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Module 12Operational Amplifiers – Part II
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Review from Operational Amplifiers I:
Negative input
Positive inputOutput
VPOS
–VNEG
Power SupplyVoltages
Anatomy of an “Op-Amp”
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Dependent Source Model
• rin is on the order of several Megohms:
• Av is on the order of 105 to 106
These features motivate the Ideal Op-Amp approximation
v+
v–
Av(v+ – v–)
Equivalent model for the circuit inside an op-amp
rin
vOUT
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Dependent Source Model
• vOUT must lie between Vpos and –Vneg
VPOS
–VNEG
vOUT
Vpos
–Vneg
Upper Limit
Lower Limit
Range
• Otherwise, the op-amp becomes saturated.
• Saturated op-amp vOUT = Vpos or –Vneg limit
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The Ideal Op-Amp ApproximationVPOS
–VNEG
rin =
v+
v–
rin = Av = Very Large
vOUT
This model greatly simplifies op-amp analysis
–Vneg < VOUT < Vpos
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A Consequence of Infinite rin
VPOS
VNEG
rin = i+ = 0
i = 0
Currents i+ and i to (or from) input terminals are zero
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A Consequence of Large Av
VPOS
–VNEG
If vOUT lies between Vpos and –Vneg …
(v+ v–) 0
Defines the Linear Region of operation
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Example: The Non-Inverting Amplifier Revisited
Use the Ideal Op-Amp approximation:
v– = vOUT R1 + R2
R1 Via voltage division (works because i = 0)
vIN
R2
R1
vOUT R1 + R2
R1
vOUT vIN
=i = 0
v vIN When vOUT in linear region:
–vneg< vOUT < vPos
vIN= vOUT R1 + R2
R1 R1 + R2
R1
vOUT vIN
= Done!
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+ –
vIN
vOUT R2
R1
vOUT vIN
=
Example: The Inverting Amplifier Revisited
R2
R1 i = 0
Use the Ideal Op-Amp approximation:
v 0v+ = 0
i1
i1 = vIN v
R1
vIN
R1
=
i2
i1 = i2 Via KCL (with i = 0)
Ohm’s Law
vOUT = i2 R2 = i1 R2vOUT =
R2
R1
vIN Done!
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+ –
The Summation Amplifier
vOUT
+
–+_
R2
R1
+_v1
v2
RF
Another Example:
Use the Ideal Op-Amp Approximation…
i2
i1 iF
KCL: i1 + i2 = iF
i1 = v1
R1
i2 = v2
R2
vOUT = iFRF
iF = + v1
R1
v2
R2
vOUT = RF
R1
RF
R2
v2v1 +
Output is weighted, inverted sum of inputs
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Can extend result to arbitrary number of input resistors:
Output is weighted, inverted sum of inputs:
vOUT
+
–
R2
R1
+_v2
RF
i2
i1 iF
...
+_v1
+_v3 +_vn
R3
Rn
vOUT = RF
R1
RF
R2
v2v1 +RF
R3
RF
Rn
v3+ vn+ …+
iF = i1 + i2 + i3+ … + in
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Difference AmplifierAnother Example:
vOUT
+
–
+_
R1
R2+_v2
v1
R2
R1
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vOUT
+
–
+_
R1
R2+_v2
v1
R2
R1
Use Superposition:
Set v2 to zero
i+ = 0 v+ = 0 v = 0
We have an inverting amplifier
vOUT =R2
R1
v1 1st Partial result for vOUT
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vOUT
+
–
+_
R1
R2+_
v1
R2
R1
Use Superposition, con’t:
Set v1 to zero
We have an non-inverting amplifier2nd Partial result for vOUT
v+ = v2 R1 + R2
R2 Via voltage division
R1 + R2
R1
vOUT v+= = v2 R1 + R2
R1 R1 + R2
R2 = v2
R2
R1
v2
i+
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vOUT
+
–
+_
R1
R2+_v2
v1 R1
Add together the 2nd and 1st partial results:
R2
= v2 R2
R1
vOUT R2
R1
v1
= (v2 v1) R2
R1
vOUT Amplifies difference between v2 and v1
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Summary
• Ideal Op-Amp Approximation simplifies circuit analysis
• “Ideal” implies rin = and v+ = v in the linear region
• Summation Amplifier
vOUT = RF
R1
RF
R2
v2v1 +RF
R3
RF
Rn
v3+ vn+ …+
= (v2 v1) R2
R1
vOUT
• Difference Amplifier
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End of This Module
Homework