ECA notes

1. ADITYA
Electronic Circuit Analysis
K. JAYARAM KUMAR
Assistant Professor
Department of Electronics and Communication
Engineering

2. ADITYA
AC analysis of Differential amplifier using BJT
2
2
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE

3. ADITYA
AC analysis of
Differential Amplifier using BJT
K.JAYARAM KUMAR
Assistant Professor
Department of Electronics and Communication Engineering

4. ADITYA
Learning Outcomes
At the end of this lecture, Student will be able to:
LO 1: Learns the AC analysis of differential amplifiers
using BJT.
4
4
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE

5. ADITYA
5
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER
a) Dual Input Balanced Output(DIBO)
b) Dual Input Unbalanced Output (DIUBO)
c) Single Input Balanced Output(SIBO)
d) Single Input Unbalanced Output(SIUBO)
 Configurations of Differential Amplifier:

6. ADITYA
6
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
DIBO DIUBO
 Configurations (Contd.)

7. ADITYA
7
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
SIBO SIUBO
 Configurations (Contd.)

8. ADITYA
8
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
• Differential Amplifier is capable of amplifying DC as well as AC
input signals.
• DC Analysis: We have to find operating collector current (ICQ)
and operating collector to emitter voltage (VCEQ).
• AC Analysis: We have to find voltage gain (Ad), Input Resistance
(Ri) and Output Resistance (Ro)
 Analysis:

9. ADITYA
9
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis:
• To perform AC analysis to drive the expression for the voltage
gain (Ad), the input resistance (Ri) and output resistance (Ro) of
the differential amplifier.
• The dc voltages are reduced to zero.
• AC equivalent of CE configuration is used.

10. ADITYA
10
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis:
DIBO
• AC equivalent circuit is drawn using r-parameters

11. ADITYA
11
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis:
DIBO
Fig.4 AC Equivalent Circuit
AC equivalent circuit is drawn using
r-parameters as shown in Fig.

12. ADITYA
12
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis:
• Since the two dc emitter currents are equal.
• Therefore, resistance r'e1 and r'e2 are also equal and designated by r'e.
• This voltage across each collector resistance is shown 180° out of
phase with respect to the input voltages v1 and v2.
• The polarity of the output voltage is shown in figure. The collector C2
is assumed to be more positive with respect to collector C1 even
though both are negative with respect to ground.

13. ADITYA
13
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis:
(a) Voltage Gain (Ad)
Applying KVL in two loops 1 & 2. We get
E
e
e
e
e
s
b R
i
i
r
i
R
i
V )
( 2
1
'
1
1
1
1 



E
e
e
e
e
s
b R
i
i
r
i
R
i
V )
( 2
1
'
2
2
2
2 



Substituting current relations,
ac
e
b
i
i

1
1 
ac
e
b
i
i

2
2
& 
……(1)
……(2)

14. ADITYA
14
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis-Voltage Gain (Ad)
E
e
e
e
e
ac
s
e R
i
i
r
i
R
i
V )
( 2
1
'
1
1
1
1 




E
e
e
e
e
ac
s
e R
i
i
r
i
R
i
V )
( 2
1
'
2
2
2
2 




Again, assuming RS1/βac and RS2/βac are very
small in comparison with RE and re' and
therefore neglecting these terms,
E
e
E
e
e R
i
R
r
i
V 2
1
1 )
'
( 


E
e
E
e
e R
i
R
r
i
V 1
2
2 )
'
( 


……….(3)
……….(4)

15. ADITYA
15
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis-Voltage Gain (Ad)
Solving these two equations (3) & (4), ie1 and ie2 can be calculated.





















2
1
2
1
'
'
e
e
E
e
E
E
E
e
i
i
R
r
R
R
R
r
V
V
E
e
E
E
E
e
E
e
E
e
e
R
r
R
R
R
r
R
r
V
R
V
i
i







'
'
'
2
1
1
1
To find ie1,
2
2
2
1
1
)
(
)
'
(
)
'
(
E
E
e
E
E
e
e
R
R
r
R
V
R
r
V
i






To find ie2,
E
e
E
E
E
e
E
E
e
e
e
R
r
R
R
R
r
V
R
V
R
r
i
i







'
'
'
2
1
2
2
⇒ 𝑖𝑒2 =
𝑉2(𝑟𝑒′ + 𝑅𝐸) − 𝑉1𝑅𝐸
(𝑟𝑒′ + 𝑅𝐸)2 − (𝑅𝐸)2
….(5) ….(6)
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16. ADITYA
16
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis-Voltage Gain (Ad)
Substituting ie1 & ie2 in the above expression from eq.(5)&(6)
𝑉
𝑜 = 𝑅𝐶
𝑉1(𝑟𝑒′ + 𝑅𝐸) − 𝑉2𝑅𝐸
(𝑟𝑒′ + 𝑅𝐸)2 − (𝑅𝐸)2
−
𝑉2(𝑟𝑒′ + 𝑅𝐸) − 𝑉1𝑅𝐸
(𝑟𝑒′ + 𝑅𝐸)2 − (𝑅𝐸)2
The output voltage VO is given by
Vo = VC2 - VC1
VO = (- RC iC2 ) - (-RC iC1)
VO = RC (iC1 - iC2)
VO = RC (ie1 - ie2)

17. ADITYA
17
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis-Voltage Gain (Ad)
















 2
2
2
2
1
2
1
'
2
'
)
(
)
)(
'
(
E
E
e
E
e
E
E
e
C
o
R
R
r
R
r
V
V
R
V
V
R
r
R
V













)
2
'
(
'
)
2
'
)(
( 2
1
E
e
e
E
e
C
o
R
r
r
R
r
V
V
R
V
)
(
'
2
1 V
V
r
R
V
e
C
o 


'
e
C
d
o
d
r
R
V
V
A 



18. ADITYA
18
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis-Voltage Gain (Ad)
• Thus a differential amplifier amplifies the difference between
two input signals.
• Defining the difference of input signals as vd = v1 - v2
• The voltage gain of the dual input balanced output
differential amplifier can be given by
'
e
C
d
o
d
r
R
V
V
A 


19. ADITYA
19
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis:
(b) Differential Input Resistance (Ri)
Differential input resistance is defined as the equivalent resistance that
would be measured at either input terminals with the other terminal
grounded.
0
1
1
1
1
2 


V
b
s
i
i
V
R
R
)
( 1
1
1
ac
e
i
i
V
R




20. ADITYA
20
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis- Differential Input Resistance (Ri)















2
2
2
1
1
1
)
(
)
'
(
)
'
(
E
E
e
E
E
e
ac
i
R
R
r
R
V
R
r
V
V
R

)
'
(
)
'
2
'
(
1
2
1
1
E
e
E
e
e
ac
i
R
r
V
R
r
r
V
R





0
2 
V

1
1
1
e
ac
i
i
V
R




Substituting ie1 value in above equation, we get

21. ADITYA
21
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis- Differential Input Resistance (Ri)
The factor of 2 arises because the re' of each transistor is in series.
⇒ 𝑅𝑖1 =
𝛽𝑎𝑐 ⋅ 𝑟𝑒′(𝑟𝑒′ + 2𝑅𝐸)
(𝑟𝑒′ + 𝑅𝐸)
Generally RE >> re’, we get
'
2
1 e
ac
i r
R 


Similarly '
2
2
2 e
ac
s
i r
R
R 


'
2
2
1 e
ac
i
i r
R
R 




22. ADITYA
22
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
 A.C. Analysis:
(b) Differential Output Resistance (R0)
• Output resistance is defined as the equivalent resistance that would be
measured at output terminal with respect to ground.
• Therefore, the output resistance RO1 measured between collector C1 and
ground is equal to that of the collector resistance RC.
• Similarly the output resistance RO2 measured at C2 with respect to ground
is equal to that of the collector resistor RC.
RO1 = RO2 = RC

23. ADITYA
23
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
Configuration Voltage Gain Input
Resistance
Output
Resistance
DIBO
DIUBO
SIBO
SIUBO
'
e
C
d
r
R
A 
'
2 e
C
d
r
R
A 
'
e
C
d
r
R
A 
'
2 e
C
d
r
R
A 
C
R
C
R
C
R
C
R
'
2 e
ac
i r
R 

'
2 e
ac
i r
R 

'
2 e
ac
i r
R 

'
2 e
ac
i r
R 

 AC Analysis Comparison:

24. ADITYA
24
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
Problem:The following specifications are given for the dual input, balanced-
output differential amplifier: RC = 2.2 kΩ, RE = 4.7 kΩ, Rin1 = Rin2 = 50Ω, +VCC=
10V, -VEE = -10 V, βdc =100 and VBE = 0.715V. Determine the
a) voltage gain. b) Input resistance c)Output resistance.
Solution:
mA
I
mV
r
E
e
25

mA
mV
988
.
0
25

)
(
2 
in
E
BE
EE
E
CQ
R
R
V
V
I
I





)
100
50
(
)
7
.
4
(
2
7
.
0
10




k
The ac emitter resistance
mA
988
.
0


 3
.
25
'
e
C
d
r
R
A 
Differential Gain



3
.
25
2
.
2 k
96
.
86


25. ADITYA
25
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE
DIFFERENTIAL AMPLIFIER Contd..
Problem:The following specifications are given for the dual input, balanced-
output differential amplifier: RC = 2.2 kΩ, RE = 4.7 kΩ, Rin1 = Rin2 = 50Ω, +VCC=
10V, -VEE = -10 V, βdc =100 and VBE = 0.715V. Determine the
a) voltage gain. b) Input resistance c)Output resistance.
Solution:
Input Resistance


 k
R
R o
o 2
.
2
2
1
'
2 e
ac
i r
R 
 3
.
25
100
2 

 
 k
06
.
5
Output Resistance

26. ADITYA
Understands about the AC analysis of differential
amplifiers using BJT.
Summary
26
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE

27. ADITYA
Understands the Design and analysis of multistage
amplifiers using BJT & FET and Differential amplifier
using BJT.
Summary
27
Electronic Circuit Analysis K.Jayaram Kumar, Assistant Professor,
ECE