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Problem for Problems for Discrete Transistor Amplifiers
ECE 65, Winter2013, F. Najmabadi
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (2/42)
Exercise 1: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).
This is a common collector amplifier (emitter follower) . o Input at the base, output at the emitter.
It has a emitter-degeneration bias with a voltage divider.
When Rsig & vsig are not given, it implies that vsig = vi and Rsig = 0
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (3/42)
Bias circuit Thevenin form of the Voltage divider
V 95.49k 18k 22
k 22
k 90.9k 18||k 22
=×+
=
==
BB
B
V
R
Real circuit
Exercise 1 (cont’d): β = 200, VA = 150 V.
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (4/42)
Bias Calculations
A 3.20/ mA 05.4
107.0)1/(109.9 4.95
4.95 :KVLBE33
µβ
β
==≈=
+++×=
++=−
CB
CE
EE
EEBEBB
IIII
IIRIVRI
V 7.0 and 0 V, 7.0: Activein is BJT Assume
≥>= CECBE VIV
V 0.7 V 5 10104 9
9 :KVLCE
0
33
=>=××+=
+=−−
DCE
CE
EECE
VVV
RIV
V 95.4k 9.9 == BBB VR
k 28.1 r
k 0.3710x 4
150r
mA/V 156 1026104.05
3o
3
3
===
==≈
=××
==
mB
T
-C
A
-
-
T
Cm
gIVIVVIg
βπ
Thevenin form of the Voltage divider
Exercise 1 (cont’d): β = 200, VA = 150 V.
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (5/42)
11511
15115196510156)||||(
965k100||k1||k8.38)||||()||||(1
)||||(
3
≈+
=
=××=
Ω==+
=
−
i
o
LEom
LEo
LEom
LEom
i
o
vv
RRrgRRr
RRrgRRrg
vv
k 28.1 k, 8.38 mA/V, 156 === πom rrg
[ ][ ] k 9.9k194k3.1 ||k9.9
)||||( || ≈+=
+=
i
LEoBi
RRRrrRR βπ
Ω==+
= 4.6||||
||ββππ rR
RRrRR E
sigBEo
Exercise 1 (cont’d): β = 200, VA = 150 V.
Signal circuit (IVC = 0)
Amplifier Parameters Emitter Follower
1==×+
=i
o
i
o
sigi
i
sig
o
vv
vv
RRR
vv
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (6/42)
Ω==
4.6k9.9
o
i
RR
k 28.1k 8.38mA/V 156
===
π
o
m
rrg
Hz 39.31047.0)10100(6.4 2
1
)( 21
Hz 2.341047.0)0109.9( 2
1
)( 21
632
22
631
11
=×××+
=
+=
=××+×
=
+=
−
−
π
π
π
π
p
cLop
p
csigip
f
CRRf
f
CRRf
Hz 6.374.32.34 21 =+=+≈ ppp fff
Exercise 1 (cont’d): β = 200, VA = 150 V.
Amplifier Cut-off frequency 2 caps: 2 poles
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (7/42)
This is a common emitter amplifier with RE . o Input at the base, output at the collector.
It has a emitter-degeneration bias with a voltage divider.
Exercise 2: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (8/42)
Bias calculations Thevenin form of the Voltage divider
A 2.14/ mA 84.2
5107.0)1/(100.5 2.22
2.22 :KVLBE3
µβ
β
==≈=
+++×=
++=−
CB
CE
EE
EEBEBB
IIII
IIRIVRI
V 7.0 and 0 V, 7.0: Activein is BJT Assume
≥>= CECBE VIV
V 0.7 V 5.10 )51010(1084.215
15 :KVLCE
0
33
=>=+××+=
++=−−
DCE
CE
EECECC
VVV
RIVRI
k 83.1 r
k 8.5210 84.2
150r
mA/V 10.9 1026102.84
3o
3
3
===
=×
=≈
=××
==
mB
T
-C
A
-
-
T
Cm
gIVIVVIg
βπ
Exercise 2 (cont’d): β = 200, VA = 150 V
Bias circuit Caps open
V 22.215k 9.5k 34
k 9.5
k 0.5k 9.5||k 34
=×+
=
==
BB
B
V
R
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (9/42)
k 83.1 k, 8.52 mA/V, 10.9
===
πo
m
rrg
Amplifier Parameters CE amplifier with RE
k 8.4104k ||k 0.5)/1](/)||[(1
||
==
++
+=
i
EoLC
EBi
RrRrRR
RrRRπ
πβ
64.1
0.990kk 100||k 1||)/1](/)||[(1
)||(
−=
==+++
−=
i
o
LC
EoLCEm
LCm
i
o
vv
RRrRrRRRg
RRgvv
π
k 0.1
||1 ||
=≈
+++≈
Co
sigBE
EoCo
RR
RRRrRrRR
π
β
59.1)64.1(100800,4
800,4−=−×
+=×
+=
i
o
sigi
i
sig
o
vv
RRR
vv
Exercise 2 (cont’d): β = 200, VA = 150 V
Signal circuit (IVC = 0)
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (10/42)
k 0.1 k 8.4 == oi RR
Amplifier Cut-off frequency (2 caps: 2 poles)
Hz 8.1510100)1010010( 2
1
)( 21
Hz 91.6107.4)100108.4( 2
1
)( 21
9332
22
631
11
=×××+
=
+=
=××+×
=
+=
−
−
π
π
π
π
p
cLop
p
csigip
f
CRRf
f
CRRf
Hz 7.228.159.6 21 =+=+≈ ppp fff
k 83.1r k 8.52rmA/V 10.9
o ===
π
mg
Exercise 2 (cont’d): β = 200, VA = 150 V
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (11/42)
Exercise 3: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).
This is a PNP common emitter amplifier (no RE ). o Input at the base, output at the collector (RE is shorted
out for signal because of the by-pass capacitor)
It has a emitter-degeneration bias with two voltage sources (RB = ∞)
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (12/42)
Exercise 3 (cont’d). β = 200, VA = 150 V.
Bias circuit (Signal = 0, Caps open)
A 0.5/ mA 0.1
100103.2 3 :KVLBE 3
µβ ==≈=
++×=−
CB
CE
BEBE
IIII
IVI
V 7.0 and 0 V, 7.0: Activein is BJT Assume
≥>= ECCEB VIV
V 0.7 V 4.1 10106.46
3103.2103.23 :KVLCE
0
33
33
=>=××−=
−×++×=−−
DEC
EC
CECE
VVV
IVI
k 26.5 r
k 15010150r
mA/V 38.5 1026
10
3-o
3-
-3
===
==≈
=×
==
mB
T
C
A
T
Cm
gIVIVVIg
βπ
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (13/42)
3.85)k 100||k 3.2||k 150(1038.5
)||||(
3 −=×−=
−=
−
i
o
LComi
o
vv
RRrgvv
k 26.5 k, 150 mA/V, 38.5 === πom rrg
Amplifier Parameters 1) CE amplifier with no RE 2) No voltage divider biasing: no RB
RB is open circuit, RB = ∞
k 27.2k 3.2||k 150 ||
===
o
oCo
RrRR
k 26.5 || === ππ rrRR Bi
Exercise 3 (cont’d). β = 200, VA = 150 V.
3.85−=≈×+
=i
o
i
o
sigi
i
sig
o
vv
vv
RRR
vv
isig RR <<
Signal circuit (IVC = 0)
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (14/42)
Signal circuit (IVC = 0)
k 26.5r k 150r
mA/V 38.5
o
===
π
mg
Hz 1451298.150321 =++=++≈ pppp ffff
Exercise 3 (cont’d). β = 200, VA = 150 V.
k 27.2 k 26.5
==
o
i
RR
Amplifier Cut-off frequency 1) 2 caps: 2 poles 2) no coupling capacitors at the input
0)( 2
1
11 =
+=
csigip CRR
fπ
Hz 1291047)5.26||10(2.3 2
1
]/)||(/1[|| 21
633
3
=×××
=
+=
−π
βπ
p
EsigBmEp
f
CRRgRf
Hz 8.1510100)1010(2.3 2
1)( 2
1
9532
22
=××+×
=
+=
−π
π
p
cLop
f
CRRf
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (15/42)
This is a MOS common source amplifier (no RS ). o Input at the gate, output at the drain (RS is shorted
out for the signal because of the by-pass capacitor)
It has a source-degeneration bias with a voltage divider.
Exercise 4: Find the amplifier parameters of this circuit. (µnCox = 100 µA/V2, (W/L) = 6/0.1, Vt = 0.5 V, λ = 0.1 V-1. Assume capacitors are large and ignore channel width modulation in biasing. )
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (16/42)
V 353.18.110033
100V 0 G =+
=⇒=kk
kIG
Assume Saturation
232 10x3 5.0 OVOVoxnD VVL
WCI −== µ
V 803.0=+= tOVGS VVV
Saturation ⇒> OVDS VV
mA 275.010x3 23 == −OVD VI
V 550.0=−= GSGS VVV
V 249.18.1 =−= DDD IRV
V 699.0=−= SDDS VVV
V 303.0=OVV
DStOVDSGSG IRVVIRVV ++=+=
233 103 102 5.0353.1 OVOV VV −×××++=
0853.06 2 =−+ OVOV VV
GS-KVL:
Exercise 4 (cont’d): µnCox = 100 µA/V2, (W/L) = 6/0.1, Vt = 0.5 V, λ = 0.1 V-1.
Bias circuit (Signal = 0, Caps open)
k 3.3610 752.01.0
11r
mA/V 1.82 0.303
10 0.275 2 2
3o
3
=××
==
=××
==
-D
-
OV
Dm
I
VIg
λ
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (17/42)
32.3)k 50||k 2||k 3.36(101.82
)||||(
3 −=×−=
−=
−
i
o
LDomi
o
vv
RRrgvv
Amplifier Parameters This is a CS amplifier with no RS
k 90.1k 2||k 3.36 ||
===
o
Doo
RRrR
k 8.24== Gi RR
k 8.24k33||k100 ==GR
Exercise 4 (cont’d): µnCox = 100 µA/V2, (W/L) = 6/0.1, Vt = 0.5 V, λ = 0.1 V-1.
Signal circuit (IVC = 0)
19.3k 1k 8.24
k 8.24−=×
+=×
+=
i
o
i
o
sigi
i
sig
o
vv
vv
RRR
vv
Exercise 4 (cont’d):
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (18/42)
Hz 1684.757.307.61321 =++=++≈ pppp ffff
k 9.1 k 8.24
==
o
i
RR
Amplifier Cut-off frequency 1) 3 caps: 3 poles
Hz 7.6110100)1010(24.8 2
1
0)( 2
1
9331
11
=××+×
=
=+
=
−π
π
p
csigip
f
CRRf
Hz 4.75107.4]579||10[2 2
1)]/()||[(|| 2
1
633
3
=×××
=
+=
−π
π
p
SomLDoSp
f
CrgRRrRf
Hz 7.3010100)105010(1.9 2
1
)( 21
9432
22
=×××+×
=
+=
−π
π
p
cLop
f
CRRf
k 3.36mA/V 1.82
==
o
m
rg
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (19/42)
Exercise 5: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).
This is a common collector amplifier (emitter follower). o Input at the base, output at the emitter.
It is biased with a current source! o No emitter degeneration: no RE (RE = ∞) and no RB (RB = ∞)
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (20/42)
Exercise 5 (cont’d): β = 200, VA = 150 V.
V 72.0100 3
−=++=
E
EBEB
VVVI
V 7.0 and 0 V, 7.0: Activein is BJT Assume
≥>= CECBE VIV
k 21.1 r
k 9.34104.3
150r
mA/V 165 1026104.3
3o
3
3
===
=×
=≈
=××
==
mB
T
-C
A
-
-
T
Cm
gIVIVVIg
βπ
A 5.21/mA 3.4
µβ ==≈=
CB
CE
IIII
V 0.7 V 72.44 0 =>=−= DECE VVV
Bias circuit (Signal = 0, Caps open)
EV
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (21/42)
k 21.1 k, 9.34 mA/V, 165 === πom rrg
1104.281
104.28
104.28)||||(
k9.25k100||||k9.34||||)||||(1
)||||(
3
3
3
≈×+
×=
×=
=∞=+
=
i
o
LEom
LEo
LEom
LEom
i
o
vv
RRrgRRr
RRrgRRrg
vv
Ω=+
≈ 111||
||β
π sigBEo
RRrRR
[ ]M 2.5k9.25201k2.1
)||||)(1( || =×+=
++=
i
LEoBi
RRRrrRR βπ
Exercise 5 (cont’d): β = 200, VA = 150 V.
Signal circuit (IVC short, ICS open))
Amplifier Parameters 1) Emitter Follower 2) Biased with an ICS (RE = ∞ ) 3) No voltage divider biasing:
RB is open circuit, RB = ∞
1=≈×+
=i
o
i
o
sigi
i
sig
o
vv
vv
RRR
vv
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (22/42)
Signal circuit
k 21.1rk 9.34rmA/V 165
o
===
π
mg
Hz 34.31047.0)11110100( 2
1][ 2
1
632
22
=××+×
=
+=
−π
π
p
coLp
f
CRRf
Ω==
111 M 2.5
o
i
RR
Exercise 5 (cont’d): β = 200, VA = 150 V.
Hz 34.32 == pp ff
Amplifier Cut-off frequency 1) 1 cap: 1 pole 2) no coupling capacitors at the input
0)( 2
1
11 =
+=
csigip CRR
fπ
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (23/42)
Exercise 6: Find the bias point and the amplifier parameters of the circuit below. (µpCox (W/L) = 400 µA/V2, Vtp = − 4 V, λ = 0.01 V-1. Ignore channel width modulation in biasing).
This is a PMOS common drain amplifier. o Input at the gate, output at the source.
It has a source-degeneration bias with two voltage sources.
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (24/42)
Exercise 6 (Cont’d): µpCox (W/L) = 400 µA/V2, Vtp = − 4 V,λ = 0.01 V-1
Assume Saturation:
092
09104005.010
||101013
2
264
44
=−+
=−+×××
++×=+×=−
OVOV
OVOV
tpOVDSGD
VVVV
VVIVISG-KVL:
5.0 2OVoxpD V
LWCI µ=
V 9.5=+= tOVSG VVV
Saturation ⇒> OVDS VV
mA 71.0104005.0 26 =××= −OVD VI
V 9.5 0 =→−= SSSG VVV
V 9.10)5( =−−= SSD VV
V 9.1=OVV
k 141100.7101.0
11
mA/V 0.747 0.303
100.71 2 2
3
3
=××
==
=××
==
−
−
Do
OV
Dm
Ir
VIg
λ
SV
Bias circuit (Signal = 0, Caps open)
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (25/42)
Signal circuit
87.038.61
38.66.38)||||(
k54.8k100||k10||k 141||||
88.0)||||(1
)||||(
=+
=
===
=+
=
i
o
LSom
LSo
LSom
LSom
i
o
vv
RRrgRRr
RRrgRRrg
vv
Amplifier Parameters Common collector Amp.
Exercise 6 (Cont’d): µpCox (W/L) = 400 µA/V2, Vtp = − 4 V,λ = 0.01 V-1
k 2.1k 34.1||k 101 ||
===
∞==
mSo
Gi
gRR
RR
87.0==×+
=i
o
i
o
sigi
i
sig
o
vv
vv
RRR
vv
k 141 mA/V, 0.747 == om rg
Amplifier Cut-off frequency 1) 1 cap: 1 pole 2) no coupling capacitors at the input
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (26/42)
Signal circuit
Exercise 6 (Cont’d): µpCox (W/L) = 400 µA/V2, Vtp = − 4 V,λ = 0.01 V-1
k 141mA/V 0.747
==
o
m
rg
k 2.1
=∞=
o
i
RR
Hz 39.31047.0)102.110( 2
1
][ 21
635
2
=×××+
=
+=
−π
π
p
coLp
f
CRRf
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (27/42)
Exercise 7: Find the bias point and the amplifier parameters of the circuit below. (µnCox (W/L) = 800 µA/V2, Vt = 1 V, λ = 0.01 V-1. Ignore channel width modulation in biasing).
This is a NMOS common gate amplifier. o Input at the source, output at the drain.
It has a source-degeneration bias with a voltage divider.
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (28/42)
Bias circuit
054
05108005.010
10106
2
264
44
=−+
=−+×××
×++=×+==−
OVOV
OVOV
DtOVDGSG
VVVV
IVVIVVGS-KVL:
Assume Saturation: 5.0 2OVoxnD V
LWCI µ=
V 0.2=+= tOVGS VVV
Saturation ⇒> OVDS VV
mA 40.0108005.0 26 =××= −OVD VI
V 74415101015 44
=−−=×++×=
DS
DDSD
VIVI
V 0.1=OVV
Exercise 7 (Cont’d): µnCox (W/L) = 800 µA/V2, Vt = 1 V, λ = 0.01 V-1.
GV
V 615M8.1M2.1
M2.1=×
+=GV
DS-KVL:
k 250100.401.0
11r
mA/V 0.800 1
100.4 2 2
3o
3
=××
==
=××
==
−
−
D
OV
Dm
I
VIg
λ
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (29/42)
69.7)||k 10||k 250(108.0
)||||(
3 =∞×=
=
−
i
o
LSomi
o
vv
RRrgvv
Exercise 7 (Cont’d): µnCox (W/L) = 800 µA/V2, Vt = 1 V, λ = 0.01 V-1.
Signal circuit (IVC short)
Amplifier Parameters 1) CG amplifier 2) Biased with two voltage sources (RG = 0)
11.469.7k 1k 15.1
k 15.1=×
+=×
+=
i
o
sigi
i
sig
o
vv
RRR
vv
k 250mA/V 0.800
==
o
m
rg
k 15.1k 1.30||k 10
/)||(1 ||
==
+=
i
m
oLDSi
Rg
rRRRR
k 77.9k 432||k 10
)]||(1([ ||
==
+=
o
sigSmoDo
RRRgrRR
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (30/42)
Hz 74021 =+≈ ppp fffk 250mA/V 0.80
==
o
m
rg
Exercise 7 (Cont’d): µnCox (W/L) = 800 µA/V2, Vt = 1 V, λ = 0.01 V-1.
Signal circuit (IVC short) Amplifier Cut-off frequency
2 caps: 2 poles
Hz 74010100)1010(1.15 2
1
0)( 2
1
9331
11
=××+×
=
=+
=
−π
π
p
csigip
f
CRRf
010100)10(9.77 2
1
)( 21
932
22
=××∞+×
=
+=
−π
π
p
cLop
f
CRRf
k 77.9 k 15.1
==
o
i
RR
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (31/42)
Exercise 8: Design a common emitter amplifier with RE with vo / vi = 5 when driving a 100 kΩ load and a bias point of IC = 4 mA and VCE = 6 V. The circuit should have a cut-off frequency of 50~Hz for Rsig = 100 Ω. A 16 V power supply is available. Use a Si BJT with β = 200, β min = 100, VA = 150 V, ignore Early effect in bias calculations.
Prototype of the circuit is shown.
Problem specification gives
We should find RC , RE , RB1 , RB2 , Cc1 , and Cc2
k 100 , 100 V, 16 =Ω== LsigCC RRV
* See Exercise 4 of BJT lecture slides for bias design
mA, 4 V, 6 k, 100 , 100 V, 16 ===Ω== CCELsigCC IVRRV
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (32/42)
Exercise 8 (Cont’d):
k 30.1
k 5.37104
150
mA/V 154 1026104
3
3
3
===
=×
=≈
=××
==
mB
Tπ
-C
Ao
-
-
T
Cm
gIVr
IVr
VIg
β
1. Bias point (IC = 4 mA & VCE = 6 V) gives the SSM parameters as well as a relationship between RC and RE through CE-KVL:
k 5.2104
616)(
3 =×
−=+
+≈−++=
−EC
ECCCECC
EECECCCC
RR
RRIVVIRVIRV
2. Amp gain of vo / vi = 5 also gives another relationship between RC and RE:
CE-KVL:
5)/1](/)||[(1
)||( =+++
=πrRrRRRg
RRgvv
EoLCEm
LCm
i
o
Solution can be simplified by noting RC + RE = 2.5k and thus RC < 2.5k: 1) Since RC << RL = 100 k, then RC || RL ≈ RC 2) Since RC << ro = 37.5 k, we can drop the 3rd term in the denominator of the gain formula:
5.32555
51
+=+=
=+
≈
Em
EC
Em
Cm
i
o
Rg
RR
RgRg
vv
k 09.2 411
k 5.25.325
=Ω=
=++
C
E
EE
RR
RR
A 0.20/ µβ == CB II
Exercise 8 (Cont’d):
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (33/42)
3102.73.631316
)/1](/)||[(1)||( −×++
=+++
=πrRrRRRg
RRgvv
EoLCEm
LCm
i
o
In fact, because gm terms are large:
Note that the (open-loop gain) is independent of BJT parameters!
E
C
Em
Cm
Em
Cm
i
o
RR
RgRg
RgRg
vv
=≈+
≈1
For discrete BJT amplifiers (RC << ro ), dropping the third term is a very good approximation because gm terms are actually large:
k 09.2 , 411 =Ω= CE RR
k 30.1 k, 5.37 mA/V, 154 === πom rrg
A. 20 mA, 4 V, 6 k, 100 , 100 V, 16 µ====Ω== BCCELsigCC IIVRRV
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (34/42)
Exercise 8 (Cont’d): A. 20 mA, 4 V, 6 k, 100 , 100 V, 16 µ====Ω== BCCELsigCC IIVRRV
1. Bias point (IC = 4 mA & VCE = 6 V) and requirements of good bias give RB1 and RB2 through BE-KVL. However, it is easier to find RB and VBB (Thevenin equivalent) first.
2. Resistors RB1 and RB2 are found from:
BE-KVL:
Dividing the above equations gives RB1 :
k 89.4k 4.27
2
1
==
B
B
RR
k 09.2 , 411 =Ω= CE RR
k 4.154111011.04111011.0)1( 0.1
)1(
min
min
=××=××=+≤
+<<
B
EB
EB
RRR
RRβ
β
V 43.2114 10 4 0.710 4.15 1020 336
=××++×××=
++=
BB
--BB
EEBEBBBB
VV
RIVRIV
152.016
2.43
k 15.4||
21
2
21
2121
==+
=
=+
==
BB
B
CC
BB
BB
BBBBB
RRR
VV
RRRRRRR
The 2nd condition for stable bias is RE IE ≥ 1 V:
V 1 64.1114 10 4 3 >=××= -EE RI
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (35/42)
Exercise 8 (Cont’d):
Coupling capacitors Cc1 and Cc2 are found from specification of cut-off frequency (50 Hz). We need to compute Ri and Ro first.
Can choose either capacitor and compute the other one. A good design guideline is to keep capacitors small.
k 15.4 ,k 89.4 ,k 4.27 ,k 09.2 , 411 21 ====Ω= BBBCE RRRRR
k 95.3k 81.6 ||k 15.4)/1](/)||[(1
||
==
++
+=
i
EoLC
EBi
RrRrRR
RrRRπ
πβ
k 09.2M 1.74 ||
||1 ||
=≈=
+++=
CCo
sigBE
EoCo
RRR
RRRrRrRR
π
β
k 30.1 k, 5.37 mA/V, 154 === πom rrg
Amplifier Cut-off frequency (2 caps: 2 poles)
23
22
13
11
10102 21
)( 21
1005.4 21
)( 21
ccLop
ccsigip
CCRRf
CCRRf
××=
+=
××=
+=
ππ
ππ
5010102 2
11005.4 2
1
Hz 50
23
13
21
=××
+××
=+≈
cc
ppp
CC
fff
ππ
A. 20 mA, 4 V, 6 k, 100 , 100 V, 16 µ====Ω== BCCELsigCC IIVRRV
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (36/42)
Exercise 8 (Cont’d):
We can choose either capacitor and compute the other one. A good design guideline is to keep capacitors small. Since the coefficient of CC1 is 25 times smaller than that CC1, a reasonable choice is to assign 80% of fp to CC1 :
k 15.4 ,k 89.4 ,k 4.27 ,k 09.2 , 411 21 ====Ω= BBBCE RRRRRk 30.1 k, 5.37 mA/V, 154 === πom rrg
nF 156 502.010102 2
1
F 982.0 508.01005.4 2
1
Hz 50
12
3
11
3
21
=→×=××
=→×=××
=+≈
cc
cc
ppp
CC
CC
fff
π
µπ
Design Commercial Value Value (5%)
nF 156F 982.0
411k 09.2k 89.4k 4.27
2
1
2
1
==
Ω====
C
C
E
C
B
B
CCRRRR
µnF 160
F 1 390
k 2k 7.4
k 27
1
2
1
==
Ω====
Cw
C
E
C
B
B
CCRRRR
µ
A. 20 mA, 4 V, 6 k, 100 , 100 V, 16 µ====Ω== BCCELsigCC IIVRRV
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (37/42)
Exercise 9: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).
This is a two-stage amplifier. o Stage 1 is a common emitter amplifier with RE o Stage 2 is a common collector amplifier (emitter follower).
Both stages have source-degeneration bias with a voltage divider. o Each stage is independently biased because of the 470 nF
coupling capacitor between stages.
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (38/42)
Stage 1 Bias Thevenin form of the Voltage divider
V 37.215k 2.6k 33
k 2.6
k 22.5k 2.6||k 33
1
1
=×+
=
==
BB
B
V
R
Exercise 9 (cont’d) : β = 200, VA = 150 V
Bias circuit Caps open
A 9.15/ mA 18.3
5007.0)1/(1022.5 2.37
:KVLBE
111
11
1113
111111
µβ
β
==≈=
+++×=
++=−
CB
CE
EE
EEBEBBBB
IIII
IIRIVRIV
V 7.0 and 0 V, 7.0: Activein is BJT Assume
111 ≥>= CECBE VIV
V 0.7 V 05.7 )500102(1018.315
15 :KVLCE
01
331
11111
=>=+×××+=
++=−−
DCE
CE
EECECC
VVV
RIVRI
k 64.1
k 2.4710 18.3
150
mA/V 122 1026103.18
1
1
11
31
11
3
31
1
===
=×
==
=××
==
mB
Tπ
-C
Ao
-
-
T
Cm
gIVr
IVr
VIg
β
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (39/42)
Stage 2 Bias Thevenin form of the Voltage divider
V 25.815k 18k 22
k 22
k 90.9k 22||k 18
2
2
=×+
=
==
BB
B
V
R
Exercise 9 (cont’d) : β = 200, VA = 150 V
Bias circuit Caps open
A 36/ mA 20.7
107.0)1/(109.9 8.25
8.25 :KVLBE
222
22
23
223
22222
µβ
β
==≈=
+++×=
++=−
CB
CE
EE
EEBEBB
IIII
IIRIVRI
V 7.0 and 0 V, 7.0: Activein is BJT Assume
222 ≥>= CECBE VIV
V 0.7 V 80.7 101020.715
15 :KVLCE
02
332
222
=>=××+=
+=−−
DCE
CE
EECE
VVV
RIV
k 722.0
k 8.2010 20.7
150
mA/V 277 1026107.20
2
2
22
32
22
3
32
2
===
=×
==
=××
==
mB
Tπ
-C
Ao
-
-
T
Cm
gIVr
IVr
VIg
β
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (40/42)
Exercise 9 (cont’d) : β = 200, VA = 150 V
12621
26226294510277)||||(
)||||(1)||||(
2
2
32222
2222
2222
2
2
≈+
=
=××=
+=
−
i
o
LEom
LEom
LEom
i
o
vv
RRrgRRrg
RRrgvv
[ ][ ] k 41.9k 189722 ||k 90.9
)||||( ||
2
2222222
=+=+=
i
LEoBi
RRRrrRR βπ
Stage 2: Emitter Follower For Gain & Ri , start from the load side because we need to know RL
k 41.9 21 == iL RR
k 100 2 == LL RR
k 90.9k 722.0
k 8.20mA/V 277
2
2
2
2
====
B
π
o
m
Rrrg
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (41/42)
Exercise 9 (cont’d) : β = 200, VA = 150 V
k 95.497.2k ||k 22.5)/1](/)||[(1
||
1
11111
11111
==
++
+=
i
EoLC
EBi
RrRrRR
RrRRπ
πβ
25.30.62
201046.0500101221
1065.110122
046.0)/1](/)||[(k 1.65k 41.9||k 2||
)/1](/)||[(1)||(
3
33
1
1
11111
11
1111111
111
1
1
−=−=+××+×××
−=
=+==
+++−=
−
−
i
o
EoLC
LC
EoLCEm
LCm
i
o
vv
rRrRRRR
rRrRRRgRRg
vv
π
π
Stage 1: CE amplifier with RE k 41.9 21 == iL RR
k 22.5k 64.1k 2.47mA/V 122
1
1
1
1
====
B
π
o
m
Rrrg
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (42/42)
Exercise 9 (cont’d) : β = 200, VA = 150 V
Stage 2: Emitter Follower
For Ro , start from the source side because we need to know Rsig
k 0.2 12 == osig RR
100 1 Ω== sigsig RR
k 2
||1 ||
11
1111
11111
=≈
+++=
Co
sigBE
EoCo
RR
RRRrRrRR
π
β
Stage 1: CE amplifier with RE
Ω==+
= 8.119.11||k 1||
||2
22222 β
π sigBEo
RRrRR
k 22.5k 64.1k 2.47mA/V 122
1
1
1
1
====
B
π
o
m
Rrrg
k 90.9k 722.0
k 8.20mA/V 277
2
2
2
2
====
B
π
o
m
Rrrg
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (43/42)
Exercise 9 (cont’d) : β = 200, VA = 150 V
k 0.2 k 95.4
25.3/ k 41.9
100
1
1
11
21
1
==
−===
Ω==
o
i
io
iL
sigsig
RR
vvRRRR
k 22.5k 64.1k 2.47mA/V 122
1
1
1
1
====
B
π
o
m
Rrrg
k 90.9k 722.0
k 8.20mA/V 277
2
2
2
2
====
B
π
o
m
Rrrg
Ω==
===
==
8.11 k 41.9
1/ k 100
k 0.2
2
2
22
2
12
o
i
io
LL
osig
RR
vvRRRR
19.31)25.3(98.0
11.8 k, 4.95
2
2
1
1
21
−=×−×=××+
=
Ω====
i
o
i
o
sigi
i
sig
o
ooii
vv
vv
RRR
vv
RRRR
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (44/42)
Exercise 9 (cont’d) : β = 200, VA = 150 V
k 0.2 k 41.9 11.8
k 4.95
1
2
==
Ω==
o
i
o
i
RRRR
Hz 3.527.299.1571.6321 =++=++≈ pppp ffff
Amplifier Cut-off frequency 3 caps: 3 poles
Hz 71.6)( 2
1
11 =
+=
csigip CRR
fπ
Hz 9.15)( 2
1 2
2 =+
=cLo
p CRRf
π
Hz 7.2910470)(2
1
)(21
912
1,,
=××+
=
+=
−
−
oipj
cjjojipj
RRf
CRRf
π
π
Coupling cap at the input:
Coupling cap at the output:
Coupling cap between stages:
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (45/42)
Exercise 10: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).
This is a two-stage amplifier. Both stages are common emitter amplifier with RE.
Both stages have source-degeneration bias with two power supplies (no voltage divider). o There are NO coupling capacitors and the second stage is
biased from VC1.
A 3.33/mA 65.6
510)1/(106.3 510106.351.3
3 5107.0106.31083.3106.3 15
3 510)(106.3 15
222
22
223
223
22333
22213
µβ
β
==≈=
++×=+×=
−++×+×××=
−+++×=−
CB
CE
EEEB
EB
EBEBC
IIII
IIIIII
IVII
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (46/42)
Exercise 10 (cont’d) : β = 200, VA = 150 V
A 2.19/mA 83.3
3 6007.0)1/( 100 03 600 100 0
111
11
111
111
µβ
β
==≈=
−+++=−++=
CB
CE
EE
EBEB
IIII
IIIVI
V 7.0 and 0 V, 7.0: Activein are sBJTboth Assume
111 ≥>= CECBE VIV
BE1-KVL:
BE2-KVL:
CE2-KVL:
0.7 V 63.43 510105.1 15
02
2223
=>=−++×=
DCE
ECEC
VVIVI
CE1-KVL:
0.7 V 79.13 600)(106.3 15
01
11213
=>=−+++×=
DCE
ECEBC
VVIVII
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (47/42)
Exercise 10 (cont’d) : β = 200, VA = 150 V
k 36.1
k 2.3910 83.3
150
mA/V 147 1026103.83
1
1
11
31
11
3
31
1
===
=×
==
=××
==
mB
Tπ
-C
Ao
-
-
T
Cm
gIVr
IVr
VIg
β
V 63.4A 2.19
mA 83.3
1
1
1
===
CE
B
C
VII
µV 79.1
A 3.33mA 65.6
2
2
2
===
CE
B
C
VII
µ
k 782.0
k 6.2210 65.6
150
mA/V 256 10261065.6
2
2
22
32
22
3
32
2
===
=×
==
=××
==
mB
Tπ
-C
Ao
-
-
T
Cm
gIVr
IVr
VIg
β
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (48/42)
Exercise 10 (cont’d) : β = 200, VA = 150 V
For Gain & Ri , start from the load side because we need to know RL
k 8.92 21 == iL RR
k 100 2 == LL RR
∞====
2
2
2
2
k 782.0k 6.22mA/V 256
B
π
o
m
Rrrg
k 8.9292.8k ||)/1](/)||[(1
||
2
22222
22222
=∞=
++
+=
i
EoLC
EBi
RrRrRR
RrRRπ
πβ
88.2132379
108.05101056211048.110562
108.0)/1](/)||[(k 1.48k 100||k 5.1||
)/1](/)||[(1)||(
3
33
1
1
22222
22
2222222
222
2
2
−=−=+××+×××
−=
=+==
+++−=
−
−
i
o
EoLC
LC
EoLCEm
LCm
i
o
vv
rRrRRRR
rRrRRRgRRg
vv
π
π
Stage 2: CE amplifier with RE
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (49/42)
Exercise 10 (cont’d) : β = 200, VA = 150 V
k 108k 108 ||)/1](/)||[(1
||
1
11111
11111
=∞=
++
+=
i
EoLC
EBi
RrRrRR
RrRRπ
πβ
71.53.89
510127.0600101471
1047.310147
127.0)/1](/)||[(k47.3k 8.92||k 6.3||
)/1](/)||[(1)||(
3
33
1
1
11111
11
1111111
111
1
1
−=−=+××+×××
−=
=+==
+++−=
−
−
i
o
EoLC
LC
EoLCEm
LCm
i
o
vv
rRrRRRR
rRrRRRgRRg
vv
π
π
Stage 1: CE amplifier with RE k 8.92 21 == iL RR
∞====
1
1
1
1
k 36.1k 2.39mA/V 147
B
π
o
m
Rrrg
∞====
2
2
2
2
k 782.0k 6.22mA/V 256
B
π
o
m
Rrrg
∞====
1
1
1
1
k 36.1k 2.39mA/V 147
B
π
o
m
Rrrg
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (50/42)
Exercise 10 (cont’d) : β = 200, VA = 150 V
For Ro , start from the source side because we need to know Rsig
k 6.3 12 == osig RR
100 1 Ω== sigsig RR
k 6.3
||1 ||
11
1111
11111
=≈
+++=
Co
sigBE
EoCo
RR
RRRrRrRR
π
β
Stage 1: CE amplifier with RE
k 5.1
||1 ||
21
2222
22222
=≈
+++=
Co
sigBE
EoCo
RR
RRRrRrRR
π
β
Stage 2: CE amplifier with RE
∞====
2
2
2
2
k 782.0k 6.22mA/V 256
B
π
o
m
Rrrg
∞====
1
1
1
1
k 36.1k 2.39mA/V 147
B
π
o
m
Rrrg
F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (51/42)
Exercise 10 (cont’d) : β = 200, VA = 150 V
k 6.3 k 108
71.5/ k 8.92
100
1
1
11
21
1
==
−===
Ω==
o
i
io
iL
sigsig
RR
vvRRRR
k 5.1 k 8.92
88.2/ k 100
k 6.3
2
2
22
2
12
==
−===
==
o
i
io
LL
osig
RR
vvRRRR
4.16)88.2()71.5(1
k 5.1 k, 108
2
2
1
1
21
=−×−×=××+
=
====
i
o
i
o
sigi
i
sig
o
ooii
vv
vv
RRR
vv
RRRR
Amplifier Cut-off frequency: only Coupling cap at the output:
Hz 7.15)( 2
1 2
=+
=cLo
p CRRf
π