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Lecture 06 - 1ELE2110A © 2008
Week 6: Small Signal Analysis for Single Stage BJT amplifiers
ELE 2110A Electronic Circuits
Lecture 06 - 2ELE2110A © 2008
Topics to cover …Family of single-stage BJT amplifiers
Small signal analysis – Common-Emitter Amplifier
Common-Emitter Amplifier with Emitter R
Common-Base Amplifier
Reading Assignment: Chap 14.1 – 14.5 of Jaeger and Blalock , orChap 5.7 of Sedra & Smith
Lecture 06 - 3ELE2110A © 2008
Signal Injection
Transistor is a VCCS:
To cause changes in current, vBE must be changed
– Base or emitter terminals can be used as i/pterminal
– Collector is not used as an i/p terminal because even if Early voltage is considered, collector voltage has negligible effect on terminal currents
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
=TV
BEvSICi exp
Lecture 06 - 4ELE2110A © 2008
Signal Extraction
Large changes in iC or iE create large voltage drops across collector and emitter resistors
– Collector or emitter can be used as o/pterminal
– Base terminal is not used as o/p terminal due to small iB
Lecture 06 - 5ELE2110A © 2008
BJT Amplifier FamilyConstraints for signal injection and extraction yield three families of amplifiers
Input Output
– Common-Emitter (C-E): Base Collector– Common-Base (C-B): Emitter Collector– Common-Collector (C-C): Base Emitter
– All circuit examples in the following slides use the four-resistor bias circuits to establish Q-point for the various amplifiers.
Lecture 06 - 6ELE2110A © 2008
Topics to cover …Family of single-stage BJT amplifiers
Small signal analysis – Common-Emitter Amplifier
Common-Emitter Amplifier with Emitter R
Common-Base Amplifier
Lecture 06 - 7ELE2110A © 2008
Using Small Signal Models in BJT ckt Analysis
Steps for using small-signal models:
Determine the DC operating point of the BJT– in particular, the collector current
Calculate small-signal model parameters gm, rπ, & re for this DC operating pointEliminate DC sources– Replace DC voltage sources with short circuits– Replace DC current sources with open circuits
Replace BJT with an equivalent small-signal model– Choose most convenient one depending on surrounding circuitry
Analyze
Lecture 06 - 8ELE2110A © 2008
Common Emitter Amplifier
Biased using the four-resistor networkAC input is injected to base through a coupling capacitorAC output is taken from collectorEmitter is ac-grounded
Lecture 06 - 9ELE2110A © 2008
DC operating point
All capacitors in original amplifier circuits are replaced by open circuits, disconnecting vI, RI, and R3 from circuit
Q-point can be found from dc equivalent circuit by using DC model for BJT
Lecture 06 - 10ELE2110A © 2008
AC EquivalentReplace all ∞ capacitors by short circuits, ∞ inductors by open circuits
Set all independent DC sources to 0, i.e., replace DC voltage sources by short circuits and DC current sources by open circuits
Replace transistor by small-signal model
Use small-signal AC equivalent to analyze AC characteristics of amplifier
Lecture 06 - 11ELE2110A © 2008
Simplified AC Equivalent
kΩ100kΩ3.43
kΩ30kΩ1021
==
==
RCRR
RRBR
Lecture 06 - 12ELE2110A © 2008
Small Signal Performance Parameters
Voltage gainInput resistanceOutput resistance
and sometimesCurrent gainInput signal range
In AC analysis, we are interested to find
Lecture 06 - 13ELE2110A © 2008
Voltage Gain
3RCRorLR =
LRmgbevov
bvcv
vtA −===
Terminal voltage gain between base and collector is:
Overall voltage gain from source vito output voltage across R3 is:
( )⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
+−=∴
===
π
πrBRIR
rBRLRmgvA
ivbev
vtAiv
bev
bevov
ivov
vA
Lecture 06 - 14ELE2110A © 2008
Input Resistance
The total resistance looking into the amplifier at coupling capacitor C1represents total resistance of the amplifier presented to signal source
ππ
πrRRrBRR
rBR
21xixv
in
)(xixv
===
=
Lecture 06 - 15ELE2110A © 2008
Output Resistance
Output resistance is the total equivalent resistance looking into the output of the amplifier at coupling capacitor C3
To find Rout, input source is set to 0 and test source is applied at output
CRorCRR
mgorC
R
≅==∴
++=
xixv
out
bevxvxvxi But vbe=0.
As ro>> RC.
Lecture 06 - 16ELE2110A © 2008
Example 1
Problem: Find voltage gain, input and output resistances.Given data: β= 65, VA =50 VAssumptions: Active-region operation, VBE=0.7 V, small signal operating conditions.
Analysis: To find the Q-point, dc equivalent circuit is constructed.
µA24566µA24165
µA71.3
==
===∴
BIEIBICI
BI
5)4106.1()1(510 =×+++ BIBEVBI β
V67.30)5()4106.1(4105
=∴
=−−×−−−
CEVEICEVCI
Active region of operation is correct.
Lecture 06 - 17ELE2110A © 2008
Example 1 (Cont.)Next we construct the ac equivalent and simplify it
0.84in
in)3out( −=+
−==⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
RIR
RRRmg
ivov
vA
S31064.9 −×==TVC
Img
kΩ64.6==
CI
TVr
βπ
kΩ223=+
=C
ICE
VA
Vor
kΩ23.6in == πrBRR
kΩ57.9out == orCRR
Lecture 06 - 18ELE2110A © 2008
Topics to cover …Family of single-stage BJT amplifiers
Small signal analysis – Common-Emitter Amplifier
Common-Emitter Amplifier with Emitter R
Common-Base Amplifier
Lecture 06 - 19ELE2110A © 2008
Common-Emitter Amplifier with Emitter R
AC equivalent:
A resistor exists at the emitter of the ac equivalent circuit
Also called emitter degeneratedCE amplifier
RE provides negative feedback
AC ground
OutputInput
Lecture 06 - 20ELE2110A © 2008
Negative Feedback due to RE
Assume iC for any reasoniE (=iC/α) vE
vBE
iC : counter act the original change
More discussion on negative feedback later in the course
Lecture 06 - 21ELE2110A © 2008
Overall Voltage Gain
Thus, the task breaks down into two steps:Find the terminal voltage gainFind the input resistance Rin.
Overall voltage gain from vi to vocan be expressed as:
b
ovt v
vA ≡
)||(||
inBI
inB
i
b
RRRRR
vv
+=
Define terminal voltage gain as:
And it can be observed that:
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
=≡ivb
v
bvov
ivov
vA
Lecture 06 - 22ELE2110A © 2008
Terminal Voltage GainUse hybrid-π model (neglecting ro):
for gmrπ = β >> 1
( )E
Eb
RriiRirv)1(
)1(++=
++=β
β
π
π
Lo iRv β−=
βπ =mgr
ERr
LR
bvov
vtA)1( ++
−=≡βπ
β
ERmgL
Rmg
ERrmgr
LRrmg
vtA+
−≅++
−=1)1( ππ
π
Using , we have
Effect of RE:For RE=0,– Upper limit of Avt
For gmRE>>1,– Avt becomes less dependent on gm
which varies widely
Increasing RE decreases voltage gain!
LRmgvtA −≅
ERLRvtA /−=
Lecture 06 - 23ELE2110A © 2008
Input ResistanceTo find Rin to find Vb/i:
)1()1(
)1(
ERmgrERrmgrERr
ib
vinR
+≅
++=
++==
πππ
βπ
( )Eb Rriv )1( ++= βπ
for gmrπ = β >> 1
Increasing RE increases input resistance!
Lecture 06 - 24ELE2110A © 2008
Output Resistance
0ixi0i ==⇒=∴ β ∞==∴xixv
outR
To find Rout:Set vi to zeroApply a test source at outputRout = vx/ix
E)iR(βev 1+=BIth RRR //=
KVL at left mesh:
( )
( )
0
0)1(
0
=⇒
=+++
=++
e
ethE
e
eth
v
vrRR
vvrRi
π
π
β
Lecture 06 - 25ELE2110A © 2008
Output ResistanceNow, we also include ro in our analysis:
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
+++≅≡∴πrthRER
EβRor
xixv
outR 1
KVL along loop 1:
evoβi)rx(ievrvxv +−=+=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛ += ERπrthRxiev //
Voltage at E:
Current division at Emitter:πrth
RERER
xii++
−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
++++
+=∴E
RπrthRxi
πrthRERE
Rxixiorxv //β
Put 2nd and 3rd equations into 1st equation:
for large value of ro
Lecture 06 - 26ELE2110A © 2008
Output Resistance
πβ rmg=Assuming and , withthRERr >>+ )( π ERor >>
orR )1(out += β
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
+++≅π
βrthRER
ERorR 1out
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
+=++≅ )//(11out ERrmgorrERERrmg
orR πππ
Effect of RE:For RE = 0, Rout = ro
For RE = ∞, upper limit ofRout is obtained:
Increasing RE increases Rout!
Lecture 06 - 27ELE2110A © 2008
Current Gain
Terminal current gain:
Overall current gain:
β−=≡iLi
itA
ii
inB
Bit
s
L
s
Li RR
RAii
ii
iiA
+==≡
Lecture 06 - 28ELE2110A © 2008
Input Signal Range
321
AC
beT
C
DCC
TbeCVv
CC
vVII
VvIeIi Tbe
+=
+≅= )/1(/
Remember, in deriving the linear small signal model, we assumed vbe << VT :
Typically, vbe is required to be less than 5mV!
Lecture 06 - 29ELE2110A © 2008
Input Signal Range
πβπππ rERERmg
bv
ERrb
vrirbev
/1)1( ++=
++==
V)1(005.01005.0 ERmgERERmg
bv +≅++≤
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
β
1>>ERmgIf , vb can be increased beyond 5 mV limit.
( )Eb Rriv )1( ++= βπ
V025.0
1
=>>∴
>>≅=
TVEREITV
EREI
TVER
CI
ERmgCondition for gmRE >>1 :
Increasing RE increases input signal range!
Lecture 06 - 30ELE2110A © 2008
Summary of Emitter Degenerated Common-Emitter Amplifier
Terminal voltage gain: Inverting and Large
Overall voltage gain: Inverting and Large
Input resistance: Large
Output resistance: Large
Terminal Current gain: Large
Effects of the resistor at Emitter:– Voltage gain decreased, but more stabilized (less sensitive to gm)– Input signal range increased– Input and output resistance increased
)1( ERmgr +π
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ + )(1 ERrmgor π
ERmgL
Rmg
+−
1
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
++−
in//RB
RI
Rin//R
BR
ERmgL
Rmg
1
β−
Lecture 06 - 31ELE2110A © 2008
Topics to cover …Family of single-stage BJT amplifiers
Small signal analysis – Common-Emitter Amplifier
Common-Emitter Amplifier with Emitter R
Common-Base Amplifier
Lecture 06 - 32ELE2110A © 2008
Common-Base Circuits
AC/Small-signal equivalent:
73 || RRRL =
Output
Input
AC ground
Load
SignalSource
Lecture 06 - 33ELE2110A © 2008
Terminal Voltage Gain
Non-inverting!Magnitude same as the CE amplifier with RE=0.
LRmgevov
ACBvt +=≡
Apply test source to input (E)And use BJT small signal model:
Lecture 06 - 34ELE2110A © 2008
Input Resistance
mgmgr
mgr
r
iev
RCBin
1)1//(1
≅=+
== ππ
π
eme vg
rvi +=π
KCL at emitter:
Rin is small (as gm is usually large)!
(gm=IC/VT)
Lecture 06 - 35ELE2110A © 2008
Overall Voltage Gain
IRmgLRmg
RIR
R
IRRmgL
RmginRR
IR
inRRvtA
ivev
evov
ivov
ACBv
+≅
++=
+===
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
1
4
4)//
4(1
)//4
(
//4
Overall voltage gain is
For ,
This is the upper bound.
For ,
1<<IRmg
LRmgACBv +=
1>>IRmg
IR
LR
ACBv +=
IRR >>4for
For large voltage gain, a very small RI is required!Not a good candidate for voltage amplifier
Lecture 06 - 36ELE2110A © 2008
ExampleProblem: Find overall voltage gain.Given data: β=100, Q-point values: IC=245uA,VCE=3.64V, gm=9.8mS, rπ=10.2kΩ, rο=219kΩ.Assumptions: Small-signal operating conditions.Analysis:
kΩ1873
Ω102/1
==
=≅
RRLRmgRCB
in
176=+= LRmgACBvt
59.84
4)
4(1
+=++
=⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
RIR
R
RIRmgACB
vtACBv
Lecture 06 - 37ELE2110A © 2008
Input Signal Range
For small-signal operation,
Relative size of gm and RI determine signal-handling limit.
V)1(005.0 IRmgb
v +≤
)1(ebviv IRmg+≅ for R4 >> RI.
⎟⎟⎠
⎞⎜⎜⎝
⎛++
=+
=4
4
44
4
)//(1////
RRR
RRgv
RRRRRvv
IIm
i
inI
inieb
Lecture 06 - 38ELE2110A © 2008
Output Resistance
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
++=∴π
βrthR
thRorRCB
out 1
πβ rmg=
⎟⎟⎠
⎞⎜⎜⎝
⎛ +≅ )//(1 thRrmgorRCBout π
Using
Rout here is equivalent to the Rout of CE amplifier with RE=Rth and resistance at base equal to zero.
Rout is large.
Lecture 06 - 39ELE2110A © 2008
Current Gain
Terminal current gain:
Current gain from source to load:
1+≅== αeili
ACBit
i
inR for RitARinR
RitA
iei
eili
ili
ACBi >>=≅
+===
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
1
Lecture 06 - 40ELE2110A © 2008
Summary of Common Base Amplifier
Terminal voltage gain: Non-inverting and LargeInput resistance: LowOutput resistance: HighCurrent gain: close to UnityInput range: determined by gmRth
Excellent for use as a current buffer
Lecture 06 - 41ELE2110A © 2008
Current Buffer
RsRL
Io
Ii
Rs Ri RLRoA=1
Current buffer
VoIi
Io
iiLS
So II
RRRI <<+
=
LOSiii
OL
Oi
iS
So
RRRRIAIRR
RAIRR
RI
>><<=≅
+⎟⎟⎠
⎞⎜⎜⎝
⎛+
=
and for
• Requirement of a current buffer:- Low input resistance- High output resistance- Unity current gain
sL RR >>
Only a small portion of source current is delivered to load!
• With a buffer:
• Without a buffer:
for