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EE301/2011 Operational Amplifier 1
Chapter 2 :Operational-Amplifier
• Introduction of Operation Amplifier (Op-Amp)
• Understand the Differential Amplifier
• Comparison of ideal and non-ideal Op-Amp
• Non-ideal Op-Amp consideration
EE301/2011 Operational Amplifier 2
Vd
+
Vo
Rin~inf Rout~0
Input 1
Input 2
Output
+Vcc
-Vcc
Operational Amplifier (Op-Amp)
• The circuit symbol for an op-amp is shown to the right, where:
• Input 1: non-inverting input• Input 2: inverting input• Vo: output• +Vcc: positive power
supply• -Vcc: negative power
supply
The power supply pins ( and ) can be labeled in different ways Despite different labeling, the function remains the same
• Very high differential gain
• High input impedance
• Low output impedance
• Provide voltage changes (amplitude and polarity)
• Used in oscillator, filter and instrumentation
• Accumulate a very high gain by multiple stages
EE301/2011 Operational Amplifier 3
EE301/2011 Operational Amplifier 4
IC Product
+
1
2
3
4
8
7
6
5
OFFSET NULL
-IN
+IN
V
N.C.
V+
OUTPUT
OFFSET NULL
+
1
2
3
4
8
7
6
5
OUTPUT A
-IN A
+IN A
V
V+
OUTPUT B
-IN B
+IN B+
DIP-741 Dual op-amp 1458 device
Op- Amp Block Diagram
EE301/2011 Operational Amplifier 5
• The input stage is a differential amplifier. The differential amplifier used as an input stage provides differential inputs and a frequency response down to d.c. Special techniques are used to provide the high input impedance necessary for the operational amplifier.
EE301/2011 Operational Amplifier 6
Differential Amplifier
• The second stage is a high-gain voltage amplifier. This stage may be made from several transistors to provide high gain. A typical operational amplifier could have a voltage gain of 200,000. Most of this gain comes from the voltage amplifier stage.
EE301/2011 Operational Amplifier 7
More Stages of Gain Amplifier
• The final stage of the OP AMP is an output amplifier. The output amplifier provides low output impedance. The actual circuit used could be an emitter follower. The output stage should allow the operational amplifier to deliver several milliamperes to a load. The operational amplifier can be used with three different input conditions (modes). With differential inputs (first mode), both input terminals are used and two input signals which are 180 degrees out of phase with each other are used.
EE301/2011 Operational Amplifier 8
Push-pull amplifier
• Push-pull amplifier An electronic circuit in which two transistors (or vacuum tubes) are used, one as a source of current and one as a sink, to amplify a signal. One device “pushes’’ current out into the load, while the other “pulls” current from it when necessary. A common example is the complementary-symmetry push-pull output stage widely used to drive loudspeakers (see illustration), where an npn transistor can source (push) current from a positive power supply into the load, or a pnp transistor can sink (pull) it into the negative power supply.
EE301/2011 Operational Amplifier 9
EE301/2011 Operational Amplifier 10
Single-Ended Input
+
Vo
~ Vi
+
Vo
~ Vi
• + terminal : Source• – terminal : Ground• 0o phase change
• + terminal : Ground• – terminal : Source• 180o phase change
EE301/2011 Operational Amplifier 11
Double-Ended Input
~ V1
+
Vo
~ V2
+
Vo~
Vd
• Differential input
•
• 0o phase shift change
between Vo and Vd
VVVd
Qu: What Vo should be if,
V1
V2
(A) (B)Ans: (A or B) ?
EE301/2011 Operational Amplifier 12
Distortion
Vd
+
Vo
+Vcc=+5V
Vcc= 5V
0
+5V
5V
The output voltage never excess the DC voltage supply of the Op-Amp
EE301/2011 Operational Amplifier 13
Common-Mode Operation
+
Vo
Vi ~
• Same voltage source is applied at both terminals
• Ideally, two input are equally amplified
• Output voltage is ideally zero due to differential voltage is zero
• Practically, a small output signal can still be measured
Note for differential circuits:Opposite inputs : highly amplifiedCommon inputs : slightly amplified
Common-Mode Rejection
EE301/2011 Operational Amplifier 14
Common-Mode Rejection Ratio (CMRR)
Differential voltage input :
VVVd
Common voltage input :
)(2
1 VVVc
Output voltage :
ccddo VGVGV
Gd : Differential gainGc : Common mode gain
)dB(log20CMRR 10c
d
c
d
G
G
G
G
Common-mode rejection ratio:
Note:When Gd >> Gc or CMRR Vo = GdVd
+
Noninverting Input
Inverting Input
Output
EE301/2011 Operational Amplifier 15
CMRR ExampleWhat is the CMRR?
Solution :
dBCMRR and
V (2) From
V (1) From
V V
V V
40)10/1000log(20101000
607007060
806006080
702
4010060
2
20100
60401008020100
21
21
cd
cdo
cdo
cc
dd
GG
GGV
GGV
VV
VV
+
100V
20V80600V
+
100V
40V60700V
NB: This method is Not work! Why?
(1) (2)
EE301/2011 Operational Amplifier 16
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically, Gd = 20,000 to 200,000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically, Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
Vo'Rload
outload
loadoload RR
RVV
EE301/2011 Operational Amplifier 17
Frequency-Gain Relation• Ideally, signals are amplified
from DC to the highest AC frequency
• Practically, bandwidth is limited
• 741 family op-amp have an limit bandwidth of few KHz.
• Unity Gain frequency f1: the gain at unity
• Cutoff frequency fc: the gain drop by 3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0.707Gd
fc0
1
GB Product : f1 = Gd fc
20log(0.707)=3dB
EE301/2011 Operational Amplifier 18
GB ProductExample: Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV
Sol:
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 / Gd = 10 MHz / 20 V/mV
= 10 106 / 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0.707Gd
fc0
1
10MHz
? Hz
EE301/2011 Operational Amplifier 19
Ideal Vs Practical Op-AmpIdeal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin >1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)/2 Common-Mode signal
CMRR 10-100dB
+
~
AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
EE301/2011 Operational Amplifier 20
Ideal Op-Amp ApplicationsAnalysis Method :Two ideal Op-Amp Properties:
(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit:(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V
(3) Set V+ = V and solve for the desired closed-loop gain
EE301/2011 Operational Amplifier 21
Noninverting Amplifier(1) Kirchhoff node equation at V+
yields,
(2) Kirchhoff node equation at V yields,
(3) Setting V+ = V– yields
or
+Vin
Vo
RaRf
iVV
00
f
o
a R
VV
R
V
0
f
oi
a
i
R
VV
R
V
a
f
i
o
R
R
V
V1
EE301/2011 Operational Amplifier 22
+
vi
Ra
vov-
v+
Rf
+
vi
Ra
vov-
v+
Rf
R2R1
+
vi vov-
v+
Rf
+
vi
vov-
v+
Rf
R2R1
Noninverting amplifier Noninverting input with voltage divider
ia
fo v
RR
R
R
Rv ))(1(
21
2
Voltage follower io vv
ia
fo v
R
Rv )1(
Less than unity gain
io vRR
Rv
21
2
EE301/2011 Operational Amplifier 23
Inverting Amplifier(1) Kirchhoff node equation at V+
yields,
(2) Kirchhoff node equation at V yields,
(3) Setting V+ = V– yields
0V
0_
f
o
a
in
R
VV
R
VV
a
f
in
o
R
R
V
V
Notice: The closed-loop gain Vo/Vin is dependent upon the ratio of two resistors, and is independent of the open-loop gain. This is caused by the use of feedback output
voltage to subtract from the input voltage.
+
~
Rf
Ra
Vin
Vo
EE301/2011 Operational Amplifier 24
Multiple Inputs(1) Kirchhoff node equation at V+
yields,
(2) Kirchhoff node equation at V yields,
(3) Setting V+ = V– yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
EE301/2011 Operational Amplifier 25
Inverting IntegratorNow replace resistors Ra and Rf by complex
components Za and Zf, respectively, therefore
Supposing(i) The feedback component is a capacitor C,
i.e.,
(ii) The input component is a resistor R, Za = RTherefore, the closed-loop gain (Vo/Vin) become:
where
What happens if Za = 1/jC whereas, Zf = R? Inverting differentiator
ina
fo V
Z
ZV
CjZ f
1
dttvRC
tv io )(1
)(
tjii eVtv )(
+
~
Zf
Za
Vin
Vo
+
~
R
Vin
Vo
C
EE301/2011 Operational Amplifier 26
Op-Amp IntegratorExample:
(a) Determine the rate of changeof the output voltage.
(b) Draw the output waveform.
Solution:
+
R
VoVi
0+5V
10 k
0.01FC
Vo(max)=10 V
100s
(a) Rate of change of the output voltage
smV/
F k
V
50
)01.0)(10(
5
RC
V
t
V io
(b) In 100 s, the voltage decrease
VμssmV/ 5)100)(50( oV
0+5V
0
-5V
-10V
Vi
Vo
EE301/2011 Operational Amplifier 27
Op-Amp Differentiator
RCdt
dVv i
o
+
RC
VoVi
0to t1 t2
0
to t1 t2
EE301/2011 Operational Amplifier 28
Non-ideal case (Inverting Amplifier)
+
~
Rf
Ra
Vin
Vo
3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Equivalent CircuitRf
RaVin
Vo
+
+
R RV
-AV
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
EE301/2011 Operational Amplifier 29
Close-Loop GainRf
RaVin
Vo
+
+
R R
-AV
Ra Rf
R
V
V
Vin Vo
Applied KCL at V– terminal,
0
f
o
a
in
R
VV
R
V
R
VV
By using the open loop gain,
AVVo
0f
o
f
oo
a
o
a
in
AR
V
R
V
AR
V
AR
V
R
V
fa
aafafo
a
in
RRAR
RARRRRRRRV
R
V
The Close-Loop Gain, Av
RARRRRRRR
RAR
V
VA
aafaf
f
in
ov
EE301/2011 Operational Amplifier 30
Close-Loop Gain
When the open loop gain is very large, the above equation become,
a
fv R
RA
~
Note : The close-loop gain now reduce to the same form as an ideal case
EE301/2011 Operational Amplifier 31
Input ImpedanceRf
Ra
Vin Vo
+
+
R
-AV
R'
RV
Rf
+
R
-AV
if
V
Input Impedance can be regarded as,RRRR ain //
where R is the equivalent impedance of the red box circuit, that is
fi
VR
However, with the below circuit,
A
RR
i
VR
RRiAVV
of
f
off
1
)()(
EE301/2011 Operational Amplifier 32
Input ImpedanceFinally, we find the input impedance as,
1
11
of
ain RR
A
RRR
RARR
RRRRR
of
ofain )1(
)(
Since, RARR of )1( , Rin become,
)1(
)(~
A
RRRR of
ain
Again with )1( ARR of
ain RR ~
Note: The op-amp can provide an impedance isolated from input to output
EE301/2011 Operational Amplifier 33
Output ImpedanceOnly source-free output impedance would be considered, i.e. Vi is assumed to be 0
Rf
Ra
Vo
+
R
-AV
R
V
io
V
V
Rf
Ra R+
R
-AV
V
i2 i1
(a) (b)
Firstly, with figure (a),
ofafa
ao
af
a VRRRRRR
RRVV
RRR
RRV
//
//
By using KCL, io = i1+ i2
o
o
faf
oo R
AVV
RRR
Vi
)(
//
By substitute the equation from Fig. (a),
RRARRRRRRR
RRRRRRR
i
V
R
afafao
fafao
o
o
out
)1())(1(
)(
is impedance,output The
R and A comparably large,
a
faoout AR
RRRR
)(~