Bjt+and+jfet+frequency+response

BJT and JFET Frequency
Response

Effects of Frequency on Operation of Circuits
• The frequency of a signal can affect the response of circuits.
• The reactance of capacitors increases when the signal frequency decreases,
and its reactance decreases when the signal frequency increases.
• The reactance of inductors and winding of transformers increases when the
signal frequency increases, and its reactance decreases when the signal
frequency decreases.
• Devices such as BJTs, FETs, resistors, and even copper wires have intrinsic
capacitances, whose reactance at high frequencies could change the response
of circuits.
• The change in the reactance of inductors and capacitors could affect the gain
of amplifiers at relatively low and high frequencies.
• At low frequencies, capacitors can no longer be treated as short circuits,
because their reactance becomes large enough to affect the signal.
• At high frequencies, the reactance of intrinsic capacitance of devices
becomes low enough, that signals could effectively pass through them,
resulting to changes in the response of the circuit.
• At low frequencies, reactance of primary of transformers become low,
resulting to poor low frequency response. Change in magnetic flux at low
frequencies become low.
• At high frequencies, the stray capacitance of transformer windings reduces
the gain of amplifiers.

• Increase in the number of stages could also affect the frequency response of a
circuit.
• In general, the gain of amplifier circuits decreases at low and high
frequencies.
• The cutoff frequencies are the frequencies when the power delivered to the
load of the circuit becomes half the power delivered to the load at middle
frequencies.
Voltage gain
Frequency
0.707 AVmid
AVmid
f1 f2Bandwidth
Avmid = voltage gain of amplifier at middle frequencies
0.707 Avmid = voltage gain of amplifier at lower cutoff frequency and higher cutoff frequency
(when output power is half the output power at middle frequencies)
f1 = low cutoff frequency PO(HPF) = output power at higher cutoff frequency Vi = input voltage
f2= high cutoff frequency PO(LPF) = output power at lower cutoff frequency
Pomid= output power at middle frequencies
    Omid
2
vmid
2
vmid
O(LPF)(HPF) P0.5
Ro
ViA
5.0
Ro
Vi0.707A
PPo 
Bandwidth = f2-f1

Frequency Response of Amplifier Circuits
• f1 and f2 are called half power, corner, cutoff, band, break, or -3db
frequencies.
• f1 is the low cutoff frequency and f2 is the high cutoff frequency.
• When the amplitude of a signal is 0.707 of its original amplitude, its power
becomes half of its original power.
PHP = PMF / 2 = power at half power frequency
where: PHP = Power at half power point (f1 or f2)
PMF = Power at middle frequencies
• The bandwidth of the signal is equal to f2 – f1
B = f2 – f1 = bandwidth

• The 180 degrees phase shift of most amplifiers (Common emitter, common
source) is only true at middle frequencies.
• At low frequencies, the phase shift is more than 180 degrees.
• At high frequencies, the phase shift is less than 180 degrees.
Phase shift
between Vo
and Vi
Frequency
1800
2700
f1 f2
Phase shift between Vo and Vi
900

• The graph of the frequency response of amplifier circuits can be plotted
with a normalized gain. (gain is divided by the gain at middle frequencies.)
Frequency
0.707 AVmid
1 AVmid
f1 f2
Normalized Gain
in Ratio
frequencymiddleatgainvoltageA
ffrequencyatgainvoltageA:where
A
A
GainNormalized
Vmid
V
Vmid
V



Normalized Plot of Voltage Gain Versus Frequency

• A decibel plot of the gain can be made using the following formula:
Voltage
gain
Frequency
0.707 AVmid
1 AVmid
f1 f2
Normalized Gain in db
frequencymiddleatgainvoltageA
ffrequencyatgainvoltageA:where
dbingainnormalized
A
A
20log
A
A
Vmid
V
Vmid
V
Vmid
V


db
Decibel plot of Normalized Voltage Gain Versus Frequency
0 db
-3 db
-6 db
-9 db

Capacitor Coupled Amplifier Circuit Frequency Response
• For capacitor coupled (also called RC-coupled) amplifiers:
– The drop in gain at low frequencies is due to the increasing reactance of
the coupling capacitors (Cc), and bypass capacitors (Cb, CE, and Cs).
– The drop in gain at high frequencies is due to the parasitic capacitance
of network and active devices, and frequency dependence of the gain of
BJTs, FETs, or vacuum tubes.
Voltage
gain
Frequency
0.707 AVmid
AVmid
f1 f2
Drop in gain is due to increase in reactance
of coupling and bypass capacitors
Avmid = voltage gain of amplifier at middle frequencies
0.707 Avmid = voltage gain of amplifier at lower cutoff frequency and higher cutoff frequency
(when output power is half the output power at middle frequencies)
f1 = low cutoff frequency
f2= high cutoff frequency
Bandwidth

Transformer Coupled Amplifier Circuit Frequency Response
• For transformer coupled amplifier circuits:
– The drop in gain at low frequencies is caused by “shorting effect” of the
input terminals (primary) of the transformer at low frequencies. The reactance
of the primary of a transformer becomes very low at low frequencies and
becomes zero at 0 hertz.
– At low frequencies, change in magnetic flux becomes low, resulting to lower
output voltage.
– The drop in gain at high frequencies is due to the stray capacitance at the
primary and secondary of a transformer, and frequency dependence of gain of
devices. At high frequencies, the reactance of the stray capacitances becomes
low enough) that high frequency signals are also “shorted out”.
Voltage
gain
Frequency
0.707 AVmid
AVmid
f1 f2
Drop in gain is due to “shorting” effect
of primary of transformer
at low frequencies.
Drop in gain is due to stray
capacitance at primary and
secondary of transformer and
other components, and
frequency dependence of gain
of active devices.
Bandwidth

Direct Coupled Amplifier Circuit Frequency Response
• For direct coupled amplifier circuits:
– There are no coupling or bypass capacitors, or transformers to cause a
drop in the gain at low frequencies. The gain at low frequencies is
typically the same as that at middle frequencies.
– The drop in gain at high frequencies is due to stray capacitance of the
circuit and the frequency dependence of the gain of active devices.
Voltage
gain
Frequency
0.707 AVmid
AVmid
f2
Drop in gain is due to stray
capacitance of the circuit, and
the frequency dependence
of the gain of active devices.
Bandwidth

Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit
• A capacitor coupled circuit which acts as a high pass filter is shown below.
• At middle and high frequencies, the capacitor C can be considered a short
circuit because its reactance becomes low enough that the voltage appearing
across RL is almost equal to Vi (input voltage of combination of C and RL).
• At low frequencies, the coupling capacitor C could no longer be treated as a
short circuit because its reactance becomes high enough that the voltage
appearing at the load (RL) becomes significantly lower than Vi.
• R can represent any resistance or resistance combination in a circuit.
• At low frequencies, the RC combination of the coupling capacitor (C) and the
resistance (R) determines the frequency response of the amplifier circuit.
• The reactance of the coupling capacitor C can be computed as:
R
Cc
IR
Vi =
Input voltage
to RC network
Capacitor Coupled Circuit Which Acts As A High Pass Filter
Vo = Output
voltage
(Farad)CcofecapacitancC
(Hz)signaloffrequencyf:where
ffrequencyatCcofreactance
πfC2
1
Xc




1
Vi
Vo
RloadtheacrossvoltageViVo
sfrequenciehighatCcofreactance0
fC2
1
Xc
L




• At high and middle frequencies, Xc becomes low enough that it can be
assumed to be zero (0) and Cc is assumed to be a short circuit.
– The voltage across R (Vo) can be assumed to be equal to the input voltage
of the RC network (Vi).
• If the frequency is equal to zero (0) such as when the signal is a DC voltage,
the reactance of Cc is equal to infinity, and the capacitor Cc can be assumed to
be an open circuit.
– The voltage across R (Vo) is equal to zero (0).
• Between the two extremes, the ratio between Vo and Vi will vary between
zero and one (1).
0
Vi
Vo
Rloadtheacrossvoltage0Vo
hz0fwhenCcofreactance
fC2
1
Xc
L





2
ViVowhensfrequenciemiddleatRatdissipatedpower
sfrequenciemiddleatPower
R
Vi
R
Vi
2
1
R
1
2
Vi
R
Vo
P
XcRwhentageoutput vol0.707ViVo
0.707Vi
2
Vi
R
RR
Vi
R
XcR
Vi
)(R)I(Vo
Xc,RWhenbelow.shownasXcRwhenoccursthisand
s,frequenciemiddleatpoweroutputtheofthathalfisRatpoweroutputthe
(f1),frequencycutofflowthetoequalissignaltheoffrequencyWhen the
R
XcR
Vi
)(R)I(Vo
2222
2
R
2
R
222
2




















































• The magnitude of the output voltage can be computed as:

(Hz)frequencycutofflower
RC2
1
f1
f1C2
1
XcR




• When the frequency is equal to the low cutoff frequency (f1), R=XC and f1
can be computed as follows:
• The normalized voltage gain at lower cutoff frequency (f1) can be
computed as:
• The normalized voltage gain at middle frequencies (fmid) can be computed
as:
3db
A
0.707A
log20db
A
A
Vmid
Vmid
Vmid
cutoffVlower

db0
A
A
log20db
A
A
Vmid
Vmid
Vmid
Vmid


 
(Hz)frequencycutofflower
RC2
1
f1:where
(unitless)ffrequencyatgainvoltage
f
f1
j1
1
Av
fCR2
1
j1
1
R
jXc
1
1
jXcR
R
jXcRI
RI
Vi
Vo
Av














• At frequency f, the voltage gain can be computed as:
• In magnitude and phase form, the voltage gain at any frequency can be
computed as:
fi/f/Tan
f
f1
1
1
Av 1
2










(db)
f
f1
1
1
log20
Vi
Vo
log20Av
2
db








• In db (logarithmic form), the voltage gain at frequency f can be computed
as:
• When f=f1= lower cutoff frequency,
db3-
f1
f1
1
1
log20
Vi
Vo
log20Av
2
db 















































22/12
2
db
f
f1
1log10
f
f1
1-20log
(db)
f
f1
1
1
log20
Vi
Vo
log20Av
• The voltage gain at frequency f can be written as:
• When f<<f1, the above equation can be approximated by:
• If we forget the condition f<<f1 and plot the right side of the above
equation, the following points can be used. The plot is a straight line when
plotted in a log scale.
db20-20log10-f1/10fAt
db12-20log4-f1/4fAt
db6-20log2-f1/2fAt
db020log1-f1fAt




f1)f(when
f
f1
log20
f
f1
-20logAv
2/12
db 





















f
f1
log20
f
f1
-20logX
2/12























• Using the points in the preceding slide, a bode plot can be made as shown below.
• A Bode plot is a piecewise linear plot of the asymptotes and associated
breakpoints.
• A Bode plot for the low frequency region is shown below.
• One octave is equivalent to a change in frequency by a factor of two (2).
• One octave results to a 6 db change in the normalized gain.
• One decade is a change in frequency by a factor of 10.
• One decade results to a 20 db change in the normalized gain.
Frequency (log scale)
0.707 AV
1 AV
f1
Normalized Gain
in db
Bode Plot for Low Frequency Region
0 db
-3 db
-6 db
-9 db
-12 db
-15 db
-18 db
-21 db
f1/2f1/4f1/10
f
f1
log20X 





 (db)
f
f1
1
1
log20Av
2
db








Actual Response Curve
Asymptote
Asymptote

• The plot in the preceding slide shows two asymptotes. One for f<< f1 (6 db /
octave), and the other for f >> f1 (horizontal line – 0 db).
• The plot of the line corresponding to f << f1 results to frequency response of
6db per octave (6 db drop in gain for every reduction in the frequency by a
factor of 2). The plot also corresponds to a response of 20 db per decade.
• The decibel plot of voltage gain (Av) can be made by using the information
on the asymptotes and knowing that at f= f1, Avdb = -3db.

• Example: For the RC network shown below, Determine the break frequency
(cutoff frequency), sketch the asymptotes and the frequency response curve.
0.707 AV
1 AV
f1 =
99.47Normalized Gain
in db
Bode Plot for Low Frequency Region
0 db
-3 db
-6 db
-9 db
-12 db
-15 db
-18 db
-21 db
f1/2=
49.74
f1/4=
24.87
f1/10=
9.947
RL =
8 kohm
Cc = 0.2 microfarad
IRL
Vi =
Input voltage
to RC network
Vo = Output
voltage Hz99.47
)0x12.0)(000,8(2
1
CR2
1
f1 6
L

 

Asymptote
Asymptote
-3db point

Low Frequency Analysis of Capacitor Coupled BJT Amplifier
• A capacitor coupled (also called RC coupled) BJT amplifier circuit is shown
below.
• At middle and high frequencies, the capacitors Cc, Cs, and Ce can be
considered short circuits because their reactance become low enough, that there
are no significant voltage drops across the capacitors.
• At low frequencies, the coupling capacitors Cc, Cs, and Ce could no longer be
treated as short circuits because their reactance become high enough that the
there are significant voltage drops across the capacitors.
Rc
Q1
Vcc
RB2
Cc
Vo = Output
voltage
RL
Vi
Vs
Rsig
Cs
RB1
RE
Ce
Zi Zo
Vs = Signal source
Rsig = internal resistance of signal source
Cs =coupling capacitor for Vs
Cc= coupling capacitor for RL
Ce= bypass capacitor for RE

• The frequency analysis of high pass RC network can be used for capacitor
coupled BJT amplifier circuits. The values of R and C are taken from the
equivalent resistances and capacitances in the BJT amplifier circuit.
• For the portion of the circuit involving the coupling capacitor Cs, the
equivalent circuit is shown below.
– Equivalent circuit assumes that the input impedance of the amplifier (Zi) is
purely resistive and is equal to Ri.
Cs
Zi = Ri
Vi
Equivalent Circuit of Vs, Cs and Zi
Ii
Cs
Zi = Ri
Vi
Equivalent Circuit of Vs, Cs and Zi
Ii
RB1//RB2 hie = re
re
Zi = Ri
Vs
Rsig
Vs
Rsig

CsjXRsigRi
RiVs
Vi


• The value of the input impedance (resistance) of the amplifier can be computed
as:
Zi = Ri = RB1 // RB2 // hie
= RB1 // RB2 // re
• The voltage Vi can be computed using voltage divider rule.
• The voltage Vi at middle frequencies (Cs can be considered short circuit) can
be computed as:
• The lower cutoff frequency can be computed as:
RsigRi
RiVs
Vi mid


Csinvolving
circuittheofportionfor thefrequencyoffcutlower
Ri)Cs(Rsig2
1
fLs 




• For the portion of the circuit involving the coupling capacitor Cc, the
– Equivalent circuit assumes that the output impedance of the transistor is
purely resistive and is equal to ro.
Cc
Zo= Ro
VRL
Equivalent Circuit of Circuit Portion Involving Cc
IRL
Rc RL
ib ro
Cc
Zo= Ro = Rc // ro
VRL
IRL
Rc // ro
RL

• The value of the output impedance (resistance) of the amplifier can be
computed as:
Zo = Ro = RC // ro
Ccinvolving
)CR(Ro2
1
f
CL
LC 




• For the portion of the circuit involving the bypass capacitor Ce, the equivalent
circuit is shown below.
– The equivalent circuit uses the re model.
• The resistance (Re) seen looking into RE from the output side can be computed
as:
Ce(Rs’/ + re
Equivalent Circuit of Portion of Circuit Involving RE and CE
RE
(Ampere)currentquiescentEmitter(Ampere)currentDCEmitterI
(ohms)
I
10X26
r
//RRsig//RRs':Where
(ohms)sideoutputthefromRintolookingseenimpedancer
Rs'
//RRe
E
E
3-
e
B2B1
EeE





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









 eE r
Rs'
//RRe


• The lower cut-off frequency of the portion of the circuit involving the
bypass capacitor Ce can be computed as:
• The voltage gain of the amplifier without considering the effects of the
voltage source resistance Rsig can be computed as:
• At middle frequencies, RE is shorted out because the reactance of Ce is
very low. Voltage gain can be computed as:
• At low frequencies, the reactance of Ce becomes high and RE should be
considered in the computation of the voltage gain.
side.outputthefromRintolookingresistanceequivalentRe:where
CeRe2
1
f
E
LE



)considerednot(rosfrequencielowatamplifiertheofgainvoltage
Rr
Rc//R
Vi
V
Vi
Vo
Av
Ee
LRL



sfrequenciemiddleatamplifiertheofgainvoltage
r
ro//Rc//R
Vi
V
Vi
Vo
Av
e
LRL


• Overall, the effects of the capacitors Cs, Cc, and Ce must be considered in
determining the lower cutoff frequency of the amplifier.
• The highest lower cutoff frequency among the three cutoff frequencies will
have the greatest impact on the lower cutoff frequency of the amplifier.
• If the cutoff frequencies due to the capacitors are relatively far apart, the
highest lower cutoff frequency will essentially determine the lower cutoff
frequency of the amplifier.
• If the highest lower cutoff frequency is relatively close to another lower cutoff
frequency, or if there are more than one lower cutoff frequencies, the lower
cutoff frequency of the amplifier will be higher than the highest lower cutoff
frequency due to the capacitors.
fLT = overall lower cutoff frequency of amplifier
fLT > fLS
fLT > fLC
Flt > fLE

• Example: A voltage divider BJT amplifier circuit has the parameters listed below.
Determine the low cutoff frequency of the amplifier and sketch the low frequency
response.
Cs=12 uF Rsig = 2 kohm RL= 2 kohm RB1=50 kohm RB2= 10 kohm
Cc=2 uF RE = 2 kohm RC= 4 kohm =100 Vcc= 20 volts
CE=20 uF Assume that output resistance of transistor to be infinite.
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LE
S
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E



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
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

Low Frequency Response of JFET Common Source Amplifier
CcID = Drain current
(ac)
VGS
VDSIG (Gate
Current)
= 0
Drain (D)
Source
Gate (G)
VO=
Output voltage
RG1
Vi =
Input
voltage
CG
RD
VDD
Zi Zo
CsRs
• The analysis of low frequency response of FET amplifiers is similar to that of
BJT amplifiers.
• At middle and high frequencies, the capacitors Cc, Cs, and CG can be
considered short circuits because their reactance become low enough, that
there are no significant voltage drops across the capacitors.
• At low frequencies, the coupling capacitors Cc, Cs, and CG could no longer
be treated as short circuits because their reactance become high enough that
the there are significant voltage drops across the capacitors.
Vs =
Source
voltage
Rsig
Ii
RL
RG2

• The frequency analysis of high pass RC network can be used for capacitor
coupled FET amplifier circuits. The values of R and C are taken from the
equivalent resistances and capacitances in the FET amplifier circuit.
• For the portion of the circuit involving the coupling capacitor CG, the
– Equivalent circuit assumes that the input impedance of the amplifier (Zi) is
purely resistive and is equal to Ri.
CG
Zi = Ri
= RG1 // RG2
Vi
Equivalent Circuit of Vs, CG and Zi
Ii
CG
Vi
Equivalent Circuit of Vs, CG and Zi
Ii
RG1 // RG2
Vs
Rsig
Vs
Rsig
Zi = Ri
= RG1 // RG2
Zi = Ri
= RG1 // RG2

CGjXRsigRi
RiVs
Vi


• The value of the input impedance (resistance) of the amplifier can be computed
as:
Zi = Ri = RG1 // RG2 Zi = RG2 if RG1 is not present (RG1 = infinity)
• The voltage Vi can be computed using voltage divider rule.
• The voltage Vi at middle frequencies (when CG can be considered as short
circuit) can be computed as:
• The lower cutoff frequency (half power frequency) can be computed as:
RsigRi
RiVs
Vi mid


G
G
LG
Cinvolving
Ri)C(Rsig2
1
f 




• For the portion of the circuit involving the coupling capacitor Cc, the
– Equivalent circuit assumes that the output impedance of the transistor is
purely resistive and is equal to Ro.
rd
Drain (D) ID
RD
Zo = Ro
RLgmVgs
+
- -
-
-
+ +
gmVgs
Ird IRD IRL
Cc

• The value of the output impedance (resistance) of the amplifier can be
computed as:
Zo = Ro = RD // rd Zo = Ro = RD if rd is equal to infinity
Ccinvolving
)CR(Ro2
1
f
CL
LC 




• For the portion of the circuit involving the bypass capacitor Cs, the equivalent
circuit is shown below.
– The resistance (Req) seen looking into Rs from the output side can be
computed as:
Cs
Equivalent Circuit of Portion of Circuit Involving RS and CS
RS
gm
1
Rs//Req
bewillaboveequationtheinfinity,rwhen
(ohms)sideoutputthefromRintolookingseenimpedance
//RRr
rgm1
Rs1
Rs
Req
d
S
LDd
d











Req
System

• The low cut-off frequency of the portion of the circuit involving the bypass
capacitor Cs can be computed as:
side.outputthefromRintolookingresistanceequivalentReq:where
CsReq2π
1
f
S
Ls


• Overall, the effects of the capacitors CG, Cc, and CS must be considered in
determining the low cutoff frequency of the amplifier.
• The highest lower cutoff frequency among the three cutoff frequencies will
have the greatest impact on the low cutoff frequency of the amplifier.
• If the cutoff frequencies due to the capacitors are relatively far apart, the
highest low cutoff frequency will essentially determine the low cutoff
frequency of the amplifier.
• If the highest lower cutoff frequency is relatively close to another lower cutoff
frequency, or if there are more than one lower cutoff frequency, the low cutoff
frequency of the amplifier will be higher than the highest lower cutoff
frequency due to the capacitors.
involving Cs

• Example: Given a common source FET amplifier with the following
parameters, determine the lower cutoff frequency of the amplifier.
CG =0.02F Cc = 0.6 F Cs = 2 F
Rsig = 12 K RG= 1 Mohm RD= 5 K Rs= 1 K RL = 2 K
IDSS= 9 mA Vp= -7 volts rd= infinity VDD = 20 volts
Since RG1 is not present, configuration is self bias FET.
 
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sfrequenciemidlleatgainvoltage
9.1
2,0005,000
000)(5,000)(2,
)10x(1.33)//Rgm(R
Vi
Vo
Avmid 3-
LD









Av / Avmid (db)
Normalized
Voltage
gain
Frequency
0.707 AVmid
1 AVmid
fLSfLC
Low Frequency Response (Normalized Voltage Gain Versus Frequency
0 db
-3 db
-5 db
-25 db
-20 db
-15 db
-10 db
1 10 100 1K 10K 100K 1M 10M 100M
fLG fHi fHof
- 20 db / decade

High Frequency Response of Low Pass RC Network
• At the high frequency end, the frequency response of a low pass RC network
shown below is determined by the decrease in the reactance of the capacitor as
frequency of operation increases.
• Because of the decrease in the capacitance, there is a “shorting” effect across
the terminals of the capacitor at high frequencies, and the voltage drop across
the capacitor decreases as frequency increases.
R
IR
Vi =
Input voltage
to RC network
Vo = Output
voltage
0.707 AV
1 AV
f2
Normalized Gain
in db
Bode Plot for High Frequency Region
0 db
-3 db
-6 db / octave
C
Av = Vo / Vi

High Frequency Response of Low Pass RC Network
• The voltage gain of the low pass RC network can be computed as:
 
sfrequenciemidat
gainvoltagethetimes0.707isgaingewhen voltafrequency
(Hz)frequencycutoffhigh
RC2
1
f2:where
(unitless)ffrequencyatgainvoltage
f2
f
j1
1
Av
f
f2
1
j1
1
fRC2
1
1
j1
1
R
1
R
1
fC2
1
R
j1
1
fC2
1
R
j1
1
jXc
R
1
1
jXcR
jXc-
jXcRI
(-jXc)I
Vi
Vo
Av

































• The above equation results to plot that drops off at 6db per octave with
increasing frequency.

Miller Effect Capacitance
• When the frequencies being processed by an amplifier are high, the frequency response
of the amplifier is affected by:
• Interelectrode (between terminals) capacitance internal to the active device
• Wiring capacitance between leads of the network
• The coupling and bypass capacitors are considered short circuits at mid and high
frequencies because their reactance levels are very low.
• The diagram below shows the existence of a “feedback” capacitance whose reactance
becomes significantly low at high frequencies, that it affects the performance of an
amplifier.
• The input and output capacitance are increased by a capacitance level sensitive to the
interelectrode (between terminals) capacitance (Cf) between the input and output
terminals of the device and the gain of the amplifier.
• Because of Cf, an equivalent capacitance, called Miller capacitance, is produced at the
input and output.
Vo
+
-
Vi
+
-
Zi
Av =Vo / Vi
CfI2
I1
Ii
Ri

• The value of the Miller effect input capacitance can be computed as:
CMi
fMi
CM
ff
21
21
X
1
Ri
1
Zi
1
ecapacitancinputeffectMillerCAv)(1C
ecapacitancinputeffectMillerofReactanceX
fC2Av)(1
1
CAv)(1
1
Av)(1
Xcf
Av)(1
Xcf
1
Ri
1
Zi
1
Xcf
Av)(1
Ri
1
Zi
1
Xcf
Av)Vi(1
Ri
Vi
Zi
Vi
IIIi
Xcf
Av)Vi(1
Xcf
AvViVi
Xcf
VoVi
I
Ri
Vi
I
Zi
Vi
Ii























• The equivalent circuit due to the Miller Effect Capacitance is shown below.
• Above results show that for any inverting amplifier (negative AV), the input
capacitance will be increased by a Miller effect capacitance, which is a
function of the gain of the amplifier and the interelectrode (parasitic)
capacitance between the input and output terminals of the active device.
• If the voltage gain is negative (with phase reversal), Miller Effect capacitance
(CM) is positive and higher than the interelectrode capacitance.
• If the voltage gain is positive (no phase reversal) and greater than 1, Miller
Effect capacitance (CM) is negative.
+
-
Vi
Zi
Ii
CM i= (1-AV)CfRi

• At high frequencies, the voltage gain Av is a function of the Miller effect
capacitance (CM).
• There is difficulty in solving the value of the Miller effect capacitance (CM)
since it is a function of the gain AV which in turn is a function of the Miller
effect capacitance.
– In general, the midband value of the voltage gain is used for AV, to get
the worst case scenario for the Miller effect capacitance, since the
highest value of Av is the midband value.
• The Miller effect also increases the level of the output capacitance, and it
must also be considered in determining the high cutoff frequency.
• The diagram below shows the “feedback” capacitor as seen in the output side
of the amplifier.
Vo
+
-
Vi
+
-
Zo
Av =Vo / Vi
Cf I2
I1
Io
Ro

• The Miller effect output capacitance can be determined as follows:
ecapacitancoutputeffectMillerofReactance
C
1
C
Av
1
1
1
Av
1
1
Xcf
Io
Vo
Xcf
Av
1
1
Vo
Io
Xcf
Av
1
1Vo
Xcf
Av
Vo
Vo
Xcf
ViVo
Io
:byedapproximatbecanIo
largelysufficientusuallyisRobecausesmallryusually veis
Ro
Vo
but
Xcf
ViVo
Ro
Vo
Io
Xcf
ViVo
I
Ro
Vo
I
Zo
Vo
Io
IIIo
Mo
f
21
21





































ecapacitancoutputeffectMillerCfC
:byedapproximatbecan
ecapacitancoutputeffectMillerthe1,angreater thmuchisAvWhen
ecapacitancoutputeffectMillerCf
Av
1
1C
Mo
Mo







BJT High Frequency Response
• At the high frequency end, the high cutoff frequency (-3 db) of BJT circuits is
affected by:
– Network capacitance (parasitic and induced)
– Frequency dependence of the current gain hfe
• At high frequencies, the high cutoff frequency of a BJT circuit is affected by:
– the interelectrode capacitance between the base and emitter, base and
collector, and collector and emitter.
– Wiring capacitance at the input and output of the BJT.
• At high frequencies, the reactance of the interelectrode and wiring capacitance
become significantly low, resulting to a “shorting” effect across the
capacitances.
• The “shorting” effect at the input and output of an amplifier causes a reduction
in the gain of the amplifier.
• For common emitter BJT circuits, Miller effect capacitance will affect the high
frequency response of the circuit, since it is an inverting amplifier.

• The figure below shows the RC network which affects the frequency response of
BJT circuits at high frequencies.
Cc
IRC
Cwo
IB C
E
B
VO= Output
voltage
RE
Vi =
Input voltage
CS
RC
VCC
RB2
RB1
IRB2
IRB1
Ce
Zi ZoZix Zox
IRL
CwiVs =
Source
voltage
Rsig
Ii
Cbc
Cce
Cbe
Cbe = capacitance between the base and emitter of transistor
Cce = capacitance between collector and emitter of transistor
Cbc = capacitance between base and collector of transistor
Cwi = wiring capacitance at input of amplifier
Cwo = wiring capacitance at output of amplifier

• The figure below shows the ac equivalent circuit of the BJT amplifier in the
preceding slide.
• At mid and high frequencies, Cs, Cc, and Ce are assumed to be short circuits
because their impedances are very low.
• The input capacitance Ci includes the input wiring capacitance (Cwi), the
transistor capacitance Cbe, and the input Miller capacitance CMi.
• The output capacitance Co includes the output wiring capacitance (Cwo), the
transistor parasitic capacitance Cce, and the output Miller capacitance CMo.
• Typically, Cbe is the largest of the parasitic capacitances while Cce is the
smallest
E
Ci
IcIb
Vo=Vce
Ri  re
= re
ro
Zix Zox
Vi
 Ib
E
RL
B
Ii
CoRC
RiRB1// RB2
Vs
Rsig
Ci = Cwi + Cbe + CMi Co = Cwo + Cce + CMo
Thi Tho

• The Thevenin equivalent circuit of the ac equivalent circuit of the BJT amplifier
is shown below.
• For the input side, the -3db high cutoff frequency can be computed as:
Vo=Vce
Vi
Co
Ri
RThi = Rsig // RB1// RB2 // Ri
VThi
Thi Tho
VTho
Ci
RTho= Rc // RL// ro
circuitofecapacitancinputAv)C-(1CCCCCC
sideinputatresistanceequivalentThevenin
1)re(//R//R//RsigRi//R//R//RsigR
frequency)(-3dbsideinputfor thefrequencyoffcuthigher
CiR2
1
f
bcbewiMibewii
B2B1B2B1THi
Thi
Hi







• At the high frequency end, the reactance of capacitance Ci will decrease as
frequency increases, resulting to reduction in the total impedance at the input
side.
– This will result to lower voltage across Ci, resulting to lower base current,
and lower voltage gain.
• For the output side, the -3db high cutoff frequency can be computed as:
• At the high frequency end, the reactance of capacitance Co will decrease as
frequency increases, resulting to reduction in the total impedance at the output
side.
– This will result to lower output voltage Vo, resulting to lower voltage and
power gain.
circuitofecapacitancoutputC
Av
1
-1CCCCCC
sideoutputatresistanceequivalentThevenin//roR//RcR
frequency)(-3dbsideoutputfor thefrequencyoffcuthigher
CoR2
1
f
bccewoMocewoo
LTHo
Tho
Ho











• The Hybrid or Giacolletohigh frequency equivalent circuit for common
emitter is shown below.
• The resistance rb includes the base contact resistance (due to actual connection
to the base) , base bulk resistance (resistance from external base terminal to the
active region of transistor), and base spreading resistance (actual resistance
within the active region of transistor).
• The resistances r, ro, and ru are the resistances between the indicated terminals
when the BJT is in the active region.
• Cbe and Cbc are the capacitances between the indicated terminals.
E
Cu = Cbc
Ic
Ib
ro =
1 / hoe
Zix ZoxE
B
C = Cber  re
rb
C
ru
 Ib =
hfe Ib
Hybrid  High Frequency Equivalent Circuit (Common Emitter)

• At the high frequency end, hfe of a BJT will be reduced as frequency increases.
• The variation of hfe (or ) with frequency can approximately be computed as:
   
 
rtransistoofcurrentemitterDCI
I
26mV
r
))(r(hferr
CuCr2
11
f
CuCr2
1
hfe
1
CuCr2
1
ff
sheet)specsatgivenusuallyone(thefrequencymiddleathfehfe:where
ffrequencyathfe
f
f
j1
hfe
hfe
E
E
e
emidemid
mid
β
mid
hfeβ
midmid
β
mid



















e
e

• Since re is a function of the DC emitter current IE, and f is a function of re, f is a
function of the bias condition of the circuit.
• hfe will drop off from its midband value with a 6 db / octave slope.
• For the common base configuration:
• It has improved frequency response compared to common emitter configuration.
• Miller effect capacitance is not present because of its non-inverting
characteristics.
• f is higher than f.
f
Normalized hfe
in db
Bode Plot for hfe () in the High Frequency Region
0 db
-3 db
-6 db / octave (for f)
hfe / hfe mid

• The relationship of f (-3db high cutoff frequency for ) and f db high
cutoff frequency for ) is shown below.
• The upper cutoff frequency of the entire system (upper limit for the bandwidth)
is lower than the lowest upper cutoff frequency (lowest among fHi, fHo, and f)
• The lowest upper cutoff frequency has the greatest impact on the bandwidth of
the system. It defines a limit for the bandwidth of the system.
• The lower is the upper cut off frequency, the greater is its effect on the
bandwidth of the entire system.
 forfrequencyoffcutdb3)1(ff 
f
Normalized hfb
in db
Bode Plot for hfb () in the High Frequency Region (Common Base)
0 db
-3 db
-6 db / octave (for f)
hfb / hfb mid f

• The gain-bandwidth product of a transistor is defined by the following condition:
 
 
productbandwidthgain
CuCr2
1
f
CuCr2
11
)())(f(f
bandwidth
f
f
bandwidthfandgainsinceproduct,bandwidthgain))(f())(fhfe(f
1
f
f
hfe
f
f
1
hfe
:ascomputedis)f(fwhenhfeofmagnitudetheand
fbydenotedis0dbtoequalishfeat whichfrequencyThe
db01log20
f
f
j1
hfe
log20hfeand1
f
f
j1
hfe
hfe
T
mid
midβmidT
mid
T
β
βmidβmidβmidT
β
T
mid
2
β
T
mid
T
T,db
β
mid
db
β
mid

































e
e

• Example: Given a common emitter BJT amplifier with the following parameters,
determine the following:
a. High cutoff frequency for the input of the circuit (fHi)
b. High cutoff frequency for the output of the circuit (fHo)
c. High cutoff frequency for f
d. Gain bandwidth product (fT)
 e. Sketch the frequency response for the low and high frequency range
 Specs similar to example on BJT low frequency response:
Cs=12 uF Rsig = 2 kohm RL= 2 kohm RB1=50 kohm RB2= 10 kohm
Cc=2 uF RE = 2 kohm RC= 4 kohm Vcc= 20 volts
CE=20 uF  = hfemid = 100 ro = infinite
Additional specs:
C = Cbe= 35 pF Cu = Cbc= 3 pF Cce = 1 pF
Cwi = 5 pF Cwo = 6 pF

response)frequencylow(foramplifierofimpedanceinputohms1,596.08RiZi
Rs)gconsiderin(notsfrequenciemidatgainvoltage538.67Av
742.19r
responsefrequencyhighforRi2.974,1)742.19(100)(βr
response,frequencylowonexampleprevioustheFrom
mid
e
e




CeinvolvingcircuitoffrequencycutofflowerHz225.822f
CcinvolvingcircuitoffrequencycutofflowerHz263.13f
Csinvolvingcircuittheofportionfor thefrequencyoffcutlowerHz3.688f
Vs)ofresistance(internalRsgconsideringainvoltage9.9872Avs
LE
LC
LS
mid




ohms678.887
2.974,1
1
000,10
1
000,50
1
000,2
1
1
Ri//R//R//RsigR
pF614.452pF3(-67.538))-(1pF35pF5
CbcAv)-(1CbeCwiCi
:responsefrequencyhighFor the
B2B1Thi







Hz359,884,11
)10x)(10.04433.333,1(2
1
)(Co)R(2
1
f
pF044.01pF3
538.67
1
1pF1pF6
Cbc
Avmid
1
1CceCwoCCceCwoCo
ohms33.333,1
2,0004,000
000)(4,000)(2,
R//RcR
Hz981,729
)10x614.45)(2678.887(2
1
)(Ci)R(2
1
f
12-
Tho
Ho
Mo
LTho
12-
Thi
Hi


















   
 
1hfewhenfrequencyHz800,092,215)892,150,2)(001())(fhfe(f
Hz892,150,2
x103x103519.472)(2
1
(100)
1
CuCr2
1
hfe
1
CuCr2
1
ff
βmidT
1212
mid
hfeβ










  e

• In the low frequency region, the lower cutoff frequency due to the emitter capacitor
(fLE) has the highest value. Consequently, it has the greatest impact on the bandwidth of
the system, among the three lower cutoff frequencies.
• In the high frequency region, the high cutoff frequency due to the input capacitors and
resistors (fHi) has the lowest value. Consequently, it has the greatest impact on the
bandwidth of the system, among the three high cutoff frequencies.
Av / Avmid (db)
Normalized
Voltage
gain
Frequency
0.707 AVmid
1 AVmid
fLEfLC
Full Frequency Response (Normalized Voltage Gain Versus Frequency)
0 db
-3 db
-5 db
-25 db
-20 db
-15 db
-10 db
1 10 100 1K 10K 100K 1M 10M 100M
fLS fHi fHof
Bandwidth
- 20 db / decade
(-6 db / octave)
+20 db / decade
(6 db/octave)

FET High Frequency Response
• The high frequency response analysis for FET is similar to that of BJT.
• At the high frequency end, the high cutoff frequency (-3 db) of FET circuits is
affected by the network capacitance (parasitic and induced).
• The capacitances that affect the high frequency response of the circuit are
composed of:
– the interelectrode capacitance between the gate and source, gate and drain,
and drain and source.
– Wiring capacitance at the input and output of the circuit.
• At high frequencies, the reactance of the interelectrode and wiring capacitance
become significantly low, resulting to a “shorting” effect across the
capacitances.
• The “shorting” effect at the input and output of an amplifier causes a reduction
in the gain of the amplifier.
• For common source FET circuits, the Miller effect will be present, since it is an
inverting amplifier.

• The figure below shows the RC network which affects the frequency response of
FET circuits at high frequencies.
Cc
IRD
Cwo
IG D
S
G
VO= Output
voltage
RS
Vi =
Input voltage
CG
RD
VCC
RG2
RG1
IRG2
IRG1
CS
Zi ZoZix Zox
IRL
CwiVs =
Source
voltage
Rsig
Ii
Cgd
Cds
Cgs
Cgs = capacitance between the gate and source of transistor
Cds = capacitance between drain and source of transistor
Cgd = capacitance between gate and drain of transistor
Cwi = wiring capacitance at input of amplifier
Cwo = wiring capacitance at output of amplifier
Common Source FET Amplifier Circuit

• The figure below shows the ac equivalent circuit of the FET amplifier.
• At mid and high frequencies, CG, CS, and Cc are assumed to be short circuits because
their impedances are very low.
• The input capacitance Ci includes the input wiring capacitance (Cwi), the transistor
capacitance Cgs, and the input Miller capacitance CMi.
• The output capacitance Co includes the output wiring capacitance (Cwo), the
transistor parasitic capacitance Cds, and the output Miller capacitance CMo.
• Typically, Cgs and Cgd are higher than Cds.
• At high frequencies, Ci will approach a short-circuit and Vgs will drop, resulting to
reduction in voltage gain.
• At high frequencies, Co will approach a short-circuit and Vo will drop, resulting to
reduction in voltage gain.
S
Ci
Id
Vo=Vdsrd
Zix Zox
Vi = Vgs
gm Vgs
S
RL
G
Ii
CoRD
RG1// RG2
Vs
Rsig
Ci = Cwi + Cgs + CMi Co = Cwo + Cds + CMo
Thi Tho
D
IRL

• The Thevenin equivalent circuit of the ac equivalent circuit of the FET amplifier
is shown below.
• For the input side, the -3db high cutoff frequency can be computed as:
Vo=Vds
Vi
Co
RThi = Rsig // RG1// RG2
VThi
Thi Tho
VTho
Ci
RTho= RD // RL// rd
 
scenariocaseworstget thetoAvforusedisAvmidwhere
sideinputatecapacitanceffectMillerCgdAv1C
circuitofecapacitancinputCCgsCC
sideinputatresistanceequivalentTheveninR//R//RsigR
frequency)(-3dbsideinputfor thefrequencyoffcuthigh
CiR2
1
f
Mi
Miwii
G2G1THi
Thi
Hi






• For the output side, the -3db high cutoff frequency can be computed as:
scenariocaseworstget thetoAvforusedisAvmid
sideoutputat theecapacitanceffectMillerCgd
Av
1
1C
circuitofecapacitancoutputCCdsCC
sideoutputatresistanceequivalentThevenin//rdR//RR
frequency)(-3dbsideoutputfor thefrequencyoffcuthigh
CoR2
1
f
Mo
Mowoo
LDTHo
Tho
Ho












• Example: Given a common source FET amplifier with the following parameters,
determine the following:
a. High cutoff frequency for the input of the circuit (fHi)
b. High cutoff frequency for the output of the circuit (fHo)
c. Sketch the frequency response for the low and high frequency range
 Specs similar to example on FET low frequency response:
CG =0.02mF Cc = 0.6 mF Cs = 2 mF
Rsig = 12 K RG= 1 Mohm RD= 5 K Rs= 1 K RL = 2 K
IDSS= 9 mA Vp= -7 volts rd= infinity VDD = 20 volts
Since RG1 is not present, configuration is self bias FET.
Additional specs:
Cgd= 3 pF Cgs = 5 pF Cds = 1 pF
Cwi = 5 pF Cwo = 6 pF

CsinvolvingcircuittoduefrequencycutofflowerHz851f
CcinvolvingcircuittoduefrequencycutofflowerHz37.89f
Cinvolvingcircuittoduefrequencycutofflower7.86Hzf
Rs)gconsiderin(notsfrequenciemidatgainvoltage9.1Av
response,frequenclowonexampleprevioustheFrom
LS
LC
GLG
mid




Hz800,717
)10x8.71)(857,11(2
1
)(Ci)R(2
1
f
ohms857,11
000,000,1000,12
)000,000,1)(000,12(
R//RsigR
pF7.18pF3(-1.9))-(1pF5pF5
CgdAv)-(1CgsCwiCi
:responsefrequencyhighFor the
12-
Thi
Hi
GThi









Hz605,621,9
)10x579.1)(157.428,1(2
1
)(Co)R(2
1
f
pF579.11pF3
9.1
1
1pF1pF6
Cgd
Avmid
1
1CdsCwoCCdsCwoCo
ohms57.428,1
2,0005,000
000)(5,000)(2,
R//RR
12-
Tho
Ho
Mo
LDTho



















• In the low frequency region, the lower cutoff frequency due to the source capacitor (fLS)
has the highest value. Consequently, it has the greatest impact on the bandwidth of the
system, among the three lower cutoff frequencies.
• In the high frequency region, the high cutoff frequency due to the input capacitors and
resistors (fHi) has the lowest value. Consequently, it has the greatest impact on the
bandwidth of the system, among the two high cutoff frequencies.
Av / Avmid (db)
Normalized
Voltage
gain
Frequency
0.707 AVmid
1 AVmid
fLSfLC
Full Frequency Response (Normalized Voltage Gain Versus Frequency)
0 db
-3 db
-5 db
-25 db
-20 db
-15 db
-10 db
1 10 100 1K 10K 100K 1M 10M 100M
fLG fHi fHo
Bandwidth
-20 db / decade
(-6 db / octave)
+20 db / decade
(6 db / octave)

Frequency Response of Multistage (Cascaded) Amplifiers
• If there are several stages in a cascaded amplifier system, the overall bandwidth
of the system will be lower than the individual bandwidth of each stage.
• In the high frequency region:
• The output capacitance Co must now include the wiring capacitance (Cwi),
parasitic capacitance (Cbe or Cgs), and input Miller capacitance (CMi) of the
next stage.
• The input capacitance Ci must now include the wiring capacitance (Cwo),
parasitic capacitance (Cce or Cds), and input Miller capacitance (CMO) of the
preceding stage.
• The lower cutoff frequency of the entire system will be determined primarily by
the stage having the highest lower cutoff frequency.
• The upper cutoff frequency of the entire system will be determined primarily by
the stage having the lowest higher cutoff frequency.
• For n stages having the same voltage gain and lower cutoff frequency (f1), the
overall lower cutoff frequency (f1’) can be computed as:
stagesofnumbern
stageeachoffrequencycutofflowerf1:where
amplifierentiretheoffrequencycutoffloweroverall
12
f1
f1'
1/n






Frequency Response of Multistage (Cascaded) Amplifiers
• For n stages having the same voltage gain and higher cutoff frequency (f2),
the overall higher cutoff frequency (f2’) can be computed as:
stagesofnumbern
stageeachoffrequencycutoffhigherf2:where
amplifierentiretheoffrequencycutoffhigheroverall12f2f2' 1/n




Square Wave Testing
• A square wave signal can be used to test the frequency response of single
stage or multistage amplifier.
• If an amplifier has poor low frequency response or poor high frequency
response, the output of the amplifier having a square wave input will be
distorted (not exactly a square wave at the output).
• A square wave is composed of a fundamental frequency and harmonics
which are all sine waves.
• If an amplifier has poor low or high frequency response, some low or high
frequencies will not be amplified effectively and the output waveform will be
distorted.

Square Wave Testing
• The figures below show the effect of poor frequency response of an
amplifier using a square wave input.
V
t
V
t
t
t
VV
No distortion
(Good Frequency Response)
Poor High Frequency
Response
Poor Low Frequency
ResponsePoor High and Low
Frequency Response
tilt
long rise time
tilt
V
t
t
V
Very Poor High Frequency
Response
Very Poor Low Frequency
Response
tilt
very long rise time

Square Wave Testing
• The high cutoff frequency can be determined from the output waveform by
measuring the rise time of the waveform.
• Rise time is between the point when the amplitude of the waveform is 10 %
of its highest value up to the point when the amplitude is 90 % of its
highest value.
• The high cutoff frequency can be computed as:
• The lower cutoff frequency can be determined from the output waveform
by measuring the tilt of the waveform.
(seconds)timerisetr:where
)ftoequalelyapproximatis(bandwidthamplifierofBandwidth
tr
0.35
fBW
(Hz)amplifieroffrequencycutoffupper
tr
0.35
f
HH
H



(Hz)wavesquareoffrequencyfs
(unitless)tilt
V
V'-V
P:where
(Hz)amplifieroffrequencycutofflowerfs
P
fLO




t
V tilt
V
V’
Rise time (tr)

Square Wave Testing
• Example: The output waveform of an amplifier with a 4 Khz square wave
input has the following characteristics:
Rise time = 15 microseconds Maximum amplitude (V) = 40 millivolts
Minimum voltage of tilt (V’) = 30 millivolts
Determine: high cutoff frequency, bandwidth, low cutoff frequency.
Hz23,333.33
15x10
0.35
fBW
Hz23,333.33
15x10
0.35
tr
0.35
f
6-
H
6-
H


frequencycutofflowerHz318(4,000)
0.25
fs
P
f
(unitless)tilt0.25
10x40
10x30-10x40
V
V'-V
P
LO
3-
-3-3




Bjt+and+jfet+frequency+response

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