1. Chapter 5
Differential
amplifier
By
-ISMET-
Edited by Nazirah Mohamat Kasim & Shahilah Nordin

2. • An arrangement of transistors which allows
the difference between two signals source to
be amplified.
• The output is proportional to the difference
between these two inputs.
• The best direct coupled stages available to
the IC designer.

3. • Differential amplifiers +VCC
are used in low and
high frequency RC RC
amplifiers, analog
modulators and digital Vi1
V01 V02
Vi2
logic states. Q1 Q2
• The basic differential
amplifier is shown in RT
Figure 5.1:- -VEE
Figure 5.1

4. +VCC
For the dc analysis of
differential amplifier RC RC
circuit, all inputs are V01 V02
set to zero as shown in Q1 Q2
Figure 5.2. IT/2
IT/2
RT IT
-VEE
Figure 5.2

5. From Figure 5.2,
IT
IC 
2
and V B  0
therefore, V E  V B - V BE  0  0.7V   0.7V
V EE - 0.7V
Emitter current, I T 
RT
IT
Collector voltage, V C  V CC - RC
2

6. Three possible input signal combinations for differential
amplifier:
Differential
mode
Single-Ended common
Mode mode

7. •Single ended mode – an active signal is applied to only
one input while the other is grounded.
•Differential mode – two opposite polarity active signals
are applied to the amplifier.
•Common mode
-Two signals of the same amplitude, frequency and phase are
applied to the differential amplifier.
- The output of the amplifier is ideally zero when measured the
difference between the output terminals.

8. Definition
The input is applied and the
output is measured at one
of the output terminal.
The input is in differential
mode and the output is
measured between the
differences of two output
terminals.

9. +VCC
The input signal is
applied to one RC RC
input with the V0
other input is Q1 Q2 Vi2=
+
connected to the Vi1
0
ground, as shown -
RT
in Figure 5.3: -VEE
Figure 5.3

10. The ac equivalent circuit of Figure 5.3 is
shown in Figure 5.4.
Vo 1 Vo 2
bIb RC RC bIb r?
r?
+
Vi 1
-
RT
Figure 5.4

11. b VT
Assume: Ib1  Ib2  Ib , r 
I CQ
and β 1  β 2  β
Assume RT is very large and using KVL we obtain
V i1
- 2I b r  0
V i1
 2I b r
The output at the collector will be
V o
 - b IbR C
Therefore,
Voltage gain,
VO RC
Av  
Vi  VT 
2
I


 CQ 

12. The mode of this +VCC
operation is
assumed by letting RC RC
a single sine wave
V0
signal is connected
between the two Q1 Q2
+ +
inputs. Vi1 Vi2
Refer Figure 5.5 - -
RT
(Double ended -VEE
differential
amplifier circuit) Figure 5.5

13. The ac equivalent circuit of Figure 5.5
is shown in Figure 5.6.
Vo 1 Vo 2
bIb RC RC bIb r?
r? +
+
Vi 1 Vi 2
- -
RT
Figure 5.6

14. Figure 5.6 shows an ac connection of a differential amplifier. We
separate the input signals as Vi1 and Vi2, with separate outputs
resulting as VO1 and VO2 with the transistors replaced with its ac
equivalent circuit.
Similarly as for the single ended operation,
Assume R T is very large and it becomes open circuit
V i1
- V i 2  Ib r  Ib r
V i1
- Vi2  V d
Therefore, V d  2I b r
The output at the collector terminal,
V o
 - b Ib R C
Therefore,
Differenti al voltage gain,
VO RC
Av  
Vd V 
2 T 
I 
 CQ 

15. Refer Figure 5.6,
Vd
Input impedance Z i(diff ) 
Ii
V d
 2I b r
and Ii  Ib
Therefore,
Z i(diff)  2r 

16. The same input signal is applied to the
+VCC
two input terminals of differential
amplifier with the same magnitude and
phase. RC RC
An ac connection showing common
input to both transistors is shown in V01
Figure 5.7 VCM
Q1 Q2
Ideal differential amplifier
-The output voltage Vo is
expected to be zero.
RT
-because the difference
-VEE
between the two outputs at
each collector is opposite to
each other and they are Figure 5.7
cancels out each other.

17. However
…
practically there is +VCC
an output at the
collectors but the RC
value is small.
To V01
we analyze,
Q1
treat the amplifier +
VCM 2RT
as symmetry, that is -
RT is made to be
parallel and be
-VEE
replaced with 2RT as
shown in Figure 5.8
Figure 5.8

18. To determine the common mode gain, the ac
equivalent circuit of figure 5.8 is drawn
as given in Figure 5.9
VCM Vo
r βIb
RC
2RT
From the circuit of Figure 5.9,
V CM
 V i  Ib r   IE 2R T
V 
 b Ib  T  2R T 
I 
Figure 5.9  CQ 
and V o
 - b IbR C
Therefore, the common mode gain, A CM of the
differenti al amplifier,
VO RC RC
A CM
    
Vi VT 2R
 2R T T
I CQ

19. The ratio of the magnitude of its differential gain, Ad to
the magnitude of its common mode gain, ACM.
Ad
CMRR 
A CM
The value of the CMRR is often given in dB,
Ad
CMRR (dB)  20 log 10
A CM

20. Example 1.
The circuit given in Figure 5.10 has the following parameters:
hfe1 = hfe2 = 120, VT = 26mV, VBE1 = VBE2 = 0.7V
Calculate: +10V
a) ICQ1, ICQ2 and VCE1
b) Differential gain, Ad 3.9kΩ 3.9kΩ
c) Common mode gain, ACM
V01
d) CMRR in dB Vi1 Vi2
Q1 Q2
e) Differential input impedance, Zi(diff)
f) Output impedance, Zo
5.6kΩ
-10V
Figure 5.10

21. Solution
DC analysis:
Since VB = 0, VE = -0.7V
Using KVL around loop A:
VE – IERE + 10V = 0
10  V E 10  (-0.7)
Then IE    1.66mA
RT 5.6k

22. Solution IE
a) I CQ1 , I CQ2   0.83mA
2
V CE1  V CC - I C1 R C1 - V E  0
 10  (0.83m)(3. 9k)  (  0.7)
 7.46V
b) From the ac equivalent circuit of differential mode:
VO RC VT 26mV
A d
  ; but   31.32 Ω
Vd V  IC 0.83mA
2 T
 I


 C 
3.9k
Then A d     62.26
2(31.32)

23. Solution
c) From the ac equivalent circuit of common mode:
VO RC
A CM
  
Vi VT
 2R T
IC
3.9k
    0.35
31.32  2(5.6k)
d) Ad 62.26
CMRR(dB)  20log 10
 20log 10
 45dB
A CM 0.35
e)  VT 
Z i(diff)  2 
 I   2(120)(31. 32)  7.5k Ω
 C 
f) ZO  R C  3.9k Ω

24. Good differential amplifier should have high CMRR
Ad
 CMRR 
A CM 
High CMRR means the differential amplifier circuit has the
ability to reject common mode signal (noise).
Ideally, ACM = 0 not amplify the noise signal.

25. In practical way there is still small signal at the output of
common mode signal.
to  CMRR, we have A CM 
 RC
 A CM 
VT
 2R T 
IC
To reduce common mode gain, A CM  we can increase R T 
Popular method to increase  R T is by using :
Constant Current Source
Practical current source is a current supply with a resistance.

26. An ideal current source, R=∞
Whereas practical current source resistance, R is very large.
An ideal current source provides a constant current
regardless of the load connected to it.
Practical Current
Source
IT RT Figure 5.11

27. Constant Current Source
circuit can be built using:
FET devices
BJT devices
Combination of FET & BJT devices
Three popular Constant-
Current Sources for
differential amplifier
Bipolar Transistor Constant Current Source
Transistor / Zener Constant Current Source
Current Mirror circuit.

28. BJT Constant Current Source
Figure 5.12a shows that an npn transistor together with
resistors operate as a constant current source.
IT
C3
B3
E3
R1 R2 RE
-VEE RT
Figure 5.12a

29. BJT Constant Current Source
Figure 5.12b and Figure 5.12c shows simplified circuit from Figure
5.12a
IT
C3 IT
C3
B3 RTH
Q3 B3
E3
E3
VTH
R1 R2 RE LOOP1 RE
VEE
VEE VEE
RT RT
Figure 5.12b Figure 5.12c
R1
V TH    VEE  R TH  R 1 // R 2
R1  R 2
By using KVL at loop1 in Figure 5.12c
IT  IC3  IE3
- V TH  I B R TH  V BE 3  I E 3 R E  0
V TH  V EE - V BE3 V TH  V EE - V BE3
IE3  IT 
R TH R TH
 RE  RE
b1 b1

30. BJT Constant Current Source
Figure 5.12d and Figure 5.12e shows ac equivalent circuit for circuit
in Figure 5.12a
b3 rbb' c3
rce3
ib bi b b3 rbb' r e3 c3
r rce3
R1 R2 e3 ie
ib bi b
ie R1//R2 RE
RE
RT RT
Figure 5.12d Figure 5.12c
R T  rce 3 1  b    R b  rce 3 1  b 
where
Re R b   R 1 // R 2   rbb '  r
 
Re  R b Re  RE

31. Example 2
If R1 = 4.7kΩ, R2: 4.7kΩ, RE = 2.2kΩ and VEE = 20V in figure
5.12:
IT
Figure 5.12
R1 R2 RE
-VEE
Calculate current, I.
Answer: 4.23mA

32. Zener Constant Current Source
•From BJT constant current source, R2 is replaced with
zener diode as in Figure 5.13.
IT
C3
B3
+ Q3
VBE3
Figure 5.13a + _ E3
VZ loop1
R1 - RE IE3
•Using KVL at loop 1:
-VEE
- V Z  V BE3  IE3 R E  0
V Z  V BE3  IE3 R E I T  IC3  IE3
V Z - V BE3 V Z - V BE3
IE3  IT 
RE RE

33. Zener Constant Current Source
Figure 5.13b and Figure 5.13c shows ac equivalent circuit for circuit
in Figure 5.13a
b3 rbb' c3 rce3
rbb' r e3 c3
ib bi b b3
r rce
R1 ib ie
e3 bi b
ie Re
RE
RT RT
Figure 5.13b Figure 5.13c
R T  rce 3 1  b    R b  rce 3 1  b 
where
Re R b  rbb '  r
 
Re  R b Re  RE

34. Example 3
If R1 = 3.3kΩ, RE: 2.2kΩ, and VEE = 15V in figure 5.13:
IT
+ Figure 5.13
VZ
R1 - RE
-VEE
Calculate current, I.
Answer: 2.64mA

35. Current Mirror Circuit
Used primarily in IC.
Circuit in Figure 5.14 is an example of constant current
source circuit that can be used for differential amplifier.
For current mirror circuit,
+VCC
I C  I C3  I C4
IX IB  IB3  IB4
IE  IE3  IE4
RX IT=IC IC
I X  I C  2I B But IB 
b
so,
IC 2IB IC
I X  IC  2
Q4 Q3 b
IB + IB  b 2
 IC 
 

IE VBE IE  b 
- RT  b 
IC  IT  
 IX  IX

 b  2
Therefore I T  IC  I X
Figure 5.14a

36. Current Mirror Circuit
Figure 5.14b shows ac equivalent circuit for circuit in Figure 5.14a
c4 b4 b3 c3
Ib4 rbb'4 rbb'3
ib3
rce3
RX rce3 bi b 4 bi b3
r 4 r 4
e3 e3
RT
Figure 5.13b
R T  rce 3

37. Example 4
Given Rx = 1.5kΩ, VCC = 18V in figure 5.14, calculate the
mirrored current, I in the circuit.
Answer: 11.53mA +18V
IX
Solution
Using KVL around the loop:
RX I
18 – IXRX – VBE = 0 Q1 Q2
+
IX = 11.53mA VBE
-
Figure 5.14

38. Example 5
For the differential amplifier of Figure 5.15, determine the
following: +15V
a) ICQ, VCEQ 4.7kΩ 4.7kΩ
b) ACM
V0
c) Ad Vi1
Q1 Q2
Vi2
d) CMRR in dB
e) Differential input impedance, Zi(diff) 1.5kΩ
Q3
f) Output impedance, Zo
Given:rce = 1/1μS 1.2kΩ
β1 = β2 = β3 = 150
-15V
VT = 26mV Figure 5.15

39. Solution
a) DC analysis:
V Z  V BE
IE3   3.67mA
RE
IE3
ICQ   1.84mA
2
V CEQ
 V C - VE
 V C - ICQ R C - V E
 7.05V
VT
and  14.1 Ω
ICQ

40. Solution
b)
VO -RC RC
A CM
  -
Vi VT 2R T
 2R T
IC
Where R T  rce 3 1  b 
Then R T  80.635 MΩ
A CM  - 29.14  10
-6
Therefore

41. Solution
VO RC
c) A d
   -166.67
Vd  VT 
2
 I 

 C 
d) Ad
CMRR(dB)  20log 10
 135.147dB
A CM
e) Z i(diff)  2r   4.23k Ω
f) Z O  R C  4.7k Ω

43. TUTORIAL 5

44. Question 1
Figure 5.16 shows an emitter coupled pair differential amplifier
with Vi1 and Vi2 as the input and VO as the output. Q1 and Q2
have the following parameters:
β1 = β2 = 150, VT = 26mV, VBE1 = VBE2 = 0.7V +15V
Calculate:
4.7kΩ 4.7kΩ
a) ICQ1, ICQ2
b) Differential gain, Ad Vi1
V01
Vi2
c) Common mode gain, ACM Q1 Q2
d) Differential input impedance, Zi(diff)
e) Output impedance, Zo 4.9kΩ
-15V
Figure 5.16

45. Question 2
Figure 5.17 is a differential amplifier with β1=β2= β3=100,
VBE=0.7V, VT=26mV and 1/ro =40μS. Both diodes are made of
silicon. Determine: +15V
a) ICQ1, ICQ2 15kΩ 15kΩ
b) Ad, ACM and CMRR in dB V0
Vi1 Vi2
c) Zi(diff) and Zo Q1 Q2
2kΩ
Q3
470kΩ
-15V
Figure 5.17

46. Question 3
Figure 5.18 shows an emitter coupled pair differential amplifier
with Vi1 and Vi2 as the input and VO as the output. Determine:
+12V
a) The quiescent point of the
1.5k Ω 1.5k Ω
differential amplifier for transistor Q1
b) CMRR in dB Vi1
V0
Vi2
Q1 Q2
c) Output resistance (Ro)
d) Differential input resistance, Ri(diff)
1.5k Ω
Q3
Q4
-12V
Figure 5.18

47. The End
See you in the next lesson!
Don’t forget to do revision.
-ISMET-
edited by Nazirah Mohamat Kasim &
Shahilah Nordin