bjt and fet frequency response
DESCRIPTION
BJT and Fet Frequebcy responseTRANSCRIPT
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BJT and JFET Frequency Response
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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 gainof 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.
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Effects of Frequency on Operation of Circuits 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 f2BandwidthAvmid = voltage gain of amplifier at middle frequencies0.707 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 voltagef2= high cutoff frequency PO(LPF) = output power at lower cutoff frequency Pomid= output power at middle frequencies
Omid
2vmid
2vmid
O(LPF)(HPF) P 0.5Ro
Vi A5.0Ro
Vi 0.707APPo
Bandwidth = f2-f1
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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
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Effects of Frequency on Operation of Circuits 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 shiftbetween Voand Vi
Frequency
18002700
f1 f2
Phase shift between Vo and Vi
900
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Frequency Response of Amplifier Circuits 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 Gainin Ratio
frequency middleat gain voltageA ffrequency at gain voltageA :where
AAGain Normalized
Vmid
V
Vmid
V
Normalized Plot of Voltage Gain Versus Frequency
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Frequency Response of Amplifier Circuits A decibel plot of the gain can be made using the following formula:
Voltage gain
Frequency0.707 AVmid
1 AVmid
f1 f2
Normalized Gain in db
frequency middleat gain voltageA ffrequency at gain voltageA :where
dbin gain normalizedAA20log
AA
Vmid
V
Vmid
V
Vmid
V
db
Decibel plot of Normalized Voltage Gain Versus Frequency
0 db-3 db-6 db-9 db
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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 capacitanceof 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 frequencies0.707 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 frequencyf2= high cutoff frequency
Bandwidth
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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 transformerat low frequencies.
Drop in gain is due to stray capacitance at primary and secondary of transformer and other components, and frequency dependence of gainof active devices.
Bandwidth
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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
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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
IRVi =Input voltage to RC network
Capacitor Coupled Circuit Which Acts As A High Pass Filter
Vo = Outputvoltage
(Farad) Cc of ecapacitanc C (Hz) signal offrequency f :where
ffrequency at Cc of reactancefC21Xc
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Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit
1ViVo
R load theacross voltageViVo
sfrequenciehigh at Cc of reactance0fC2
1Xc
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).
0ViVo
R load theacross voltage0Vo
hz 0f when Cc of reactancefC2
1Xc
L
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Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit
2ViVo when sfrequencie middleat Rat dissipatedpower
sfrequencie middleat Power R
Vi R
Vi21
R1
2Vi
RVoP
XcR when tageoutput vol 0.707ViVo
0.707Vi2
ViRRR
ViRXcR
Vi)(R)I(Vo
Xc,R When below.shown as Xc R when occurs thisand s,frequencie middleat power output theof that half is Rat power output the
(f1),frequency cutoff low the toequal is signal theoffrequency When the
RXcR
Vi)(R)I(Vo
2222
2R
2R
222
2
The magnitude of the output voltage can be computed as:
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Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit
(Hz)frequency cutofflower RC21f1
f1C21XcR
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:
3dbA
0.707Alog20dbA
AVmid
Vmid
Vmid
cutoffVlower
db 0AAlog20db
AA
Vmid
Vmid
Vmid
Vmid
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Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit
(Hz)frequency cutofflower RC21 f1 :where
(unitless) ffrequency at gain voltage
ff1j1
1Av
fCR21j1
1
RjXc1
1jXcR
RjXcR I
R IViVoAv
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
ff11
1Av 12
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Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit
(db)
ff11
1log20ViVolog20Av
2db
In db (logarithmic form), the voltage gain at frequency f can be computed as:
When f=f1= lower cutoff frequency,
db 3-
f1f11
1log20ViVolog20Av
2db
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Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit
22/12
2db
ff11log10
ff11-20log
(db)
ff11
1log20ViVolog20Av
The voltage gain at frequency f can be written as:
When f
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Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit 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
f1Normalized 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
ff1log20X
(db)
ff11
1log20Av2
db
Actual Response Curve
Asymptote
Asymptote
- Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit The plot in the preceding slide shows two asymptotes. One for f> f1 (horizontal line 0 db). The plot of the line corresponding to f
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Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit
Example: For the RC network shown below, Determine the break frequency (cutoff frequency), sketch the asymptotes and the frequency response curve.
Frequency (log scale)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
IRLVi =Input voltage to RC network
Vo = Outputvoltage Hz 99.47
)0x12.0)(000,8(21
CR21f1 6
L
Asymptote
Asymptote
-3db point
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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 = Outputvoltage
RL
ViVs
Rsig
Cs
RB1
RE Ce
Zi Zo
Vs = Signal sourceRsig = internal resistance of signal sourceCs =coupling capacitor for VsCc= coupling capacitor for RLCe= bypass capacitor for RE
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Low Frequency Analysis of Capacitor Coupled BJT Amplifier 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 = RiVi
Equivalent Circuit of Vs, Cs and Zi
IiCs
Zi = Ri
Vi
Equivalent Circuit of Vs, Cs and Zi
Ii
RB1//RB2 hie = rere
Zi = Ri
Vs
Rsig
Vs
Rsig
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Low Frequency Analysis of Capacitor Coupled BJT Amplifier
CsjXRsigRiRi VsVi
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:RsigRiRi VsVi mid
Cs involving
circuit theofportion for thefrequency offcut lower Ri)Cs(Rsig2
1fLs
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Low Frequency Analysis of Capacitor Coupled BJT Amplifier For the portion of the circuit involving the coupling capacitor Cc, the
equivalent circuit is shown below. 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 RLib ro
Cc
Zo= Ro = Rc // ro
VRL
IRLRc // ro
RL
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Low Frequency Analysis of Capacitor Coupled BJT Amplifier The value of the output impedance (resistance) of the amplifier can be
computed as:Zo = Ro = RC // ro
The lower cutoff frequency can be computed as:
Cc involving
circuit theofportion for thefrequency offcut lower )CR(Ro2
1fCL
LC
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Low Frequency Analysis of Capacitor Coupled BJT Amplifier 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)current quiescent Emitter (Ampere)current DCEmitter I
(ohms) I
10 X 26r
//RRsig//RRs' :Where
(ohms) sideoutput thefrom R into lookingseen impedancerRs'//RRe
E
E
3-
e
B2B1
EeE
eE rRs'//RRe
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Low Frequency Analysis of Capacitor Coupled BJT Amplifier
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.output thefrom R into looking resistance equivalent Re :where
circuit theofportion for thefrequency offcut lower CeRe2
1f
E
LE
)considerednot (ro sfrequencie lowat amplifier theofgain voltageRr
Rc//RViV
ViVoAv
Ee
LRL
sfrequencie middleat amplifier theofgain voltager
ro//Rc//RViV
ViVoAv
e
LRL
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Low Frequency Analysis of Capacitor Coupled BJT Amplifier 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 lowercutoff frequency of the amplifier will be higher than the highest lower cutoff frequency due to the capacitors.
fLT = overall lower cutoff frequency of amplifierfLT > fLSfLT > fLCFlt > fLE
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Low Frequency Analysis of Capacitor Coupled BJT Amplifier 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 kohmCc=2 uF RE = 2 kohm RC= 4 kohm =100 Vcc= 20 voltsCE=20 uF Assume that output resistance of transistor to be infinite.
742.1910 x 1.317
10 x 26re
CurrentQuiescent Emitter
A10 x 1.3172000
7.0333.3R
VVRVI
ground torelative baseat voltageDC
volts3.33350,00010,000
0)(20)(10,00R R
R V V
done. becan ionsapproximat following theand Rohms 20,0000)(100)(2,00 R
3
3
3
E
BEB
E
REE
B1B2
B2CCB
B2 E
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Low Frequency Analysis of Capacitor Coupled BJT Amplifier
0.4442,0001,596.08
1,596.08RsigRi
RiVsVi
amplifier of impedanceinput ohms 1,596.08 2.974,1
1000,101
000,501
1r////RRRiZi
Rs) gconsiderin(not sfrequencie midat gain voltage538.67 742.19
000,2000,4)000,2)(000,4(
rRc//R
ViVoAv
:sfrequencie middleAt
2.974,1)742.19(100)(r
eB2B1
e
Lmid
e
-
Low Frequency Analysis of Capacitor Coupled BJT Amplifier
Cc involvingcircuit offrequency cutofflower
Hz 13.26310 X 2 )000,2(4,0002
1)CcR(Rc2
1f
:Cc of effects thegConsiderin
Cs involvingcircuit theofportion for thefrequency offcut lower Hz 3.688 10 X 21,596.08)1(2,0002
1Ri)Cs(Rsig2
1f
amplifier of impedanceinput ohms 1,596.08 2.974,1
1000,101
000,501
1r////RRRiZi
:Cs of effects thegConsiderin
Vs) of resistance (internal Rs gconsideringain voltage9.9872
0.444)(-67.538)(VsVi
ViVo
VsVoAvs
6-L
LC
6-LS
eB2B1
mid
-
Low Frequency Analysis of Capacitor Coupled BJT Amplifier
circuit.amplifier whole theoffrequency cutoff affect thetly predominan willfrequency cutofflower its Cc, and Cs todue
frequecies cutoff the tocomparedhigh relatively isC involvingcircuit theofportion theoffrequency cutofflower theBecause
Hz 225.82210 X 20 5.239)3(2
1CRe2
1f
ohms 5.2393742.91100
1612.903//2000re
'R//RRe
ohms 1612.903
000,101
000,501
20001
1//RRsig//RRs'
:C of effects thegConsiderin
E
6-E
LE
SE
B2B1
E
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Low Frequency Response of JFET Common Source Amplifier
CcID = Drain current(ac)
VGSVDSIG (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
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Low Frequency Response of JFET Common Source Amplifier 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 is shown below. Equivalent circuit assumes that the input impedance of the amplifier (Zi) is
purely resistive and is equal to Ri.
CG
Zi = Ri= RG1 // RG2Vi
Equivalent Circuit of Vs, CG and Zi
IiCG
Vi
Equivalent Circuit of Vs, CG and Zi
Ii
RG1 // RG2Vs
Rsig
Vs
Rsig
Zi = Ri= RG1 // RG2
Zi = Ri= RG1 // RG2
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Low Frequency Response of JFET Common Source Amplifier
CGjXRsigRiRi VsVi
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:
RsigRiRi VsVi mid
G
GLG
C involving
circuit theofportion for thefrequency offcut lower Ri)C(Rsig2
1f
-
Low Frequency Response of JFET Common Source Amplifier For the portion of the circuit involving the coupling capacitor Cc, the
equivalent circuit is shown below. 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
-
Low Frequency Response of JFET Common Source Amplifier 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
The lower cutoff frequency can be computed as:
Cc involving
circuit theofportion for thefrequency offcut lower )CR(Ro2
1fCL
LC
-
Low Frequency Response of JFET Common Source Amplifier 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
gm1Rs//Req
be willaboveequation theinfinity,rwhen
(ohms) sideoutput thefrom R into lookingseen impedance
//RRrr gm1Rs1
RsReq
d
S
LDd
d
Req
System
-
Low Frequency Response of JFET Common Source Amplifier
The low cut-off frequency of the portion of the circuit involving the bypass capacitor Cs can be computed as:
side.output thefrom R into looking resistance equivalent Req :where
circuit theofportion for thefrequency offcut lower Cs Req 2
1f
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
-
Low Frequency Response of JFET Common Source Amplifier Example: Given a common source FET amplifier with the following
parameters, determine the lower cutoff frequency of the amplifier.CG =0.02F Cc = 0.6 F Cs = 2 F Rsig = 12 K RG= 1 Mohm RD= 5 K Rs= 1 K RL = 2 KIDSS= 9 mA Vp= -7 volts rd= infinity VDD = 20 volts Since RG1 is not present, configuration is self bias FET.
negative) more goes V when reachedfirst is value(ThisA 10 x 2.9806I Choose
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GSDSSDQD
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22
-
Low Frequency Response of JFET Common Source Amplifier
3volts volts2.9806)1000)(10 x -(2.9806))(R-(IV 3S DGSQ
ctancetransconduSiemens x1033.19-3-1
9) x10(9 2
VV1
VI 2gm 3-
-3
P
GS
P
DSS
Hz 7.86 10 x )0.02000,000,1(12,0002
1)CR(Rsig2
1Ri)C(Rsig2
1f 6-GGG
LG
Hz 89.730.6x10 )000,2(5,0002
1)CR(R2
1)CR(Ro2
1f 6-CLDCL
LC
amplifier theoffrequency cutoff low on theimpact highest thehasit s,frequencie cutofflower three theoflargest theis f Since
Hz 185)x10(429.18)(22
1Cs Req 2
1f
LS
6-Ls
ohms 18.29410 x 33.1/11000
)10 x 33.1/1)(1000(gm1Rs//Req infinity,rd Since 3-
-3
-
Low Frequency Response of JFET Common Source Amplifier
sfrequencie midlleat gain voltage
9.12,0005,000
000)(5,000)(2,)10 x (1.33)//Rgm(RViVo Avmid 3-LD
Av / Avmid (db)Normalized Voltage gain
Frequency0.707 AVmid
1 AVmidfLSfLC
Normalized Gain in db
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 100MfLG 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 = Outputvoltage
Frequency (log scale)
0.707 AV
1 AV
f2Normalized Gain in db
Bode Plot for High Frequency Region
0 db
-3 db-6 db / octave
CAv = Vo / Vi
-
High Frequency Response of Low Pass RC Network The voltage gain of the low pass RC network can be computed as:
sfrequencie midat gain voltage the times0.707 isgain gewhen voltafrequency
(Hz)frequency cutoffhigh RC21 f2 :where
(unitless) ffrequency at gain voltage
f2fj1
1Av
ff21j1
1
fRC211j1
1
R1R1
fC21Rj1
1
fC21Rj1
1
jXcR1
1jXcR
jXc-jXcR I
(-jXc) IViVoAv
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
-
Miller Effect Capacitance The value of the Miller effect input capacitance can be computed as:
CMi
fMi
CMff
21
21
X1
Ri1
Zi1
ecapacitancinput effect Miller C Av) (1C
ecapacitancinput effect Miller of ReactanceXfC2 Av) (1
1C Av) (1
1Av) (1
XcfAv) (1
Xcf1
Ri1
Zi1
XcfAv) (1
Ri1
Zi1
XcfAv) Vi(1
RiVi
ZiVi
IIIiXcf
Av) Vi(1Xcf
Av ViViXcf
VoViI RiViI
ZiViIi
-
Miller Effect Capacitance 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
IiCM i= (1-AV)Cf Ri
-
Miller Effect Capacitance 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 sideof the amplifier.
Vo
+
-Vi
+
-Zo
Av =Vo / Vi
Cf I2
I1
Io
Ro
-
Miller Effect Capacitance The Miller effect output capacitance can be determined as follows:
ecapacitancoutput effect Miller of ReactanceC 1
C Av11
1
Av11
XcfIoVo
XcfAv11
VoIo
XcfAv11Vo
XcfAvVoVo
XcfViVoIo
:by edapproximat becan Io
largely sufficientusually is Ro because smallry usually ve is RoVobut
XcfViVo
RoVoIo
XcfViVoI
RoVoI
ZoVoIo
IIIo
Mof
21
21
-
Miller Effect Capacitance
ecapacitancoutput effect Miller CfC
:by edapproximat becan ecapacitancoutput effect Miller the1,an greater thmuch is AvWhen
ecapacitancoutput effect Miller Cf Av11C
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.
-
BJT High Frequency Response The figure below shows the RC network which affects the frequency response of
BJT circuits at high frequencies.
CcIRC
Cwo
IB C
E
B
VO= Output voltage
REVi =
Input voltage
CS
RC
VCC
RB2
RB1
IRB2
IRB1
Ce
Zi ZoZix Zox
IRL
CwiVs = Source voltage
Rsig
Ii Cbc
CceCbe
Cbe = capacitance between the base and emitter of transistorCce = capacitance between collector and emitter of transistorCbc = capacitance between base and collector of transistorCwi = wiring capacitance at input of amplifierCwo = wiring capacitance at output of amplifier
-
BJT High Frequency Response 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
BIi
CoRCRiRB1// RB2Vs
Rsig
Ci = Cwi + Cbe + CMi Co = Cwo + Cce + CMo
Thi Tho
-
BJT High Frequency Response 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=VceVi
Co
Ri
RThi = Rsig // RB1// RB2 // Ri
VThi
Thi Tho
VTho
CiRTho= Rc // RL// ro
circuit of ecapacitancinput Av)C-(1 C C C C C C
sideinput at resistance equivalentThevenin 1)re( // R// R // Rsig Ri // R// R // RsigR
frequency) (-3db sideinput for thefrequency offcut higher Ci R 2
1f
bcbewiMibewii
B2 B1B2 B1THi
ThiHi
-
BJT High Frequency Response 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.
circuit of ecapacitancoutput C Av1-1 C C C C C C
sideoutput at resistance equivalentThevenin //roR // RcR
frequency) (-3db sideoutput for thefrequency offcut higher Co R 2
1f
bccewoMocewoo
LTHo
ThoHo
-
BJT High Frequency Response The Hybrid or Giacolletohigh 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 terminalswhen the BJT is in the active region.
Cbe and Cbc are the capacitances between the indicated terminals.
E
Cu = Cbc
IcIb
ro = 1 / hoe
Zix ZoxE
B
C = Cber re
rbC
ru
Ib =hfe Ib
Hybrid High Frequency Equivalent Circuit (Common Emitter)
-
BJT High Frequency Response 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:
rtransistoofcurrent emitter DC I
I26mVr
))(r(hfe r r
CuC r 2
11 f
CuC r 21
hfe1
CuC r 21 f f
sheet) specsat given usually one (thefrequency middleat hfe hfe:where
ffrequency at hfe
ffj1
hfehfe
E
Ee
e midemid
mid
midhfe
mid mid
mid
e
e
-
BJT High Frequency Response 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.
Frequency (log scale)
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
-
BJT High Frequency Response 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.
for frequency offcut db 3)1(ff
Frequency (log scale)
f
Normalized hfbin 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
-
BJT High Frequency Response The gain-bandwidth product of a transistor is defined by the following condition:
productbandwidth gain CuC r 2
1f
CuC r 211)( ))(f( f
bandwidth ff
bandwidthf andgain since product,bandwidth gain ))(f( ))(fhfe(f
1
ff
hfe
ff1
hfe:as computed is )f(f when hfe of magnitude theand
fby denoted is 0db toequal ishfeat which frequency The
db 01log20
ffj1
hfelog20hfe and 1
ffj1
hfe hfe
T
midmidmidT
mid
T
midmidmidT
T
mid
2
T
mid
T
T, db
middb
mid
e
e
-
BJT High Frequency Response 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 fd. 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 kohmCc=2 uF RE = 2 kohm RC= 4 kohm Vcc= 20 voltsCE=20 uF = hfemid = 100 ro = infinite
Additional specs:C = Cbe= 35 pF Cu = Cbc= 3 pF Cce = 1 pFCwi = 5 pF Cwo = 6 pF
-
BJT High Frequency Response
response)frequency low(for amplifier of impedanceinput ohms 1,596.08 RiZiRs) gconsiderin(not sfrequencie midat gain voltage538.67 Av
742.19rresponsefrequency high for Ri2.974,1)742.19(100)(r
response,frequency lowon example previous theFrom
mid
e
e
Ce involvingcircuit offrequency cutofflower Hz 225.822fCc involvingcircuit offrequency cutofflower Hz 263.13 f
Cs involvingcircuit theofportion for thefrequency offcut lower Hz 3.688 fVs) of resistance (internal Rs gconsideringain voltage9.9872 Avs
LE
LC
LS
mid
ohms 678.887
2.974,11
000,101
000,501
000,21
1
Ri // R // R // RsigRpF 614.452pF 3 (-67.538))-(1 pF 35 pF 5
Cbc Av)-(1 Cbe Cwi Ci:responsefrequency high For the
B2B1Thi
-
BJT High Frequency Response
Hz 359,884,11)10 x )(10.04433.333,1(2
1)(Co)R(2
1f
pF 044.01pF 3538.67
11pF 1pF 6
CbcAvmid
11 Cce CwoC Cce Cwo Co
ohms 33.333,12,0004,000
000)(4,000)(2, R // Rc R
Hz 981,729)10 x 614.45)(2678.887(2
1)(Ci)R(2
1f
12-Tho
Ho
Mo
LTho
12-Thi
Hi
1hfewhen frequency Hz 800,092,215) 892,150,2)(001() )(fhfe(f
Hz 892,150,2 x103 x1035 19.472)( 2
1(100)
1
CuC r 21
hfe1
CuC r 21 f f
midT
1212
midhfe
e
-
BJT High Frequency Response 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
Frequency0.707 AVmid
1 AVmidfLEfLC
Normalized Gain in db
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 100MfLS 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.
-
FET High Frequency Response The figure below shows the RC network which affects the frequency response of
FET circuits at high frequencies.
CcIRD
Cwo
IG D
S
G
VO= Output voltage
RSVi =
Input voltage
CG
RD
VCC
RG2
RG1
IRG2
IRG1
CS
Zi ZoZix Zox
IRL
CwiVs = Source voltage
Rsig
Ii Cgd
CdsCgs
Cgs = capacitance between the gate and source of transistorCds = capacitance between drain and source of transistorCgd = capacitance between gate and drain of transistorCwi = wiring capacitance at input of amplifierCwo = wiring capacitance at output of amplifier
Common Source FET Amplifier Circuit
-
FET High Frequency Response 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 = Vgsgm Vgs
S
RL
GIi
CoRDRG1// RG2
Vs
Rsig
Ci = Cwi + Cgs + CMi Co = Cwo + Cds + CMo
Thi Tho
D
IRL
-
FET High Frequency Response 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=VdsVi
CoRThi = Rsig // RG1// RG2
VThi
Thi Tho
VTho
CiRTho= RD // RL// rd
scenario case worst get the toAvfor used is Avmid where
sideinput at ecapacitanceffect Miller Cgd Av1C
circuit of ecapacitancinput C Cgs C C
sideinput at resistance equivalentThevenin R// R // RsigR
frequency) (-3db sideinput for thefrequency offcut high Ci R 2
1f
Mi
Miwii
G2 G1THi
ThiHi
-
FET High Frequency Response For the output side, the -3db high cutoff frequency can be computed as:
scenario case worst get the toAvfor used is Avmid
sideoutput at the ecapacitanceffect Miller Cgd Av11C
circuit of ecapacitancoutput C Cds C C
sideoutput at resistance equivalentThevenin //rdR // RR
frequency) (-3db sideoutput for thefrequency offcut high Co R 2
1f
Mo
Mowoo
LDTHo
ThoHo
-
FET High Frequency Response 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 KIDSS= 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 pFCwi = 5 pF Cwo = 6 pF
-
FET High Frequency Response
Cs involvingcircuit toduefrequency cutofflower Hz 851fCc involvingcircuit toduefrequency cutofflower Hz 37.89f
C involvingcircuit toduefrequency cutofflower 7.86HzfRs) gconsiderin(not sfrequencie midat gain voltage9.1 Av
response, frequenc lowon example previous theFrom
LS
LC
GLG
mid
Hz 800,717)10 x 8.71)(857,11(2
1)(Ci)R(2
1f
ohms 857,11000,000,1000,12
)000,000,1)(000,12(
R // RsigRpF 7.18pF 3 (-1.9))-(1 pF 5 pF 5
Cgd Av)-(1 Cgs Cwi Ci:responsefrequency high For the
12-Thi
Hi
GThi
-
FET High Frequency Response
Hz 605,621,9)10 x 579.1)(1 57.428,1(2
1)(Co)R(2
1f
pF 579.11pF 39.1
11pF 1pF 6
CgdAvmid
11 Cds CwoC Cds Cwo Co
ohms 57.428,12,0005,000
000)(5,000)(2, R // R R
12-Tho
Ho
Mo
LDTho
-
FET High Frequency Response 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
Frequency0.707 AVmid
1 AVmidfLSfLC
Normalized Gain in db
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 100MfLG 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:
stages ofnumber n stageeach offrequency cutofflower f1 :where
amplifier entire theoffrequency cutofflower overall12
f1f1'1/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:
stages ofnumber n stageeach offrequency cutoffhigher f2 :where
amplifier entire theoffrequency cutoffhigher overall12 f2f2' 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 bedistorted (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
tt
VV
No distortion (Good Frequency Response)
Poor High FrequencyResponse
Poor Low FrequencyResponsePoor High and Low
Frequency Response
tilt
long rise time
tilt
V
t
t
V
Very Poor High FrequencyResponse
Very Poor Low FrequencyResponse
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) timerise tr :where
)f toequalely approximat is (bandwidthamplifier ofBandwidth tr
0.35fBW
(Hz)amplifier offrequency cutoffupper tr
0.35f
HH
H
(Hz) wavesquare offrequency fs
(unitless) tilt V
V'-V P :where
(Hz)amplifier offrequency cutofflower fsPfLO
t
V tilt
VV
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 millivoltsMinimum voltage of tilt (V) = 30 millivolts
Determine: high cutoff frequency, bandwidth, low cutoff frequency.
Hz 23,333.3315x10
0.35fBW
Hz 23,333.3315x10
0.35tr
0.35f
6-H
6-H
frequency cutofflower Hz 318(4,000)0.25fsPf
(unitless) tilt 0.2510 x 40
10 x 30-10 x 40 V
V'-V P
LO
3-
-3-3