16. op amps - stanford university · no current into op-amp inputs no voltage difference between...

M. Horowitz, J. Plummer, R. Howe 1

E40M

Op Amps

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

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)

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

∴

M. Horowitz, J. Plummer, R. Howe 5

OP AMPS

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

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

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

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

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( )

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

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

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

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

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

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

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−

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?

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.

M. Horowitz, J. Plummer, R. Howe 20

USEFUL OP AMPS CIRCUITS

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

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

⎛

⎝⎜⎜

⎞

⎠⎟⎟

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

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

M. Horowitz, J. Plummer, R. Howe 25

OP AMP FILTERS

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

⎛

⎝⎜⎜

⎞

⎠⎟⎟

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

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

M. Horowitz, J. Plummer, R. Howe 29

More Examples

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

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

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