lecture 1 op-amp introduction of operation amplifier (op- amp) analysis of ideal op-amp applications...
TRANSCRIPT
Lecture 1 Op-Amp
• Introduction of Operation Amplifier (Op-Amp)
• Analysis of ideal Op-Amp applications
• Comparison of ideal and non-ideal Op-Amp
• Non-ideal Op-Amp consideration
Vd
+
Vo
Rin~inf Rout~0
Input 1
Input 2
Output
+Vcc
-Vcc
Operational Amplifier (Op-Amp)• 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
ddo VGV
510say large,y ver
normally gain aldifferenti : dG
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
Single-Ended Input
+
Vo
~ Vi
+
Vo
~ Vi
• + terminal : Source• – terminal : Ground• 0o phase change
• + terminal : Ground• – terminal : Source• 180o phase change
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) ?
Distortion
Vd
+
Vo
+Vcc=+5V
Vcc= 5V
0
+5V
5V
The output voltage never excess the DC voltage supply of the Op-Amp
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
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
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)
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
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
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
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
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
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
+
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
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
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
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
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
Op-Amp Differentiator
RCdt
dVv i
o
+
RC
VoVi
0to t1 t2
0
to t1 t2
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
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
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
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
)()(
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
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
)(~