ECE 342 – Jose Schutt-Aine 1 ECE 342 Solid-State Devices & Circuits 18. Operational Amplifiers...

ECE 342 – Jose Schutt-Aine 1

ECE 342Solid-State Devices & Circuits

18. Operational Amplifiers

Jose E. Schutt-AineElectrical & Computer Engineering

University of [email protected]


Operational Amplifiers

General terminal configuration with bias

• Universal importance (e.g. amplification from microphone to loudspeakers)


Common configuration with bias implied but not shown


2 1( )outv A v v

Gain=A

• Signaling1. Differential input stage2. Difference between input is amplified


2. Zero output impedance


Also, op amps are dc (or direct coupled) amplifiers since they are expected to amplify signals with frequency as low as DC.

1. Infinite input impedance

• Ideal Op Amp

4. Infinite CMRR or zero common-mode gain

3. Infinite open-loop gain Ainf

5. Infinite bandwidth


Differential & Common-Mode Signals

- Differential input signal vID=v2-v1

- Common-mode input signal vICm=0.5(v1+v2)

1 2ID

ICm

vv v

2 2ID

ICm

vv v


Ideally, vICM should be zero to achieve high CMRR.

• Amplifier will amplify the difference between the two input signals



Practical ConsiderationsThe output voltage swing of an op amp is limited by the DC power supply. Since op amp can exhibit high gain, power supply voltage fluctuations must be minimized use decoupling capacitors from power supply


Inverting Configuration

We introduce RF (or R2) to reduce gain (from inf)

• When RF is connected to terminal 1, we talk about negative feedback. If RF is tied to terminal 2, we have positive feedback

Terminal 2 is tied to ground



Need to evaluate vo/vI

Assume ideal Op-Amp

2 1( ) 0ov v vA

Since gain is infinite:

v1 is virtual ground

1 2, 0Thus v v

11

1 1

I Iv v vi

R R

Note: A is open-loop gain


Since input impedance of OP amp is infinite, current through RF is i1

11

Ivi iR

1

0I o o I

F F

v v v vi

R R R

1 1

o I Fo I

F

v v Rv v

R R R



1

o F

I

v RG

v R

Closed-Loop gain

Observe that the closed-loop gain is the ratio of external components we can make the closed-loop as accurate as we want. Gain is smaller but more accurate.



2 1 1 /o ov v A v v v A

We assumed that the OP-amp was ideal. If we assume that the gain A is finite = A


11 1

( / ) /in o in ov v A v v Ai

R R


1

/i o Fo oo F

v v A Rv vv iR

A A R

Still assume infinite input impedance


1

1 1

/

1 / / 1o F F

I F F

v R R ARG

v R R A R A R

1 1F

F

ARG

R A R

Closed-loop gain

forinverting configuration


11

1

/

/

oR

oo

v A RvR

vi v AG

The reflected impedance of RF is given by


1

1R

RR

A

G

since 11 (1 )

F

A RA

G R

1F

R

RR

A

small



Since the reflected impedance is so small, v1 is thus very small and the inverting terminal is said to be a virtual ground in this configuration

1

, FRas A GR

We see that

, 0Ras A R Note: To minimize the closed-loop gain (G) on the value of the open-loop gain (A), make 1+RF/R1 << A


11 1/I I

iI

v vR R

i v R

Input and Output Impedances

- If high gain is required, input impedance will be low- Output impedance is zero



Find closed-loop gain for A=103, A=104 and A=105 assuming R1=1 kW and RF=100 kW. Assuming vI=0.1 V, find v1.

Example

A |G| v1

Using formulas

Note: Since output of inverting configuration is at terminal of VCVS, output impedance of closed-loop amp is zero.

103 90.83 -9.08 mV104 99.00 -0.99 mV105 99.90 -0.1 mV


Non-Inverting Configuration

0oID

vv

A Assume gain is


1 20 IDv v v virtual short

Infinite input impedance


1 0i

1 10o

F a

v v vTherefore

R R

1 a Fo

a

v R Rv

R


Virtual short


2 1Iv v v

a FI o

a

R Rv v

R

1o F

I a

v RG

v R

1 F

a

RG

R


The Buffer Stage

0, 1 1FF

a

RIf R G

R

Although voltage gain is low, current gain can be quite high. Buffer stage can be used to interface between processors and switches.


The Voltage Follower

- Unity gain amplifier- 100% negative feedback

inR

0outR


No feedback with feedback descriptionA(f) Ani(f) GainAMBOa AMBni Midband gainf2oa f2ni 3 dB freq ptGBWoa GBWni Gain-BW prod

Frequency Response – Non-Inverting

1MBoa

MBniMBoa

AA

A F

a

a F

RF

R R



1,MBoaIf A F

11a F F

MBnia a

R R RA

F R R

( )1ni

AA f

AF

2

( )1 /

MBoa

oa

AA f

jf f



2

1( )

1 11

MBoani

MBoa

MBoa oa

AA f

fA F jA F f

1MBoa

MBniMBoa

AA

A F

2 2 1ni oa MBoaf f A F


2

( )1 /

MBnini

ni

AA f

jf f


2 2 11

MBoani MBni ni oa MBoa

MBoa

AGBW A f f A F

A F

2 2ni MBni ni MBoa oa oaGBW A f A f GBW

Gain-Bandwidth product is constant


Midband voltage gain is reduced from AMBoa to AMBni

The upper 3-dB frequency will be greater than that of the op amp by the same factor of gain reduction.

If the low-frequency gain of the op amp is AMBoa= 200,000 and with resistors AMBni = 40, the gain is reduced by a factor of 5,000. If the basic 3-db frequency is 5 Hz, the noninverting 3dB frequency will be 25 kHz.



1(1 )F

iF

ARA

R A R

Frequency Response – Inverting OP Amp

2

Using1 /

MBoa

oa

AA

jf f

1 1 2(1 ) /MBoa F

iMBoa F F oa

A RA

R A R j R R f f


1 1Using ( 1) MBoa MBoaR A R A


1

2 1 1

1

11 /[ ]

Fi

oa MBoa F

RA

fR jf A R R R

and neglecting FR


1

FMBi

RA

R


2 2 1 11 /[ ]i oa MBoa Ff f A R R R

12 2

1 1

1 FMBi i oa MBoa

F

R RA f f A

R R R


2 21

1 FMBi i oa MBoa

F

RA f f A

R R


2 2MBi i oa MBoaA f f A

if RF >> R1

2

( )1 /

MBii

i

AA f

jf f



if RF >> R1 gain-bandwidth is constant

if RF ~ R1 ,

2 21

F

i MBi i MBoa oaF

RGBW A f A f

R R

1

F

i oaF

RGBW GBW

R R



12

1 1

1 oaFi oa

F MBi F

GBW RRf GBW

R R A R R


10200

0.05MBoA

ExampleDesign an amplifier to couple a microphone to a resistive load. The microphone generates a peak output of 50 mV for a typical voice input level and has a 10-kW output impedance. The output voltage across the 2-kW load is to have a peak value of 10 V. the bandwidth of the voltage gain should be at least 40 kHz. If the GBW of the op amp used is 3106 Hz, calculate the bandwidth of the final design

6

2

3 1015

200ni oa

niMBni MBni

GBW GBWf kHz

A A

The midband voltage gain is:

For a single noninverting stage with this gain, the upper corner frequency is:


1

21

1 10FR

R 2

22

1 20FR

R

This value of BW will not work need 2 stages

Example (cont’)

Pick first stage gain 10; second stage gain 20. We must then have:

and


6

2 2

3 10150

20nif kHz

6

2 1

3 10300

10nif kHz

Choose R21 and R22 arbitrarily and use above equations to extract RF1 and RF2; we get:

RF1= 18 kW, RF2 = 38 kW

Next, find 3-dB bandwidth of each stage by dividing respective gains into GBWoa or GBWni

Example (cont’)


5 5

10 20( )

1 /3 10 1 /1.5 10oA jf jf

2 22 210 10

2 1 19 10 2.25 10

o of f

2 126of kHz

The overall gain is:

At 3dB point, magnitude squared of denominator must be 2

From which

Example (cont’)

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ECE 342 – Jose Schutt-Aine 1 ECE 342 Solid-State Devices & Circuits 18. Operational Amplifiers...

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