Download Week 6: Small Signal Analysis for Single Stage BJT amplifiers

ELE 2110A Electronic Circuits
Week 6: Small Signal Analysis for
Single Stage BJT amplifiers
ELE2110A © 2008
Lecture 06 - 1
Topics to cover …
z
Family of single-stage BJT amplifiers
z
Small signal analysis
– Common-Emitter Amplifier
z
Common-Emitter Amplifier with Emitter R
z
Common-Base Amplifier
Reading Assignment:
Chap 14.1 – 14.5 of Jaeger and Blalock , or
Chap 5.7 of Sedra & Smith
ELE2110A © 2008
Lecture 06 - 2
Signal Injection
v
i = I exp BE
C
S
V
T
⎛
⎜
⎜
⎜⎜
⎝
⎞
⎟
⎟
⎟⎟
⎠
z
Transistor is a VCCS:
z
To cause changes in current, vBE must be
changed
– Base or emitter terminals can be used as i/p
terminal
– Collector is not used as an i/p terminal because
even if Early voltage is considered, collector
voltage has negligible effect on terminal
currents
ELE2110A © 2008
Lecture 06 - 3
Signal Extraction
z
Large changes in iC or iE create large
voltage drops across collector and emitter
resistors
– Collector or emitter can be used as o/p
terminal
– Base terminal is not used as o/p terminal due to
small iB
ELE2110A © 2008
Lecture 06 - 4
BJT Amplifier Family
z
Constraints for signal injection and extraction yield three
families of amplifiers
Input
Output
– Common-Emitter (C-E):
– Common-Base (C-B):
– Common-Collector (C-C):
Base
Emitter
Base
Collector
Collector
Emitter
– All circuit examples in the following slides use the four-resistor
bias circuits to establish Q-point for the various amplifiers.
ELE2110A © 2008
Lecture 06 - 5
Topics to cover …
z
Family of single-stage BJT amplifiers
z
Small signal analysis
– Common-Emitter Amplifier
z
Common-Emitter Amplifier with Emitter R
z
Common-Base Amplifier
ELE2110A © 2008
Lecture 06 - 6
Using Small Signal Models in BJT ckt Analysis
Steps for using small-signal models:
z
Determine the DC operating point of the BJT
– in particular, the collector current
z
z
Calculate small-signal model parameters gm, rπ, & re for
this DC operating point
Eliminate DC sources
– Replace DC voltage sources with short circuits
– Replace DC current sources with open circuits
z
Replace BJT with an equivalent small-signal model
– Choose most convenient one depending on surrounding circuitry
z
Analyze
ELE2110A © 2008
Lecture 06 - 7
Common Emitter Amplifier
z
z
z
z
Biased using the four-resistor network
AC input is injected to base through a coupling capacitor
AC output is taken from collector
Emitter is ac-grounded
ELE2110A © 2008
Lecture 06 - 8
DC operating point
z
All capacitors in original amplifier circuits are replaced by
open circuits, disconnecting vI, RI, and R3 from circuit
z
Q-point can be found from dc equivalent circuit by using
DC model for BJT
ELE2110A © 2008
Lecture 06 - 9
AC Equivalent
z
ELE2110A © 2008
Replace all ∞ capacitors by short
circuits, ∞ inductors by open circuits
z
Set all independent DC sources to 0,
i.e., replace DC voltage sources by
short circuits and DC current sources
by open circuits
z
Replace transistor by small-signal
model
z
Use small-signal AC equivalent to
analyze AC characteristics of amplifier
Lecture 06 - 10
Simplified AC Equivalent
R
B
= R R = 10 k Ω 30 k Ω
1 2
R = R
ELE2110A © 2008
C
R
3
= 4 . 3 k Ω 100 k Ω
Lecture 06 - 11
Small Signal Performance Parameters
In AC analysis, we are interested to find
z
z
z
Voltage gain
Input resistance
Output resistance
and sometimes
z Current gain
z Input signal range
ELE2110A © 2008
Lecture 06 - 12
Voltage Gain
Terminal voltage gain
between base and collector is:
v
v
Avt = v c = v o = − g m R
L
b
be
Overall voltage gain from source vi
to output voltage across R3 is:
⎞
⎛v
v
⎟
be ⎟ = A ⎜⎜ be
vt ⎜ v
v ⎟
i ⎠
i
⎝
⎡
⎤
R rπ
⎢
⎥
B
⎥
∴ Av = − g m R ⎢⎢
L ⎢ R + R r ⎥⎥
B π ⎦
I
⎣
v o ⎛⎜ v o
Av = v = ⎜ v
i ⎜⎝ be
⎞⎛
⎟⎜
⎟⎜
⎟⎜
⎠⎝
(
R
L
⎞
⎟
⎟
⎟
⎠
)
= ro R R
C 3
ELE2110A © 2008
Lecture 06 - 13
Input Resistance
z
The total resistance looking into the
amplifier at coupling capacitor C1
represents total resistance of the
amplifier presented to signal source
v x = i x ( R rπ )
B
v
R = x = R rπ = R R rπ
B
1 2
in
ix
ELE2110A © 2008
Lecture 06 - 14
Output Resistance
z
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
vx vx
+
+ gm v
ix =
But vbe=0.
be
R
ro
C
vx
= R ro ≅ R As ro>> RC.
∴ R out =
C
C
ix
z
ELE2110A © 2008
Lecture 06 - 15
Example 1
Analysis: To find the Q-point, dc
equivalent circuit is constructed.
105 I +V + (β +1)I (1.6×104) = 5
B BE
B
z
z
z
Problem: Find voltage gain, input
and output resistances.
Given data: β= 65, VA =50 V
Assumptions: Active-region
operation, VBE=0.7 V, small signal
operating conditions.
ELE2110A © 2008
∴ I = 3 .71µA
B
I = 65 I = 241µA
B
C
I = 66 I = 245µA
E
B
5 −104 I −V − (1.6 ×104 ) I − (−5) = 0
E
C CE
∴V = 3.67V
CE
Active region of operation is correct.
Lecture 06 - 16
Example 1 (Cont.)
Next we construct the ac
equivalent and simplify it
R = R rπ = 6.23kΩ
in B
I
gm = C = 9.64×10− 3S
V
T
βV
rπ = T = 6.64kΩ
I
C
V +V
ro = A CE = 223kΩ
I
C
ELE2110A © 2008
R out = R
vo
Av =
v
i
r = 9 . 57 k Ω
C o
⎡
⎤
R
⎢
⎥
in
⎢
⎥ = − 84 .0
= − gm ( Rout R ) ⎢
3 ⎢ R + R ⎥⎥
⎢⎣ I
in ⎥⎦
Lecture 06 - 17
Topics to cover …
z
Family of single-stage BJT amplifiers
z
Small signal analysis
– Common-Emitter Amplifier
z
Common-Emitter Amplifier with Emitter R
z
Common-Base Amplifier
ELE2110A © 2008
Lecture 06 - 18
Common-Emitter Amplifier with Emitter R
AC equivalent:
Input
AC ground
Output
A resistor exists at the emitter of
the ac equivalent circuit
Also called emitter degenerated
CE amplifier
RE provides negative feedback
ELE2110A © 2008
Lecture 06 - 19
Negative Feedback due to RE
z
z
z
z
z
z
ELE2110A © 2008
Assume iC Ç for any reason
iE (=iC/α) Ç
vE Ç
vBE È
iC È : counter act the original
change
More discussion on negative
feedback later in the course
Lecture 06 - 20
Overall Voltage Gain
Overall voltage gain from vi to vo
can be expressed as:
vo
Av ≡ =
v
i
Define terminal voltage gain as:
And it can be observed that:
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞⎛
⎟⎜
⎟⎜
⎟⎜
⎟⎜
⎟⎜
⎠⎝
⎞
vo vb ⎟⎟
⎟
v v ⎟⎟
b i⎠
vo
Avt ≡
vb
vb
RB || Rin
=
vi RI + ( RB || Rin )
Thus, the task breaks down into two steps:
z Find the terminal voltage gain
z Find the input resistance Rin.
ELE2110A © 2008
Lecture 06 - 21
Terminal Voltage Gain
Use hybrid-π model (neglecting ro):
Using rπ g m = β , we have
g m rπ R
gm R
L
L
Avt = −
≅−
1+ g m R
rπ + ( g m rπ +1)R
E
E
for gmrπ = β >> 1
vb = irπ + ( β + 1)iRE
= i(rπ + ( β + 1) RE )
vo = − βiRL
βR
vo
L
Avt ≡ = −
v
rπ + (β +1)R
E
b
ELE2110A © 2008
Effect of RE:
z For RE=0,
Avt ≅ − gm R
L
– Upper limit of Avt
z
For gmRE>>1,
Avt = − R / R
L E
– Avt becomes less dependent on gm
which varies widely
z
Increasing RE decreases voltage gain!
Lecture 06 - 22
Input Resistance
To find Rin Æ to find Vb/i:
vb = i (rπ + ( β + 1) RE )
v
R = b = rπ + (β +1)R
in
E
i
= rπ + ( gmrπ +1)R
E
≅ rπ (1+ gm R )
E
for gmrπ = β >> 1
z
Increasing RE increases input resistance!
ELE2110A © 2008
Lecture 06 - 23
Output Resistance
To find Rout:
z Set vi to zero
z Apply a test source at output
z Rout = vx/ix
Rth = RI // RB
ve = (β +1 )iR
E
KVL at left mesh:
i (Rth + rπ ) + ve = 0
ve
(Rth + rπ ) + ve = 0
( β + 1) RE
⇒ ve = 0
∴
i = 0 ⇒ ix = βi = 0
ELE2110A © 2008
vx
∴Rout =
=∞
ix
Lecture 06 - 24
Output Resistance
Now, we also include ro in our analysis:
KVL along loop 1:
vx = vr + ve = (ix − βi)ro + ve
Voltage at E:
⎛⎛
⎞
⎞
ve = ix ⎜⎜ ⎜⎜ R + rπ ⎟⎟ // R ⎟⎟
E⎠
⎠
⎝ ⎝ th
Current division at Emitter:
i = −ix
Put 2nd and 3rd equations into 1st equation:
⎛
⎜
⎜
⎜
⎜
⎜
⎝
∴ v x = ro i x + β i x
R
R
⎞
⎟
⎟
⎟
⎟
⎟
⎠
⎛⎛
⎞
E
+ i x ⎜⎜ ⎜⎜ R + rπ ⎟⎟ // R
E
⎠
⎝ ⎝ th
+ R + rπ
E
th
⎞
βR
⎟
E
⎟
∴Rout ≡ ≅ ro 1+
⎟
R
R
r
+
+
ix
E th π ⎟⎠
vx
ELE2110A © 2008
⎛
⎜
⎜
⎜
⎜
⎝
R
E
R + R + rπ
E th
⎞
⎟
⎟
⎠
for large value of ro
Lecture 06 - 25
Output Resistance
βR
⎛
⎜
⎜
⎜
⎜
⎝
E
Rout ≅ ro 1+
R + R + rπ
E th
Assuming (rπ + RE ) >> Rth
and
⎞
⎟
⎟
⎟
⎟
⎠
ro >> R , with β = gmrπ
E
gmrπ R ⎞⎟
E ⎟ = r ⎛⎜⎜1+ g (r // R ) ⎞⎟⎟
Rout ≅ ro 1+
m π
E ⎠
R + rπ ⎟⎟ o ⎝
E
⎠
⎛
⎜
⎜
⎜⎜
⎝
Effect of RE:
z For RE = 0, Rout = ro
z For RE = ∞, upper limit of
Rout is obtained:
Rout = (β +1)ro
z
Increasing RE increases Rout!
ELE2110A © 2008
Lecture 06 - 26
Current Gain
ii
z
Terminal current gain:
z
Overall current gain:
ELE2110A © 2008
i
A ≡ L = −β
it i
Ai ≡
RB
iL iL i
=
= Ait
RB + Rin
is
i is
Lecture 06 - 27
Input Signal Range
Remember, in deriving the linear small signal model,
we assumed vbe << VT :
iC = I C e vbe / VT ≅ I C (1 + vbe / VT )
IC
= IC +
v
{ V be
DC
1T23
AC
Typically, vbe is required to be less than 5mV!
ELE2110A © 2008
Lecture 06 - 28
Input Signal Range
vb = i (rπ + ( β + 1) RE )
v
v
b
b
v = irπ = rπ
=
be
rπ + (β +1)R
1+ g m R + R / rπ
E
E
E
⎛
⎜
⎜
⎜
⎜
⎝
⎛
⎜
⎜
⎜
⎜
⎝
R
⎞ ⎞⎟
⎟
⎟⎟
⎟⎟
⎟⎟
⎠⎠
v ≤ 0.005 1+ g m R + E ≅ 0.005(1+ gm R )V
E
E β
b
If gm R >> 1 , vb can be increased beyond 5 mV limit.
E
I R
I R
E
C
≅ E E >>1
gm R =
E
V
V
T
T
∴I R >>V = 0.025V
E E
T
z
Condition for gmRE >>1 :
z
Increasing RE increases input signal range!
ELE2110A © 2008
Lecture 06 - 29
Summary of Emitter Degenerated
Common-Emitter Amplifier
z
Terminal voltage gain: Inverting and Large
−
gm R
L
1+ g m R
E
⎡
⎢
⎢
⎢
⎢
⎣⎢
⎤
R //R
⎥
B in ⎥
L
−
⎥
1+ g m R RI + RB //Rin ⎥⎦⎥
E
gm R
z
Overall voltage gain: Inverting and Large
z
Input resistance: Large
z
Output resistance: Large ro ⎜⎜1+ gm (rπ RE ) ⎟⎟
⎠
⎝
z
Terminal Current gain: Large
z
Effects of the resistor at Emitter:
rπ (1+ gm R )
E
⎞
⎛
−β
– Voltage gain decreased, but more stabilized (less sensitive to gm)
– Input signal range increased
– Input and output resistance increased
ELE2110A © 2008
Lecture 06 - 30
Topics to cover …
z
Family of single-stage BJT amplifiers
z
Small signal analysis
– Common-Emitter Amplifier
z
Common-Emitter Amplifier with Emitter R
z
Common-Base Amplifier
ELE2110A © 2008
Lecture 06 - 31
Common-Base Circuits
Output
AC/Small-signal equivalent:
Load
AC ground
Input
ELE2110A © 2008
RL = R3 || R7
Signal
Source
Lecture 06 - 32
Terminal Voltage Gain
CB ≡ vo = + g R
Avt
m L
ve
z
Apply test source to input (E)
And use BJT small signal model:
ELE2110A © 2008
z
Non-inverting!
Magnitude same as the CE
amplifier with RE=0.
Lecture 06 - 33
Input Resistance
KCL at emitter:
ve
i = + g m ve
rπ
v
CB = e = rπ
= rπ //( 1 ) ≅ 1
Rin
g
g
i rπ g m +1
m
m
z
Rin is small (as gm is usually large)!
(gm=IC/VT)
ELE2110A © 2008
Lecture 06 - 34
Overall Voltage Gain
For gm R << 1 ,
I
CB
Av = + gm RL
This is the upper bound.
For gm R >>1 ,
I
R
CB
Av = + L
R
I
Overall voltage gain is
⎤
⎡
⎞
⎛
⎞⎛
R // R
⎢
⎜ v ⎟⎜ v ⎟
v
in ⎥⎥
CB = o = ⎜ o ⎟⎜ e ⎟ = A ⎢
4
Av
⎜
⎟
⎥
⎢
v ⎜⎜⎝ ve ⎟⎟⎠⎜⎜ v ⎟⎟ vt ⎢ R + ( R // R ) ⎥
⎢⎣ I
i
⎝ i ⎠
4 in ⎥⎦
⎛
⎞
⎜
⎟
R
gm R
⎜
L
4 ⎟⎟
=
⎜
1+ g m (R // R ) ⎜⎜ R + R ⎟⎟
4 I ⎝ I
4⎠
gm R
L
≅
for R4 >> RI
1+ g m R
I
ELE2110A © 2008
z
z
For large voltage gain, a
very small RI is required!
Not a good candidate for
voltage amplifier
Lecture 06 - 35
Example
z
z
z
z
Problem: 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:
CB
Rin ≅1/ gm =102Ω
R = R R =18kΩ
L 3 7
ACB
vt = + gm RL = 176
⎛
⎞
CB
⎜
⎟
R
A
⎜
vt
4 ⎟⎟ = +8.59
=
ACB
⎜
v
1+ gm ( R R ) ⎜⎜ R + R ⎟⎟
4⎠
I 4 ⎝ I
ELE2110A © 2008
Lecture 06 - 36
Input Signal Range
⎛ R4 ⎞
R4 // Rin
vi
⎜⎜
⎟⎟
veb = vi
=
RI + R4 // Rin 1 + g m ( RI // R4 ) ⎝ RI + R4 ⎠
v ≅ v (1+ gm R )
I
i eb
For small-signal operation,
for R4 >> RI.
v ≤ 0.005(1+ gm R )V
I
b
Relative size of gm and RI determine signal-handling limit.
ELE2110A © 2008
Lecture 06 - 37
Output Resistance
Rout here is equivalent to the Rout of
CE amplifier with RE=Rth and resistance
at base equal to zero.
z
⎛
⎜
⎜
⎜
⎜
⎝
∴RCB
out = ro 1+
βR
th
R + rπ
th
⎞
⎟
⎟
⎟
⎟
⎠
Using β = gmrπ
⎞
⎛
CB
Rout ≅ ro ⎜⎜⎝1+ gm (rπ // Rth ) ⎟⎟⎠
z
ELE2110A © 2008
Rout is large.
Lecture 06 - 38
Current Gain
i
i
CB = l = α ≅ +1
Ait
ie
z
Terminal current gain:
z
Current gain from source to load:
i ⎛⎜ i ⎞⎟⎛⎜ i ⎞⎟
R
CB = l = ⎜ l ⎟⎜ e ⎟ = A
≅ A =1 for R >> R
Ai
⎟
⎟⎜
⎜
it
in
R + R it
i ⎜⎝ ie ⎟⎠⎜⎜ i ⎟⎟
in
⎝
⎠
ELE2110A © 2008
Lecture 06 - 39
Summary of Common Base Amplifier
z
Terminal voltage gain: Non-inverting and Large
Input resistance: Low
Output resistance: High
Current gain: close to Unity
Input range: determined by gmRth
z
Excellent for use as a current buffer
z
z
z
z
ELE2110A © 2008
Lecture 06 - 40
Current Buffer
• Without a buffer:
Io
Rs
Ii
Io =
RL
RS
I i << I i for RL >> Rs
RS + RL
Only a small portion of source
current is delivered to load!
• With a buffer:
Current buffer
Io
Ii
Rs
Ri
A=1
Ro
RL
Vo
⎛ RS
⎞
RO
⎜
Io = ⎜
I i ⎟⎟ A
⎝ RS + Ri ⎠ RL + RO
≅ AI i = I i for Ri << RS and RO >> RL
• Requirement of a current buffer:
- Low input resistance
- High output resistance
- Unity current gain
ELE2110A © 2008
Lecture 06 - 41