Transcript Operational Amplifiers

Operational Amplifiers
Dr. Holbert
February 11, 2008
Lect8
EEE 202
1
Op Amps
• Op Amp is short for operational amplifier
• Amplifiers provide gains in voltage or
current
• Op amps can convert current to voltage
Lect8
EEE 202
2
Applications of Op Amps
• Op amps can be configured in many
different ways using resistors and other
components
• Most configurations use feedback
• Op amps can provide a buffer between
two circuits
• Op amps can be used to implement
integrators and differentiators
• Lowpass and bandpass filters
Lect8
EEE 202
3
The Op Amp Symbol
High Supply
Non-inverting input
Inverting input
+
Output
–
Ground
Low Supply
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EEE 202
4
The Op Amp Model
• An operational amplifier is modeled as a
voltage-controlled voltage source.
Non-inverting input
v+
+
Rin
Inverting input
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v–
vo
+
–
–
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A(v+ – v– )
5
Typical vs. Ideal Op Amps
Typical Op Amp:
• The input resistance
(impedance) Rin is
very large (practically
infinite).
• The voltage gain A is
very large (practically
infinite).
Lect8
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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6
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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7
The Basic Inverting Amplifier
R2
R1
Vin
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+
–
–
+
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+
Vout
–
8
Solving the Amplifier Circuit
Apply KCL at the inverting (–) input:
R2
Vin
Vout
i2
R1
V–
i1
Vin  V Vin
i1 

R1
R1
–
i–
Vout  V Vout
i2 

R2
R2
i1 + i2 + i– =0
Lect8
i  0
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9
Solve for Vout
• From KCL
i1  i2  i  0
• Thus, the amplifier
gain is
Vout
R2

Vin
R1
Vin Vout

00
R1
R2
Vin
Vout

R1
R2
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EEE 202
10
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 (e.g.,
nodal), to solve for the output voltage in
terms of the input(s)
Lect8
EEE 202
11
Where is the Feedback?
R2
R1
Vin
Lect8
+
–
–
+
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+
Vout
–
12
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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EEE 202
13
The Non-Inverting Amplifier
+
+
–
vin
+
–
R2
vout
R1
–
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14
KCL at the Inverting Input
i  0
+
vin
+
–
–
+
i2
vout
i–
i1
R2
R1
–
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EEE 202
 v  vin
i1 

R1
R1
vout  v
i2 
R2
vout  vin

R2
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Solve for vout
i1  i2  i  0
 vin vout  vin

0
R1
R2
 R2 

vout  vin 1 
R1 

• Hence, the non-inverting amplifier has a
gained output (> unity) relative to the
resistance ratio
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16
A Mixer Circuit
R1
v1
R2
+
–
v2
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+
–
Rf
–
+
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+
vout
–
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KCL at the Inverting Input
v1  v v1
i1 

R1
R1
v1
i  0
R1
i1
R2 i
2
+
–
i–
v2
+
–
–
+
vout  v vout
if 

Rf
Rf
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Rf
if
+
vout
–
v2  v v2
i2 

R2
R2
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Solve for vout
i1  i2  i f  i  0
v1 v2 vout


0
R1 R2 R f
vout  
Rf
R1
v1 
Rf
R2
v2
• So, the mixer circuit output is a (negative)
combination of the input voltages
Lect8
EEE 202
19
Class Examples
• Drill Problem P4-1
Lect8
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