BJT and JFET FrEquEncy instrumentation

rEsponsE
A number of modern VOMs and DMMs have a dB scale designed to provide an indication
of power ratios referenced to a standard level of 1 mW at 600 V. Since the reading is accu-
rate only if the load has a characteristic impedance of 600 V, the 1 mW, 600 V reference
level is normally printed somewhere on the face of the meter, as shown in Fig. 7. The dB
scale is usually calibrated to the lowest ac scale of the meter. In other words, when making
the dB measurement, choose the lowest ac voltage scale, but read the dB scale. If a higher
voltage scale is chosen, a correction factor must be used, which is sometimes printed on
the face of the meter but is always available in the meter manual. If the impedance is other
than 600 V or not purely resistive, other correction factors must be used that are normally
included in the meter manual. Using the basic power equation P 5 V2>R reveals that 1 mW
across a 600 V load is the same as applying 0.775 V rms across a 600 V load; that is,
V = 2PR = 2(1 mW)(600 V) = 0.775 V. The result is that an analog display will
have 0 dB [defining the reference point of l mW, dB 5 10 log10 P2>P1 5 10 log10 (1mW/1
mW(ref) 5 0 dB] and 0.775 V rms on the same pointer projection, as shown in Fig. 7. A
voltage of 2.5 V across a 600 V load results in a dB level of dB 5 20 log10 V2>V1 5
20 log10 25 V>0.775 5 10.17 dB, resulting in 2.5 V and 10.17 dB appearing along the
same pointer projection. A voltage of less than 0.775 V, such as 0.5 V, results in a dB level of
dB 5 20 log10 V2>V1 5 20 log10 0.5 V>0.775 V 5 23.8 dB, also shown on the scale in
Fig. 7. Although a reading of 10 dB reveals that the power level is 10 times the reference,
don’t assume that a reading of 5 dB means that the output level is 5 mW. The 10 : 1 ratio
is a special one in logarithmic use. For the 5 dB level, the power level must be found using
the antilogarithm (3.126), which reveals that the power level associated with 5 dB is about
3.1 times the reference or 3.1 mW. A conversion table is usually provided in the manual
for such conversions.

1.5
1 2.0

.5 2.5

4 5 6 7
2 3 8
0 1 1 mW, 600  9
2 10
4
3
0

3V
6 AC
8 11
12 B
+D

–D
B

Fig. 7
Defining the relationship between a dB scale referenced to
1 mW, 600 V and a 3 V rms voltage scale.

4 generaL FrequenCy COnsideratiOns

l
The frequency of the applied signal can have a pronounced effect on the response of a
single-stage or multistage network. The analysis thus far has been for the midfrequency
spectrum. At low frequencies, we shall find that the coupling and bypass capacitors can no
longer be replaced by the short-circuit approximation because of the increase in reactance
of these elements. The frequency-dependent parameters of the small-signal equivalent cir-
cuits and the stray capacitive elements associated with the active device and the network
will limit the high-frequency response of the system. An increase in the number of stages
of a cascaded system will also limit both the high- and low-frequency responses.

Low-Frequency range
To demonstrate how the larger coupling and bypass capacitors of a network will affect the
frequency response of a system, the reactance of a 1-mF (typical value for such applica-
tions) capacitor is tabulated in Table 4 for a wide range of frequencies.

570
 TABLE 4 BJT and JFET FrEquEncy
1 rEsponsE
Variation in XC = with frequency for a 1-mF
2pfC
capacitor

f XC

10 Hz 15.91 kV
100 Hz 1.59 kV Range of possible
s
1 kHz 159 V effect
10 kHz 15.9 V
100 kHz 1.59 V Range of lesser
1 MHz 0.159 V concern
10 MHz 15.9 mV s ( short-circuit
100 MHz 1.59 mV equivalence)

Two regions have been defined in Table 4. For the range of 10 Hz to 10 kHz the mag-
nitude of the reactance is sufficiently large that it may have an impact on the response of
the system. However, for much higher frequencies it appears as though the capacitor is
behaving much like the short-circuit equivalent it is designed to match.
Clearly, therefore,
the larger capacitors of a system will have an important impact on the response of a
system in the low-frequency range and can be ignored for the high-frequency region.

high-Frequency range
For the smaller capacitors that come into play due to the parasitic capacitances of the
device or network, the frequency range of concern will be the higher frequencies. Consider
a 5-pF capacitor, typical of a parasitic capacitance of a transistor or the level of capacitance
introduced simply by the wiring of the network, and the level of reactance that results for
the same frequency range appearing in Table 4. The results appear in Table 5 and clearly
reveal that at low frequencies they have a very large impedance matching the desired
open-circuit equivalence. However, at higher frequencies they are approaching a short-
circuit equivalence that can severely affect the response of a network.

TABLE 5
1
Variation in XC = with frequency for a
2pfC
5 pF capacitor

f XC

10 Hz 3,183 MV Range of lesser

100 Hz 318.3 MV concern
1 kHz 31.83 MV s ( open-circuit
10 kHz 3.183 MV equivalent)
100 kHz 318.3 kV
1 MHz 31.83 kV Range of possible
10 MHz 3.183 kV s effect
100 MHz 318.3 V

Clearly, therefore,
the smaller capacitors of a system will have an important impact on the response of a
system in the high-frequency range and can be ignored for the low-frequency region.

mid-Frequency range
In the mid-frequency range the effect of the capacitive elements is largely ignored and the
amplifier considered ideal and composed simply of resistive elements and controlled
sources.

571
 BJT and JFET FrEquEncy The result is that
rEsponsE the effect of the capacitive elements in an amplifier are ignored for the mid-frequency
range when important quantities such as the gain and impedance levels are determined.

typical Frequency response

The magnitudes of the gain response curves of an RC-coupled, direct-coupled, and transformer-
coupled amplifier system are provided in Fig. 8. Note that the horizontal scale is a logarith-
mic scale to permit a plot extending from the low- to the high-frequency regions. For each
plot, a low-, a high-, and a mid-frequency region has been defined. In addition, the primary
reasons for the drop in gain at low and high frequencies have also been indicated within
the parentheses. For the RC-coupled amplifier, the drop at low frequencies is due to the
increasing reactance of CC, Cs, or CE, whereas its upper frequency limit is determined by
either the parasitic capacitive elements of the network or the frequency dependence of the
gain of the active device. An explanation of the drop in gain for the transformer-coupled
system requires a basic understanding of “transformer action” and the transformer equiva-
lent circuit. For the moment, let us say that it is simply due to the “shorting effect” (across
the input terminals of the transformer) of the magnetizing inductive reactance at low fre-
quencies (XL = 2pfL). The gain must obviously be zero at f 5 0 because at this point
there is no longer a changing flux established through the core to induce a secondary or
output voltage. As indicated in Fig. 8, the high-frequency response is controlled primarily
by the stray capacitance between the turns of the primary and secondary windings. For
the direct-coupled amplifier, there are no coupling or bypass capacitors to cause a drop in
gain at low frequencies. As the figure indicates, it is a flat response to the upper cutoff

|V |
| Av | = | o |
| Vi |
Bandwidth (Parasitic capacitances
of network and active
(CC Cs or CE)
,
devices and frequency
Avmid dependence of the gain of
0.707Avmid the transistor, FET, or tube)

Mid-frequency
Low- High-frequency
frequency
10 fL 100 1000 10,000 100,000 fH 1 MHz 10 MHz f (log scale)

(a)
|V |
| Av | = | o |
| Vi | Bandwidth
Avmid
(Transformer)
0.707Avmid
(Transformer)
Low- Mid-frequency
frequency High-frequency

10 fL 100 1000 10,000 fH 100,000 f (log scale)

(b)
|V |
| Av | = | o |
| Vi | (Parasitic capacitances
Bandwidth of network and active
Avmid devices and frequency
0.707Avmid dependence of the gain of
the transistor, FET, or tube)

10 ( fL ) 100 1000 10,000 fH 100,000 1 MHz f (log scale)

(c)

Fig. 8
Gain versus frequency: (a) RC-coupled amplifiers; (b) transformer-coupled amplifiers; (c) direct-coupled amplifiers.

572
frequency, which is determined by either the parasitic capacitances of the circuit or the BJT and JFET FrEquEncy
frequency dependence of the gain of the active device. rEsponsE
For each system of Fig. 8, there is a band of frequencies in which the magnitude of the
gain is either equal or relatively close to the midband value. To fix the frequency boundar-
ies of relatively high gain, 0.707Avmid was chosen to be the gain at the cutoff levels. The
corresponding frequencies f1 and f2 are generally called the corner, cutoff, band, break, or
half-power frequencies. The multiplier 0.707 was chosen because at this level the output
power is half the midband power output, that is, at midfrequencies,
0 V2o 0 0 AvmidVi 0 2
Pomid = =
Ro Ro
and at the half-power frequencies,
0 0.707 AvmidVi 0 2 0 AvmidVi 0 2
PoHPF = = 0.5
Ro Ro

and PoHPF = 0.5 Pomid (18)

The bandwidth (or passband) of each system is determined by fH and fL, that is,

bandwidth (BW) = fH - fL (19)

with fH and fL defined in each curve of Fig. 8.

5 nOrmaLizatiOn prOCess
l
For applications of a communication nature (audio, video) a decibel plot versus frequency
is normally provided rather than the gain versus frequency plot of Fig. 8. In other words,
when you pick up a specification sheet on a particular amplifier or system, the plot will
typically be of dB versus frequency rather than gain versus frequency.
To obtain such a dB plot the curve is first normalized—a process whereby the vertical
parameter is divided by a specific level or quantity sensitive to a combination or variables
of the system. For this area of investigation, it is usually the midband or maximum gain for
the frequency range of interest.
For example, in Fig. 9 the curve of Fig. 8a is normalized by dividing the output voltage
gain at each frequency by the midband level. Note that the curve has the same shape, but
the band frequencies are now defined by simply the 0.707 level and not linked to the actual
midband level. It clearly reveals that
The band frequencies define a level where the gain or quantity of interest will be 70.7%
or its maximum value.

Av
Av
mid

1
0.707

10 fL 100 1000 10,000 100,000 fH 1 MHz 10 MHz f (log scale)

Fig. 9
Normalized gain versus frequency plot.

Consider also that the plot of Fig. 9 is not sensitive to the actual level of the midband
gain. The midband gain could be 50, 100, or even 200, and the resulting plot of Fig. 9 would
be the same. The plot of Fig. 9 is now defining frequencies where the relative gain is defined
rather than concerning itself with the “actual gain.”
The next example will demonstrate the normalization process for a typical amplifier
response.

573
 BJT and JFET FrEquEncy
rEsponsE eXampLe 9 Given the frequency response of Fig. 10:
a. Find the cutoff frequencies fL and fH using the measurements provided.
b. Find the bandwidth of the response.
c. Sketch the normalized response.
Av

128
90.5

0
100 fL 1000 10,000 fH 100,000 1 MHz f (log scale)
1/ " 7/16"
4
1" 1"
d1 d2

Fig. 10
Gain plot for Example 8.

Solution:
d1 1>4
a. For fL: = = 0.25
d2 1
10d1>d2 = 100.25 = 1.7783
Value = 10 x * 10 d1>d2 = 102 * 1.7783 = 177.83 Hz
d1 7>16
For fH: = = 0.438
d2 1
10d1>d2 = 100.438 = 2.7416
Value = 10 x * 10 d1>d2 = 104 * 2.7416 = 27,416 Hz
b. The bandwidth:
BW = fH - fL = 27,416 Hz - 177.83 Hz  27.24 KHz
c. The normalized response is determined by simply dividing each level of Fig. 10 by the
midband level of 128, as shown in Fig. 11. The result is a maximum value of 1 and
cutoff levels of 0.707.
Av
Avmid

128
=1
128
90.5
= 0.707
128

0
100 fL 1000 10,000 fH 100,000 1 MHz f (log scale)
= 177.83 Hz = 27,416 Hz

Fig. 11
Narmalized plot of Fig. 10.

A decibel plot of Fig. 11 can be obtained by applying Eq. (13) in the following manner:

Av Av
` = 20 log10 (20)
Avmid dB Avmid

574
 At midband frequencies, 20 log10 1 = 0, and at the cutoff frequencies, 20 log10 BJT and JFET FrEquEncy
1>12 = -3 dB. Both values are clearly indicated in the resulting decibel plot of Fig. 12. rEsponsE
The smaller the fraction ratio, the more negative is the decibel level.

Av
Av
mid (dB)
10 100 1000 10,000 100,000 1 MHz 10 MHz
fL fH f (log scale)
0 dB
− 3 dB
− 6 dB
− 9 dB
− 12 dB

Fig. 12
Decibel plot of the normalized gain versus frequency plot of Fig. 9.

For the greater part of the discussion to follow, a decibel plot will be made only for the
low- and high-frequency regions. Keep Fig. 12 in mind, therefore, to permit a visualization
of the broad system response.
Most amplifiers introduce a 180° phase shift between input and output signals. This
fact must now be expanded to indicate that this is the case only in the midband region. At
low frequencies, there is a phase shift such that Vo lags Vi by an increased angle. At high
frequencies, the phase shift drops below 180°. Figure 13 is a standard phase plot for an
RC-coupled amplifier.

< (Vo leads Vi)

10 fL 100 1000 10,000 100,000 fH 1 MHz 10 MHz f (log scale)

Fig. 13
Phase plot for an RC-coupled amplifier system.

6 LOw-FrequenCy anaLysis—bOde pLOt

l
In the low-frequency region of the single-stage BJT or FET amplifier, it is the RC combi-
nations formed by the network capacitors CC, CE, and Cs and the network resistive param-
eters that determine the cutoff frequencies. In fact, an RC network similar to Fig. 14 can be
established for each capacitive element, and the frequency at which the output voltage
drops to 0.707 of its maximum value can be determined. Once the cutoff frequencies due to
Fig. 14
each capacitor are determined, they can be compared to establish which will determine the RC combination that
low-cutoff frequency for the system. will define a low-cutoff
Consider, for example, the voltage-divider BJT network of Fig. 15 that was analyzed in frequency.
detail in the Section 6 of the chapter “BJT AC Analysis”. The analysis resulted in an input
impedance of
Zi = Ri = R1  R2  bre
and an equivalent circuit at the input as shown in Fig. 16.
For the mid-frequency range the capacitor Cs is assumed to be an equivalent short-circuit
state, and Vb = Vi. The result is a high midband gain for the amplifier that is not affected
by the coupling or bypass capacitors. However, as we lower the applied frequency the re-
actance of the capacitor will increase and take an increasing share of the applied voltage Vi.
Neglecting the effects of the coupling capacitor CC and bypass capacitor CE for the moment,
if the voltage Vb should decrease, it will result in the same decrease in overall gain Vo>Vi.

575
 BJT and JFET FrEquEncy VCC
rEsponsE
Io

RC
R1
C Vo
Ii CC
B
Vi
Cs Zo
network
Zi
E + Cs
+
R2
RE CE Vi Vb Ri

– –

Fig. 15 Fig. 16
Voltage-divider bias configuration. Equivalent input circuit for the
network of Fig. 15.

Less of the applied voltage is reaching the base of the transistor reducing the output voltage
+ +
Vo. In fact if the Vb should drop to 0.707 of the peak possible value of Vi the overall gain
Vi R Vo will drop the same amount. In total, therefore, if we find the frequency that will result in Vb
being only 0.707 Vi, we will have the low-cutoff frequency for the full amplifier response.
– – Finding this frequency will now be examined by analyzing the generic RC network
of Fig. 14 introduced above. Once the results are obtained it can be applied to any RC
Fig. 17 combination that may develop due to the other coupling capacitors or bypass capacitors.
RC circuit of Fig. 14 at At high frequencies, the reactance of the capacitor of Fig. 14 is
very high frequencies.
1
 0V
XC =
2pfC
and the short-circuit equivalent can be substituted for the capacitor as shown in Fig. 17.
The result is that Vo  Vi at high frequencies. At f = 0 Hz,
1 1
XC = = = V
2pfC 2p(0)C
and the open-circuit approximation can be applied as shown in Fig. 18, with the result that
Vo = 0 V.
Fig. 18 Between the two extremes, the ratio Av = Vo >Vi will vary as shown in Fig. 19 As the
RC circuit of Fig. 14 at frequency increases, the capacitive reactance decreases, and more of the input voltage ap-
f 5 0 Hz. pears across the output terminals.

A v = Vo / Vi
1

0.707

0
fL f

Fig. 19
Low-frequency response for the RC circuit of Fig. 14.

The output and input voltages are related by the voltage-divider rule in the following
manner:
RVi
Vo 
R  XC

576
where the boldface roman characters represent magnitude and angle of each quantity. BJT and JFET FrEquEncy
The magnitude of Vo is determined as follows: rEsponsE

RVi
Vo =
2R2 + X2C
For the special case where XC 5 R,
RVi RVi RVi RVi 1
Vo = = = = = Vi
2R +2
X2C 2R + R2 2
22R 2 12R 12

Vo 1
and 0 Av 0 = = = 0.707 0 XC = R (21)
Vi 12

the level of which is indicated on Fig. 19. In other words, at the frequency for which XC 5
R, the output will be 70.7% of the input for the network of Fig. 14.
The frequency at which this occurs is determined from
1
XC = = R
2pfLC

1
and fL = (22)
2pRC

In terms of logs,
1
Gv = 20 log10 Av = 20 log10 = -3 dB
12
whereas at Av = Vo >Vi = 1 or Vo = Vi (the maximum value),
Gv = 20 log10 1 = 20(0) = 0 dB
In Fig. 8, we recognize that there is a 3-dB drop in gain from the midband level when
f = fL. In a moment, we will find that an RC network will determine the low-frequency
cutoff for a BJT transistor and fL will be determined by Eq. (22).
If the gain equation is written as
Vo R 1 1 1
Av = = = = =
Vi R - jXC 1 - j(XC>R) 1 - j(1>vCR) 1 - j(1>2pfCR)
we obtain, using the frequency defined above,

1
Av = (23)
1 - j( fL >f )

In the magnitude and phase form,

Vo 1
Av = = ltan-1( fL >f ) (24)
Vi 21 + ( fL >f )2
5 3
magnitude of Av phaseby which
Vo leads Vi

For the magnitude when f = fL,

1 1
0 Av 0 = = = 0.707 1 -3 dB
21 + (1) 2 12
In the logarithmic form, the gain in dB is

1
Av(dB) = 20 log10 (25)
21 + ( fL >f )2

577
 BJT and JFET FrEquEncy Expanding Eq. (25):
rEsponsE
fL 2 1>2
Av(dB) = -20 log10 c 1 + a b d
f
fL 2
= - 1 12 2 (20) log10 c 1 + a b d
f
fL 2
= -10 log10 c 1 + a b d
f
2
For frequencies where f V fL or ( fL >f ) W 1, the equation above can be approximated by
fL 2
Av(dB) = -10 log10 a b
f
and finally,

fL
Av(dB) = -20 log10 (26)
f
f V fL

Ignoring the condition f V fL for a moment, we find that a plot of Eq. (26) on a fre-
quency log scale yields a result very useful for future decibel plots.
fL
At f = fL: = 1 and -20 log10 1 = 0 dB
f
fL
At f = 12 fL: = 2 and -20 log10 2  -6 dB
f
fL
At f = 14 fL: = 4 and -20 log10 4  -12 dB
f
1 fL
At f = 10 fL: = 10 and -20 log10 10 = -20 dB
f
A plot of these points is indicated in Fig. 20 from 0.1 fL to fL with a dark blue straight
line. In the same figure, a straight line is also drawn for the condition of 0 dB for f W fL.
As stated earlier, the straight-line segments (asymptotes) are only accurate for 0 dB when
f W fL and the sloped line when fL W f. We know, however, that when f = fL, there is
a 3-dB drop from the midband level. Employing this information in association with the
straight-line segments permits a fairly accurate plot of the frequency response as indicated
in the same figure.
The piecewise linear plot of the asymptotes and associated breakpoints is called a Bode
plot of the magnitude versus frequency.

Av(dB)

fL /10 fL /4 fL /2 fL 2fL 3fL 5fL 10fL

fL /f

Fig. 20
Bode plot for the low-frequency region.

578
 The approach was developed by Professor Hendrik Bode in the 1940s (Fig. 21). BJT and JFET FrEquEncy
The calculations above and the curve itself demonstrate clearly that: rEsponsE

A change in frequency by a factor of two, equivalent to one octave, results in a 6-dB

change in the ratio, as shown by the change in gain from fL>2 to fL.
As noted by the change in gain from fL >2 to fL:
For a 10:1 change in frequency, equivalent to one decade, there is a 20-dB change in
the ratio, as demonstrated between the frequencies of fL>10 and fL.
Therefore, a decibel plot can easily be obtained for a function having the format of Eq.
(26). First, simply find fL from the circuit parameters and then sketch two asymptotes—one
along the 0-dB line and the other drawn through fL sloped at 6 dB/octave or 20 dB/decade.
Then, find the 3-dB point corresponding to fL and sketch the curve.
The gain at any frequency can be determined from the frequency plot in the following American (Madison, WI; Summit,
manner: NJ; Cambridge, MA)
(1905–1981)
Vo V.P. at Bell Laboratories
Av(dB) = 20 log10
Vi Professor of Systems Engineering,
Av(dB) Vo Harvard University
but = log10
20 Vi In his early years at Bell Laborato-
ries, Hendrik Bode was involved with
Vo electric filter and equalizer design.
and Av = = 10Av(dB)>20 (27)
Vi He then transferred to the Mathemat-
ics Research Group, where he spe-
For example, if Av(dB) = -3 dB, cialized in research pertaining to
electrical networks theory and its
Vo
Av = = 10(-3>20) = 10(-0.15)  0.707 as expected application to long distance commu-
Vi nication facilities. In 1948 he was
The quantity 10-0.15 is determined using the 10 x function found on most scientific calculators. awarded the Presidential Certificate
The phase angle of u is determined from of Merit for his work in electronic
fire control devices. In addition to
the publication of the book Network
fL
u = tan-1 (28) Analysis and Feedback Amplifier
f Design in 1945, which is considered
a classic in its field, he has been
from Eq. (24). granted 25 patents in electrical engi-
For frequencies f V fL, neering and systems design. Upon
retirement, Bode was elected Gordon
fL
u = tan-1 S 90 McKay Professor of Systems Engi-
f neering at Harvard University. He
For instance, if fL = 100f, was a fellow of the IEEE and Ameri-
can Academy of Arts and Sciences.
fL
u = tan-1 = tan-1(100) = 89.4 Fig. 21
f
Hendrik Wade Bode.
For f = fL, (Courtesy of AT&T Archives and
History Center.)
fL
u = tan-1 = tan-11 = 45
f
For f W fL,
fL
u = tan-1 S 0
f
For instance, if f = 100fL,
fL
u = tan-1 = tan-1 0.01 = 0.573
f
A plot of u = tan-1( fL >f ) is provided in Fig. 22. If we add the additional 180° phase
shift introduced by an amplifier, the phase plot of Fig. 13 is obtained. The magnitude and
phase response for an RC combination have now been established. In Section 7, each ca-
pacitor of importance in the low-frequency region will be redrawn in an RC format and the
cutoff frequency for each determined to establish the low-frequency response for the BJT
amplifier.

579
 BJT and JFET FrEquEncy
rEsponsE

0.1fL 0.2fL 0.3fL 0.5fL fL 2fL 3fL 5fL 10fL

Fig. 22
Phase response for the RC circuit of Fig. 14.

eXampLe 10 For the network of Fig. 23:

a. Determine the break frequency.
b. Sketch the asymptotes and locate the 23-dB point.
c. Sketch the frequency response curve.
d. Find the gain at Av(dB) 5 26 dB.

Solution:
Fig. 23
1 1
Example 10. a. fL = =
2pRC (6.28)(5 * 10 V)(0.1 * 10-6 F)
3

 318.5 Hz
b. and c. See Fig. 24.
Vo
d. Eq. (27): Av = = 10 Av(dB)>20
Vi
= 10(-6>20) = 10-0.3 = 0.501
and Vo 5 0.501 Vi or approximately 50% of Vi.

Av(dB)

fL /10 fL /2 fL 2fL 3fL 5fL 10fL

Av

Fig. 24
Frequency response for the RC circuit of Fig. 23.

7 LOw-FrequenCy respOnse—bjt
ampLiFier with RL
l
The analysis of this section will employ the loaded (RL ) voltage-divider BJT bias configu-
ration introduced earlier in Section 6. For the network of Fig. 25, the capacitors Cs, CC, and
CE will determine the low-frequency response. We will now examine the impact of each
independently in the order listed.

580
Cs Because Cs is normally connected between the applied source and the active device, BJT and JFET FrEquEncy
the general form of the RC configuration is established by the network of Fig. 26, match- rEsponsE
ing that of Fig. 16 with Ri = R1  R2  bre.

VCC

RC
R1 CC
Vo
Cs
Vi Vi
+ Cs + System

RL Vb Ri = R1 R2  βre
Vb R2 Ri
RE CE –
Zi
–

Fig. 25 Fig. 26
Loaded BJT amplifier with capacitors that affect the low- Determining the effect of Cs on the low-
frequency response. frequency response.

Applying the voltage-divider rule:

RiVi
Vb = (29)
Ri - jXCs

The cutoff frequency defined by Cs can be determined by manipulating the above equa-
tion into a standard form or simply using the results of Section 6. As a verification of the
results of Section 6 the manipulation process is defined in detail below. For future RC
networks, the results of Section 6 will simply be applied.
Rewriting Eq. (29):
Vb Ri 1
= =
Vi Ri - jXCs Xcs
1 - j
Ri
The factor
Xcs 1 1 1
= a ba b =
Ri 2pfCs Ri 2pfRiCs

Defining 1
fLs = (30)
2pRiCs

we have Vb 1
Av = = (31)
Vi 1 - j( fLs >f )

At fLs the voltage Vb will be 70.7% of the mid band value assuming Cs is the only capacitive
element controlling the low-frequency response.
For the network of Fig. 25, when we analyze the effects of Cs we must make the as-
sumption that CE and CC are performing their designed function or the analysis becomes
too unwieldy, that is, that the magnitudes of the reactances of CE and CC permit employing
a short-circuit equivalent in comparison to the magnitude of the other series impedances.

581
 BJT and JFET FrEquEncy CC Because the coupling capacitor is normally connected between the output of the
rEsponsE active device and the applied load, the RC configuration that determines the low-cutoff
frequency due to CC appears in Fig. 27. The total series resistance is now Ro + RL, and the
cutoff frequency due to CC is determined by

1
fLC = (32)
2p(Ro + RL)CC

Ignoring the effects of Cs and CE, we find that the output voltage Vo will be 70.7% of its
midband value at fLC. For the network of Fig. 25, the ac equivalent network for the output
section with Vi = 0 V appears in Fig. 28. The resulting value for Ro in Eq. (32) is then
simply

Ro = RC  ro (33)

CC +
C
System RL Vo + CC +
Ro
ro Vc RC RL Vo
–
– Ro –
Thévenin

Fig. 27 Fig. 28
Determining the effect of CC on the Localized ac equivalent for CC with
low-frequency response. Vi 5 0 V.

CE To determine fLE, the network “seen” by CE must be determined as shown in Fig. 29.
Once the level of Re is established, the cutoff frequency due to CE can be determined using
the following equation:

1
fLE = (34)
2pReCE

For the network of Fig. 25, the ac equivalent as “seen” by CE appears in Fig. 30 as derived
from Fig. 38. The value of Re is therefore determined by

R1  R2
Re = RE  a + re b (35)
b

The effect of CE on the gain is best described in a quantitative manner by recalling that
the gain for the configuration of Fig. 31 is given by
-RC
Av =
re + RE

RC

Vo

E Vi

System CE R1 R2 Re
Re + re RE CE RE
β

Fig. 29 Fig. 30 Fig. 31

Determining the effect of CE on the Localized ac equivalent of CE. Network employed to describe the
low-frequency response. effect of CE on the amplifier gain.

582
The maximum gain is obviously available where RE is 0 V. At low frequencies, with the BJT and JFET FrEquEncy
bypass capacitor CE in its “open-circuit” equivalent state, all of RE appears in the gain rEsponsE
equation above, resulting in the minimum gain. As the frequency increases, the reactance
of the capacitor CE will decrease, reducing the parallel impedance of RE and CE until the
resistor RE is effectively “shorted out” by CE. The result is a maximum or midband gain
determined by Av = -RC>re. At fLE the gain will be 3 dB below the midband value deter-
mined with RE “shorted out.”
Before continuing, keep in mind that Cs, CC, and CE will affect only the low-frequency
response. At the midband frequency level, the short-circuit equivalents for the capacitors
can be inserted. Although each will affect the gain Av = Vo>Vi in a similar frequency
range, the highest low-frequency cutoff determined by Cs, CC, or CE will have the greatest
impact because it will be the last encountered before the midband level. If the frequencies
are relatively far apart, the highest cutoff frequency will essentially determine the lower
cutoff frequency for the entire system. If there are two or more “high” cutoff frequencies,
the effect will be to raise the lower cutoff frequency and reduce the resulting bandwidth
of the system. In other words, there is an interaction between capacitive elements that can
affect the resulting low-cutoff frequency. However, if the cutoff frequencies established by
each capacitor are sufficiently separated, the effect of one on the other can be ignored with
a high degree of accuracy—a fact that will be demonstrated by the printouts to appear in
the following example.

eXampLe 11 Determine the cutoff frequencies for the network of Fig. 25 using the fol-
lowing parameters:
Cs = 10 mF, CE = 20 mF, CC = 1 mF
R1 = 40 kV, R2 = 10 kV, RE = 2 kV, RC = 4 kV,
RL = 2.2 kV
b = 100, ro = V, VCC = 20 V

Solution: To determine re for dc conditions, we first apply the test equation:

bRE = (100)(2 kV) = 200 kV W 10R2 = 100 kV
Since satisfied the dc base voltage is determined by
R2VCC 10 kV(20 V) 200 V
VB  = = = 4V
R2 + R1 10 kV + 40 kV 50
VE 4 V - 0.7 V 3.3 V
with IE = = = = 1.65 mA
RE 2 kV 2 kV
26 mV
so that re =  15.76 
1.65 mA
and bre = 100(15.76 V) = 1576 V = 1.576 k

Vo -RC  RL (4 kV)  (2.2 kV)

midband gain Av = = = -  -90
Vi re 15.76 V

Cs Ri = R1  R2  bre = 40 kV  10 kV  1.576 kV  1.32 kV

1 1
fLS = =
2pRiCs (6.28)(1.32 kV)(10 mF)
fLS  12.06 Hz

1
CC fLC = with Ro = RC  ro  RC
2p(Ro + RL)CC
1
=
(6.28)(4 kV + 2.2 kV)(1 mF)
 25.68 Hz

583
 BJT and JFET FrEquEncy R1  R2
rEsponsE CE Re = RE  a + re b
b
40 kV  10 kV
= 2 kV  a + 15.76 V b
100
8 kV
= 2 kV  a + 15.76 V b
100
= 2 kV  (80 V + 15.76 V)
= 2 kV  95.76 V
= 91.38 V
1 1 106
fLE = = =  87.13 Hz
2pReCE (6.28)(91.38 V)(20 mF) 11,477.73
Since fLE W fLC or fLS the bypass capacitor CE is determining the lower cutoff fre-
quency of the amplifier.

8 impaCt OF RS On the bjt

LOw-FrequenCy respOnse
l
In this section we will investigate the impact of the source resistance on the various cutoff
frequencies. In Fig. 32 a signal source and associated resistance have been added to the
configuration of Fig. 25. The gain will now be between the output voltage Vo and the sig-
nal source Vs.

Cs The equivalent circuit at the input is now as shown in Fig. 33, with Ri continuing to be
equal to R1  R2  bre.

VCC

RC
R1 CC
Vo
Cs

+ +
Cs
Rs Rs
RL
Vb R2 + Vb System
+ Ri
RE CE Vs
Vs
Zi – –
– –

Fig. 32 Fig. 33
Determining the effect of Rs on the low-frequency response of a Determining the effect of Cs on the low-
BJT amplifier. frequency response.

Using the results of the last section it would appear we could simply find the total sum of
the series resistors and plug it into Eq. (22). Doing so would result in the following equation
for the cutoff frequency:

1
fLs = (36)
2p(Ri + Rs)Cs

584
However, it would be best to validate our assumption by first applying the voltage-divider BJT and JFET FrEquEncy
rule in the following manner: rEsponsE

RiVs
Vb = (37)
Rs + Ri - jXCs

The cutoff frequency defined by Cs can be determined by manipulating the above equa-
tion into a standard form, as demonstrated below.
Rewriting Eq. (37):
Vb Ri 1
= =
Vs Rs + Ri - jXCs Rs Xcs
1 + - j
Ri Ri
1 1
= =
Rs XCs 1 Rs XCs
a1 + b 1 - j a1 + b a1 - j b
Ri £ Ri ° Rs ¢ § Ri Ri + Rs
1 +
Ri
The factor
Xcs 1 1 1
= a ba b =
Ri + Rs 2pfCs Ri + Rs 2pf(Ri + Rs)Cs
1
Defining fLs =
2p(Ri + Rs)Cs
Vb 1
we have =
Vs 1 1
a Rs b a 1 - b
1 + Ri 1 - jfLs>f
Vb Ri 1 Rs
and finally Av = = c dc d
Vs Ri + Rs 1 - j( fLs>f ) +
For the midband frequencies, the input network will appear as shown in Fig. 34. +
Vs Ri Vb
Vb Ri
so that Avmid = = (38) –
Vs Ri + Rs
–
Av 1
and =
Avmid 1 - j( fLs>f )
Fig. 34
Noting the similarities with Eq. (23) the cutoff frequency is defined by fLs above and Determining the effect of Rs on the
gain Avs.
1
fLs = (39)
2p(Rs + Ri)Cs

as assumed in the derivation of Eq. (36).

At fLs, the voltage Vo will be 70.7% of the midband value determined by Eq. (38), as-
suming the Cs is the only capacitive element controlling the low-frequency response.
CC Reviewing the analysis of Section 7 for the coupling capacitor CC, we find that the
derivation of the equation for the cutoff frequency remains the same. That is,

1
fLC = (40)
2p(Ro + RL)CC

CE Again, following the analysis of Section 7 for the same capacitor, we find that Rs will
affect the resistance level substituted into the cutoff equation so that

1
fLE = (41)
2pReCE

585
 BJT and JFET FrEquEncy Rs
rEsponsE with Re = RE  a + re b and Rs = Rs  R1  R2
b
In total, therefore, the introduction of the resistance Rs reduced the cutoff frequency de-
fined by Cs and raised the cutoff frequency defined by CE. The cutoff frequency defined by
CC remained the same. It is also important to note that the gain can be severely affected by
the loss in signal voltage across the source resistance. This last factor will be demonstrated
in the next example.

eXampLe 12
a. Repeat the analysis of Example 11 but with a source resistance Rs of 1 kV. The gain of
interest will now be Vo>Vs rather than Vo>Vi. Compare results.
b. Sketch the frequency response using a Bode plot.
c. Verify the results using PSpice.
Solution: a. The dc conditions remain the same:
re = 15.76 V and bre = 1.576 kV

Vo -RC  RL
midband gain Av = =  -90 as before
Vi re
The input impedance is given by
Zi = Ri = R1  R2  bre
= 40 kV  10 kV  1.576 kV
 1.32 kV
Rs and from Fig. 35,
+ RiVs
1 kΩ Vb =
+ Ri + Rs
Vs Ri 1.32 kΩ Vb Vb Ri 1.32 kV
or = = = 0.569
– Vs Ri + Rs 1.32 kV + 1 kV

– so that Avs =
Vo
= # = (-90)(0.569)
Vo Vb
Vs Vi Vs
= 251.21
Fig. 35
Determining the effect of Rs on
the gain Avs. Cs Ri = R1  R2  bre = 40 kV  10 kV  1.576 kV  1.32 kV
1 1
fLS = =
2p(Rs + Ri)Cs (6.28)(1 kV + 1.32 kV)(10 mF)
fLS  6.86 Hz vs. 12.06 Hz without Rs

1
CC fLC =
2p(RC + RL)CC
1
=
(6.28)(4 kV + 2.2 kV)(1 mF)
 25.68 Hz as before

CE Rs = Rs  R1  R2 = 1 kV  40 kV  10 kV  0.889 kV
Rs 0.889 kV
Re = RE g a + re b = 2 kV g a + 15.76 V b
b 100
= 2 kV  (8.89 V + 15.76 V) = 2 kV  24.65 V  24.35 V
1 1 106
fLE = = =
2pReCE (6.28)(24.35 V)(20 mF) 3058.36
 327 Hz vs. 87.13 Hz without Rs.

586
 The net result is a severe reduction in overall gain (almost 43%) but a correspond- BJT and JFET FrEquEncy
ing reduction in the lower cutoff frequency. Recall that the highest of the low cutoff rEsponsE
frequencies will determine the overall low cutoff frequency for the amplifier. The
results point out that the internal series resistance can have a very strong impact on the
midband gain, but on the other side of the coin it can improve the overall bandwidth. In
this case it is clear that the loss in gain far outweighs any gain in bandwidth.
b. It was mentioned earlier that dB plots are usually normalized by dividing the voltage
gain Av by the magnitude of the midband gain. For Fig. 32, the magnitude of the mid-
band gain is 51.21, and naturally the ratio 0 Av >Avmid 0 will be 1 in the midband region.
The result is a 0-dB asymptote in the midband region as shown in Fig. 36. Defining fLE
as our lower cutoff frequency fL, we can draw an asymptote at 26 dB/octave as shown
in Fig. 36 to form the Bode plot and our envelope for the actual response. At f1, the
actual curve is 23 dB down from the midband level as defined by the 0.707Avmid level,
permitting a sketch of the actual frequency response curve as shown in Fig. 36. A
26-dB/octave asymptote was drawn at each frequency defined in the analysis above to
demonstrate clearly that it is fLE for this network that will determine the 23-dB point. It
is not until about 224 dB that fLC begins to affect the shape of the envelope. The mag-
nitude plot shows that the slope of the resultant asymptote is the sum of the asymptotes
having the same sloping direction in the same frequency interval. Note in Fig. 36 that
the slope has dropped to 212 dB/octave for frequencies less than fLC and could drop to
218 dB/octave if the three defined cutoff frequencies of Fig. 36 were closer together.
Using Eq. (9), the cutoff frequency for the low-frequency region is about 325 Hz.

fL

fL (low-cutoff
frequency)

Fig. 36
Low-frequency plot for the network of Example 12.

c. The PSpice solution can be found in Section 15.

Keep in mind as we proceed to the next section that the analysis of this section is not limited
to the networks of Figs. 25 and 32. For any transistor configuration it is simply necessary to iso-
late each RC combination formed by a capacitive element and determine the break frequencies.
The resulting frequencies will then determine whether there is a strong interaction between ca-
pacitive elements in determining the overall response and which element will have the greatest
effect on establishing the lower cutoff frequency. In fact, the analysis of the next section will
parallel this section as we determine the low-cutoff frequencies for the FET amplifier.

9 LOw-FrequenCy respOnse—Fet ampLiFier

l
The analysis of the FET amplifier in the low-frequency region will be quite similar
to that of the BJT amplifier of Section 7. There are again three capacitors of primary
concern as appearing in the network of Fig. 37: CG, CC, and CS. Although Fig. 37

587
 BJT and JFET FrEquEncy
rEsponsE

Fig. 37
Capacitive elements that affect the low-frequency response of a JFET amplifier.

will be used to establish the fundamental equations, the procedure and conclusions can
be applied to any FET configuration.

CG For the coupling capacitor between the source and the active device, the ac equivalent
network is as shown in Fig. 38. The cutoff frequency determined by CG is

1
fLG = (42)
2p(Rsig + Ri)CG

Vi Vg

Rsig CG
+
Vs System
Ri
–

Fig. 38
Determining the effect of CG on the low-frequency response.

which is an exact match of Eq. (39). For the network of Fig. 37,

Ri = RG (43)

Typically, RG W Rsig, and the lower cutoff frequency is determined primarily by RG and
CG. The fact that RG is so large permits a relatively low level of CG while maintaining a
low cutoff frequency level for fLG.

CC For the coupling capacitor between the active device and the load the network of
CC Fig. 39 results, which is also an exact match of Fig. 27. The resulting cutoff frequency is
System RL
Ro 1
fLC = (44)
2p(Ro + RL)CC

For the network of Fig. 37,

Fig. 39
Determining the effect of CC on the
Ro = RD  rd (45)
low-frequency response.

588
CS For the source capacitor CS, the resistance level of importance is defined by Fig. 40. BJT and JFET FrEquEncy
The cutoff frequency is defined by rEsponsE

1
fLS = (46)
2pReqCS

For Fig. 37, the resulting value of Req is System CS

Reg
RS
Req = (47)
1 + RS (1 + gmrd)>(rd + RD  RL)
Fig. 40
which for rd   V becomes Determining the effect of CS on the
low-frequency response.
1
Req = RS  (48)
gm
rd V

eXampLe 13
a. Determine the lower cutoff frequency for the network of Fig. 37 using the following
parameters:
CG = 0.01 mF, CC = 0.5 mF, CS = 2 mF
Rsig = 10 kV, RG = 1 MV, RD = 4.7 kV, RS = 1 kV, RL = 2.2 kV
IDSS = 8 mA, VP = -4 V, rd =  V, VDD = 20 V
b. Sketch the frequency response using a Bode plot.
c. Verify the results of part (b) using PSpice.
d. Perform a complete analysis of the network of Fig. 37 using Multisim.

Solution:
a. DC analysis: Plotting the transfer curve of ID = IDSS(1 - VGS >VP)2 and superimpos-
ing the curve defined by VGS = -ID RS results in an intersection at VGSQ = -2 V and
IDQ = 2 mA. In addition,
2IDSS 2(8 mA)
gm0 = = = 4 mS
ƒVP ƒ 4V
VGSQ -2 V
gm = gm0 a 1 - b = 4 mS a 1 - b = 2 mS
VP -4 V

1 1
CG Eq. (36): fLG = =  15.8 Hz
2p(Rsig + Ri)CG 2p(10 kV + 1 MV)(0.01 mF)

1 1
CC Eq. (38): fLC = =  46.13 Hz
2p(Ro + RL)CC 2p(4.7 kV + 2.2 kV)(0.5 mF)

1 1
CS Req = RS  = 1 kV  = 1 kV  0.5 kV = 333.33 V
gm 2 mS
1 1
Eq. (40): fLS = = = 238.73 Hz
2pReqCS 2p(333.33 V)(2 mF)
Because fLs is the largest of the three cutoff frequencies, it defines the low-cutoff
frequency for the network of Fig. 37.
b. The midband gain of the system is determined by
Vo
Avmid = = -gm(RD  RL) = -(2 mS)(4.7 kV  2.2 kV)
Vi
= -(2 mS)(1.499 kV)
 23

589
 BJT and JFET FrEquEncy Using the midband gain to normalize the response for the network of Fig. 37 results in the
rEsponsE frequency plot of Fig. 41.

Fig. 41
Low-frequency response for the JFET configuration of Example 13.

c. and d. The computer solutions can be found in Section 15.

10 miLLer eFFeCt CapaCitanCe

l
In the high-frequency region, the capacitive elements of importance are the interelectrode
(between-terminals) capacitances internal to the active device and the wiring capacitance
between leads of the network. The large capacitors of the network that controlled the low-
frequency response are all replaced by their short-circuit equivalent due to their very low
reactance levels.
For inverting amplifiers (phase shift of 180° between input and output, resulting in
a negative value for Av), the input and output capacitance is increased by a capacitance
level sensitive to the interelectrode capacitance between the input and output terminals
of the device and the gain of the amplifier. In Fig. 42, this “feedback” capacitance is
defined by Cf.

Fig. 42
Network employed in the derivation of an equation for the
Miller input capacitance.

590
 Applying Kirchhoff’s current law gives BJT and JFET FrEquEncy
rEsponsE
Ii = I1 + I2
Using Ohm’s law yields
Vi Vi
Ii = , I1 =
Zi Ri
Vi - Vo Vi - AvVi (1 - Av)Vi
and I2 = = =
XCf XCf XCf
Substituting, we obtain
Vi Vi (1 - Av)Vi
= +
Zi Ri XCf
1 1 1
and = +
Zi Ri XCf >(1 - Av)
XCf 1
but = = XCM
1 - Av v(1 - Av)Cf
t
CM
1 1 1
and = +
Zi Ri XCM
establishing the equivalent network of Fig. 43. The result is an equivalent input imped-
ance to the amplifier of Fig. 44 that includes the Ri with the addition of a feedback
capacitor magnified by the gain of the amplifier. Any interelectrode capacitance at the
input terminals to the amplifier will simply be added in parallel with the elements of
Fig. 43.

Ii

+
Zi
Vi Ri CM = (1 − Aυ ) Cf

Fig. 43
Demonstrating the effect of the Miller effect capacitance.

In general, therefore, the Miller effect input capacitance is defined by

CMi = (1 - Av)Cf (49)

This shows us that:

For any inverting amplifier, the input capacitance will be increased by a Miller effect
capacitance sensitive to the gain of the amplifier and the interelectrode (parasitic)
capacitance between the input and output terminals of the active device.
The dilemma of an equation such as Eq. (49) is that at high frequencies the gain
Av will be a function of the level of CMi. However, because the maximum gain is the
midband value, using the midband value will result in the highest level of CMi and the
worst-case scenario. In general, therefore, the midband value is typically employed for
Av in Eq. (49).
The reason for the constraint that the amplifier be of the inverting variety is now more
apparent when one examines Eq. (49). A positive value for Av would result in a negative
capacitance (for Av 7 1).
The Miller effect will also increase the level of output capacitance, which must also be
considered when the high-frequency cutoff is determined. In Fig. 44, the parameters of

591
 BJT and JFET FrEquEncy
rEsponsE

Fig. 44
Network employed in the derivation of an equation for the
Miller output capacitance.

importance to determine the output Miller effect are in place. Applying Kirchhoff’s current
law results in
Io = I1 + I2
Vo Vo - Vi
with I1 = and I2 =
Ro XCf
The resistance Ro is usually sufficiently large to permit ignoring the first term of the equa-
tion compared to the second term and assuming that
Vo - Vi
Io 
XCf
Substituting Vi = Vo >Av from Av = Vo >Vi results in
Vo - Vo >Av Vo(1 - 1>Av)
Io = =
XCf XCf
Io 1 - 1>Av
and =
Vo XCf
Vo XCf 1 1
or = = =
Io 1 - 1>Av vCf (1 - 1>Av) vCMo
resulting in the following equation for the Miller output capacitance:

1
CMo = a 1 - bC (50)
Av f

For the usual situation where Av W 1, Eq. (50) reduces to

CMo  Cf 0 Av 0 W1
(51)

Examples of the use of Eq. (50) appear in the next two sections as we investigate the
high-frequency responses of BJT and FET amplifiers.
For noninverting amplifiers such as the common-base and emitter-follower configura-
tions, the Miller effect capacitance is not a contributing concern for high-frequency
applications.

11 high-FrequenCy respOnse—bjt ampLiFier

l
At the high-frequency end, there are two factors that define the 23-dB cutoff point:
the network capacitance (parasitic and introduced) and the frequency dependence
of hfe(B).

592
network parameters BJT and JFET FrEquEncy
rEsponsE
In the high-frequency region, the RC network of concern has the configuration appearing
in Fig. 45. At increasing frequencies, the reactance XC will decrease in magnitude, result-
ing in a shorting effect across the output and a decrease in gain. The derivation leading to R
the corner frequency for this RC configuration follows along similar lines to that encoun-
+ +
tered for the low-frequency region. The most significant difference is in the following Vi C Vo
general form of Av: – –
1
Av = (52) Fig. 45
1 + j( f>fH)
RC combination that will define
a high-cutoff frequency.
This results in a magnitude plot such as shown in Fig. 46 that drops off at 6 dB/octave with
increasing frequency. Note that fH is in the denominator of the frequency ratio rather than
the numerator as occurred for fL in Eq. (23).

fH
f (log scale)

–3 dB
–6 dB/octave

Fig. 46
Asymptotic plot as defined by Eq. (52).

In Fig. 47, the various parasitic capacitances (Cbe, Cbc, Cce) of the transistor are included
with the wiring capacitances (CWi, CWo) introduced during construction. The high-fre-
quency equivalent model for the network of Fig. 47 appears in Fig. 48. Note the absence of
the capacitors Cs, CC, and CE, which are all assumed to be in the short-circuit state at these
frequencies. The capacitance Ci includes the input wiring capacitance CWi, the transition
capacitance Cbe, and the Miller capacitance CMi. The capacitance Co includes the output
wiring capacitance CWo, the parasitic capacitance Cce, and the output Miller capacitance
CMo. In general, the capacitance Cbe is the largest of the parasitic capacitances, with Cce
the smallest. In fact, most specification sheets simply provide the levels of Cbe and Cbc and
do not include Cce unless it will affect the response of a particular type of transistor in a
specific area of application.

VCC

RC
R1 Cbc
CC
Vo
Cs
Vi
Cce

Rs CWo RL
R2 Cbe
+ CWi

RE CE
Vs
–

Fig. 47
Network of Fig. 25 with the capacitors that affect the high-frequency response.

593
 BJT and JFET FrEquEncy Ci = CWi + Cbe + CMi Co = CWo + Cce + CMo
rEsponsE Ib
Rs
Vo
+
+ Thi Tho
Vs Vi R1 R2 Ri Ci β Ib ro RC RL Co
–
–

Fig. 48
High-frequency ac equivalent model for the network of Fig. 47.

Determining the Thévenin equivalent circuit for the input and output networks of Fig.
48 results in the configurations of Fig. 49. For the input network, the 23-dB frequency is
defined by

1
fHi = (53)
2pRThiCi

with RThi = Rs  R1  R2  Ri (54)

and Ci = CWi + Cbe + CMi = CWi + Cbe + (1 - Av)Cbc (55)

RThi = Rs R1 R2  Ri RTho = RC RL  ro

+ +
EThi Ci ETho Co
– –

(a) (b)

Fig. 49
Thévenin circuits for the input and output networks of the network of Fig. 48.

At very high frequencies, the effect of Ci is to reduce the total impedance of the parallel
combination of R1, R2, Ri, and Ci in Fig. 48. The result is a reduced level of voltage across
Ci, a reduction in Ib, and a gain for the system.
For the output network,

1
fHo = (56)
2pRThoCo

with RTho = RC  RL  ro (57)

and Co = CWo + Cce + CMo (58)

or Co = CWo + Cce + (1 - 1>Av)Cbc

For Av large (typical): 1 W 1>Av

and Co  CWo + Cce + Cbc (59)

At very high frequencies, the capacitive reactance of Co will decrease and consequently
reduce the total impedance of the output parallel branches of Fig. 48. The net result is that
Vo will also decline toward zero as the reactance XC becomes smaller. The frequencies

594
 fHi and fHo will each define a 26-dB/octave asymptote such as depicted in Fig. 46. If the BJT and JFET FrEquEncy
parasitic capacitors were the only elements to determine the high-cutoff frequency, the rEsponsE
lowest frequency would be the determining factor. However, the decrease in hfe (or b)
with frequency must also be considered as to whether its break frequency is lower than
fHi or fHo.

hfe (or B) variation

The variation of hfe (or b) with frequency will approach, with some degree of accuracy, the
following relationship:

hfemid
hfe = (60)
1 + j(f>fb)

The use of hfe rather than b in some of this descriptive material is due primarily to the
fact that manufacturers typically use the hybrid parameters when covering this issue in their
specification sheets and so on.
The only undefined quantity, fb, is determined by a set of parameters employed in the
hybrid p or Giacoletto model of Fig. 50. The resistance rb includes the base contact, base
bulk, and base spreading resistance. The first is due to the actual connection to the base.
The second includes the resistance from the external terminal to the active region of the
transistors, and the last is the actual resistance within the active base region. The resis-
tances rp, ro, and ru are the resistances between the indicated terminals when the device
is in the active region. The same is true for the capacitances Cbc and Cbe, although the
former is a transition capacitance, whereas the latter is a diffusion capacitance. A more
detailed explanation of the frequency dependence of each can be found in a number of
readily available texts.

ru
rb
B b' C
Cu(Cbc)
+ I'b
Ib

V r C(Cbe) β Ib ro

–
E E

Fig. 50
Giacoletto (or hybrid p) high-frequency transistor small-signal ac equivalent circuit.

If we remove the base resistance rb, the base-to-collector resistance ru, and all the
parasitic capacitances, the result is an ac equivalent circuit that matches the small-signal
equivalent for the common-emitter configuration. The base-to-emitter resistance rp is bre
and the output resistance ro is simply a value provided through the hybrid parameter hoe.
The controlled source is also bIb. However, if we include the resistance ru (usually quite
large) between base and collector, there is a feedback loop between output and input cir-
cuits to match the contribution of hre for the hybrid equivalent circuit. The feedback term is
normally inconsequential for most applications, but if a particular application puts it at the
forefront, then the model of Fig. 50 will bring it into play. The resistance ru is a result of
the fact that the base current is somewhat sensitive to the collector-to-base voltage. Because
the base-to-emitter voltage is linearly related to the base current through Ohm’s law and the
output voltage is equal to the difference between the base-to-emitter voltage and collector-
to-base voltage, we can conclude that the base current is sensitive to the changes in output
voltage as revealed by the hybrid parameter hre.

595
 BJT and JFET FrEquEncy In terms of these parameters,
rEsponsE
1
fb (often appearing as fhfe ) = (61)
2prp(Cp + Cu)
or, because rp = bre = hfemidre,

1 1
fb = (62)
hfemid 2pre(Cp + Cu)

or
Equation (62) clearly reveals that because re is a function of the network design:
fB is a function of the bias configuration.
The basic format of Eq. (60) is exactly the same as Eq. (52) if we extract the multiply-
ing factor hfemid, revealing that hfe will drop off from its midband value with a 6-dB/octave
slope as shown in Fig. 51. The same figure has a plot of hfb (or a) versus frequency. Note
the small change in hf b for the chosen frequency range, revealing that the common-base
configuration displays improved high-frequency characteristics over the common-emitter
configuration. Recall also the absence of the Miller effect capacitance due to the noninvert-
ing characteristics of the common-base configuration. For this very reason, common-base
high-frequency parameters rather than common-emitter parameters are often specified for a
transistor—especially those designed specifically to operate in the high-frequency regions.

hfe hfe, hfb

hfe
mid
40 dB
Midband value for hfe
(hfe = hfe mid) 1.0
−3 dB hfe = 0.707 hfe mid
0.707 30 dB
hfe

20 dB
−6
dB
/
oc

10 dB
tav
es
lo
pe

Midband value for hfb hfe = 1

0 dB
−3 dB
hfb

−10 dB

fβ , ( fh ) 5fβ fT fh , ( f α )
−20 dB fe
fb

0.1 MHz 1.0 MHz 10.0 MHz 100.0 MHz 1 kMHz 10 kMHz f (log scale)

Fig. 51
hfe and hf b versus frequency in the high-frequency region.

The following equation permits a direct conversion for determining fb if fa and a are
specified:

fb = fa(1 - a) (63)

gain-bandwidth product
There is a Figure of Merit applied to amplifiers called the Gain-Bandwidth Product
(GBP) that is commonly used to initiate the design process of an amplifier. It provides

596
important information about the relationship between the gain of the amplifier and the BJT and JFET FrEquEncy
expected operating frequency range. rEsponsE
In Fig. 52 the frequency response of an amplifier with a gain of 100, a low cutoff fre-
quency of 250 Hz, and an upper cutoff frequency of 1 MHz has been plotted on a linear
scale rather than the typical log scale. Note that because a linear scale was chosen for the
horizontal axis it is impossible to show the low cutoff frequency, and the curve appears as
essentially a straight vertical line at f 5 0 Hz. Because f 5 0 Hz represents a dc situation,

Av
dB Bode plot
Av = 100
40 dB
37 dB –3 dB –6 dB
Actual response –6 dB/octave
30 dB Av = 70.71

BW = fH – fL ≅ fH
20 dB/decade
20 dB
10 MHz
10 dB
50 kHz Octave

0 dB
100 kHz 500 kHz 1 MHz 2 MHz 3 MHz f
10 kHz ( fH)
Linear scale
250 Hz

Fig. 52
Plotting the dB gain of an amplifier in a linear-frequency plot.

the gain at the low end of an amplifier is often called the DC gain.
Note also that the use of a linear horizontal axis results in a very slow decline in gain
with frequency past the breakpoint. It would take many pages to show the full frequency
plot at the high end.
It is also clear from Fig. 52 that the bandwidth is essentially defined by the upper cutoff
frequency because the low cutoff frequency is so small in comparison.
If Fig. 52 were plotted using a log scale for the horizontal axis, the plot of Fig. 53 would
result.
The low end is expanded and the frequency response at the upper end is complete with
a boundary defined by the 20-dB drop per decade slope. The upper breakpoint frequency
is labeled fH with the lower breakpoint frequency labeled fL.
At Av = Avmid = 100 the bandwidth as shown in Fig. 53 is approximately 1 MHz.

AvdB

Av = 100
40 dB
37 dB BW
BW = fH – fL = 1 MHz – 250 Hz ≅ 1 MHz
30 dB
–20 dB/decade
Av = 10
–20 dB/decade
20 dB
BW
BW = fH – fL = 10 MHz – 25 Hz ≅ 10 MHz
10 dB

Av = 1
0 dB
1 10 100 1 kHz 10 kHz 100 kHz 1 MHz 10 MHz 100 MHz f
( fH) ( f T) (log scale)
25 Hz 250 Hz
( fL )

Fig. 53
Finding the bandwidth at two different gain levels.

597
 BJT and JFET FrEquEncy The gain-bandwidth product is
rEsponsE
GBP = Avmid BW (64)

which for this example is

GBP = (100)(1 MHz) = 100 MHz
At Av = 10, 20 log1010 = 20 and the bandwidth as shown in Fig. 53 is approximately
10 MHz.
The resulting gain-bandwidth product is now
GBP = (10)(10 MHz) = 100 MHz
In fact, at any level of gain the product of the two remains a constant.
At Av 5 1 or Av  dB = 0 bandwidth is defined as fT in Fig. 53.
In general,
the frequency fT is called the unity-gain frequency and is always equal to the product of
the midband gain of an amplifier and the bandwidth at any level of gain.
That is,

fT = Avmid fH (Hz) (65a)

The result is that the expected bandwidth of an amplifier for any level of gain can be
found quite directly. Consider an amplifier with a given fT of 120 MHz. At a gain of 80
the expected fH or bandwidth is fT >Avmid = 120 MHz>80 = 1.5 MHz. At a gain of 60 the
bandwidth is 120 MHz/60 5 2 MHz and so on—a very useful tool.
For transistors themselves, where a voltage gain has not been defined by a configu-
ration, specification sheets will provide a value of fT that relates to the transistor only.
That is,

fT = hfemid fb (Hz) (65b)

The dB plot would appear as shown in Fig. 49.

The general equation for the hfe variation with frequency is defined by Eq. 60. For the
amplifier it is defined by

Avmid
Av = (66)
1 + j( f>fH)

Note that in each case the frequency fH defines the corner frequency.
Substituting Eq. (62) for fb in Eq. (65) gives
1
fT = hfemid
2phfemidre(Cp + Cu)

1
and fT  (67)
2pre(Cp + Cu)

eXampLe 14 Use the network of Fig. 47 with the same parameters as in Example 12, that
is,
Rs = 1 kV, R1 = 40 kV, R2 = 10 kV, RE = 2 kV, RC = 4 kV, RL = 2.2 kV
Cs = 10 mF, CC = 1 mF, CE = 20 mF
hfe = 100, ro =  V, VCC = 20 V
with the addition of
Cp(Cbe) = 36 pF, Cu(Cbc) = 4 pF, Cce = 1 pF, CWi = 6 pF, CWo = 8 pF
a. Determine fHi and fHo.
b. Find fb and fT.

598
c. Sketch the frequency response for the low- and high-frequency regions using the BJT and JFET FrEquEncy
results of Example 12 and the results of parts (a) and (b). rEsponsE
d. Obtain the PSpice response for the full frequency spectrum and compare with the
results of part (c).

Solution:
a. From Example 12:
Ri = 1.32 kV, Avmid(amplifier:not including effects of Rs) = -90
and RThi = Rs  R1  R2  Ri = 1 kV  40 kV  10 kV  1.32 kV
 0.531 kV
with Ci = CWi + Cbe + (1 - Av)Cbc
= 6 pF + 36 pF + [1 - (-90)]4 pF
= 406 pF
1 1
fHi = =
2pRThiCi 2p(0.531 kV)(406 pF)
= 738.24 kHz
RTho = RC  RL = 4 kV  2.2 kV = 1.419 kV
1
Co = CWo + Cce + CMo = 8 pF + 1 pF + a 1 - b 4 pF
-90
= 13.04 pF
1 1
fHo = =
2pRThoCo 2p(1.419 kV)(13.04 pF)
= 8.6 MHz

Av
Av
mid dB

fHi fβ
fL fL fL
1 s 10 C 100 E 1 kHz 10 kHz 100 kHz 1 MHz 10 MHz 100 MHz
0
fHo f (log scale)

–3
f1 f2
–5 BW

−6 dB/octave

–10

–15

+20 dB/decade −12 dB/octave

–20

–25

Fig. 54
Full frequency response for the network of Fig. 47.

599
 BJT and JFET FrEquEncy b. Applying Eq. (63) gives
rEsponsE
1
fb =
2phfemidre(Cbe + Cbc)
1 1
= =
2p(100)(15.76 V)(36 pF + 4 pF) 2p(100)(15.76 V)(40 pF)
= 2.52 MHz
fT = h femid fb = (100)(2.52 MHz)
= 252 MHz
c. See Fig. 54. The corner frequency fHi will determine the high cutoff frequency
and the bandwidth of the amplifier. The upper cutoff frequency is very close to
600 kHz.
d. The PSpice analysis will appear in Section 15.

12 high-FrequenCy respOnse—Fet ampLiFier

l
The analysis of the high-frequency response of the FET amplifier will proceed in a very
similar manner to that encountered for the BJT amplifier. As shown in Fig. 55, there are
interelectrode and wiring capacitances that will determine the high-frequency characteris-
tics of the amplifier. The capacitors Cgs and Cgd typically vary from 1 pF to 10 pF, whereas
the capacitance Cds is usually quite a bit smaller, ranging from 0.1 pF to 1 pF.
Because the network of Fig. 55 is an inverting amplifier, a Miller effect capacitance will
appear in the high-frequency ac equivalent network appearing in Fig. 56. At high frequen-
cies, Ci will approach a short-circuit equivalent and Vgs will drop in value and reduce the
overall gain. At frequencies where Co approaches its short-circuit equivalent, the parallel
output voltage Vo will drop in magnitude.

VDD

RD
CC
Cgd Vo

Vi Vg
Cds
Rsig CG CW o RL
+
Vs RG CW i Cgs
CS
– RS

Fig. 55
Capacitive elements that affect the high-frequency response of a JFET amplifier.

Rsig
Vi =V g
Vo
+
+
Vs RG Ci Vgs gm Vgs rd RD RL Co
–
Thi – Tho

Fig. 56
High-frequency ac equivalent circuit for Fig. 55.

600
 The cutoff frequencies defined by the input and output circuits can be obtained by first find- BJT and JFET FrEquEncy
ing the Thévenin equivalent circuits for each section as shown in Fig. 57. For the input circuit, rEsponsE

1
fHi = (68)
2pRThiCi

RThi RTho

EThi ETho

Fig. 57
The Thévenin equivalent circuits for: (a) the input circuit and (b) the output circuit.

and RThi = Rsig  RG (69)

with Ci = CWi + Cgs + CMi (70)

and CMi = (1 - Av)Cgd (71)

for the output circuit,

1
fHo = (72)
2pRThoCo

with RTho = RD  RL  rd (73)

and Co = CWo + Cds + CMo (74)

1
and CMo = a 1 - bC (75)
Av gd

eXampLe 15
a. Determine the high-cutoff frequencies for the network of Fig. 55 using the same
parameters as Example 13:
CG = 0.01 mF, CC = 0.5 mF, CS = 2 mF
Rsig = 10 kV, RG = 1 MV, RD = 4.7 kV, RS = 1 kV, RL = 2.2 kV
IDSS = 8 mA, VP = -4 V, rd =  V, VDD = 20 V
with the addition of
Cgd = 2 pF, Cgs = 4 pF, Cds = 0.5 pF, CWi = 5 pF, CWo = 6 pF
b. Obtain a PSpice response for the full frequency range and note whether it supports the
conclusions of Example 13 and the calculations above.

601
 BJT and JFET FrEquEncy Solution:
rEsponsE
a. RThi = Rsig  RG = 10 kV  1 MV = 9.9 kV
From Example 13, Av = -3. We have
Ci = CWi + Cgs + (1 - Av)Cgd
= 5 pF + 4 pF + (1 + 3)2 pF
= 9 pF + 8 pF
= 17 pF
1
fH1 =
2pRThiCi
1
= = 945.67 kHz
2p(9.9 kV)(17 pF)
RTho = RD  RL
= 4.7 kV  2.2 kV
 1.5 kV
1
Co = CWo + Cds + CMo = 6 pF + 0.5 pF + a 1 - b 2 pF = 9.17 pF
-3
1
fHo = = 11.57 MHz
2p(1.5 kV)(9.17 pF)
The results above clearly indicate that the input capacitance with its Miller effect
capacitance will determine the upper cutoff frequency. This is typically the case due to
the smaller value of Cds and the resistance levels encountered in the output circuit.
b. The PSpice analysis will appear in Section 15.

Even though the analysis of the last few sections has been limited to two configurations,
the general procedure for determining the cutoff frequencies should support the analysis
of any other transistor configuration. Keep in mind that the Miller capacitance is limited to
inverting amplifiers and that fa is significantly greater than fb if the common-base configu-
ration is encountered. There is a great deal more literature on the analysis of single-stage
amplifiers that goes beyond the coverage of this chapter. However, the content of this
chapter should provide a firm foundation for any analysis of frequency effects.

13 muLtistage FrequenCy eFFeCts

l
For a second transistor stage connected directly to the output of a first stage, there will be
a significant change in the overall frequency response. In the high-frequency region, the
output capacitance Co must now include the wiring capacitance (CW1), parasitic capaci-
tance (Cbe), and Miller capacitance (CMi) of the following stage. Furthermore, there will be
additional low-frequency cutoff levels due to the second stage, which will further reduce
the overall gain of the system in this region. For each additional stage, the upper cutoff
frequency will be determined primarily by the stage having the lowest cutoff frequency.
The low-frequency cutoff is primarily determined by that stage having the highest low-
frequency cutoff frequency. Obviously, therefore, one poorly designed stage can offset an
otherwise well-designed cascaded system.
The effect of increasing the number of identical stages can be clearly demonstrated
by considering the situations indicated in Fig. 58. In each case, the upper and lower cut-
off frequencies of each of the cascaded stages are identical. For a single stage, the cutoff
frequencies are fL and fH as indicated. For two identical stages in cascade, the drop-off
rate in the high- and low-frequency regions has increased to 212 dB/octave or 240 dB/
decade. At fL and fH, therefore, the decibel drop is now 26 dB rather than the defined band
frequency gain level of 23 dB. The 23-dB point has shifted to f L and f H  as indicated,
with a resulting drop in the bandwidth. A 218-dB/octave or 260-dB/decade slope will
result for a three-stage system of identical stages with the indicated reduction in bandwidth
( f L and f H ).

602