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Chapter 6
The Operational
Amplifier
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The operational amplifier oropamp for short,
finds daily usage in a large variety of
electronic applications.
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op amps have three principal terminals:
the outputthe non-inverting input
theinverting
input
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Ideal Op Amp Rules
No current ever flows into either
input terminal. There is no voltage difference
between the two input terminals.
The op amp actsto make this
happen!
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Apply KVL, Ohms
law, and the ideal op
amp rules to find
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vout Rf
R1
v in
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Example: vin(t)=5sin3tmV,Rf=47 k,R1=4.7 k
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vout
(t) = -50sin3tmV
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vout 1Rf
R1
v in
To solve, use KVL, KCL,and op amp rules.
Suggested circuit variables to
perform the circuit analysis
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Example: vin(t)=5sin3tmV,Rf=47 k,R1=4.7 k
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vout
(t) = 55sin3tmV
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vout(t) =vin(t)
this design allows connection of apractical voltage source to a load without
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vout Rf
R(v1 v2 v3)
This amplifier performs the operationof adding.
It also introduces a gain ofRf/R
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vout
Rf
R
R2
R1
(v1 v
2) This voltage is notaffected by thecircuit on the right.
Op amps can be combined in stages to create the
desired relationship between the outputs and the inputs.
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Zener diode: i=0 if v
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With a reference voltage source Vref, we candrive a constant currentIs=Vref/ Rrefthrough
any loadRL.
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The op amp can be modeled as a dependent
voltage source, with the following components
as shown: input resistanceRi output resistanceRo open loop gainA
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Example:vin(t)=5sin3tmV,
Rf=47 k,
R1=4.7 k
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For a 741op amp (A=200,000, Ri=2M, Ro=75
vout(t) = -49.997 sin 3t mV.
An ideal op amp produces vout(t) = -50 sin 3t mV.[Analyze the detailed op amp model using nodal analysis.]
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When A=,Ro=0 , and Ri= , the op amp
behaves according to the ideal op amp rules.
(vd=0 and iin=0)
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Whenv1= v2= vCM, the output should be zero,
but real op amps produce a small common
mode voltage voCM. ACM=| voCM / vCM |
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The enormous but unpredictable gain of the op amp is made
usable through negative feedback.
When vingoes up, vdgoes down, and the op amp reacts by
lowering voutuntil the unwanted non-zero vdis pushed backto zero.
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this feedback
resistor allows the
output to affect the
input terminal.
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An op amp requires power supplies.
Usually, equal and opposite
voltages are connect to the V+
andV-terminals.
Typical values are 5 to 24 volts.
The power supply ground must be
the same as the signal ground.
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in this example +18V is connected to V+
and -18 V is connected to V-
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vout=10vin,but only up to the18 V supplies
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Non-zero output offsets can be removed:
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Slew rate is the maximum V/s for output.
examples: input (green) and output (red)
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Op amps in open loop can be used to make
decisions. In this case, is vin>2.5 V?
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Design a circuit that provides a logic 1 5 V output if a
certain voltage signal drops below 3 V, and zero volts
otherwise.
Answer:
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This device allows precise amplification of
small voltage differences:
vout=K(v+-v-)
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