9. BJT and FET Frequency Response
9. BJT and FET Frequency Response
Logarithms
Decibels
P
G = log 2
P1
(bel)
P2
V22 /R 2
V
P2
G dB = 10 log
= 10 log
= 20 log 2 (R1 = R 2 )
G dB = 10 log
(decibel or dB)
2
P1
V1
V1 /R1
P1
P2: output power level
P1: reference or input power level
In electronic communication or audio equipment, the reference power level is generally accepted to be
1 mW associated with resistance of 600  (chosen because of its characteristic impedance of audio
transmission lines).
P
G dBm = 10 log 2 600 (dBm)
1 mW
A cascade system:
A VT = A V1 A V2 ..... A Vn → G V = 20log A VT = 20log A V1 + 20log A V2 + ..... + 20log A Vn
(dB)
= G V1 + G V2 + ..... + G Vn (dB)
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9. BJT and FET Frequency Response
Frequency Response
The corresponding frequencies f1 and f2 are generally called the corner, cutoff, band, break, or halfpower frequencies (HPF).
Bandwidth (BW) = f2 – f1
Po mid =
2
Vo
Ro
=
A Vmid Vi
2
PoHPF =
Ro
0.707A Vmid Vi
Ro
2
= 0.5
A Vmid Vi
Ro
2
= 0.5 Po mid
AV
AV
dB = 20 log
A Vmid
A Vmid
Low-Frequency Analysis
XC =
Vo =
1
1
=
ωC 2fC
R
1
R+
j2 fC
Vi → A V =
Vo
=
Vi
1
1
→ AV =
XC
1− j
X
1+  C
R
 R



2
X
1
= 0.707 → C = 1
R
2
1
1
→ XC =
= R → f1 =
2 f1C
2RC
f = f1 → A V =
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9. BJT and FET Frequency Response
AV =
Vo
=
Vi
1
1
1
1
f 
=
=
=
 tan −1  1 
XC
1
f1
2
f 
1− j
1− j
1− j
f 
1+  1 
2fCR
f
R
f 
f
0.001f1
0.01 f1
0.1 f1
f1
10 f1
100 f1
1000 f1
f1 =
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Av
10-3
10-2
10-1
0.707
1
1
1

90
89
84
45
6
1
0
Av
dB
-60
-40
-20
-3
0
0
0
1
= 318.5 Hz
2(5kΩ )(0.1F)
9-3
9. BJT and FET Frequency Response
Low-Frequency Response - BJT Amplifier
CS
When we analyze the effects of CS, we must make the assumption that CE and CC are too unwieldy, that
the magnitude of reactances of CE and CC permits employing a short-circuit equivalent.
R i = Zi = R1 R 2 βre
1
f LS =
2Π(R s + R i )CS
CC
R o = Zo = ro R C
1
f LC =
2Π(R o + R L )CC
CE
R s' = R s R1 R 2
 R s'

Re = RE 
+ re 
 β



1
fLE =
2 R e C E
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9. BJT and FET Frequency Response
Midband Frequency Response – BJT Amplifier
At the midband frequency level the short-circuit equivalents for the capacitors (CS, CC, and CE) can be
inserted.
R R
V
A Vmid = o = − L C
Vi
re
Sketch the lower cutoff frequency response using the following parameters:
CS = 10 F, CE = 20 F, CC = 1 F
RS = 1 k, R1 = 40 k, R2 = 10 k, RE = 2 k, RC = 4 k, RL = 2.2 k
 = 100, ro =  k, VCC = 20 V
Solution:
26mV
= 16.39 Ω
1.586mA
R i = 40kΩ 10kΩ (100x16.39 ) = 1.36 kΩ
I E = 1.586mA → re =
f LS =
1
= 6.74 Hz
2Π(1kΩ + 1.36kΩ )10 μF
R o = 4kΩ
f LC =
1
= 25.67 Hz
2Π(4kΩ + 2.2kΩ )1F
R s' = 1kΩ 40kΩ 10kΩ = 0.889 kΩ
 889 

R e = 2kΩ 
+ 16.39   = 24.96 Ω
 100

1
fLE =
= 318.82 Hz
2 (24.96Ω )(20 F)
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A
A
V
mid
Vs
=−
=
mid
2.2kΩ 4kΩ
16.39Ω
= −86.59
Z
i A
= −49.8
R + Z Vmid
s
i
9-5
9. BJT and FET Frequency Response
Low-Frequency Response - FET Amplifier
Ri = RG
R o = R D rd
1
f LG =
2Π R sig + R i CG
1
f LC =
2Π(R o + R L )CC
(
)
Midband Frequency Response – FET Amplifier
Vo
= −g m (R L R D )
Vi
Sketch the lower cutoff frequency response using the following parameters:
CG = 0.01 F, CC = 0.5 F, CS = 2 F
Rsig = 10 k, RG = 1 M, RD = 4.7 k, RS = 1 k, RL = 2.2 k
IDSS = 8 mA, VP = -4V, rd =  
1+
R s' =
R D R L + Rs
rd
gm
R eq = R S R s'
f LS =
1
2R eq CS
A Vmid =
Solution:
VGS = −2V → g m = 2 mS
f L G = 15.76 Hz
f L C = 46.13 Hz
f L S = 238.73 Hz
A Vmid = −3
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9. BJT and FET Frequency Response
Miller Effect Capacitance
Input Capacitance:
V − Vo Vi − A V Vi (1 − A V )Vi
Vi
I2 = i
=
=
Ri
X Cf
X Cf
X Cf
1
1
1
I i = I1 + I 2 →
=
+
X Cf
Zi R i
Ii =
Vi
Zi
I1 =
(1 − A V )
but
→
X Cf
(1 − A V )
=
1
1
=
= X CM
i
2f (1 − A V )C f 2fC Mi
1
1
1
=
+
→ Z i = R i X CM
i
Z i R i X CM
i
with C Mi = (1 − A V )C f
Output Capacitance:
V
Io = o
Zo
I1 =
Io = I1 + I 2 →
Vo
V − Vi
I2 = o
=
Ro
XCf
1
1
=
+
Zo R o
Vo −
Vo
AV
XCf

1 
1 −
Vo
A V 

=
XCf
1
XCf

1 
1 −

 AV 
but
XCf
1
=
=
1
= XCM
o
2fC M o


1 
1 
1 −
 2f 1 −
Cf
 AV 
 AV 
1
1
1
→
=
+
→ Zo = R o X C M
o
Zo R o X C M
o

1 
Cf
with C M o = 1 −
 AV 
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9. BJT and FET Frequency Response
High-Frequency Analysis
1
V
1
1
j2 fC
Vo =
Vi → A V = o =
→ AV =
1
2
Vi 1 + j R
+R
 R 


j2 fC
XC
1+ 

 XC 
1
R
= 0.707 →
=1
XC
2
1
1
→ XC =
= R → f2 =
2f 2C
2RC
f = f2 → AV =
AV =
Vo
=
Vi
1
1+ j
R
XC
=
1
=
1 + j2 fCR
1
1+ j
f
f2
=
 f 
 − tan −1  
2
 f2 
 f 
1 +  
 f2 
1
f
0.001f2
0.01 f2
0.1 f2
f2
10 f2
100 f2
1000 f2
Av
1
1
1
0.707
0.1
0.01
0.001

0
-1
-6
-45
-84
-89
-90
Av
dB
0
0
0
-3
-20
-40
-60
High-Frequency Response - BJT Amplifier
The various parasitic capacitances (Cbe, Cbc, Cce) of the transistor have been included with the wiring
capacitances (CWi, and CWo) introduced during construction. The capacitors CS, CC, CE are assumed to
be in the short-circuit state at these frequencies.
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9. BJT and FET Frequency Response
The input network:
R i = βre
R TH1 = R s R 1 R 2 R i
C i = C Wi + C be + C Mi = C Wi + C be + (1 − A V )C bc
f Hi =
1
2R TH1 C i
The output network:
R TH 2 = R C R L ro

1
Co = C Wo + Cce + C M o = C Wo + Cce + 1 −
 AV
1
fHo =
2R TH 2 Co

C bc

Sketch the higher cutoff frequency response using the following parameters:
CS = 10 F, CE = 20 F, CC = 1 F
RS = 1 k, R1 = 40 k, R2 = 10 k, RE = 2 k, RC = 4 k, RL = 2.2 k
 = 100, ro =  k, VCC = 20 V
Cbe = 36 pF, Cbc = 4 pF, Cce = 1 pF, CWi = 6 pF, CWo = 8 pF
Solution:
R TH1 = 1KΩ 40KΩ 10KΩ (100x16,39 Ω) = 0.576 KΩ
C i = 6pF + 36pF + (1 − (− 86.59 ))4pF = 392.36 pF
f Hi =
1
= 704.23 kHz
2Π(0.576KΩ )(392.36pF )
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R TH 2 = 4kΩ 2.2kΩ = 1.419 kΩ
1 

C o = 8pF + 1pF + 1 −
4pF = 13.05 pF
 − 86.59 
1
f Ho =
= 8.59 MHz
2(1.419kΩ )(13.05pF )
9-9
9. BJT and FET Frequency Response
High-Frequency Response - FET Amplifier
The input network:
The output network:
R TH 2 = R D R L rd
R TH1 = R sig R G
C i = C Wi + C gs + C Mi = C Wi + C gs + (1 − A V )C gd
1
f Hi =
2R TH1 C i

1
C o = C Wo + C ds + C M o = C Wo + C ds + 1 −
 AV
1
f Ho =
2R TH 2 C o

C gd

Sketch the higher cutoff frequency response
using the following parameters:
CG = 0.01 F, CC = 0.5 F, CS = 2 F
Rsig = 10 k, RG = 1 M, RD = 4.7 k,
RS = 1 k, RL = 2.2 k
IDSS = 8 mA, VP = -4V, rd =  
Cgd = 2 pF, Cgs = 4 pF, Cds = 0.5 pF,
CWi = 5 pF, CWo = 6 pF
Solution:
f H i = 945.67 kHz
f H o = 11.57 MHz
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