Alexandria High Institute for Engineering & Technology
Frequency Response of BJT &
JFET
Course ECE241- Chapter (2)
Sameh Selim
)1)
Logarithms
To clarify the relationship between the variables of a logarithmic
function, consider the mathematical equations:
b is the base, x the power.
If b = 10 and x = 2,
a = bx = (10)2 = 100
but x = logb a = log10 100 = 2
For the electrical/electronics industry, the base in the logarithmic
equation is limited to 10 and the number e = 2.71828…
Common logarithms: logarithms taken to the base 10.
Natural logarithms: logarithms taken to the base e.
The two are related by
In calculators, the common logarithm is denoted by "log" key and the
natural logarithm by the "ln" key.
Example 1: Using the calculator, determine the logarithm of the
following numbers to the indicated base:
a) Log10 106
b) logee3
c) log10 10-2
)2)
d) loge e-1
Example 2: Using the calculator, determine the logarithm of the
following numbers:
a) Log10 64
b) loge 64
c) log10 1600
d) log10 8000
N.B. the logarithm of a number does not increase in the same linear
fashion as the number.
The following table shows how the logarithm of a number increases
only as the exponent of the number.
If the antilogarithm of a number is desired, the 10x or ex calculator
functions are employed.
Example 3: Using a calculator, determine the antilogarithm of the
following expressions:
a) 1.6 = log10 a.
b) 0.04 = loge a.
Some properties of common logarithms:
)3)
Example 4: Using calculator, determine the logarithm of the
following numbers:
(
)
a) log10(0.5)
b)
(
)
c)
The use of log scales can significantly expand the range of variation
of a particular variable on the graph.
The semilog scale graph paper appears as in Figure below:
)4)
DECIBELS
The term "bel" (B) (derived from Alexander Graham Bell) was defined
by the following equation to relate power levels P1 and P2
Practically, so the decibel (dB) was defined such that:
For a specified terminal (output) power (P2) there must be a
reference power level (P1). The reference level is generally accepted
to be 1mW, with an associated resistance of 600  (the characteristic
impedance of audio transmission lines).
There exists a second equation for decibels. For the system shown,
Vi equal to some value V1,
. If Vi is changed to some other
level, V2, then
. To determine the resulting difference in
decibels between the power levels,
One of the advantages of the logarithmic relationship is the manner
in which it can be applied to cascaded stages.
)5)
For example, the magnitude of the overall voltage gain of cascaded
system is given by
Applying the proper logarithmic relationship results in
In words, the equation states that the decibel gain of cascaded
system is simply the sum of the decibel gain of each stage, i.e.
Example 5: Find the magnitude gain corresponding to a decibel gain
of 100.
Check
Example 6: The I/p power to a device is 10,000 W at a voltage of
1000 V. The O/P power is 500 W, while the o/p impedance is 20.
a) Find the power gain in decibels. ( -13.01 dB )
b) Find the voltage gain in decibels. ( - 20 dB )
c) Explain why parts (a) and (b) agree or disagree.( Ri = 100  Ro)
Example 7:
An amplifier rated at 40-W O/P is connected to a 10- speaker.
a) Calculate the I/P power required for full power O/P if the power
gain is 25 dB. ( 126.5 mW)
b)
c) Calculate the I/P voltage for rated O/P if the amplifier voltage gain
is 40 dB. ( 200 mV )
)6)
General Frequency considerations The magnitudes of the gain
response of an RC – coupled amplifier system is provided in the
following Figure. Note that the horizontal scale is a logarithmic scale
to permit a plot extending from the low- to the high-frequency
regions.
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 boundaries of relatively high gain, 0.707 Avmid was chosen
to be the gain at the cutoff levels. The corresponding frequencies f1
and f2 are generally called the corner, cutoff, break, or half-power
frequencies. The multiplier 0.707 was chosen because at this level
the O/P power is half the midband power O/P, that is, at
midfrequencies,
and at the half-power frequencies,
and POHPF = 0.5 Pomid
(16)
The bandwidth (or passband) of the system is determined by f1 and
f2, that is,
Bandwidth (BW) = f2 – f1
(17)
)7)
For applications of communications nature (audio, video), a decibel
plot of the voltage gain vs frequency is more useful than that
appearing in the above Figure: the curve is generally normalized as in
the following Figure.
In this Figure, the gain at each frequency is divided by the midband
value. Obviously the midband value is then 1 as indicated. At halfpower frequencies, the resulting level is 0.707 = √ . A decibel plot
can now be obtained by applying
At midband frequencies, 20 log10 1 = 0, and at the cutoff frequencies,
20 log10 √ = -3 dB. Both values are clearly indicated in the
resulting decibel plot, as shown in the following Figure,
Most amplifiers introduce a 180o phase shift between I/P and O/P
signals. In fact, 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 will drop below 180o. the
following Figure is a standard phase plot for an RC –coupled
amplifier.
)8)
Low–Frequency Analysis-Bode Plot:
In the low-frequency region of the single-stage BJT or FET amplifier,
it is the R-C combination formed by the network capacitors Cc, CE and
Cs and the network resistive parameters that determine the cutoff
frequencies.
The series RC combination, as shown, and the development of a
procedure that will result in a plot of frequency response with
minimum of time and effort.
At high frequencies,
and the short-circuit equivalent can be
substituted for the capacitor. The result is
that Vo  Vi at high frequencies.
At f = 0 Hz,
and the open-circuit approximation can be applied, with the result
that Vo = 0 V.
Between the two extremes, the ratio Av = Vo /Vi will vary as shown in
)9)
the following Figure. As the frequency increases, the capacitive
reactance decreases and more of the input voltage appears across
the O/P terminals.
The O/P and I/P voltages are related by the voltage-divider rule as
With magnitude of Vo determined by
√
For the special case where XC = R,
√
√
√
√
√
And
The level of which is indicated in on Figure above. In other words, at
the frequency of which XC = R, the O/P will be 70.7% of the I/P for the
RC network shown before.
The frequency at which this occurs is determined from
and in terms of logs,
)11)
While at Av = Vo/Vi = 1 or Vo = Vi (the maximum value)
If the gain equation is written as
For the magnitude when f = f1
In logarithmic form, the gain in dB is
For frequencies For where f  f1 or (f1/f)2  1, the equation above
can be approximated by
And finally,
)11)
Ignoring the condition f  f1 for the moment, a plot of the last
equation on a frequency log scale will yield a result of a very useful
nature for future decibel plots.
The piecewise linear plot of the asymptotes and associated
breakpoints is called a Bode Plot of magnitude vs frequency.
The piecewise linear plot of the asymptotes and associated
breakpoints is called a Bode plot of the magnitude versus frequency.
Notes:
 A change in frequency by a factor of 2, i.e. one octave, results in 6-dB
change in ratio, as shown by the gain change from f1/2 to f1.
 A change in frequency by a factor of 10, i.e. one decade, results in 20dB change in ratio, as shown by the gain change from f1/10 to f1.
)12)
Steps for Bode plot:
1) Find f1 from the circuit parameters.
2) Sketch 2 asymptotes: one along the 0-dB and the other drawn
thro' f1 sloped at 6 dB/octave or 20 dB/decade.
3) Find the 3-dB point corresponding to f1 and sketch the curve.
Example 8: For the network shown in Figure:
1. Determine the break frequency.
2. Sketch the asymptotes and locate the -3 dB point.
3. Sketch the frequency response curve.
The Frequency response of the gain Av(dB) for the R-C circuit is shown
in Figure. The gain at any frequency can be determined from the
frequency plot in the following manner:
)13)
Check: Av(dB) = -3 dB, Av = 0.707, and
Av(dB) = - 1 dB, (at f = 2f1), Av = 0.891
The phase angle is determined from
f  f1,  = 90o
f1 = 100 f,  = 89.4o
f = f1,  = 45o
f  f1,  = 0o
f = 100 f1,  = 0.573o
The phase response for
the R-C circuit is shown.
Low-Frequency Response (BJT) Amplifier
In the analysis of the voltage-divider BJT, it will simply be necessary
to find the appropriate equivalent resistance for the RC combination.
Capacitors Cs, Cc, CE will determine the low-frequency response. We
will examine the impact of each independently.
Effect of Cs: the general form of the R-C configuration is established by
the network of the following Figure.
The total resistance is Rs + Ri.
)14)
The cutoff frequency:
At mid or high frequencies: The reactance of the capacitor will be short
circuit approximate. The voltage Vi is related to Vs by:
At fLs, the voltage Vi will be 70.7% of the
value determined by this eqn., assuming
that Cs is the only capacitive element
controlling the low frequency response.
The ac equivalent for the i/p section of
BJT amplifier:
The value of Ri is determined by:
Ri = R1//R2//re
The voltage Vi applied to the i/p of the active device can be calculated
using the voltage-divider rule:
Effect of CC: Since the coupling capacitor is normally connected
between the O/P of the active device and the applied load, the R-C
configuration that determines the low cutoff frequency due to CC
appears in the following Figure.
The total series resistance is now RO +
RL and the cutoff frequency due to CC is
determined by:
(
)
)15)
Ignoring the effects of CS and CE the O/P
voltage Vo will be 70.7% of its midband
value at fLC. For the network of the loaded
BJT amplifier, the ac equivalent network
for the O/P section with Vi = 0 V appears
in the following Figure.
The resulting value of Ro in the equation of fLC is then simply
Ro = RC//ro
Effect of CE: To determine fLE, the network seen by CE must be
determined as shown in the Figure below.
Once the level of Re is established, the
cutoff frequency due to CE can be
determined using the following
equation:
For the BJT network, the ac equivalent as seen by CE appears in the
following Figure.
The value of Re is therefore determined by
(
)
Where
The effect of CE on the gain is best described
by recalling that the gain for the configuration
)16)
of the shown Figure is given by:
The maximum gain is obviously available where RE is zero ohms.
The higher fL will be the predominant factor in determining the lowfrequency response for the complete system.
Example: For the network shown in Figure with the following
parameters:
Cs = 10 F, CE = 20 F, Cc = 1 F, Rs = 1 K, R1 = 40 K, R2 = 10 K,
RE = 2 K, Rc = 4 K, RL = 2.2 K,  = 100, ro =  , Vcc = 20 V
a) Determine re
b) Find AVmid = Vo/Vi
c) Calculate Zi.
d) Find AVmid = Vo/Vs.
e) Determine fLs, fLc,
and fLE.
f) Determine the lower
cutoff frequency.
g) Sketch the
asymptotes of the
Bode plot defined by the cutoff frequencies.
h) Sketch the low-frequency response for the amplifier using results
of (f).
______________________________________________________
a) Determine re for dc conditions:
RE = (100)(2 K) = 200 K >> 10 R2 = 100 K
, check VB = 4 V
, IE = 1.65 mA
, re  15.76 
)17)
And re = 100 (15.76 ) = 1576  = 1.576 K
, check  -90
b) Midband gain
c)
= 40K//10K//1.576K
 1.32 K
d) From the Figure shown:
So that
(
)(
)
e) Effect of Cs
Ri = R1//R2//re , check Ri = 1.32 K
(
)
, check  6.86 Hz
Effect of CC
(
)
, check  25.68 Hz
Effect of CE
= 1 K// 40 K // 10 K = 0.889 K
)  24.35 
(
=(
)(
)(
f) fL = 327 Hz
)18)
)
 327 Hz
g) , h) The low –frequency plot for the network:
Low-Frequency Response (FET) Amplifier
The analysis is quite similar to that of the BJT amplifier. There are
again capacitors CG, CC, and CS. The following Figure is used to
establish the fundamental equations.
Effect of CG: The ac equivalent network for the coupling capacitor
between the source and the active device is shown in the following
Figure.
The cutoff frequency determined by CG will
be:
(
For the above network,
)
)19)
Typically RG>> Rsig and the lower cutoff frequency will be determined
primarily by RG and CG.
Effect of CC: For the coupling capacitor between the active device
and the load the network of following Figure will result.
The resulting cutoff frequency is
(
)
For the given network,
Effect of CS: For the source capacitor CS the resistance level of
importance is defined by the following Figure.
The cutoff frequency will be defined by
For the above network, the resulting value of Req
(
) (
Which for rd    becomes
)
Example:
(a) Determine the lower cutoff frequency for the network of the
above Fig, using the following parameters:
CG = 0.01 F, CC = 0.5 F, Cs = 2 F
Rsig = 10 k, RG = 1 M, RD = 4.7 k, Rs= 1 k,
RL = 2.2 k, gm = 2 mS
IDSS = 8 mA, VP = -4 V, rd = , VDD = 20 V.
(b) Sketch the frequency response using Bode plot.
)21)
(a)
Effect of CG:
Using Rsig=10 K, Ri = RG = 1 M, check fLG  15.8 Hz
Effect of CC:
Using RO = RD//rd = 4.7 K, RL = 2.2 K, check fLC 46.13 Hz
Effect of CC:
Using Req = RS//(1/gm), check Req = 333.33 ,and fLs = 238.73 Hz
(b) The midband gain is determined by:
(
), check AVmid  -3.
Using the midband gain to normalize the response will result in the
frequency plot of Figure.
)21)
Miller Effect Capacitance:
In high-frequency region, the capacitive elements of importance are
the inter-electrode (between terminals) capacitances internal to the
active device and the wiring capacitance between leads of the
network.
For inverting amplifiers, the I/P and O/P capacitance is increased by a
capacitance level sensitive to the inter-electrode capacitance between
the I/P and O/P terminals of the device and the gain of the amplifier.
In Figure, this "feedback" capacitance is defined by Cf.
Miller I/P Capacitance:
Applying KCL:
Ii = I 1 + I2
Using Ohm's law:
Substituting, we obtain
Establishing the equivalent network shows that the I/P impedance
includes Ri with the addition of a feedback capacitor magnified by the
)22)
gain of the amplifier. In general the Miller effect I/P capacitance is
defined by
CMi = (1 - Av)Cf
This shows that: For any inverting amplifier, the I/P capacitance will
be increased by a Miller effect capacitance sensitive to the gain of the
amplifier and the inter-electrode (parasitic) capacitance between the
I/P and O/P terminals of the active device.
The reason for the constraint that the amplifier be of the inverting
type is now more apparent when you examine the equation of CMI. A
positive Av would result in a negative capacitance (for Av > 1).
Miller O/P Capacitance:
Applying KCL:
Io = I1 + I2
Using Ohm's law:
The resistance Ro is usually sufficiently large to permit ignoring I1,
, substituting Vi = Vo/Av will result in
Or
Resulting in the Miller O/P
capacitance:
(
)
)23)
For the usual situation where Av >> 1, this equation reduces to
CMO = Cf
High-Frequency Response (BJT) Amplifier:
Network Parameters:
In the high-frequency region, the RC network of concern has the
configuration appearing in Figure, the general form of Av:
(
)
This results in a magnitude plot that drops off at 6 dB/ octave with
increasing frequency.
The high-frequency model for the network of the following Figure
appears in the next Figure.
(BJT network with capacitors that affect the high frequency response)
)24)
In fact, most specification sheets provide levels of Cbe and Cbc and do
not include Cce.
Determining the Thevenin equivalent circuit for the I/P and O/P
networks of the above Figure will result in the configuration of the
following Figures.
For the I/P network,
the -3 dB frequency is defined by
With
(
And
)
At very high frequencies, the effect of Ci is to reduce the total
impedance of the parallel combination of R1, R2, Ri and Ci, which results
in a reduced level of voltage across Ci, a reduction in Ib, and gain of the
system.
)25)
For the O/P network,
And
At very high frequencies, Co will decrease reducing the total
impedance of the O/P parallel branches of the equivalent model. The
net result is that Vo will also decline toward zero as the reactance XC
becomes smaller. The frequencies fHi and fHo will each define a -6
dB/octave asymptote.
)26)
Example
For the network shown in the Figure with the following parameters:
Rs = 1 K, R1 = 40 K, R2 = 10 K, RE = 2 K, RC = 4 K, RL = 2.2 K,
CS = 10 F, CC = 1 F, CE = 20 F,
Cbe = 36 pF, Cbc = 4 pF, CCe = 1 pF, CWi = 6 pF, CWo = 8 pF,
 = 100, ro = , VCC = 20 V.
a) Find the I/P resistance Ri
b) Find the voltage gain Avmid.
c) Determine the Thevenin equivalent I/P resistance Rthi.
d) Determine the Thevenin equivalent O/P resistance Rtho.
e) Find the input capacitance Ci.
f) Find the output capacitor Co.
g) Determine fHi and fHo.
a) Get VB, check VB = 4 V, get re = 15.76 .
re = 1.576 K,
Ri = R1//R2//re , check Ri = 1.32 K
b)
, check  -90
c) Rthi = Rs// R1 // R2 // Ri, check  0.531 K
d) Rtho = RC// RL, check Rtho = 1.419 K
)27)
e)
f)
(
)
, check Ci = 406 pF
= 8 pF + 1 pF + [1 – (-1/90)]4 pF = 13.04 pF
g)
, check fHi = 738.24 KHz
h)
, check fHo = 8.6 MHz
In general, the lowest of the upper-cutoff frequencies defines a
maximum possible bandwidth for the system.
)28)
High-Frequency Response FET Amplifier
The shown network is an inverting amplifier, so Miller effect
capacitance will appear in the high-frequency ac equivalent network.
Thevenin's i/p circuit
For the i/p circuit fHi = 1/2RthiCi
Rthi = Rsig//RG
Ci = Cwi + Cgs + CMi
CMi = (1 – Av)Cgd
For the o/p circuit fHo = 1/2RthoCo
Rtho = RD//RL//rd
Co = Cwo + Cds + CMo
CMo = (1 – 1/Av)Cgd
)29)
Thevenin's o/p circuit
Example:
Determine the high cutoff frequency for the network of the above
Fig, using the following parameters:
CG = 0.01 F, CC = 0.5 F, Cs = 2 F
Rsig = 10 k, RG = 1 M, RD = 4.7 k, Rs= 1 k,
RL = 2.2 k, gm = 2 mS
IDSS = 8 mA, VP = -4 V, rd = , VDD = 20 V.
Cgd = 2 pF, Cgs = 4 pF, Cds = 0.5 pF, Cwi = 5 pF, Cwo = 6 pF.
Av = -gm(RD//RL), check Av = -3.
Rthi = Rsig//RG = 9.9 K.
Ci = Cwi + Cgs + (1-Av)Cgd, check = 17 pF.
fHi = 1/(2RthiCi) = 945.67 KHz
Rtho = RD//RL  1.5 K
Co = Cwo + Cds + CMO , check = 9.17 pF
fHo = 1/(2)(1.5K)(9.17pF) = 11.57 MHz
From which it is noticed that the i/p capacitance with its Miller
capacitance will determine the upper cutoff frequency.
)31)