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op-amps (v.1a) 1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

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Page 1: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

op-amps (v.1a) 1

CENG4480_A2 Op Amps and Analog interfacing

Analog interfacing techniques

Page 2: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

op-amps (v.1a) 2

Computer interfacing Introduction

To learn how to connect the computer to various physical devices.

Some diagrams of this manuscript are taken from the following references:

[1] S.E. Derenzo, Interfacing -- A laboratory approach using the microcomputer for instrumentation, data analysis and control prentice hall.

[2] D.A. Protopapas, Microcomputer hardware design, Prentice hall

[3] G C Loveday, Designing electronic hardware, Addison Wesley

Page 3: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

op-amps (v.1a) 3

Topics include:

Overall interfacing schemesAnalog interface circuits, active filtersAnalog/digital conversionssensors, controllersControl techniquesAdvanced examples

Page 4: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

op-amps (v.1a) 4

Overall view: a typical data acquisition and control system

Sensor filter A/D

Computer

D/APowercircuit

Mechanicaldevice

Timere.g.

8253

Sample&

Hold

Digital controlcircuit

Op-amp

Page 5: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

op-amps (v.1a) 5

Analog interface example1 Audio recording systems

Audio recording systemsAudio signal is 20~20KHzSampling at 40KHz, 16-bit is Hi-fiStereo ADC require to sample at 80KHz.Calculate storage requirement for one hour?Audio recording standards

Audio CD Mini-disk MD MP3

Page 6: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

op-amps (v.1a) 6

Analog interface example2 Surround sound audio systems

A common two channels audio CD Calculate storage size for one hour of recording

of a CD. 44.1KHz*2bytes*60sec*60min*2 channels=633.6Mbytes

Calculate the play time of a CD. 700M/(2bytes*44.1KHz*2channels*60sec)=61.4 minutes

6 Channels: Front R/L,Rear R/L, Middle, Sub woofer

44.1KHz, Calculate the sampling frequency.

Page 7: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

op-amps (v.1a) 7

Analog interface example3Play stations and Wii

Play station 3, Analog hand held controller (http://ryangenno.tripod.com/images/PlayStation3-system.gif)

Wii, http://www.onlinekosten.de/news/bilder/wii_controller.jpg

Driving wheel http://www.bizrate.com/gamecontrollers/logitech-driving-force-driving-force-wheel--pid11297651/

Page 8: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

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Operational Amplifier choices (op amp)

Why use op amp?What kinds of inputs/outputs do you want?What frequency responses do you want?

Page 9: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

op-amps (v.1a) 9

Factors for choosing an amplifier Source load requirements:

DC ( static or slow changing, without decoupling capacitors)

AC( for fast changing signal, use decoupling capacitors)

Input range, biased – absolute min, max voltage

Output range, biased – absolute min, max voltage

Frequency: range, allowed attenuation in dB

Noise tolerance Power – output current/output

impedance.

DC-direct current amplifier

AC-alternating current amplifier

Op-

amp

DC

Source Load

Op-

amp

AC

Source Load

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Discussions: what kind of amplifiers should we use? It is an art.

Condenser microphone(+/-10mV)Audio amplifier from MP3 player to speakerUltrasonic sensors (+/-1mV) to ADC (analog to

digital converter) (0-5V)Accelerometers (+/-5V), or (+/-500mV)Temperature sensors to ADC (010mv)Moving coil microphone with 50Hz noise (+/-

0.5mV)

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Suggestions

Condenser microphone(+/-100mV): AC, bandwidth=audio range(20-20KHz, voltage gain ~=100,low power.

Moving coil microphone with 50Hz noise (+/-0.5mV): AC, bandwidth=audio range, voltage gain ~=2000,low power.

Audio amplifier from MP3 player to speaker: AC, high power, voltage gain ~=5, power gain~=100, bandwidth=audio range

Ultrasonic sensors (+/-1mV) to ADC (0-5V): DC-shift needed, voltage gain~=5000. bandwidth=tuned narrow band

Accelerometers (+/-500mV): DC-shift needed, voltage gain ~=10, Temperature sensors to ADC (010mv): DC, voltage gain~=100.

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Operational amplifiers (op-amps)

ideal op-amps inverting amplifier non-inverting amplifier voltage follower current-to-voltage amplifier summing amplifier full-wave rectifier instrumental amplifier

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General concept about OP amps

Controllable gainFor DC or AC amplifierNot too high frequency responses

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Ideal Vs. realistic op-amp

Ideal Realistic RinA= infinite 105->108

Zin= infinite 106(bipolar input) 1012(FET input)

output offset existsV0=A(V+-V-)

V-

V+ +

_

+

2

3

6LM741

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Inverting amplifier

Gain(G) = -R2/R1

For min. output offset, set R3 = R1 // R2

Rin=R1

+

_

R2

R1

R3

V0V1

Virtual-ground,V2

A+

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Non-inverting amplifier

_V0

V1

R1

R2

Gain(G) 1 + (R2/R1)

For min. offset output , set R1//R2=Rsource

High input resistance

+

V2

A

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Differential amplifier

V0=(R2/R1)(V2-V1)

Minimum output offset R1 //R2 =R3 //R4

_V0

R2

+

R1V1

R3V2

R4

A

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Exercise 1

A temperature sensor has an offset of 100mV (produces an output of 100mV at 0 °C-degrees Celsius), and the gradient is 10 mV per °C. The temperature to be measured is ranging from 0 to 50 °C.

The required ADC input range is 0 to 9Volts. Given that the power supply is +/-9V, design a differential amplifier

for this application.

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Voltage follower (Unit voltage gain, high current gain, high input impedance)

Gain=1,Rin=high

For minimum output offset R=Rsource

_V0=V1

V1 +

R

A

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Current to voltage converter: application to photo detector – no loading effect for the light detector

_V0

IR

V0=I R

I should not be too large otherwise offset voltage will be too high.

+A

PhotodiodeLight detector

See http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/photdet.html#c1

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Summing amplifier

_V0

R

+

V0 = -{(V1/R1)+(V2/R2)+(V3/R3)}R

R1

R2

R3

I1

I1+I2+I3

V1

V2

V3

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Op-amp characteristics

Input and output offset voltages It is affected by power supply variations,

temperature, and unequal resistance paths. Some op-amps have offset setting inputs. Unequal resistance paths and bias currents on

inverting and non-inverting inputs

Temperature variations -- read data sheet for operating temperatures

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Op-amp dynamic response

Slew rate -- the maximum rate of output change (V/us) for a large input step change. A741 slew rate=0.5V/ s. Fast slew rate is

important in video circuits , fast data acquisition etc.

Gain bandwidth product higher gain --> lower frequency response lower gain --> higher frequency response

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Common mode gain

If the two inputs (V+,V-) are connected together and is given Vc, output is found to be Vo.

ideal differential amplifier only amplifies the voltage difference between its two inputs, so Vo should be 0.

But in practice it is not.This deficiency can be measured by the Common

mode gain=Vo/Vc.

Page 25: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

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Diagram of gain bandwidth product, from [1]

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Instrumental amplifier To make a better DC amplifier from op-amps

Diagram of instrumental amplifier, from [1]

Applications: DSO input amplifiers, amplifiers in medical measurement systems

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op-amps (v.1a) 27

Instrumental amplifier

It has all the advantages of an amplifier.Gain(G)=V0/(V+-V-)

=(R4/R3)[1+(2R2/R1)] (typically 10 to 1000)

Even V+=V-= Vc , there is a slight output because of the Common Mode Gain=Gc=V0/Vc

Therefore, V0= G(V+-V-)+GcVc

To measure this imperfection, Common Mode rejection ratio (CMRR)=G/Gc (typically 103 to 107, or 60 to 140 dB)is used , the bigger the better.

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Comparing amplifiers, from [1]

Op Inv. Noninv. Diff. Instu. Amp Amp Amp Amp AmpHigh Rin Yes No Yes No Yes

Diff’tial Yes No No Yes Yes input

Defined No Yes Yes Yes Yesgain

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Operational amplifier selection techniques and keywords National semiconductor is the main manufacturer: See

http://www.national.com/appinfo/milaero/analog/highp.html General Purpose: LM741* High Slew Rate:50V/ ms --> 2000V/ ms (how fast the output can be changed) Follower (high speed):50MHz Low Supply Current: 1.5mA --> 20 µA/Amp Low offset voltage: 100 µV Low Noise Low Input Bias Current: 50pA -->10pA High Power : 0.2A --> 2A Low Drift: 2.5 mV/ _C --> 1.0 mV/ _C Dual/Quad High Power Bandwidth High Power Bandwidth : 300KHz - 230Mhz

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Practical op-amp usage examples

Example 1: Working with one power supply

Example 2: Active filters Example 3: Sample and holdExample 4: Example 4: Voltage

Comparator and schmit triggerExample 5: Power amplifier

Page 31: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

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Example 1: Single power supply for op-amps

Small systems usually have 1 power supplyOutput V0 is biased at E/2 rather than 0.

E.g. Inverting amplifier. GainR2/R1

+

_

R2

R1

R3Vo

V1

A

R=20K R=20K E Volts Power supply0-VoltE/2

E/2 E/2

Vo

E

0-Volt

V+V-+

Page 32: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

op-amps (v.1a) 32

Typical AC amplifier designCondenser microphone amplifier circuit, and the diagram showing theoutput swing around the biased 2.5V. The capacitors isolate the stages.Condenser MIC output impedance is 75 OhmsWhat is the input impedance of the amplifier?

Page 33: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

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Example 2: Active filters (analog and using op-amps)

Applications: accept or reject certain signals with specific frequencies. High-pass, low-pass, band-pass etc. E.g. reject noise extract signal after demodulation reject unwanted side effect signals

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Types

2-1: Low pass2-2: High pass2-3: Band stop (notch): e.g. noise removal2-4: Band pass

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definition of power gain in decibel (dB)

Output power is P2, input power is P1

Power Gain in dB=10 log10 (P1/P2)

OrOutput voltage is V2, input voltage is V1

Assume load R is the same, power=V2/RPower Gain in dB=10 log10 (V1

2/ V22)

Power Gain in dB=20 log10 (V1/ V2)

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Important terms for filters: Formulas are not important, remember the frequency-gain curve and concepts

Pass band-- range of frequency that are passed unfiltered

Stop band -- range of frequency that are rejected.Corner frequency -- where amplitude dropped by

2-1/2=0.707= -3dBI.e. in dB: 20log(0.707) = -3dBSettling time -- time required to rise within 10%

of the final value after a step input.

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2-1 Low pass

Only low frequency signal can passone-pole: attenuates slower 20dB/decadetwo-pole: attenuates faster 40dB/decadeApplications:

remove high freq. Noise, remove high freq. before sampling to avoid

aliasing noise

Page 38: Op-amps (v.1a)1 CENG4480_A2 Op Amps and Analog interfacing Analog interfacing techniques

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Diagram for low-pass one pole filter, from [1]for simplicity make R2/R1=1,

-R2/R1

|G| =-------------------

{1+(f/fc)2}1/2

gain

3dB

fc Freq.

20 dB/decadedrop

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op-amps (v.1a) 39

Formula, also

22

11 2

2

1 2

2

2

1

2

2 21

// 2 1,

1 2

1( )

// 221 2[ ( )]

2

[1 2 ]

1put

2

1

1 1 1,since ( _ 2)

1 11

cc

c

c

c

c

X RGain X

R j fc

RX R fcj fc

GainR fcR R

j fc

RGain

R j fcR

fcR

RGain

fR j

f

RGain see appendix

R ja af

f

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op-amps (v.1a) 40

2-1aLow pass, one pole filter formulas for simplicity make R2/R1=1

-R2/R1

|G| =-------------------

{1+(f/fc)2}1/2

Corner frequency= fc=1/(2R2C)

The gain drops 6dB/octave or 20 dB/decade

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op-amps (v.1a) 41

Diagram for Low-pass two-pole filter, from [1] for simplicity make R3/(R2+R1)=1

- R3/(R2+R1)

|G|= ------------------ 1+(f/fc)2

gain

6dB

fc Freq.

40 dB/decadedrop

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op-amps (v.1a) 42

2-1bLow-pass two-pole filter formulas for simplicity make R3/(R2+R1)= 1

- R3/(R2+R1)

|G|= ------------------ 1+(f/fc)2

Corner frequency=fc

fc=(R1//R2)/2C1 =(2 R3C2)-1 when gain G drops at -6dB.

G is dropping at 40dB/decade

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Matlab, lp42.m %lp42.m, ceg3480 matlab demo %low pass filter-one pole clear f=[0:100:2000] N=length(f) fc=1000 for i=1:N % -----gain1 , low pass one pole , for simplifcity make (R2/R1)=1 gv1(i)=-1/sqrt(1+(f(i)/fc)^2); %db=10*log_base10(power1/power2); or dB=20*log_base10(voltage1/voltage2; gain1_db(i)=20*log10(abs(gv1(i))); % -----gain2 , low pass two pole , for simplifcity make {R3/(R1+R2)}=1 gv2(i)=-1/(1+(f(i)/fc)^2); %db=10*log_base10(power1/power2); or dB=20*log_base10(voltage1/voltage2; gain2_db(i)=20*log10(abs(gv2(i))); end % %======================================================== figure(1) clf limit_y=min(gain2_db) semilogx(f,gain1_db,'g-.') hold on semilogx(f,gain2_db,'b--') %------------------------ semilogx([fc,fc],[0,limit_y],'k-.') legend('one pole','two pole','fc=1000',4); %------------------------------------------------------- semilogx(f,ones(N)*-3,'r') text(f(N),-3,'- 3d B line') %------------------------ semilogx(f,ones(N)*-6,'r') text(f(N),-6,'- 6d B line') ylabel('low pass filter gain in d B') xlabel('freq Hz f in log scale')

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Plotting the comparison of the low pass filters (one-pole, two-pole)

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2-2High pass

Only high frequency signal can passone-pole: attenuates slower 20dB/decadetwo-pole: attenuates faster 40dB/decadeApplications:

Remove low freq. Noise (50Hz main) Remove DC offset drift.

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op-amps (v.1a) 46

2-2aDiagram for high-pass one-pole filter, from [1]For simplicity make R1=R2

(f/fc)

|G|=------------------- {1+(f/fc)2}1/2

fc=1/{2 (R1C)}

gain3dB

Freq.

20 dB/decadedrop

high freq. CutoffunintentionallyCreated by Op-amp

fc

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op-amps (v.1a) 47

High-pass one-pole filter formulas

(f/fc)

|G|=------------------- {1+(f/fc)2}1/2

Corner frequency= fc=1/{2 (R1C)}

At low f , |G|=f/fc; at high f , |G|=R2/R11

Since op-amp has a certain gain-bandwidth, so at high frequency the gain drops. So all op-amp high-pass filters are actually band-pass.

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Matlab hp52.m %hp52.m ceg3480 matlab demo %high pass filter-one pole clear f=[500:100:100000] N=length(f) fc=1000 for i=1:N % -------------------gain3 , high pass ,one pole gv3(i)=-(f(i)/fc)/sqrt(1+(f(i)/fc)^2); %db=10*log_base10(power1/power2); or dB=20*log_base10(voltage1/voltage2; gain3_db(i)=20*log10(abs(gv3(i))); end % %======================================================== figure(1) clf limit_y=min(gain3_db) semilogx(f,gain3_db,'k-.') hold on %------------------------ semilogx([fc,fc],[0,limit_y],'g-.') legend('one pole','fc=1000',4); %------------------------------------------------------- semilogx(f,ones(N)*-3,'r') text(f(N),-3,'- 3d B line') ylabel('high pass filter gain in d B') xlabel('f freq in log scale')

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high pass one pole filter

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2-3Band stop (notch) filter

Suppresses a narrow frequency band of signal

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2-3Band pass filter

Passes a frequency band of signal.

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Diagram for Notch filter (band-stop), from [1]

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op-amps (v.1a) 53

Notch filter (band-stop) formulas

Rejects a narrow band of frequencies and passes all others. Say reject the 60Hz main noise for noise removal.

High Q,Fc1=(4 RC)-1

Low Q, Fc2=( RC)-1

Voltagegain

frequency

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op-amps (v.1a) 54

Example 3: Sample and hold amplifier

For a fast changing signal, if you want to know the voltage level of a snap shot (e.g. using a slow AD converter to view a short pulse), you need a sample and hold device, e.g. AD582, AD389 etc.

At Sample(S), V0=V1; at Hold(H) the output is held at the level just before switching to H. It is like taking a photograph of a signal.

Some AD converter has this circuit incorporated inside.

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Diagram for Sample and hold amplifier, from [1]

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Example 4: Voltage Comparator with hysteresis and schmit trigger

E.g. in IR motor speed encoder V1=IR receiver input

Vref

V1comparator

Comparator gives bad resultUnstable region when V1 and Vref are closed

V0

0V

0V=

V+

V-

V-

V+

IR receiverSignal with noise

Better outputUsing Schmit trigger

Schmit trigger

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op-amps (v.1a) 57

Diagram for hysteresis (non-inverting schmit trigger), see P.420, S. Franco, Design with operational

amplifiers and analog integrated circuits, McGraw Hill.

OutputVoltage

Input voltage

Switch overvoltage

t

VTH

VTL

V1 V0Voltage

VTH -VTL=(Vohigh –Volow)(R1/R2)=2V

Vref =0

VTL

=-1VVTH

=1V

10V

-10V

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Example 4: Schmit trigger using dual-power supply non-inverting op amp : A small amount of hysteresis is used to stabilize the output when V1 is near to Vref.(set R1/R2=0.1)

To make Vo from low to high V1 >VTH, at V1=VTH

(VTH-0)/R1=(0-Vlow)/R2

V1>VTH= -R1/R2(Vlow)=1V

V1>VTH= -(0.1)(-10)=1V

To make Vo from high to low V1 <VTL

(Vhigh-0)/R2=(0-VTL)/R1

V1<VTL= -R1/R2(VHigh)= -1V

If R2 >> R1, hysteresis is small. For schmit trigger devices R1 0.1*R2, e.g. R1=1K, R2=10K, so

hysteresis is good enough to reject noise.

0V= Vo

The op-amp uses V+,V_ power supplies.Output is clamped to Vlow

or Vhigh, setVref =0 to make the math easier

Vhigh=10V

Vlow=-10V

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op-amps (v.1a) 59

Example 5: Power amplifier

Most op-amps can drive outputs with low currents, we need transistors to raise the power to drive heavy loads, e.g. mechanical relays, motors or speakers.

V0=V1-1.2 Volts;

TIP3055 type transistors can drive current up to 15A.

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op-amps (v.1a) 60

Power amplifier, from [1]

From: http://www.st.com/stonline/books/pdf/docs/4136.pdf

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From : http://www.fairchildsemi.com/ds/TI/TIP41C.pdf

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Summary

Studied Basic digital data acquisition systems The configuration of operational amplifier

circuits and their applications

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Appendix

To be discussed in class

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Appendix 1Exercise—problem setting

A magnetic sensor is used to detect the magnetic flux density (in K Gauss) of an environment. The resistance of the sensor is proportional to the flux density detected with a gradient of 4 KΩ per K Gauss, and when there is no magnetic flux the resistance is 2K. The range of the magnetic flux density in the environment is between 0 and 5 K Gauss, and the flux density changes at a rate of not more than 10 K Gauss per second. The smallest change in flux density detectable is required to be 5 Gauss. The system uses an Analogue-to-Digital Converter (ADC) to convert the magnetic flux density into digital data, and the power supply used for this system is 5 Volt.

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Exercise — Question Draw the circuit diagram of the bridge circuit and the operational

amplifier circuit needed to transform the flux density detected to an output voltage. The output voltage is proportional to the flux density. When the flux density is 0 the output is 0 Volt; when the flux density is 5 K Gauss the output is 5 Volts.

It is found that the temperature of the environment affects the resistance of the magnetic sensor at a rate of RT (in KΩ per degree Celsius). Discuss how you can use two or more sensors to reduce the effect of temperature change. Draw the circuit of your scheme and explain with the help of formulas of how the system can be freed from the effect of temperature change.

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Appendix 2, To prove

1/(1+ja)=1/(1+ja)*[(1-ja)/(1-ja)] = (1-ja)/(12-(ja)2)=(1-ja)/(1+a2), since j2= -1 =1/(1+a2)+(-ja)/(1+a2)=real + imaginary so |1/(1+ja)|={real2+ imaginary2}1/2

={[1 /(1+a2)]2+[(-ja)/(1+a2)]2}1/2

={[1+a2]2-(1+a2)a2/[1+a2]2}1/2

={[1+2a2+a4-a2-a4/[1+a2]2}1/2

={[1+a2]/[1+a2]2}1/2

=1/[1+a2]1/2, proved! 2

1 1

1 1ja a

2

1 1

1 1ja a

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Solution for Exercise 1

Gain =Vout/Vin=9V/(10mV*50 °C )=18, set R2/R1=R4/R3=18 How to solve the offset problem.

Sensor V2 Offset of 100mV at V1, 9*Rb/(Ra+Rb)=100mV (make R4>> Ra) why? Add a small variable resistor Rc between 9V & Ra for offset trimming.

_V0

R2

+

R1

V1

R3

Sensor

R4

A

Ra

Rb

9V

0V-9V

9V

V2