Bipolar Differential Amplifiers: Qualitative Analysis
Common Mode
Differential Mode
Different modes of operation of the BJT differential pair: (a) The differential pair with a common-mode input signal vCM. (b) The differential pair with a
“large” differential input signal. (c) The differential pair with a large differential input signal of polarity opposite to that in (b). (d) The differential
pair with a small differential input signal vi. Note that we have assumed the bias current source I to be ideal (i.e., it has an infinite output
resistance) and thus I remains constant with the change in vCM.
Bipolar Differential Amplifiers: Large Signal Analysis
The exponential relationship applied to each
of the two transistors may be written as:
I
I
iE1 = S e( vB1 −vE ) / VT and iE 2 = S e( vB 2 −vE ) / VT
α
α
These two equations can be combined to obtain
iE1
= e( vB1 −vB 2 ) / VT
iE 2
which can be manipulated to yield
iE1
iE 2
1
1
and
=
=
( vB 2 − vB 1 ) / VT
( vB1 − vB 2 ) / VT
iE1 + iE 2 1 + e
iE1 + iE 2 1 + e
From the circuit we have
iE1 + iE 2 = I
Which may be used to obtain the following
expressions for iE1 and iE 2
iE1 =
iC1
I
( vB 2 − vB1 ) / VT
and iE 2 =
I
( vB 1 − vB 2 ) / VT
1+ e
1+ e
and iC 2 may be obtained by multiplying
iE1 and iE 2 by α which is almost unity and
plotted as shown in the figure
Bipolar Differential Amplifiers: Linearization
For Your Information
What is the role of the
degeneration resistance (Re)?
The transfer characteristics of the BJT differential pair (a) can be linearized (b) (i.e., the linear range of operation can be extended) by including resistances in the emitters.
Bipolar Differential Amplifiers: DC Analysis (Example 1)
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Problem:
Find the Q-points of transistors in the shown
differential amplifier.
Given data: VCC=VEE=15 V, REE=RC=75kΩ, β =100
Analysis:
⎛⎜15 − 0 .7 ⎞⎟ V
−V
⎠
BE
EE
I =
= ⎝
= 95 .3μA
E
3
2R
2(75 ×10 )Ω
EE
100
I =α I =
I = 94.4μ A
E 101 E
C
V
94.4μ A
I =
=
= 0.944μ A
B β
100
IC
V = 15 − I R = 7.92V
C
C C
= V −V = 7.92V -(-0.7V) = 8.62V
V
CE C E
Due to symmetry, both
transistors are biased at
Q-point (94.4 μA, 8.62V)
Bipolar Common-mode Input Voltage Range
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Problem:
Find the max. VIC before saturation in the shown
differential amplifier.
Given data: VCC=VEE=15 V, REE=RC=75kΩ, β =100
Analysis:
We want to find max. VIC while the C-B
junction is reverse biased.
V
=V
− I R −V ≥ 0
CB CC C C
IC
V −V +V
I = α IC BE EE
C
2 REE
⎛
⎞
RC ⎜⎝VEE −VBE ⎟⎠
1−α
2REE
VCC
∴V ≤ V
IC CC
RC
1+α
2REE
For symmetrical power supplies (VEE=VCC) ,
VEE >> VBE, and RC = REE,
V CC
V
≤
= 5V
IC
3
Bipolar Differential Amplifiers: DC Analysis (Example 2)
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Problem: Find vE ,vc1 , and vc2 in the shown differential amplifier.
Given data: VCC=VEE= 5 V, REE=RC=1kΩ, α ≈, |VBE|=0.7 V
Analysis:
We can assume Q1 to be off and Q2 on
vE
vC1
= +0.7 V
= −5 + R
C
* I C1
= −5 + 0 = −5 V
I E 2 = I E = (5 − 0.7) /1 = 4.3 mA
I C 2 = α I E 2 = 4.3 mA
vC 2 = −5 + RC * I C 2
= −5 + 4.3 = −0.7 V
Differential-mode Gain and Input Resistance
v
v
id
v =
− ve v = − id − ve
3 2
4
2
( g +1/ r )(v + v ) =1/ R ve
π 3 4
m
EE
∴ve (1/ R + 2/ r + 2 g ) = 0 → ve = 0
EE
m
π
Emitter node in differential amplifier represents
virtual ground for differential-mode input signals.
v
v
v = − id
4
2
∴v = id
3 2
Output signal voltages are:
v
v
v
v = + gm R id
= − gm R id
c2
C 2
c1
C 2
∴v = − gm R v
C id
od
Differential-mode Gain and Input Resistance (contd.)
Differential-mode gain for balanced output, v od = v c1 − v c2 is:
v
A = od
= − gm R
C
dd v
id v = 0
ic
If either vc1 or vc2 is used alone as output, output is said to be single-ended.
A
gm R
C
c1
A =
=−
= dd
dd1 v
2
2
id v = 0
ic
v
A
gm R
C
c2
A
=
=
= − dd
dd 2 v
2
2
id v = 0
ic
v
Differential-mode input resistance is small-signal resistance presented to
differential-mode input voltage between the two transistor bases.
(v / 2)
i = id
b1
rπ
If vid =0, R
od
= 2( R ro ) ≅ 2R
C
C
∴R = v / i = 2rπ
id id b1
. For single-ended outputs, R
od
≅R
C
Common-mode Gain and Input Resistance
Both arms of differential amplifier are symmetrical.
So terminal currents and collector voltages are equal.
Characteristics of differential pair with common-mode
input are similar to those of a C-E amplifier with large
emitter resistor.
v
ic
i =
b r + 2(β +1)R
EE
π
Output voltages are:
−β RC
v = v = −β i R =
v
c1 c2
b C r + 2(β +1)R
ic
EE
π
ve = 2(β +1)i R
b EE
2(β +1) REE
=
v ≅v
rπ + 2(β +1)REE ic ic
Common-mode Gain and Input Resistance (contd.)
Common-mode gain is given by:
β RC
voc
RC
=−
≅−
A =
cc v
2REE
rπ + 2(β +1)REE
ic v =0
id
For RC=REE, common-mode gain =0.5. Thus, common-mode output voltage
and Acc is 0 if REE is infinite. This result is obtained since output resistances of
transistors are neglected. A more accurate expression is:
⎛
⎜
⎜
⎜
⎜
⎝
1
1
Acc ≅ R
−
C β r 2R
o
EE
v
od
=v
c1
−v
c2
⎞
⎟
⎟
⎟
⎟
⎠
= 0 Therefore, common-mode conversion gain is found to be 0.
v
rπ + 2(β +1)REE rπ
ic
R =
=
= + (β +1)R
EE
ic 2i
2
2
b
Common-Mode Rejection ratio (CMRR)
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Represents ability of amplifier to amplify desired
differential-mode input signal and reject undesired
common-mode input signal.
For differential output, common-mode gain of balanced
amplifier is zero, CMRR is infinite. For single-ended output,
A
A
/2
dm
dd
=
= ⎛
CMRR =
A cm
A cc
2 ⎜⎜
1
1
1
−
2g m R EE
⎜ β ro g m
⎝
⎞
⎟
⎟
⎟
⎠
≅ gm R
EE
Analysis of Differential Amplifiers Using Half-Circuits
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Half-circuits are constructed by first
drawing the differential amplifier in a fully
symmetrical form- power supplies are
split into two equal halves in parallel,
emitter resistor is separated into two
equal resistors in parallel.
None of the currents or voltages in the
circuit are changed.
For differential mode signals, points on
the line of symmetry are virtual grounds
connected to ground for ac analysis
For common-mode signals, points on line
of symmetry are replaced by open
circuits.
Bipolar Differential-mode Half-circuits
Direct analysis of the half-circuits yield:
v
v
= − gm R id
c1
C 2
v
v
= + gm R id
c2
C 2
v = v − v = −g R v
m C id
od c1 c2
v
Applying rules for drawing halfcircuits, the two power supply
lines and emitter become ac
grounds. The half-circuit
represents a C-E amplifier stage.
A = od
= − gm R
C
dd v
id v = 0
ic
A
gm R
C
c1
A =
=−
= dd
dd1 v
2
2
id v = 0
ic
v
R = v / i = 2rπ
id id b1
R = 2( R ro )
C
od
Bipolar Common-mode Half-circuits
Direct analysis of the half-circuits yield:
−β RC
v = v = −β i R =
v
c1 c2
b C r + 2(β +1)R
ic
EE
π
voc
β RC
RC
=−
≅−
A =
cc v
2REE
rπ + 2(β +1)REE
ic v =0
id
v = v −v =0
od c1 c2
Applying rules for drawing halfcircuits, the points at the line of
symmetry are open circuited.
The half-circuit represents a C-E
amplifier stage with an emitter
resistance.
v
rπ + 2(β +1)REE rπ
ic
R =
=
= + (β +1) R
EE
ic 2i
2
2
b
A
A /2
dm
CMRR =
= dd ≅ g m R
EE
Acm
Acc
Biasing with Electronic Current Sources
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Differential amplifiers are biased using electronic
current sources to stabilize the operating point and
increase effective value of REE to improve CMRR
Electronic current source has a Q-point current of ISS
and an output resistance of RSS as shown.
DC model of the electronic current source is a dc
current source, ISS while ac model is a resistance RSS.
V
I
=I − 0
DC SS R
SS