op-amp building blocks and applications - engineeringrhabash/elg4139ln212.pdf · op-amp building...
TRANSCRIPT
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Lecture 2
Op-Amp Building Blocks and Applications
• Instrumentation Amplifiers
• Filters
• Integrators
• Differentiators
• Frequency-Gain Relation
• Non-Linear Op-Amp Applications
• DC Imperfections
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Check List for Selecting Op-Amps
• Power Supply
• Gain Bandwidth
• Cost
• Voltage Offset
• Stability
• Output Current
• Noise
• Special Functions
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Types of Op-Amps
• Precision Amplifiers:
– Low noise
– Low Offset Voltage
– Low Input Bias Current.
• Low Power/Low Bandwidth
• Video Amplifiers
• Audio Amplifiers
• High Voltage/High Current
• Differential/Instrumentation Amplifiers
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Instrumentation Amplifier (IA) Applications
• Used in Environments with High Noise.
• Offers high Input Impedance.
• Low Bias Currents.
• Gain Determined by One Resistor.
• Commercial IA is based on Standard 3-OP Amps.
• Requires Several Matched Resistors.
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Instrumentation Amplifier
Combines 2 non-inverting amplifiers
with the difference amplifier to
provide higher gain and higher input
resistance.
)b
va(v
3
4v R
R
o
bv
2i)
1i(2
2iav RRR
12
2v
1v
iR
)2
v1
(v
1
21
3
4v
R
R
R
R
o
Ideal input resistance is infinite
because input current to both op
amps is zero. The CMRR is
determined only by Op-Amp 3.
NOTE
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6
Figure
8.14,
8.15
Instrumentation Amplifier Input (a) output (b)
Stages of IA
1
2
21
21
R
R
R
R
vv
vA Fout
V
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Instrumentation Amplifier: Example
• Problem: Determine Vo
• Given Data: R1 = 15 kW, R2 = 150 kW, R3 = 15 kW, R4 = 30 kW V1 = 2.5 V,
V2 = 2.25 V
• Assumptions: Ideal op amp. Hence, v-= v+ and i-= i+= 0.
• Analysis: Using dc values,
Adm
R
4R
3
1R
2R1
30kW
15kW1
150kW
15kW
22
Vo Adm
(V1
V2
)22(2.52.25)5.50V
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8
Figure
8.20
Op-amp Circuits Employing Complex Impedances
S
F
S
out
S
F
S
out
Z
Zj
V
V
Z
Zj
V
V
1)(
)(
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Figure 8.21,
8.23
Active Low-Pass Filter
Normalized Response of Active Low-Pass Filter
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The Active Low-Pass Filter
Use a phasor approach to gain analysis of
this inverting amplifier. Let s = j.
Av ˜ v o
( j)
˜ v ( j)
Z2
( j )
Z1( j )
Z1
j R1
Z2( j)
R2
1
jC
R2
1
jC
R
2jCR
21
Av R
2R1
1
(1 jCR2
)
R2
R1
ej
(1j
c
)
c 2fc 1R
2C
fc 12R
2C
fc is called the high frequency “cutoff” of
the low-pass filter. ELG4139 10
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Active Low-Pass Filter
• At frequencies below fc (fH in the figure), the amplifier is an inverting amplifier with gain set by the ratio of resistors R2 and R1.
• At frequencies above fc, the amplifier response “rolls off” at -20dB/decade.
• Notice that cutoff frequency and gain can be independently set.
Av R
2R1
ej
(1j
c
)
R
2
R1
12
c
2
ej
ejtan1( /
c)
R
2
R1
1
c
2e
j[ tan1( /c
)]
magnitude phase
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Active Low-Pass Filter: Example
• Problem: Design an active low-pass filter
• Given Data: Av= 40 dB, Rin= 5 kW, fH = 2 kHz
• Assumptions: Ideal op amp, specified gain represents the desired low-frequency gain.
• Analysis:
Input resistance is controlled by R1 and voltage gain is set by R2 / R1.
The cutoff frequency is then set by C.
The closest standard capacitor value of 160 pF lowers cutoff frequency to 1.99 kHz.
100dB20/dB4010 vA
W k51 in
RR
Av
R
2R1
R2
100R1
500kW
C 12f
HR
2
12(2kHz)(500kW)
159pF
and
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13
Figure
8.24,
8.25
Active High-Pass Filter
Normalized Response of Active High-Pass Filter
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14
Figure
8.26,
8.27
Active Band-Pass Filter
Normalized Amplitude Response of Active Band-Pass Filter
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Cascaded Amplifiers
• Connecting several amplifiers in cascade (output of one stage connected to the input of the next) can meet design specifications not met by a single amplifier.
• Each amplifer stage is built using an op amp with parameters A, Rid, Ro, called open loop parameters, that describe the op amp with no external elements.
• Av, Rin, Rout are closed loop parameters that can be used to describe each closed-loop op amp stage with its feedback network, as well as the overall composite (cascaded) amplifier.
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Two-Port Model for a 3-Stage Cascade Amplifier
• Each amplifier in the 3-stage cascaded amplifier is replaced by its 2-port
model.
vo AvA
vs
RinB
RoutA
RinB
AvB
RinC
RoutB
RinC
AvC
vCA
vBA
vAAvA
svov
Since Rout= 0
Rin= RinA and Rout= RoutC = 0
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Figure
8.30
Op-Amp Integrator
t
S
FS
out dttvCR
tv )(1
)(
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Inverting Integrator
Now 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 = R
Therefore, the closed-loop gain (Vo/Vin) become:
where
What happens if Za = 1/jC whereas, Zf = R?
Inverting differentiator
in
a
f
o VZ
ZV
CjZ
f
1
dttvRC
tv io )(1
)(
tj
ii eVtv )(
+
~
Zf
Za
Vin
Vo
+
~
R
Vin
Vo
C
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Op-Amp Integrator
Example:
(a) Determine the rate of change
of the output voltage.
(b) Draw the output waveform.
Solution:
(a) Rate of change of the output voltage
smV/
F k
V
50
)01.0)(10(
5
W
RC
V
t
V io
(b) In 100 s, the voltage decrease
VμssmV/ 5)100)(50( oV
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Figure 8.35
Op-Amp Differentiator
dt
tdvCRtv S
SFout
)()(
+
R
C
VoVi
0to t1 t2
0
to t1 t2ELG4139 20
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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
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Building Filters: Gain and Bandwidth Gain Bandwidth Product = Gain x Bandwidth
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Gain-Bandwidth Product
Example: 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
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Ideal Versus Practical Op-Amp
Ideal Actual
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin >1MW
Output Impedance Zout 0 W 10-100 W
Output Voltage Vout Depends only
on Vd = (V+V)
Differential
mode signal
Depends slightly
on average input
Vc = (V++V)/2
Common-Mode
signal
CMRR 10-100dB
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Analysis
Ideal Op-Amp Main 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
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+
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
i
a
f
o vRR
R
R
Rv ))(1(
21
2
Voltage follower
io vv
i
a
f
o vR
Rv )1(
Less than unity gain
io vRR
Rv
21
2
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Non-Linear Op-Amp Applications
• Applications using saturation
– Comparators
– Comparator with hysteresis (Schmitt Trigger)
– Oscillators.
• Applications using active feedback components
– Log, antilog, squaring etc. amplifiers
– Precision rectifier
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Comparators A comparator is a device that compares the magnitude of two inputs and
gives an output that indicates which of the two is larger!
VOUT
VIN
Ideal response VOUT = A0VIN
Practical response (clipped)
If A0 is large, practical response can be approximated as :
VIN > 0 V+ > V- VOUT = +VSAT
VIN < 0 V+ < V- VOUT = -VSAT
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Hysteresis
• A comparator with hysteresis has a ‘safety margin’.
• One of two thresholds is used depending on the current
output state.
V
time
Upper threshold
Lower threshold
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Schmitt Trigger
• The Schmitt trigger is an op-amp comparator circuit
featuring hysteresis.
• The inverting variety is the most commonly used.
21
1
RR
RVVVV OUTIN
Switching occurs when:
But,
21
1
RR
RVV
VV
SATTHRESH
SATOUT
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Schmitt Trigger Op-amp Circuit
• The open-loop comparator from the previous two slides
is very susceptible to noise on the input
– Noise may cause it to jump erratically from + rail to –
rail voltages
• The Schmitt Trigger circuit (see at the left) solves this
problem by using positive feedback
– It is a comparator circuit in which the reference voltage is
derived from a divided fraction of the output voltage, and
fed back as positive feedback.
– The output is forced to either VPOS or VNEG when the
input exceeds the magnitude of the reference voltage
– The circuit will remember its state even if the input
comes back to zero (has memory)
• The transfer characteristic of the Schmitt Trigger is
shown at the left
– Note that the circuit functions as an inverter with
hysteresis
– Switches from + to – rail when vIN > VPOS(R1/(R1 + R2))
– Switches from – to + rail when vIN< VNEG(R1/(R1 + R2))
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Asymmetrical Thresholds
• We do not need always
want the threshold levels to
be symmetrical around 0 V.
• More general configuration
features an arbitrary
reference level.
Using Kirchoff’s current law:
012
R
VV
R
VV REFOUT
21
21
1212 RR
RRV
R
V
R
V
R
V
R
V REFOUT
21
2
21
1
RR
RV
RR
RVV REFOUT
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Realising VREF
21
2
21
1
RR
RV
RR
RVV REFSATTHRESH
211 || rrR 21
2
rr
rVV SREF
Providing and
But,
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DC Imperfections Three DC Imperfections of Real Op-Amps
Input Bias Current; Input Offset Current; and Input Offset Voltage (output voltage
may not be zero for zero input voltage)
𝐼𝐵 =𝐼𝐵++𝐼𝐵_
2
𝐼𝑜𝑓𝑓 = 𝐼𝐵+ − 𝐼𝐵−
Bias Current
Offset Current
Bias Current: All op-amps draw a small constant DC bias currents at their inputs.
Typical value for a 741 is around 100 nA. This is only notable when very high
impedance sources are used. In such cases, an alternative op-amp with lower bias
current should be used.
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Offset Voltage
• When both input voltages are equal, the output should be zero. Actually it probably won’t be due to an offset voltage between the inputs. Typically, this is around 2 mV.
• Offset voltage is automatically compensated by a negative feedback network. It can be a problem for precision comparator applications.
• Both the offset voltage and bias current are DC. A.C.
operation is not affected by them (they just add an offset)
Negative feedback reduces the effect of both. Steps can be
taken to reduce them (further reading)
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Example: Find the worst-Case DC Output Voltage of an
Inverting Amplifier assuming vin = 0. The maximum bias
current of the Op-Amp is 100 nA. The maximum offset
current is 40 nA, and the maximum offset voltage is 2 mV.
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First, Consider the Offset Voltage
𝑉0,𝑣𝑜𝑓𝑓 = − 1 +𝑅2
𝑅1𝑉𝑜𝑓𝑓
𝑉0,𝑜𝑓𝑓 = −11𝑉𝑜𝑓𝑓
𝑉0,𝑜𝑓𝑓 = −22 and + 22mV ELG4139
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Second, Bias Current Sources
• 𝑉0,𝑏𝑖𝑎𝑠 = −𝑅2𝐼2 − 𝑅1𝐼1
• 𝐼1 = 0
• 𝐼2 = −𝐼𝐵
• 𝑉0,𝑏𝑖𝑎𝑠 = 𝑅2𝐼𝐵
ELG4139
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Third, Offset Current Source
• 𝑉0,𝑖𝑜𝑓𝑓 = 𝑅2𝐼𝑜𝑓𝑓
2= -2 and 2 mV
• 𝑉0 = 𝑉0,𝑣𝑜𝑓𝑓 + 𝑉𝑜,𝑏𝑖𝑎𝑠 + 𝑉𝑜,𝑖𝑜𝑓𝑓
• 𝑉0 = 22 + 10 +2=34mV
• 𝑉0 = −22 + 0 −2=-24mV
ELG4139
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