Upload
austin-barrett
View
212
Download
0
Embed Size (px)
Citation preview
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
DMT 231 / 3ELECTRONICS II
Lecture IV AC Analysis II
[BJT Common-Emitter Amplifier]
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
RC
RB
vs
vO
VBB
VC
C
Example I
Given : = 100, VCC = 12V
VBE = 0.7V, RC = 6k,
RB = 50k, and VBB = 1.2V
Calculate the small-signal voltage gain.
Small-signal hybrid- equivalent circuit
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
1.
2.
3.
4.
5.
6.
AR
VVI
B
onBEBBBQ 10
50
7.02.1)(
mAAII BQCQ 1)10(100
VRIVV CCQCCCEQ 6)6)(1(12
kI
Vr
CQ
T 6.21
)026.0)(100(
VmAV
Ig
T
CQm /5.38
026.0
1
4.11
B
Cms
ov Rr
rRg
V
VA
Example I: Solutions
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
Basic Common-Emitter Amplifier Circuit
vs
RS
R1
R2
RC
CCvO
VCCExample II
Given : = 100, VCC = 12V
VBE(on) = 0.7V, RS = 0.5k,
RC = 6k, R1 = 93.7k, R2 = 6.3k
and VA = 100V.
Calculate the small-signal voltage gain.
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
Como RrVgV
sS
VRrRR
rRRV
21
21
Co
S
ms
ov Rr
RrRR
rRRg
V
VA
21
21
Coo RrR
rRRRi 21
B C
E
R1 \\ R2Vs
RS
RCrOr gmV
Vo
Ri Ro
Example II: SolutionSmall-signal equivalent circuit
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
• The basic common-emitter circuit used in previous analysis causes a serious defect :– If BJT with VBE(on) = 0.7 V is used, IB = 9.5 μA & IC = 0.95 mA– But, if new BJT with VBE(on) = 0.6 V is used, IB = 26 μA & BJT goes
into saturation; which is not acceptable Previous circuit is not practical
– So, the emitter resistor is included: Q-point is stabilized against variations in β, as will the voltage gain, AV
• Assumptions– CC acts as a short circuit– Early voltage = ∞ ==> ro neglected due to open circuit
Basic Common-Emitter Amplifier
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
Common-Emitter Amplifier with Emitter Resistor
CE amplifier with emitter resistor Small-signal equivalent circuit (with current gain parameter, β)
inside transistor
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
Common-Emitter Amplifier
with Emitter Resistor• ac output voltage
• Input voltage loop
• Input resistance, Rib
• Input resistance to amplifier, Ri
• Voltage divider equation of Vin to Vs
Remember: Assume VA is infinite, ro is neglected
Cbo RIV
Ebbbin RIIrIV
Eb
inib Rr
I
VR 1
ibi RRRR 21
sSi
iin V
RR
RV
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
Common-Emitter Amplifier
with Emitter Resistor
Cont..
• So, small-signal voltage gain, AV
• If Ri >> Rs and (1 + β)RE >> rπ
Remember: Assume VA is infinite, ro is neglected
Si
i
E
Cv
sib
inC
s
Cb
s
ov
RR
R
Rr
RA
VR
VR
V
RI
V
VA
1
1
E
C
E
Cv R
R
R
RA
1
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
Example IIIGiven : = 100, VBE(on) = 0.7V,
VT = 26 mV and VA = ∞.Determine:
(i) Base-emitter input resistance, r
(ii) transconductance, gm
(iii) small-signal transistor output resistance, ro
(iv) Input resistance to the base, Rib
(v) Input resistance to the amplifier, Ri
(vi) Small-signal voltage gain, Av.
Common-Emitter Amplifier with Emitter Resistor
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
RS
R1
R2 RE
RC
vs
vO
CC
VCC
CE
B C
E
Vo
Vs RC
RS
r roR1|| R2 gmV
Emitter bypass capacitor, CE provides a short circuit
to ground for the ac signals
Common-Emitter Amplifier
with Emitter Bypass Capacitor
Small-signal hybrid-π equivalent circuit
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
Example IVGiven : = 100, VBE(on) = 0.7V,
VT = 26 mV and VA = 100.Determine:
(i) Quiescent value of base current, IBQ
(ii) Quiescent value of collector current, ICQ
(iii) Quiescent value of collector-emitter voltage, VCEQ
(iv) Base-emitter input resistance, r
(v) transconductance, gm
(vi) small-signal transistor output resistance, ro
(vii) Input resistance seen by the signal source, Rin
(viii) Output resistance looking back into the output terminal, Ro
(ix) Small-signal voltage gain, Av.
Common-Emitter Amplifier with Emitter Bypass
Capacitor
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
AC LOAD LINE ANALYSIS
• DC load line – Visualized the relationship between Q-point & transistor
characteristics
• AC load line – Visualized the relationship between small-signal response
& transistor characteristics
– Occurs when capacitors added in transistor circuit
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
AC LOAD LINE ANALYSIS
Common-emitter amplifier with emitter bypass capacitor
Note: The next DC & AC load line analysis will be based on this circuit
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
• KVL on C-E loop
AC LOAD LINE ANALYSIS - DC Load Line
21
21
21
21
21
1 Slope
)(So,
11
1, when point,-QFor
)(1
1 when ,)(
1
)(
EEC
EECCQCEQ
EECCCCE
CEEECCECC
EEECECC
RRR
-
RRRIVVV
RRIRIVVV
IIVRRIVRI
VRRIVRIV
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
• KVL on C-E loop
AC LOAD LINE ANALYSIS - AC Load Line
1
11
1
1- Slope
)(
Assuming
0
EC
ECcEcCcce
ec
EeceCc
RR
RRiRiRiv
ii
RivRi
AC equivalent circuit
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
AC and DC Load Lines
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
RS
R1
R2 RE
RC
RL
vs
vO
CC1
CC2
VCC
vO
vs R1 R2
RS
RE
RC RL
AC LOAD LINE ANALYSIS
Common-emitter amplifier with emitter resistor
AC equivalent circuit
Note: The DC & AC load line analysis will be based on these circuits
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
AC LOAD LINE ANALYSIS - DC Load Line
+ VCE
0
+ IC
ICQ
VCEQ
Q
CCV
EC
CC
RR
V
EC RR
1
Slope
• KVL at C-E loop
EC
ECCQCEQCC
CEQECCQCC
ECCECC
EECECCCC
RR
RRIVV
VRRIV
RIVRI
RIVRIV
1- Slope
assume point, -Q For
)(
)(
11
1
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
+ VCE0
+ IC
ICQ
VCEQ
Q
CCV
EC
CC
RR
V
i
v
LC
CEQ
RR
Vi
)( LCCQ RRIv )( LCCQCEQoffcut RRIVv
LC
CEQCQcsat RR
VIi
AC LOAD LINE ANALYSIS - AC Load Line
SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II
Self-ReadingChapter 6, page 399 - 402: Basic
Common-Emitter Amplifier Circuit
Chapter 6, page 402 - 409: Circuit with Emitter Resistor
Chapter 6, page 409 - 413: Circuit with Emitter Bypass Capacitor
Chapter 6, page 415 - 418: AC Load Line Analysis
Donald A. Neamen, MICROELECTRONICS Circuit Analysis and Design, Third Edition