# CHAPTER 3 DIFFERENTIAL AMPLIFIERS

```ELECTRONICS-1
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CHAPTER 3
DIFFERENTIAL AMPLIFIERS
Introduction: the common and the differential mode components of two voltages
differential mode component:
vd
v1
common mode component:
vc
vd = v1 - v2
v  v2
vc = 1
2
v2
vd
2
vd
v 2  vc 
2
(Remark: the same definitions are valid for DC as well.)
with these:
v1  v c 
With these the linear system of equations describing the linear operation of differential amplifiers:
vod = Avdd vind + Avdc vinc
voc = Avcd vind + Avcc vinc
vin1
vin2
vo1
vo2
Target: as high Avdd as possible. Quality factors for this:
- discrimination factor
A
D  vdd
A vcc
- CMRR = Common Mode Rejection Ratio
A
CMRR  vdd
Avdc
Single ended differential amplifier (symmetrical input, asymmetrical output):
vin1
Vin2
vo = Avd vind + Avc vinc
Here, the CMRR definition:
CMRR 
Elec1-Part3-Diffamp.doc
Avd
Avc
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Basic connection of the differential amplifier
VS+
RC
RC
RL
vin1
vin2
vod
RB
RB
RE
a) Pure differential mode input (vinc = 0 and
vin1 = -vin2)
The equal magnitude but opposite phase current changes
cancel each-other on RE, therefore the potential of the
common E point does not change - virtual ground.
Because of the symmetry the electrical middle point of
the load resistance neither does change - this also a
virtually grounded point. From the aspect of one of the
transistors the situation on AC is the following:
According to this the differentialdifferential voltage gain by means
of the gain formula of the CE
stage (g22 0):
vod/2
RC
VS-
Avdd   g 21 ( RC x
Rt/2
vind/2
RL
)
2
Due to the symmetry, the
differential input does not result a
common output voltage:
Avcc = 0
b) Pure common input: (vind = 0, vin1 = vin2 = vinc)
Now the equal current changes flow together through RE, therefore from the aspect of one
transistor the situation is such as if RE had double value. Due to the symmetry the collectors
remain equipotential, no current flows through R L as if it were an open circuit. Therefore the
equivalent circuit for one transistor now:
The voltage gain referring to the common output due to the
common input can be calculated as the gain of the CE stage
with a 'degenerated' transistor:
RC
voc
vinc
2RE
g 21 RC
1  g 21 2 RE
Because of the symmetry there is no differential mode output
voltage:
Avdc = 0.
Avcc  
The quality factors:
R

 R Cx L
2
D  1  g 21R E 
 RC


CMRR = ∞
Elec1-Part3-Diffamp.doc






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(Practice)
Numeric example #1
Problem: vin1 = 1,005 V and vin2 = 0,995 V
How much are the voltages vo1 and vo2 ?
Solution: (observe DC input coupling)
a) Op. point:
IC0 = 1 m A,
VC0 = 7.5 V (collector potentials)
To this belongs
g21 = 38 mS.
VS+ = 15V
RC
7,5k
RC
7,5k
RL =10k
10k
vo1
b) Input voltage components:
vind = 0,01 V
vinc = 1 V
vo2
c) Gains (g22 = 0) :
vin2
vin1
RE
7,5k
RL
) = -114
2
vod = -1,14 V
Avdd = - g21(RC x
With this:
VS- = -15,6V
RC
 0.5
2 RE
voc = - 0,5 V
Avcc  
With this:
d) Output voltages:
- Regarding only the change of the collector potentials:
vo1 = - 0.5*1,14 - 0,5 = - 1,07 V
vo2 = + 0.5*1,14 - 0,5 = 0,07 V
- Voltages that can be measured between the output points and ground:
Vo1 = 7,5 – 0.5*1,14 - 0,5 = 6,43 V (VC0 + 0.5v0d + v0C)
Vo2 = 7,5 + 0.5*1,14 - 0,5 = 7,57 V (VC0 – 0.5v0d + v0C)
End of class 6.
Common mode rejection in case of asymmetry
RC1
RC2
vodc
Conditions: R g  0, R L   and g 22  0 .
Notations:
RC = RC1 – RC2
g21 = g21(1) - g21(2)
RC = 0.5(RC1+RC2) g21 = 0.5(g21(1)+g21(2))
iC1=g21(1)vB
vB
iC2=g21(2)vB
vinc
RE
vB
vinc
The differential mode output voltage generated
by the pure common mode input voltages:
vodc = iC2 RC2 - iC1 RC1 =
= vB (g21 – 0.5*g21)(RC - 0.5*RC) - (g21 + 0.5*g21)(RC + 0.5*RC) =
= -vB (RC g21 + RC g21)
Expressing vB by vinc:
vB = vinc - 2g21 vB RE
Elec1-Part3-Diffamp.doc
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from where:
vinc
1  2 g 21 RE
The common to differential gain:
 g
RC 

g 21 RC  21 
g 21
RC 
v0 dc

Avdc 

vinc
1  2 g 21 R E
Since -g21 RC = Avdd , the common mode rejection:
A
1  2 g 21 R E
CMRR  vdd 
g 21 RC
Avdc

g 21
RC
vB 
Elec1-Part3-Diffamp.doc
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Versions of circuit arrangement
a) For the sake of large
CMRR (and D) the realisation
of a large RE
b) Using E-degeneration to stabilise
the gain (by reducing it)
RE
RE
2I0
g 21
1  g 21 RE
d) Differential amplifier with
symmetric output
*

Here: g 21
c) Trans-conductance amplifier
IC1
IC2
RC
RC
∞ RL
RL
RE
∞
CC
∞
RL
vin
I0
vin1
I0
VS-
vin
;
RE
v
 I 0  in
RE
vin2
RE
The connection can be retraced to the original one (see dashed lines).
I C1  I 0 
So:
IC2
Avd = - 0.5Avdd = 0.5g21(RC x RL)
One of the collectors is used as a current output. If both are used, then the output signal
is the difference of the two collector
v
currents ( 2 in ): this is decided by the (not
RE
shown) connection receiving the currents.
Elec1-Part3-Diffamp.doc
vo = Avd vind + Avc vinc
g 21 RC xRL 
1  2 g 21 RE
The notion of inverting
and non-inverting inputs.
Avc  Avcc  
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e) Two-stage differential amplifier with direct coupling
RC1
RC1
0V
IE1 -0.6V IE1
IN1
IN2
2IE2
IE2
VS+
RE2
RL
IE2
RC2
0V 2I
E1
IB
v0
RE1
VS-
Supposing that IB ≈ 0, find the necessary value of RC2 to get direct output coupling.
I E1 
1  0.6  VS

2
R E1
I E2 
1 I E1 R C1  0.6

2
R E2
R C2 
 VS
I E2
Elec1-Part3-Diffamp.doc
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