Transcript
Chapter 5
Differential amplifier
By -ISMET-
Edited by Nazirah Mohamat Kasim & Shahilah Nordin
• 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.
• Differential amplifiers are used in low and high frequency amplifiers, analog modulators and digital logic states.
• The basic differential amplifier is shown in Figure 5.1:-
Figure 5.1
For the dc analysis of differential amplifier circuit, all inputs are set to zero as shown in Figure 5.2.
Figure 5.2
From Figure 5.2,
CT
CCC
T
EET
BEBE
B
TC
R 2
I-V Vvoltage, Collector
R
0.7V-VI current, Emitter
0.7V0.7V0V-V Vtherefore,
0 Vand2
II
Three possible input signal combinations for differential amplifier:
Single-Ended Mode
Differential mode
common mode
•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.
Definition
Single ended output
The input is in differential mode and the output is measured between the
differences of two output terminals.
The input is applied and the output is measured at
one of the output terminal.
Double
ended
output
The input signal is applied to one input with the other input is connected to the ground, as shown in Figure 5.3:
Figure 5.3
The ac equivalent circuit of Figure 5.3 is shown in Figure 5.4.
Figure 5.4
Vi1
Vo 1 Vo 2
RT
RC IbIb RC
-
+r?
r?
Assume:
CQ
T
C
i
O
Co
b1i
b1i
21
bb2b1
IV
2
R
V
V Av
gain, Voltage
Therefore,
IbR- V
be willcollector the at output The
r2I V
0r2I- V
obtain weKVL using and largevery is RT Assume
βββ and
III
CQ
T
I
Vr,
The mode of this operation is assumed by letting a single sine wave signal is connected between the two inputs. Refer Figure 5.5 (Double ended differential amplifier circuit) Figure 5.5
The ac equivalent circuit of Figure 5.5 is shown in Figure 5.6.
Figure 5.6
Vi1 Vi2
Vo 1 Vo 2
RT
RC IbIb RC
+
--
+r?
r?
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.
CQ
T
C
d
O
Cbo
bd
d2i1i
bb2i1i
T
IV
2
R
V
V Av
gain, voltage alDifferenti
Therefore,
RI- V
terminal, collector the at output The
r2I VTherefore,
VV- V
rIrIV- V
circuit open becomes it and largevery is R Assume
operation, ended single the for asSimilarly
Refer Figure 5.6,
2rZ
Therefore,
II and
r2I V
I
V)Z impedance Input
i(diff)
bi
bd
i
di(diff
The same input signal is applied to the two input terminals of differential amplifier with the same magnitude and phase.An ac connection showing common input to both transistors is shown in Figure 5.7
Ideal differential amplifier-The output voltage Vo is expected to be zero.-because the difference between the two outputs at each collector is opposite to each other and they are cancels out each other.
Figure 5.7
RC
Q2Q1
RC
RT
VCMV01
-VEE
+VCC
However…practically there is
an output at the collectors but the value is small.
we treat the amplifier as symmetry, that is RT is made to be parallel and be replaced with 2RT as shown in Figure 5.8
To analyze,
RC
Q1
2RT
V01
-VEE
+VCC
VCM-
+
Figure 5.8
To determine the common mode gain, the ac equivalent circuit of figure 5.8 is drawn as given in Figure 5.9
T
C
T
C
i
OCM
CM
Co
TCQ
Tb
TEbiCM
2R
R
2R
R
V
V A
amplifier, aldifferenti
the of Again, mode common the Therefore,
IbR- Vand
2RI
VI
2RIrIV V
5.9, Figure of circuit the From
CQ
T
IV
Figure 5.9
RC2RT
r bβIVCM Vo
The ratio of the magnitude of its differential gain, Ad to the magnitude of its common mode gain, ACM.
A
A(dB) CMRR
dB, in given often is CMRR the of value The
A
ACMRR
CM
d
CM
d
10log20
Example 1.The circuit given in Figure 5.10 has the following parameters:
hfe1 = hfe2 = 120, VT = 26mV, VBE1 = VBE2 = 0.7V
Calculate:a) ICQ1, ICQ2 and VCE1
b) Differential gain, Ad
c) Common mode gain, ACM
d) CMRR in dBe) Differential input impedance, Zi(diff)
f) Output impedance, Zo
Figure 5.10
SolutionDC analysis:
Since VB = 0, VE = -0.7V
Using KVL around loop A: VE – IERE + 10V = 0
Then 1.66mA5.6k
(-0.7)10
R
V10I
T
EE
Solution a)
b) From the ac equivalent circuit of differential mode:
7.46V
0.7)(9k)(0.83m)(3.10
0V-RI-VV
0.83mA2
II ,I
EC1C1CCCE1
ECQ2CQ1
62.262(31.32)
3.9k AThen
31.32Ω0.83mA
26mV
I
V but ;
IV
2
R
V
V A
d
C
T
C
T
C
d
Od
Solution c) From the ac equivalent circuit of common mode:
d)
e)
f)
45dB0.35
62.2620log
A
A20logCMRR(dB) 10
CM
d10
7.5kΩ32)2(120)(31.I
V2Z
C
Ti(diff)
3.9kΩRZ CO
0.352(5.6k)31.32
3.9k
2RIV
R
V
V A
TC
T
C
i
OCM
Good differential amplifier should have high CMRR
High CMRR means the differential amplifier circuit has the ability to reject common mode signal (noise).
Ideally, ACM = 0 not amplify the noise signal.
CM
d
A
ACMRR
In practical way there is still small signal at the output of common mode signal.
to we have
To reduce common mode gain, we can increase Popular method to increase is by using :Constant Current Source
Practical current source is a current supply with a resistance.
CMA TR
TR
CMRR, CMA
TC
T
CCM
2RIV
RA
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
RTIT Figure 5.11
Constant Current Source Constant Current Source circuit can be built using:circuit can be built using:
FET devices
BJT devices
Combination of FET & BJT devices
Three popular Constant-Current Three popular Constant-Current Sources for differential amplifierSources for differential amplifier
Bipolar Transistor Constant Current Source
Transistor / Zener Constant Current Source
Current Mirror circuit.
BJT Constant Current SourceBJT Constant Current Source Figure 5.12a shows that an npn transistor together with
resistors operate as a constant current source.
RER2R1
-VEE
ITC3
B3
E3
RT
Figure 5.12a
BJT Constant Current SourceBJT Constant Current Source Figure 5.12b and Figure 5.12c shows simplified circuit from Figure
5.12a
By using KVL at loop1 in Figure 5.12c
EE21
1TH V
RR
RV
RER2R1
VEE
ITC3
B3
E3
VEERT
Q3
RE
VEE
ITC3
B3
E3
RTH
VTH
RT
LOOP1
21TH R//RR
Figure 5.12b Figure 5.12c
ETH
BE3EETH
TH
R1
RV-VV
V-
3E
E3E3BETHB
I
0RIVRI 3E3CT III
ETH
BE3EETH
R1
RV-VV
TI
BJT Constant Current SourceBJT Constant Current SourceFigure 5.12d and Figure 5.12e shows ac equivalent circuit for circuit
in Figure 5.12a
where
1rR1r 3ceb3ceTR
bR
e
e
R
R
Ee RR
rrR//RR 'bb21b
Figure 5.12d Figure 5.12c
rbb'
birce3
rR1 R2
RE
b3 c3
e3
RT
ib
ie
rbb'
bi
r
R1//R2 RE
b3 c3e3
RT
ib ie
rce3
Example 2If R1 = 4.7kΩ, R2: 4.7kΩ, RE = 2.2kΩ and VEE = 20V in figure 5.12:
Calculate current, I.Answer: 4.23mA
RER2R1
-VEE
IT
Figure 5.12
Zener Constant Current SourceZener Constant Current Source•From BJT constant current source, R2 is replaced with zener diode as in Figure 5.13.
•Using KVL at loop 1:
E
BE3ZE3
EE3BE3Z
EE3BE3Z
R
V-V I
RIVV
0RIVV-
RER1
-VEE
IT
+
-VZ
Q3
C3
B3
E3VBE3
+
_IE3
loop1
E3C3T III
Figure 5.13a
E
BE3ZT R
V-V I
Figure 5.13b and Figure 5.13c shows ac equivalent circuit for circuit in Figure 5.13a
where
11 33 cebce rRrTR
bR
e
e
R
R
Ee RR
rrR 'bbb
Figure 5.13b Figure 5.13c
Zener Constant Current SourceZener Constant Current Source
rbb'
birce
rR1
RE
b3 c3
e3
RT
ib
ie
rbb'
bi
r
Re
b3 c3e3
ib ie
RT
rce3
Example 3If R1 = 3.3kΩ, RE: 2.2kΩ, and VEE = 15V in figure 5.13:
Calculate current, I.Answer: 2.64mA
RER1
-VEE
IT
+
-VZ
Figure 5.13
Current Mirror CircuitCurrent Mirror CircuitUsed primarily in IC.
Circuit in Figure 5.14 is an example of constant current source circuit that can be used for differential amplifier.
XCT
C
CX
BBCX
E4E3E
B4B3B
C4C3C
III Therefore
I
I I
so,
I But 2III
III
III
III
circuit, mirror current For
XXTC
C
C
II2
II
2
I2
I
RX
Q4 Q3
VBE
+VCC
+
-
IT=IC
IX
RT
IC 2IB
IB IB IEIE
Figure 5.14a
Figure 5.14b shows ac equivalent circuit for circuit in Figure 5.14a
3cerTR
Figure 5.13b
Current Mirror CircuitCurrent Mirror Circuit
rbb'3
3birce3
4r
RX
b3 c3
e3
RT
ib3rbb'4
4r
Ib4
4birce3
b4c4
e3
Example 4Given Rx = 1.5kΩ, VCC = 18V in figure 5.14, calculate the mirrored current, I in the circuit.Answer: 11.53mA
Using KVL around the loop:18 – IXRX – VBE = 0
IX = 11.53mA
Solution
R X
Q 1 Q 2
V BE
+18V
+
-
I
IX
Figure 5.14
Example 5For the differential amplifier of Figure 5.15, determine the following:
a) ICQ, VCEQ
b) ACM
c) Ad
d) CMRR in dBe) Differential input impedance, Zi(diff)
f) Output impedance, Zo
Given:rce = 1/1μS
β1 = β2 = β3 = 150
VT = 26mV
Q 2Q 1
V i1
V 0 V i2
+15V
4.7kΩ 4.7kΩ
Q 3
-15V
1.2kΩ
1.5kΩ
Figure 5.15
Solutiona) DC analysis:
14.1ΩI
V and
7.05V
V-RI- V
V- V V
1.84mA2
II
3.67mAR
VVI
CQ
T
ECCQC
ECCEQ
E3CQ
E
BEZE3
Solutionb)
6-CM
T
T
T
C
TC
T
C
i
OCM
1029.14- ATherefore
MΩ 80.635 R Then
R Where
2R
R-
2RIV
R-
V
V A
13cer
Solutionc)
d)
e)
f)
-166.67
IV
2
R
V
V A
C
T
C
d
Od
135.147dBA
A20logCMRR(dB)
CM
d10
4.23kΩ2rZ i(diff)
4.7kΩRZ CO
TUTORIAL 5
Question 1Figure 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
Calculate:a) ICQ1, ICQ2
b) Differential gain, Ad
c) Common mode gain, ACM
d) Differential input impedance, Zi(diff)
e) Output impedance, Zo
Figure 5.16
Question 2Figure 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:
a) ICQ1, ICQ2
b) Ad, ACM and CMRR in dB
c) Zi(diff) and ZoQ 2Q 1
V i1
V 0 V i2
+15V
15kΩ 15kΩ
Q 3
-15V
470kΩ
2kΩ
Figure 5.17
Q 2Q 1
V i1
V 0 V i2
+12V
1.5k Ω 1.5k Ω
Q 3
-12V
1.5k Ω
Q 4
Question 3Figure 5.18 shows an emitter coupled pair differential amplifier with Vi1 and Vi2 as the input and VO as the output. Determine:
a) The quiescent point of the differential amplifier for transistor Q1
b) CMRR in dBc) Output resistance (Ro)
d) Differential input resistance, Ri(diff)
Figure 5.18
The End See you in the next lesson!Don’t forget to do revision.
-ISMET- edited by Nazirah Mohamat Kasim & Shahilah Nordin
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