in) =
lOmV; RA = 10k0; .RB = 1000; G
= 10,000 (at low/middle freqencies).
From these values we can
calculate that:
H = RB + (RA + RB)
= 100 + (10,000 + 100)
= 100 + 10,100
= 0.00990099.
Alternatively, 1/H = 101.
By this means we can calculate
signal voltage values all around the
circuit. Without all the calculation
details, we have written the results
on Fig.5, so if you don't want to
bother with calculations, just look
at the diagram. Fig.5 shows that at
low and middle frequencies, the
gain stages of the amplifier give an
open loop gain G = 10,000.
(a) An input of lOmV gives an output of 999.9mV, so the closed loop
gain T (measured from input X to
output Y) is T = 100 (approx).
(b) The feedback network divides
the output by 101 to give a feedback
voltage FB = 9.9mV (approx),
which is subtracted from V(in) to
give an error signal E = 0.099mV
(approx). This is multiplied by
10,000 to give an output V(out) =
999.9mV.
At high frequencies we must ex
pect the gain G of the amplifier to
be less than 10,000 (because of the
shunting effect of stray capacitances, the Miller capacitance and
the falloff in hre at high frequencies as discussed earlier). At some
high frequency, the open loop gain
G could be down to half; ie, G =
5000. For this condition, as illustrated by Fig.6:
(c) The same input V(in) = lOmV
gives an output of 990mV, so the
closed loop gain (measured from X
to Y) is approximately 99.
(d) The feedback divider fraction is
not subject to frequency, so H still
divides the new output by 101, giving a feedback voltage FB = 9.8mV
(approx), which is subtracted from
V(in) to give an error signal E =
0.198mV (approx), which multiplied
by the reduced value G = 5000
gives the new output V(out) =
990mV.
(e) We observe (with joy) that even
though the gain stages only had half
gain (G = 5000 at high frequencies), we still found an overall gain
only 1 % down (T = 99) at high frequencies; ie, T = 990mV + 10mV
= 99.
Conclusion
From all the above we conclude
that negative feedback automatically compensates for at least one
amplifier shortcoming; the fact that
the open loop gain falls at high frequencies. What about those other
deficiencies, like distortion, hum
and noise?.
Can it be that negative feedback
will cure those ills too? That will be
our topic for next month's episode.~
]ULY 1988
73
