Op-Amps Microprocessor Interface. Operational Amplifier (Op-Amp) Very high differential gain High...
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Op-AmpsOp-Amps
Microprocessor Interface Microprocessor Interface
Vd
+
Vo
Rin~inf Rout~0
Input 1
Input 2
Output
+Vcc
-Vcc
Operational Amplifier Operational Amplifier (Op-Amp)(Op-Amp)
Very high differential Very high differential gaingain
High input impedanceHigh input impedance Low output impedanceLow output impedance Provide voltage changes Provide voltage changes
(amplitude and polarity)(amplitude and polarity) Used in oscillator, filter Used in oscillator, filter
and instrumentationand instrumentation Accumulate a very high Accumulate a very high
gain by multiple stagesgain by multiple stages
ddo VGV
510say large,y ver
normally gain aldifferenti : dG
IC ProductIC 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
DistortionDistortion
Vd
+
Vo
+Vcc=+5V
Vcc= 5V
0
+5V
5V
The output voltage never excess the DC voltage supply of the Op-Amp
Op-Amp PropertiesOp-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 Frequency-Gain RelationRelation
• 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
GainBandwidth ProductGainBandwidth 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 Op-Amp Ideal Op-Amp ApplicationsApplicationsAnalysis Method :Analysis Method :Two ideal Op-Amp Properties:Two ideal Op-Amp Properties:
(1)(1) The voltage between The voltage between VV++ and and VV is zero is zero VV++ = = VV
(2)(2) The current into both The current into both VV++ and and VV termainals is zero termainals is zero
For ideal Op-Amp circuit:For ideal Op-Amp circuit:(1)(1) Write the kirchhoff node equation at the Write the kirchhoff node equation at the
noninverting terminal noninverting terminal VV++
(2)(2) Write the kirchhoff node eqaution at the inverting Write the kirchhoff node eqaution at the inverting terminal terminal VV
(3)(3) Set Set VV++ = = VV and solve for the desired closed-loop and solve for the desired closed-loop gaingain
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 AmplifierInverting Amplifier(1)(1) Kirchhoff node equation at Kirchhoff node equation at VV++
yields,yields,
(2)(2) Kirchhoff node equation at Kirchhoff node equation at VV yields, yields,
(3)(3) Setting Setting VV++ = = VV–– yields 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 InputsMultiple Inputs(1)(1) Kirchhoff node equation at Kirchhoff node equation at VV++
yields, yields,
(2)(2) Kirchhoff node equation at Kirchhoff node equation at VV yields, yields,
(3)(3) Setting Setting VV++ = = VV–– yields 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 IntegratorInverting IntegratorNow replace resistors Ra and Rf by complex
components Za and Zf, respectively, therefore
Supposing• The feedback component is a capacitor C,
i.e.,
• 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 IntegratorOp-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 DifferentiatorOp-Amp Differentiator
RCdt
dVv i
o
+
RC
VoVi
0to t1 t2
0
to t1 t2
Slew RateSlew Rate
FIGURE 10-17 The rate of change of a linear, or ramp, signal is the change in voltage FIGURE 10-17 The rate of change of a linear, or ramp, signal is the change in voltage divided by the change in timedivided by the change in time
The maximum possible rate at which an amplifier’s output voltage can change, in volts per second, is called its slew rate.