page 39
I-
V-
I+
V+
V+ = V- |
I+ = I- = 0 |
5.1.2 Simple Applications
• Considering that the Op-amp was originally designed to allow simple mathematical operations in circuit form, the following circuits tend to bemathematical in nature.
5.1.2.1 - Inverting Amplifier
• A typical op-amp application is the inverting amplifier.
page 40
R2
R1
VI
-
V- Vo
+
V+
R3
R1 and R2 are effectively a voltage divider,
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– V ) ----------------- |
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+ V |
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+ R |
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The circuit tries to keep both op-amp inputs equal. And, the V+ input is grounded.
V- = V+ = 0 |
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R2 |
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1 – R |
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0 = VI R------------------ |
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R1 + R2 – R2 |
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Amplifier Gain |
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R-----1 |
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5.1.2.2 - Non-Inverting Amplifier
• We can also make non-inverting amplifiers using the following circuit,
page 41
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V- |
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Use R1 and R2 as a voltage divider, |
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If we consider the the two inputs to have an equivalent input voltages, |
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V- = V+ |
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VO |
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R1 |
+ R2 |
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Gain |
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------ |
----------------- |
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5.1.2.3 - Integrator
• The integrating amplifier is a very powerful application,
page 42
1MΩ
If
C
VI |
R1 |
II |
- |
Vo |
+
R2
In the circuit the 1MΩ resistor is used to prevent output drift. Recall that in this configuration,
V- = V+ = 0
We can then find the currents,
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VI |
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d |
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----- |
f |
= C----V |
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dt |
Finally,
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IV- |
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– If = 0 = |
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– C----Vo |
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R1 |
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∫ VIdt |
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1 |
C |
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5.1.2.4 - Differentiator
• The following device is one form of differentiator using an inductor,
page 43
L |
RL |
If
II RI
VI
-
+ 
Vo
We may begin by realizing that the two op-amp inputs are at zero volts.
V- = V+ = 0
Next the input, and feedback currents may be found and summed,
II |
= |
VI |
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If = |
Vo |
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---- |
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--------------------- |
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RI |
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d |
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L---- + RL |
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VI |
Vo |
dt |
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∑ |
I = II – If |
= 0 |
= |
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---- |
– --------------------- |
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RI |
d |
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L---- + RL |
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dt |
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L---- |
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V |
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= |
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L |
dt |
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----- |
+ -------- |
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o |
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I |
R |
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R |
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I |
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I |
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• A second type of circuit uses a capacitor to find the differential,