Transcript Operational Amplifiers (4.1-4.3)

Operational Amplifiers (4.1-4.3)
Dr. Holbert
April 3, 2006
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Op Amps
• Op Amp is short for operational amplifier.
• An operational amplifier is modeled as a
voltage controlled voltage source.
• An operational amplifier has a very high
input impedance and a very high gain.
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Use of Op Amps
• Op amps can be configured in many
different ways using resistors and other
components.
• Most configurations use feedback.
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Applications of Op Amps
• Amplifiers provide gains in voltage or
current.
• Op amps can convert current to voltage.
• Op amps can provide a buffer between two
circuits.
• Op amps can be used to implement
integrators and differentiators.
• Lowpass and bandpass filters.
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The Op Amp Symbol
High Supply
Non-inverting input
Inverting input
+
Output
Ground
Low Supply
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The Op Amp Model
Non-inverting
input
v+
+
Rin
Inverting input
v-
vo
+
–
–
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A(v+ -v- )
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Typical Op Amp
• The input resistance Rin is very large
(practically infinite).
• The voltage gain A is very large (practically
infinite).
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“Ideal” Op Amp
• The input resistance is infinite.
• The gain is infinite.
• The op amp is in a negative feedback
configuration.
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The Basic Inverting Amplifier
R2
R1
Vin
+
–
–
+
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Vout
–
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Consequences of the Ideal
• Infinite input resistance means the current
into the inverting input is zero:
i- = 0
• Infinite gain means the difference between
v+ and v- is zero:
v+ - v- = 0
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Solving the Amplifier Circuit
Apply KCL at the inverting input:
R2
i2
R1
–
i1
i-
i1 + i2 + i-=0
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KCL
i  0
vin  v vin
i1 

R1
R1
vout  v vout
i2 

R2
R2
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Solve for vout
vin
vout

R1
R2
Amplifier gain:
vout
R2

vin
R1
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Recap
• The ideal op-amp model leads to the
following conditions:
i- = 0 = i+
v+ = v• These conditions are used, along with KCL
and other analysis techniques, to solve for
the output voltage in terms of the input(s).
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Where is the Feedback?
R2
R1
Vin
+
–
–
+
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Vout
–
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Review
• To solve an op-amp circuit, we usually
apply KCL at one or both of the inputs.
• We then invoke the consequences of the
ideal model.
– The op amp will provide whatever output
voltage is necessary to make both input
voltages equal.
• We solve for the op-amp output voltage.
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The Non-Inverting Amplifier
+
+
–
vin
+
–
R2
vout
R1
–
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KCL at the Inverting Input
+
+
–
vin
+
–
ii1
i2
R2
vout
R1
–
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KCL
i  0
 v  vin
i1 

R1
R1
vout  v vout  vin
i2 

R2
R2
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Solve for Vout
 vin vout  vin

0
R1
R2
vout
 R2 

 vin 1 
R1 

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A Mixer Circuit
R1
v1
R2
+
–
v2
+
–
Rf
–
+
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vout
–
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KCL at the Inverting Input
R1 i1
v1
R2 i
2
+
–
iv2
Rf
if
+
–
–
+
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vout
–
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KCL
v1  v v1
i1 

R1
R1
v2  v v2
i2 

R2
R2
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KCL
i  0
vout  v vout
if 

Rf
Rf
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Solve for Vout
v1 v2 vout


0
R1 R2 R f
vout  
Rf
R1
v1 
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Rf
R2
v2
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Class Example
• Learning Extension E4.1
• Learning Extension E4.2
• Learning Extension E4.3
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