session 4: analog circuits bjt biasing single stage...
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1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
BJT: Bipolar Junction Transistor
3
D
A C
iD
VD
p n
F : Forward
R : ReverseF
R
p n p n p n
21E C
B
21E C
B
Qn
E
C
B
B
E
C
Qp
VBE
VBC
1 : F
2 : R
1 : F
2 : F
1 : R
2 : F1 : R
2 : R
SaturationActive
ReversedCut-Off
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
BJT: Bipolar Junction Transistor
4
D
A C
iD
VD
p n
B Qn
E
C+
-
∝ /
ΔΔ 2 2 !"/#
∝ $%/ ΔΔ 2 %& & 2 !"/#
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
BJT Configurations
5
BQn
E
C
+
-
&
&+
-
+
-B
Qn
E C
&
&+
-B
Qn
E
C
+
-
&+
-
CE: Common Emitter CB: Common Base CC: Common Collector
Output Characteristic
Input Characteristic
Transfer Characteristic
vs. for different '&
& vs. & for different ' or
& vs. or for different '&
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
CE: Common Emitter
6
BQn
E
C
+
-
&
&+
-
Output Characteristic
Input Characteristic
&
&
( 010+,20+,30+,
&
&
' . 0.5'5602'6002'6202'
'3
'3
Early voltage
Saturation
Active
Cut-Off0.2'
& 4 (5$% 1 6 &'3 "
Active:
'& 4 0'& 7 0
very little
dependence to
output voltage
'& ↑→ : ↑→ ; ↑→ (& ↑
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
CE: Transistor Model
7
BQn
E
C
+
-
&
&+
-
Output Characteristic
&
& <S
atu
ratio
n
Cut-Off
&
&
0.2'
(&=>
'&=>
?=>
SOA: Safe Operating Area
(
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
CE: Output Characteristic
8
BQn
E
C
+
-
&
&+
-
&
& <S
atu
ratio
n
Cut-Off
(
4 (′5$%
Ideally linear:
& (5$% 1 6 &'3 "
4 (5; $%
; 4 & 4(&( 4
A
BC 4 ;D 4 &EF$#SPICE
BGH 4 ;& 4 &EIFIJ4 ;D1 6 &'3 "
B,H 4 K&KEIFIJ4 A
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
CB: Common Base
10
Output Characteristic
Input Characteristic
&
&
( 012,22,32,
0.4'
Saturation
Active
Cut-Off
& 4 MActive:
'& 4 0'& 7 0
very little
dependence to
output voltage
+
-B
Qn
E C
&
&+
-
M 4 ;; 6 1
ΔΔ 2 %
7
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
;(′
Large Signal Model
11
BQn
E
C
+
-
&
&+
- &
& <S
atu
ratio
n
(
2
1
E
C
Bideal
ideal
E
C
B
0.7
0.4
(′
&B
E
C
(5 ;(
Active
E C
B
M(
Saturation
B
E
C
(5 '&N=0.2~0.3'
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Voltage Amplifier
12
B
Qn
E
C
+
-
&
&+
-+
-I
&P&
Q+
-
'&& Q 4 '&& P&&& 4 ; 4 (N
R 1"
I
& Q
I'&&
cut-off
act
ive saturation
0.1v
10v
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Biasing: VBE= cte
13
+
-I
&P&
Q+
-
'&&
I
& Q
I'&&
cut-off
act
ive
saturation
0.1v
10v
5'
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Biasing: IB= cte
14
+
-
I&P&
Q+
-
'&&( 4 (#(& 4 ;(
5
(#∞
+
-
I&P&
Q+
-
'&&
∞
5
P ( 4 '&& 'P
(& 4 12, ; 4 100(# 4 10+,
→ P& ~5UΩ
'&& 4 10'
For max swing: '&~5'→ P& ~9.7UΩFor max gain: '&~0.3'
→ P4 930UΩ'&& 4 10'(& 4 12,
What is the problem?
Replace it with transistor with ; 4 250(& 4 2.52, '& 4 10 5 Y 2.5 4 2.5 Z '&N=
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Biasing: IB= cte
15
+
-
I&P&
Q+
-
'&&
∞
5
P
P 4 '& '( 4 ; '& '(&
(& 4 12, ; 4 100'&& 4 10'For max swing: '&~5'
→ P4 430UΩ
; ↑ → (& ↑→ '& ↓
→ ( ↓→ (& ↓
Negative feedback!
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Biasing: IE= cte
16
+
-I
&P&
Q+
-
'&&
Q
I
'&& 10v
5'
P+
-I
&P&
Q+
-
'&&
5 (
( 4 ' 'P
' ≫ 'independent of '(
'~2.7'
12,P&
10'
P2.7' 2'
4 2UΩ
6'4UΩ 4
I&
P&
0.7' 6.7'
Q+
-
+
-
10'5UΩ
Where is the
trade-off?
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Biasing: IE= cte
17
Q
'&& ( 4 (&;Assume:
12,
P&10'
P
2.7'2'
4 2UΩ
6'4 4UΩ
P
P∞
I (
(( ≫ ((&
P
P(
('&& 10'
'' 4 PP 6 P '&&
P
'&&P&10'
''
' 4 ' 'Q'&
(
( 4 'P (& 4 M( ((& '& 4 '&& (&P&
Q
'&&
12,
P&10'
P
2.7'2'4 2UΩ
6'4 4UΩ
P
P∞
I 10+,
0.12, 4 73UΩ
4 27UΩThis is for design
how about analysis
?
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Biasing: IE= cte
18
' 4 2'Assume: ( ≫ (
P
P(
('&&
'' 4 2.7'Q
'&&
12,
P&10'
P
2.7'2'
2UΩ
6'4UΩ
P
P∞
I 10+,
0.12, 73UΩ
27UΩ( 4 12, ( 4 0.012,
( 4 0.12,
What if ; was 10!
P
P(
('&&
''&& PP 6 P
P ∥ P Q
'&&
(&P&10'
P 1M (&
(&/;KVL
PP 6 P '&& 4 (&P ∥ P; 6 '^ 6 1M P(& → (& 4 ⋯
For the above numbers: (& 4 21.01 Y 2U 6 0.01 Y 19.7U 4 0.922,
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Biasing: IE= cte
19
(P 6 ' 4 '^ 6 (PQ
'&&
(
P&
PP
P
similarD(
( '&&P 6 P ( 4 PP (
'&&P&
PP
P
(
( 4 '&& '^P 6 P
(P 4 (P'&&
P&P( < <
( 4 '&& '^P( ( 4 (5 $%/ 1
+
- ( 4 (5 $%/ 1
(( 4(5(5 4
Only in Integrated Circuits!
,`,`area
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Biasing: Example 01
20
(′3.9(′ 6 0.7 4 5.60.9 (′; "
Q1
Q2
4.7kΩ
5.6kΩ 3.9kΩ
56kΩ 3.3kΩ
150kΩ
15V ; 4 ; 4 100 'ab 4 0.7< <
(& [mA]
'& [V]4'
0.072,
3.3'12,
9.4' 10.1'1.262,
1.262,5.9'
0.012,1 1.266.1 4.2
Q1
41kΩ
150||564.08
3.3kΩ
(
(/;
KVL
5.6kΩ
15V
( 4 4.083.3 6 40.8/101 4 0.912,
3'
Q2
3.9kΩ
15V
4.7kΩ
(′/;0.9
(′ 4 1.132,
9.89'
KVL
10.59'5.31'
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Biasing: Example 02
21
; 4 ; 4 100 'ab 4 0.7< <
(& [mA]
'& [V]
1.492,
1.05 1.491.4 9.3Q2
8.2kΩ
Q1
470Ω
Vcc=10V
0.7'
1.4' (& 4 1.492,
1.052,
( 4 1.052, 0.01√
0.015 √
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Biasing: Example 03
22
; 4 ; 4 100'ab 4 0.7 < <
(& [mA]
'& [V]
2 1.72.4 3.56
Vcc=9V
47kΩ
Q2
2.2 kΩ3.3 kΩ
Q1
1 kΩ
0.7'0.7'
0.72,
0.72,7.46'1.4'
2.32,
0.007 √
0.023 ××××
1.7'1.72,
1.72,5.26'2.4'
22,
8.3 330 4 0.747
47kΩ
Q2
2.2 kΩ3.3 kΩ
Q1
1 kΩ
9V
0.747U
6 0.7
8.3 3.3U
⇒ 4 1.65
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Biasing: Example 04
23
; 4 ; 4 100 'ab 4 0.7< <
(& [mA]
'& [V]
1.1 1.11.5 3.11
Vcc=10V
Q2
3.9 kΩ
18 kΩ
Q1
1 kΩ
1.8'0.12, 15 kΩ
68 kΩ
P3.3'
1.1'1.12,
2.6'5.71'
P 4 8.2UΩFind bias points if
Q2
8.2 kΩ
P
2.6'3.3'
' 4 10'
0.2'
'& 4 10 1.1 Y 8.2 2.64 1.622.8' 0.882, < <
(& [mA]
'& [V]
1.1 0.881.5 0.2
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Linear BJT Amplifier
24
Vin
R
VCC
Vo
Qn
& 4 (5$% 1 6 &'3 " (5
$%
' 4 262' '3 4 200'
4 ' 6 e sinij
Q 4 '&& P& 4 '&& P(5$ ek lmn o (&
4 '&& P(& 1 6 e' sinij 6e2' sin
ij 6 ⋯ '&& P(& P(& e' sinij'Q ^
, 4 ^ 4P&(&' 4 pP& p 4 (&'
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
BJT Small Signal Model (h-π)
25
Trance-Conductance:
BQn
E
C
+
-
&&+
-
A qr pr q#r+
-
Input resistance:
qr ≡ KKOutput resistance:
4 KK 4 ; K&K
& 4 (5 %$ 1 1 6 &'3 "
≅ (5%$ 1 6 &'3 "(&
4 ; (5' %$
4 ;'(&4
;p 4 ;q
q ≡ K&K& 4 K&K& 4 (&'3
4 '3(&
p ≡ K&K 4 (5' %$ 4 (&' 4
1q
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
N
Example 01 -CE
26
(&P&
Q+
-
10'
∞5
P
P 4 10 0.70.012,
(& 4 12,; 4 100Assume
Design for
r 4 N qr ∥ Pqr ∥ P 6 PN
PN1U
'3~∞and maximum swing
Find ,, P, P^v
AC:
Qwindow
for
10'cutoff
saturation0.2'~0? 'Q 4 5 P& 4 5UΩ4 930UΩDC:
PN1UP
^P&qr p
r
r+-
~N qrqr 6 PN# 4 prP&
, 4 #N 4 pP&qrqr 6 PN 4
;P&qr 6 PN4P&q 6 PN ;⁄ 4
Collector resistance
Emitter’s circuit resistance
if , 4 pP&PN → 0:
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Q
Example 02 -CE
27
Q12,
P&10'
P
2.7'2'2UΩ
6'4UΩ
P
P∞10+,
0.12, 73UΩ
27UΩ
y 6 Py rqr 6 r 6 Prqr 1 6 ; 4 0
PN ∞
Find ,, P, P^v
PNP P P
P&
P ∥ P
y Py
yPy
P
P&qr+
-
r pr
r qrz
r 1qr 6 p
'|
4 rqr 1 6 ;
Q~20UΩ
1UΩ
y 4 P ∥ PP ∥ P 6 PN Py 4 P ∥ P ∥ PN'|:
r 4 y qrPy 6 qr 6 P 1 6 ;
^ 4 prP&
,~ 4 ^y 4;P&Py 6 qr 6 P 1 6 ;
4 P&Py 6 qr; 6 P 1 6 ;;, 4 ^N 4
yN ∙^y 4
2021 ∙
43.5100 6 2 Y 101100
4 1.8
; 4 100Assume '3~∞
AC
circuit
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Example 02 -CE
28
QP&
10'
P2UΩ
4UΩ
P
P 73UΩ
27UΩ
PN1UΩ
, 4 1.8
How we can increase gain?
Q
QP&
10'
P2UΩ
4UΩ
P
P 73UΩ
27UΩ
PN1UΩ Bypass capacitance
H
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Example 03 -CC
29
Q
10'
P
P
6 P5 rqr 6 r 6 Prqr 6 pr 4 0
PN
P5
P
qr+
-
r pr
r qrz
r 1qr 6 p
'|'|:
r 4 P5qr 6 1 6 P p 6 1qr
, 4 ^
Design a buffer window for
9.3'saturation
0'? ' 4 5
P 4 5UΩ
~10
cutoff
(& 4 12,
P 4 10 5.70.012 4 430UΩ
PNP P
AC circuit
^^ 4 r 1
qr 6 p P
41qr 6 p P
P5qr 6 1 6 P p 6 1qr4 PP5 6 qr1 6 ; 6 P
~1
^
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Example 03 -CC
30
4 qr 6 P 1 6 ;
P5
P
qr+
-
r pr
r qrz
r 1qr 6 p
P 4r 6 P rqr 6 prrqr
PN
P
^
^ 4PP5 6 qr1 6 ; 6 P^
P PN
P ^P
P 4 qr 6 P 1 6 ;
PN
P ^P^v
P^v 4 P ∥ q
q
4 P ∥ P5 6 qr1 6 ;
q
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Example 04 –Multi-stage Amplifier
31
P&!10'
2UΩ
73UΩ
27UΩ
1UΩP& Q
Design an amplifier:
(& 4 12,, 1000; 4 100' 4 10'P5 4 1UΩ
930UΩ
P&! 4 10 312 4 7UΩP& 4 10 512 4 5UΩ
AC circuit
1UΩ
27 ∥ 73
20UΩ
< <
<<
7UΩ 930UΩ 5UΩP! ^ , 4 ^ 4
⋅ ⋅
^P
420 ∥ qr!
20 ∥ qr! 6 1 4 0.69
4 p! 7 ∥ 930 ∥ qr 4 73^ 4 p5 4 200
, 4 10143
< <7UΩ 5UΩ
1UΩ^
buffer
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Example 05 –CE , CB, CC
32
10'
2
73
27PN
Q
AC circuit
<<
<<
7<
73
27 2
4430
512,2.723
7.38 12,
4 12,5.7 5
20PN
7 2< 4 Q
< <1.5
< 4 Q
PN ≪?
5
5430
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Common Base
33
PPN
NQPNP&
PQP& B
E
C
qr+
- prE
B
rC
NE
B
C
≡
≡E
B
C
qr
prr
+
-
E C
qrr+
-
≡B
prpr
≡ E C
qrr+
-
B
prq
E C
qr+
-
B
pr
q 4 q ∥ qr 4 q ;; 6 1 ≅ q
NPN
PQP&
, 4 ^N 4P&P 6 qr; 6 1 6 PN
P 4 P 6 qr; 6 1P^ 4 P
P P^
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Example 05 –CE , CB, CC
34
< <1.5
< 4 Q
N 4 p 1.5 ∥ P
5 P
4 1 4 6p 4 ∥ qr 6 ;5" 4 160^ 4
55 6 qr;
≅ 1
, 4 ^N 4N ⋅
⋅^ 4 160
P
P 4 qr! 4 2.5P^ 4 4 ∥ 4
6 qr; 6 1 4 63Ω
P^
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
neglecting
base
current
CascodeAmplifier , CE-CB
35
Q(
18'1.5
AC circuit:
6.8
5.6
1
4.7<
<
( 4 186.8 6 5.6 6 4.7 4 1.13
5
11.2
4.34.3
10.511.5 Q
1.5
<
<
4 p! q 4 1
q
^ 41.5qr;
4 245
, 4 245
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Some Notes:
36
qr pr q#r+-
E
BC
qr pr q#r+-
E
B C
qr pr q#r+-
E
B C
qr pr q#r+-
< < q≡ ≡
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
Summary
37
NPN
PQP&
^N 4P&P 6 qr; 6 1 6 PN
P 4 P 6 qr; 6 1P^ 4 P
P^ PN
P
^ 4PP5 6 qr1 6 ; 6 P
^P
P 4 qr 6 P 1 6 ;
P^
P^ 4 P ∥ P5 6 qr1 6 ;
QPNP
P&P
PP^
^N 4P&PN 6 qr; 6 1 6 P
P 4 qr 6 P 1 6 ;
P^ 4 P&
1. Lab
2. Power
3. Ser/Parl
4. Small sig.
5. Applic
? VA
38
PN
P
^ 4PP5 6 qr1 6 ; 6 P
^P
P 4 qr 6 P 1 6 ;
P^
P^ 4 P ∥ P5 6 qr1 6 ;
q PN
P ∥ q ^PP^
q 4 ∞^ 4
P ∥ qP5 6 qr1 6 ; 6 P ∥ qP 4 qr 6 P ∥ q 1 6 ;
P^ 4 P ∥ q ∥ P5 6 qr1 6 ;
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