Op Amps M. Horowitz, J. Plummer, R. Howe 1 Reading A&L: - - PowerPoint PPT Presentation

• M. Horowitz, J. Plummer, R. Howe

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E40M

Op Amps

• M. Horowitz, J. Plummer, R. Howe

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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

• M. Horowitz, J. Plummer, R. Howe

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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)

• M. Horowitz, J. Plummer, R. Howe

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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

∴

• M. Horowitz, J. Plummer, R. Howe

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OP AMPS

• M. Horowitz, J. Plummer, R. Howe

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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 10 K to 1 M

• Output voltage can be + or –

– Often can swing between +Vdd and -Vdd supplies – Huh?

LM741

• M. Horowitz, J. Plummer, R. Howe

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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

• M. Horowitz, J. Plummer, R. Howe

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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

• M. Horowitz, J. Plummer, R. Howe

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What You Will Actually Do

• Use the USB supply

– Just change the reference voltage

+

• 5 V

+

• R

R

• M. Horowitz, J. Plummer, R. Howe

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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 small

so ip ≈ 0 and in ≈ 0.

vo = A v+ − v−

( ) = A vp − vn

( )

• M. Horowitz, J. Plummer, R. Howe

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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

• M. Horowitz, J. Plummer, R. Howe

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What Does This Do?

+

• vin +
• 2.5 V

+

• +
• vout

Vcc = 5 V vout

1 2 3 1 2 3 4 5 4 5

• M. Horowitz, J. Plummer, R. Howe

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Same Circuit Different Reference

+

• vin +
• 2.5 V

+

• +
• vout

Vdd = 2.5 V vout

• 2
• 1
• 2
• 1

1 2 1 2

• M. Horowitz, J. Plummer, R. Howe

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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

• utput, and the value you want the output to have.
• Let’s see how to do this

• M. Horowitz, J. Plummer, R. Howe

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Connect Vout to Vin-

+

• vin +
• +
• vout

Vcc = 5 V

• Vcc = -5 V

vout = A(V+ − V−) = A(vin − vout) ∴ A +1

( )vout = Avin

∴vout = A A +1

( )

vin ≅ vin

• M. Horowitz, J. Plummer, R. Howe

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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 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

• M. Horowitz, J. Plummer, R. Howe

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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−

• M. Horowitz, J. Plummer, R. Howe

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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?

• M. Horowitz, J. Plummer, R. Howe

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Ideal Op Amps

No current into op-amp inputs No voltage difference between

• p-amp input terminals

The Two Golden Rules for circuits with ideal op-amps*

* when used in negative feedback amplifiers

1. 2.

• M. Horowitz, J. Plummer, R. Howe

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USEFUL OP AMPS CIRCUITS

• M. Horowitz, J. Plummer, R. Howe

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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 makes the input difference 0

– Assume V+ = V- – Find Vout such that KCL holds

• We’ll do some examples

E40M Lecture 19

• M. Horowitz, J. Plummer, R. Howe

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Non-inverting Amplifier

• ip = 0 so vp = vs
• V+ = V- so vn = vp = vs

i1 i2

i1 = i2 so vo − vs R1 = vs R2

∴ vo R1 = vs 1 R1 + 1 R2 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ∴vo = vs R1+R2 R2 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟

• M. Horowitz, J. Plummer, R. Howe

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Inverting Amplifier

At node vn

vn − vs Rs + vn − vo Rf +in = 0

But vn =vp = 0 and in = 0, so

− vs Rs − vo Rf = 0 or vo = −vs Rf Rs

• M. Horowitz, J. Plummer, R. Howe

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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 = − vo Rf so vo = −isRf

• M. Horowitz, J. Plummer, R. Howe

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OP AMP FILTERS

• M. Horowitz, J. Plummer, R. Howe

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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 = −vs Rf Rs

Zs = Rs

Zf = 1 1 Rf + j∗2πFCf

∴vo = −vs Zf Zs = − 1 1 Rf + j∗2πFCf Rs = −vs Rf Rs 1 1+ j∗2πFRfCf ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟

• M. Horowitz, J. Plummer, R. Howe

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Sketching the Bode Plot

F [Hz]

20 log10 |Vo/Vs| Rs = 1 kΩ, Rf = 100 kΩ, Cf = 160 nF

vo vs − Rf Rs 1 1+ j∗2πFRfCf ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟

Vo Vs

= ⎯

Fc = 1/(2pRfCf) = 10 Hz

0.1 1 10 100 103 104 105 20 40 60

• 20

• M. Horowitz, J. Plummer, R. Howe

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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

• M. Horowitz, J. Plummer, R. Howe

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More Examples

• M. Horowitz, J. Plummer, R. Howe

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Summing Amplifier

Output voltage is a scaled sum of the input voltages: i2 i1 i

3

i1+i2 = i3 so v1 R1 + v2 R2 = − vo R

KCL at the summing point (or summing node):

vo = − Rf R1 v1+ Rf R2 v2 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟

• ip = in = 0
• vn = vp = 0

• M. Horowitz, J. Plummer, R. Howe

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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 − vo Rf = 0

vn = v2 so v2 − v1 Rs = vo − v2 Rf

∴ vo Rf = v2 − v1 Rs + v2 Rf

∴vo = −v1 Rf Rs + v2 Rf +Rs Rs

• M. Horowitz, J. Plummer, R. Howe

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Differential Amplifier 1.0

v1− vn R1 = vn − vo R2 v1− v2 R4 R3 +R4 R1 = v2 R4 R3 +R4 − vo R2 ∴ vo R2 = − v1 R1 + v2 R1 R4 R3 +R4 + R1 R2 R4 R3 +R4 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟

But if R3 = R1 and R4 = R2

vo = v2 − v1

( )

R2 R1