cascade and cascode configurations
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EE2603-03 1
3. Cascade and Cascode configurations (12.2 - 12.5)
• Cascaded amplifiers • Cascode amplifiers • Darlington pair • Feedback pair
EE2603-03 2
Cascaded amplifiers Cascaded amplifiers consists of two or more stages of the basic BJT and/or FET circuits each of which utilize individual power supply VCC
Cascaded amplifiers are used to obtain higher voltage amplification, or matching of the input impedance with the transducer and matching the output impedances with the loading of the following stage.
Cascade amplifiers
+Vcc
-Vcc or "0"
Load RL
Transducer ROI1 I2 I3
Rin
Ro
Vin
Vo
EE2603-03 3
0.2K15k
VCC=12V
8.2k
b=100
47k
Vin
Rin
Iin
0.2K15k
VCC=12V
8.2k
b=100
47k VO
RO2Rin2
10k
IL
Example:1
.amplifierCascadeBJTBJTfolowingtheofii
A,vvA,R,RFind
1in
Li
1in
2ovT1in2o −==
Ω=====
+⎟⎠
⎞⎜⎝
⎛ +=+⎟⎟⎠
⎞⎜⎜⎝
⎛+
β==
===+
×=+
=
7.3mA03.7mV26
ImV26rmA03.7
k313.0V2.2)mA(I
7.0k2.0100
k37.11)mA(I7.0RR
I9.2V
k37.1115//47R,V9.24715
1512RR
RVV
CeC
CEB
CBB
B21
1CCBB
(a) Find re for both BJT
EE2603-03 4
(b) Find Rin, Ro, Av, and Ai of the cascade amplifier
32.9037.0345.0
037.036.0//2.8// 2
1
11 −==−=−==
kkk
rRR
vvA
einC
ino
vVoltage gain 1
kkrRivR eBinin
in 36.07.3100//37.11//1 =×=== βInput resistance 1
Output resistance 2
Voltage gain 2
Voltage gain overall ( ) ( ) 24.11358.12132.9AAvv
vv
vvA 2v1v
1o
2o
1in
1o
1in
2ovT =−×−=×=×==
0.2K15k
VCC=12V
8.2k
b=100
47k
Vin
Rin
Iin
0.2K15k
VCC=12V
8.2k
b=100
47k VO
RO2Rin2
10k
IL
k2.8RivR Co
oo ===
8.121037.0
10//2.8//
2−=−=−==
kkk
rRR
vvA
eLC
ino
v
87.4010
36.024.1135=
×−==
×⎟⎠⎞⎜
⎝⎛
===k
kRRA
R
Rvv
Rv
Rv
ii
ALinv
L
inino
inin
Lo
inL
iCurrent gain
EE2603-03 5
1k1M
1k
1k
12V
VO2
RO2Rin2
1k1M
1k
12V
Rin
VinIDSS=8mAVP=-8V
Example:2
.amplifierCascadeFETFETfolowingtheofvvA,R,RFindin
2ovTin2o −=
(a) Find gm of the FET ( )
( )
( ) ( )( )
mS235.1806.31
882
VV
1VI2
g
neglectedisitV8VbeyondisV94.20VButV94.20k1mA94.20orV06.3k1mA06.3V
94.20or06.32
88.17242
25657624I
064I24II16I64I8
I8648
8I
18VV
1II
IRIVbutk1R,mA8I,V8V
P
GS
P
DSSm
PGS
GS
D
D2DD
2DD
2D
2D
2
P
GSDSSD
DSDGSSDSSp
=⎟⎠
⎞⎜⎝
⎛−
−−
−
×−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−
−=∴
∴−=−=
−=×−−=×−=∴
=±
=−±−−
=
=+−⇒−+=
+−=⎟⎟⎠
⎞⎜⎜⎝
⎛
−
−−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=∴
−=−===−=
EE2603-03 6
1k1M
1k
1k
12V
VO2
RO2Rin2
1k1M
1k
12V
Rin
Vin IDSS=8mAVP=-8V
( ) 235.1k1mS235.1M1//k1gvv
A m1in
1o1v −=×−=−==Voltage gain 1
Ω== MRR Gin 1Input resistance 1
Ω== kRR Do 12Output resistance 2
Voltage gain 2
Voltage gain overall ( ) ( ) 76.06175.0235.1AAvv
vv
vv
A 2v1v1o
2o
1in
1o
1in
2ovT =−×−=×=×==
( ) 6175.0k5.0mS235.1k1//k1gvv
A m1in
1o2v −=×−=−==
(b) Find AVT of the FET-FET cascade amplifier
EE2603-03 7
0.2K15k
VCC=12V
8.2k
b=100
47k
Vin
Rin
Iin
VDD=12V
RS=2.4k
IDSS=8mAVP=8V
RL=2.4kVo
Ro
Rin
RG=1M
Example:3
.amplifierCascadeFETBJTfolowingtheofvvA,R,RFind
1in
2ovT1in2o −=
Ω=====
+⎟⎠
⎞⎜⎝
⎛ +=+⎟⎟⎠
⎞⎜⎜⎝
⎛+
β==
===+
×=+
=
7.3mA03.7mV26
ImV26rmA03.7
k313.0V2.2)mA(I
7.0k2.0100
k37.11)mA(I7.0RR
I9.2V
k37.1115//47R,V9.24715
1512RR
RVV
CeC
CEB
CBB
B21
1CCBB(a) Find re of the BJT
EE2603-03 8
( )
( )
( ) ( )
mSVV
VI
g
neglectedVVbeyondVVButVkmAorVkmAV
orI
IIIII
II
VVII
IRIVbutkRmAIVV
PGS
P
DSSm
PGS
GS
D
DDDDD
DD
PGS
DSSD
DSDGSSDSSp
94.0825.41
8821
281.15
1.154.229.625.44.277.1
29.677.152.11
04.264.4652.11
6.147421534.46
0644.4676.54.3876.5648
4.28648
84.2
181
4.24.2,8,8
22
222
=⎟⎠
⎞⎜⎝
⎛−
−−
−
×−=⎟
⎟⎠
⎞⎜⎜⎝
⎛−
−=∴
∴−=−=
−=×−−=×−=∴
=±
=−±−−
=
=+−⇒−+=
+−=⎟⎟⎠
⎞⎜⎜⎝
⎛
−
−−=⎟
⎟⎠
⎞⎜⎜⎝
⎛−=∴
−=−===−=
VDD
RS=2.4k
IDSS=8mAVP=8V
RL=2.4k
Vin
Vo
Ro
Rin
RG=1M
(b) Find gm of the FET
EE2603-03 9
0.2K15k
VCC=12V
8.2k
b=100
47k
Vin
Rin
Iin
VDD=12V
RS=2.4k
IDSS=8mAVP=8V
RL=2.4kVo
Ro
Rin
RG=1M
6.221037.0
1//2.8// 2
1
11 −=−=−==
kMk
rRR
vvA
einC
ino
vVoltage gain 1
kkrRivR eBinin
in 36.07.3100//37.11//1 =×=== βInput resistance 1
( ) Ω==== kkkkkgRR mSo 74.006.1//4.294.01//4.2/1//2
( ) 53.013.213.1
)4.2//4.2(94.01)4.2//4.2(94.0
//1)//(
2
22 ==
+=
+==
kkkk
RRgRRg
vvA
LSmLSm
ino
v
Output resistance 2
Voltage gain 2
Voltage gain overall 45.11753.06.221AAvv
vv
vvA 2v1v
1o
2o
1in
1o
1in
2ovT −=×−=×=×==
Ω== 1M R R Gin2
(c) Find AVT of the BJT-FET cascade amplifier
EE2603-03 10
Cascode amplifiers
Cascode amplifiers
+Vcc
-Vcc or "0"
Load RL
Transducer RO
Rin
Ro
Vin
VoI1
I1
Cascode amplifiers consists of two or more stages of the basic BJT and/or FET circuits each of which utilize single power supply VCC for both stages
Cascaded amplifiers are used to obtain higher voltage amplification, or matching of the input impedance with the transducer and matching the output impedances with the loading of the following stage.
EE2603-03 11
1.1k4.7k
b=200Vin
Rin
IinVB1
VE1
VC1
5.6k
VCC=18V
1.8k
b=200
6.8k VO
RO2
10k
IL
VC2
VB2
VE2
IC1=IC2
Example:4 – (page 549 text)
amplifier.CascodeBJTBJTfolowingtheofii
A,vvA,R,RFind
in
Li
in
o2vTino2 −==
Ω===⇒==
=−
=⇒=++
×=
8.69.3
268.3
8.31.1
7.09.49.48.66.57.4
7.418
2121
11
mAmVrrmAII
mAk
VVIVkkk
kV
eeCC
CB
(a) Find re
32.224k0068.0k10//k8.1
rR//R
A)Cwith(CBisstageSeconde
LC2VB ===∴
1rrAloadasrwithCEisstageFirste
e1V2e −=−=∴
32.22432.2241212 −=×−=×== VVin
oVT AA
VVA
Ω=×== kkkrRR eBin 887.00068.0200//6.5//7.4// 1β Ω== k8.1RR 2C2o
(b) Find Ro2, Rin, AVT, Ai
89.1910
887.032.224−=
×−==
kk
RRAAL
inVTi
EE2603-03 12
Example:4 – (page 301 text) amplifier.CascodeBJTFETfolowingtheof
ii
A,vvA,R,RFind
in
Li
in
o2vTino2 −==
VCC=20V
1.1k
b=160
91k VO
RO2
10k
IL
VC2
VB2
VE2
ID=IC
RS=1.2k
IDSS=6mAVP=6V
Vin
Rin
RG=18k
Iin
(a) Find VGS and gm of FET
( )
( )
( ) ( )( )
mS586.1624.11
662
VV
1VI2
g
V24.1VisanswerCorrectneglectedisitV6VbeyondisV768.15VBut
V768.15k2.1mA89.153.3orV24.1k2.1mA78.33.3V
mA89.15ormA78.388.2
43.1732.2888.2
2.49880232.28I
049.86I32.28I44.1I32.22I44.149.86I6
I2.13.36366
6I2.13.3
16VV
1II
V3.3k91k18k1820whereV
I2.13.3RIVVbutk2.1R,mA6I,V6V
P
GS
P
DSSm
GS
PGS
GS
D
D2DD
2DD
2D
2D
2
P
GSDSSD
G
DSDGGSSDSSp
=⎟⎠
⎞⎜⎝
⎛−
−−
−
×−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−
−=∴
−=
∴−=−=
−=×−−=×−=∴
=±
=−±−−
=
=+−⇒−+=
+−−=⎟⎟⎠
⎞⎜⎜⎝
⎛
−
−−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=∴
=+
×=
−=−===−=
EE2603-03 13
(c) Find Ro2, Rin, AVT, Ai
(b) Find re
Ω==⇒== 88.6mA78.3mV26rmA1.2II eCD
Ω== kkkRin 1591//18Ω== kRR Co 1.12
144k0068.0k10//k1.1
rR//R
A)Cwith(CBisstageSeconde
LC2VB ===∴
0109.088.610586.1rgAloadasrwithCSisstageFirst 32em1V2e −=××−=−=∴ −
56.11440109.0AAVV
A 2V1Vin
2oVT −=×−=×==
34.2k10
k1556.1RRA
AL
inVTi −=
×−==
VCC=20V
1.1k
b=160
91k VO
RO2
10k
IL
VC2
VB2
VE2
ID=IC
RS=1.2k
IDSS=6mAVP=6V
Vin
Rin
RG=18k
Iin
EE2603-03 14
Darlington pair Type 2N999 – NPN Darlington-connected Silicon
Transistor Package
Parameter Test conditions Min Max
VBE IC=100mA 1.8V
hFE (βD) IC=10mA IC=100mA
4000 7000
70,000
BEEEB
BDBDE
EDB
BECCB
VRIV,II)1(I
,RR
VVI
+=
β=+β=
β+
−=
Q2
Q1
E
C
BQD
E
CB
Q2
Q1
E
C
B
VCC
IB
IE
RE
RB
IC
DC analysis of the Darlington pair
EE2603-03 15
Q2
Q1
E
C
B
18V
IB
IE
3.3MW
IC
RB
bD = 8000VBE = 1.6V
390W
Vin
VoRin
Ro
mA48.20A56.28001I8001I)1(I BBDE =µ×==+β=
amplifier.DarlingtonfolowingtheofV,V,I,IFind BEEB
V6.96.18VVV BEEB =+=+=
A56.23908000M3.3
6.118RR
VVI
EDB
BECCB µ=
Ω×+Ω
−=
β+
−=
V8390mA48.20RIV EEE =Ω×==
EE2603-03 16
RE
riRB
vo
Rin Ro
vin
iin ib CB
EbDib
ioIQ2
Q1
E
C
B
18V
iin
IC
RB
bD = 8000VBE = 1.6VVin
Vo
Rin
Ro
RE
ioI
EDiBb
inBin
EDib
inEbDibin
i
oinb
Rr//RiV//RRbut
RriVRiriVand
rVVi
β+==
β+=⇒β+=−
=
AC analysis of the Darlington pair
( )( )
( ) [ ]E
EDB
E
EDiB
EDi
ED
E
inv
E
inino
inin
Eo
in
oi R
R//RR
Rr//RRr
RRRA
RRV/V
R/VR/V
iiA
β=
β+×⎥⎥⎦
⎤
⎢⎢⎣
⎡
β+
β==
×===
[ ] )Rrif(1Rr
RRriRi
vvA EDi
EDi
ED
EDib
EbD
in
ov β<<=
β+
β=
β+
×β==
D
iE
bD
ibE
bD
oE
o
oE
oo
r//Riri
//Ri
v//Ri
v//RIvR
β=
β−
−=
β−=
−==
EE2603-03 17
Q2
Q1
E
C
B
18V
IB
IE
3.3MW
IC
RB
bD = 8000VBE = 1.6V
390W
Vin
VoRin
Ro
ri = 5kW
amplifier.DarlingtonfolowingtheofiiA,
vvA,R,RFind
in
oi
in
ovino ==
( ) Ω==×+=β+== M605.1M125.3//M3.3k39.08000k5//M3.3Rr//RiV//RR EDiBb
inBin
4107k39.0
k1605998.0RRA
iiA
E
inv
in
oi =
×===
1998.0k3125k5
k3125k39.08000k5
k39.08000Rr
RA
EDi
EDv ≈=
+=
×+
×=
β+
β=
Ω=Ω==β
= 625.0625.0//k39.08000
k5//k39.0r//RRD
iEo
EE2603-03 18
Feedback pair Q2
Q1
C
E
BbD
C
E
B
VCC
IB
IE
RC
RB
IC
Q2
Q1
C
E
B
IB2
[ ]
( )7.0RIVVRIVRIV
III1I,II
,RR
VVI
ECCC1BECCCCBBB
B212B22B2EB12B
21DCDB
1EBCCB
−−=−−==
ββ=β≈+β=β=
β×β=ββ+
−=
DC analysis of the Feedback pair
EE2603-03 19
amplifier.pairFeedbackfolowingtheofV,V,I,IFind BCEB
18V
IB
IE
RC
RB
IC
Q2
Q1
C
E
B
IB2
2MW
75Wb1=140b2=180
( ) mA1.112A45.425200IIII1I BDB212B22B2E =µ×=β=ββ=β≈+β=
V6.975mA1.11218RIVV CCCCC =Ω×−=−=
[ ]2520180140A45.47525200M2
7.018RR
VVI D
CDB
1EBCCB =×=βµ=
Ω×+
−=
β+
−=
V9.8M2A45.4RIV BBB =Ω×µ==
EE2603-03 20
RC
ri1RB vo
Rin Ro
vin
iin ib C1B1
E1 b1ib
IRC I
C2B2
E2 b2ib2
ib2ri2 -b2ib2
Q2
Q1
E
C
B
VCC
iin
RB
b1 = 140
Vin
Vo
Rin
Ro
RE
iob2 = 180RC
( ) ( ) ( )
( ) C21BC211iBb
inBin
C211ibC2b11ibb12b2C1ibin
R//RRr//RiV//RRbut
RriR1iriiiRriV
ββ=ββ+==
ββ+=+ββ+=β+β−+=
AC analysis of the Feedback pair
( )( )
( ) [ ]
( ) ( )C
C21B
C21
C21B21
C
C21B
C21i
C21
C
inv
C
inino
inin
Co
in
oi
RR//R
RR//R
RR//R
RrR
RRA
RRV/V
R/VR/V
iiA
ββ=
ββ
ββββ=
ββ×⎥⎦
⎤⎢⎣
⎡
ββ+
ββ=
−=
×−=
−==
[ ] )Rrif(1Rr
RRriRi
vvA C21i
C21i
C21
C21ib
Cb21
in
ov ββ<<=
ββ+
ββ=
ββ+
×ββ==
Ai of the Feedback pair
Rin of the Feedback pair
Av of the Feedback pair
EE2603-03 21
RC
ri1RB vo
Ro
vin=0
ib C1B1
E1 b1ib
I
C2B2
E2 b2ib2
ib2ri2 -b2ib2
IRC
( )( )
21
1iC
b21
1ibC
b21
oC
RC
o
RC
oinoo
b21RC2b1RC
b1b12RCb12b2RC
r//Ri
ri//R
iv//R
IIv//
Iv
I)0Vat(vR
iI1iIiiIiiII
ββ=
ββ−
−=
ββ−=
−=
==
ββ−=−ββ−=
β+β−×β+=β+β+=
Ro of the Feedback pair
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