16. Op Amps - Stanford University · No current into op-amp inputs No voltage difference between...
Transcript of 16. Op Amps - Stanford University · No current into op-amp inputs No voltage difference between...
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M. Horowitz, J. Plummer, R. Howe 1
E40M
Op Amps
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M. Horowitz, J. Plummer, R. Howe 2
Reading
A&L: Chapter 15, pp. 863-866.
Reader, Chapter 8
• 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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M. Horowitz, J. Plummer, R. Howe 3
How to Measure Small Voltages?
• Arduino input has full-scale around 5V– It produces a 10 bit answer (1024)– This means a LSB (least significant bit) is 5mV
• Need to make the signal bigger before input to Arduino– So, we will use an amplifier
• Many ways to build amplifiers– One often uses a standard building block for amplifiers
• Called an Operational Amplifier, or Op-Amp• A circuit with very high gain at low frequencies (< 10 kHz)
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M. Horowitz, J. Plummer, R. Howe 4
Electrical Picture
• Signal amplitude ≈ 1 mV
• Noise level will be significant
• will need to amplify and filter
• We’ll use filtering ideas from the last two lectures
∴
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M. Horowitz, J. Plummer, R. Howe 5
OP AMPS
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M. Horowitz, J. Plummer, R. Howe 6
Op Amp
• Is a common building block– It is a high-gain amplifier
• Output voltage isA (V+ ⎯ V-)Gain, A, is 10 K to 1 M
• Output voltage can be + or –– Often can swing between
+Vdd and -Vdd supplies– Huh?
LM741
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M. Horowitz, J. Plummer, R. Howe 7
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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M. Horowitz, J. Plummer, R. Howe 8
Moving the Reference
+
-+
-
vout
Vdd = 5 V
2.5 V
+
--
+-
+-
vout
Vdd = 2.5 V
The voltages are all the same, only the reference voltage has moved
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M. Horowitz, J. Plummer, R. Howe 9
What You Will Actually Do
• Use the USB supply– Just change the reference voltage
+
-5 V +
-
R
R
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M. Horowitz, J. Plummer, R. Howe 10
Op Amp Behavior
• Relationship between output voltage and input voltage:
A is the op-amp gain (or open-loop gain), and is huge 10K-1M
• The input currents are very, very smallso ip ≈ 0 and in ≈ 0.
vo = A v+ − v−( ) = A vp − vn( )
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M. Horowitz, J. Plummer, R. Howe 11
Since the Output Swing is Limited
• The high gain only exists for a small range of input voltages– If the input difference is too large, the output “saturates”
• Goes to the max positive or negative value possible• Close to supply voltages
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M. Horowitz, J. Plummer, R. Howe 12
What Does This Do?
+
-vin +
-
2.5 V+-
+
-
voutVcc = 5 V
vout
1
2
3
1 2 3
4
5
4 5
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M. Horowitz, J. Plummer, R. Howe 13
Same Circuit Different Reference
+
-vin +
-
2.5 V+-
+
-
vout
Vdd = 2.5 V vout
-2
-1
0
-2 -1 0
1
2
1 2
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M. Horowitz, J. Plummer, R. Howe 14
How To Get A Useful Amplifier
• The gain of the op amp is too high to make a useful 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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M. Horowitz, J. Plummer, R. Howe 15
Connect Vout to Vin-
+
-vin +
- +
-
vout
Vcc = 5 V
-Vcc = -5 V
vout = A(V+ −V−) = A(vin − vout)
∴ A +1( )vout = Avin
∴vout =AA +1( )
vin ≅ vin
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M. Horowitz, J. Plummer, R. Howe 16
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 0 V to 0.1 V– 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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M. Horowitz, J. Plummer, R. Howe 17
Feedback in an Op-amp Circuit
• As the output rises– The input difference decreases– So A* DVin also decreases
• The system is stable when– A* DVin is exactly equal to Vout
• If A is large (106) for any Vout
– Say in the range of ± 10V– Dvin will be very, very small– Can approximate that by saying Dvin will be driven to 0– Output will be set so vin+ ≈ vin−
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M. Horowitz, J. Plummer, R. Howe 18
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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M. Horowitz, J. Plummer, R. Howe 19
Ideal Op Amps
No current into op-amp inputs
No voltage difference betweenop-amp input terminals
The Two Golden Rules for circuits with ideal op-amps*
* when used in negative feedback amplifiers
1.
2.
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M. Horowitz, J. Plummer, R. Howe 20
USEFUL OP AMPS CIRCUITS
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M. Horowitz, J. Plummer, R. Howe 21
Approach To Solve All Op-amp Circuits
• First check to make sure the feedback is negative– If not, STOP!
• Find the output voltage that makes the input difference 0
– Assume V+ = V-
– Find Vout such that KCL holds
• We’ll do some examples
E40M Lecture 19
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M. Horowitz, J. Plummer, R. Howe 22
Non-inverting Amplifier
• ip = 0 so vp = vs
• V+ = V- so vn = vp = vs
i1
i2
i1 = i2 so vo − vs
R1=
vsR2
∴voR1
= vs1R1
+1R2
⎛
⎝⎜⎜
⎞
⎠⎟⎟
∴vo = vsR1+R2R2
⎛
⎝⎜⎜
⎞
⎠⎟⎟
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M. Horowitz, J. Plummer, R. Howe 23
Inverting Amplifier
At node vn
vn − vsRs
+vn − voRf
+ in = 0
But vn =vp = 0 and in = 0, so
−vsRs
−voRf
= 0 or vo = −vsRfRs
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M. Horowitz, J. Plummer, R. Howe 24
Current-to-Voltage Converter
KCL at the vn node:
i2
i1
iR
• ip = in = 0
• vn = vp = 0
• So iR = 0 as well
i1 = is = i2 = −voRf
so vo = −isRf
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M. Horowitz, J. Plummer, R. Howe 25
OP AMP FILTERS
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M. Horowitz, J. Plummer, R. Howe 26
Adding Capacitors
Sinusoidal voltage
Cf • Suppose we add a capacitor in the feedback
• We can treat this exactly as we did the earlier circuits by using impedances.
• Our earlier analysis showed
vo = −vsRfRs
Zs =Rs Zf =1
1Rf
+ j∗2πFCf
∴vo = −vsZfZs
= −
11Rf
+ j∗2πFCf
Rs= −vs
RfRs
11+ j∗2πFRfCf
⎛
⎝⎜⎜
⎞
⎠⎟⎟
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M. Horowitz, J. Plummer, R. Howe 27
Sketching the Bode Plot
F [Hz]
20 log10 |Vo/Vs|
Rs = 1 kΩ, Rf = 100 kΩ, Cf = 160 nF
vovs
−RfRs
11+ j∗2πFRfCf
⎛
⎝⎜⎜
⎞
⎠⎟⎟
Vo
Vs= ⎯
Fc = 1/(2pRfCf) = 10 Hz
0.1 1 10 100 103 104 105
20
40
60
0
-20
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M. Horowitz, J. Plummer, R. Howe 28
Learning Objectives
• Understand how living things use electricity
• Understand what an op amp is:– The inputs take no current– The output is 106 times larger than the difference in input
voltages
• The two Golden Rules of op amps in negative feedback– Input currents are 0; Vin- = Vin+
• Be able to use feedback to control the gain of the op amp– For inverting and non-inverting amplifiers
• Understand op amp filters and differential amplifiers
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M. Horowitz, J. Plummer, R. Howe 29
More Examples
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M. Horowitz, J. Plummer, R. Howe 30
Summing Amplifier
Output voltage is a scaled sum of the input voltages:
i2
i1 i3
i1+ i2 = i3 so v1R1
+v2R2
= −voR
KCL at the summing point (or summing node):
vo = −RfR1v1+
RfR2v2
⎛
⎝⎜⎜
⎞
⎠⎟⎟
• ip = in = 0
• vn = vp = 0
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M. Horowitz, J. Plummer, R. Howe 31
A Subtracting (Difference) Amplifier?
• Take an inverting amplifier and put a 2nd voltage on the other input?
• Not quite what we wanted. We’d like vo a (v1 – v2).
v2
v1
i1+ i2 = 0 so vn − v1
Rs+
vn − voRf
= 0
vn = v2 so v2 − v1
Rs=
vo − v2Rf
∴voRf
=v2 − v1Rs
+v2Rf
∴vo = −v1RfRs
+ v2Rf +RsRs
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M. Horowitz, J. Plummer, R. Howe 32
Differential Amplifier 1.0
v1− vnR1
=vn − voR2
v1− v2R4
R3 +R4R1
=
v2R4
R3 +R4− vo
R2
∴voR2
= −v1R1
+v2R1
R4R3 +R4
+R1R2
R4R3 +R4
⎛
⎝⎜⎜
⎞
⎠⎟⎟
But if R3 = R1 and R4 = R2
vo = v2 − v1( )R2R1