Lecture 19
Op Amps
Mark Horowitz
Stanford University
horowitz@ee.stanford.edu
Copyright © 2016 by Mark Horowitz
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Reading
A&L: Chapter 15, pp. 863-866.
Reader, Chapter 8 which will be out soon
•
•
•
Noninverting Amp
– http://www.electronics-tutorials.ws/opamp/opamp_3.html
Inverting Amp
– http://www.electronics-tutorials.ws/opamp/opamp_2.html
Summing Amp
– http://www.electronics-tutorials.ws/opamp/opamp_4.html
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Roadmap
To measure our heart’s electrical signal, we need to measure
small voltages. We introduce a new component, an Op Amp,
that can amplify small signals and make them much larger.
The problem with an op-amp is that the gain is too large,
around a million.
Fortunately we can use feedback around this component to
build an amplifier with the gain that we need. In this lecture we
will go over the basics of op-amp circuits. In the next lecture we
will show how to combine impedance ideas with op amps to
create filters that have a peak gain greater than one.
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Learning Objectives
•
Understand what an op amp is:
– The inputs take no current
– The output is 106 times larger than the difference in input
•
The “golden rules” of op amps
– Input current is 0; Vin- = Vin+
•
Be able to use feedback to control the gain of the op amp
– For inverting and non-inverting amplifiers
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PREVIOUSLY IN E40M
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Bode Plot
•
Plot of Vout/Vin
– Where both axis use logarithmic scales
– Y is in dB
– X is in log(F)
•
Easy to estimate shape, since every region is a straight line
– k
– slope = 0
– k*F
- slope = 1
– k/F
- slope =-1
– k/F2
- slope = -2
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Example Bode Plot
10
100
1
10
0.1
1
0.01
0.001
0.1
0.0001
0.01
0.00001
1
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10000
1
1000000
E40M Lecture 19
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10000
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Frequency Domain Analysis
•
Analyze the circuit by what it does to each sinewave
– What is the output/input ratio for each frequency sinewave
•
This is called the transfer function:
𝑉𝑜𝑢𝑡 (𝐹)
𝐻 𝐹 =
𝑉𝑖𝑛 (𝐹)
•
Notice that this is a property of only the circuit, and not the input
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Want to Measure our Heart’s Electrical Signal
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OP AMPS
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Op Amp
•
Is a common building block
– It is a high-gain amplifier
•
Output voltage is
– A (V+ - V-)
– Gain, A, is 10K to a 1M
•
Output voltage can be + or –
– Often runs from +Vdd and -Vdd supply
– Huh?
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Op-Amp Power Supply
•
Up to now we had one supply voltage, Vdd
– All voltages were between Vdd and Gnd
– Generally measured relative to Gnd
• So all voltages were positive.
•
A sinewave goes positive and negative
– And most input signals do that too
•
It is convenient to have a reference where
– The output can be positive and negative
– Can do that by changing what we call the reference
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Moving the Reference
Vdd = 2.5 V
Vdd = 5 V
+
+
-
+
-
vout
-
+ vout
+ 2.5 V
-
-
The voltages are all the same, only the reference voltage has moved
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Op Amp Behavior
•
Relationship between output voltage and input voltage:
𝒗𝒐 = 𝑨 𝑽𝒊𝒏+ − 𝑽𝒊𝒏− = 𝑨 𝒗𝒑 − 𝒗𝒏
– A is the op-amp gain (or open-loop gain), and is 10K-1M
•
The input currents are generally small
So ip ≈ 0 and in ≈ 0.
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Since the Output Swing is Limited
•
Inside the op-amp are transistors
– Which drive the output voltage
• Using energy from the power supplies
– If the input difference is too large, the output “saturates”
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What Does This Do?
vout
Vdd = 5 V
5
+
4
vin +
-
+
+
-
vout
2.5 V
2
1
-
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Same Circuit Different Reference
vout
Vdd = 2.5 V
+
vin +
-
2
+
+
-
vout
1
0
-1
2.5 V
-2
-
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-1
0
2
1
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How To Get A Useful Amplifier
•
The gain of the op amp is to high to make a good amplifier
– We need to do something to make it useful
•
We will use analog feedback to fix this problem
– Feedback makes the input the error between the value of the
output, and the value you want the output to have.
•
Let’s see how to do this
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Connect Vout to Vin-
vout  A(Vin   Vin  )  A(vin  vout )
Vdd = 2.5 V
+
vin +
-
+
-Vdd = -2.5 V
vout
-
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What Is Going On
•
We solved the equation to find the answer
– But how does the op-amp get this answer?
•
Think about what happens when the input increases in voltage
– From 0V to 0.1V
– Initially the output can’t change
• There is capacitance at every node
– The op-amp thinks it needs to create a huge output voltage
• So it drives current into the output
• Which charges the capacitor
• Causing the output to increase
– This then decreases the input difference
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Feedback in an Op-amp
•
As the output rises
– The input difference decreases
– So A * DVin also decreases
•
The system is stable when
– A * DVin is exactly Vout
•
If A is large (like a million) for any Vout
– Say in the range +/- 10V
– DVin will be very, very small
– Can approximate that by saying DVin will be driven to 0
– 𝑽𝒊𝒏+ ≈ 𝑽𝒊𝒏−
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BUT
•
This is only true if you connect the output feedback
– To the negative terminal of the amplifier
•
What happens if you connect it to the positive terminal?
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Ideal Op Amp Model
(when using negative feedback)
The Two Golden Rules for circuits with ideal op-amps*
No voltage difference between
op-amp input terminals
No current into op-amp inputs
* when used in negative feedback amplifiers
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USEFUL OP AMPS CIRCUITS
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Approach To Solve All Op-amp Circuits
•
First check to make sure the feedback is negative
– If not stop!
•
Find the output voltage that will make the input difference 0
– Assume Vin- is equal to Vin+
• Find Vout such that KCL holds
– Solve for Vin- as a function of Vout, Vin- =kVout
• Then set Vout = Vin+/k
•
Both methods always work, but usually one is easier
– When in doubt, use the first one
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Non-inverting Amplifier
At node vn
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Inverting Amplifier
vn = vp = 0
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Summing Amplifier
KCL at the summing point (or summing node):
Output voltage is a scaled sum of the input voltages:
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Current-to-Voltage Converter
KCL at the vn node:
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DIFFERENTIAL AMPLIFIERS
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What Happens If We:
•
Take an inverting amplifier and drive the other input?
v1
v2
•
Not quite what we want.
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Differential Amplifier 1.0
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Learning Objectives
•
Understand what an op amp is:
– The inputs take no current
– The output is 106 times larger than the difference in input
•
The “golden rules” of op amps
– Input current is 0; Vin- = Vin+
•
Be able to use feedback to control the gain of the op amp
– For inverting and non-inverting amplifiers
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